2<\/sub>O<\/p>\nQuestion 5.
\nThe ratio of number of sigma (\u03c3) bond and pi (\u03c0) bonds in 2 – butynal is
\na) 8\/3
\nb) 5\/3
\nc) 8\/2
\nd) 9\/2
\nAnswer:
\na) 8\/3<\/p>\n
<\/p>\n
Question 6.
\nWhich one of the following is the likely bond angles of sulphur tetrafluo-ride molecule?
\na) 120\u00b0, 80\u00b0
\nb) 109\u00b028\u2019
\nc) 90\u00b0
\nd) 89\u00b0, 117\u00b0
\nAnswer:
\nd) 89\u00b0, 117\u00b0<\/p>\n
Question 7.
\nAssertion:
\nOxygen molecule is paramagnetic.
\nReason :
\nIt has two unpaired electron in its bonding molecular orbital
\na) both assertion and reason are true and reason is the correct explanation of assertion.
\nb) both assertion and reason are true but reason is not the correct explanation of assertion.
\nc) assertion is true but reason is false.
\nd) both assertion and reason are false.
\nAnswer:
\nc) assertion is true but reason is false.<\/p>\n
Question 8.
\nAccording to Valence bond theory, a bond between two atoms is formed when
\na) fully filled atomic orbitals overlap
\nb) half filled atomic orbitals overlap
\nc) non – bonding atomic orbitals overlap
\nd) empty atomic orbitals overlap
\nAnswer:
\nb) half filled atomic orbitals overlap<\/p>\n
Question 9.
\nIn ClF3<\/sub>, NF3<\/sub> and BF3<\/sub> molecules the chlorine, nitrogen and boron atoms are
\na) sp3<\/sup> hybridised
\nb) sp3<\/sup>, sp3<\/sup> and sp2<\/sup> respectively
\nc) sp3<\/sup> hybridised
\nd) sp3<\/sup>d, sp3<\/sup> and sp hybridised respectively
\nAnswer:
\nd) sp3<\/sup>d, sp3<\/sup> and sp hybridised respectively<\/p>\nQuestion 10.
\nWhen one s and three p orbitals hybridise,
\na) four equivalent orbitals at 90\u00b0 to each other will be formed
\nb) four equivalent orbitals at 109\u00b028\u2019 to each other will be formed
\nc) four equivalent orbitals, that are lying the same plane will be formed
\nd) none of these
\nAnswer:
\nb) four equivalent orbitals at 109\u00b028\u2019 to each other will be formed<\/p>\n
<\/p>\n
Question 11.
\nWhich of these represents the correct order of their increasing bond order.
\na) C2<\/sub>+<\/sup> < C2<\/sub>2-<\/sup> < O2<\/sub>2-<\/sup> < O2<\/sub>
\nb) C2<\/sub>2-<\/sup> < C2<\/sub>+<\/sup> < O2<\/sub> < O2<\/sub>2-<\/sup>
\nc) O2<\/sub>2-<\/sup> < O2<\/sub> < C2<\/sub>2-<\/sup> < C2<\/sub>+<\/sup>
\nd) O2<\/sub>2-<\/sup> < C2<\/sub>+<\/sup> < O2<\/sub> < C2<\/sub>2-<\/sup>
\nAnswer:
\nd) O2<\/sub>2-<\/sup> < C2<\/sub>+<\/sup> < O2<\/sub> < C2<\/sub>2-<\/sup><\/p>\nQuestion 12.
\nHybridisation of central atom in PCl5<\/sub>\u00a0involves the mixing of orbitals.
\na) s, Px<\/sub>, Py<\/sub>,\u00a0dx2<\/sup><\/sub>,\u00a0dx2<\/sup> – y2<\/sup><\/sub>
\nb) s, px<\/sub>, py<\/sub>, pxy<\/sub>, dx2<\/sup> – y2<\/sup><\/sub>
\nc) s, px<\/sub>, py<\/sub>, pz<\/sub>, dx2<\/sup> – y2<\/sup><\/sub>
\nd) s, px<\/sub>, Py<\/sub>, dxy<\/sub>, dx2<\/sup> – y2<\/sup><\/sub>
\nAnswer:
\nc) s, px<\/sub>, py<\/sub>, pz<\/sub>, dx2<\/sup> – y2<\/sup><\/sub><\/p>\nQuestion 13.
\nThe correct order of O – O bond length in hydrogen peroxide, ozone and oxygen is
\na) H2<\/sub>O2<\/sub> > O3<\/sub> > O2<\/sub>
\nb) O2<\/sub> > O3<\/sub> > H2<\/sub>O2<\/sub>
\nc) O2<\/sub> > H2<\/sub>O2<\/sub> > O3<\/sub>
\nd) O3<\/sub> > O2<\/sub> > H2<\/sub>O2<\/sub>
\nAnswer:
\nb) O2<\/sub> > O3<\/sub> > H2<\/sub>O2<\/sub><\/p>\nQuestion 14.
\nWhich one of the following is diamagnetic?
\na) O2<\/sub>
\nb) O2<\/sub>2-<\/sup>
\nc) O2<\/sub>+<\/sup>
\nd) None of these
\nAnswer:
\nb) O2<\/sub>2-<\/sup><\/p>\nQuestion 15.
\nBond order of a species is 2.5 and the number of electrons in its bonding molecular orbital is formed to be 8. The no. of electrons in its antibonding molecular orbital is
\na) three
\nb) four
\nc) Zero
\nd) can not be calculated from the given information
\nAnswer:
\na) three<\/p>\n
<\/p>\n
Question 16.
\nShape and hybridisation of IF5<\/sub> are
\na) Trigonal bipyramidal, sp3<\/sup>d2<\/sup>
\nb) Trigonal bipyramidal, sp3<\/sup>d
\nc) Square pyramidal, sp3<\/sup>d2<\/sup>
\nd) Octahedral, sp3<\/sup>d2<\/sup>
\nAnswer:
\nc) Square pyramidal, sp3<\/sup>d2<\/sup><\/p>\nQuestion 17.
\nPick out the incorrect statement from the following:
\na) sp3<\/sup> hybrid orbitals are equivalent and are at an angle of 109\u00b028\u2019 with each other
\nb) dsp2<\/sup> hybrid orbitals are equivalent and bond angle between any two of them is 90\u00b0
\nc) All five sp3<\/sup>d hybrid orbitals are not equivalent out of these five sp3<\/sup>d hybrid orbitals, three are at an angle of 120\u00b0 remaining two are perpendicular to the plane containing the other three
\nd) none of these
\nAnswer:
\nc) All five sp3<\/sup>d hybrid orbitals are not equivalent out of these five sp3<\/sup>d hybrid orbitals, three are at an angle of 120\u00b0 remaining two are perpendicular to the plane containing the other three<\/p>\nQuestion 18.
\nThe molecules having same hybridisation, shape and number of lone pairs of electrons are
\na) SeF4<\/sub>, XeO2<\/sub>F2<\/sub>
\nb) SF4<\/sub>, XeF2<\/sub>
\nc) XeOF4<\/sub>, TeF4<\/sub>
\nd) SeCl4<\/sub>, XeF4<\/sub>
\nAnswer:
\na) SeF4<\/sub>, XeO2<\/sub>F2<\/sub><\/p>\nQuestion 19.
\nIn which of the following molecules \/ ions BF3<\/sub>, NO2<\/sub>–<\/sup>\u00a0H2<\/sub>0 the central atom is sp2<\/sup> hybridised?
\na) NH2<\/sub>–<\/sup> and H2<\/sub>O
\nb) NO2<\/sub>–<\/sup> and H2<\/sub>O
\nc) BF3<\/sub> and NO2<\/sub>–<\/sup>
\nd) BF3<\/sub> and NH2<\/sub>–<\/sup>
\nAnswer:
\nc) BF3<\/sub> and NO2<\/sub>–<\/sup><\/p>\nQuestion 20.
\nSome of the following properties of two species, NO3<\/sub>–<\/sup> and H3<\/sub>O+<\/sup> are described below. Which one of them is correct?
\na) dissimilar in hybridisation for the central atom with different structure
\nb) isostructural with same hybridisation for the central atom.
\nc) different hybridisation for the central atom with same structure.
\nd) none of these
\nAnswer:
\na) dissimilar in hybridisation for the central atom with different structure<\/p>\n<\/p>\n
Question 21.
\nThe types of hybridisation on the five-carbon atom from right to left in the, 2,3 pentadiene.
\na) sp3<\/sup>, sp2<\/sup>, sp, sp2<\/sup>, sp3<\/sup>
\nb) sp3<\/sup>, sp, sp, sp, sp3<\/sup>
\nc) sp2<\/sup>, sp, sp2<\/sup>, sp2<\/sup>, sp3<\/sup>
\nd) sp3<\/sup>, sp3<\/sup>, sp2<\/sup>, sp3<\/sup>, sp3<\/sup>
\nAnswer:
\na) sp3<\/sup>, sp2<\/sup>, sp, sp2<\/sup>, sp3<\/sup><\/p>\nQuestion 22.
\nXeF2<\/sub> is isostructural with
\na) SbCl2<\/sub>
\nb) BaCl2<\/sub>
\nc) TeF2<\/sub>
\nd) ICl2<\/sub>–<\/sup>
\nAnswer:
\nd) ICl2<\/sub>–<\/sup><\/p>\nQuestion 23.
\nThe percentage of s-character of the hybrid orbitals in methane, ethane, ethene, and ethyne are respectively
\na) 25, 25, 33.3, 50
\nb) 50, 50, 33.3, 25
\nc) 50, 25, 33.3, 50
\nd) 50, 25, 25, 50
\nAnswer:
\na) 25, 25, 33.3, 50<\/p>\n
Question 24.
\nOf the following molecules, which have shape similar to carbondioxide?
\na) SnCl2<\/sub>
\nb) NO2<\/sub>
\nc) C2<\/sub>H2<\/sub>
\nd) All of these
\nAnswer:
\nc) C2<\/sub>H2<\/sub><\/p>\nQuestion 25.
\nAccording to VSEPR theory, the repulsion between different parts of electrons obey the order
\na) 1. p – 1. p > b. p – b. p > 1. p – b. p
\nb) b. p – b. p > b. p – 1. p > 1. p – b. p
\nc) 1. p – 1. p > b. p – 1. p > b. p – b. p
\nd) b. p – b. p > 1. p – 1. p > b. p – 1. p
\nAnswer:
\nc) 1. p – 1. p > b. p – 1. p > b. p – b. p<\/p>\n
<\/p>\n
Question 26.
\nShape of ClF3<\/sub> is
\na) Planar triangular
\nb) Pyramidal
\nc) “T” Shaped
\nd) none of these
\nAnswer:
\nc) “T” Shaped<\/p>\nQuestion 27.
\nNon – Zero dipole moment is shown by
\na) CO2<\/sub>
\nb) p – dichlorobenzene
\nc) carbontetrachloride
\nd) water
\nAnswer:
\nd) water<\/p>\nQuestion 28.
\nWhich of the following conditions is not correct for resonating structures?
\na) the contributing structure must have the same number of unpaired electrons
\nb) the contributing structures should have similar energies
\nc) the resonance hybrid should have higher energy than any of the contributing structure.
\nd) none of these
\nAnswer:
\nc) the resonance hybrid should have higher energy than any of the contributing structure.<\/p>\n
Question 29.
\nAmong the following, the compound that contains, ionic, covalent, and Coordinate linkage is
\na) NH4<\/sub>Cl
\nb) NH3<\/sub>
\nc) NaCl
\nd) none of these
\nAnswer:
\na) NH4<\/sub>Cl<\/p>\nQuestion 30.
\nCaO and NaCl have the same crystal structure and approximately the same radii. If U is the lattice energy of NaCl, the approximate lattice energy of CaO is
\na) U
\nb) 2U
\nc) U \/2
\nd) 4U
\nAnswer:
\nd) 4U<\/p>\n
<\/p>\n
II. Write brief answer to the following questions:<\/span><\/p>\nQuestion 31.
\nDefine the following:
\ni) Bond order
\nii) Hybridisation
\niii) \u03c3 – bond
\nAnswer:
\ni) Bond order:
\nThe number of bonds formed between the two bonded atoms in a molecule is called the bond order.<\/p>\n
ii) Hybridisation:
\nHybridisation is the process of mixing of atomic orbitals of the same atom with comparable energy to form an equal number of new equivalent orbitals with the same energy.<\/p>\n
iii) \u03c3 – bond:
\nWhen two atomic orbitals overlap linearly along the axis, the resultant bond is called a sigma (\u03c3) bond.<\/p>\n
Question 32.
\nWhat is a pi – bond?
\nAnswer:
\nWhen two atomic orbitals overlaps sideways, the resultant covalent bond is called a pi (\u03c0) bond. When we consider x – axis as molecular axis, the py<\/sub> – py<\/sub> and pz<\/sub> – pz<\/sub> overlaps will result in the formation of a \u03c0 – bond.<\/p>\nQuestion 33.
\nIn CH4<\/sub>, NH3<\/sub> and H2<\/sub>O, the central atom undergoes sp3 hybridization – yet their bond angles are different. Why?
\nAnswer:
\nAccording to VSEPR theory, as H2<\/sub>O has two lone pairs so it repels the bond pairs much more and makes bond angle shorter of 104.5 degrees, and as NH3<\/sub> has one lone pair that repels the three bond pairs but not much effectively and strongly as two lone pairs of water repel one bond pair.<\/p>\n