JEE Chemistry Structure of the Atom MCQs Set 03

Answer: (c) 4

Explanation:
1. The element with \(Z = 27\) is cobalt, \([Ar]\,3d^7 4s^2\).
2. \(\ce{Co^{3+}}\) loses the two 4s electrons and one 3d electron, giving \(3d^6\).
3. The \(3d^6\) configuration has 4 unpaired electrons.

Teacher's Note:
1. Remove the 4s electrons before the 3d electrons when forming a cation.
2. Fill the five d orbitals singly first and then pair up.

Answer: (c) \(\frac{2.5h}{\pi}\)

Explanation:
1. By Bohr's postulate \(mvr = \frac{nh}{2\pi}\).
2. For \(n = 5\) the angular momentum is \(\frac{5h}{2\pi} = \frac{2.5h}{\pi}\).

Teacher's Note:
1. Angular momentum is quantised in whole multiples of \(\frac{h}{2\pi}\).
2. It depends on \(n\) only and not on the element.

Answer: (b) 2s: [↑]; 2p: [↑↓] [↑] [↑]

Explanation:
1. The Aufbau principle says that orbitals of lower energy are filled before those of higher energy.
2. In (b) the 2s orbital has only one electron while electrons are already present in 2p, so Aufbau is violated.
3. Diagram (a) violates Hund's rule instead, while (c) and (d) follow all the rules.

Teacher's Note:
1. The 2s orbital must be completely filled before 2p is occupied.
2. Do not confuse Aufbau with Hund's rule.

Answer: (a) \(3.3\times 10^{-33}\ \text{m}\)

Explanation:
1. \(v = 30\ \text{m/min} = 0.5\ \text{m s}^{-1}\) and \(m = 0.4\ \text{kg}\).
2. \(\lambda = \frac{h}{mv} = \frac{6.626\times 10^{-34}}{0.4\times 0.5}\).
3. \(\lambda = 3.3\times 10^{-33}\ \text{m}\).

Teacher's Note:
1. Convert the speed to \(\text{m s}^{-1}\) and the mass to kg.
2. The wavelength of a large object is far too small to observe.

Answer: (a) \(1.06\times 10^{16}\ \text{Hz}\)

Explanation:
1. Use \(KE = h(\nu - \nu_0)\) and the condition \(h(3.2\times 10^{16} - \nu_0) = 4h(1.6\times 10^{16} - \nu_0)\).
2. So \(3.2 - \nu_0 = 6.4 - 4\nu_0\) in units of \(10^{16}\), giving \(3\nu_0 = 3.2\times 10^{16}\).
3. \(\nu_0 = 1.06\times 10^{16}\ \text{Hz}\).

Teacher's Note:
1. Kinetic energy is photon energy minus work function.
2. Write the equation for both cases and divide.

Answer: (d) both (b) and (c)

Explanation:
1. Let the number of protons be \(Z\). The ion has \(Z - 3\) electrons and \(56 - Z\) neutrons.
2. \(56 - Z = 1.304(Z - 3)\) gives \(Z = 26\).
3. So there are 26 protons, 23 electrons and 30 neutrons, and (b) and (c) are correct.

Teacher's Note:
1. A positive ion has fewer electrons than protons.
2. Check your answer by adding protons and neutrons to get the mass number.

Answer: (a) Gamma rays

Explanation:
1. In the electromagnetic spectrum frequency increases in the order microwaves, ultraviolet, X-rays, gamma rays.
2. Gamma rays have the highest frequency and the shortest wavelength.

Teacher's Note:
1. Frequency is inversely proportional to wavelength.
2. Energy of a photon is proportional to its frequency.

Answer: (a) \(122.4\ \text{eV}\)

Explanation:
1. The third ionisation energy removes the last electron from \(\ce{Li^{2+}}\), a hydrogen-like ion with \(Z = 3\).
2. \(IE = 13.6\times Z^2 = 13.6\times 9 = 122.4\ \text{eV}\), and ionisation energy is positive.

Teacher's Note:
1. The ionisation energy of a hydrogen-like species is \(13.6\,Z^2/n^2\ \text{eV}\).
2. The energy of the electron in the orbit is negative.

Answer: (b) 3

Explanation:
1. The 4th excited state is \(n = 5\) and the 1st excited state is \(n = 2\).
2. Lines ending at \(n = 2\) belong to the Balmer series, which lies in the visible region.
3. The lines are 5 to 2, 4 to 2 and 3 to 2, so there are 3 lines.

Teacher's Note:
1. The ground state is \(n = 1\), so the nth excited state is \(n + 1\).
2. Balmer lines end on \(n = 2\).

Answer: (a) \(\frac{\pi a_0}{4}\)

Explanation:
1. The radius is proportional to \(n^2\), so \(r_4 = 16r_1 = a_0\) and \(r_2 = 4r_1 = \frac{a_0}{4}\).
2. For the nth orbit \(2\pi r = n\lambda\).
3. \(\lambda = \frac{2\pi r_2}{2} = \frac{\pi a_0}{4}\).

Teacher's Note:
1. The circumference of the orbit holds a whole number of wavelengths.
2. Compare orbit radii using \(r \propto n^2\).

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

Explanation:
1. Here \(\Delta p = \Delta x\), and the minimum uncertainty gives \(\Delta x\,\Delta p = \frac{h}{4\pi}\).
2. So \((\Delta p)^2 = \frac{h}{4\pi}\) and \(\Delta p = \frac12\sqrt{\frac h\pi}\).
3. \(\Delta v = \frac{\Delta p}{m} = \frac{1}{2m}\sqrt{\frac h\pi}\).

Teacher's Note:
1. Heisenberg's principle states \(\Delta x\,\Delta p \ge \frac{h}{4\pi}\).
2. Divide by mass to get the uncertainty in velocity.

Answer: (c) 1, 1

Explanation:
1. Radial nodes \(= n - l - 1 = 3 - 1 - 1 = 1\).
2. Angular nodes \(= l = 1\).

Teacher's Note:
1. Total nodes equal \(n - 1\).
2. For 3p this is 2, split as 1 radial and 1 angular.

Answer: (c) zero

Explanation:
1. The configuration of Cr is \([Ar]\,3d^5 4s^1\) because of the extra stability of a half filled d subshell.
2. The outermost orbit is the fourth shell, which holds one electron in the 4s orbital, so \(l = 0\).
3. The orbital angular momentum \(\sqrt{l(l+1)}\frac{h}{2\pi}\) is zero.

Teacher's Note:
1. Chromium and copper are exceptional configurations.
2. An s electron has no orbital angular momentum.

Answer: (c) \(\frac 1R\)

Explanation:
1. The shortest wavelength is the series limit, from \(n_2 = \infty\) to \(n_1 = 3\) for the Paschen series.
2. \(\frac{1}{\lambda} = RZ^2\left(\frac{1}{3^2} - 0\right) = R\times 9\times\frac19 = R\).
3. So \(\lambda = \frac 1R\).

Teacher's Note:
1. Paschen series ends on \(n = 3\).
2. The shortest wavelength corresponds to the transition from infinity.