2<\/sup>
\n(b) 2l + 1
\n(c) 4l + 2
\n(d) none of these
\nAnswer:
\n(c) 4l + 2<\/p>\nQuestion 12.
\nFor d-electrons, the orbit angular momentum is
\n(a) \\(\\frac{\\sqrt{2} h}{2 \\pi}\\)<\/p>\n
(b) \\(\\frac{\\sqrt{2 h}}{2 \\pi}\\)<\/p>\n
(c) \\(\\frac{\\sqrt{2 \\times 4} \\mathrm{~h}}{2 \\pi}\\)<\/p>\n
(d) \\(\\frac{\\sqrt{6} \\mathrm{~h}}{2 \\pi}\\)
\nAnswer:
\n(d) \\(\\frac{\\sqrt{6} \\mathrm{~h}}{2 \\pi}\\)<\/p>\n
Question 13.
\nWhat is the maximum number electrons that car be associated with following set of quantum numbers? n = 3, l = 1 and m = -1
\n(a) 4
\n(b) 6
\n(c) 2
\n(d) 10
\nAnswer:
\n(c) 2<\/p>\n
Question 14.
\nAssertion:
\nThe number of radials and angular nodes for 3p orbital are I, 1 respectively.
\nReason:
\nThe number of radials and angular nodes depends only one the quantum number.
\n(a) Both assertion and reason are true and the reason is the correct explanation of the assertion
\n(b) Both assertion and reason are true but the reason is not the correct explanation of the assertion
\n(c) Assertion is true but the reason is false
\n(d) Both assertion and reason are false
\nAnswer:
\n(c) Assertion is true but the reason is false<\/p>\n
Question 15.
\nThe total number of orbitals associated with the principal quantum number n = 3 is
\n(a) 9
\n(b) 8
\n(c) 5
\n(d) 7
\nAnswer:
\n(a) 9<\/p>\n
<\/p>\n
Question 16.
\nIf n = 6, the sequence for filling electrons will be,
\n(a) ns \u2192 (n – 2)f \u2192 (n – 1)d \u2192 np
\n(b) ns \u2192 (n – 1 )d \u2192 (n – 2)f \u2192 np
\n(c) ns \u2192 {n – 2)f \u2192 np \u2192 (n – 1 )d
\n(d) none of these are correct
\nAnswer:
\n(a) ns \u2192 (n – 2)f \u2192 (n – 1)d \u2192 np<\/p>\n
Question 17.
\nConsider the following sets of quantum numbers:
\n
\nWhich of the following sets of quantum numbers is not possible?
\n(a) (i), (ii) and (iv)
\n(b) (ii), (iv) and (v)
\n(c) (i) and (iii)
\n(d) (ii), (iii) and (iv)
\nAnswer:
\n(b) (ii), (iv) and (v)<\/p>\n
Question 18.
\nHow many electrons in an atom with atomic number 105 can have (n + l) = 8?
\n(a) 30
\n(b) 17
\n(c) 15
\n(d) unpredictable
\nAnswer:
\n(b) 17<\/p>\n
Question 19.
\nElectron density in the yz plane of 3dx2<\/sup> – y2<\/sup><\/sub> orbital is
\n(a) zero
\n(b) 0.50
\n(c) 0.75
\n(d) 0.90
\nAnswer:
\n(a) zero<\/p>\nQuestion 20.
\nIf uncertainty in position and momentum are equal, then minimum uncertainty in velocity is
\n(a) \\(\\frac{1}{m} \\sqrt{\\frac{h}{\\pi}}\\)<\/p>\n
(b) \\(\\sqrt{\\frac{\\mathrm{h}}{\\pi}}\\)<\/p>\n
(c) \\(\\frac{1}{2 m} \\sqrt{\\frac{h}{\\pi}}\\)<\/p>\n
(d) \\(\\frac{\\mathrm{h}}{4 \\pi}\\)
\nAnswer:
\n(c) \\(\\frac{1}{2 m} \\sqrt{\\frac{h}{\\pi}}\\)<\/p>\n
<\/p>\n
Question 21.
\nA macroscopic particle of mass 100 g and moving at a velocity of 100 cm s-1<\/sup> will have a de Broglie wavelength of
\n(a) 6.6 \u00d7 10-29<\/sup> cm
\n(b) 6.6 \u00d7 10-30<\/sup> cm
\n(c) 6.6 \u00d7 10-31<\/sup> cm
\n(d) 6.6 \u00d7 10-32<\/sup> cm
\nAnswer:
\n(c) 6.6 \u00d7 10-31<\/sup> cm<\/p>\nQuestion 22.
\nThe ratio of de Brogue wavelengths of a deuterium atom to that of an \u03b1 – particle, when the velocity of the former is five times greater than that of later, is
\n(a) 4
\n(b) 0.2
\n(c) 2.5
\n(d) 0.4
\nAnswer:
\n(d) 0.4<\/p>\n
Question 23.
\nThe energy of an electron in the 3rd orbit of a hydrogen atom is -E. The energy of an electron in the first orbit will be
\n(a)-3E
\n(b) -E\/3
\n(c) -E\/9
\n(d) -9E
\nAnswer:
\n(d) -9E<\/p>\n
Question 24.
\nTime independent Schnodinger wave equation is
\n(a) H\u03c8 = E\u03c8
\n(b) \u22062<\/sup>\u03c8 + 8\u03c02<\/sup>m(E + V)\u03c8
\n(c) \\(\\frac{\\partial^{2} \\psi}{\\partial x^{2}}+\\frac{\\partial^{2} \\psi}{\\partial y^{2}}+\\frac{\\partial^{2} \\psi}{\\partial z^{2}}+\\frac{2 m}{h^{2}}(\\mathrm{E}-\\mathrm{V}) \\psi=0\\)
\n(d) all of these
\nAnswer:
\n(a) H\u03c8 = E\u03c8<\/p>\nQuestion 25.
\nWhich of the following does not represent the mathematical expression for the Heisenberg uncertainty principle?
\n(a) \u2206x . \u2206p \u2265 \\(\\frac{h}{4}\\)
\n(b) \u2206x . \u2206v \u2265 \\(\\frac{h}{4 \\pi m}\\)
\n(c) \u2206E . \u2206t \u2265 \\(\\frac{h}{4 \\pi}\\)
\n(d) \u2206E . \u2206x \u2265 \\(\\frac{h}{4 \\pi}\\)
\nAnswer:
\n(d) \u2206E . \u2206x \u2265 \\(\\frac{h}{4 \\pi}\\)<\/p>\n
<\/p>\n
II. Write brief answers to the following questions:<\/span><\/p>\nQuestion 26.
\nWhich quantum number reveals information about the shape, energy, orientation, and size of orbitals?
\nAnswer:
\nMagnetic quantum numbers reveal information about the shape, energy, orientation, and size of orbitals.<\/p>\n
Question 27.
\nHow many orbitals are possible for n = 4?
\nAnswer:
\nWhen n = 0, l = 0, 1,2 and 3. Hence, there are four subshells namely, s, p, d and f
\nl = 0, m1<\/sub> = 0; one 4s orbital, l = 1, m = -1, 0, +1; three 4p orbitals,
\nl = 2, m1<\/sub> = – 2, -1, 0, +1, +2; five 4d orbitals and
\nl = 3, m1<\/sub> = – 3, -2, -1, 0, 1, 2, 3; seven 4f orbitals. Hence, the number of possible orbitals when n = 4 are sixteen.<\/p>\nQuestion 28.
\nHow many radial nodes for 25, 4p, 5d and 4f orbitals exhibit? How many angular nodes?
\nAnswer:
\nThe number of radial nodes is equal to (n – l – 1) and angular nodes is l.<\/p>\n
\n\n\nOrbital<\/td>\n | N<\/td>\n | l<\/td>\n | Radial node (n – l \u2013 1)<\/td>\n | Angular node, l<\/td>\n<\/tr>\n |
\n2s<\/td>\n | 2<\/td>\n | 0<\/td>\n | 1<\/td>\n | 0<\/td>\n<\/tr>\n |
\n4p<\/td>\n | 4<\/td>\n | 1<\/td>\n | 2<\/td>\n | 1<\/td>\n<\/tr>\n |
\n5d<\/td>\n | 5<\/td>\n | 2<\/td>\n | 2<\/td>\n | 2<\/td>\n<\/tr>\n |
\n4f<\/td>\n | 4<\/td>\n | 3<\/td>\n | 0<\/td>\n | 3<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n The number of radial nodes for 2s, 4p, 5d, and 4f orbitals are respectively 1,2,2 and 0 and the number of angular nodes for 2s, 4p, 5d, and 4f orbitals respectively are 0, 1, 2, and 3.<\/p>\n Question 29. \nThe stabilization of a half-filled d – orbital is more pronounced than that of the p-orbital. Why? \nAnswer: \nThe exactly half-filled orbitals have greater stability. The reason for their stability are –<\/p>\n \n- symmetry<\/li>\n
- exchange energy.<\/li>\n<\/ol>\n
(1) Symmetry: The half-filled orbitals are more symmetrical than partially filled orbitals and this symmetry leads to greater stability.<\/p>\n (2) Exchange energy: The electrons with the same spin in the different orbitals of the same sub-shell can exchange their position. Each such exchange releases energy and this is known as exchange energy. Greater the number of exchanges, the greater the exchange energy, and hence greater the stability. In d-orbital, 10 exchanges are possible but in p-orbital 6 exchanges are possible. So, d – orbital with 5 unpaired electrons (10 exchanges)n i.e. half-filled is more stable than p – orbital with 3 unpaired electrons (6 exchanges).<\/p>\n Question 30. \nConsider the following electronic arrangements for the d5<\/sup> configuration. \n \n(i) Which of these represents the ground state? \n(ii) Which configuration has the maximum exchange energy? \nAnswer: \n(i) The ground state electronic configuration is \n<\/p>\n(ii) The configuration has the maximum exchange energy is \n<\/p>\n <\/p>\n Question 31. \nState and explain Paull\u2019s exclusion principle. \nAnswer: \nPauli’s exclusion principle states that “No two electrons in an atom can have the same set of values of all four quantum numbers”. \nIllustration: H(Z = 1) 1s1<\/sup>. \nOne electron is present in hydrogen atom, the four quantum numbers are n = 1, l = 0, m = 0 and s = + \\(\\frac {1}{2}\\). For helium Z = 2. He: 1s2<\/sup>. In this one electron has the quantum number same as that of hydrogen, n = 1,l = 0, m = 0 and s = +\u00bd For other electron, fourth quantum number is different, i.e. n = 1, l = 0, m = 0 and s = – \u00bd.<\/p>\nQuestion 32. \nDefine orbital. What are \u2018n\u2019 and ‘l’ values for 3px and 4d x2<\/sup> – y2<\/sup><\/sub> electron? \nAnswer: \nOrbital is a three-dimensional space in which the probability of finding the electron is maximum. The values of \u2018n\u2019 and \u2018l\u2019 for 3px orbital are n = 3 and l = 1, 4d x2<\/sup> – y2<\/sup><\/sub> orbital are n = 4 and l = 2.<\/p>\nQuestion 33. \nExplain briefly the time-independent Schrodinger wave equation. \nAnswer: \nErwin Schrodinger expressed the wave nature of electrons in terms of a differential equation. This equation determines the change of wave function in space depending on the field of force in which the electron moves. The time-independent Schrodinger equation can be expressed as \nH\u03c8 = E\u03c8, where H is called Hamiltonian operator, \u03c8 is the wave function and is a function of position coordinates of the particle and is denoted as \u03c8(x, y, z), E is the energy of the system. \n \nThe above Schrodinger wave equation does not contain time as a variable and is referred to as time-independent Schrodinger wave equation. This equation can be solved only for certain values of E, the total energy, i.e., the energy of the system is quantized. The permitted total energy values are called eigenvalues and corresponding wave functions represent the atomic orbitals.<\/p>\n Question 34. \nCalculate the uncertainty in position of an electron, if \u2206v = 0.1 % and \u03c5 = 2.2 \u00d7 106<\/sup> ms-1<\/sup>. \nAnswer: \nHeisenberg’s Uncertainty Principle is \u2206x. \u2206v > h\/4\u03c0m. \nGiven: \n\u2206v = 0.1%, \n\u03c5 = 22 \u00d7 106<\/sup> ms-1<\/sup>. \nh = 6.626 \u00d7 10-34<\/sup> kgm2<\/sup>s-1<\/sup> . \nm = 9.1 X 10 21 kg. \n\u2206v = \\(\\frac{0.1 \\times 2.2 \\times 10^{6} \\mathrm{~ms}^{-1}}{100}\\)<\/p>\n= 2.2 \u00d7 103<\/sup> ms-1<\/sup> \nUncertainty in position, \n\u2206x \u2265 \\(\\frac{h}{4 \\pi m}\\) \n\u2206x \u2265 \\(\\frac{6.626 \\times 10^{-34} \\mathrm{kgm}^{2} \\mathrm{~s}^{-1}}{9.1 \\times 10^{-31} \\mathrm{~kg} . \\times 2.2 \\times 10^{3} \\mathrm{~ms}^{-1}}\\)<\/p>\n\u2206x \u2265 2.64 \u00d7 10-8<\/sup> m.<\/p>\nQuestion 35. \nDetermine the values of all the four quantum numbers of the 8th electron in the O – atom and 15th electron in the Cl atom and the last electron in Chromium. \nAnswer: \n(1) O (Z = 8) 1s2<\/sup> 2s2<\/sup> 2px<\/sub>2 <\/sup>2py<\/sub>1<\/sup> 2pz<\/sub>1<\/sup> \nFour quantum numbers for 2px<\/sub>1<\/sup> electron in oxygen atom: \nn = principal quantum number = 2 \nl = azimuthal quantum number =1 \nm = magnetic quantum number =+1 \ns = spin quantum number = +\\(\\frac {1}{2}\\)<\/p>\n(2) Cl (Z = 17) 1s2<\/sup> 2s2<\/sup> 2p6<\/sup> 3s2<\/sup> 3px<\/sub>2<\/sup> 3py<\/sub>2<\/sup> 3pz<\/sub>1<\/sup> \nFour quantum numbers for 15th<\/sup> electron in chlorine atom: \nn = 3, l = 1, m = 0, s = + \u00bd<\/p>\n(3) Cr (Z = 24) 1s2<\/sup> 2s2<\/sup> 2p2<\/sup> 3s2<\/sup> 3p2<\/sup> 3d2<\/sup> 4s1<\/sup> \nn = 3, l = 2, m = +2, s = + \u00bd<\/p>\n<\/p>\n Question 36. \nThe quantum mechanical treatment of the hydrogen atom gives the energy value: \nEn<\/sub> = –\\(\\frac{-13.6}{n^{2}}\\) eV\/atom \n(i) Use this expression to find \u2206E between n = 3 and n = 4. \n(ii) Calculate the wavelength corresponding to the above transition. \nAnswer: \nEnergy of the electron in the nth orbit is \nEn<\/sub> = –\\(\\frac{-13.6}{n^{2}}\\)eV\/atom.<\/p>\nWhen n = 3, \nE3<\/sub> = \\(\\frac{-13.6 \\mathrm{eV} \/ \\text { atom }=-1.51 \\mathrm{eV} \/ \\text { atom }}{9}\\)<\/p>\nWhen n = 4, \nE4<\/sub> = \\(\\frac{-13.6 \\mathrm{eV} \/ \\text { atom }=-0.85 \\mathrm{eV} \/ \\text { atom }}{16}\\)<\/p>\n\u2206E = (E4<\/sub> – E3<\/sub>) = (-0.85) – (-1.51) = 0.66 eV\/atom \nWavelength corresponding to this transition,<\/p>\n<\/p>\n Question 37. \nHow fast must a 54g tennis ball travel in order to have a de Broglie wavelength that is equal to that of a photon of green light 5400 \u00c5? \nAnswer: \nde Broglie wavelength, \n\u03bb = \\(\\frac{h}{m v}\\) \nGiven: \nde Broglie wavelength, \u03bb = 5400 \u00c5 and mass, m = 54 g. \nVelocity of the tennis ball \nv = \\(\\frac{h}{m \\lambda}\\)<\/p>\n v = \\(\\frac{6.626 \\times 10^{-34} J s}{54 \\times 10^{-3} k g \\times 5400 \\times 10^{-10} m}\\)<\/p>\n v = 2.27 \u00d7 10-26<\/sup> ms-1<\/sup><\/p>\nQuestion 38. \nFor each of the following, give the sub level designation, the allowable m values and the number of orbitals, \n(i) n = 1, l = 2 \n(ii) n = 5, l = 3 \n(iii) n = 7, l = 0 \nAnswer: \n<\/p>\n Question 39. \nGive the electronic configuration of Mn2+<\/sup> and Cr3+<\/sup>. \nAnswer: \n1. Mn (Z = 25) \nMn \u2192 Mn2+<\/sup> + 2e–<\/sup> \nMn2+<\/sup> electronic configuration is 1s 1s2<\/sup> 2s2<\/sup> 2p6<\/sup> 3s2<\/sup> 3p6<\/sup> 3d5<\/sup><\/p>\n2. Cr (Z = 24) \nCr \u2192 Cr3+<\/sup> + 3e–<\/sup> \nCr3+<\/sup> electronic configuration is Is2<\/sup> 2s2<\/sup> 2p6<\/sup> 3s2<\/sup>3p6<\/sup> 3d3<\/sup><\/p>\nQuestion 40. \nDescribe the Aufbau principle. \nAnswer: \nAufbau Principle states that \u201cIn the ground state of the atoms, the orbitals are filled in the order of their increasing energies\u201d. That is the electrons first occupy the lowest energy orbital available to them. Once the lower energy orbitals are completely filled, then the electrons enter the next higher energy orbitals. The order of filling of various orbitals as per the Aufbau principle is given in the figure. which is in accordance with the (n + l) rule. \n<\/p>\n <\/p>\n Question 41. \nAn atom of an element contains 35 electrons and 45 neutrons. Deduce \n(i) the number of protons \n(ii) the electronic configuration for the element \n(iii) All the four quantum numbers for the last electron. \nAnswer: \n(i) Atomic Number of the element, z = No. of protons or No. of electrons. = 35 \nMass number of the element, A = No. of protons + No of neutrons = 35 + 45 = 80 \nNumber of Protons = 80 – 45 = 35.<\/p>\n (ii) Electronic configuration of the element (z = 35) \n1s2<\/sup> 2s2<\/sup> 2p6<\/sup> 3s2<\/sup> 3p6<\/sup> 4s2<\/sup> 3d10<\/sup> 4p5<\/sup><\/p>\n(iii) Quantum number for the last electron (4pz<\/sub>), \nn = 4, l = 1, m = +1, or -1 s = +1\/2<\/p>\nQuestion 42. \nShow that the circumference of the Bohr orbit for the hydrogen atom is an integral multiple of the de Broglie wavelength associated with the electron revolving around the nucleus. \nAnswer: \nAccording to the de Broglie concept, the electron that revolves around the nucleus exhibits both particle and wave character. In order for the electron wave to exist in phase, the circumference of the orbit should be an integral multiple of the wavelength of the electron wave, Otherwise, the electron wave is out of phase. \nCircumference of the orbit = n\u03bb \n2\u03c0r = n\u03bb \n2\u03c0r = \\(\\frac{n h}{m v}\\) \nRearranging, \nmvr = \\(\\frac{n h}{m v}\\) \nAngular momentum = \\(\\frac{n h}{2 \\pi}\\) \nThe above equation was already predicted by Bohr. \nHence, de Broglie and Bohr\u2019s concepts are in agreement with each other.<\/p>\n Question 43. \nCalculate the energy required for the process. \nHe+<\/sup>(g)<\/sub> \u2192 He2+<\/sup>(g)<\/sub> + e–<\/sup> \nThe ionization energy for the H atom in its ground state is \u2014 13.6 eV\/atom. \nAnswer: \nHe+<\/sup>(g)<\/sub> \u2192 He2+<\/sup>(g)<\/sub> + e and \nEn<\/sub> = – 13.6z2<\/sup>\/n2<\/sup> \nEl<\/sub> = – \\(\\frac{13.6(2)^{2}}{(1)^{2}}\\) = -56.4 eV<\/p>\nE\u221e<\/sub> = \\(\\frac{-13.6(2)^{2}}{(\\infty)^{2}}\\) = 0<\/p>\nRequired energy for the given process is, \nE\u221e<\/sub> – El<\/sub> = 0 – (-56.4) = 56.4 eV.<\/p>\nQuestion 44. \nAn ion with mass number 37 possesses unit negative charge. It the ion contains 11.1% more neutrons than electrons. Find the symbol of the ion. \nAnswer: \nLet the number of electrons in an ion = x \nnumber of neutrons = n = x + \\(\\frac{11.1}{100}\\) eV = 1.111 x \n(As the number of neutrons is 11.1% more than the number of electrons) \nIn the neutral of the atom, a number of electrons. \ne–<\/sup> = x – 1 (as the ion carries -1 charge) \nSimilarly number of protons = P = x – 1 \nNumber of protons + number of neutrons = mass number = 37 \n(x – 1) + 1.111 x = 37 . \n2.111 x = 37 +1 \n2.111 x = 38 \nx = \\(\\frac{38}{2.111}\\) = 18.009 = 18 \n\u2234 Number of protons = atomic number – 1 = 18-1 = 17 \n\u2234 The symbol of the ion = \\(_{17}^{37} \\mathrm{Cl}\\).<\/p>\nQuestion 45. \nThe Li2+<\/sup> ion is a hydrogen-like \u00a1on that can be described by the Bohr model. Calculate the Bohr radius of the third orbit and calculate the energy of an electron in 4th orbit. \nAnswer: \n<\/p>\n<\/p>\n Question 46. \nProtons can be accelerated in particle accelerators. Calculate the wavelength (in \u00c5) of such accelerated proton moving at 2.85 \u00d7 108<\/sup>ms-1<\/sup> (mass of proton is 1.673 \u00d7 10-27<\/sup> kg). \nAnswer: \nGiven: \nvelocity, v = 2.85 \u00d7 108<\/sup>ms-1<\/sup>. \nmass, m = 1.673 \u00d7 10-27<\/sup> kg \n\u03bb = \\(\\frac{h}{m v}\\)<\/p>\n\u03bb = \\(\\frac{6.626 \\times 10^{-34} \\mathrm{kgms}^{-1}}{1.673 \\times 10-27 \\mathrm{~kg} \\times 2-85 \\times 10^{8} \\mathrm{~ms}^{-1}}\\)<\/p>\n \u03bb = 1.389 \u00d7 10-8<\/sup> \u00c5<\/p>\nQuestion 47. \nWhat is the de Broglie wavelength (in cm) of a 160g cricket ball travelling at 140 Km hr-1<\/sup>. \nAnswer: \nGiven: \nvelocity, v = 140 km\/hr. = \\(\\frac{140 \\times 10^{3}}{60 \\times 60 \\mathrm{~ms}^{-1}}\\)<\/p>\nmass, m = 160g = 160 \u00d7 10-3<\/sup> kg |