JEE Chemistry Structure of the Atom MCQs Set 09

Answer: (c) 486 nm

Explanation:
1. \(\frac1\lambda \propto \frac14 - \frac1{n^2}\). For 5 to 2: \(\frac14 - \frac1{25} = \frac{21}{100}\). For 4 to 2: \(\frac14 - \frac1{16} = \frac{3}{16}\).
2. \(\lambda_{4 \to 2} = 434 \times \frac{21/100}{3/16} = 434 \times 1.12 = 486\) nm.

Teacher's Note:
1. Wavelength is inversely proportional to the gap.
2. The 4 to 2 line is the blue-green Balmer line.

Answer: (a) Thomson's atomic model

Explanation:
1. This is the plum-pudding (watermelon) model.
2. It was proposed by J. J. Thomson.
3. Rutherford later replaced it with the nuclear model.

Teacher's Note:
1. Thomson's model was disproved by the gold foil experiment.
2. Chadwick discovered the neutron.

Answer: (c) The electronic configuration of the N atom is 1s2 2s2 2p\(^1_x\) 2p\(^1_y\) 2p\(^1_z\) written with the p\(_z\) electron having opposite spin to the other two

Explanation:
1. For an s orbital \(\ell\) = 0, so the angular momentum \(\sqrt{\ell(\ell+1)}\,\hbar\) is zero (a is correct).
2. An orbital needs n, \(\ell\), m and an electron also needs s (b is correct). d\(_{z^2}\) has m = 0 (d is correct).
3. In nitrogen the three 2p electrons must have parallel spins (Hund's rule), so an arrangement with one opposite spin is wrong (c).

Teacher's Note:
1. Hund's rule requires parallel spins in a half-filled subshell.
2. The wrong statement is the one that breaks it.

Answer: (a) \(\frac{2\pi k e^2}{h}\)

Explanation:
1. \(v = \frac{2\pi Z k e^2}{nh}\).
2. For \(\ce{He+}\), Z = 2 and n = 2.
3. \(v = \frac{2\pi \times 2 k e^2}{2h} = \frac{2\pi k e^2}{h}\).

Teacher's Note:
1. This is the Bohr velocity formula \(v = \frac{2\pi k Z e^2}{nh}\).
2. For hydrogen with n = 1 it equals \(2.18 \times 10^6\) m/s.

Answer: (c) \(10.9 \times 10^{5}\) m/s

Explanation:
1. \(v = 2.18 \times 10^{6}\,\frac{Z}{n}\) m/s.
2. For Z = 1 and n = 2: \(v = 1.09 \times 10^{6}\) m/s.
3. This equals \(10.9 \times 10^{5}\) m/s.

Teacher's Note:
1. Velocity in the first orbit is \(2.18 \times 10^{6}\) m/s and halves for n = 2.
2. Express the value with the required power of ten.

Answer: (a) \(\frac{2hc}{m_e}\left[\frac1{\lambda_2} - \frac1{\lambda_1}\right]\)

Explanation:
1. Photoelectric equation: \(\frac{hc}{\lambda_1} = \phi + \frac12 m_e V_1^2\) and \(\frac{hc}{\lambda_2} = \phi + \frac12 m_e V_2^2\).
2. Subtracting, \(\frac12 m_e (V_2^2 - V_1^2) = hc\left(\frac1{\lambda_2} - \frac1{\lambda_1}\right)\).
3. \(V_2^2 - V_1^2 = \frac{2hc}{m_e}\left[\frac1{\lambda_2} - \frac1{\lambda_1}\right]\).

Teacher's Note:
1. The work function is the same for both, so it cancels on subtraction.
2. Photon energy is \(\frac{hc}{\lambda}\).

Answer: (a) 820.8 nm

Explanation:
1. The shortest wavelength is the series limit: \(\frac1\lambda = R\left(\frac1{3^2} - 0\right) = \frac{R}{9}\).
2. \(\lambda = \frac{9}{R} = \frac{9}{1.097 \times 10^{7}}\) m \(= 820\) nm.
3. This matches 820.8 nm.

Teacher's Note:
1. Paschen series ends at n = 3.
2. The series limit uses \(n_2 = \infty\).

Answer: (b) B

Explanation:
1. He (1s2), Li (1s2 2s1) and Be (1s2 2s2) contain only s electrons, which are spherically symmetric.
2. B (1s2 2s2 2p1) has a p electron, whose density is dumbbell shaped.
3. So B has non-spherical electron density.

Teacher's Note:
1. s orbitals are spherical and p orbitals are directional.
2. A single p electron breaks spherical symmetry.

Answer: (d) 10

Explanation:
1. n = 3 and \(\ell\) = 2 denote the 3d subshell.
2. It has \(2\ell + 1 = 5\) orbitals.
3. Maximum electrons = 5 x 2 = 10.

Teacher's Note:
1. Maximum electrons in a subshell = \(4\ell + 2\).
2. For \(\ell\) = 2 this gives 10.

Answer: (c) 36

Explanation:
1. In ground-state atoms the subshells filled up to n = 4 follow Aufbau: 1s to 3d and then 4s, 4p.
2. The 4d and 4f subshells are filled only after 5s (so an atom whose highest n is 4 stops at krypton).
3. Maximum number of electrons = 36.

Teacher's Note:
1. The 4d and 4f subshells come after 5s in energy.
2. Krypton (Z = 36) completes the n = 4 level of occupation.

Answer: (d) two quantum mechanical spin states which have no classical analogue

Explanation:
1. Electron spin is a purely quantum mechanical property with no true classical counterpart.
2. The +1/2 and -1/2 labels denote two possible spin states, and the picture of a spinning ball is only a crude model.
3. So (d) is the correct statement.

Teacher's Note:
1. Do not think of the electron as a literally rotating sphere.
2. Spin states are described only by quantum mechanics.

Answer: (b) \(2.1 \times 10^{-26}\) m/s

Explanation:
1. \(\Delta v = \frac{h}{4\pi m \Delta x}\).
2. \(m = 0.25 \times 10^{-3}\) kg, so \(4\pi m \Delta x = 4\pi \times 2.5 \times 10^{-4} \times 10^{-5} = 3.14 \times 10^{-8}\).
3. \(\Delta v = \frac{6.6 \times 10^{-34}}{3.14 \times 10^{-8}} = 2.1 \times 10^{-26}\) m/s.

Teacher's Note:
1. Convert grams to kilograms.
2. The tiny result shows that uncertainty is negligible for macroscopic objects.

Answer: (d) n only

Explanation:
1. In the hydrogen atom (a one-electron system) \(E_n = -\frac{13.6}{n^2}\) eV.
2. The energy depends only on the principal quantum number n.

Teacher's Note:
1. In multi-electron atoms the energy depends on n and \(\ell\).
2. Hydrogen is a special one-electron case.

Answer: (b) n = 2 to n = 1

Explanation:
1. For \(\ce{He+}\), \(\Delta E \propto Z^2\left(\frac14 - \frac1{16}\right) = 4 \times \frac{3}{16} = \frac34\).
2. For H, the transition 2 to 1 gives \(1 - \frac14 = \frac34\).
3. The energies match, so the frequencies are equal.

Teacher's Note:
1. Compare Z squared times the bracket for both species.
2. 3/4 of 13.6 eV is 10.2 eV for both.

Answer: (c) (d) < (b) < (c) < (a)

Explanation:
1. The subshells are (a) 4p, (b) 4s, (c) 3d, (d) 3p.
2. (n + \(\ell\)) values: 4p = 5, 4s = 4, 3d = 5, 3p = 4.
3. Lower (n + \(\ell\)) has lower energy, and ties are broken by lower n: 3p < 4s < 3d < 4p.

Teacher's Note:
1. Convert each pair to a subshell name first.
2. Order: (d) < (b) < (c) < (a).

Answer: (a) \(1.214 \times 10^{-7}\) m

Explanation:
1. \(\Delta E = 2.178 \times 10^{-18}\left(1 - \frac14\right) = 1.6335 \times 10^{-18}\) J.
2. \(\lambda = \frac{hc}{\Delta E} = \frac{6.62 \times 10^{-34} \times 3 \times 10^{8}}{1.6335 \times 10^{-18}}\).
3. \(\lambda = 1.216 \times 10^{-7}\) m, which matches \(1.214 \times 10^{-7}\) m (121.6 nm, the Lyman alpha line).

Teacher's Note:
1. Use \(\Delta E = hc/\lambda\).
2. This is the first line of the Lyman series, in the ultraviolet.

Answer: (c) 5

Explanation:
1. \(\Delta E = 13.6 Z^2\left(\frac14 - \frac19\right) = 13.6 Z^2 \times \frac{5}{36} = 1.889 Z^2\).
2. \(1.889 Z^2 = 47.2\), so \(Z^2 = 25\).
3. Z = 5.

Teacher's Note:
1. Use the hydrogen-like energy formula for two levels.
2. Solve for \(Z^2\) and take the square root.

Answer: (c) 4

Explanation:
1. Final energy \(= -13.6 + 12.75 = -0.85\) eV.
2. \(-\frac{13.6}{n^2} = -0.85\) gives \(n^2 = 16\).
3. So n = 4.

Teacher's Note:
1. Add the photon energy to the initial energy.
2. The result must match a level energy exactly.

Answer: (b) 2.55 eV

Explanation:
1. The third excited state is n = 4 and the first excited state is n = 2.
2. \(\Delta E = 13.6\left(\frac14 - \frac1{16}\right) = 13.6 \times \frac{3}{16}\).
3. \(\Delta E = 2.55\) eV.

Teacher's Note:
1. Excited state number is n - 1.
2. This is the second line of the Balmer series (blue-green).

Answer: (b) The atom will be ionised and the electron will have a kinetic energy of 0.4 eV

Explanation:
1. The ionisation energy of hydrogen is 13.6 eV, and 14 eV is more than that.
2. The photon ionises the atom, and the extra energy goes to the electron as kinetic energy.
3. K.E. = 14 - 13.6 = 0.4 eV.

Teacher's Note:
1. A photon with energy above the ionisation energy is absorbed.
2. Excess energy becomes kinetic energy of the free electron.

Answer: (c) 10.2 eV of the electron's energy would be absorbed by the H atom and it would come out with 0.6 eV energy

Explanation:
1. A colliding electron can transfer part of its kinetic energy to the atom, but only in amounts equal to an allowed energy gap.
2. The smallest gap is 10.2 eV (n = 1 to 2), and 10.8 eV exceeds it.
3. The atom absorbs 10.2 eV and the electron leaves with 10.8 - 10.2 = 0.6 eV.

Teacher's Note:
1. Unlike a photon, an electron need not give up all of its energy.
2. The atom accepts only quantised amounts.

Answer: (c) 27/5

Explanation:
1. In units of 13.6 eV: \(E_2 - E_1 = 1 - \frac14 = \frac34\) and \(E_3 - E_2 = \frac14 - \frac19 = \frac{5}{36}\).
2. Ratio \(= \frac{3/4}{5/36} = \frac{3}{4} \times \frac{36}{5} = \frac{27}{5}\).

Teacher's Note:
1. Level spacings shrink rapidly with increasing n.
2. Compute both differences with a common factor.

Answer: (a) A-(iv), B-(ii), C-(i), D-(iii)

Explanation:
1. A: \(E_1\) of \(\ce{He+}\) = -13.6 x 4 = -54.4 eV (iv).
2. B: PE = 2E = 2(-13.6) = -27.2 eV (ii).
3. C: second excited state n = 3, K.E. = \(\frac{13.6 \times 4}{9} = +6.04\) eV (i).
4. D: ionisation energy = 54.4 eV = \(54.4 \times 1.602 \times 10^{-19} = 8.72 \times 10^{-18}\) J (iii).

Teacher's Note:
1. Kinetic energy is positive and equals minus the total energy.
2. Convert eV to joule with \(1.602 \times 10^{-19}\).

Answer: (c) \(\nu_3 = \nu_2 + \nu_1\)

Explanation:
1. The energy for 3 to 1 equals the sum of the energies for 3 to 2 and 2 to 1: \(E_{3 \to 1} = E_{3 \to 2} + E_{2 \to 1}\), so (a) is wrong.
2. Since \(E = h\nu\), \(\nu_3 = \nu_2 + \nu_1\), so (c) is correct.
3. In terms of wavelength, \(\frac1{\lambda_3} = \frac1{\lambda_1} + \frac1{\lambda_2}\), so (b) is wrong.

Teacher's Note:
1. Energies and frequencies add, wavelengths do not.
2. Check with the numbers: 12.09 = 1.89 + 10.2 eV.

Answer: (d) all of them

Explanation:
1. Each single electronic jump emits exactly one photon (one quantum) with energy equal to the gap.
2. This is true for 4 to 2, 3 to 1 and 4 to 1.
3. So all of them emit one quantum.

Teacher's Note:
1. One transition equals one photon.
2. A multi-step cascade would emit several photons.