JEE Chemistry Structure of the Atom MCQs Set 05

Answer: (d) \(2.188 \times 10^{3}\) km/s

Explanation:
1. \(v = 2.188 \times 10^{6}\,\frac{Z}{n}\) m/s.
2. For \(\ce{Be^{3+}}\), Z = 4 and the third excited state has n = 4.
3. \(v = 2.188 \times 10^{6} \times \frac44 = 2.188 \times 10^{6}\) m/s = \(2.188 \times 10^{3}\) km/s.

Teacher's Note:
1. The third excited state is n = 4.
2. Convert m/s to km/s by dividing by 1000.

Answer: (d) 2.12 Å

Explanation:
1. \(r \propto n^2\).
2. \(r_2 = 0.530 \times 4 = 2.12\) Å.

Teacher's Note:
1. The first excited state is n = 2.
2. Radius scales as n squared.

Answer: (a) 1 : 1

Explanation:
1. Radius depends on n and Z only (\(r \propto n^2/Z\), neglecting the small reduced-mass effect).
2. Both H and D have Z = 1.
3. So the ratio is 1 : 1.

Teacher's Note:
1. Isotopes have the same nuclear charge.
2. The extra neutron does not change the Bohr radius appreciably.

Answer: (b) 1 : 1/2 : 1/3

Explanation:
1. \(v \propto \frac1n\).
2. \(V_1 : V_2 : V_3 = 1 : \frac12 : \frac13\).

Teacher's Note:
1. Speed falls as 1/n.
2. Compare with radius, which grows as n squared.

Answer: (d) -82 kJ/mol

Explanation:
1. \(E_n \propto \frac{1}{n^2}\).
2. \(\frac{E_4}{E_2} = \frac{4}{16} = \frac14\).
3. \(E_4 = \frac{-328}{4} = -82\) kJ/mol.

Teacher's Note:
1. Energy rises towards zero as n increases.
2. Ratio of energies is the inverse ratio of \(n^2\).

Answer: (a) -1.51 eV

Explanation:
1. The second excited state has n = 3.
2. \(E_3 = \frac{-13.6}{9} = -1.51\) eV.

Teacher's Note:
1. Ground state is n = 1, first excited n = 2, second excited n = 3.
2. Divide -13.6 eV by \(n^2\).

Answer: (c) -6.8 eV

Explanation:
1. In a Bohr orbit, total energy \(E = \frac{U}{2}\). The first excited state of H is n = 2, \(E_2 = \frac{-13.6}{4} = -3.4\) eV.
2. (Check: for \(\ce{He+}\), n = 2, \(E = -13.6\) eV, so U = -27.2 eV, which agrees.)
3. Double the total energy = 2 x (-3.4) = -6.8 eV.

Teacher's Note:
1. The potential energy data are consistent with the Bohr relation U = 2E.
2. The question asks for hydrogen, not helium.

Answer: (a) \(E_n\) for A = \(Z^2 \times E_n\) for H

Explanation:
1. For one-electron species \(E_n = -13.6\,\frac{Z^2}{n^2}\) eV.
2. So the energy at a given n is \(Z^2\) times the hydrogen value.

Teacher's Note:
1. One-electron ions follow the Bohr formula with Z squared.
2. Examples: \(\ce{He+}\), \(\ce{Li^{2+}}\) and \(\ce{Be^{3+}}\).

Answer: (a) the kinetic energy decreases

Explanation:
1. Kinetic energy \(= +\frac{13.6 Z^2}{n^2}\) eV, so it decreases as n increases.
2. Total energy rises (becomes less negative) and potential energy also rises.
3. Only the kinetic energy decreases.

Teacher's Note:
1. Moving outward, the electron slows down but its energy increases.
2. Potential and total energy are negative and increase towards zero.

Answer: (c) from n = 1 to n = 3 (instead of \(\infty\) to n = 1, which is an emission)

Explanation:
1. Energy absorbed: 1 to 2 = 10.2 eV; 2 to 3 = 1.89 eV; 1 to 3 = 12.09 eV; 3 to 5 = 0.97 eV.
2. The largest value is for 1 to 3.

Teacher's Note:
1. Absorption energy is large when the lower level is n = 1.
2. Higher levels are closer together.

Answer: (d) n = 5 to n = 2

Explanation:
1. Emission requires a drop to a lower level, so (a) and (b) (absorptions) are ruled out.
2. Energy for 3 to 1 is 12.09 eV; energy for 5 to 2 is \(13.6(\frac14 - \frac1{25}) = 2.86\) eV.
3. The smaller energy gives the lower frequency, so 5 to 2.

Teacher's Note:
1. Low frequency means a small energy gap.
2. Check whether each option is emission or absorption first.

Answer: (d) 16.53 eV

Explanation:
1. \(\Delta E = 13.6 Z^2\left(\frac1{3^2} - \frac1{4^2}\right) = 13.6 \times 25 \times \frac{7}{144}\).
2. \(= 340 \times 0.04861 = 16.53\) eV.

Teacher's Note:
1. Use the hydrogen-like energy formula with Z = 5.
2. Subtract the energies of the two levels.

Answer: (d) 9/4

Explanation:
1. First excited state: n = 2; second excited state: n = 3.
2. \(\frac{E_2}{E_3} = \frac{1/4}{1/9} = \frac94\).

Teacher's Note:
1. Excited state number is n - 1.
2. Energy varies as \(1/n^2\).

Answer: (c) 3

Explanation:
1. \(n^2 = \frac{313.6}{34.84} = 9\).
2. So n = 3.

Teacher's Note:
1. Rearrange the formula for \(n^2\) and take the square root.
2. The result must be a whole number.

Answer: (b) \(E_1\)(H) = \(E_2\)(\(\ce{He+}\)) = \(E_3\)(\(\ce{Li^{2+}}\)) = \(E_4\)(\(\ce{Be^{3+}}\))

Explanation:
1. \(E_n = -13.6\,\frac{Z^2}{n^2}\) eV.
2. The ratio Z/n is 1 for H (n = 1), \(\ce{He+}\) (n = 2), \(\ce{Li^{2+}}\) (n = 3) and \(\ce{Be^{3+}}\) (n = 4).
3. So all four energies are equal to -13.6 eV.

Teacher's Note:
1. Check Z/n for each case.
2. Equal Z/n means equal energy.

Answer: (a) \((E_2 - E_1) > (E_3 - E_2) > (E_4 - E_3)\)

Explanation:
1. For Z = 1 in units of 13.6 eV: \(E_2 - E_1 = \frac34\), \(E_3 - E_2 = \frac{5}{36}\), \(E_4 - E_3 = \frac{7}{144}\).
2. The gaps decrease as n increases.
3. The same order holds for any Z.

Teacher's Note:
1. Energy levels crowd closer together at larger n.
2. They converge towards zero energy.

Answer: (d) 1.5 eV

Explanation:
1. The second excited state is n = 3.
2. \(E_3 = -\frac{13.6}{9} = -1.51\) eV.
3. Energy to ionise = 1.51 eV (about 1.5 eV).

Teacher's Note:
1. Ionisation from an excited level needs less energy than from the ground state.
2. Use \(E = 13.6/n^2\) with n = 3.

Answer: (c) 3.4

Explanation:
1. The first excited state is n = 2.
2. \(E_2 = -\frac{13.6}{4} = -3.4\) eV.
3. The energy required to remove the electron is 3.4 eV.

Teacher's Note:
1. An excited atom is easier to ionise than a ground-state atom.
2. Ionisation energy equals the magnitude of the level energy.

Answer: (a) the kinetic energy of the first orbit

Explanation:
1. Ionisation potential of \(\ce{He+}\) is the energy needed to remove the electron from n = 1: 54.4 eV.
2. In a Bohr orbit the kinetic energy equals minus the total energy, i.e. +54.4 eV for n = 1.
3. So it equals the kinetic energy of the first orbit.

Teacher's Note:
1. In Bohr orbits, K.E. = |E|.
2. Ionisation energy is the magnitude of the ground-state energy.

Answer: (d) Li\(^{+}\)

Explanation:
1. The spectrum depends on the electronic structure.
2. He has 2 electrons, as does \(\ce{Li+}\), so they are isoelectronic.
3. H, Na and \(\ce{He+}\) do not have 2 electrons, so \(\ce{Li+}\) is the closest.

Teacher's Note:
1. Isoelectronic species have similar spectral patterns.
2. \(\ce{He+}\) is only hydrogen-like, with one electron.

Answer: (c) \(n_4 \to n_1\)

Explanation:
1. Shortest wavelength means largest energy.
2. The gaps are: 2 to 1 = 10.2 eV, 3 to 1 = 12.09 eV, 4 to 1 = 12.75 eV, 4 to 3 = 0.66 eV.
3. The largest is 4 to 1.

Teacher's Note:
1. Wavelength is inversely proportional to energy.
2. Transitions down to n = 1 are in the ultraviolet.

Answer: (d) Lyman

Explanation:
1. The Lyman series ends at n = 1, where the energy gaps are largest.
2. Balmer ends at 2, Brackett at 4 and Pfund at 5, all with smaller gaps.

Teacher's Note:
1. The lower the final level, the higher the photon energy.
2. Lyman lies in the ultraviolet.

Answer: (a) \(\ce{Al^{3+}}\)

Explanation:
1. \(\ce{Mg^{2+}}\) has 12 - 2 = 10 electrons.
2. \(\ce{Al^{3+}}\) has 13 - 3 = 10 electrons, Na has 11, \(\ce{Mg+}\) has 11 and F has 9.
3. Only \(\ce{Al^{3+}}\) is isoelectronic.

Teacher's Note:
1. Similar spectra arise from similar electronic structure.
2. Count electrons for each species.

Answer: (b) 0.25

Explanation:
1. The minimum wavelength is the series limit (\(n_2 = \infty\)). \(\frac1\lambda = R\frac1{n_1^2}\).
2. Lyman: \(\lambda = \frac1R\); Balmer: \(\lambda = \frac4R\).
3. Ratio = 1/4 = 0.25.

Teacher's Note:
1. The series limit gives the shortest wavelength.
2. Balmer limit is at 364.7 nm and Lyman limit at 91.2 nm.

Answer: (d) \(\lambda_2 = \lambda_1/4\)

Explanation:
1. \(\frac1\lambda \propto Z^2\) for the same pair of levels.
2. \(\frac{\lambda_2}{\lambda_1} = \frac{Z_H^2}{Z_{He}^2} = \frac14\).

Teacher's Note:
1. Energy gaps scale as \(Z^2\).
2. Larger energy gap means shorter wavelength.

Answer: (a) 1 : 9

Explanation:
1. \(\frac1\lambda \propto Z^2\) for the same transition.
2. \(\frac{\lambda_{Li}}{\lambda_D} = \frac{1^2}{3^2} = \frac19\).

Teacher's Note:
1. Deuterium has Z = 1 like hydrogen.
2. Lithium ion lines are at 9 times shorter wavelength.