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Students get through the TN Board 12th Chemistry Important Questions Chapter 11 Hydroxy Compounds and Ethers which is useful for their exam preparation.

TN State Board 12th Chemistry Important Questions Chapter 11 Hydroxy Compounds and Ethers

Answer the following questions.

Question 1.
Classify the following as primary, secondary and tertiary alcohols.


Answer:
Primary alcohols: (i), (ii), (iii)
Secondary alcohols: (iv), (v)
Tertiary alcohol: (vi)

Question 2.
Identify allyl alcohols in the above (Q.No. 1) examples.
Answer:
(ii) and (vi).

Question 3.
Name the following compounds according to IUPAC system:
Answer:

Question 4.
Write IUPAC names of the following:
Answer:

Question 5.
Write the structures of the compounds whose IUPAC names are as follows:
Answer:

Question 6.
Complete the following reactions:


Answer:

Question 7.
Using suitable Grignard reagent and a carbonyl compound, how will you prepare
(i) phenyl methanol,
(ii) butan-2-ol,
(iii) 2-methylhexan-2-ol,
(iv) propan-2-ol.
Answer:
(i) phenyl methanol:

(ii) butan-2-ol:

(iii) 2-methylhexan-2-ol:

(iv) propan-2-ol:

Question 8.
Give equation for the preparation of 2-methyl-2-propanol using a suitbale Grignard reagent and a carbonyl compound.
Answer:
Tertiary containing at least two identical groups can be prepared by the addition of a Grignard reagent to an ester other the formic ester followed by acid hydrolysis.

Question 9.
Name the reducing agents that will convert a carbonyl compound to an alcohol. Give an example for each.
Answer:
Raney Ni, / Sodium analgen in H₂O (Na—Hg/H₂O), H₂ in the presence of Pt.

where ‘R’ is an alkyl group.

Question 10.
Why is LiAlH₄ is used to reduce crotonaldehyde?
Answer:
Li crotonaldehyde is an unsaturated aldehyde (CH₃CH=CHCHO). Lithium aluminium hydride selectively reduces ‘CHO’ group to ‘CH₂OH’ group without affecting the C=C.

Question 11.
Name the product formed when sodium borohydride is used as a reducing agent for the reduction of a compound containing both carbonyl and carboxyl group, (or) Identify the product:
Answer:

Question 12.
Complete the following reactions:


Answer:

Question 13.
How is glycerol prepared from triglycerides or fats?
Answer:
The alkaline hydrolysis of these fats gives glycerol and the reaction is known as saponification.

Question 14.
Complete the following equations:

Answer:

Question 15.
Give mechanism of reaction between alcohols and halogen acids.
Answer:
Primary alcohols react by S_N2 mechanism while secondary and tertiary alcohol react by S_N1 mechanism.

Question 16.
How will you distinguish between

Answer:

(ii) Secondary alcohol will give a blue colour as shown above whereas a tertiary alcohol will not produce any colour.

Question 17.
Give the structures of the products you would expect when each of the following alcohol reacts with (a) HCl-ZnCl₂, (b) HBr, (c) SOCl₂ (i) Butan-l-ol and (ii) 2-methylbutan-2-ol.
Answer:
(a) with HCl-ZnCl₂ (Lucas reagent):
[2-methylbutan-2-ol being a tertiary alcohol reacts with Lucas reagent to produce turbidity immediately due to formation of insoluble tert-alkyl-halide, while butan-l-ol being a primary alcohol does not react with Lucas reagent at room temperature.]

(b) with HBr: Both alcohols produce the corresponding alkyl bromides.

(c) with SOCl₂: Both alcohols react to form their corresponding alkyl chlorides.

Question 18.
Arrange the following compounds in increasing order of boiling points, pentan-1 -ol, butan-1-ol, butan-2-ol, ethanol, propan-1-ol, methanol. Give reason for your answer.
Methanol < ethanol < propan-1-ol < butan-2- ol < butan-1-ol < pentan-1-ol.
Answer:
Boiling point increases regularly as the molecular mass increases due to the corresponding increase in their vander waals forces of attraction. The boiling point increases in the order.
methanol, ethanol, propan-1-ol, butan-1-ol and pentan-1-ol.
Among isomeric alcohol, 2° alcohols have lower boiling point than 1° alcohols due to corresponding decrease in the extent of hydrogen bonding because of steric hindrance. Thus the boiling point of butan-2-ol is lower than butan-1-ol.

Question 19.
Explain why propanol has higher boiling point than that of hydrocarbon butane?
Answer:
The molecuels of butane are held together by weak vander waals forces of attraction while these of propanol are held together by stronger hydrogen bonding.

Therefore boiling point of propanol is higher than that of butane.

Question 20.
Write the mechanism for the reaction

Answer:
The reaction follows S_N1 mechanism.
Step 1: Formation of protonated alcohol.

Step 2: Protonated alcohols forms carbocation.

Step 3: Elimination of a proton to form alkane.

Question 21.
Give mechanism for the formation of prop-1 – ene from propan-1-ol.
Answer:
The reaction is

Mechanism:
Step 1: Formation of protonated alcohol.

Step 2: Formation of carbocation.

Step 3: Elimination of a proton.

Question 22.
State Saytzeff rule.
Answer:
During intramolecular dehydration, if there is a possibility to form a carbon-carbon double bond at different locations, the preferred location is the one that gives the more (highly) substituted alkene i.e., the stable alkene.

Question 23.
Name the reagents in the following reactions:
(i) oxidation of a primary alcohol to a carboxylic acid.
(ii) oxidation of a primary alcohol to an aldehyde.
(iii) benzyl alcohol to benzoic acid.
(iv) butan-2-one to butan-2-ol.
Answer:
(i) Acidified potassium dichromate or neutral or acidic or alkaline potassium permanganate (followed by hydrolysis with dilute H₂SO₄).
(ii) Pyridinium chloro chromate(PCC) in CH₂Cl₂ or pyridinium dichromate (PDC) in CH₂Cl₂.
(iii) Acidified or alkaline KMnO₄ (followed by hydrolysis with dil H₂SO₄).
(iv) Ni/H₂ or NaBH₄ or LiAlH₄.

Question 24.
What happen when,
(i) vapours of ethanol is passed over heated copper at 573 K.
(ii) vapours of propan-2-ol is passed over heated copper at 573 K.
(iii) vapours of 2-methyl propan-2-ol is passed over heated copper at 573 K. Give equation.
Answer:
(i) Ethanal is formed.

(ii) Propanone is formed.

(iii) 2-methyl prop-1-ene is formed.

Question 25.
What is nitroglycerine? How it is formed? Give equation.
Answer:
Glycerotrinitrate (1,2,3, trinitroxy propane) is nitroglycerine. It is formed by nitration of glycerine in the presence of conc. H₂SO₄.

Question 26.
How is acrolein formed from glycerol?
Answer:
When glycerol is heated with dehydrating agents like conc. H₂SO₄, KHSO_4, or P₂O₅ acrolein is formed.

Question 27.
Mention the cause for the acidic nature of alcohols.
Answer:
The acidic character of alcohols is due to the electronegative oxygen atom which withdraws electrons of the O—H bond towards itself. As a result, the O—H bond becomes weak and hence a proton can be easily abstracted by a base, i.e., ionization of alcohol to yield alkoxide ion and H+ ions are facilitated.
ROH ⇌ RO^– + H⁺

Question 28.
Mention the uses of (i) methanol, (ii) ethanol, (iii) ethylene glycol, (iv) glycerol.
Answer:
(i) Uses of methanol:
(a) Methanol is used as a solvent for paints, varnishes, shellac, gums, cement, etc.
(b) In the manufacture of dyes, drugs, perfumes, and formaldehyde.

(ii) Uses of ethanol:
(a) Ethanol is used as an important beverage.
(b) It is also used in the preparation of

• Paints and varnishes.
• Organic compounds like ether, chloroform, iodoform, etc.,
• Dyes, transparent soaps.

(c) As a substitute for petrol under the name power alcohol used as fuel for airplane.
(d) It is used as a preservative for biological specimens.

(iii) Uses of ethylene glycol:
(a) Ethylene glycol is used as an antifreeze in an automobile radiators.
(b) Its dinitrate is used as an explosive with DNG.

(iv) Uses of glycerol:
(a) Glycerol is used as a sweetening agent in confectionery and beverages.
(b) It is used in the manufacture of cosmetics and transparent soaps.
(c) It is used in making printing inks and stamp pad ink and lubricant for watches and clocks.
(d) It is used in the manufacture of explosives like dynamite and cordite by mixing it with china clay.

Question 29.
Alcohols are easily protonated in comparison to phenols. Explain why.
Answer:
In phenols, the lone pair of electrons on the oxygen atom is delocalized over the benzene ring due to resonance and hence, not easily available for protonation. In contrast, in alcohols, the lone pair of electrons on the oxygen atom is localized due to the absence of resonance and hence are easily available for protonation.

Question 30.
Identify the major product obtained in the following reaction:

Answer:

Question 31.
Identify the product of the reaction:

Answer:
In the presence of a weak nucleophile such as ethanol, neopentyl bromide ionizes to form first a 1° carbocation which rearranges to form a more stable 3° carbocation. This is then attacked by the weak nucleophile ethanol followed by a less proton to yield ethyl neopentyl ether.

Question 32.
Show how 2-methyl propan-1-ol is prepared by the reaction of a suitable Grignard reagent.
Answer:

This part that has been enclosed in the box comes from methanal while the remaining part comes from the Grignard reagent. Therefore the Grignard reagent should be isopropyl magnesium bromide. Thus,

Question 33.
Write the structure of the products of the following reactions.

Answer:

(ii) NaBH₄ (sodium bromohydride) is a reducing agent. It reduces aldehydes and ketones but not esters.

(ii) NaBH₄ reduces, -CHO group to CH₂OH group. Thus,

Question 34.
What is esterification? Give an example.
Answer:
Formation of an ester from an alcohol and an organic acid in the presence of a catalyst is known as esterification. In the reaction, the ‘OH’ of the carboxylic acid and ‘H’ of the alcoholic group are removed as water.
Eg:

Question 35.
Why are alcohols are weaker acids than water?
Answer:
In water, ‘O’ atom is attached to two ‘H’ atom, but in alcohols, ‘O’ atom is attached to one ‘H’ atom and one ‘R’ group (alkyl group). Since ‘R’ group has an electron-donating inductive effect (+I), it increases the electron density in the O—H bond compared to that O—H bond in water. In other words, —O—H bond in alcohols is stronger than the —O—H bond in water. As a result, proton removal from alcohol is more difficult than in water. Hence, alcohols are weaker acids than water.

Question 36.
The acidic nature of alcohols is in the order 1° alcohol > 2° alcohol > 3° alcohol. Explain.
Answer:

Alkyl (R) groups are electron releasing (i.e., +I effect) the electron density in the O—H bond in tertiary alcohols is maximum followed by secondary alcohol, while in primary alcohols, it is minimum. This means that O—H bond in tertiary alcohol is the strongest and hence most difficult to break followed by O—H bond in the secondary alcohols, while the O—H bond in primary alcohols is the weakest and hence most easy to break. Thus the acidic strength is in the order.
Primary > Secondary > Tertiary

Question 37.
Arrange the following compounds in increasing order against each.
(i) CH₃CH₂OH, CF₃CH₂OH. CCl₃CH₂OH (acid strength)
(ii) 2-methyl-2-propanol, 1-butanol, 2-butanol (reactivity towards sodium)
Answer:
(i) Due to -I effect of halogens, the electron density in the O—H bond decreases. As a result, the O—H bond strength becomes weak and proton removal is easier compared to CH₃CH₂OH. Further, since fluorine has a stronger -I effect, than chlorine, CF₃CH₂COOH is a stronger acid than CCl₃CH₂COOH. Hence, CF₃CH₂OH is the strongest acid while CH2CH2OH is the weakest acid. Hence increasing order of acid strength is
CH₃CH₂OH < CCl₃CH₂OH < CF₃CH₂OH.
(ii) It is an acid-base reaction since alcohols are acidic and sodium is a strong base. As such, the reactivity of these alcohols towards sodium increases as the acidic character of alcohols increases. Since the acidic character increases in the order 3° < 2° < 1°. Therefore the reactivity towards alcohol increases in the same order.

Question 38.
How will you convert phenol to (i) benzene, (ii) aniline, (iii) phenylacetate, (iv) phenyl benzoate, (v) anisole?
Answer:
(i) Benzene: Phenol is converted to benzene on heating with zinc dust. In this reaction, the hydroxyl group which is attached to the aromatic ring is eliminated.

(ii) Aniline: Phenol on heating with ammonia in presence of anhydrous ZnCl₂ gives aniline.

(iii) Phenyl acetate:

(iv) Phenyl benzoate:

(v) Anisole:

Question 39.
Identify A and B in the following:

Answer:

(ii) A= HNO₂ (273 -278) K
B = H₂O

Question 40.
What is picric acid? How it is prepared? 2,4,6 trinitrophenol is called picric acid.
Answer:
Nitration with cone. HNO₃ and conc.H₂SO₄ gives picric acid.

Question 41.
Give equation for the following:
(i) nitration of phenol,
(ii) sulphonation of phenol,
(iii) nitrosation of phenol.
Answer:
(i) Nitration of phenol: Phenol can be nitrated using 20% nitric acid even at room temperature, a mixture of ortho and para nitro phenols are formed.

(ii) Sulphonation of phenol: Phenol reacts with cone. H₂SO₄ at 280 K to form o-phenol sulphonic acid as the major product. When the reaction is carried out at 373 K the major product is p-phenol sulphonic acid.

(iii) Nitrosation of phenol: phenol can be readily nitrosated at low temperature with HNO₂.

Question 42.
What happens when phenol is treated with
(i) acidified potassium dichromate,
(ii) hydrogen in the presence of nickel as a catalyst,
(iii) bromine water,
(iv) bromine in the presence of CCl₄.
Answer:
Phenol treated with
(i) acidified potassium dichromate (oxidation):
Phenol undergoes oxidation with air or acidified K₂CrO₇ with conc. H₂SO₄ to form 1, 4-benzoquinone.

(ii) hydrogen in the presence of nickel as catalyst (reduction): Phenol on catalytic hydrogenation gives cyclohexanol.

(iii) bromine water (halogenation): phenol reacts with bromine water to give a white precipitate of 2,4,6-tribromo phenol.

(iv) bromine in the presence of CCl₄: If the reaction is carried out in CS₂ or CCl₄ at 278 K, a mixture of ortho and para bromo phenols are formed.

Question 43.
Explain why the phenolic group is ortho para directing.
Answer:
The lone pair of electrons on the oxygen atom of the phenolic group enter into conjugation with the benzene ring due +M effect of the phenolic group. As a result, the ortho and para position become electron-rich compounds to the meta position. Hence the electrophilic attack the ortho and para position.

Question 44.
Write a short note on Reimer-Tiemann reaction.
Answer:
On treating phenol with CHCl₃ / NaOH, a -CHO group is introduced at the ortho position.
This reaction proceeds through the formation of substituted benzal chloride intermediate.

Question 45.
How will you distinguish between alcohol and phenol?
Answer:

1. Phenol reacts with benzene diazonium chloride to form a red-orange dye, but ethanol has no reaction with it.
2. Phenol gives purple coloration with neutral ferric chloride solution, alcohols do not give such coloration with FeCl₃.
3. Phenol reacts with NaOH to give sodium phenoxide. Ethyl alcohol does not react with NaOH.

Question 46.
Mention the uses of phenol.
Answer:

1. About half of the world production of phenol is used for making phenol-formaldehyde resin. (Bakelite).
2. Phenol is a starting material for the preparation of
   (a) drugs such as phenacetin, salol, aspirin, etc.
   (b) phenolphthalein indicator.
   (c) explosive like picric acid.
3. It is used as an antiseptic-carbolic lotion and carbolic soaps.

Question 47.
Give the IUPAC names of the following ethers.


Answer:
(i) 1-ethoxy-2-methyl propane
(ii) 2-chloro-l-methoxy ethane
(iii) 4-nitroanisole
(iv) 1-methoxy propane
(v) 4-ethoxy-1, 1, dimethyl cyclohexane
(vi) ethoxy benzene

Question 48.
Ethers are soluble in water. Give reason.
Answer:
The oxygen of ether can also form hydrogen bonds with water and hence they are miscible with water. Ethers dissolve a wide range of polar and non-polar substances.

Question 49.
Write the names of reagents and equations for the preparation of the following ethers by Williamson’s synthesis.
(i) 1 -propoxypropane
(ii) Ethoxy benzene
(iii) 2-methyl-2-methoxy propane
(iv) 1-methoxyethane
Answer:
(i)

Question 50.
Give the mechanism for the conversion of (i) ethanol to ethoxyethane, (ii) dehydration of 1-propanol to 1-proproxy propane.
Answer:
(i)

[Acid catalyzed 1° alcohol to ethers occurs via S_N2 reaction, involving a nucleophilic attack by the alcohol molecule on the protonated alcohol molecule.]

Question 51.
Explain why the boiling point of ethers is slightly higher than that of alkanes and lower than that of alcohols of comparable molecular mass.
Answer:
Due to weak dipole-dipole interactions, the boiling point of ethers is only slightly higher than those of u-alkanes having comparable molecular masses.
The boiling points of ethers are less than that of alcohols is that ethers do not form hydrogen bonds.

Question 52.
Mention the uses of (i) Diethyl ether and (ii) Anisole.
Answer:
Uses of Diethyl ether:

1. Diethyl ether is used as a surgical anesthetic agent in surgery.
2. It is a good solvent for organic reactions and extraction.
3. It is used as a volatile starting fluid for diesel and gasoline engines.
4. It is used as a refrigerant.

Uses of anisole:

1. Anisole is a precursor to the synthesis of perfumes and insecticide pheromones
2. It is used as a pharmaceutical agent.

Question 53.
Discuss the mechanism of the reaction
CH₃OCH₂CH₃ + HI → CH₃I + CH₃CH₂OH.
Answer:
This reaction is an S_N2 reaction. The cleavage of ethers by halogen acids occurs by the following mechanism.
Step 1: Ethers being Lewis acids, undergo protonation to form oxonium ions.

This reaction takes place with HBr or HI because they are sufficiently acidic.
Step 2: Iodide ion is a good nucleophile. Due to steric hindrance, it attacks the smaller alkyl group of oxonium ion formed in step 1 and displaces the alcohol molecule by S_N2 mechanism as shown below:

Step 3: when excess HI is used, ethanol formed reacts with another molecule of HI to form ethyl iodide.

Question 54.
Explain why ‘OCH₃’ in the anisole group is ortho-para directing.
Answer:
The lone pair of electrons on the oxygen atom of the alkoxy group enters into the conjugation with the benzene ring due to its +M effect.

As a result, the ortho and para position become rich in electron density compared to the meta position. Hence, the electrophile attacks ortho and para position.

Question 55.
Briefly account of dipolar nature of C—O bonds in ethers.
Answer:
Because of the greater electronegativity of oxygen and carbon, the C—O bond is slightly polar thus have a dipole moment. Since the two C—O bonds in ethers are inclined to each other at an angle of 110°, the two dipoles do not cancel each other. As a result, ethers have a dipole moment of 1.15 to 1.3D

Question 56.
Give the equation for the reaction between diethyl ether with an excess of oxygen.
Answer:

Question 57.
Give examples for (i) phthalein fusion and (ii) coupling reaction.
Answer:
(i) Phthalein fusion: On heating phenol with phthalic anhydride in presence of conc. H₂SO₄, phenolphthalein is obtained.

(ii) Coupling reaction: Phenol couples with benzene diazonium chloride in an alkaline solution to form p-hydroxy azobenzene (a red-orange dye).

Choose the correct answer:

1. The IUPAC name of the compound is:

(a) 2-chloro-5- hydroxyhexane
(b) 2-hydroxy-5-chlorohexane
(c) 5-chlorohexan-2-ol
(d) 2-chlorohexan-5-ol
Answer:
(c)

2. The IUPAC name of the compound is:

(a) 1-methoxy-1-methyl ethane
(b) 2-methoxy-2-methyl ethane
(c) 2-methoxypropane
(d) isopropyl-methyl ether
Answer:
(c)

3. Arrange the following compounds in increasing order of boiling point:
I – propan-1-ol
II – butan-1-ol
III – butan-2-ol
IV – pentan-1-ol
(a) I, III, II, IV
(b) I, II, III, IV
(c) IV, III, II, I
(d) IV, II, III, I
Answer:
(a)
Hint: The boiling point increases as the molecular mass increases. Further among isomeric alcohols 1° alcohols have higher boiling points than 2° alcohols.

4. Acid catalysed hydration of alkenes except ethene, leads to the formation of:
(a) primary alcohol.
(b) secondary or tertiary alcohol.
(c) a mixture of primary and secondary alcohol.
(d) a mixture of secondary and tertiary alcohol.
Answer:
(b)
Hint: Hydration of ethene gives a 1° alcohol while all others give a either a 2° alcohol or 3° alcohol.

5. Which of the following Grignard reagent is suitable for the preparation of 3-methyl-2- butanol?
(a) 2-butanone + methyl magnesium bromide
(b) Acetone + ethyl magnesium bromide
(c) Acetaldehyde + isopropyl magnesium bromide
(d) Ethyl propionate + methyl magnesium bromide
Answer:
(c)
Hint:

6. Which of the following esters shown, after reduction with LiAlH₄ and aqueous workup, will yield two molecules of only a single alcohol?
(a) C₆H₅COOC₆H₅
(b) CH₃CH₂COOCH₂CH₃
(c) CH₃COOCH₃
(d) C₆H₅COOCH₂C₆H₅
Answer:
(d)
Hint: Acyl and arylalkyl groups contain the same number of carbon atoms.

7. Which of the following will exhibit highest boiling point?

Answer:
(b)
Hint: Among isomeric alcohols, the alcohol with no branching has the highest boiling point.

8. Consider the reactions:

The mechanisms of reactions (i) and (ii) are respectively:
(a) S_N1 and S_N2
(b) S_N1 and S_N1
(c) S_N2 and S_N2
(d) S_N2 and S_N1
Answer:
(c)

9.

which of the following compounds will be formed?

Answer:
(a)
Hint: Nucleophilic attack occurs on the smaller alkyl group.

10. The reaction of with RMgX leads to the formation of:

Answer:
(d)
Hint:

11.

X is:
(a) bromine in benzene
(b) bromine in water
(c) potassium bromide solution
(d) bromine in CCl₄ at 0° C
Answer:
(b)
Hint:

12. Phenol can be distinguished from ethanol by the following reagent except.
(a) sodium
(b) I₂ / NaOH
(c) neutral FeCl₃
(d) Br₂ / H₂O
Answer:
(a)
Hint: Na reacts with both ethapol and phenol.

13. An ether is more volatile than an alcohol having the same molecular formula. This is due to:
(a) intermolecular hydrogen bonding in alcohols
(b) dipolar character of ethers
(c) alcohols having resonance structures
(d) intermolecular hydrogen bonding in ethers
Answer:
(a)

14. Mark the correct increasing order of reactivity of the following compounds with HBr / HI.

(a) I < II < III
(b) II < I < III
(c) II < III < I
(d) III < II < I
Answer:
(c)
Hint: All the three benzyl alcohols react with HBr or HI through the formation of intermediate carbocation. Obviously more stable the carbocation, more reactive is the alcohol. Now, electron withdrawing group i.e, NO₂, Cl, etc., decrease the stability of carbocations. Since, NO₂ is a stronger, electron withdrawing group, then Cl, stability of carbocation increases in the order.

Therefore, the reactivity of the benzyl alcohols increases in the same orders, i.e., II < III < I.

15. Which of the following reagents can be used to oxidize primary alcohols to aldehydes?
1. CrO₃ in acidic medium,
2. KMnO₄ in acidic medium,
3. Pyridinium chlorochromate,
4. Heat in the presence of Cu at 573 K.
(a) 1, 3, 4
(b) 1, 3
(c) 1, 4
(d) none of the above
Answer:
(a)
Hint: KMnO₄ will oxidise the aldehyde formed to carboxylic acid.

16. Assertion: Bond angles in ethers is slightly more than the tetrahedral angle.
Reason: There are repulsions between two bulky ‘R’ groups.
(a) Assertion and reason both are correct, and reason is the correct explanation of assertion.
(b) Assertion and reason both are correct, but reason is not the correct explanation of assertion.
(c) Assertion is wrong, but reason is correct.
(d) Both assertion and reason are wrong.
Answer:
(c)
Hint: Correct assertion: Bond angles in ethers are slightly more than the tetrahedral bond angle.

17. Assertion: Bromination of benzene does not require Lewis acid is catalyst.
Reason: Lewis acid polarises the bromine molecule in Friedel craft’s reaction.
(a) Assertion and reason both are correct, and reason is the correct explanation of assertion.
(b) Assertion and reason both are correct, but reason is not the correct explanation of assertion.
(c) Assertion is true, but reason is wrong.
(d) Both assertion and reason are wrong.
Answer:
(b)

18. Match the entries in column I with appropriate entries in column II and choose the correct option using the codes given:



Answer:
(b)

19. Match the entries in column I with appropriate entries in column II and choose the correct option using the codes given:


Answer:
(d)

20. In the following sequence of reactions,

(a) CH₃CH₂CH₂OH
(b) CH₃CH(OH)CH₃
(c) CH₃CH₂CH₂CH₂OH
(d) (CH₃)₃CCH₂OH
Answer:
(b)
Hint: Since alkenes give Markovnikoff addition products, therefore 1° alcohols (a), (c) and (d) cannot be obtained by hydration. Thus, ‘Z’ is 2-propanol.
i.e.

21. Consider the following reaction:

The compound ‘Z’ is:
(a) CH₃CH₂OCH₂CH₃
(b) CH₃CH₂OSO₃H
(c) CH₃CH₂OH
(d) CH₂=CH₂
Answer:
(c)
Hint:

22. Formation of tert-butyl ether by reaction of sodium tert-butoxide and methyl bromide involves:
(a) elimination reaction
(b) electrophilic addition reaction
(c) nucleophilic addition reaction
(d) nucleophilic substitution reaction
Answer:
(d)

23.

Answer:
(a)

24. The red coloured compound formed during Victor-Meyer’s test is:

Answer:
(b)
Hint:

25. Following compounds are given:
(i) CH₃CH₂OH
(ii) CH₃COCH₃
(iii) CH₃CH(OH)CH₃
(iv) CH₃OH
Which of the above compounds on being warmed with iodine solution andNaOH, will give iodoform?
(a) (i), (iii) and (iv)
(b) only (i)
(c) (i), (ii) and (iii)
(d) (i) and (ii)
Answer:
(c)
Hint: Compounds containing CH₃CO group or CH₃CH(OH) group on warming with I₂ and NaOH will form iodoform. Hence, (i), (ii) and (iii) will give iodoform.

Students get through the TN Board 12th Chemistry Important Questions Chapter 10 Surface Chemistry which is useful for their exam preparation.

TN State Board 12th Chemistry Important Questions Chapter 10 Surface Chemistry

Answer the following questions.

Question 1.
Give three examples of adsorption.
Answer:

1. Charcoal adsorbs ammonia.
2. Silica gel adsorbs water.
3. Charcoal adsorbs colorants from sugar.

Question 2.
Distinguish between adsorption and absorption.
Answer:
Adsorption is a surface phenomenon while absorption is a bulk phenomenon, i.e., the adsorbate molecules are distributed throughout the adsorbent.

Question 3.
What are adsorbent and adsorbates? Give examples.
Answer:
Adsorbent is the material on which adsorption take place. Adsorbed substance is called adsorbate.
Examples for adsorbates: The gaseous molecules like He, Ne, O₂, N₂, SO₂ and NH₃ and solutions of NaCl or KCl.
Examples for adsorbents: Silica gel, metals like Ni, Cu, Pt, Ag and Pd and certain colloids.

Question 4.
What is an interface?
Answer:
The surface of separation of two phases where the concentration of adsorbed molecules is high is known as interface.

Question 5.
Mention the characteristics of adsorption.
Answer:

1. Adsorption can occur in all interfacial surfaces i.e., the adsorption can occur in between gas-solid, liquid solid, liquid- liquid, solid- solid and gas-liquid.
2. Adsorption is always accompanied by decrease in free energy. When ΔG reaches zero, the equilibrium is attained.
3. Adsorption is a spontaneous process.
4. When molecules are adsorbed, there is always a decrease in randomness of the molecules.
   We know, ΔG = ΔH – T ΔS where ΔG is change in free energy.
   ΔH is change in enthalpy and ΔS is change in entropy.
   Hence, ΔH = ΔG + TΔS Adsorption is exothermic as there is an interaction between adsorbate and adsorbent.

Question 6.
What do you understand by the term ‘sorption’?
Answer:
The term represents simultaneous adsorption and absorption.

Question 7.
What is the term used for sorption of gases on metal surfaces?
Answer:
Occlusion.

Question 8.
Give three examples for chemical adsorption or chemisorption.
Answer:

1. Adsorption of O₂ on tungsten.
2. Adsorption of H₂ on nickel.
3. Adsorption of ethyl alcohol vapours on nickel.

Question 9.
Give two examples for physical adsorption.
Answer:

1. Adsorption of N₂ on mica.
2. Adsorption of gases on charcol.

Question 10.
Explain the nature of the force of attraction that exists between an adsorbent and an adsorbate in (i) physical adsorption and (ii) chemical adsorption.
Answer:
The forces of attraction that exists between adsorbate and adsorbent in
(i) Physical adsorption are (a) vander waals force of attraction, (b) dipole-dipole interaction and (c) dispersion forces.
(ii) In chemisorption gas molecules are held to the surface by formation of chemical bonds. Since strong bonds are formed, nearly 400 kJ mole is given out as heat of adsorption.

Question 11.
Why is adsorption exothermic?
Answer:
When adsorption occurs, entropy decreases, i.e., ΔS = negative. As ΔG = ΔH – TΔS, for a spontaneous process, AG must be negative. This can only be so, if ΔH is negative, i.e., the process is exothermic.

Question 12.
How do size of particles of adsorbent, pressure of gas and prevailing temperature influence the extent of adsorption of a gas on a solid?
Answer:

1. The smaller the size of the particles of the adsorbent, greater is the surface area, greater is the adsorption.
2. At constant temperature, adsorption first increases with increase in pressure and then attains equilibrium at high pressures.
3. In physical adsorption, it decreases with increase in temperatures, but in chemisorption first, it increases and then decreases.

Question 13.
What are adsorption isotherms and isobars? Explain.
Answer:
A plot between the amount of adsorbate adsorbed and pressure or concentration of adsorbate at constant temperature is called adsoiption isotherm.
The plot of the amount of adsorption verses temperature at constant pressure is called adsorption isobar. The adsorption isobars of physisorption and chemisorption as shown in figure.

In physical adsorption, \(\frac{x}{m}\) decreases with increases in T, But in chemical adsorption, \(\frac{x}{m}\) increases with rise in temperature and then decreases. The increase illustrates the requirement of activation of the surface for adsorption is due to fact that the formation of activated complex requires certain energy.
The decrease at high temperature is due to desorption, as the kinetic energy of the adsorbate increases.

Question 14.
Derive Freundlich adsorption isotherm.
Answer:
A plot between the amount of adsorbate adsorbed and the pressure or concentration of adsorbate at constant temperature is called adsorption isotherms.
In order to explain these isotherms, various equations were suggested as follows: Freundlich adsorption isotherm.
According to Freundlinch,
\(\frac{x}{m}\) = Kp^{\(\frac{1}{n}\)}
where V is the amount of adsorbate or adsorbed on ‘m’ gm of adsorbent at a pressure of p. K and n are constants.
Value n is always less than unity.
This equation is applicable for adsoiption of gases on solid surfaces. The same equation becomes \(\frac{x}{m}\) = Kc^{\(\frac{1}{n}\)} when used for adsorption in solutions with c as concentration.
This equation quantitively predict the effect of pressure (or concentration) on the adsorption of gases (or adsorbates) at a constant temperature.
Taking log on both sides of the equation

Hence the intercept represents the value of log K and the slope \(\frac{b}{q}\) gives \(\frac{1}{n}\).
This equation explains the increase of \(\frac{x}{m}\) with an increase in pressure. But experimental values show the deviation at low pressure.

Question 15.
Explain the application of adsorption (i) in the use of gas masks, (ii) in the use of softening hard water, (iii) in decolourisation of sugar.
Answer:
(i) Gas masks are devices containing suitable adsorbents so that poisonous gases present in the atmosphere are preferentially adsorbed and the air for breathing is purified.
Activated charcoal is one of the best adsorbents.
(ii) Permutit is employed for this process which adsorbs Ca²⁺ and Mg²⁺ ions in its surface, there is an ion exchange as shown below it occurs on the surface.

Exhausted permutit is regenerated by adding a solution of common salt.

(iii) Sugar prepared from molasses is decolourised to remove coloured impurities by adding animal charcoal which acts as decolourising material.

Question 16.
What are adsorption indicators?
Answer:
In the precipitation titrations, the endpoint is indicated by an external indicator that changes its colour after getting adsorbed on precipitate. It is used to indicate the endpoint of the titration.

Question 17.
Distinguish between homogeneous catalysis and heterogeneous catalysis with suitable examples.
Answer:

Question 18.
Name the catalysts used in the following:
(i) Haber’s process in the manufacture of ammonia.
(ii) Oxidation of ammonia.
(iii) Hydrogenation of alkenes.
(iv) Decomposition of H₂O₂.
(v) Formation of acetophenone from benzene and ethanoyl chloride. Also, give equations for the reactions.
Answer:
(i) The catalyst used is Iron,

(ii) The catalyst is platinum gauze.

(iii) The catalyst is finely divided nickel.

(iv) The catalyst is platinum

(v) The catalyst is anhydrous aluminium chloride

Question 19.
Mention the characteristics of catalysts.
Answer:

1. For a chemical reaction, the catalyst is needed in very small quantities. Generally, a pinch of catalyst is enough for a reaction in bulk.
2. There may be some physical changes, but the catalyst remains unchanged in mass and chemical composition in a chemical reaction.
3. A catalyst itself can not initiate a reaction. It means it can not start a reaction that is not taking place. But, if the reaction is taking place at a slow rate it can increase its rate.
4. A solid catalyst will be more effective if it is taken in a finely divided form.
5. A catalyst can catalyse a particular type of reaction, hence they are said to be specific in nature.
6. In an equilibrium reaction, the presence of catalyst reduces the time for attainment of equilibrium and here it does not affect the position of equilibrium and the value of the equilibrium constant.
7. A catalyst is highly effective at a particular temperature called as optimum temperature.
8. The presence of a catalyst generally does not change the nature of products
   eg: 2SO₂ + O₂ → 2SO₃
   This reaction is slow in the absence of a catalyst, but fast in the presence if Pt catalyst.

Question 20.
What are promoters? Explain with an example.
Answer:
In a catalysed reaction the presence of a certain substance increases the activity of a catalyst. Such a substance is called a promoter. For example in the Haber’s process of manufacture ammonia, the activity of the iron catalyst is increased by the presence of molybdenum. Hence molybdenum is called a promoter. In the same way, Al₂O₃ can also be used as a promoter to increase the activity of the iron catalyst.

Question 21.
Explain the action of anhydrous aluminium chloride in Friedel-Crafts reaction.
Answer:
The mechanism of Friedel crafts reaction is given below

The action of catalyst is explained as follows:

It is an intermediate.

Question 22.
Give the mechanism of the following reactions based on intermediate formation theory.
Answer:

1. Thermal decomposition of KClO₃ in the presence of MnO₂.
   Steps in the reaction
   2KClO₃ → 2KCl + 3O₂
   can be given as
   2KClO₃ + 6MnO₂ → 6MnO₃ + 2KCl
   It is an intermediate,
   6MnO₃ → 6MnO₂ + 3O₂
2. The reaction between H2 and O2 in the presence of copper.
   2Cu + \(\frac{1}{2}\) O₂ → Cu₂O It is an intermediate.
   Cu₂O + H₂ → H₂O + 2Cu
3. Oxidation of HCl by air in the presence of CuCl₂.
   2CuCl₂ → Cl₂ + Cu₂Cl₂
   2Cu₂Cl₂ + O₂ → 2Cu₂OCl₂
   It is an intermediate.
   2Cu₂OCl₂ + 4HCl → 2H₂O + 4CuCl₂
   This theory describes
   (a) the specificity of a catalyst and
   (b) the increase in the rate of the reaction with an increase in the concentration of a catalyst.

Question 23.
Mention the limitations of the intermediate compound formation theory of catalysis.
Answer:

1. The intermediate compound theory fails to explain the action of catalytic poison and activators (promotors).
2. This theory is unable to explain the mechanism of heterogeneous catalysed reactions.

Question 24.
Explain the salient features of the adsorption theory of catalysis.
Answer:
According to this theory, the reactants are adsorbed on the catalyst surface to form an activated complex which subsequently decomposes and gives the product.
The various steps involved in a heterogeneous catalysed reaction are given as follows:

1. Reactant molecules diffuse from bulk to the catalyst surface.
2. The reactant molecules are adsorbed on the surface of the catalyst.
3. The adsorbed reactant molecules are activated and form an activated complex which is decomposed to form the products.
4. The product molecules are desorbed.
5. The product diffuses away from the surface of the catalyst.

Question 25.
What are active centres? How does the presence of active centres influence the rate of catalysed reaction?
Answer:
The surface of a catalyst is not smooth. It bears steps, cracks and comers. Hence the atoms on such locations of the surface are coordinatively unsaturated. So, they have much residual force of attraction. Such sites are called active centres. So, the surface carries high surface free energy.
The presence of such active centres increases the rate of reaction by adsorbing and activating the reactants.
The adsorption theory explains the following

1. Increase in the activity of a catalyst by increasing the surface area. An increase in the surface area of metals and metal oxides by reducing the particle size increases the rate of the reaction.
2. The action of catalytic poison occurs when the poison blocks the active centres of the catalyst.
3. A promoter or activator increases the number of active centres on the surfaces.

Question 26.
What is the role of a promoter in heterogeneous catalysis?
Answer:
A promoter increases the number of active centres on the surfaces.

Question 27.
What are catalytic poisons? How do they affect a catalysed reaction?
Answer:
Substances that destroy the activity of catalysts are known as catalytic poisons. They affect the catalytic reaction by occupying the active centres of the catalyst as a result active centres will not available for adsorption by reactants.

Question 28.
Give some examples of enzyme catalyses. Some common examples for enzyme catalysis.
Answer:

1. The peptide glycyl L-glutamyl L-lyrosin is hydrolysed by an enzyme called pepsin.
2. The enzyme diastase hydrolyses starch into maltose.
   2(C₆H₁₀O₅)_n + nH₂O → nC₂H₂₂O₁₁
3. The yeast contains the enzyme zymase which converts glucose into ethanol.
   C₆H₁₂O₆ + H₂O → 2C₂H₅OH + 2CO₂
4. The enzyme micoderma aceti oxidises alcohol into acetic acid.
   C₂H₅OH + O₂ → CH₃COOH + H₂O
5. The enzyme urease present in soya beens hydrolyses the urea.
   

Question 29.
Briefly outline the characteristics of enzyme catalysed reaction.
Answer:

1. An enzyme catalyst is specific in nature, i.e., one catalyst cannot catalyse more than one reaction.
2. The enzyme catalysed reactions are very fast compared to other catalysed reactions.
3. The rate of an enzyme-catalysed reaction depends on temperature. At an optimum temperature, the rate of the reaction becomes maximum. Beyond the optimum temperate, the rate decrease.
4. The enzyme activity is maximum at a particular pH called optimum pH. The favourable pH range for enzymatic reactions is 5-7.
5. The enzymatic activity is increased by the presence of containing substances, known as coenzymes.
6. The activity of enzymes can also be inhibited or poisoned due to the presence of certain substances.

Question 30.
Explain phase transfer catalysis with an example.
Answer:
Substitution of Cl^– and CN^– in the following reaction.

R-Cl = 1-chlorooctane
R-CN = 1-cyanooctane
By direct heating of two-phase mixture of organic 1 -chlorooctane with aqueous sodium cyanide for several days, 1-cyanooctane is not obtained. However, if a small amount of quaternary ammonium salt like tetraalkyl ammonium chloride is added, a rapid transition of 1 -cyanooctane occurs in about 100% yield after 1 or 2 hours. In this reaction, the tetraalkyl ammonium cation, which has hydrophobic and hydrophilic ends, transports CN from the aqueous phase to the organic phase using its hydrophilic end and facilitates the reaction with 1-chloroocatne as shown below:

So phase transfer catalyst speeds up the reaction by transporting one reactant from one phase to another.

Question 31.
Name the dispersed phase and dispersed medium in the following: fog, dust, sharing cream, milk, paints.
Answer:

Question 32.
Give the name of the colloid and one example for (i) a gas dispersed in solid, (ii) a liquid dispersed in a solid, (iii) a solid dispersed in a solid.
Answer:
(i) A gas dispersed in a solid is known as a solid foam. Pumice stone, rubber, bread are examples.
(ii) A liquid dispersed in a solid is known as a gel. Butter and cheese are examples.
(iii) A solid dispersed in another solid is known as a solid sol. Pearls, coloured glass, alloys are examples of solid sols.

Question 33.
Briefly outline the principle involved in the preparation of colloidal solution by dispersion methods.
Answer:
In the dispersion method, the particles whose size is higher than that of colloidal particles are disintegrated to that of the colloidal particles subjecting it to a mechanical disintegration in a colloid mill or by electro-dispersion.

Question 34.
Explain how colloidal particles of metals are prepared by electro dispersion method.
Answer:
An electrical arc is struck between electrodes dispersed in the water surrounded by ice. When a current of 1 amp /100V is passed an arc produced forms vapours of metal which immediately condense to form a colloidal solution. By this method colloidal solution of many metals like copper, silver, gold, platinum, etc. can be prepared Alkali hydroxide is added as a stabilising agent for the colloidal solution.

Question 35.
Explain peptisation with an example.
Answer:
The conversion of a freshly precipitated substance into a colloid by the addition of an electrolyte (peptising agent) is known as peptisation. For example, freshly precipitated silver chloride is converted into a colloidal solution by the addition of HCl.

Question 36.
Explain the principle involved in the preparation of colloids by the condensation method.
Answer:
Particles whose size are lesser than the colloidal particles are brought together to the size of colloidal particles.

Question 37.
What is the principle involved in the purification of a colloidal solution by dialysis?
Answer:
The separation of crystalloids (electrolytes) from colloids is based upon the principle that particles of crystalloid pass through the animal membrane (bladder) or parchment paper, or cellophane sheet whereas those of colloid do not.

Question 38.
Explain why? (i) colloidal particles are not visible to the naked eye, (ii) colloidal particles pass through ordinary filter paper, (iii) the sky appears blue, (iv) finest gold sol is red in colour.
Answer:
(i) No particle is visible to the naked eye if its diameter is less than half the wavelength of light used. The shortest wavelength of light is about 4000Å or 400 mμ. Hence no particle of diameter less than 200 mμ can be seen. The size of the colloidal particles is less than 200 mμ.
(ii) The size of the colloidal particles are smaller than the pores in the filter paper and hence, they pass through the filter paper.
(iii) This is due to the Tyndall effect. The colloidal particles absorb light energy and then scatter in all directions.
(iv) The colour of the colloidal solution depends as the wavelength of scattered radiation by dispersed particles. The wavelength of light further depends on the size and nature of particles. The finest gold sol is red in colour. As the size of particles increase, it becomes purple, then blue and finally golden yellow.

Question 39.
Explain the Tyndall effect.
Answer:
Colloids have optical properties. When a homogeneous solution is seen in the direction of light, it appears clear but it appears dark, in a perpendicular direction.
But when light passes through a colloidal solution, it is scattered in all directions. The colloidal particles absorb a portion of light and the remaining portion is scattered from the surface of the colloid. Hence the path of light is made clear. This phenomenon is known as the Tyndall effect.

Question 40.
Give an account of the Brownian movement.
Answer:
The zig-zag random motion of colloidal particles, when viewed through ultramicroscope, is known as Brownian movement. The reason for Brownian movement is that the colloidal sol particles are continuously bombarded with the molecules of the dispersion medium and hence they follow a zigzag, random, continuous movement.
Brownian movement enables us,
I. to calculate Avogadro number.
II. to confirm kinetic theory which considers the ceaseless rapid movement of molecules that increases with an increase in temperature.
III. to understand the stability of colloids: As the particles are in continuous rapid movement they do not come close and hence not get condensed. That is Brownian movement does not allow the particles to be acted on by force of gravity.

Question 41.
Explain the term electrophorosis.
Answer:
The movement of colloidal particles towards cathode or anode depending on the charge of the particles under the influence of electric current is known as electrophorosis.
This experiment is used to detect the charge on the colloidal particles. If the colloidal particles move towards the cathode, it is negatively charged. If they are positively charged, they migrate towards anode.

Question 42.
Give examples for positively charged and negatively charged colloids.
Answer:

Question 43.
Account for the charge on colloidal particles.
Answer:
The colloidal particles, in whichever method is prepared, contain traces of electrolytes. It preferentially adsorbs either positive or negative ions from the solution. If the particles adsorb positive ions from the solution, it becomes a positively charged colloid. If it adsorbs a negative ion from the solution, it becomes a negatively charged colloid.

Question 44.
Account for the stability of a colloid.
Answer:
Since the colloidal particles are either positive or negatively charged, they do not come close together. They repel each other. Thus, the charge on the colloidal particles is responsible for its stability.

Question 45.
Explain the term coagulation or flocculation.
Answer:
The conversion of a colloid to a precipitate is known as coagulation or precipitation. Coagulation results in the removal of charges on the colloid by any one of the following:

1. Addition of electrolytes
2. electrophoresis
3. mixing oppositely charged sols
4. boiling.

These methods help remove charges from the particles. Thus, the particles come closer, form an aggregate of particles and finally settles down as a precipitate due to gravity.

Question 46.
The peptising agent is added to convert a precipitate into a colloidal solution. Give reason.
Answer:
Ions (either positive or negative) of the peptising agent (electrolyte) are adsorbed on the particles of the precipitate. They repel and hit each other breaking the particles of the precipitate into colloidal size.

Question 47.
Explain what is observed (i) when a beam of light is passed through a colloidal solution.
(ii) an electrolyte, NaCl is added to ferric hydroxide solution,
(iii) electric current is passed through a colloidal solution.
Answer:
(i) Scattering of light by colloidal particles take place and the path of light becomes visible.
(ii) The positively charged colloidal particles of Fe(OH)₃ gets coagulated by the oppositely charged Cl^– ions provided by NaCl.
(iii) On passing electric current, colloidal particles move towards the oppositely charged electrode where they lose their charge Cl^– get coagulated.

Question 48.
What happens when gelatin is mixed with gold sol?
Answer:
Gold sol is lyophobic. Gelatin forms lyophilic sol and acts as a protective colloid. On adding gelatin to gold sol, the latter becomes more stable.

Question 49.
A colloid is formed by adding FeCl₃ in excess of hot water. What will happen if an excess ferric chloride is added to this colloid?
Answer:
On adding FeCl₃ to water, a colloidal sol of hydrated ferric oxide is formed. This is a positively charged colloid as it adsorbs Fe⁺³ ions on its surface. On adding NaCl, Cl^– ions bring about coagulation.

Question 50.
How does the boiling of a colloid, help the coagulation of a colloidal solution?
Answer:
Boiling vessels in increased collisions between the colloidal particles. The sol particles combine and settle down as a precipitate.

Question 51.
Explain the protective action of lyophilic colloids.
Answer:
The addition of a lyophobic colloid into lyophilic colloid protects the latter from coagulation. Hence, lyophilic colloids act as protective colloids.

Question 52.
Define gold number. Mention its use.
Answer:
‘Gold number’ as a measure of protecting the power of a colloid. The gold number is defined as the number of milligrams of hydrophilic colloid that will just prevent the precipitation of 10ml of gold sol on the addition of 1ml of 10% NaCl solution. Smaller the gold number greater the protective power.

Question 53.
What are emulsions? Mention various types of emulsions.
Answer:
Emulsions are a colloidal solutions in which a liquid is dispersed in another liquid.
Generally, there are two types of emulsions.

1. Oil in Water (O/W)
2. Water in Oil (W/O)

Question 54.
Give one example each for (i) oil in water and (ii) water in oil emulsions.
Answer:
(i) In Oil in Water (O/W) emulsions, the dispersed phase is oil and the dispersion medium is water.
eg: Milk is an emulsion of liquid fat dispersed in water and vanishing cream.
(ii) Emulsion of Water in Oil (W/O) in which water is the dispersed phase and oil is the dispersion medium.
eg: Cod liver oil, butter and cold cream.

Question 55.
What is emulsification?
Answer:
The process of preparation of emulsion by the dispersal of one liquid in another liquid is called Emulsification.

Question 56.
What is an emulsifier or emulsification agent?
Answer:
It is a substance added to have a stable emulsion. The type of emulsion formed depends upon the nature of the emulsifying agent. For example, the presence of soluble soaps as an emulsifying agent favours the formation of oil in water type of emulsions, whereas insoluble soaps (containing non alkali metal atoms) favours the formation of water in oil emulsions.

Question 57.
Mention the properties of the emulsion.
Answer:

1. Emulsions exhibit all the properties like the Tyndall effect, Brownian movement, Electrophoresis, Coagulation on the addition of electrolytes.
2. Emulsions can be separated into their constituent liquids by boiling, freezing, centrifuging, electrostatic precipitation by adding large amounts of the electrolytes to precipitate out the dispersed phase or by chemical destruction of emulsifying agent.
3. Emulsions can be diluted by adding any amount of dispersion medium.

Choose the correct answer:

1. Which of the following is true with respect of adsorption?
(a) ΔG < 0, ΔS > 0, ΔH < 0
(b) ΔG < 0, ΔS < 0, ΔH < 0
(c) ΔG > 0, ΔS > 0, ΔH < 0
(d) ΔG < 0, ΔS < 0, ΔH > 0
Answer:
(b)

2. Which of the following statements is incorrect regarding physisorption?
(a) It occurs because of vander Waals forces.
(b) More easily liquefiable gases are adsorbed readily.
(c) Under high pressure, it results in multimolecular layer on the adsorbent surface.
(d) Enthalpy of adsorption in low and positive.
Answer:
(d)
Hint: Enthalpy of adsorption is always negative.

3. Physical adsorption of a gaseous species may change to chemical adsorption with
(a) decrease in temperature
(b) increase in temperature
(c) increase in surface area of the adsorbent
(d) decrease in surface area of the adsorbent
Answer:
(b)
Hint: At high temperatures, gaseous species may dissociate into atoms that are chemisorbed, eg: N₂ an iron surface at ≥ 773 K.

4. In physisorption, adsorbent does not show specificity for any particular gas because:
(a) involved van der Waals forces are universal
(b) gases involved behave like ideal gases
(c) enthalpy of adsorption is low
(d) it is a reversible gas
Answer:
(a)

5. Which of the following is an example of adsorption?
(a) water on silica gel
(b) water on calcium carbide
(c) hydrogen on finely divided nickel
(d) oxygen on the metal surface
Answer:
(c)

6. On the basis of data given below predict which of the following gases show least adsorption on a definite amount of charcoal

(a) CO₂
(b) SO₂
(c) CH₄
(d) H₂
Answer:
(d)
Hint: Higher the critical temperature, the greater is the adsorption. H₂ has the least critical temperature and hence least adsorbed.

7. Which of the following are the characteristics of chemisorption?
(i) High heat of adsorption
(ii) Irreversibility
(iii) Low activation energy
(a) (i) and (ii) only
(b) (i) and (iii) only
(c) (ii) and (iii) only
(d) (i), (ii) and (iii)
Answer:
(a)

8. The plot of \(\frac{x}{m}\) (along Y-axis) versus log C (along X-axis) in the Freundlich adsorption isotherm is a horizontal line parallel to X-axis when,
(a) n = 0
(b) n = 1
(c) n = ∞
(d) such a plot is impossible
Answer:
(c)
Hint:

i.e., constant parallel to log C axis.
[Note: This is the equation for adsorption of solids in solution.]

9. Match the following:

Which of the following is the correct option?

Answer:
(b)

10. Which of the following will show the Tyndall effect?
(a) Aqueous solution of soap below critical miscelle concentration.
(b) Aqueous solution of soap above critical miscelle concentration.
(c) Aqueous solution of sodium chloride
(d) Aqueous solution of sugar
Answer:
(b)

11. Method by which lyophobic sol can be protected:
(a) addition of oppositely charged sol
(b) addition of an electrolyte
(c) addition of lyophilic sol
(d) by boiling
Answer:
(c)

12. Freshly prepared precipitate sometimes gets converted to colloidal solution by:
(a) coagulation
(b) electrolysis
(c) diffusion
(d) peptisation
Answer:
(d)

13. Which of the following electrolytes will have maximum coagulating value for AgI / Ag⁺ Sol?
(a) Na₂S
(b) Na₃PO₄
(c) Na₂SO₄
(d) NaCl
Answer:
(b)
Hint: AgI / Ag⁺ is a positively charged sol. Electrolytes whose negative ion (anion) has the least negative charge will require the maximum amount and will have the maximum coagulating value. Hence, Cl^– in NaCl.

14. A colloidal system having a solid substance as a dispersed phase and liquid as dispersion medium is classified as:
(a) solid sol
(b) gel
(c) emulsion
(d) sol
Answer:
(d)

15. Which of the following process is responsible for the formation of the delta at places where the river meets the sea?
(a) Emulsification
(b) Colloid formation
(c) Coagulation
(d) Peptisation
Answer:
(c)
Hint: Formation of the delta-shaped heap of sand clay etc where the river falls into the sea due to coagulation of sand/clay by NaCl present in seawater.

16. When an excess of a very dilute aqueous solution of KI is added to a very dilute aqueous solution of silver nitrate, the colloidal particles of silver iodide are associated with which of the following Helmholtz double layer?

Answer:
(d)
Hint:
As excess KI has been added, I^– ions are adsorbed on AgI forming a fixed layer (giving negative charge). It then attracts the counter ion (K⁺) from the medium forming a second layer, (diffused layer).

17. The ratio of a number of moles of AgNO₃, Pb(NO₃)₂ and Fe(NO₃)₃ required for coagulation of a definite amount of colloidal sol of silver iodide prepared by mixing AgNO₃ with an excess of KI will be:
(a) 1:2:3
(b) 3:2:1
(c) 6:3:2
(d) 2:3:6
Answer:
(c)
Hint:
With excess KI, colloidal particles will be [AgI]I^–.

Molar ratio required for coagulation of same amount of [AgI]I^– is 1 : \(\frac{1}{2}\) : \(\frac{1}{3}\) = 6 : 3 : 2.

18. Gelatin is mostly used in making ice creams in order to:
(a) prevent the formation of colloidal sol.
(b) enrich fragrance.
(c) prevent crystallisation and stabilise the mix.
(d) modify the taste.
Answer:
(c)
Hint: Gelatin is a protective colloid. It stabilises the mix and prevents crystallisation.

19. Which of the following has a minimum gold number?
(a) Starch
(b) Sodium oleate
(c) Gum arabic
(d) Gelatin
Answer:
(d)

20. Match the following:

Which of the following is the correct option?

Answer:
(a)

Students get through the TN Board 12th Chemistry Important Questions Chapter 9 Electro Chemistry which is useful for their exam preparation.

TN State Board 12th Chemistry Important Questions Chapter 9 Electro Chemistry

Answer the following questions.

Question 1.
Define the following terms:
(i) Resistivity,
(ii) conductivity,
(iii) cell constant,
(iv) specific conductance,
(iv) molar conductivity,
(vi) equivalent conductance.
Answer:
(i) Resistivity is defined as the resistance of an electrolyte confined between two electrodes having unit cross-sectional area and are separated by unit distance.
(ii) The reciprocal of specific resistance \(\left(\frac{1}{\rho}\right)\) is called specific conductance or conductivity.
(iii) The ratio between the distance between the two electrodes in a conductivity cell to their area of cross-section is called cell constant.
(iv) The specific conductance is defined as the conductance of a cube of an electrolytic solution of unit dimension.
(v) Molar conductance is defined by the conducting power of all the ions produced by 1 mole of the electrolyte.
(vi) Equivalent conductance is defined as the conductance of ‘V’ m³ of the electrolyte solution containing one gram equivalent of electrolyte in a conductivity cell in which the electrodes are 1 meter apart.

Question 2.
Give the relationship between resistance and specific resistance of an electrolytic conductor.
Answer:
R = ρ × \(\frac{l}{a}\)
where ‘R’ is the resistance, ρ is the specific resistance or resistivity, ‘l’ is the distance between the electrodes in a conductivity cell, and ‘a’ is the area of cross-section of electrodes.

Question 3.
Mention the units of (i) resistance, (ii) specific resistance, (iii) conductance, (iv) specific conductance, (v) molar conductance and (vi) equivalent conductance.
Answer:

Question 4.
Give the relationship between molar conductance and specific conductance of an electrolyte.
Answer:

Question 5.
Give the relationship between equivalent conductance and specific conductance of an electrolyte.
Answer:

Question 6.
Mention the factors which influence electrolytic conductance.
Answer:

1. A solvent of a higher dielectric constant shows high conductance in solution.
2. Conductance is inversely proportional to the viscosity of the medium, i.e., conductivity increases with the decrease in viscosity.
3. If the temperature of the electrolytic solution increases, conductance also increases. An increase in temperature increases the kinetic energy of the ions and decreases the attractive force between the oppositely charged ions and hence conductivity increases.
4. Molar conductance of a solution increases with an increase in dilution. This is because, for a strong electrolyte, interionic forces of attraction decrease with dilution. For a weak electrolyte, the degree of dissociation increases with dilution.

Question 7.
Explain how does the molar conductance of the electrolytes varies with the concentration of the electrolyte.
Answer:
The molar conductance of a strong electrolyte increases on dilution. The variation of molar conductance with dilution is given by an empirical formula

where Λ_m⁰ is called limiting molar conductivity and C is the concentration of the electrolyte.

For strong electrolytes such as KCl, NaCl, etc., the plot, Λ_m vs √C, gives a straight line as shown in the graph. It is also observed that the plot is not a linear one for weak electrolytes.
For strong electrolyte, at a high concentration, the number of constituent ions of the electrolyte in a given volume is high and hence the attractive force between the oppositely charged ions is also high. Moreover, the ions also experienced a viscous drag due to greater solvation. These factors attribute to the low molar conductivity at high concentrations. When the dilution increases, the ions are far apart and the attractive forces decrease. At infinite dilution the ions are so far apart, the interaction between them becomes insignificant, and hence, the molar conductivity increases and reaches a maximum value at infinite dilution.
For a weak electrolyte, at high concentration, the plot is almost parallel to the concentration axis with a slight increase in conductivity as the dilution increases. When the concentration approaches zero, there is a sudden increase in the molar conductance and the curve is almost parallel to Λ_m⁰ axis. This is due to the fact that the dissociation of the weak electrolyte increases with the increase in dilution (Ostwald dilution law). Λ_m⁰ values for strong electrolytes can be obtained by extrapolating the straight line. But the same procedure is not applicable for weak electrolytes, as the plot is not a linear one, A°m values of the weak electrolytes can be determined using Kohlraush’s law.

Question 8.
Flow much charge is required for the following reactions?
(i) 1 mole of Al³⁺ to Al
(ii) 1 mole of Cu²⁺ to Cu
(iii) 1 mole of MnO₄^– to Mn²⁺
Answer:
(i) Al³⁺ +3e → Al
Reduction of 1 mole of Al³⁺ requires 3 moles of electrons.
∴ Charge on 3 moles of electrons
= 3 × 96500 C
= 289500 coulomb
(ii) Cu²⁺ + 2e → Cu
Reduction of 1 mole of Cu²⁺ requires 2 mole of electrons
= 2 × 96500 C
= 193000 coulomb
(iii) MnO₄^– + 8H⁺ + 5e → Mn²⁺ + 4H₂O
Reduction of MnO₄^– ions require 5 mole of electrons
= 5 × 96500 C
= 482500 Coulomb

Question 9.
A solution of CuSO₄ is electrolyzed for 10 minutes with a current of 1.5 amperes. What mass of copper deposited at the cathode?
Answer:

∴ Mass of copper deposited at the cathode = 0.296 g

Question 10.
How many hours does it take to reduce 3 mole of Fe³⁺ to Fe²⁺. With 2 ampere of current? (F = 96500 coulomb)
Answer:
The required equation is Fe³⁺ + e → Fe²⁺
3 mole of Fe³⁺ requires 3 mole of electrons.
∴ Required charge = 3 × 96500 coloumbs
= 289500 coloumbs
charge = current × time
289500 = 2 × t
t = \(\frac{289500}{2}\)
= 1.475 × 10⁵ sec
= 40 hour.

Question 11.
In an electrolysis experiment, a current was passed for 5 hours through two cells connected in series. The first cell contains a solution of gold salt and the second cell contains copper sulfate solution.
If the oxidation number of gold is +3, find the amount of copper deposited on the cathode in the second cell. Also, calculate the magnitude of the current in amperes.
Answer:

Let Z be the electrochemical equivalent of copper

Question 12.
The specific conductance of a decinormal solution of KCl is 0.00112 ohm^-1 cm^-1. The resistance of a cell containing the solution was found to be 56 ohms. What is the value of the cell constant?
Answer:
Specific conductance = \(\frac{Cell constant}{Resistance}\)
Cell constant = Specific conductance × Resistance
= 0.0112 × 56
= 0.6272 cm^-1

Question 13.
1.0 N solution of salt surrounding two platinum electrodes 2.1 cm apart and 4.2 sq.cm in area was found to offer a resistance of 50 ohms. Calculate the equivalent conductivity of the solution.
Answer:
Given: l = 2.1 cm; a = 4.2 sq.cm; R = 50 ohm

Question 14.
The specific conductivity of 0.02 M KCl solution at 25° C is 2.768 × 10^-3 ohm^-1 cm^-1. The resistance of this solution at 25° C, when measured with a particular cell, was 250.2 ohm. The resistance of 0.01 M CuSO₄ solution at 25° C measured with the same cell was 8331 ohm. Calculate the molar conductivity of the copper sulfate solution.
Answer:
Cell constant = Specific conductivity of KCl × Resistance = 2.768 × 10^-3 × 250.2
For 0.01 M CuSO₄

Question 15.
At 291 K, the molar conductivities at infinite dilution of NH₄Cl, NH₄OH, and NaCl are 129.8, 217.8, and 108.9 S cm² respectively. If the molar conductivity of a cent normal solution of NH₄OH is 9.33 S cm², what is the percentage dissociation of NH₄OH at this dilution? Also, calculate the dissociation constant of NH₄OH.
Answer:

∴ degree of dissociation (α)

Calculation of dissociation constant

Question 16.
The conductivity of a saturated solution of AgCl at 288 K. is found to be 1.382 × 10^-6 Scm^-1. Find its solubility. Given ionic conductances of Ag⁺ and Cl^– ion at infinite dilution are 61.9 S cm² mol^-1 and 76.3 S cm² mol^-1 respectively.
Answer:

Question 17.
Define (i) electrode potential and (ii) emf of a cell.
Answer:
(i) Electrode potential: The potential difference set up between the metal and its ions in the solution is called the electrode potential, (or) electrode potential may be defined as the tendency of an electrode to lose or gain electrons when it is in contact with a solution of its own ions.
(ii) emf of a cell; The difference between the electrode potentials of the two half cells in a galvanic cell is known as the cell potential (or) It is the driving forces that permit the flow of electrons from anode to cathode.

Question 18.
Explain the terms (i) oxidation potential and (ii) reduction potential of an electrode.
Answer:
(i) Oxidation potential: when an electrode is negatively charged with respect to the solution, i.e., it acts as an anode, the potential difference is called oxidation potential.
M ⇌ Mⁿ⁺ + ne
It refers to the tendency for oxidation to occur at the electrode.
(ii) Reduction potential: When an electrode is positively charged with respect to solution, i.e., it acts as the cathode, the potential difference developed is called reduction potential
Mⁿ⁺ + ne ⇌ M
It refers to the tendency for reduction to occur at the electrode.

Question 19.
What is a reference electrode? Mention its use.
Answer:
A reference electrode is an electrode whose electrode potential value is known. It is used to determine, electrode potentials of various electrodes.

Question 20.
Mention the characteristics of electrode chemical series.
Answer:

1. The negative sign of the standard reduction potential indicates that the electrode when joined with SHE, acts as an anode and oxidation occurs at this electrode. Similarly, the sign of standard reduction potential indicates that the electrode when joined with SHE acts as and reduction occurs on this electrode.
2. Greater the value of standard reduction potential move easily the substance (element or ions) is reduced or in other words, stronger oxidizing agent it is. Similarly, the lower the value of standard reduction potential of the substance greater is the extent of oxidation half-reaction and the substance acts as a reducing agent.
3. Reactivity of metals: A metal that has a high negative value (or small positive value) of standard reduction potential loses electrons is said to be chemically active. The chemical reactivity decreases from top to bottom of the series.

Question 21.
Given the standard reduction potentials

Arrange these metals in their increasing order of reducing character.
Answer:
Ag < Hg < Cr < Mg < K.
The lesser the reduction potentials greater is their reducing character.

Question 22.
Write down the electrode reactions and the net cell reaction for the following cells. Which electrode would be the positive terminal in each cell.

Answer:

Question 23.
The standard EMF of the cell:
Ni | N₁²⁺ || Cu²⁺ | Cu is 0.59 V. The standard reduction potential of the copper electrode is 0.34 V. Calculate the standard reduction potential of the nickel electrode.
Answer:

Question 24.
A cell is set up between copper and silver electrodes as Cu | Cu²⁺ || Ag⁺ | Ag.
Given: . Calculate the emf of the cell.
Answer:

Question 25.
Two half cells are Al³⁺ | Al and Mg²⁺ | Mg. The reduction potentials of these half cells are -1.66 and -2.36 V. Construct the galvanic cell, write the cell reaction and calculate the emf of the cell.
Answer:
For the construction of a galvanic cell,
= (Electrode with lesser value of reduction potential – Electrode with higher value of reduction potential )

Question 26.
A cell is prepared by dipping a copper rod in 1M CuSO₄ solution and a nickel rod in 1M NiSO₄ solution. The standard reduction potentials of copper and nickel electrodes are 0.34 V and – 0.25 V respectively.
(a) How will you represent the cell?
(b) What will be the cell reaction?
(c) What will be the standard emf of the cell?
(d) Which electrode will be positive?
Answer:
The cell is represented as

(d) The cathode is a copper electrode. Because reduction occurs at this electrode.

Question 27.
Predict whether zinc and silver react with 1M H₂SO₄ to give hydrogen gas or not. Given the standard reduction potentials of zinc and silver electrodes are -0.76 and 0.80 V respectively.

Answer:
The cell is

The emf of the cell is positive. Hence the cell reaction is spontaneous. Thus, zinc will liberate hydrogen gas from 1M H₂SO₄.
Similarly,

The cell is

The emf of the cell is negative. Hence the reaction will not occur, i.e., Ag will not displace hydrogen from 1M H₂SO₄.

Question 28.
Can a nickel sulphate be used to stir a solution of copper sulphate. Support your answer with a reason.
[OR]
Can a solution of 1M copper sulphate be stored in a vessel made of nickel.
Given:

Answer:
The reaction,
Ni + CuSO₄ → NiSO₄ + Cu
should not occur, if CuSO₄ solution to be stored in a nickel vessel.
The cell would be Ni | Ni²⁺ || Cu²⁺ | Cu

The emf is positive. This means copper sulfate reacts with nickel, Hence CuSO₄ cannot be stored in a nickel vessel.

Question 29.
Give the relationship between the electrode potential and the concentration of the electrode.
Answer:
This relationship is given by Nernst equation. For an electrode reaction,
Mⁿ⁺ + ne → M
the electrode potential is given by the relationship,

Question 30.
For general cell reactions

Write the Nernst equation.
Answer:

where is the number of electrons involved in the cell reaction?

Question 31.
Calculate the electrode potential of copper electrode, when a copper wire is dipped in 0.1 M CuSO₄ solution E_Cu²⁺/Cu⁰ = 0.34.
Answer:
The electrode reaction is Cu²⁺ + 2e → Cu
Applying Nernst equation

[Note: Decreasing concentration of the electrolyte decreases the electrode potential]

Question 32.
The following reaction takes place in a galvanic cell.

Calculate the emf of the cell.
Answer:

Question 33.
Give the relationship between the free energy charge of the cell reaction and the emf of the cell.
Answer:
ΔG° = -n FE_cell

Question 34.
Give the relationship between the free energy change for the cell reaction and the equilibrium constant.
Answer:
ΔG° = – 2.303 RT log K_c

Question 35.
(a) Calculate the standard free energy change and maximum work obtainable by a galvanic cell, where the cell reaction

(b) Also calculate the equilibrium constant for the reaction.
Answer:
(a) The cell is

Thus the maximum work that can be obtainable from the cell = 212.3 kJ mol^-1

Question 36.
Describe the nature of anode, cathode, the cell reaction, and salient features of the Leclanche cell.
Answer:
Anode: Zinc container
Cathode: Graphite rod in contact with Mn02
Electrolyte: ammonium chloride and zinc chloride in water.
Emf of the cell is about 1.5 V.
Cell reaction:
Oxidation at anode:
Zn(s) → Zn²⁺(aq) + 2e^– … (1)
Reduction at cathode:
2NH₄⁺(aq) + 2e^– → 2NH₃(aq) + H₂(g) … (2)
The hydrogen gas is oxidised to water by MnO₂
H₂(g) + 2MnO₂(s) → Mn₂O₃(s) + H₂O (l) …(3)
Equation (1) + (2) + (3) gives the overall redox reaction
Zn(s) + 2NH₄⁺ (aq) + 2MnO₂(s) → Zn²⁺(aq) + Mn₂O₃(s) + H₂O (l) + 2NH₃ ….. (4)
The ammonia produced at the cathode combines with Zn²⁺ to form a complex ion [Zn(NH₃)₄]²⁺(aq). As the reaction proceeds, the concentration of NH₄⁺ will decrease and the aqueous NH₃ will increase which lead to the decrease in the emf of cell.

Question 37.
Describe the nature of anode, cathode the cell reaction, and salient features of the mercury button cell.
Answer:
Anode: Zinc amalgamated with mercury.
Cathode: HgO mixed with graphite.
Electrolyte: Paste of KOH and ZnO.

Cell emf: about 1.35V.
Uses: It has a higher capacity and longer life. Used in pacemakers, electronic watches, cameras etc…

Question 38.
Explain the functioning of lead storage batteries.
[OR]
Explain the chemical reaction involved in the process of charging and recharging in a lead storage battery.
Answer:
Anode: Spongy lead
Cathode: Lead plate bearing PbO₂
Electrolyte: 38% by mass of H₂SO₄ with density 1.2g / mL.
Oxidation occurs at the anode:
Pb(s) → Pb²⁺(aq) + 2e^– … (1)
The Pb²⁺ ions combine with SO₄^-2 to from PbSO₄ precipitate.
Pb²⁺(aq) + SO₄² (aq) → PbSO₄(s) …….. (2)
Reduction occurs at the cathode:
PbO₂(s) + 4H⁺(aq) + 2e^– → Pb²⁺(aq) + 2H₂O (l) ……. (3)
The Pb²⁺ ions also combine with SO₄^2- ions from sulphuric acid to form PbS04 precipitate.
Pb²⁺(aq) + SO₄^2-(aq) → PbSO …… (4)
Overall reaction:
Equation (1) + (2) + (3) + (4) ⇒
Pb(s) + PbO₂ (s) + 4H⁺(aq) + 2SO₄^2-(aq) → 2PbSO₄(s) + 2H₂O (l)
The emf of a single cell is about 2 V. Usually six such cells are combined in series to produced 12 volt.
The emf of the cell depends on the concentration of H₂SO₄. AS the cell reaction uses SO₂² ions, the concentration H₂SO₄ decreases. When the cell potential falls to about 1.8V, the cell has to be recharged.

Question 39.
Discuss the nature of the anode, cathode, and the cell reaction of the Lithium-ion battery. How does it act as a secondary cell?
Answer:
Anode: Porus graphite.
Cathode: transition metal oxide such as CoO₂. Electrolyte: Lithium salt in an organic solvent At the anode oxidation occurs:
Li(s) → Li⁺(aq) + e^–
At the cathode reduction occurs:
Li⁺ + CoO₂ (s) + e^– → Li CoO₂ (s)
Overall reactions:
Li(s) + CoO₂ → Li CoO₂ (s)
Both electrodes allow Li⁺ ions to move in and out of their structures.
During discharge, the Li⁺ ions produced at the anode move towards the cathode through the non-aqueous electrolyte. When a potential greater than the emf produced by the cell is applied across the electrode, the cell reaction is reversed and now the Li⁺ ions move from cathode to anode where they become embedded on the porous electrode. This is known as intercalation.
Uses: Used in cellular phones, laptop computer digital camera, etc…

Question 40.
Mention the various methods of protecting metals from corrosion.
Answer:
This can be achieved by the following methods.

1. Coating metal surface by paint.
2. Galvanizing by coating with another metal such as zinc, zinc is a stronger oxidizing agent than iron and hence it can be more easily corroded than iron, i.e., instead of iron, the zinc is oxidized.
3. Catholic protection: In this technique, unlike galvanizing the entire surface of the metal to be protected need not be covered with a protecting metal instead, metals such as Mg or zinc which is corroded more easily than iron can be used as a sacrificial anode and the iron material acts as a cathode. So iron is protected, but Mg or Zn is corroded.

Passivation: The metal is treated with strong oxidizing agents such as concentrated HNO₃. As a result, a protective oxide layer is formed on the surface of the metal.
Alloy formation: The oxidizing tendency of iron can be reduced by forming its alloy with other more anodic metals.
eg: stainless steel – an alloy of Fe and Cr.

Choose the correct answer:

1. In the electrolytic cell, the flow of electrons is from:
(a) cathode to anode in the solution
(b) cathode to the anode through external supply
(c) cathode to the anode through the internal supply
(d) anode to the cathode through the internal supply
Answer:
(b)
Hint: In electrolytic cells, electrons enter the cathode and leave at the anode. Thus in the external supply, electrons flow from cathode to anode. In the solution, only ions flow and not the electrons.

2. Which of the statements about solutions of electrolytes is not correct?
(a) the conductivity of the solution depends upon the size of the ions.
(b) conductivity depends upon the viscosity of the solution.
(c) conductivity does not depend upon the solvation of ions present in the solution.
(d) the conductivity of the solution increases with temperature.
Answer:
(c)
Hint: Greater the solvation of ions, lesser is the conductivity. Hence, (c) is an incorrect statement.

3. A dilute solution of Na₂SO₄ is electrolyzed using platinum electrodes. The products at the cathode and anode are:
(a) O₂ and H₂
(b) SO₂, Na
(c) O₂, Na
(d) S₂O₈^2-, H₂
Answer:
(a)
Hint: H20 is more readily reduced than Na+. It is more readily oxidised than S04 2. Hence, the electrode reactions are:
at cathode: 2H₂O + 2e → H₂ + 2OH^–
at anode: 2H₂O → O₂ + 4H⁺ + 4e.

4. The quantity of charge required to obtain one mole of aluminum from Al₂O₃ is:
(a) 1F
(b) 6F
(c) 3F
(d) 2F
Answer:
(c)
Hint: Al₂O₃ → 2Al.
i.e., 2Al₂O₃ + 6e → 2Al or
Al₂O₃ + 3e → Al.
Hence, to obtain one mol of Al, the charge required is 3F.

5. Electrolysis of dilute aqueous solution of NaCl solution was carried out by passing 10 milliampere current. The time required to liberate 0.01 mole of H₂ gas at the cathode is (1 Faraday = 96500 C mol^-1):
(a) 9.65 × 10⁴ sec
(b) 19.3 × 10⁴ sec
(c) 28.95 × 10⁴ sec
(d) 38.6 × 10⁴ sec
Answer:
(b)
Hint:

Thus,
0.5 mol of H₂ is liberated by IF = 96500 C 0.01 mol of H₂ will be liberated by charge

6. Al₂O₃ is reduced by electrolysis at low potentials and high currents. If 4.0 × 10⁴ amperes of current is passed through molten Al₂O₃ for 6 hours, what mass of aluminium is produced? (Assume 100% current efficiency. At mass of Al = 27 g mol^-1):
(a) 8.1 × 10⁴ g
(b) 2.4 × 10⁵ g
(c) 1.3 × 10⁴ g
(d) 9.0 × 10³ g
Answer:
(a)
Hint: Al³⁺ + 3e → Al
1 mol of Al = 27 g of Al is deposited by 3F or 3 × 96500 C of electricity.
Quantity of electricity passed

7. What current is to be passed for 0.02 A for deposition of a certain weight of metal which is equal to electrochemical equivalent?
(a) 4 A
(b) 100 A
(c) 200 A
(d) 2 A
Answer:
(a)
Hint: Electrochemical equivalent is the weight deposited by 1 coulomb.
Q = I × t
1 = I × 0.25 or I = 4A

8. Acertain current liberates 0.504 g of hydrogen in 2 hours. How many grams of oxygen can be liberated by the same current in the same time?
(a) 2.0 g
(b) 0.4 g
(c) 4.0 g
(d) 8.0 g
Answer:
(c)
Hint: H₂O → H₂ + \(\frac{1}{2}\) O₂
Thus if one mole of H₂ is liberated, O₂ liberated = \(\frac{1}{2}\) mole

9. Resistance of 0.2 M solution of an electrolyte is 50Ω. The specific conductance of the solution is 1.3 Sm^-1. If the resistance of 0.4M solution of the same electrolyte is 260Ω the molar conductivity is:
(a) 6250 Sm² mol^-1
(b) 6.25 × 10^-4 Sm² mol^-1
(c) 625 × 10^-4 Sm² mol^-1
(d) 62.5 Sm² mol^-1
Answer:
(b)
Hint:
specific conductance = observed conductance × cell constant
K = \(\frac{1}{R}\) × cell constant
or cell constant = K × R
= 1.3 Sm^-1 × 50Ω
= 65 m^-1
For 0.4 M solution, specific conductance

10. An increase in equivalent conductance of a strong electrolyte with dilution is mainly due to:
(a) increase in the number of ions
(b) increase in ionic mobility
(c) 100% dissociation of electrolyte at normal dilution
(d) increase in both i.e., number of ions and ionic mobility of ion.
Answer:
(b)
Hint: A strong electrolyte is completely ionized at all concentrations. Hence, the number of ions remains the same. However, on dilution, interionic forces decrease, and hence ionic mobility increases. Therefore, equivalent conductance increases.

11. In the electrolysis of which solution, OH^– ions are discharged in preference to Cl^– ions?
(a) dilute NaCl
(b) very dilute NaCl
(c) fused NaCl
(d) solid NaCl
Answer:
(b)
Hint: In very dilute NaCl solution, the following reactions take place on electrolysis.

12. The charge in coulombs on lg ion of N^-3 is:
(a) 28.9 × 10⁵ C
(b) 2.89 × 10⁵ C
(c) 2890 C
(d) 28900 C
Answer:
(b)
Hint: Charge on 1 gm ion of N^-3
= 3 × 1.6 × 10^-19 C
One g ion = 6.023 × 10²³ ion
Charge on 1 gm ion of N^-3
= 3 × 1.6 × 10^-19 × 6.023 × 10²³
= 2.89 × 10⁵ C

13. A solution of copper sulfate is electrolyzed for 10 minutes with a current of 1.5 amperes. The mass of copper deposited at the cathode is: (atomic mass of Cu = 63.5 g mol^-1)
(a) 29.6 g
(b) 0.296 g
(c) 2.96 g
(d) 1.564 g
Answer:
(b)
Hint: Current = 1.5 amperes
Time = 10 minutes
= 10 × 60 = 600 s
Quantity of electricity passed = i × t
= 1.5 × 600 = 900 C
The reaction occurring at cathode is
Cu²⁺ + 2e → Cu
i.e., 2F or 2 × 96500 C deposit
= 1 mol = 63.5 g of Cu
900 C will deposit = \(\frac{63.5}{2 \times 96500}\) × 900 = 0.296 g

14. The limiting molar conductance of HCl, CH₃COONa, and NaCl are respectively 425, 90, and 125 mho cm² mol^-1 at 25° C. The molar conductivity of 0.1 M CH₃COOH solution is 7.8 mho cm² mol^-1 at the same temperature. The degree of dissociation of 0.1 M acetic acid at this temperature is:
(a) 0.1
(b) 0.02
(c) 0.15
(d) 0.03
Answer:
(b)

15. Degree of dissociation of pure water is 1.9 × 10^-9. Molar conductances of H⁺ and OH^– ions at infinite dilution are 200 S cm² mol^-1 and 350 Scm² mol^-1 respectively.
(a) 3.8 × 10^-7 Scm² mol^-1
(b) 5.7 × 10^-7 Scm² mol^-1
(c) 9.5 × 10^-7 Scm² mol^-1
(d) 1.045 × 10^-6 Scm² mol^-1
Answer:
(b)
Hint:

16. When measured against a standard calomel electrode, an electrode is found to have a standard reduction potential of 0.100 V. If the standard reduction potential of the calomel electrode is 0.244 V, the standard reduction potential of the same electrode against the standard hydrogen electrode will be:
(a) – 0.144 V
(b) + 0.100 V
(c) – 0.344 V
(d) – 0.100 V
Answer:
(b)
Hint: Standard electrode potential is a fixed quantity for a given electrode. Hence, its standard electrode potential against SHE will remain the same. Viz., (0.100 V). (emf of the cell formed will change)

17. On the basis of the following E° values, the strongest oxidising agent is:


Answer:
(c)
Hint: The strongest oxidizing agent is a substance that is reduced move easily, i.e., it has the highest reduction potential. Reduction potentials of [Fe(CN)₆]^-3 and Fe³⁺ are 0.355 V and 0.77 V. Hence, Fe³⁺ is reduced more easily and hence is the strongest oxidizing agent.

18. Standard electrode potentials for Sn⁴⁺ / Sn²⁺ couple is +0.15 V and that for Cr³⁺/Cr couple is – 0.74 V. These two couples in their standard states are connected to make a cell. The cell potential will be:
(a) +1.83 V
(b) +1.19 V
(c) +0.89 V
(d) +0.18 V
Answer:
(c)
Hint:

19. Cr₂O₇^2- + I^– → I₂ + Cr³⁺;

(a) 0.54 V
(b) – 0.54 V
(c) +0.18 V
(d) -0.18 V
Answer:
(a)
Hint: In the given reaction, I^– is oxidized to I₂ and Cr₂O₇²⁺ ions have been reduced to Cr³⁺. The cell is

20. Given the standard electrode potentials K⁺/K = -2.93 V; Ag⁺ / Ag = 0.80 V; Mg₂²⁺ / Hg = 0.79; Mg²⁺/Mg = -2.37 V; Cr²⁺/Cr = -0.74 V.
Arrange these metals in their increasing order of reducing power.
(a) Ag < Hg < Cr < Mg < K
(b) K < Mg < Cr < Hg < Ag
(c) Hg < Cr < Mg < K < Ag
(d) Mg < Cr < Hg < Ag < K
Answer:
(a)
Hint: Lower the reduction potential, more easily it is oxidized and hence greater is its reducing power.

21. The reduction potential of hydrogen half cell will be negative if:
(a) pH₂ = 1 atm and [H⁺] = 1.0 M
(b) pH₂ = 2 atm and [H⁺] = 1.0 M
(c) pH₂ = 2 atm and [H⁺] = 2.0 M
(d) pH₂ = 1 atm and [H⁺] = 2.0 M
Answer:
(b)
Hint: For hydrogen electrode,

22. 2Hg → Hg²⁺; E° = 0.855 V
Hg → Hg²⁺; E° = 0.799 V
Equilibrium constant for the reaction
Hg + Hg²⁺ ⇌ Hg₂²⁺ at 27° C is:
(a) 89
(b) 82.3
(c) 79
(d) none of these
Answer:
(c)
Hint:
E° for the reaction = 0.855 – 0.799 = 0.056 V

23. Standard free energies of formation of (in kJ mol^-1) at 298 K are -237.3, -394.2, and -8.2 for H₂O(l), CO₂(g), and pentane (g) respectively. The value of E°ell for the pentane-oxygen fuel cell is:
(a) 1.968 V
(b) 2.0968 V
(c) 1.0968 V
(d) 0.968 V
Answer:
(c)
Hint: The balanced equation for pentane oxygen cell reaction will be:

Students get through the TN Board 12th Chemistry Important Questions Chapter 8 Ionic Equilibrium which is useful for their exam preparation.

TN State Board 12th Chemistry Important Questions Chapter 8 Ionic Equilibrium

Answer the following questions.

Question 1.
State the formula and the name of the conjugate base of each of the following acids.
(a) H₃O⁺,
(b) HSO₄^–,
(c) HF,
(d) NH₄⁺,
(e) CH₃NH₃⁺,
(f) CH₃PO₄,
(g) CH₃COOH,
(h) H₂PO₄^–,
(i) HS^–,
(j) HCO₃^–.
Answer:
(a) H₂O – water
(b) SO₄^-2 – sulphate ion
(c) F^– – fluoride ion
(d) NH₃ – ammonia
(e) CH₃NH₂ – methyl amine
(f) H₂PO₄^– – dihydrogen phosphate ion
(g) CH₃COO^– – acetate ion
(h) HPO₄^-2 – hydrogen phosphate ion
(i) S^-2 – sulphide ion
(j) CO₃^-2 – carbonate ion

Question 2.
State the formula and name of the conjugate acid of each of the following:
(a) OH^–,
(b) CH₃COOH,
(c) HPO₄^-2,
(d) CO₃^-2,
(e) NH₃,
(f) CH₃NH₂,
(g) HS^–,
(h) CH₃COO^–
(i) H₂PO₄^–,
(j) Cl^–.
Answer:
(a) H₃O – water
(b) CH₃COOH₂⁺ – protonated acetic acid
(c) H₂PO₄^– – dihydrogen phosphate ion
(d) HCO₃^– – bicarbonate ion
(e) NH₄⁺ – ammonium ion
(f) CH₃NH₃⁺ – protonated methyl amine
(g) H₂S – hydrogen sulphide
(h) CH₃COO – acetic acid
(i) H₃PO₄ – phosphoric acid
(j) HCl – hydrochloric acid

Question 3.
Which of the following behave both as Bronsted acids as well as Bronsted bases?
NH₃, HSO₄^–, H₂SO₄, HCO₃^–; H₃PO₄, HS^–, H₂O.
Answer:
NH₃, HSO₄^– , HCO₃^– ; HS^– and H₂O behave both as Bronsted acids and Bronsted bases.

Question 4.
Classify Lewis acids and bases from the following:
H₃O⁺, CH₃NH₂, BF₃, OH^–, Cu²⁺, S^2-, CH₃COO^–.
Answer:
Lewis acids: H₃O⁺, BF₃, Cu²⁺.
Lewis bases: CH₃NH₂, OH^–, S^2-, CH₃COO^–.

Question 5.
Classify the following species into Lewis acids and Lewis bases and show how these act as Lewis acid/base.
(a) OH^–, (b) F^–, (c) H⁺, (d) BCl₃
Answer:
(a) OH^– is a Lewis base because it acts as an electron-pair donor.
(b) F^– is also a Lewis base.
(c) H⁺ is a Lewis acid because it acts as an electron pair acceptor.
(d) BCl₃ is an electron-deficient compound electron pair acceptor, hence Lewis acid.

Question 6.
Explain how the [H⁺] concentration or [OH^–] concentration, decides a solution as acidic, neutral, or alkaline.
Answer:

Question 7.
Calculate the hydrogen and hydroxyl ion concentrations in (i) 0.01M HNO₃, (ii) 0.001M KOH solution at 298 K.
Answer:
(i) HNO₃ is a strong acid. It ionizes as

(ii) KOH is a strong base. It ionizes as

Question 8.
Assuming complete dissociation, calculate the pH of the following solutions:
(i) 0.003M HCl,
(ii) 0.005M NaOH,
(iii) 0.002M HBr,
(iv) 0.002M KOH.
Answer:
(i) HCl → H⁺ + Cl^–
[H⁺] = [HCl] = 0.003M
pH = – log [H⁺]
= – log [0.003] = 2.52

(ii) NaOH → Na⁺ + OH^–
[OH^–] = [NaOH] = 0.005M
pOH = – log[OH^–]
= – log [0.005] = 2.301
pH = 14 – pOH
= 14 – 2.301 = 11.699

(iii) HBr → H⁺ + Br^–
[H⁺] = [HBr]
= 0.002M
pH = – log [H⁺] (or)
pH = – log [0.002] = 2.6989

(iv) KOH → K⁺ + OH^–
[OH^–] = [KOH]
= 0.002M
pOH = – log [OH ]
pH = – log (0.002) = 2.6929
pH = 14 – pOH (or)
= 14 – 2.6929 = 11.3011

Question 9.
Calculate the pH of the resulting mixtures: 10 ml of 0.02M Ca(OH)₂ + 25 ml of 0.1M HCl.
Answer:

Number of moles of OH^– ion = 2 x 0.002 = 0.004
Number of moles of H⁺ ion = 0.0025
Number of moles of OH^– remaining after neutralisation = 0.004 – 0.0023 = 0.0015
Molarity of OH^– ions

Question 10.
The concentration of H⁺ ions in 0.1 M solution of a weak acid is 1.0 x 10^-5 mol L^-1. Calculate the dissociation constant of the acid.
Answer:

[HA] can be taken as 0.1 as 1 x 10^-5 is very small. Hence,

Question 11.
Find whether the resulting solution is acidic, neutral, or basic when
(i) a strong acid is mixed with a strong base
(ii) a strong acid is mixed with a weak base
(iii) a weak acid is mixed with a strong base
(iv) a weak acid is mixed with a weak base.
Answer:
(i) A strong acid and a strong base give a neutral solution because both are completely ionized and the reaction goes on completion.
eg:

(ii) A strong acid and a weak base give an acidic solution, as the weak acid is hot completely ionized. The reaction does not go for completion and there is an excess of hydrogen ions in the solution.
eg:

The solution is therefore is acidic.
(iii) A weak acid and a strong base give a basic solution, as the weak acid is not completely ionized. The reaction does not go for completion and there is an excess of hydroxyl ions in the solution.

Hence the solution is basic.
(iv) A weak acid and a weak base give an acidic, neutral, or basic depending on the relative strength of acid or base. If both weak acid and weak base have equal strength, the resulting solution is neutral.

Question 12.
State Ostwald’s dilution law.
Answer:
When dilution increases, the degree of dissociation of weak electrolytes also increases.
The mathematical form is
α = \(\sqrt{\frac{\mathrm{K}_{a}}{\mathrm{C}}}\)
where ‘α’ is the degree of dissociation, K_a is the dissociation constant of the acid and ‘C’ is its concentration.

Question 13.
Explain the term ‘common ion effect’ with an example.
Answer:
The suppression of the dissociation of a weak electrolyte by the addition of an ion in common with that of the weak electrolyte is known as the common ion effect.
Consider the dissociation of acetic acid in the presence of sodium acetate or HCl.
In the absence of any added salt, the dissociation of acetic acid is represented as

The addition of acetate ions in the form of sodium acetate increases the concentration of acetate ions. This increases the value of K_a. But K_a is a constant at a given temperature. Thus to maintain the K_a value constant, H⁺ ions combine with acetate ion to form undissociated acetic acid. i.e., The degree of dissociation of acetic acid in the presence of acetate ion is less than in the absence of sodium acetate. The same happens in the presence of HCl.

Question 14.
What are buffer solutions? Give example.
Answer:

1. Buffer solutions resist changes in pH by the addition of small quantities of an acid or a base. This ability is known as buffer action.
2. Buffer solutions consist of a mixture of a weak acid and a salt of a weak acid and a strong base.
   eg: CH₃COOH + CH₃COONa. This type of buffer is known as an acid buffer.
3. Buffer solution which consists of a mixture of a weak base and a salt of a weak base and strong acid is known as a basic buffer.
   eg: NH₄OH and NH₄Cl.

Question 15.
Give the characteristics of a buffer solution.
Answer:

1. It should have a definite pH. i.e., it has reserve acidity or alkalinity.
2. Its pH does not change on standing.
3. Its pH does not change on dilution.
4. Its pH is only slightly changed by the addition of a small quantity of acid or base.

Question 16.
Give examples of acidic and basic buffers.
Answer:
Acidic buffer:

1. CH₃COOH + CH₃COONa
2. Boric acid and borax.
3. Pthalic acid and potassium acid pthalate.

Basic buffer:

1. NH₄OH + NH₄Cl
2. Glycine and glycine hydrochloride.
3. Aniline and aniline hydrochloride.

Question 17.
Explain buffer action with a suitable example.
Answer:
The buffer action in a solution containing CH₃COOH and CH₃COONa is explained as follows:
The dissociation of the buffer components occurs as below.


If an acid is added to this mixture, it will be consumed by the conjugate base CH₃COO^– to form the undissociated weak acid i.e., the increase in the concentration of H⁺ does not reduce the pH significantly.

If a base is added, it will be neutralized by H₃O⁺, and the acetic acid is dissociated to maintain the equilibrium. Hence the pH is not significantly altered.

Question 18.
Explain the terms ‘buffer capacity” and ‘buffer index”.
Answer:
The buffer capacity is the ability of the buffer to resist changes in pH by the addition of small quantities of acid or base. It is measured in terms of buffer capacity. The quantitative measure of buffer capacity is the buffer index. It is defined as the number of equivalents of acid or base to one liter of the buffer solution to change its pH by unity.
β = \(\frac{d \mathrm{~B}}{d(\mathrm{pH})}\)
where P = buffer index, dB = number of gram equivalent of acid/base added to one liter of buffer solution, d(pH) = The change in pH after the addition of an acid or base.

Question 19.
Derive Henderson equation for the determination of pH of an acid buffer.
Answer:
An acid buffer consists of a mixture of a weak acid and salt with strong acid. The ionization of weak acid, HA is given by
HA ⇌ H⁺ + A^–
The dissociation constant of the weak acid is given by,

It can be assumed that the concentration of [A^–] ions from the complete ionization of the salt compared to the ionization of the weak acid. Hence, [A^–] concentration may be considered as the concentration of the salt.
So,

Taking logarithms on both sides and reversing the sign,

This is known as Henderson’s equation.

Question 20.
Write Henderson’s equation for a basic buffer.
Answer:

Question 21.
Suppose it is required to make a buffer solution of pH = 4. Using acetic acid and sodium acetate. How much sodium acetate to be added to 1 liter of N/10 acetic acid? The dissociation constant of acetic acid is 1.8 × 10^-5.
Answer:
Applying Henderson’s equation,
The molecular mass of CH₃COONa = 82
Amount of salt = 0.018 × 82 = 1.476 g.

Question 22.
Calculate the pH of a buffer solution containing 0.15 moles of NH₄OH and 0.25 moles of NH₄Cl. (K_b for NH₄OH = 1.8 × 10^-5)
Answer:

Question 23.
Calculate the pH of 0.1M ammonia solution. Calculate the pH after 50 ml of this solution is treated with 25 ml of 0.1M HCl. The dissociation constant of ammonia, K_b = 1.77 × 10^-5.
Answer:
For NH₄OH

When NH₃ is added, neutralization occurs

(M_NH4Cl) number of mole of NH₄Cl = 0.0025
(M_NH4OH) Remaining mole of NH₄OH = 0.005 – 0.0025 = 0.0025
According to Henderson’s equation,

Question 24.
Define hydrolysis.
Answer:
The reaction between cation or anion of the salt with water to produce the acid and base from which the salt formed is known as hydrolysis.
salt + H₂O = acid + base

Question 25.
Define the term degree of hydrolysis.
Answer:
It is the fraction of one mole of the salt that undergoes hydrolysis is known as the degree of hydrolysis.

Question 26.
Define the term hydrolysis constant (K_h), with an example.
Answer:
The equilibrium constant for the reaction, salt + H₂O ⇌ acid + base, is known as the hydrolysis constant (K_h). Consider the hydrolysis of acetate ion.

Question 27.
The pK_a of acetic acid and pK_b of ammonium hydroxide are 4.76 and 4.75 respectively. Calculate the pH of the ammonium acetate solution.
Answer:
Ammonium acetate is the salt of a weak acid and weak base.

Question 28.
The ionization constant of nitrous acid is 4.5 × 10^-4. Calculate the pH of 0.04 M sodium nitrite solution and its degree of hydrolysis.
Answer:
Sodium nitrite is a salt of a weak acid and a strong base.

Substituting these values in equation (1)
pH = \(\frac{1}{2}\) [14 + 3.346 – 1.3979]
= 7.974
Degree of hydrolysis (h)

Question 29.
Calculate the degree of hydrolysis and pH of a 0.1M sodium acetate solution. The hydrolysis constant of sodium acetate is 5.6 × 10^-10.
Answer:

Question 30.
Calculate the pH of an aqueous solution of 1.0M ammonium formate assuming complete dissociation. pK_a for formic acid is 3.8 and pK_b of ammonia = 4.8.
Answer:
Ammonium formate is a salt of a weak acid and weak base.

Question 31.
A 0.02 M solution of pyridinium hydrochloride has pH = 3.44. Calculate the ionization constant of pyridine.
Answer:
Pyridinium hydrochloride is a salt of a weak base and strong acid.

Question 32.
Define solubility product.
Answer:
It is defined as the product of the concentration of ions in a saturated solution of an electrolyte at a given temperature.

Question 33.
Give expressions for the solubility product of the following: (i) BaSO₄, (ii) H₂S, (iii) Sb₂S₃, (iv) AlI₃.
Answer:

Question 34.
Mention the criteria of precipitation of an electrolyte.
Answer:
The ionic product should exceed the value of the solubility product, for precipitation to occur.

Question 35.
How will you decide whether a solution is saturated or unsaturated, from the ionic product values?
Answer:

1. When the ionic product is less than the solubility product, the solution is unsaturated.
2. When an ionic product is equal to the solubility product, the solution is saturated.

Question 36.
Obtain a relationship between solubility and solubility product of the following: (i) AgCl, (ii) Ag₂CO₃, (iii) CaF₂, (iv) Cr(OH)₃.
Answer:

Question 37.
Equal volumes of 0.002 M solutions of sodium iodate and cupric chlorate are mixed together. Will it lead to the precipitation of copper iodate? K_s for cupric iodate = 7.4 x 10^-8.
Answer:
Given, [NaIO₃] = 0.002 M
[Cu(ClO₃)₂] = 0.002 M
∴ [IO₃^–] = 0.002 M
[Cu⁺²] = 0.002 M
When equal volumes of these solutions are mixed, their concentrations will be halved.
i.e., [IO₃^–] = 0.001 M
[Cu⁺²] = 0.001 M
Cu(IO₃)₂ ⇌ Cu⁺² + 2IO₃^–
K_s = [Cu⁺²][IO₃^–]²
= 0.001 x (0.001)2 = 10^-9
Since IP is less than solubility products precipitation will not occur.

Question 38.
What is the minimum volume of water required to dissolve 1 gm of calcium sulfate at 298 K. The K_s for calcium sulfate is 9.1 x 10^-6.
Answer:
Let the solubility of CaSO₄ at 298 K be ‘s’ mol L^-1.

Minimum volume of water required to dissolve
1 g of CaSO₄ = \(\frac{1000}{0.41}\)
= 2439 mL = 2.439 L

Question 39.
The solubility product of BaS04 is 1.5 x 10^-9 Find its solubility (i) in pure water and (ii) in 0.1M BaCl₂, solution.
Answer:
(i)

Let ‘s’ be the solubility in mol L^-1. Then
K_s = s²
(or) 1.5 x 10⁷ = s²
s = \(\sqrt{1.5 \times 10^{-9}}\)
= 3.87 x 10^-5 mol L⁷

(ii) Let (s’) be the solubility of BaSO₄ in 0.1M BaCl₂ solution.
Total [Ba²⁺] = [Ba²⁺] from BaSO₄ from [Ba²⁺] from BaCl₂.

Question 40.
Calculate the pH at which Mg(OH)₂ begins to precipitate from a solution containing 0.10 Mg²⁺ ions.
Answer:

Choose the correct answer:

1. What is the conjugate base of OH?
(a) O₂
(b) H₂O
(c) O^–
(d) O^2-
Answer:
(d)
Hint:

2. C₂H₅ONa is …………. of C₂H₅OH.
(a) strong acid
(b) weak acid
(c) strong base
(d) weak base
Answer:
(c)
Hint: C₂H₅OH is a strong base. a weak acid, C₂H₅O^– is a strong base

3. Among the following, the one which can act as Bronsted acid as well as Bronsted base is:
(a) H₃PO₄
(b) AlCl₃
(c) CH₃COO^–
(d) H₂O
Answer:
(d)
Hint:

4. At 25° C, the dissociation constant of a base, BOH is 1.0 x 10^-12. The concentration of hydroxyl ions in 0.01M aqueous solution of the base would be:
(a) 1.0 x 10^-6 mol lit^-1
(b) 1.0 x 10^-7 mol lit^-1
(c) 2.0 x 10^-6 mol lit^-1
(d) 1.0 x 10^-5 mol lit^-1
Answer:
(b)
Hint:

5. Which of the following is the correct statement?
(a) HCO₃^– is the conjugate base of CO₃^2-
(b) NH₂^– is the conjugate acid of NH₃
(c) H₂SO₄ is the conjugate acid of HSO₄^–
(d) NH₃ is the conjugate base of NH₂^–
Answer:
(c)
Hint:

6. Three reactions involving H₂PO₄^– are given below:

In which of the above H₂PO₄^– act as an acid?
(a) (iii) only
(b) (i) only
(c) (ii) only
(d) (i) and (ii)
Answer:
(c)

7. Which one of the following will decrease the pH of 50 ml of 0.01M hydrochloric acid?
(a) Addition of 50 ml of 0.01 M HCl
(b) Addition of 50 ml of 0.002 M HCl
(c) Addition of 150 ml of 0.002 M HCl
(d) Addition of 5 ml of 1 M HCl
Answer:
(d)
Hint: For 0.001 M HCl, [H⁺] = 10^-2M; pH = 2

8. If pK_a for fluoride ion at 25° C, is 10.83, the ionisation constant of hydrofluoric acid at this temperature is:
(a) 1.74 x 10^-5
(b) 3.52 x 10^-3
(c) 6.75 x 10^-4
(d) 5.38 x 10^-2
Answer:
(c)
Hint:
pK_a + pK_b = 14
pK_a = 14 – 10.83 = 3.17
pK_a= – log Ka = – log 3.17
log K_a= -3.17 = 4.83
(or) K_a= 6.76 x 10^-4

9. The pH of a solution obtained by mixing 100 ml of a solution of pH = 3 with 400 ml of a solution of pH = 4 is:
(a) 3 – log 2.8
(b) 1 – log 2.8
(c) 4 – log 2.8
(d) 5 – log 2.8
Answer:
(c)
Hint:

10. The pH of 0.1M aqueous solution of a weak acid, HA is 3. What is its degree of dissociation?
(a) 1%
(b) 10%
(c) 50%
(d) 25%
Answer:
(a)
Hint: HA ⇌ H⁺ + A^–
[H⁺] = Cα
0.1 x α = 10^-3
α = 10^-2 i.e., 1%

11. The pK_a of acetic acid is 4.74. The concentration of acetic acid is 0.01 M. The pH of acetic acid is:
(a) 3.37
(b) 4.37
(c) 4.74
(d) 0.474
Answer:
(a)

12. How many times 1M CH₃COOH solution should be diluted so that pH of the solution is doubled?
(a) 20 times
(b) 200 times
(c) 5.55 x 10² times
(d) 5.55 x 10⁴ times
Answer:
(d)
Hint: pH of a weak acid

13. 40 ml of 0.1M ammonia is mixed with 20 ml 0.1M HCl. What is the pH of the mixture? (pK_b for ammonia is 4.74).
(a) 4.74
(b) 2.26
(c) 9.26
(d) 5.00
Answer:
(c)
Hint: 40 ml of 0.1M NH₃ solution
= 40 x 0.1 milli = 4 millimol
20 ml of 0.1 M HCl = 20 x 0.1 = 2 millimol
2 millimol of HCl neutralise 2 millimol of NH₄OH, to form 2 millimol of NH₄OH.
NH₄OH left = 2 millimol
Total volume = 60 ml

14. The pH of a solution formed on mixing 20 ml of 0.05M H₂SO₄ with 5 ml of 0.45 M NaOH at 298 K is:
(a) 6
(b) 2
(c) 12
(d) 1
Answer:
(c)
Hint:
2NaOH + H₂SO₄ → Na₂SO₄ + 2H₂O
20 ml of 0.05M H₂SO₄
= 20 ml of 0.05 millimol
= 1.0 millimol

5 ml of 0.45M NaOH
= 5 x 0.45 millimol
= 2.25 millimol

2 millimol of NaOH will react with 1 millimol of H₂SO₄.
∴ NaOH left in solution = 0.5 millimol Volume of the solution = 25 ml
∴ [OH^–] = \(\frac{0.5}{25}\) = 0.01 M = 10^-2 M
[H⁺] = 10^-12;
pH = 12

15. Which of the following salts will have the highest pH in water?
(a) KCl
(b) NaCl
(c) Na₂CO₃
(d) CuSO₄
Answer:
(c)
Hint: KCl and NaCl, being salts of strong acid and strong base do not get hydrolysed. Their aqueous solutions are neutral and hence their pH = 7.
Na₂CO₃ being a salt of weak acid and a strong base, gives NaOH as a product of hydrolysis. Hence, its aqueous solution is alkaline and has pH >7.
CuSO₄ being a salt of strong acid and weak base, gives H₂SO₄, as a product of hydrolysis H₂SO₄ is a strong acid and its pH < 7.

16. The H₃O⁺ ion concentration of a solution of pH 6.58 is:
(a) antilog (- 6.58)
(c) antilog (- 5.58)
Answer:
(a)
Hint: pH = – log [H₃O⁺];
log [H₃O⁺] = -pH = – 6.58
[H₃O⁺] = antilog (-6.58)
= 2.63 x 10^-7 g ion/lit

17. 30 CC of \(\frac{M}{4}\) HCl, 20 CC of \(\frac{M}{2}\) HNO₃, and 40 CC of \(\frac{M}{4}\) NaOH solutions are mixed and the volume is made upto 1 dm³. The pH of the resulting solution is:
(a) 2
(b) 1
(c) 3
(d) 8
Answer:
(a)
Hint:
Total millimoles of H^–
= (30 x \(\frac{1}{3}\)) + (20 x \(\frac{1}{2}\))
= 10 + 10 = 20
Total millimol of OH^–
= 40 x \(\frac{1}{4}\) = 10
∴ H⁺ ions left after the neutralisation = 10 millimoles Volume of the solution
= 1 dm3
∴ Molarity of H⁺ ions
= \(\frac{10}{1000}\) = 10^-2M
pH = – log [H⁺]
= – log [10^-2] = 2

18. The pH of 0.1 M solutions of the following salts increases in the order:
(a) NaCl < NH₄Cl < NaCN < HCl
(b) HCl < NH₄Cl < NaCl < NaCN
(c) NaCN < NH₄Cl < NaCl < HCl
(d) HCl < NaCl < NaCN < NH₄Cl
Answer:
(b)
Hint: NaCl = neutral (pH = 7); NH₄Cl = slightly acidic (pH < 7); NaCN = basic (pH > 7); HCl = strongly acidic (pH << 7).

19. A weak acid HX has the dissociation constant 1 x 10^-5. It forms NaX on reaction with alkali. The degree of hydrolysis of 0.1 M solution of NaX is:
(a) 0.0001%
(b) 0.01 %
(c) 0.1%
(d) 0.15 %
Answer:
(b)
Hint: Hydrolysis reaction is
X^– + H₂O ⇌ HX + OH^–.
For a salt of weak acid with strong base,

20. Among the following hydroxides, the one which has the lowest value of K_s (solubility product) at ordinary temperature is:
(a) Mg(OH)₂
(b) Ca(OH)₂
(c) Ba(OH)₂
(d) Be(OH)₂
Answer:
(d)
Hint: The solubility increases down the group due to an increase in the size of the ion and a decrease in lattice energy. Lower the solubility, lower is the K_s.

21. On adding 0.1M solution each of Ag⁺, Ba²⁺, Ca²⁺ ions in a Na₂SO₄ solution, the species first precipitated is (K_s(CaSO₄) = 10^-6, K_s(BaSO₄) = 10^-11, K_s(Ag₂SO₄) = 10^-5:
(a) Ag₂SO₄
(b) BaSO₄
(c) CaSO₄
(d) All of these
Answer:
(b)

The minimum [SO₄^2-] concentration required for precipitation in for BaSO₄.

22. The K_s of Ag₂CrO₄, AgCl, AgBr, and AgI are respectively 1.1 x 10^-12, 1.8 x 10^-6, 5.0 x 10^-13 and 8.3 x 10^-17. Which of the following salts will precipitate last of AgNO₃ solution containing equal moles of NaCl, NaBr, Nal and Na₂CrO₄.
(a) AgBr
(b) Ag₂CrO₄
(c) AgI
(d) AgCl
Answer:
(b)

As the solubility of Ag₂CrO₄ is highest, it will be precipitated last at all.

23. The Ks of Ag₂CrO₄ is 1.1 x 10 12 at 298 K.
The solubility in mol per litre of Ag₂CrO₄ in 0.1M. AgNO₃ solution is:
(a) 1.1 x 10^-11
(b) 1.1 x 10^-10
(c) 1.1 x 10^-12
(d) 1.1 x 10^-9
Answer:
(b)

24. Using Gibb’s free energy change, AG° = +63.3 kJ for the following reaction:
Ag₂CO₃ (s) ⇌ 2Ag⁺(aq) + CO₃^2-(aq)
the K_s for Ag₂CO₃ in water at 25° C is (R = 8.314 JK^-1 mol^-1)
(a) 3.2 x 10^-26
(b) 8.0 x 10^-12
(c) 2.9 x 10^-3
(d) 7.9 x 10^-2
Answer:
(b)
Hint:

25. A buffer solution is prepared in which the concentration of NH₃ is 0.30M and the concentration of NH₄⁺ is 0.20M. If the equilibrium constant, K_b for NH₃ equals 1.8 x 10^-5, what is the pH of the solution?
(a) 8.73
(b) 9.08
(c) 9.43
(d) 11.72
Answer:
(b)

= 4.74 – 0.176 = 4.56
pH = 14 – pOH = 14 – 4.56 = 9.44

26. What is the [H⁺] in mol/L of a solution that is 0.20M in CH₃COONa and 0.10M in CH₃COOH? K_b for CH₃COOH = 1.8 x 10^-5:
(a) 9.0 x 10⁶
(b) 3.5 x 10^-6
(b) 1.1 x 10^-5
(d) 1.8 x 10^-5
Answer:
(a)

27. In what volume ratio of NH₄Cl and NH₄OH solution (each 1M) should be mixed to get a buffer solution of pH, 9.80 (pK_b for NH₄OH is 4.74):
(a) 1 : 2.5
(b) 2.5 : 1
(c) 1 : 3.5
(d) 3.5 : 1
Answer:
(c)
Hint:
Suppose ‘V₁’ ml of 1M NH₄Cl is mixed with ‘V₂’ ml of 1M NH₄OH solution:
V₁ ml of 1M NH₄Cl = V₁ millimole
V₂ ml of 1M NH₄OH = V₂ millimole
pH = 9.80, Hence pOH = 14 – 9.80 = 4.20

28. Which one of the following pairs of solutions is not an acidic buffer?
(a) H₂CO₃ + Na₂CO₃
(b) H₃PO₄ + Na₃PO₄
(c) HClO₄ + NaClO₄
(d) CH₃COOH + CH₃COONa
Answer:
(c)
Hint: HClO₄ is a strong acid. Hence, it cannot be used to make an acidic buffer.

Students get through the TN Board 12th Chemistry Important Questions Chapter 7 Chemical Kinetics which is useful for their exam preparation.

TN State Board 12th Chemistry Important Questions Chapter 7 Chemical Kinetics

Answer the following questions.
Question 1.
Express the rate of the following reaction in terms of disappearance of the reactant and appearance or formation of the product: A → B
Answer:
Rate of disappearance of the reactant = \(-\frac{\Delta[\mathrm{A}]}{\Delta t}\)
where the negative sign indicates the concentration of the reactant decreases with time.
Rate of formation of the product = \(+\frac{d[\mathrm{B}]}{d t}\)
The positive sign indicates that the concentration of the product increases with time.

Question 2.
Mention the unit of rate of reaction.
Answer:

For a gaseous reaction, a unit of rate = atm sec^-1 since concentration is expressed in partial pressures.

Question 3.
Explain how will you determine experimentally determine
(i) average rate of reaction
(ii) instantaneous rate of reaction and
(iii) initial rate of reaction.
Answer:
The changes in concentration of the reactants or products are experimentally determined by withdrawing a small portion of the reaction mixture (usually 2 or 5 ml) at suitable intervals of time and then freeze to about 0°C to stop the reaction. The concentration of the reactant or product is measured by a suitable method. A plot of concentration-time is made from the graph, the average rate, instantaneous and initial rates can be determined.
(i) Average rate: Two equivalent points are taken with respect to the time at which the average rate is to be determined. The concentrations corresponding to these points are noted from the graph. The difference in concentration is divided by the time interval between equidistant points. Two equidistant points with respect to T is selected. Let these be ‘t₁,’ and ‘t₂’ respectively. The corresponding concentrations are x₁, and x₂ respectively.

(ii) Instantaneous rate of reaction: For the determination of the instantaneous rate of reaction at any time ‘t’, a tangent is the direction to the curve at the point ‘P’ corresponding to that time. The slope of this tangent gives the instantaneous rate of reaction.

Where Q is the angle of a tangent with the time axis. In a similar manner, the rate of formation of the product can be determined by measuring the slope of the tangent at a particular instant in concentration to the time curve.
(iii) Initial rate of reaction:

The rate of reaction is maximum at the time of mixing the concentration. This cannot be determined experimentally. This is determined from the concentration is time curve. The curve is extrapolated to zero time and the slope of this curve at zero time gives the initial rate of the reaction.
\(\frac{\Delta x}{\Delta t}\) = initial rate.

Question 4.
In the reaction 2A → products the concentration of A decreases from 0.5 mol L¹ to 0.4 mol L^-1 in 10 minutes. Calculate the rate during this period.
Answer:

Question 5.
The concentration of reactant changes from 0.03 M to 0.02 M in 25 minutes. Calculate the average rate using units of time both in minutes and in seconds.
Answer:

Question 6.
The decomposition of N₂O₅ is expressed by the equation.

If during a certain interval of time the rate of decomposition of N₂O₅ is 1.8 x 10^-3 mol liter 1 min^-1, what will be the rates of formation of NO₂ and O₂ during the same interval.
Answer:
The rate expression for the decomposition of N₂O₅ is

Question 7.
For each of the following reactions, express the given rate of change of concentration of the product or reactant in terms of rate of change of concentration of other reactants or products in that reaction.

Answer:
(a)

(b) The equality in this case is,


(c) The equality in this case is,

Question 8.
From the concentration of R at different times given below, calculate the average rate of the reaction.
Answer:
R → P during different intervals of time

Average rate of reaction between 0 → 5 sec

Average-rate of reaction between 5-10 sec

Similarly, average rate between 10 → 20 and 20 → 30 see can be calculated.
Note: What we understand from the above problem is that the rate of reaction is not a constant quantity. It decreases with time.

Question 9.
The decomposition of N₂O₅ is CCl₄ solution at 318K has been studied by monitoring the concentration of the N₂O₅ in the solution. Initially, the concentration of N₂O₅ in the solution is 2.33 M and after 184 minutes it is reduced to 2.08 M. The reaction takes place according to the equation
2N₂O₅ → 4NO₂ + O₂.
Calculate the average rate of the reaction in terms of hours, minutes and seconds. What is the rate of production during this period?
Answer:
Average rate of reaction (mol L^-1 min^-1)

For the given reaction

Question 10.
In hydrogenation reaction at 25°C it is observed that hydrogen gas pressure falls from 2 atm to 1.2 atm in 50 min. Calculate the rate of reaction in molarity per sec.
Answer:
R = 0.0821 lit mol^-1 deg^-1

Question 11.
Explain why the average rate cannot be used to produce the rate of reaction at any instant.
Answer:
Because the rate decreases with time as the reaction proceeds. It is not constant.

Question 12.
Bring out the difference between the rate and rate constant of a reaction?
Answer:

Question 13.
The decomposition of dimethyl ether leads to the formation of CH₄, H_2, and Co and the reaction rate in given by
rate = R [CH₃OCH₃]^3/2
The rate is followed by an increase in pressure in a closed vessel so that the rate is expressed in terms of partial pressure of dimethyl ether.
rate = R[P_CH3OCH3]^3/2
If the pressure is measured in the bar and the time in seconds then what are theTxmts of rate and rate constant?
Answer:
The rate law of the reaction

Question 14.
State the order with respect to each reactant, overall reaction, and the units of the rate constant in each of the following reactions.

Answer:
(a) Order with respect to NO and Br₂ are 2 and 1 respectively, over all order is 3.

(b) Order \(\frac{3}{2}\)

(c) Order with respect to H₂ and NO are 1 and 2 respectively.

= mol^-2 litre² sec^-1
(d) Order with respect to CO and Cl₂ are 2 and \(\frac{1}{2}\) respectively.
Over all order = 2\(\frac{1}{2}\) = \(\frac{5}{2}\)

(e) Order with respect to H₂O₂ and I^– are one respectively. Over all order 1 + 1 = 2

Question 15.
Identify the reaction order from the following rate constants.
(a) k = 3.1 × 10^-4 sec-1
(b) k = 1.12 × 10^-2
(c) k = 1.35 × 10^-2 mol^-2 let² s^-1
(d) k = 3.4 × 10^-3 mol^– let s^-1
Answer:
(a) Let the reaction be of nth order

i.e., reaction is of first order.
(b) (atm)^1-n × s^-1 = atm^-1 s^-1
1 – n = -1 : or n = 2
i.e., reaction is of second order.
(c)

i.e., reaction is of third order.
(d)

i.e., reaction is of second order.

Question 16.
Derive an expression for the rate constant for the first-order reaction.
Answer:
A reaction whose fate depends on the reactant concentration raised to the first power is called a first order reaction. Let us consider the following Cl₂ first-order reaction,
A → product
Rate law can be expressed as Rate = k [A]¹
Where, k is the first order rate constant.

Integrate the above equation between the limits of time t = 0 and time equal to t, while the concentration varies from the initial concentration [A₀] to [A] at the later time.

This equation is in a natural logarithm. To convert it into the usual logarithm with base 10, we have to multiply the term by 2.303.

Question 17.
How will you determine the first-order rate constant graphically?
Answer:
For a first order reaction, the rate constant k is given by

where [A₀] is the initial concentration of the reactant and [A] is the concentration of the reactant at time it\
Rearranging this equation in the form . y = mx + c, we get
log [A₀] – log [A] = \(\frac{kt}{2.303}\)
2.303 [log [A] – log [A]] = kt
log [A₀] – log [A] = \(\frac{k}{2.303}\) t.
Hence a plot of log [A] vs t gives a straight line with a negative slope having the volume of \(\frac{-k}{2.303}\)

Question 18.
Give example for first order reaction.
Answer:
(i) Decomposition of dinitrogen pentoxide
N₂O₅ → (g) 2NO₂ (g) + \(\frac{1}{2}\) O₂ (g)
(ii) Decomposition of thionylchloride;
SO₂Cl₂(l) → SO₂ (g) + Cl₂ (g)
(iii) Decomposition of the H₂O₂ in aqueous solution;
H₂O₂ (aq) → H₂O (l) + \(\frac{1}{2}\) O₂ (g)
(iv) Isomerisation of cyclopropane to propene.

Question 19.
Show that acid hydrolysis of an ester is a pseudo first order reaction.
Answer:
A second order reaction can be altered to a first order reaction by taking one of the reactant in large excess, such reaction is called pseudo first order reaction. Let us consider the acid hydrolysis of an ester,

If the reaction is carried out with the large excess of water, there is no significant change in the concentration of water during hydrolysis, i.e., the concentration of water remains almost constant.
Now, we can define k [H₂O] = k’ ; Therefore the above rate equation becomes
Rate = k’ [CH₃COOCH₃]
Thus it follows first-order kinetics.

Question 20.
Give examples of zero-order reactions.
Answer:
(i) Photochemical reaction between H₂ and I₂

(ii) Decomposition of N₂O on a hot platinum surface

(iii) Iodination of acetone in an acid medium is zero-order with respect to iodine.

Question 21.
Give the general rate equation for nth-order reaction involving one reactant A.
Answer:
The general rate equation for an nth order reaction involving one reactant [A],
A → product

Consider the case in which n ≠ 1, integration of above equation between [A₀] and [A] at time t = 0 and t = t respectively gives

Question 22.
Derive an expression to calculate t_1/2 for a zero-order reaction.
Answer:
Let us calculate the half life period for a zero-order reaction.

The half-life of a zero-order reaction is directly proportional to the initial concentration of the reactant.

Question 23.
Give the general expression for t_1/2 of the nth order (n ≠ 1) reaction.
Answer:
Half-life for an nth order reaction involving reactant A and n ≠ 1

Question 24.
Give the characteristics of first order reaction.
Answer:

1. A change in concentration unit will not change the numerical value of ‘k’
   k = \(\frac{2.303}{t}\) log \(\frac{a}{a-x}\)
   Where ‘a’ and (a – x) are the initial concentration of the reactant and concentration of the reactant at time Y respectively. Thus for the first-order reaction, any quantity which is proportional to the concentration can be used in place of concentration to calculate ‘k’.
2. The time taken for the completion of the same fraction of change is independent of the initial concentration.
3. Time required to calculate nth fraction of reaction t_1/2 can be calculated as When x = \(\frac{a}{n}\) ; t = t_1/n
   Substituting these values in,
   
   The above equation can be used to calculate t_3/4 of a reaction.
   
4. In a first-order reaction, the amount of reactant remaining after V half-lives can be calculated as follows:
   where [A₀] – initial concentration of the reactant, [A] = concentration of the reactant remaining after ‘n’ half-lives.

Question 25.
Calculate the half life of a first order reaction from their rate constant given below:
(a) 200 sec^-1 (b) 2 min^-1 (c) 4 year^-1
Answer:

= 3.465 x 10~3 sec
= 0.1732 year.

Question 26.
The rate constant for a first-order reaction is 60 sec^–. How much time will it take to reduce the initial concentration to its 1/16th value?
Answer:

Question 27.
The thermal decomposition of a compound is of the first order. If 50% of the sample is decomposed in 120 minutes, how long will it take for 90% of the compound to decompose?
Answer:
Half life of the reaction = 120 min.
We know that

If a= 100, x = 90, a-x = 10
= 5.77 x 10^-3 min^-1
Applying first order rate equation.

Question 28.
Sucrose decomposes in acid solution into glucose and fructose according to the first-order rate law. With t_1/2 = 3.00 hrs. What fraction of sucrose remains after 8 hours?
Answer:

Therefore fraction of reactant remaining = 0.1576

Question 29.
Explain the pressure change method in determining a first-order reaction.
Answer:
If pressure is given in gaseous reactions then we use the following kinetic equation.

where p₀ is the pressure at the initial stage p₀-x is the pressure at time ‘t’ values of p₀ and x can be calculated using the following examples.

Case I: If total pressure of the reaction mixture is given in place of pressure of reactants.
p_t = (p₀ – x + x + x + x)
Where p_t is the pressure of reaction vessel at time ‘t’.
Case II: If the pressure of the reaction vessel after a long time or infinite time is given
p_t = p₀ + p₀ +p₀

Question 30.
The decomposition of Cl₂O₇ at 400K in the gas phase to Cl₂ and O₂ is a first order reaction.
(i) After 55 seconds at 400K, the pressure of Cl₂O₇ falls from 0.062 to 0.044 atm. Calculate the rate constant.
(ii) Calculate the pressure of Cl₂O₇ after 100 seconds of decomposition at this temperature.
Answer:
(i) As pressure ∝ concentration.

(ii) Applying the first-order rate equation.

Question 31.
For the decomposition of azo isopropanol to hexane and nitrogen at 543K, the following data are obtained.

Calculate the rate constant for the reaction.
Answer:
The reaction is

Total pressure – p₀ – x + x + x
p_t = p₀ + x
or x = p_t – p₀
a ∝ p₀
∴ (a – x) ∝ p₀ – (p_t – p₀)
= 2p₀ – p_t
For the first order reaction

Question 32.
Explain Oswald dilution method for determining the order of the reaction.
Answer:
It is applied for these reactions which involve more than one reactant. In this method, the concentration of all reactants are taken in large excess than one. The changes in the concentration of the reactant that is not taken in excess will influence the rate of reaction.
The concentrations of all other reactants practically remain same during the course of reaction. Consider the reaction.
m₁A + m₂B + m₃C → product
Rate = k[A]^α [B]^β [C]^γ
When B and C are taken in excess, the rate law will reduce to
rate = k’ [A]^α
where k’ = k[B]^β [C]^γ
Two sets are taken in which the concentration of A is [A₁] and [A₂] while B and C are present in large excess,
r₁ = [A₁]^α
r₂ = [A₂]^α

From this, the value of α, i.e., the order with respect to A is obtained. The process is repeated one by one and order with respect to others is determined.
Total order of reaction = α + β + γ

Question 33.
Show that if the concentration of a reactant is doubled, the rate of the reaction is also doubled for a first-order reaction, increases four times for a second-order reaction, increases by eight times for a third-order reaction, i.e., A → product.
Answer:
(i) If the concentration of A is doubled.

Hence, for the first-order reaction, if the concentration is doubled, the rate is also doubled.
(ii) For a second-order reaction.

i.e., for a second-order reaction, if the concentration of A is doubled, the rate of the reaction increases by 4 times.
(iii) For a third-order reaction,

Hence, for a third-order reaction, if the concentration of the reactant is doubled the rate increases 8 times.

Question 34.
Compounds A and B react according to the following chemical equation
A (g) +2B(g) → 2C(g)
Answer:
The concentration of either ‘A’ or ‘B’ was changed keeping the concentration of one of the reactants constant and rates were measured as a function of initial concentration following results were obtained. Find the order with respect to A and B and write the rate law for the reaction

From experiments 1 and 3,
[B] = constant; [A] = doubled
rate is doubled.
Hence rate ∝ [A] i.e., the order with respect to A is 1.
From experiments 1, and 2,
[A] = constant; [B] doubled rate is quadrupled.
Hence, rate ∝ [B]²
i.e., order with respect by B is 2.
Combining rate = k [A] [B]².

Question 35.
The initial rate of reactions
3A + 2B + C → products at different initial concentrations are given below.

Answer:
(i) Consider equation (1) and (2)
The concentration of A and B are constant. The rate remains the same for different concentrations of C. Hence, the order with respect to C is zero.
(ii) Consider equations (3) and (1)
The concentration of A and C are constant. The concentration of B is doubled, rate increases by 2 times. Hence order with respect to B is 1.
(iii) Consider equations (4) and (1), The concentration of B and C remain constant. When the concentration of A is doubled, the rate increases by 4 times. Hence order with respect to A is 2.
Alternate method:
Suppose order with respect to A, B and C are α, β and γ respectively. Then the rate law at different concentration of A, B and C are
5.0 × 10^-3 = k [0.010]^α [0.005]^β [0.010]^γ …(1)
5.0 × 10^-3 = k [0.010]^α [0.005]^β [0.015]^γ …(2)
1.0 × 10^-3 = k [0.010]^α [0.010]^β [0.010]^γ …(3)
1.25 × 10^-3 = k [0.005]^α [0.005]^β [0.010]^γ …(4)
Dividing (1) by (2)

i.e., order with respect to C is zero.
Dividing (3) by (2)

Dividing (1) by (4) ⇒ 4 = 2^α or α = 2
Order with respect to A, B, C are 2, 1 and 0.

Question 36.
A reaction is first order in A and second order in B.
(i) Write the differential rate equation.
(ii) How is the rate affected by increasing the concentration of B, three times, keeping the concentration of A constant?
(iii) How is the rate affected by when the concentration of both A and B are doubled?
Answer:
(i) A + B → product
rate = k [A] [B]²
(ii) If the concentration of B is tripled, the rate increases by 9 times i.e.,

(iii) If the concentration of both A and B are doubled, the rate increases by 8 times.

Question 37.
The following rate data were obtained at 303K for the following reaction.
2A + B → C + D.

What is the order with respect to each reactant and the overall order of the reaction? write the rate law. Also, calculate the rate constant for the reaction and the units of the rate constant.
Answer:
From experiments 1 and 4, it is seen that [B] the concentration of B is the same but [A] has been made 4 times, the rate of the reaction has also become 4 times. Hence the order with respect to A is 1. i.e., rate ∝ [A],
From experiments 2 and 3, it may be noted that [A] is kept the same but [B] is doubled. The rate of reaction increases by 4 times, i.e., the order with respect to B is 2 combining both,
rate = k[A] [B]². This is the rate law for the reaction.
Alternatively, suppose the order with respect to A is α and the order with respect to B is β, the general rate law for the reaction is
rate = k [A]^α [B]^β
From the data given

Now,

Unit of rate constant (k)

Calculation of rate constant: Substitute the values of α and β in any of the equations.

Hence k = 6.0 mol^-1 L² min^-1.

Question 38.
In a reaction between A and B, the initial rate of reaction was measured at a different initial concentration of A and B as given below.

What is the order with respect to A and B?
Answer:
The general rate law is rate = k [A]^α [B]^β where α, β are ordered with respect to A and B respectively.
The given data is


Thus order with respect to A is 0.5 and that of B is 0.

Question 39.
The decomposition of ammonium nitrate in an aqueous solution was studied by placing the apparatus in a thermostat maintained at a particular temperature. The volume of nitrogen gas collected at different intervals of time was as follows:

The above data prove that the reaction is of the first order.
Answer:
For a first-order reaction,

In the present case, V_∞ = 35.05 cm³. The value of R at each instant of time can be calculated as follows:
Values of ‘k’ come nearly constant and hence the reaction is first order.

Question 40.
What do understand by a fraction of effective collisions? Mention its significances.
Answer:
Fraction of effective collisions (f) is given by the following expression

To understand the magnitude of collision factor (f), Let us calculate the collision factor (f ) for a reaction having an activation energy of 100 kJ mol^-1 at 300K.

Thus, out of 10¹⁸ collisions, only four collisions are sufficiently energetic to convert reactants to products.

Question 41.
Explain the importance of the proper orientation of molecules in the collision theory.
Answer:
Even if the reactant possesses sufficient activation energy, their collisions need not result in the formation of products. The molecules should orient themselves in such a way that their collision becomes fruit fill. The fraction of effective collisions (f) having orientation is given by steric factor (p). Thus, the rate of reaction is given by
rate = p x f x collision rate
(or) rate = pz\(e^{-\mathrm{E}_{a} / \mathrm{RT}}\)

Question 42.
Define activation energy of a reaction.
Answer:
Activation energy is the extra amount of energy that is supplied from outside so that colloiding molecules must produce effective collisions.

Question 43.
Arrhenius equation is given by k = A\(e^{-\mathrm{E}_{a} / \mathrm{RT}}\). Based on this equation answer the following questions.
(i) Can reactions have zero activation energy?
(ii) Can a reaction have negative activation energy?
Answer:
(i) k = A\(e^{-\mathrm{E}_{a} / \mathrm{RT}}\), when the expression becomes k = e° = A.
The rate constant k = collision frequency. This implies that every collision is an effective collision and leads to product formation which is not possible. Thus activation energy can not be zero.
(ii) When E_a is negative, the above equation

Rate constant having a higher value than collision frequency which is not possible.

Question 44.
Explain the effect of temperature on reaction rate based on Arrhenius’s theory.
Answer:
An increase in temperature increases the fraction of the total number of molecules having the necessary energy of activation which in terms increases the number of effective collisions and hence the rate.

Question 45.
Write Arrhenius equation and explain the terms? What is the significance of frequency factor A in the equation?
Answer:
The Arrhenius equation is k = A\(e^{-\mathrm{E}_{a} / \mathrm{RT}}\) Where ‘k’ is the rate constant for the reaction, A is called the pre-exponential factor, E_a is the energy of activation and R is the gas constant and T is the absolute temperature. The two quantities A and E_a are called the Arrhenius parameters.
The frequency factor gives the frequency of binary collision of the reacting molecules per liter per second. It practically remains constant and does not vary with temperature.

Question 46.
Explain how the energy of activation for a reaction is determined.
Answer:
The rate constant of a reaction is determined at two different temperatures. If R₁ is the rate constant at temperature T₁ arid R₂ is the rate constant at temperature T₂ thereby using a logarithmic form of Arrhenius equation

E_a can be calculated.

Question 47.
Describe how the energy of activation of a reaction is determined graphically.
Answer:
The Arrhenius equation is written in the form,

This equation is of the form y = mx + C i.e., equation for a straight line. Thus, if a plot of log k is \(\frac{1}{T}\) is a straight line, the validity of the equation is confirmed.
Thus the values of k at different temperatures are found and a plot of log k vs \(\frac{1}{T}\) is made.
A straight line with a negative slope is obtained from the value of the slope, E_a can be calculated.
Slope = \(\frac{-\mathrm{E}_{a}}{2.303 \mathrm{R}}\)

Question 48.
The rate constant for a reaction is 1.2 x 10^-3 sec^– at 30°C and 2.1 x 10^-3 sec^-1 at 40°C. Calculate the energy of activation.
Answer:
Given that k₁ = 1.2 x 10^-3 sec^-1
T₁ = 30 + 273 = 303K
k₂ = 2.1 x 10^-3 sec^–
T₂ = 40 + 273 = 313K
Substituting these values in the equation
Solving E_a = 44126.3J = 44.13kJ

Question 49.
The rate of particular reaction doubles when temperature changes from 27°C to 37°C. Calculate the energy of activation.
Answer:
Given that T₁ = 27°C = 273 +21 = 300K
T₂ = 37°C = 273 + 37 = 310K
k₂ = 2k₁
Substituting these values in the equation.
This as solving for E_a
E_a = 53598.6 J mol^-1 or 53.6 kJ mol^-1

Question 50.
The activation energy of a reaction is 94.14 kJ mol^-1 and the value of the rate constant at 313K. is 1.8 x 10^-1 sec^-1 . Calculate the frequency factor (A).
Answer:
Given that, E_a = 94.14 kJ mol^-1 = 9414 J mol^-1
T = 313K,
k = 1.8 x 10^-1 sec^-1 Substituting these values in
= (log 1.8 – 5)+ 15.7082
= (0.2553 – 5)+ 15.708
= 10.963J
A = Anti log (10.963 J)
= 9.194 x 10^-10 sec

Question 51.
The first order rate constant for the decomposition of ethyl iodide by the reaction
C₂H₅I (g) → C₂H₄ (g) + HI (g)
at 600 K is 1.60 x 10^-1 s^-1. Its energy of activation is 209 kJ / mol. Calculate the rate constant of the reaction at 700K.
Answer:
Given T₁ = 600 K
k₁ = 1.60 x 10^-5 sec^-1
T₂ = 700K
k₂ = ?
E_a = 299 kJ mol^-1 = 299000 J mol^-1
Substituting these values in the equation.

Question 52.
Rate constant ‘k’ of a reaction varies with temperature according to the equation
where E_a is the energy of activation for the reaction. When a graph is plotted for log k vs \(\frac{1}{T}\) a straight line with slope – 6670 K is obtained.
Calculate the energy of activation for the reaction and mention is units.
(R = 8.314 J k^-1 mol^-1)
Answer:
Slope of the line = \(\frac{-\mathrm{E}_{a}}{2.303 \mathrm{R}}\) = -6670K
E_a = 2.303 x 8.314 x 6670 = 127711.4 J mol^-1

Choose the correct answers:

1. Which of the following statements is not correct about the order of reaction?
(a) The order of a reaction can be fractional.
(b) Order of reaction is an experimental quantity.
(c) The order of the reaction is always equal to the sum of the stoichiometric coefficients of reactants in a balanced chemical equation for the reaction.
(d) The order of a reaction is the sum of the powers of molar concentrations of the reactants in the rate law expression.
Answer:
(c)

2. Which of the following statements is correct?
(a) The rate of reaction decreases with the passage of time as the concentration of the reactant decreases.
(b) The rate of reaction is the same at any time during the reaction.
(c) The rate of reaction is independent of temperature change.
(d) The rate of reaction decreases with an increase in the concentration of reactants.
Answer:
(a)

3. Which of the following expression is correct for the rate of reaction given below:

Answer:
(c)

4. Time required for 100 percent completion of a zero-order reaction is:

Answer:
(c)

5. For the reaction aA+ bB → cC,

if , then a, b and c respectively are:
(a) 3,1,2
(b) 2,1,3
(c) 1,3,2
(d) 6, 2, 3
Answer:
(c)
Hint: Dividing throughout by 3, we get

This is so for the reacting A + 3B → 2C

6. The rate of a gaseous reaction is given by the expression k [A] [B]. If the volume of the reaction vessel is suddenly reduced to 1/4th of the initial volume, the reaction rate relating to the original rate will be:
(a) 1/10
(b) 1/8
(c) 8
(d) 16
Answer:
(d)
Hint: Rate = k ab. When volume is reduced to 1/4th, concentrations will become = 4 times.
New rate = k (4a) (4b) = 16 kab = 16 times.

7. In a reaction A → B, the rate of reaction increases two times on increasing the concentration of the reactant four times, then order of reaction is:
(a) 0
(b) 2
(c) 1/2
(d) 4
Answer:
(c)
Hint: (i) r = ka^α, (ii) 2r = k(4a)^α
Dividing (ii) by (i),
4^α = 2 or 2^2α = 2 or 2α = 1 or α = 1/2.

8. The rate of the reaction 2 NO + Cl₂ → 2 NOCl is given by the rate equation: rate = k[NO]²[Cl₂]. The value of the rate constant can be increased by:
(a) increasing the temperature
(b) increasing the concentration of NO
(c) increasing the concentration of Cl₂
(d) doing all of these
Answer:
(a)
Hint: The rate constant of a reaction depends only on temperature and does not depend upon concentrations of the reactants.

9. The unit of rate constant for a zero-order reaction is:
(a) mol L^-1 s^-1
(b) L mol^-1 s^-1
(c) L² mol^-2 s^-1
(d) s^-1
Answer:
(a)

10. Rate constant of a reaction (k) is 175 litre² mol^-2 sec^-1. What is the order of reaction?
(a) first
(b) second
(c) third
(d) zero
Answer:
(c)
Hint: On the basis of given units of k, the reaction is of 3rd order.

11. The reaction A → B follows first order kinetics. The time taken for 0.8 mole of A to produce 0.6 mole of B is 1 hour. What is the time taken for conversion of 0.9 mole of A to produce 0.675 mole of B?
(a) 1 hour
(b) 0.5 hour
(c) 0.25 hour
(d) 2 hours
Answer:
(a)
Hint: The fraction of A reacted in each case is same (0.2 / 0.8 = 1/4),
(0.9 – 0.675)/ 0.90 = 0.225 / 0.90 = 1/4. Hence, time taken is same.

12. 75% of the first-order reaction was completed in 32 min. 50% of the reaction was completed in:
(a) 24 min
(b) 8 min
(c) 16 min
(d) 4 min
Answer:
(c)
Hint: 75 % of reaction is completed in two half-lives, i.e., 2 x t_1/2 = 32 min or t_1/2 = 16 min.

13. 1/[A] vs time is a straight line. The order of the reaction is:
(a) 1
(b) 2
(c) 3
(d) 0
Answer:
(b)
Hint: log [A] vs time is linear for 1st order reactions, 1/[A] vs time is linear for 2nd order reactions; 1/[A]² vs time is linear for 3rd order reactions.

14. If a graph is plotted between in k and 1/T for the first-order reaction, the slope of the straight line so obtained is given by:

Answer:
(a)

15. 10g of a radioactive isotope is reduced to 1.25g in 12 years. Therefore, half-life period of the isotope is:
(a) 24 years
(b) 4 years
(c) 3 years
(d) 8 years
Answer:
(b)

16. The half-life period of a radioactive element is 20 days. What will be the remaining mass of 100 g of it after 60 days?
(a) 25 g
(b) 50 g
(c) 125 g
(d) 20 g
Answer:
(c)
Hint: 60 days = 3 half-lives, i.e, n = 3;

17. Activation energy of a chemical reaction can be determined by:
(a) determining the rate constant at standard temperature.
(b) determining the rate constants at two temperatures.
(c) determining the probability of collision.
(d) using catalyst.
Answer:
(b)
Hint:
Knowing k₁ and k₂ at T₁ and T₂, E_a can be determined. (Arrhenius equation)

18. According to Arrhenius equation, rate constant k is equal to A\(e^{-\mathrm{E}_{a} / \mathrm{RT}}\). Which of the following options represents the graph of k vs \(\frac{1}{T}\)?

Answer:
(a)
Hint:
k = A\(e^{-\mathrm{E}_{a} / \mathrm{RT}}\). In k = ln A – \(\frac{\mathrm{E}_{a}}{\mathrm{RT}}\). Hence, graph of In k vs \(\frac{1}{T}\) will be linear with negative slope and intercept = ln A.

19. Which of the following statement is incorrect about the collison theory of chemical reaction?
(a) It considers reacting molecules or atoms to be hard spheres and ignores their structural features.
(b) Number of effective collisions determines the rate of reaction.
(c) Collision of atoms or molecules possessing sufficient threshold energy results into the product formation.
(d) Molecules should collide with sufficient threshold energy and proper orientation for the collision to be effective.
Answer:
(c)
Hint: (c) is incorrect because the formation of the product depends not only on energy but also on proper orientation at the time or collision.

20. A first-order reaction is 50% completed in 1.26 x 1014 s. How much time would it take for 100% completion?
(a) 1.26 x 1015 s
(b) 2.52 x 1014 s
(c) 2.52 x 1028 s
(d) infinite
Answer:
(d)
Hint: The whole of the substance never reacts because in every half-life, 50% of the substance reacts. Hence, the time taken for 100% completion of a reaction is infinite.

21. Compounds ‘A’ and ‘B’ react according to the following chemical equation:
A (g) + 2 B(g) → 2C (g)
The concentration of either ‘A’ or ‘B’ was changed keeping the concentrations of one of the reactants constant and rates were measured as a function of initial concentration. Following results were obtained. Choose the correct option for the rate equations for this reaction.

(a) Rate = k [A]² [B]
(b) Rate = k [A] [B]²
(c) Rate = k [A] [B]
(d) Rate = k [A]² [B]⁰
Answer:
(b)
Hint: From expts.l and 3, [B] = constant, [A] = doubled, rate is doubled. Hence, Rate ∝ [A].
From expts. 1 and 2, [A] = constant, [B] = doubled, rate quadrupled. Hence, Rate ∝ [B]².
Combining, Rate = k [A][B]².

22. The rate constant of reaction A → B is 0.6 x 10³ mole per liter per second. If the concentration of A is 5 M, then the concentration of B after 20 minutes is:
(a) 0.36 M
(b) 0.72 M
(c) 1.08M
(d) 3.60 M
Answer:
(b)
Hint: The unit mol L^-1 s^-1 of the rate constant show that it is a reaction of zero order. For a zero order reaction, x = kt. Amount of A reached (x) = (0.6 x 10^-3 mol L^-1 s^-1 ) (20 x 60 s) = 0.72 mol L^-1 [B] formed = [A] reacted = 0.72 mol L^-1

23. The half-life of a substance in a certain enzyme-catalyzed reaction is 138 s. The time required for the concentration of the substance to fall from 1.28 mg L^-1 to 0.04 mg L^-1, is:
(a) 414 s
(b) 552 s
(c) 690 s
(d) 276 s
Answer:
(c)

24. The following data is obtained during the first-order thermal decomposition of A (g) → B (g) + C (s) at constant volume and temperature.

The rate constant in min^-1 is
(a) 0.0693
(b) 6.93
(c) 0.00693
(d) 69.3
Answer:
(a)

25. t_1/4 can be taken as the time taken for the A concentration of a reactant to drop to its initial value. If the rate constant for a first-order reaction is k, then t_1/4 can be written as:
(a) 0.10/k
(b) 029/k
(c) 0.69/k
(d) 0.75/k
Answer:
(b)
Hint:

26. A plot of log t_1/2, versus log C₀ is given in the adjoining fig. The conclusion that can be drawn from this graph is:


Answer:
(b)

27. In the catalysed decomposition of benzene diazonium chloride,

the half-life period is found to be independent of the initial concentration of the reactant. After 10 minutes, the volume of N₂ gas collected is 10 L and after the reaction is complete, it is 50 L. Hence rate constant of the reaction (in min^-1) is:

Answer:
(b)

28. The activation energy of a reaction can be determined from the slope of which of the following graph?

Answer:
(a)
Hint: From the Arrhenius equation,

29. The rate of a reaction doubles when its temperature changes from 300 K to 310 K. Activation energy of such a reaction will be:
(a) 60.5 kJ mol^-1
(b) 53.6 kJ mol^-1
(c) 48.6 kJ mol^-1
(d) 58.5 kJ mol^-1
Answer:
(b)

30. In the presence of a catalyst, the activation energy of a reaction is lowered by 2 kcal at 27°C. The rate of reaction will increase by:
(a) 2 times
(b) 14 times
(c) 28 times
(d) 20 times
Answer:
(c)

31. Which one of the following is not correct?
(a) Every bimolecular collision does not result into a chemical reaction.
(b) Collision theory is not applicable to unimolecular reaction.
(c) According to collision frequency, k = PZ_AB \(e^{-\mathrm{E}_{a} / \mathrm{RT}}\) where Z_AB is collision frequency and P is steric factor.
(d) Collision theory assumes molecules to be hard spheres.
Answer:
(b)
Hint: Collision theory is applicable to unimolecular reactions also.

Students get through the TN Board 12th Chemistry Important Questions Chapter 6 Solid State which is useful for their exam preparation.

TN State Board 12th Chemistry Important Questions Chapter 6 Solid State

Answer the following questions.

Question 1.
Why are solids rigid?
Answer:
In solids, the constituent particles (atoms, molecules, or ions) are not free to move but can only oscillate about their mean position due to strong interatomic or intermolecular forces. This imparts rigidity.

Question 2.
Classify the following as amorphous or crystalline solids.
Answer:
Polyurethane, naphthalene, benzoic acid Teflon, potassium nitrate, cellophane, polyvinyl chloride fiberglass, copper.
Amorphous solids: Polyurethane, Teflon, cellophane, polyvinyl chloride, and fiberglass. Crystalline solids: Naphthalene, benzoic acid, potassium nitrate, and copper.

Question 3.
Explain why glass is considered an amorphous solid.
Answer:
Like liquids, it has a tendency to flow, though very slowly. It does not have a sharp melting point. It is isotropic.

Question 4.
Classify the following solids in different categories based on the nature of intermolecular forces operating in potassium sulfate, tin, benzene, urea, ammonia water, zinc sulfide, graphite, rubidium, argon, silicon carbide.
Answer:
Ionic solids: potassium sulfate, zinc sulfide. Covalent solids: graphite, silicon carbide. Molecular solids: benzene, urea, ammonia water, argon.
Metallic solids: rubidium, tin.

Question 5.
Ionic solids conduct electricity in the molten state but not in the solid state. Explain.
Answer:
In ionic solids, free ions are present in a molten state and they move towards the oppositely charged electrode under the influence of electric current. In a solid state, these ions are fixed in their lattice position and cannot move under the influence of electric current.

Question 6.
Briefly outline the properties of covalent solids.
Answer:

1. They have a high melting point.
2. They are poor thermal and electrical conductors.

Question 7.
What are covalent solids? Give example.
Answer:
In covalent solids, the constituents (atoms) are bound together in a three-dimensional network entirely by covalent bonds. eg; Diamond, silicon carbide, etc.

Question 8.
What are molecular solids? Give examples.
Answer:
In molecular solids, the constituents are neutral molecules. They are held together by weak Van der Waals forces. Generally, molecular solids are soft and they do not conduct electricity, eg: I₂, graphite.

Question 9.
Explain various types of molecular solids for example.
Answer:
Molecular solids are classified into (i) non-polar molecular solids, (ii) polar molecular solids, (iii) hydrogen-bonded molecular solids.

1. In nonpolar molecular solids, the constituent molecules are held together by weak dispersion or London forces, eg: naphthalene, anthracene.
2. In polar molecular solids, the constituent molecules are formed by polar covalent bonds, eg solid CO₂, solid NH₃.
3. In hydrogen-bonded molecular solids, the constituent molecules are held together by hydrogen bonds, eg: ice (H₂O), glucose, urea, etc.

Question 10.
Explain the types of the force of attraction that exist (i) nonpolar molecular solids, (ii) polar molecular solids, (iii) hydrogen-bonded molecular solids.
Answer:

1. In non-polar molecular solids, the constituent particles are held together by weak dispersion or London forces.
2. In polar molecular solids, the constituent molecules are held by covalent bonds. They are attracted to each other by relatively strong dipole-dipole attraction.
3. In hydrogen-bonded molecular solids, the molecules are held together by hydrogen bonds.

Question 11.
What are the characteristics of metallic solids? What type of attractive forces exists among the constituent particles?
Answer:
The constituents of a metallic solid are held together by a metallic bonding. The lattice points are occupied by metals ion and the electron’s charge bond encloses the metal ion.
They are hard, possess a high melting point, They conduct heat and electricity. They possess metallic luster.

Question 12
Give examples for metallic solids.
Answer:
Metal and metal alloys, eg: Cu, Fe, Zn, Ag, Au, Cu, Zn, etc.

Question 13.
Define the following: (i) Lattice point, (ii) Crystal lattice, (iii) Unit cell.
Answer:

1. Lattice point: Each lattice point represents one constituent particle of a solid. This may be an atom, molecule, or ion.
2. The crystal lattice is a three-dimensional arrangement of identical points in the space which represents how the constituent particles (atoms, molecules, or ions) are arranged in a crystal.
3. Unit cell A unit cell is the smallest portion of a space lattice which when repeated in different directions generates the entire lattice.

Question 14.
What are the characteristics of a unit cell?
Answer:
A unit cell is characterized by the three edge lengths or lattice constants a, b, and c and the angle between the edges α, β, and γ.

Question 15.
Briefly explain how constituent particles are arranged in (i) simple cubic, (ii) body-centered cubic, and (iii) face-centered cubic unit cells.
Answer:

1. In the simple cubic unit cell, each comer is occupied by identical atoms or ions or molecules. And they touch along the edges of the cube, do not touch diagonally. The coordination number of each atom is 6.
2. In a body-centered cubic unit cell, each comer is occupied by an identical particle, and in addition to that one atom occupies the body center. Those atoms which occupy the comers do not touch each other, however, they all touch the one that occupies the body center. Hence, each atom is surrounded by eight nearest neighbors and the coordination number is 8.
3. In a face-centered cubic unit cell, identical atoms lie at each comer as well as in the center of each face. Those atoms in the comers touch those in the faces but not each other. The coordination number is 2.

Question 16.
Distinguish between primitive and nonprimitive unit cells.
Answer:

Question 17.
Define coordination number? What is the coordination number of each constituent particle present at (i) simple cubic, (ii) body-centered cubic, and (iii) face-centered cubic unit cells?
Answer:
The number of constituent particles that surround an atom/ion in a crystal lattice is called the coordination number.

1. The coordination number of each atom in a simple cubic unit cell is 6.
2. The coordination number of each atom in a body-centered cubic unit cell is 8.
3. The coordination number of an atom is a face-centered cubic unit cell is 2.

Question 18.
Explain how will you calculate the number of particles in a unit cell.
Answer:

1. A point that lies at the comer of a unit cell is shared among the eight-unit cells and therefore, only one eight (\(\frac{1}{8}\)th) of each point lies within the unit cell.
2. A point along the edge is shared by 4 unit cells and only \(\frac{1}{4}\)th of it lies within any one cell.
3. A face-centered point is shared by 2 unit cells and only \(\frac{1}{2}\) of it is present in a given unit cell.
4. A body-centered point lies entirely within the unit cell and contributes one complete point to the cell.
   
   Total number of particles in an unit cell (z) = (\(\frac{1}{8}\) × occupied by corners) + \(\frac{1}{4}\) × (occupied by edge centres) + \(\frac{1}{2}\) × (occupied by 24 face centres) + one occupied by body centre.

Question 19.
Calculate the number of particles present in (i) simple cubic (ii) body-centered and (iii) face-centered unit cells.
Answer:

1. In simple cubic unit cell particles are present only at the comers of the cube and this is shared by 8 other unit cells. Hence, the contribution of the point at the comers to the given unit cell is
   ∴ Z = 8 × \(\frac{1}{8}\) = 1
   i.e, A simple cubic unit cell has a single constituent (atom/molecule/ion) unit per unit cell.
2. In the body-centered cubic unit cell, comers, as well as the center of the unit cell, are occupied.
   ∴ Z = 8 × \(\frac{1}{8}\) + 1 = 2
   Hence body-centered cubic (bcc) has 2 constituent units per unit cell.
3. In face-centered cubic unit cell (fee) comers as well as face centers are occupied.
   ∴ Z = 8 × \(\frac{1}{8}\) + 6 + 2 = 4
   Hence, the face-centered cubic unit cell has four constituent units per unit cell.
   [Note: Remember that the number of atoms per unit cell is in the same ratio as the stoichiometry of the compound. Hence, it helps to predict the formula of the compound.]

Question 20.
A compound formed by elements A and B has a cubic structure in which A atoms are at the comers and B atoms are at face centers. Derive the formula of the compound.
Answer:
As ‘A’ atoms are present at the 8 comers of the cube, therefore a number of atoms of A in the unit cell = 8 × \(\frac{1}{8}\) = 1.
As ‘B’ atoms are present at the face centers of the 6 faces of the cube, therefore, the number
of atoms of B in the unit cell = \(\frac{1}{2}\) × 6 = 3.
∴ Ratio of atoms A : B = 1 : 3
∴ The formula of the compound is AB₃.

Question 21.
A cubic solid is made up of two elements X and Y. Atoms ‘Y’ are present at the comers of the cube and atoms X at the body center. What is the formula of the compound? What are the coordination numbers of X and Y?
Answer:
As Y atoms are present at 8 comers, of the cube, the number of atoms of Y in the unit cell
= \(\frac{1}{8}\) × 8 = 1
As atoms X are present at the body center, therefore, the number of atoms of X in the unit cell = 1.
∴ Ratio of atoms X : Y = 1 : 1
Hence the formula of the compound is XY.
The coordination number of each X and Y = 8.

Question 22.
A compound formed by elements X and Y crystallizes in the cubic structure where Y atoms are at the comers of the cube and X atoms are at alternate faces. What is the formula of the compound?
Answer:
As there are 8 Y atoms are at the comers of the cube and the contribution of each
= \(\frac{1}{8}\) therefore no. of Y atoms per unit cell
= 8 × \(\frac{1}{8}\) = 1
These can only be 2 × atoms an alternate forces.
As contribution of each of them = \(\frac{1}{2}\).
Therefore number of X atom per unit cell
= 2 × \(\frac{1}{2}\) = 1
Ratio of atom X : Y = 1 : 1
Hence the formula of the compound is 1 : 1

Question 23.
Give the expression to find out the inter planar distance (d) between two successive planes of atoms.
Answer:

Where n is the number of planes,
λ is the wavelength of X-ray used,
θ is the angle of diffraction.
Using these values, the edge of the unit cell can be calculated.

Question 24.
X-rays of wavelength 1.54Å strike a crystal and are observed to be deflected at an angle of 22.5°. Assuming that n = 1, calculate the spacing between the planes of atoms that are responsible for this reflection.
Answer:
Applying Bragg equation,

Question 25.
Derive an expression to find the density of a crystal from lattice parameters.
Answer:
Using the edge length of a unit cell, we can calculate the density (ρ) of the crystal by considering a cubic unit cell as follows.

substitute (3) in (2)
mass of the unit cell = n × \(\frac{\mathrm{M}}{\mathrm{N}_{\mathrm{A}}}\) …(4)
For a cubic unit cell, all the edge lengths are equal i.e, a = b = c
Volume of the unit cell = a × a × a = a³ …(5)

Equation (6) contains four variables namely ρ, n, M, and a. If any three variables are known, the fourth one can be calculated.

Question 26.
The edge length of a unit cell is 408 pm. Its density is 10.6 g cm^-3. Predict whether the metal is body-centered, or face-centered, or simple cubic.
Answer:

The value of Z indicates that the metal has a fee structure.

Question 27.
An element crystallizes in bcc structure the edge length of its unit cell is 288 pm. The density of the crystal is 7.2g cm^-3. What is the atomic mass of the element?
Answer:

Question 28.
A metallic element exists as a body-centered cubic lattice. Each edge of the unit cell is 2.88 pm. The density of the metal is 7.2g cm^-3. How many atoms and unit cells are there in 100 g of the atom?
Answer:
Let the atomic mass of the element be M.

Question 29.
The density of a face-centered cubic clement is 6.25 g cm^-3. Calculate the length of the unit cell. (Atomic mass of the element = 60.2 AMU).
Answer:
Given Z = 4; M = 60.2 amu; ρ = 625 g cm^-3
N_A = 6.023 × 10²³
Applying the formula:

Question 30.
KBr has NaCl type structure. What is the distance between K⁺ and Br^– in KBr, if the density is 2.75 g cm^-3?
Answer:
Edge length ‘a’ of the unit cell is calculated as

Question 31.
Derive an expression to calculate the packing efficiency in a simple cubic arrangement.
Answer:
Let us calculate the packing efficiency in simple cubic arrangement,

Let us consider a cube with an edge length ‘a’ as shown in fig.
Volume of the cube with edge length a is = a × a × a = a³
Let V is the radius of the sphere. From the figure, a = 2r ⇒ r = \(\frac{a}{2}\)
∴ Volume of the sphere with radius ‘r’

In a simple cubic arrangement, the number of spheres that belongs to a unit cell is equal to one
∴ Total volume occupied by the spheres in sc unit cell = 1 × \(\left(\frac{\pi a^{3}}{6}\right)\) ……(2)
Dividing (2) by (3)

i.e., only 52.31% of the available volume is occupied by the spheres in simple cubic packing, making inefficient use of available space and hence minimizing the attractive forces.

Question 32.
Show that the packing efficiency in the face-centered cubic unit cells is 74%.
Answer:
The cubic close packing is based on the face-centered cubic unit cell. Let us calculate the packing efficiency in the fcc unit cell.


The total number of spheres belongs to a single fcc unit cell is 4

Question 33.
Give the relationship between the nearest neighbor distance (d) and the edge (a) of the unit cell of a cubic crystal.
Answer:
For simple cubic: d = a
For face centered cubic: d = \(\frac{a}{\sqrt{2}}\) = 0.707a
For body centered cubic:
d = \(\frac{\sqrt{3}}{2} a\)

Question 34.
Give the relationship between atomic radius (r), (which is d/2 for crystals of pure substances) and edge (a) of the unit cell of a cubic crystal.
Answer:
For simple cubic: r = \(\frac{a}{2}\)

Question 35.
Xenon crystallizes in face-centered cubic lattice and the edge of the unit cell is 620 pm. What is the nearest neighbor distance and what is the radius of the xenon atom?
Answer:
Here, a = 620 pm; d = 2 ; r = ?

Question 36.
CsCl has bcc arrangement and its unit cell edge length is 400 pm. Calculate the interionic distance in CsCl.
Answer:

The bcc arrangement of CsCl is shown in fig. Where black circle is Cs⁺ ion and colored circles are Cl^– ions. The aim is to find half of the body diagonal AE. If the edge of the unit cell is ‘a’.

Question 37.
If the radius of the atom is 75 pm and the lattice type is body-centered cubic, what is the edge of the unit cell?
Answer:

Question 38.
The radius of an atom of an element is 500 pm. If it crystallizes as fee lattice, what is the length of the side of the unit cell?
Answer:

Question 39.
A solid AB has a CsCl type structure. The edge length of the unit cell is 404 pm. Calculate the distance of the closest approach between A⁺ / B^– ions.
Answer:
The distance of the closest approach is equal to the distance between nearest neighbors (d). As CsCl has a bcc lattice.

Question 40.
What is the radius ratio? Mention its importance.
Answer:
In ionic solids, the ratio of the radius of the cation to the radius of an anion is known as the radius ratio.

The radius ratio of an ionic solid gives the structural arrangement of ionic solids.

Question 41.
What is the meaning of the term imperfection in solids?
Answer:
Imperfection refers to the departure from the perfect periodic arrangement of atoms, ions, or molecules in the structure of crystalline solids.

Question 42.
What are the types of lattice imperfections found in crystals?
Answer:

1. Stoichiometric defect i.e., Schottky defect and Frenkel defect.
2. Nonstoichiometric defect viz metal excess, metal deficiency, and impurity defects.

Question 43.
What are interstitials in a crystal?
Answer:
Atoms or ions that fill the normal vacant interstitial voids in a crystal are known as interstitials.

Question 44.
What is the Schottky defect?
Answer:
If an equal number of cations and anions are missing from their lattice sites, the defect is known as the Schottky defect.

Question 45.
What is Frenkel’s defect?
Answer:
When some ions (usual cation) are missing from the lattice sites and they occupy the interstitial sites. So that, the electro-neutrality, as well as stoichiometry are maintained, it is called Frenkel defect.

Question 46.
Which crystal defect lowers the density of the solid?
Answer:
Schottky defect.

Question 47.
Which crystal defect in crystals does not alter the density of a relevant solid?
Answer:
Frenkel defect.

Question 48.
Which point does defect increase the density of the solid?
Answer:
Schottky defect.

Question 49.
Which point defect in the crystal increases the density of the solid?
Answer:
Interstitial defect.

Question 50.
Name one solid in which both Frenkel and Schottky defects occur.
Answer:
Silver Bromide.

Question 51.
Why does the Frenkel defect does not change the density of AgCl crystal?
Answer:
Because in Frenkel defect, no ion is missing from the lattice sites. Therefore, no change in density occurs.

Question 52.
What are F. Centres?
Answer:
The free-electron trapped in the anion vacancy is called F.Centres.

Question 53.
What are non-stoichiometric defects?
Answer:
If as a result of imperfections in the crystal, the ratio of the cation to the anions becomes different from that indicated by their ideal chemical formula, then the defect is termed a non-stoichiometric defect.

Question 54.
Why does table salt, NaCl appear yellow in color?
Answer:
Yellow color in sodium chloride is due to metal excess defect due to which unpaired electrons occupy anionic sites. These sites are called F. Centres. These electrons absorb energy from the visible region for excitation which makes the crystal yellow.

Question 55.
Why is FeO (s) not formed in stoichiometric composition?
Answer:
In the crystal of FeO, some of Fe⁺² cations are replaced by Fe⁺³ cations. Three Fe⁺² cations are replaced by two Fe⁺³ cations to make up for the loss of positive charge. Eventually, there would be fewer atoms of the metal as compared to stoichiometric proportion.

Question 56.
How would you account for the following?

1. Frenkel defects are not found in alkali metal halides.
2. Schottky defects lower the density of the solid.
3. Impurity doped silicon is a semiconductor.

Answer:

1. This is because alkali metal ions have larger sizes that cannot fit into interstitial sites.
2. As the number of ions decreases as a result of the Schottky defect, the mass decreases whereas the volume remains the same.
3. This is due to additional electrons or the creation of holes and doping with impurity. Creation of hole results in p-type semiconductor and creation of electron results in the n-type semiconductor.

Question 57.
Briefly explain the metal excess defect.
Answer:
The metal excess defect arises due to the presence of more metal ions as compared to anions. Alkali metal halides NaCl, KCl show this type of defect. The electrical neutrality of the crystal can be maintained by the presence of anionic vacancies equal to the excess metal ions (or) by the presence of extra cation and electrons present in the interstitial position.

Question 58.
Briefly explain metal deficiency defects.
Answer:
Metal deficiency defect arises due to the presence of fewer cations than the anions. This defect is observed in a crystal in which, the cations have variable oxidation states.

Question 59.
Briefly explain the impurity defect.
Answer:
A general method of introducing defects in ionic solids is by adding impurity ions. If the impurity ions are in a different valance states from that of the host, vacancies are created in the crystal lattice of the host. For example, the addition of CdCl₂ to silver chloride yields solid solutions where the divalent cation Cd²⁺ occupies the position of Ag⁺. This will disturb the electrical neutrality of the crystal. In order to maintain the same, a proportional number of Ag⁺ ions leaves the lattice. This produces a cation vacancy in the lattice, such kinds of crystal defects are called impurity defects.

Choose the correct answer.

1. Which one of the following is a covalent crystal?
(a) Rock salt
(b) Ice
(c) Quartz
(d) Dry ice
Answer: (c)
Hint: Rock salt is an ionic solid, while dry ice and ice are molecular solids.

2. Total volume of atoms present in a face centered cubic unit cell of a metal is:
(r = atomic radius)

Answer: (b)
Hint: Total no. of atoms in fee unit cell = 4

3. The fraction of the total volume occupied by the atoms present in a simple cube is:

Answer: (b)
In a simple cube, no. of atoms / unit cell = 8 × \(\frac{1}{8}\) = 1

4. Three elements A, B and C crystallise into a cubic solid lattice. Atoms A occupies the comers, B atoms, the cubic centres, and atoms C, the edges. The formula of the compound is:
(a) ABC
(b) ABC₂
(c) ABC₃
(d) ABC₄
Answer: (c)
Hint:
Atoms A per unit cell = 8 × \(\frac{1}{8}\) = 1
Atom B per unit cell = 1
Atoms C per unit cell = 12 × \(\frac{1}{4}\) = 3
Ratio A : B : C = 1 : 1 : 3.
Hence the formula is ABC₃.

5. An alloy of copper, silver and gold is found to have copper forming the simple cubic close packed lattice. If silver atoms occupy the comers and gold atoms are present at the body centres, the formula of the alloy will be:
(a) Cu₃ Ag Au
(b) Cu Ag₃ Au
(c) Cu₄ Ag₂ Au
(d) Cu Ag Au.
Answer: (b)
Hint:
Cu atoms per unit cell (present at the comer) = 8 × \(\frac{1}{8}\) = 1
Ag atoms per unit cell (present at the face centres) = 6 × \(\frac{1}{2}\) = 3
Au atoms per unit cell (present at body centre) = 1
Hence the formula is Cu Ag₃ Au.

6. A solid compound XY has NaCl structure. If the radius of the cation is 100 pm, the radius of the anion (Y^–) will be:
(a) 275.1 pm
(b) 322.5 pm
(c) 241.5 pm
(d) 165.7 pm
Answer: (c)
Hint: NaCl has a face centered cubic structure. Cl^– ions and Na⁺ ions are present in the octahedral voids. Hence for such a solid, radius of the cation is = 0.414 × radius of the anion.

7. The number of atoms in 100g of the fcc crystal with density = 10 g/cm³, and the cell edge equal to 200 pm is equal to:
(a) 5 × 10²⁴
(b) 5 × 10²⁵
(c) 6 × 10²³
(d) 2 × 10²⁵
Answer:
(a)
Hint:

Thus 12g of the atom contain = 6.023 × 10²³ atom.
∴ 100g of the atom will contain

8. Which of the following statements is not true about amorphous solids?
(a) On heating they may become crystalline at certain temperature.
(b) They may become crystalline or keeping for long time.
(c) Amorphous solids can be moulded by heating.
(d) They are anisotropic in nature.
Answer:
(d)

9. The sharp melting point of crystalline solids is due to:
(a) a regular arrangement of constituent particles observed over a short distance in the crystal lattice.
(b) a regular arrangement of constituent particles observed over a long distance in the crystal lattice.
(c) same arrangement of constituent particles in different directions.
(d) different arrangement of constituent particles in different directions.
Answer:
(b)

10. Iodine molecules are held in the crystal lattice by:
(a) London forces
(b) dipole – dipole interaction
(c) covalent bonds
(d) columbic forces
Answer:
(a)

11. Which of the following is a network solid?
(a) SO₂ (solid)
(b) I₂
(c) diamond
(d) H₂O (ice)
Answer:
(c)

12. Which of the following solids is not an electrical conductor?
(i) Mg(s)
(ii) TiO(s)
(iii) I₂ (s)
(iv) H₂O (ice)
(a) (i) only
(b) (ii) only
(c) (iii) and (iv)
(d) (ii), (iii) and (iv)
Answer:
(c)

13. The lattice sites in a pure crystal cannot be occupied by:
(a) molecule
(b) ion
(c) electron
(d) atom
Answer:
(c)
Hint: Electrons can occupy only in the interstitial sites.

14. Cations are present in interstitial sites in:
(a) Frankel defect
(b) Schottky defect
(c) Valency defect
(d) Metal deficiency defect
Answer:
(a)

15. Schottky defect is obtained in crystal when:
(a) Some cations move from their lattice sites to interstitial sites.
(b) equal number of cations and anions are missing from the lattice.
(c) Some lattice sites are occupied by electrons.
(d) Some impurity is present in the lattice.
Answer:
(b)

16. The total number of tetrahedral void in the face centered unit cell is:
(a) 6
(b) 8
(c) 10
(d) 12
Answer:
(b)
Hint: No. of atoms per unit cell in fee = 4
No. of tetrahedral void: 2 × 4 = 8.

17. Which of the following statements is not true about hexagonal close packing?
(a) The coordination number is 12.
(b) It has 74% packing efficiency.
(c) Tetrahedral voids of the second layer are covered by spheres of the third layer.
(d) In this arrangement spheres of the fourth layer are exactly aligned with those of the first.
Answer: (d)
Hint: In hcp arrangement ABAB type and not ABC ABC type. Hence (d) is not true.

18. What is the coordination number in square close packed structure in two dimensions?
(a) 2
(b) 3
(c) 4
(d) 6
Answer:
(c)
Hint: The arrangement is AA AA type. In this arrangement each sphere is touching four other spheres touching the particular sphere, a square is formed. Hence, the coordination number is 4.

19. The correct order to packing efficiency in different type of unit cells is:
(a) fee < bee < simple cubic
(b) fee > bcc > simple cubic
(c) fee < bcc > simple cubic
(d) bcc < fee > simple cubic
Answer:
(b)
Hint: fcc = 74%, bcc = 68%; simple cubic = 52.4%

20. In cubic close packing, the unit cell has:
(a) 4 tetrahedral voids each of which is shared by adjacent unit cells. .
(b) 4 tetrahedral voids within the unit cell.
(c) 8 tetrahedral voids each of which is shared by four adjacent unit cells.
(d) 8 tetrahedral voids within the unit cells.
Answer:
(d)
Hint: ccp = hcp
No. of atoms per unit cell in fcc unit cell = 4.
∴ No. of tetrahedral voids = 2 × 4 = 8

21. KCl crystallises in the same type of lattice as does NaCl.
Given that \(r_{\mathrm{Na}^{+}} / r_{\mathrm{Cl}^{-}}\) is 0.55 and \(r_{\mathrm{K}^{+}} / r_{\mathrm{Cl}^{-}}\) is 0.74, Calculate the ratio of the side of the unit cell of KCl and NaCl.
(a) 1.123
(b) 0.891
(c) 1.414
(d) 0.414
Answer:
(a)

22. Ice crystallises in a hexagonal lattice having a volume of the unit cell is 132 × 10 cm³. It density of ice at the given temperature is 0.92 g cm^-3, then number of H₂O molecules per unit cell is:
(a) 1
(b) 2
(c) 3
(d) 4
Answer:
(d)
Hint:

24. The edge length of the face-centered cubic unit cell is 508 pm. If the radius of the cation is 110 pm, the radius of the anion will be:
(a) 144 pm
(b) 288 pm
(c) 618 pm
(d) 398 pm
Answer:
(a)
Hint: For the face-centered cubic unit cell (eg: NaCl),
the edge length = 2 × distance between cation and anion
= 2 (r₊ + r_–)
2(r₊ + r_–) = 508 pm
given r₊ = 110 pm
2(110 + r_–) = 508 pm
r_– = 144 pm

25. Which of the following statements is not correct?
(a) The fraction of the total volume occupied by the atoms in a primitive cell is 0.48
(b) Molecular solids are generally volatile.
(c) The number of carbon atoms in a unit cell of a diamond is 4.
(d) The number of brave lattices in which a crystal can be categorized it.
Answer:
(c)

26. Match the entities with a column I with appropriate entities in column II and choose the correct option using the code given.

(a) (A) – r; (B) – p, r, s; (C) – r, (D) – q
(b) (A) – p,r; (B) – s; (C) – r; (D) – q
(c) (A) – r; (B) – p,s; (C) – q; (D) – r
(d) (A) – r, (B) – s; (C) – r; (D) – p
Answer:
(a)

27. Match the entities with column I with appropriate entities in column II and choose the correct option using the code given.

a = cell edge
d = nearest neighbour distance
r = radius
(a) (A) – q; (B) – r; (C) – p; (D) – s
(b) (A) – q; (B) – p; (C) – r; (D) – 5
(c) (A) – q; (B) – s; (C) – q; (D) – p
(d) (A) – q; (B) – s; (C) – r; (D) – p
Answer:
(c)

28. Match the entities with column I with appropriate entities in column II and choose the correct option using the code given.

(a) (A) – s; (B) – q; (C) – r; (D)-p
(b) (A) – q; (B) – p; (C) – r; (D) – s
(c) (A) – r; (B) – s; (C) – q; (D) – p
(d) (A) – s; (B) – r; (C) – p; (D) – q
Answer:
(a)

29. Assertion (A): Hexagonal close packing is more closely packed than cubic close packing.
Reason (R): Hexagonal close-packing has a coordination number 12 whereas the cubic close packing has a coordination number 8.
(a) If both assertion and reason are true and the reason is the correct explanation of assertion.
(b) If both assertion and reason are true, and the reason is not the correct explanation of assertion.
(c) If the assertion is true but the reason is false.
(d) If both assertion and reason are false.
Answer:
(d)
Hint:
Correct assertion: Hexagonal close packing and cubic close packing are equally close-packed with a packing efficiency of 74%.
Correct reason: Both have the coordination number 12.

30. Assertion (A): Frankel defects are shown by silver halides.
Reason (R): Ag⁺ ion is smaller in size.
(a) If both assertion and reason are true and the reason is the correct explanation of assertion.
(b) If both assertion and reason are true, and reason is not the correct explanation of assertion.
(c) If the assertion is true but the reason is false.
(d) If both assertion and reason are false.
Answer:
(a)

Students get through the TN Board 12th Chemistry Important Questions Chapter 5 Coordination Chemistry which is useful for their exam preparation.

TN State Board 12th Chemistry Important Questions Chapter 5 Coordination Chemistry

Answer the following questions.

Question 1.
What is the coordination number of the central metal ions in the following complexes?
(i) [Cu(NH₃)₄]⁺²
(ii) [Fe(C₂O₄)₃]^-3
(iii) [Pt (en)₂Cl₂]
(iv) [Mo(CN)₈]^-4
(v) Fe[EDTA]]^–
(vi) [Pd(H₂O)₂(ONO)₂I₂]
Answer:
(i) NH₃ is a monodentate ligand.
Point of attachment with Cu⁺² = 4 x 1 = 4
C.N of Cu⁺² = 4
(ii) C₂O₄^-2 is bidentate ligand.
Point of attachment with Fe⁺³ = 3 x 2 = 6
C.N of Fe⁺³ = 6
(iii) ‘en’ is a bidentate ligand and Cl^– is a monodentate ligand.
Point of attachment with
Pt⁺² = 2 x 2 + 2 x 1 = 6
C.N of Pt⁺² = 6
(iv) CN^– is a monodentate ligand.
Point of attachment with Mo⁺⁴ = 8 x 1 = 8
C.N of Mo⁺⁴ = 8
(v) EDTA is a hexadentate ligand.
Point of attachment with Fe⁺² = 6 x 1 = 6
C.N of Fe⁺² = 6
(vi) Point of attachment with
Pd⁺⁴ = 2 x 1 + 2 x 1 + 2 x 1 = 6
C.N of Pd⁺² = 6

Question 2.
Calculate the oxidation state, of the central metal atom, in the following:
(i) [CO(NH₃)₅Cl]⁺²
(ii) K₄[Fe(CN)₆]
(iii) [Co(NO₂)₂(Py)₂(NH₃)₂] NO₃
(iv) Ni[CO]_4/
(v) [Fe(EDTA)]^–
Answer:
(i) x + 5 × (0) – 1 = + 2
x = 2 + 1 = 3
oxidation state of cobalt = +3
(ii) 4 × (+1) + x + 6 (-1) = 0
x = +6 – 4 = +2
oxidation state of Iron = + 2.
(iii) x + 2 (-1) + 2(1) + 2(0) – 1 = 0
x = 2 + 1 = 3
oxidation state of cobalt = +3
(iv) x + 4 (0) = 0 or x = 0:
oxidation State of nickel = 0
(v) x + 1 × x (-4) = -1
x = 4 – 1 = +3
oxidation state of Iron = +3

Question 3.
Give the IUPAC names of the following compounds:
(i) K₃[Al(C₂O₄)₃]
(ii) [Pt(NH₃)₄(NO₂)Cl] SO₄
(iii) K₃[Cr(CN)₆]
(iv) [CO(NH₃)₅ONO]Cl₂
(v) [Cr(NH₃)₅CO₃]Cl
(vi) [Cr(NH₃)₅ (NCS)][ZnCl₄]
(vii) K[Au(CN)₂]
Answer:
(i) Pottasium trioxalatoalurhinate (III)
(ii) Tetraamminechloridonitroplatinum (IV) sulphate.
(iii) Potassiumhexacyanochromate (III)
(iv) Pentaammine nitritocobalt (III) chloride.
(v) Pentaammine carbonate chromium (III) chloride.
(vi) Pentaammine isothiocyanato chromium (III) chloride.
(vii) Potassium dicyanoaurate (I).

Question 4.
Write the formula of the following co-ordination compounds.
(i) Tetraammine diaqua cobalt (III) chloride.
(ii) Potassium tetracyano nickelate (II).
(iii) Tris (ethane 1,2, diamine) chromium (III) chloride.
(iv) Amminebromidochloridonitrito-N- platinate (III).
(v) Dichlorido bis (ethane 1, 2, diamine) platinum (IV) nitrate.
(vi) Iron (III) hexacyanoferrate (III).
Answer:
(i) [Co(NH₃)₄(H₂O)₂]Cl₃
(ii) K₂[Ni(CN)₄]
(iii) (Cr (en)₃]Cl₃
(iv) [Pt(NH₃)Br Cl (NO₂)]‘-
(v) [Pt Cl₂ (en)₂] (NO₂)₂
(vi) Fe₄(Fe(CN)₂]₃

Question 5.
Write the IUPAC names of the following
co-ordination compounds:
(i) [Co(NH₃)₆]Cl₃
(ii) [Co(NH₃)₅Cl]Cl₂
(iii) K₃[Fe(CN)₆]
(iv) K₃[Fe (C₂O₄)₃]
(v) K₂[Pd Cl₄]
(vi) Pt [(NH₃)₂ Cl (NH₂CH₃)]Cl.
Answer:
(i) Hexaammine cobalt (III) chloride.
(ii) Pentaammine chloridocobalt (III) chloride.
(iii) Potassium hexacyano ferrate (III).
(iv) Potassium trioxalato ferrate (III).
(v) Potassium tetrachlorido palladate (II).
(vi) Diammine chlorido (methylamine) platinum (II) chloride.

Question 6.
Using IUPAC names, write the formula of the following:
(i) Tetra hydrazozincate (II)
(ii) Hexaammine cobalt (III) sulphate.
(iii) Potassium trioxalato chromate (III)
(iv) Diammine dichlorido platinum (II)
(v) Tetra bromido cuprate (II)
(vi) Pentaamine nitrito-O-cobalt (III)
(vii) Pentaamine nitrito-N-cobalt (III)
Answer:
(i) [Zn(OH)₄]^-2
(ii) [CO(NH₃)₆]₂SO₄
(iii) K₃[Cr (C₂O₄)₃]
(iv) [Pt(NH₃)₂Cl₂]
(v) [Cu Br₄]^-2
(vi) [CO(NH₃)₅ (ONO)]⁺²
(vii) [CO(NH₃)₅NO₂]⁺²

Question 7.
Specify the oxidation numbers of the metal ions in the following coordination entities.
(i) [CO(H₂O) (CN) (en)₂]⁺²
(ii) [Pt Cl₄]^-2
(iii) [Cr(NH₃)₃ Cl₃]
(iv) [Co Br₂ (en)₂]⁺
(v) K₃[Fe(CN)₆]
Answer:
(i) x + 0 + (-1) + 0 = + 2; x = +3
(ii) x – 4 = – 2 ; or x = +2
(iii) x + 0 + 3(-1) = 0; x = +3
(iv) x – 4 = -2 or x = +2
(v) 3(1) + x + 6(-1) = 0; or x = +3

Question 8.
Aqueous copper sulphate (blue in colour) gives (i) a green precipitate with aqueous potassium fluoride and (ii) a bright green solution with aqueous potassium chloride. Explain these experimental results.
Answer:
Aqueous copper sulphate exists as [Cu(H₂O)₄] SO₄. It is a labile complex. The blue colour is due to [Cu(H₂O)₄]⁺² ions.
(i) When KF is added, weak H₂O ligands are replaced by F^– ligands forming [CuF₄]^-2 ions which is a green precipitate.

(ii) When KCl is added, Cl^– ligands replaced by weak H₂O ligands forming [CuCl₄]^-2 ion which has a bright green colour.

Question 9.
Give two examples for each of the following:
(i) Cationic complex
(ii) Anionic complex
(iii) Neutral complex
Answer:
(i) [Ag(NH₃)₂]⁺, [Co(NH₃)₆]⁺³
(ii) [Co (CN)₆]^-3, [Fe(CN)₆]^-4
(iii) [Ni(CO)₄], [CO(NH₃)₃ Cl₃]

Question 10.
What are homoleptic and heteroleptic complexes? Give one example for each.
Answer:
A coordination compound in which the central metal atom / ion is co-ordinated to only one kind of ligand is called homoleptic complex. Examples (Co(NH₃)₆]⁺³, [Fe(H₂O)₆]⁺³.
A coordination compound in which the central metal atom / ion is coordinated to more than 6ne kind of ligand is called heteroleptic complex. Example [CO(NH₃)₅Cl]⁺², [Pt(NH₃)₂Cl₂].

Question 11.
Identify the following complexes as homoleptic or heteroleptic complexes.
(i) [Fe(CN)₆]^-4
(ii) [CO(NH₃)₄Cl₂]⁺
(iii) [Cr(en)₃]⁺³
(iv) [Ag(NH₂)₂]⁺
(v) [Fe(CN)₅NO]^-3
Answer:
(i) Homoleptic complex ion.
(ii) Heteroleptic complexion.
(iii) Homoleptic complexion.
(iv) Homoleptic complexion.
(v) Heteroleptic complex ion.

Question 12.
Explain the following giving an example in each case: (i) Linkage isomerism, (ii) Coordination isomerism, (iii) Ionisation isomerism, (iv) Solvate or hydrate isomerism.
Answer:
(i) Linkage isomerism This type of isomers arises when an ambidentate ligand is bonded to the central metal atom/ion through either of its two different donor atoms, eg: In the nitrite ion is bound to the central metal ion Co³⁺ through a nitrogen atom in one complex,and through oxygen atom in other complex.
[CO(NH₃)₅(NO₂)]^2-

(ii) Coordination isomerism: This type of isomerism arises in coordination compounds having both complex cations and complex anions. The interchange of one or more ligands between the cationic and the anionic coordination entities gives different isomers, eg:
[CO(NH₃)₆] [Cr(CN)₆] and [Cr(NH₃)₆] [CO(CN)₆]
[Cr(NH₃)₆] [Co(C₂O₄)₃] and [Co(NH₃)₆]
[Cr(C₂O₄)₃]
[Co(en)₃] [Cr(CN)₆] and [Cr(en)₃] [CO(CN)₆]
(iii) Ionisation isomerism: This type of isomerism arises when the coordination compounds give different ions in solution. For example, the complex having the formula [Co(NH₃)₅ BrSO₄] exists in two isomeric forms.
[CO(NH₃)₃ Br]SO₄ and [Co(NH₃)₅SO₄] Br. They give different ions in an aqueous solution.

(iv) Solvate or hydrate isomerism: This isomerism arises when free solvent molecules like H₂0, ammonia or alcohol present in the coordination entity are exchanged with the ligands outside the coordination entity.
eg: [Cr(H₂O)₆]Cl₃, [Cr(H₂O)₅Cl]Cl₂ H₂O [Cr(H₂O)₄Cl₂] Cl₂H₂O.

Question 13.
Give the ionisation isomer of the following and write their IUPAC names.
(i) [Pt(NH₃)₄ Cl₂] Br₂
(ii) [CO(NH₃)₄Cl₂] NO₂
Answer:
(i) The ionisation isomer of [Pt(NH₃)₄ Cl₂] Br₂ is [Pt(NH₃)₄ Br₂] Cl₂. Its IUPAC name is tetraammine dibromido platinum (IV) chloride.
(ii) The ionisation isomer of [CO(NH₃)₄ Cl₂] NO₂ is [CO(NH₃)₄Cl.NO₂]Cl. Its IUPAC name is tetraammine chlorido nitro cobalt (III) chloride.

Question 14.
Briefly outline the geometrical isomerism exhibited by square planar and with an example.
Answer:
Square planar complexes: In square planar complexes of the form [MA₂B₂]^n± and [MA₂BC]^n± (where A, B and C are mono dentate ligands and M is the central metal ion/ atom), Similar groups (A or B) present either on same side or on the opposite side of the central metal atom (M) give rise to two different geometrical isomers and they are called, cis and trans isomers respectively. The square planar complex of the type [M(xy)₂]^n± where xy is a bidentate ligand with two different coordinating atoms also shows cis-trans isomerism. Square planar complex of the form [MABCD]^n± also shows geometrical isomerism. In this case, by considering any one of the ligands (A, B, C or D) as a reference, the rest of the ligands can be arranged in three different ways leading to three geometrical isomers.
eg:

Question 15.
Write the structure of geometrical isomers in octahedral complexes of the type Ma₄b₂, Ma₂b₄, Ma₄bc, Ma₃b₃ where a + b are monodentate ligands. Give example for each type.
Answer:
(i) Ma₄b₂ type:


(ii) Ma₃b₃ type:

(iii) M(aa₃)₂b₂ or M(aa)₂ bc type:
where aa is a symmetrical bidentate ligand, b and c are monodentate ligands, eg: [Co(en)₂Cl₂]⁺

Question 16.
Mention the types of stereoisomerism exhibited by coordination compounds.
Answer:
Coordination compounds exhibit geometrical and is optical isomerism.

Question 17.
Explain the geometrical isomerism exhibited by the [MA₃B₃]^±n where A and B are monodentate ligands. [OR]
Explain the terms facial and meritorial isomers.
Answer:
Octahedral complex of the type [MA₃B₃]^n± also shows geometrical isomerism. If the three similar ligands (A) are present in the comers of one triangular face of the octahedron and the other three ligands (B) are present in the opposing triangular face, then the isomer is referred as a facial isomer (fac isomer). If the three similar ligands are present around the meridian which is an imaginary semicircle from one apex of the octahedral to the opposite apex as shown in the figure, the isomer is called as a meridional isomer (mer isomer). This is called meridional because each set of ligands can be regarded as lying on a meridian of an octahedron.

Question 18.
Write the formula of coordination isomers of the following. Write their IUPAC names.
(i) [Pt(NH₃)₄] [CuCl₄]
(ii) [Cr(NH₃)₆] [Co(C₂O₄)₃]
(iii) [Cr(NH₃)₆] [Cr(SCN)₆]
(iv) Co(en)₃[Cr(CN)₆]
Answer:
(i) The coordination isomer of [Pt(NH₃)₄] [CuCl₄] is [Cu(NH₃)₄] [PtCl]₄. Its IUPAC name is Tetraammine copper (II) tetrachlorido platinate (II).
(ii) The coordination isomer of [Cr(NH₃)₆] [Co(C₂O₄)₃] is [CO(NH₃)₆] [Cr(C₂O₄)₃]. Its IUPAC name is Hexaammine cobalt (III) trioxalato chromate (III).
(iii) The coordination isomer of [Cr(NH₃)₆] [Cr(SCN)₆] is [Cr(NH₃)₄(SCN)₂] [Cr(NH₃)₂(SCN)₄] Tetraammine dithiocyanatochromium (III) diammine tetrathiocyanatochromate (III).
(iv) The coordination isomer of [Co(en)₃] [Cr(CN)₆] is [Cr(en)₃]₄ [Co(CN)₆], Its ionisation isomer is Tri (ethane 1, 2, diamine) chromium (III) hexacyano cobaltate (III).

Question 19.
Discuss the bonding in metal carbonyls.
Answer:
The metal-carbon bond in metal carbonyls have both σ and π bond character. The metal – carbon sigma bond is formed by the donation of lone pair of electrons from the carbonyl carbon into a vacant orbital of the metal. The metal- carbon pi bond is formed by the donation of a pair of electrons from a filled d orbital of the metal to the vacant antibonding pi molecular orbital of carbon monoxide. The metal ligand bonding causes a synergic effect which strengthens the bond between Co and the metal.

Question 20.
Indicate the type of isomerism exhibited by the following complexes and draw structures for these isomers.
(i) KCr (H₂O)₂(C₂O₄)₃
(ii) [Co(en)₃]Cl₃
(iii) [ CO(NH₃)₅(NO₂)(NO₃)₂]
(iv) [Pt (NH₃) (H₂O)Cl₂]
Answer:
(i) Both show geometrical (cis-trans) isomerism. Optical isomerism is shown by cis form.

(ii) Two optical isomers can exist. [Co(en)₃]⁺

(iii) There are 10 possible (geometrically ionisation and linkage) isomers.
Ionisation isomers:

(iv) Geometrical (cis-trans) isomerism can exist.

Question 21.
Give evidence to show that [CO(NH₃)₅Cl]SO₄ and [Co(NH₃)₅ SO₄]Cl are ionisation isomers.
Answer:
The ionisation isomers dissolve in water to yield different ions and thus react differently with different reagents.

A white precipitate is than obtained.

Question 22.
Draw the structure of optical isomers of (i) PtCl₂ (en)₂, (ii) [Cr(NH₃)₂ Cl₂ (en)₂]⁺
Answer:

Question 23.
Write all the geometrical isomers of [Pt (NH₃) BrCl Py] and how many of these exhibit optical isomerism?
Answer:
Three isomers are possible.

Isomers of this type do not show optical isomerism. Optical isomerism rarely occurs in square planar or tetrahedral complexes and that too when they contain asymmetrical chelating ligand.

Question 24.
Draw the structure of all the isomers (geometrical and optical) of (i) [CO(NH₃)Cl (en)₂]⁺² (ii) [Co(NH₃)₂Cl₂(en)]⁺
Answer:
(i)

I and II are optical isomers.
II and III are geometrical isomers.
(ii)

I and II are geometrical isomers.
II and III are optical isomers.

Question 25.
Draw the structure of geometrical isomers of [Fe(NH₃)₂(CN)₄]^–.
Answer:

Question 26.
Out of the following two coordination entities which is optically active? (i) cis [CrCl₃(oX)₂]^-3(ii) trans [Cr Cl₂ (oX)₂]^-3 The two entities are represented as follows:
Answer:

Of these cis form is optically active.

Question 27.
Write the structures of isomers, if any and write the names of the following complexes. (i) [Cr (NH₃)₄ Cl₂]⁺, (ii) [Co(en)₃]⁺
Answer:
(i) Geometrical isomers of [Cr(NH₃)₄Cl₂]⁺

(ii) Optical isomers of OX [Co(en)₃]⁺³

Question 28.
Platinum (II) forms’square planar complexes and platinum (IV) gives octahedral complexes.. How many geometrical isomers are possible for each of the following. Draw their structures. (a) [Pt(NH₃)₃Cl]⁺, (b) [Pt(NH₃)Cl₅]^–, (c) [Pt(NH₃)₂ClNO₂], (d) [Pt(NH₃)₄Cl Br]⁺².
Answer:
(a) No isomer is possible for a square planar complex MA₃B.

(b) No isomer is possible for an octahedral complex of the type MAB₅.

(c) cis-trans isomers are possible for a square planar complex MA₂BC.

(d) cis-trans isomers are possible for an octahedral complex of the type MA₂BC.

Question 29.
Explain the term (i) inner orbital complex and (ii) outer orbital complex with an example for each.
Answer:
(i) Bonding in complexes are explained in terms of hybridisation. If the hybridisation of central metal atom/ion involves the use of (n-1) d, ns and np orbitals, then d²sp³ hybridisation is possible for octahedral complexes. Complexes formed by d²sp³ hybridisation of the central metal atom or ion is known as inner orbital complex. [Fe(CN)₆]^-4, [Fe(CN)₆]^-3, [Cr(NH₃)₆]⁺³ are examples of inner orbital complexes.
(ii) If the hybridisation of the metal ion involved the ns, np and nd orbitals and for octahedral complexes sp³d² hybridisation involved, the complex formed is known as outer orbital complex. [CoF₆]^-3 is an example for outer orbital complex.

Question 30.
[Co(CN)₆]^-3 and [CoF₆]^-3 are both octahedral complexes. Then what is the difference between the two?
Answer:
In both complexes, Co is in +3 oxidation state. It has 6 electrons in the ‘d’ subshell (d⁶).

In the formation of [Co(CN)₆]^–, the ‘d’ electrons in 3d level gets paired leaving behind two 3d orbital vacant.

The Co⁺³ ion makes use of two 3d orbitals, one 4s and two 4p orbitals for hybridisation. d² sp³ hybridisation and two 4p orbitals for hybridisation.

The CN^– ion donates the electron pairs to the d2spi hybrid orbitals and form six covalent bond. Thus [(CO(CN)₆]^-3 is an inner orbital or low spin complex.
On the other hand, in the formation of [CoF₆]^-3 the Co⁺³ ion makes use of outer 4s, 4p and 4d orbitals for hybridisation.

In the presence of F^– ion, electron pairing does not occur in 3d orbitals, and Co⁺³ ion makes use of one 4s, three 4p and two 4d orbitals for hybridisation.

The six fluoride ion donates a pair of electrons to the six sp³d² hybrid orbitals and form six-coordinate covalent bonds.

Since outer orbitals are used in the formation it is an outer orbital complex. Since more unpaired electrons are present it is also known as high spin complex.

Question 31.
Discuss the nature of bonding in the following coordination entities on the basis of valence bond theory.
(i) [Fe(CN)₆]^-4
(ii) [FeF₆]^-3
(iii) [Co(C₂O₄)₃]^-3
(iv) [NiCl₄]^-2
Answer:
(i) [Fe(CN)₆]^-4: d²sp³ hybridisation octahedral: diamagnetic.
In this complex, the oxidation state of iron is +2.
Electronic configuration of Fe = [Ar]3d₄⁶s²
Electronic configuration of Fe⁺² = [Ar]3d⁶
To accommodate six pairs of electrons from six cyanide ions, the iron (II) must make available six empty orbitals. This can be achieved by the d²sp³ hybridisation.

In the presence of CN^– ion, a strong ligand, the electrons in 3d subshell gets paired and two of the 3d one 4s and three 4p orbitals hybridise to produce six d²sp³ hybrid orbitals.
Fe⁺³ (d⁶) excited state.

Thus, the six pairs of electrons from six cyanide ion occupy the six hybridised orbitals of iron (II) ion. [Fe(CN)₆]^-4

Since, there are no unpaired electron present, it is diamagnetic and its geometry is octahedral.

(ii) [FeF₆]^-3: This complex is high spin or octahedral, since the central atom Fe(II) utilises nd orbitals for hybridisation. It is an octahedral complex involving sp³d² hybridisation. Each orbital accommodates a lone pair of electrons from six fluoride ions as shown below. [Fe⁺³(d⁵) ground state]

In the presence of F^– ions (weak ligand) the electrons in 3d orbitals do not get paired and the iron (II) utilises ns, np and nd orbitals for hybridisation. Thus six sp³d² hybrid orbitals are formed.
[Fe⁺³ excited state]

The six pairs of electrons from F ions accommodate the six vacant spid1 hybrid orbitals
[FeF₆]^-3

Thus, the complex is outer orbital complex, octahedral and paramagnetic.

(iii) [Co(C₂O₄)₃]^-3: d²sp³ hybridisation diamagnetic.
The cobalt ion is in +3 oxidation state. It has six electrons in d subshell (d⁶).
[Co⁺³(d⁶)]

Oxalate ion being a relatively strong ligand pairs of the electrons in ‘d’ subshell, learning two of the ‘d’ orbital for hybridisation. Thus Co⁺³ ion under goes d²sp³ hybridisation.
Co⁺³(ground state)

The three oxalate anions (bidentate) donate two pairs of electrons to two of the hybrid orbitals.

Thus [Co(C₂O₄)₃]^-3 is an inner orbital, octahedral, complex and diamagnetic.

(iv) [NiCl₄]^-2: In this complex Ni is in +2 oxidation state and has d⁸ configuration.

The chloride ions donate pair of electrons to each of the four sp³ hybrid orbitals.

Thus, the complex is tetrahedral and paramagnetic.

Question 32.
[NiCl₄]^-2 is paramagnetic while Ni(CO)₄ is diamagnetic though both are tetrahedral why?
Answer:
In Ni(Co)₄, Ni is in the zero oxidation state whereas in [NiCl₄]^-2, it is in +2 oxidation state. In the presence of a CO being a strong ligand, the unpaired ‘d’ electrons of Ni pair up but Cl^– being a weak ligand it is unable to pair up the unpaired electrons. Hence [NiCl₄]^-2 is paramagnetic where as Ni(CO)₄ is diamagnetic.

Question 33.
Explain [Co(NH₃)₆]⁺³ is an inner orbital complex whereas [Ni(NH₃)₆]⁺² is an outer orbital complex.
Answer:
In the presence of NH₃, the 3d electrons pair up learning two ‘d’ orbitals empty to be involved in d²sp³ hybridisation forming inner orbital complex.
In [Ni(NH₃)₆]⁺², Ni is in +2 oxidation state and has d⁸ configuration. The hybridisation involved in sp³d² and hence, it forms an outer orbital complex.

Question 34.
[Cr(NH₃)₆]⁺³ is paramagnetic while [Ni(CN)₄]^-2 is diamagnetic explain why?
Answer:
(i) Formation of [Cr(NH₃)₆]⁺³:
The oxidation state of chromium in [Cr(NH₃)₆]⁺³ ion is +3. The electronic configuration of chromium is [Ar] 3d⁵ 4s¹. The hybridisation scheme is shown below:

Ammonia being a weak ligand is unable to pair of up the electrons of Cr⁺³ and hence Cr⁺³ undergoes d²sp³ hybridisation.

Thus the complex is inner orbital, octahedral complex and paramagnetic because of the presence of three unpaired electrons.

(ii) Formation of [Ni(CN)₄]^-2: In the square planar complexes, the hybridisation involved is dsp2. In [Ni(CN)₄]^-2, nickel is in +2 oxidation state and has the electronic configuration of d⁸.

CN^– ion being a strong ligand, pairs up the 3d electrons leaving one 3d orbital vacant that is used for dsp² hybridisation.

Thus the complex is diamagnetic as there are no impaired electron.

Question 35.
A solution of [Ni(H₂O)₆]⁺² is green but a solution of [Ni(CN)₄]^-2 is colourless explain.
Answer:
In [Ni(H₂O)₆]⁺³, Ni is in +2 oxidation state with configuration d⁸. i.e., it has 2 unpaired electrons which do not pair in the presence of weak H₂O ligand. Hence d-d transition is possible and so it is coloured. The d-d transition absorbs red light and the complimentary light emitted is green.
In case of [Ni(CN)₄]⁺², Ni is in +2 oxidation state with d⁸ configuration. But in the presence of strong CN^– ion as ligand, the two unpaired electrons in 3d orbitals are paired up. Hence there are no unpaired electrdhs present and so it is colourless.

Question 36.
[Fe(CN)₆]^-4 and [Fe(H₂O)₆]⁺² are of different colours in dilute solution why?
Answer:
In both complexes, Fe is in +2 oxidation state with d⁶ configuration. In the presence of a weak ligand, they do not pair up. In the presence of strong CN^– ion, they paired up learning no unpaired electrons. Due to the difference in the number of unpaired electrons they have different colours.

Question 37.
Write down the IUPAC name of each of the following complexes and indicate the oxidation state, electronic configuration and coordination number of the central metal ion. Also, give the stereochemistry and magnetic moment of each complex.
(i) K₄[Cr(H₂O)₂(C₂O₄)₂]
(ii) [CrCl₃(Py₃)]
(iii) K₂[Mn(CN)₄]
(iv) [CO(NH₃)₅Cl]Cl₂
(v) CS[FeCl₄]
Answer:
(i) Potassiumdiaquaoxalatochromate (III) hydrate.
Coordination number = 6
Shape: Octahedral
Oxidation state of Cr = x + 0 + 2 (- 2) = -1 or x = +3

(ii) Trischloridopyridinechromium (III)
Coordination number of Cr = 6
Shape: Octahedral
Oxidation state of Cr = x – 3 + 0 + 0 = 0 or x = +3

(iii) Potassiumhexacyanomanganate (II)
Coordination number of Mn = 6.
Shape: Octahedral
Oxidation state of Mn = x – 6 = – 4 (or) x = +2

(iv) Pentaammine chloridocobalt (III) chloride.
Coordination number of Co = 6.
Shape: Octahedral
Oxidation state of Cr = x + 0 – 1 = +2 or x = +3

Magnetic moment = 0

(v) Caesiumtetrachloridoferrate (III).
Coordination number of Fe = 4.
Shape: Tetrahedral
Oxidation number of Fe = x – 4 = -1 or x = +3

Question 38.
Calculate the magnetic moment of the following coordination entities.
(i) [Cr(H₂O)₆]⁺³,
(ii) [Fe(H₂O)₆]⁺³,
(iii) [Zn(H₂O)₆]⁺²
Answer:
The oxidation states are (i) Cr(III); (ii) Fe(II) and (iii) Zn(II)
(i) E.C of Cr⁺³ = 3d³:
no ofunpaired electrons = 3
Magnetic moment = \(\sqrt{3(3+2)}\) = √15
(ii) E.C of Fe⁺² = 3d⁶:
no of unpaired electrons = 4
Magnetic moment = \(\sqrt{4(4+2)}\) = √24
(iii) E. C of Zn⁺² = 3d¹⁰:
no of unpaired electrons = 0
Magnetic moment = 0

Question 39.
What will be the correct order for the wavelength of absorption in the visible region of the following:
[Ni(NO₂)₆]^-4, [Ni(NH₃)₆]⁺², [Ni(H₂O)₆]⁺²
Answer:
As the metal ion is fixed the increasing field strength (CFSE values) of the ligands from the spectrochemical series are in the order.
H₂O<NH₃<NO₂^–
Thus the energies absorbed for excitation will be in the same order
[Ni(H₂O)₆]⁺² < [Ni(NH₃)₆]⁺² < [Ni(H₂O)₆]⁺²
As, E = \(\frac{h c}{\lambda}\), the wavelength absorbed will be in the opposite order.

Question 40.
Give the number of unpaired electrons in the following complex ions: (i) [FeF₆]^-4, (ii) [Fe(CN)₆]^-4
Answer:
(i) In [FeF₆]^-4, Fe is in +2 oxidation state. It has d⁶ configuration. As F^– is a weak ligand, the electrons in 3d orbital are not paired.
i-e.,
Hence, it has 4 unpaired electrons.

(ii) In [Fe(CN)₆]^-4 also, Fe is in +2 oxidation state and it has d⁶ configuration. Because CN^– ion is a strong ligand, the electrons in 3d orbital of Fe⁺² are paired and so there are no unpaired electrons.

Question 41.
Explain as how two complexes [Ni(CN)₄]^-2 and [Ni(CO)₄] have different structures but do not differ in their magnetic behaviour. (Ni = 28)
Answer:
(i) In Ni(CO)₄, nickel is in zero oxidation state. (Ni⁺² = 3d⁸ 4s²)

E.C of Ni in hybridised state

It is diamagnetic and has a tetrahedral geometry.

(ii) In [Ni(CN)₄]^-2, Ni is in +2 oxidation state. Electronic configuration of Ni⁺² ion is 3d⁸.

CN^– ion being a strong liquid and in its presence, the electrons in 3d orbital gets paired leaving are 3d orbital for hybridisation. Thus, it undergoes dsp² hybridisation.

The CN^– ions denote a pair of electrons to the valent dsp² hybrid orbitals. Thus, four Ni^– CN bonds are formed.

Thus the geometry is square planar and the complex is diamagnetic.

Question 42.
Using valence bond theory explain [Co(NH₃)₆]⁺³ in relation to the complex given below:
(i) type of hybridisation [Co(NH₃)₆]⁺³ : Co⁺³ = d⁶ configuration
(ii) inner or outer orbital complex
(iii) magnetic behaviour
(iv) spin only magnetic moment value.
Answer:
(i) In the presence of ammonia ligand, the electrons in 3d level gets paired up leaving two of the 3d orbitals for hybridisation. Hence, the type of hybridisation is d ²sp³

(ii) Since inner ‘d’ orbitals are used for hybridisation, it is an inner orbital complex.
(iii) The complex is diamagnetic as there are no unpaired electrons.
(iv) Since, there are no unpaired electrons the magnetic moment is zero.

Question 43.
Discuss the nature of bonding in the following coordination entities on the basis of valence bond theory.
(i) [FeCN₆]^-4
(ii) [FeF₆]^2-
(iii) [Co(C₂O₄)₃]^-3
(iv) [COF₆]^-3
Answer:
(i) Oxidation state of Fe in the complex is +2. Its electronic configuration is [Ar] 3d⁶. Iron undergoes d²sp³ hybridisation giving octahedral, inner orbital complex.

(ii) Oxidation state of iron in the complex is +3. Its electronic configuration is [Ar] 3d⁵. In the presence of F^– (weak ligand), the 3d electrons do not get paired. Hence it undergoes sp³d² hybridisation and forms an outer orbital, octahedral complex.

(iii) The oxidation state of cobalt in the complex is +3. Its electronic configuration is [Ar] 3d⁶. Since oxalate ion is relatively a strong ligand, the 3d electrons get paired up leaving two of 3d orbital for hybridisation. Thus, cobalt undergoes d²sp³ hybridisation giving octahedral, inner orbital complex.
(iv) Oxidation state of cobalt in the complex is +3. Its electronic configuration is [Ar] 3d⁶. The F ion being a weak ligand is unable to pair up the electrons in 3d level. It undergoes sp³d² hybridisation giving an octahedral, outer orbital complex.

Question 44.
What are crystal fields?
Answer:
The ligands especially anionic (or polar neutral ligands) has around them negatively charged field because of which they are called crystal fields.

Question 45.
What do you understand by the term crystal field splitting?
Answer:
When the ligands approach the central metal ion, the electrons in the ‘d’ orbital of central metal ion will be repelled by the lone pairs of the ligands. Because of these interactions, the degeneracy of ‘d’ orbitals of the metal ion is lost and they split into two sets of orbitals having different energies. This is known as crystal field splitting. The extent of crystal field splitting depends on the nature of the ligand.

Question 46.
Briefly discuss the crystal fieid splitting in octahedral complexes.
Answer:
As a result of repulsive forces between the ‘d’ electrons of the metal ion and that of ligands (negative or polar neutral), the degeneracy of the ‘d’ orbitals is lost. Since the two lobes of two e_g{d_x²-y², d_z²) orbitals lie in the path of approaching ligands these orbitals experience greater forces of repulsion than those of t_2g (d_xy, d_yz and d_xz) orbitals is decreased.
Thus, an energy difference exists between two sets of orbitals. This energy difference is called crystal field splitting energy and is represented by Δ_o. (the subscript ‘o’ stands for octahedral). It measures the crystal field strength of the ligands. The crystal field splitting occurs in such a way that the average energy of the ‘d’ orbitals do not change.

(a) Five degenerate ‘d’ orbitals of the free metal ion.
(b) Hypothetical degenerate ‘GP orbitals at higher energy levels under spherically symmetrical ligand field.
(c) Splitting of ‘d’ orbitals under the influence of ligands.
Thus, three orbitals lie at an energy i.e., 2/5 Δ_o below the average ‘d’ orbital energy and two ‘d’ orbitals lie at an energy 3/5 Δ_o above the average energy. The energy gap between t_2g and e_g sets also denoted by 10Dq.
The energy of t_2g orbitals is 4Dq less than that of hypothetical degenerate ‘d’ orbitals and that of e_g orbitals is 6Dq above that of hypothetical degenerate ‘d’ orbitals: Thus, t_2g set loses energy equal to 0.4 Δ_o or 4 Dq white e_g sets gain energy equal to 0.6 Δ_o or 6 Dq.

Question 47.
For the complex
[Fe(en)₂Cl₂]Cl (en = ethylene diamine), identify The complex is [Fe(en)₂Cl₂]Cl.
(i) The oxidation number of Fe.
(ii) The hybrid orbital and shape of the complex.
(iii) The magnetic behaviour of the complex.
(iv) The number of geometrical isomers.
(v) Whether there is an optical isomer also.
(vi) The name of the complex (At. No of Fe 26)
Answer:
(i) Oxidation number of Fe
= x + 2 x 0 + 2(-1)
= +1
x – 2 = +1
⇒ x = 3
(ii) Orbitals in Fe⁺³(d⁵)

In the presence of a strong ligand (en), the electrons in 3d orbital get paired.
Fe⁺³ excited state

Thus, Fe⁺³ ion undergoes d²sp³ hybridisation therefore the shape is octahedral.
(iii) The complex is paramagnetic due to the presence of one unpaired electron.
(iv) It exhibits cis-trans isomerism

(v) Cis isomer will show optical isomerism.

(vi) Dichlorido bis(ethane 1, 2, diamine) iron, (III) chloride.

Question 48.
Compare the following complexes with respect to their shape, magnetic behaviour and the hybrid orbitals involved.
(i) [CoF₄]^-2, (ii) [Cr(H₂O)₂(C₂O₄)₂]^–, (iii) [Ni(Co)₄], (At No: Co = 27, Cr = 24, Ni = 28).
Answer:

Question 49.
Briefly discuss the crystal field splitting in tetrahedral complexes.
Answer:
The three orbitals i.e., t_2g orbitals are close to approaching ligands. As a result of this, the t_2g electrons suffer more repulsions than e_g electrons. The energy of t_2g orbitals increases more than e_g orbitals. The splitting is shown below:

The energy gap between two sets of orbitals is designated as Δ_t. It is observed that Δ_t, is considerably less than Δ_o. It has been found that.
Δ_t = \(\frac{4}{9}\) Δ_o

Question 50.
Briefly outline crystal field splitting in square planar complexes.
Answer:
As the lobes of d_x²-y² point towards the ligands, this orbital has the highest energy. The lobes of d_xy orbital lie between the ligands but are coplanar with them this orbital is the next highest energy. The lobes of d_z² orbital point out of the plane of the complex but the belt around the centre of the orbital (which contain 1/3 of the electron density) lies in the plane. Therefore d_z² orbital is the next highest in energy. The lobes of d_x² and d_y² orbitals point out of the plane of the complex. Hence, they are least affected by the electrostatic field of the ligands. They are degenerate and lowest in energy.

Question 51.
What is crystal field splitting energy? How does the magnitude of Δ_o decide the actual configuration of ‘d’ orbitals in a coordination entity?
Answer:
When ligands approach a transition metal ion, the ‘rf orbital splits into two sets one with lower energy and the other with higher energy. The difference in energy between two sets (t_2g and e_g) of orbitals is called crystal field splitting energy Δ_o for an octahedral field.
If Δ_o < p (pairing energy), the fourth electron enters one of the e_g orbitals giving the configuration t_2g³ e_g¹, thereby forming high spin complexes. Such ligands for which Δ_o < p are called weak field ligands.
If Δ_o > p, the fourth electron pair up in one of the t_2g orbital giving the configuration t_2g⁴ e_g⁰, thus forming low spin complexes. Such ligands for which Δ_o > p are called strong field ligands.

Question 52.
What is spectrochemical series? Explain the difference between a weak field ligand and a strong field ligand.
Answer:
The arrangement of ligands in order of their increasing field strength i.e increasing crystal field splitting energy (CFSE) values is called spectrochemical series.
The ligand with a small value of Δ_o CFSE (Δ_o) is called weak field ligands whereas those with a large value of CFSE(Δ_o) are called strong field ligands.

Question 53.
How does the oxidation state of a metal ion influence or affect the crystal field splitting energy?
Answer:
Generally higher the oxidation state of the metal, greater is the crystal field splitting. For example most of the cobalt (II) complexes have low value of Δ whereas all cobalt (III) complexes have high value of Δ.

Question 54.
Briefly explain the distribution of ‘d’ electrons in t_2g and e_g orbitals in octahedral complexes in a weak ligand field.
Answer:
Under the influence of weak ligands the energy difference, Δ_o between t2g and eg sets is relatively small and hence all the five ‘d’ orbitals may be supposed to be degenerate. The distribution of electrons in t_2g and e_g sets occurs according to Hund’s rule.
When the ligands are weak, the first three electrons numbered as 1,2,3 go to t_2g set those numbered 4, 5 go to e_g set, those numbered 6, 7, 8 go to t_2g set and the remaining two electrons numbered 9 and 10 will occupy e_g set. i.e.,
In complexes of weak ligands, Δ_o is less than P(P is called pairing energy which is the energy required to pair two electrons in the same orbital). Δ_o is the octahedral crystal field splitting energy, tends to force as many electrons into t_2g set while P tends to prevent the electrons to pair up in the t_2g level.

Question 55.
The Hexa aqua manganese (II) ions contain five unpaired electrons while hexacyano ions contain only one unpaired electron. Explain using crystal field theory.
Answer:
Mn is in +2 oxidation state and has 3d⁵ configuration. H₂O is a weak ligand. In the presence of water molecules, the distribution of electrons is t_2g³ e_g² i.e., all the electrons are unpaired.
In the other hand CN^– ion is a strong ligand. The distribution of electrons is t_2g⁵ e_g⁰.
i.e., and it has one unpaired electron.

Question 56.
Why do compounds having similar geometry have a different magnetic moment?
Answer:
It is due to the presence of strong and weak ligands in complexes. If GFSE is high, the complex will show a low value of the magnetic moment and vice versa. eS: [CoF₆]^-3 and [CO(NH₃)₆]⁺³, the former is paramagnetic and the latter is diamagnetic.

Question 57.
Briefly explain how the electrons are distributed between t_2g and e_g sets in an octahedral complex in the presence of a strong field ligand.
Answer:
Under the influence of strong ligands, the energy difference between t_2g and e_g sets is relatively high and thus the distribution of ‘d’ electrons in t_2g and e_g sets does not obey Hund’s rule. The first electrons numbers 1, 2, 3, 4, 5, 6 will go to t_2g set remaining four electrons numbered 7, 8, 9, 10 will go to e_g set.

For these complexes, Δ_o is greater than p.

Question 58.
Using crystal field theory, draw an energy level diagram of the central metal atom/ion and determine the magnetic moment value of the following.
(i) [COF₆]^-3,
(ii) [Co(H₂O)₆]⁺²,
(iii) [Cu(CN)₆]^-3
Answer:
(i) F^– is a weak field ligand and cobalt is in a +3 oxidation state. It has d⁶ configuration. The electron distribution is t_2g⁴ e_g².

i.e., Number of unpaired electron = 4
Magnetic moment = \(\sqrt{4(4+2)}\) = √24
= 4.9 BM
(ii) In this complex cobalt is in +2 oxidation state. Its electronic configuration is d1. Since water is a weak ligand, electron filling follows Hund’s rule, i.e., t_2g⁵ e_g².

i.e., Number of unpaired electron = 3 Magnetic moment = \(\sqrt{3(3+2)}\) = √15
= 3.87BM
(iii) In this complex cobalt is in +3(d⁶) oxidation state and CN^– ion a strong ligand. Electron filling is not obeyed by Hund’s rule. All the six electrons get paired in t_2g set.

Since, there is no unpaired electron the complex is diamagnetic.

Question 59.
In the basis of crystal field theory explain why cobalt (III) forms a paramagnetic octahedral complex with weak field ligands whereas it forms a diamagnetic octahedral complex with strong field ligands.
Answer:
With weak field ligand, Δ_o < P, the electronic configuration (of cobalt (III)) i.e., d⁶ will be t_2g⁴ e_g² and it has 4 unpaired electrons and is paramagnetic. With strong field ligands, Δ_o > P, the electron distribution is t_2g⁶ e_g⁰. It has no unpaired electron and hence diamagnetic.

Question 60.
Differentiate between weak field and strong field coordination entity.
Answer:

Question 61.
State for a d⁶ configuration, how the actual configuration of the split ‘d’ orbitals in an octahedral field is decided by relative values of Δ_o and P.
Answer:
For d⁶ configuration, three electrons will first enter t_2g. Now, if P < Δ_o electrons will pair up in t_2g orbitals giving the configuration t_2g⁶ e_g⁰. If Δ_o< P, two electrons will first enter into e_g orbitals and one will pair up in t_2g, giving the configuration t_2g⁴ e_g².

Question 62.
Arrange the following ligands in the increasing order of crystal field splitting power.
Answer:
H₂O, OH^–, Cl^–, F^–, CN^–
Cl^– < F^– < OH^– < H₂O < CN^–

Question 63.
Using crystal field theory, show energy level diagrams, write the electronic configuration of the central metal atom/ion and determine the magnetic moment value in the following. (i) [FeF₆]^-3, (ii) [Fe(H₂O)₆]⁺²,
(iii) [Fe(CN)₆]^-4
Answer:
(i) In this complex, Fe is +3 oxidation state and Fe⁺³ has 3d⁵ configuration. Since F^– is a weak field ligand, electron filling in t_2g and e_g sets follow Hund’s rule.

i.e., t_2g⁴ e_g². The number of unpaired electron is 5. The magnetic moment value is \(\sqrt{5(5+2)}\) = √35 = 5.92 BM.
(ii) In [Fe(H₂O)₆]⁺², Fe is in +2 oxidation state and has d⁶ configuration. Since water is a weak ligand electron distribution in t_2g and e_g orbitals take place according to Hund’s rule.

i.e., t_2g⁴ e_g². The number of unpaired electrons is 4. Its magnetic moment value is \(\sqrt{4(4+2)}\) = √24 = 4.9 BM.
(iii) In [Fe(CN)₆]^-4, Fe is in +2 oxidation state and has 3d⁶ configuration. As CN is a strong ligand, electron distribution between t_2g and e_g sets take place not in accordance with Hund’s rule. All the t_2g orbitals are filled with six electrons.

Since there is no unpaired electron, the complex is diamagnetic.

Question 64.
Give the oxidation state, ‘d’ orbital occupation and coordination number of the central metal ion in the following complexes.
(i) K₃[Co(C₂O₄)₃],
(ii) (NH₄)[CoF₄],
(iii) Cis – [Cr(en)₂Cl₂]Cl,
(iv) [Mn(H₂O)₆]SO₄.
Answer:

Question 65.
With the help of crystal field theory, product the number of unpaired electrons in [Fe(CN)₆]^-4 and [Fe(H₆O)₂]⁺² complexes.
Answer:
In both complexes, Fe is present in +2 oxidation state and has d⁶ configuration. As CN^– is a strong field ligand, these electrons will pair up giving the configuration t_2g⁶ e_g⁰. As there are no unpaired electrons, the complex is paramagnetic. But H₂O is a weak ligand and electron filling follow Hund’s rule i.e., t_2g⁴ e_g². It has 4 unpaired electron and hence paramagnetic.

Question 66.
CO is stronger ligand than NH₃ for many metals. Explain giving appropriate reason.
Answer:
Ligands such as CO, CN^–, NO⁺ have empty π orbitals which overlap with the filled d orbitals (t_2g) of transition metals forming π bonds (back bonding). These π interactions increase the value of Δ_o. This account for the position
of these ligands as strong ligand. NH₃ cannot form π bonds by back bonding.

Question 67.
What is meant by the stability of a coordination compound in solution?
Answer:
The stability of a complex in a solution refers to the degree of association between two species involved in a state of equilibrium. The magnitude of the equilibrium constant for the association is a quantitative measure of stability.

For the above equilibrium, the larger the stability constant, the higher is the proportion of ML₄ that exists in the solution.

Question 68.
The stability constant of some complexes are given below:

Among CN^– and NH₃ which is a stronger ligand? Why?
Answer:
The numerical value of the stability constant is a measure of the stability of the complex. Greater the magnitude of the stability constant more stable the complex. Thus CN^– ion is a stronger ligand than ammonia.

Question 69.
The dissociation constants of [Cu(NH₃)₄]⁺² and [CO(NH₃)₆]⁺³ are 1.10 x 10^-12 and 6.2 x 10^-36 respectively. Which complex would be more stable and why?
Answer:
Smaller the values of the dissociation constant more stable the complexion in solution Hence [CO(NH₃)₆]⁺³ is more stable than [Cu(NH₃)₄]⁺² ion.

Question 70.
Calculate the overall complex dissociation equilibrium constant for the [Cu(NH₃)₄]⁺² ion given that β₄ for this complex is 2.14 x 10¹³.
Answer:
Overall stability constant (B₄)
= 2.1 x 10¹³. Overall dissociation constant is the reciprocal of overall stability constant. Hence, overall dissociation constant.

= 4.7 x 10^-14.

Question 71.
Calculate the ratio of [Ag(S₂O₃)₂]^-3 and [Ag⁺] in 0.1 M S₂O₃ solution. Given the stability or formation constant of [Ag(S₂O₃)₂]^-3 is 1.0 x 10¹³.
Answer:
The equilibrium is

Question 72.
Mention the use of coordination compounds in metallurgy.
Answer:

1. Purification of nickel by Mond’s process involves the formation of nickel carbonyl Ni(CO)₄ which yields 99.5% pure nickel and decomposition.
2. Coordination compounds are used in the extraction of silver and gold from their ores by forming a soluble cyano complex. These cyano complexes are reduced by zinc to yield pure metals.

Question 73.
Mention the use of complexes used as catalysts in organic reactions.
Answer:

1. Wilkinson’s catalyst [(PPh₃)₃RhCl] is used for the hydrogenation of alkenes.
2. Ziegler-Natta catalyst – [TiCl₄] + Al[C₂H₅]₃ is used in the polymerization of ethene.
3. In order to get a fine and uniform deposit of superior metals (Ag, Au, Pt etc.,) over base metals, coordination complexes [Ag(CN)₂]^– and [Au(CN)₂]^– etc., are used in an electrolytic bath.

Question 74.
Mention the use of complexes in photography.
Answer:
In photography, when the developed film is washed with sodium thio sulphatesolution (hypo), the negative film gets fixed. Undecomposed AgBr forms a soluble complex called sodiumdithiosulphatoargentate(I) which can be easily removed by washing the film with water.

Question 75.
Mention the use of metal complexes in biological systems.
Answer:

1. A red blood corpuscle (RBC) is composed of the heme group, which is Fe²⁺ – Porphyrin complex. It plays an important role in carrying oxygen from the lungs to tissues and carbon dioxide from tissues to the lungs.
2. Chlorophyll, a green pigment present in green plants and algae, is a coordination complex containing Mg²⁺ as a central metal ion surrounded by a modified Porphyrin ligand called a corrin ring. It plays an important role in photosynthesis, by which plants converts CO₂ and water into carbohydrates and oxygen.
3. Vitamin B₁₂ (cyanocobalamine) is the only vitamin consist of the metal ion. It is a coordination complex in which the central metal ion is Co⁺ surrounded by a Porphyrin ligand.
4. Many enzymes are known to be metal complexes, they regulate biological processes. For example, Carboxypeptidase is a protease enzyme that hydrolytic enzyme important in digestion contains a zinc ion coordinated to the protein.

Question 76.
What is Cisplatin? Mention its use.
Answer:
Cisplatin is a square planar coordination complex (cis- [Pt (NH₃)₂C₁₂]), in which two similar ligands are in adjacent positions. It is a platinum-based anticancer drug. This drug undergoes hydrolysis and reacts with DNA to produce various crosslinks. These crosslinks hinder DNA replication and transcription, which results in cell growth inhibition and ultimately cell death. It also crosslinks with cellular proteins and inhibits mitosis.

Question 77.
What are metal carbonyls? Mention their uses.
Answer:
Metal carbonyls are transition metal complexes of carbon monoxide, containing metal-carbonyl (M ← CO) bonds. They are used as catalysts.

Question 78.
What are mononuclear and polynuclear carbonyls? Give examples.
Answer:
Mono nuclear carbonyls contain only one metal atom.eg: Ni(CO)₄, ie(CO)₅, Cr(CO)₆. Polynuclear carbonyls contain two or more metal atoms. They may be homonuclear or heteronuclear.

Question 79.
What are non-bridged metal carbonyls? Give examples.
Answer:

1. Non bridged metal carbonyls contain only terminal carbonyls, eg: [Ni (Co)₄] [Fe(Co)₅] and [Cr(Co)₆]
2. Non bridged metal carbonyl which contains terminal carbonyls as well as metal- metal bonds, eg: Mn₂(Co)₁₀. It contains Mn—Mn bonds and also Mn ← 10 bonds.

Question 80.
What are bridged carbonyls? Give example.
Answer:
These metal carbonyls contain one or more bridging carbonyl ligands along with terminal carbonyl ligands and one or more Metal-Metal bonds. For example,

The structure of Fe₂(CO)₉, di-ironennea carbonyl molecule consists of three bridging CO ligands, six terminal CO groups and a single Fe-Fe bond formed by the weak coupling of the unpaired electrons present in two 3d orbitals of two Fe atoms. This bond is represented by a dotted line and is called a fractional single bond.