JEE Chemistry Structure of the Atom MCQs Set 08

Answer: (b) \(\frac{1}{2m}\sqrt{\frac{h}{\pi}}\)

Explanation:
1. \(\Delta x \cdot \Delta p = \frac{h}{4\pi}\) with \(\Delta x = \Delta p\), so \((\Delta p)^2 = \frac{h}{4\pi}\).
2. \(\Delta p = \frac12\sqrt{\frac{h}{\pi}}\).
3. \(\Delta v = \frac{\Delta p}{m} = \frac{1}{2m}\sqrt{\frac{h}{\pi}}\).

Teacher's Note:
1. Velocity uncertainty is momentum uncertainty divided by mass.
2. Take the square root carefully after equating.

Answer: (b) \(1 \times 10^{9}\) cm s\(^{-1}\)

Explanation:
1. \(\Delta v = \frac{\Delta p}{m}\).
2. \(\Delta v = \frac{1 \times 10^{-18}}{9 \times 10^{-28}} = 1.1 \times 10^{9}\) cm/s.
3. This is closest to \(1 \times 10^{9}\) cm/s.

Teacher's Note:
1. Divide the momentum uncertainty by the mass.
2. Keep both quantities in grams and centimetres.

Answer: (b) \(4\ell + 2\)

Explanation:
1. A subshell has \(2\ell + 1\) orbitals.
2. Each orbital holds 2 electrons.
3. Maximum electrons = \(2(2\ell + 1) = 4\ell + 2\).

Teacher's Note:
1. \(2n^2\) is for a whole shell.
2. \(2\ell + 1\) counts orbitals, not electrons.

Answer: (b) 16

Explanation:
1. Orbitals in a shell = \(n^2\).
2. For n = 4, there are 16 orbitals: 4s (1), 4p (3), 4d (5) and 4f (7).

Teacher's Note:
1. The shell holds \(2n^2\) = 32 electrons.
2. Do not confuse number of orbitals with number of electrons.

Answer: (a) ns \(\to\) (n-2)f \(\to\) (n-1)d \(\to\) np

Explanation:
1. Use the (n + \(\ell\)) rule: 6s = 6, 4f = 7, 5d = 8, 6p = 8.
2. 6s fills first, then 4f, and 5d and 6p tie at 8, with 5d (lower n) before 6p.
3. So the order is ns, (n-2)f, (n-1)d, np.

Teacher's Note:
1. Remember the order 6s, 4f, 5d, 6p.
2. Lower n fills first when (n + \(\ell\)) is equal.

Answer: (a) the Lyman series

Explanation:
1. The series limit wavelength is \(\frac{n_1^2}{R}\), smallest for the smallest \(n_1\).
2. Lyman has \(n_1 = 1\), so it reaches the shortest wavelength (91.2 nm).

Teacher's Note:
1. Lyman is in the ultraviolet region.
2. Shorter wavelength means higher energy.

Answer: (d) The electron behaves as both a particle and a wave

Explanation:
1. Bohr's model treats the electron as a particle in a definite circular orbit.
2. The dual particle-wave nature of the electron was introduced later by de Broglie.
3. So statement (d) is wrong for Bohr's model.

Teacher's Note:
1. Bohr's model gave fixed orbits, which conflicts with the uncertainty principle.
2. Wave nature is part of the quantum mechanical model.

Answer: (c) 14

Explanation:
1. \(\ell\) = 3 is the 4f subshell, with \(2\ell + 1 = 7\) orbitals.
2. Maximum electrons = 7 x 2 = 14.

Teacher's Note:
1. Use \(4\ell + 2\).
2. For \(\ell\) = 3 it gives 14.

Answer: (a) 5, 0, 0, +1/2

Explanation:
1. Rubidium is [Kr] 5s1.
2. The valence electron has n = 5, \(\ell\) = 0, m = 0.
3. Its spin is conventionally +1/2.

Teacher's Note:
1. Rb is an alkali metal with one electron in its outermost s orbital.
2. The period number gives n.

Answer: (b) \(3.3 \times 10^{-20}\) J

Explanation:
1. \(K.E. = h(\nu - \nu_0)\).
2. \(= 6.626 \times 10^{-34} \times (1 \times 10^{14} - 5 \times 10^{13}) = 6.626 \times 10^{-34} \times 5 \times 10^{13}\).
3. \(= 3.3 \times 10^{-20}\) J.

Teacher's Note:
1. Photoelectric effect: photon energy minus work function.
2. Work function is \(h\nu_0\).

Answer: (d) angular momentum

Explanation:
1. Bohr's quantisation condition: \(mvr = \frac{nh}{2\pi}\).
2. The left side is the angular momentum of the electron.
3. So \(\frac{nh}{2\pi}\) is the angular momentum.

Teacher's Note:
1. Bohr's second postulate quantises angular momentum.
2. Its unit is J s.

Answer: (a) For n = 1 the electron has a more negative energy than for n = 6, which means the electron is more loosely bound in the smallest allowed orbit

Explanation:
1. A more negative energy means a more tightly bound electron.
2. For n = 1 the energy is the most negative, so the electron is most tightly (not loosely) bound in the smallest orbit.
3. Statement (a) is therefore incorrect, and (b), (c) and (d) are correct.

Teacher's Note:
1. Lower (more negative) energy means greater stability.
2. The reference zero is the electron at infinite distance.

Answer: (a) 2

Explanation:
1. n = 3, \(\ell\) = 1, m = -1 identifies one particular 3p orbital.
2. An orbital can hold a maximum of 2 electrons.

Teacher's Note:
1. Fixing n, \(\ell\) and m fixes the orbital.
2. Spin gives the two electrons in it.

Answer: (a) 1

Explanation:
1. The three quantum numbers n, \(\ell\) and m define one orbital, here 3p with m = 0.
2. So the maximum number of orbitals is 1.

Teacher's Note:
1. Each (n, \(\ell\), m) combination is a single orbital.
2. Compare with the number of electrons, which would be 2.

Answer: (d) \(4.42 \times 10^{-18}\)

Explanation:
1. \(E = \frac{hc}{\lambda}\).
2. \(\lambda = 45 \times 10^{-9}\) m, so \(E = \frac{6.63 \times 10^{-34} \times 3 \times 10^{8}}{45 \times 10^{-9}}\).
3. \(E = 4.42 \times 10^{-18}\) J.

Teacher's Note:
1. Convert nanometres to metres first.
2. The answer is typical of ultraviolet photons.

Answer: (c) the same as that of the 2p\(_x\) and 2p\(_z\) orbitals

Explanation:
1. The three 2p orbitals are degenerate, with the same n and \(\ell\).
2. So 2p\(_y\) has the same energy as 2p\(_x\) and 2p\(_z\).
3. In a multi-electron atom 2s lies lower than 2p.

Teacher's Note:
1. Orbitals of the same subshell have equal energy in the absence of a field.
2. Their orientations differ, not their energies.

Answer: (c) \(\ce{ClO3^-}, \ce{SO3^{2-}}\)

Explanation:
1. Electron counts: \(\ce{ClO3^-}\) = 17 + 24 + 1 = 42; \(\ce{SO3^{2-}}\) = 16 + 24 + 2 = 42; \(\ce{CO3^{2-}}\) = 6 + 24 + 2 = 32; \(\ce{NO3^-}\) = 7 + 24 + 1 = 32.
2. The pairs in (a), (b) and (d) differ in electron count.
3. \(\ce{ClO3^-}\) and \(\ce{SO3^{2-}}\) both have 42 electrons and a trigonal pyramidal shape.

Teacher's Note:
1. Isoelectronic means equal electrons; isostructural means equal shape.
2. Both central atoms have one lone pair, giving a pyramidal shape.

Answer: (a) p-electrons in Cl (Z = 17)

Explanation:
1. \(\ce{Fe^{2+}}\) is [Ar] 3d6, so it has 6 d-electrons.
2. p-electrons in Cl (1s2 2s2 2p6 3s2 3p5) = 6 + 5 = 11.
3. d-electrons in Fe = 6, p-electrons in Ne = 6, s-electrons in Mg = 2 + 2 + 2 = 6. So Cl is the odd one out.

Teacher's Note:
1. Write the full configurations and count.
2. Do not forget the 2p electrons when counting p-electrons of Cl.

Answer: (d) \(\ce{Ni^{2+}}\)

Explanation:
1. A moment of 2.84 BM corresponds to 2 unpaired electrons (\(\sqrt{2 \times 4} = 2.83\)).
2. \(\ce{Ti^{3+}}\) d1: 1, \(\ce{Cr^{2+}}\) d4: 4, \(\ce{Co^{2+}}\) d7: 3, \(\ce{Ni^{2+}}\) d8: 2.
3. So \(\ce{Ni^{2+}}\).

Teacher's Note:
1. Use \(\sqrt{n(n+2)}\) BM.
2. Ni2+ is d8 with 2 unpaired electrons.

Answer: (d) \(\sqrt6\,\hbar\)

Explanation:
1. Angular momentum \(= \sqrt{\ell(\ell+1)}\,\hbar\).
2. For a d orbital, \(\ell\) = 2, giving \(\sqrt{2 \times 3}\,\hbar = \sqrt6\,\hbar\).

Teacher's Note:
1. Here \(\hbar = \frac{h}{2\pi}\).
2. s, p, d, f have \(\sqrt{\ell(\ell+1)}\) = 0, \(\sqrt2\), \(\sqrt6\), \(\sqrt{12}\).

Answer: (b) 3s 3p 4s 3d

Explanation:
1. (n + \(\ell\)) values: 3s = 3, 3p = 4, 4s = 4, 3d = 5.
2. 3p and 4s tie at 4, and the lower n (3p) has lower energy.
3. So the order is 3s < 3p < 4s < 3d.

Teacher's Note:
1. Apply the (n + \(\ell\)) rule with the lower n first for ties.
2. This is the orbital energy order in a neutral atom.

Answer: (c) 2 to 1

Explanation:
1. For \(\ce{He+}\) (Z = 2): \(\frac1\lambda = 4R\left(\frac14 - \frac1{16}\right) = 4R \times \frac{3}{16} = \frac{3R}{4}\).
2. For hydrogen, \(\frac1\lambda = R\left(\frac1{n_1^2} - \frac1{n_2^2}\right) = \frac{3R}{4}\) for n = 2 to 1.
3. So the matching transition is 2 to 1.

Teacher's Note:
1. \(\ce{He+}\) transitions have a factor of 4 in wave number.
2. Match the wave numbers.

Answer: (d) spin quantum number

Explanation:
1. Electrons in the same orbital share n, \(\ell\) and m.
2. By Pauli's principle they must differ in the fourth quantum number, spin.
3. So they are distinguished by the spin quantum number.

Teacher's Note:
1. No two electrons can have all four quantum numbers equal.
2. The two spin orientations are +1/2 and -1/2.

Answer: (a) d\(_{z^2}\), d\(_{x^2-y^2}\)

Explanation:
1. d\(_{z^2}\) has its large lobes along the z-axis, and d\(_{x^2-y^2}\) has lobes along the x and y axes.
2. d\(_{xy}\), d\(_{xz}\) and d\(_{yz}\) have lobes between the axes.
3. So the correct pair is d\(_{z^2}\) and d\(_{x^2-y^2}\).

Teacher's Note:
1. These two are the e\(_g\) set.
2. The other three (t\(_{2g}\)) point between the axes.

Answer: (c) 2

Explanation:
1. An orbital is defined by n, \(\ell\) and m.
2. Any single orbital holds a maximum of 2 electrons.
3. So 2 electrons (the whole 3p subshell would hold 6).

Teacher's Note:
1. An orbital holds 2 electrons, a subshell holds \(2(2\ell + 1)\).
2. Read carefully whether the question asks about an orbital or a subshell.