CBSE Class 11 Chemistry Structure of Atom MCQs Set 04

Answer: (c) 50

Explanation:
1. Speed of light, c = 3 × 10¹⁷ nm s⁻¹; frequency, ν = 6 × 10¹⁵ s⁻¹.
2. Using c = νλ, λ = c/ν.
3. λ = 3 × 10¹⁷ ÷ 6 × 10¹⁵ = 0.5 × 10² nm = 50 nm.

Teacher's Note:
1. Keep the units of c in mind: here c is already given in nm s⁻¹, so λ comes out directly in nm.
2. Planck's constant is not needed in this calculation; it is given only as a distractor.

Answer: (d) 4.42 × 10⁻¹⁸

Explanation:
1. λ = 45 nm = 45 × 10⁻⁹ m.
2. E = hν = hc/λ.
3. E = (6.63 × 10⁻³⁴ J s × 3 × 10⁸ m s⁻¹) ÷ (45 × 10⁻⁹ m) = 4.42 × 10⁻¹⁸ J.

Teacher's Note:
1. Always convert nm to metres (1 nm = 10⁻⁹ m) before using E = hc/λ in SI units.
2. A photon's energy in joules is a very small number, so any option with a positive power of 10 can be rejected at once.

Answer: (a) Lyman and Paschen

Explanation:
1. According to Rydberg's equation, 1/λ ∝ (1/n₁² − 1/n₂²).
2. The shortest wavelength (series limit) of a series corresponds to n₂ = ∞, so 1/λ_min ∝ 1/n₁², i.e. λ_min ∝ n₁².
3. Lyman (n₁ = 1) and Paschen (n₁ = 3): λ_L : λ_P = 1 : 9.
4. Brackett (n₁ = 4) and Pfund (n₁ = 5): 16 : 25; Paschen and Pfund: 9 : 25; Balmer (n₁ = 2) and Brackett: 4 : 16 = 1 : 4.
5. Only Lyman and Paschen give a ratio of 9.

Teacher's Note:
1. The series limit of any hydrogen series is proportional to n₁², which makes such ratio questions quick.
2. Remember n₁ = 1, 2, 3, 4, 5 for Lyman, Balmer, Paschen, Brackett and Pfund.

Answer: (a) Balmer series

Explanation:
1. The Balmer series of transitions in the hydrogen spectrum falls in the visible region.
2. The Lyman series falls in the ultraviolet region, while the Paschen, Brackett and Pfund series fall in the infrared region.

Teacher's Note:
1. Balmer lines end on n = 2, and four of them (red to violet) are visible to the eye.
2. Lyman is UV because transitions end on n = 1, the largest energy gap.

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

Explanation:
1. For a hydrogen-like species, 1/λ = Z²R (1/n₁² − 1/n₂²).
2. For He⁺, Z = 2: 1/λ = 4R (1/2² − 1/4²) = 4R × 3/16 = 3R/4.
3. For the H atom (Z = 1), the same value is obtained for n = 2 to n = 1: 1/λ = R (1 − 1/4) = 3R/4.

Teacher's Note:
1. For a species of atomic number Z, multiplying both n values of a hydrogen transition by Z gives the same wavelength: (2 → 1) × 2 = (4 → 2).
2. Do not forget the Z² factor in the Rydberg formula for He⁺ and Li²⁺.

Answer: (d) 24,372 cm⁻¹

Explanation:
1. \( \bar{\nu} = R_H \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right) \)
2. For the Balmer series n₁ = 2; the 4th line corresponds to n₂ = 6.
3. \( \bar{\nu} = 109677 \left( \frac{1}{4} - \frac{1}{36} \right) = 109677 \times \frac{8}{36} \)
4. \( \bar{\nu} \) = 24,372 cm⁻¹.

Teacher's Note:
1. The k^th line of a series has n₂ = n₁ + k, so the 4th Balmer line is 6 → 2.
2. A common slip is to take n₂ = 4 for the 4th line; that gives the 2nd line instead.

Answer: (d) \( \frac{4}{27R} \)

Explanation:
1. \( \frac{1}{\lambda} = R Z^2 \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right) \), with Z = 3 for Li²⁺.
2. \( \frac{1}{\lambda} = R (3)^2 \left( \frac{1}{1^2} - \frac{1}{2^2} \right) = \frac{9 \times 3R}{4} = \frac{27R}{4} \)
3. \( \lambda = \frac{4}{27R} \)

Teacher's Note:
1. The question asks for λ, not 1/λ, so take the reciprocal at the end.
2. Li²⁺ is a one-electron species, so Bohr's formula with Z² = 9 applies.

Answer: (c) spectrum of any atom or ion containing one electron only

Explanation:
1. Bohr's model is applicable only to the hydrogen atom and hydrogen-like species, i.e. one-electron systems.
2. Thus Bohr's model can explain the spectrum of any atom or ion containing one electron only (H, He⁺, Li²⁺, Be³⁺, etc.), not just the hydrogen atom.

Teacher's Note:
1. Hydrogen-like ions such as He⁺ and Li²⁺ are also covered by Bohr's model; only the formula picks up a Z².
2. Bohr's model fails for multi-electron atoms and for molecules like H₂.

Answer: (a) \( \nu = \frac{\Delta E}{h} = \frac{E_2 - E_1}{h} \)

Explanation:
1. The frequency of radiation absorbed or emitted when a transition occurs between two stationary states that differ in energy by ΔE is given by
2. \( \nu = \frac{\Delta E}{h} = \frac{E_2 - E_1}{h} \)
3. where E₁ and E₂ are the energies of the lower and higher allowed energy states respectively. This expression is known as Bohr's frequency rule.

Teacher's Note:
1. Bohr's frequency rule is simply ΔE = hν applied to a jump between two orbits.
2. Check dimensions: only energy divided by h gives a frequency.

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

Explanation:
1. According to Bohr, the angular momentum of the electron is quantised: \( mvr = \frac{nh}{2\pi} \).
2. Hence an electron can move only in those orbits for which its angular momentum is an integral multiple of h/2π.

Teacher's Note:
1. h/2π is written as ħ (h-bar); Bohr's condition is mvr = nħ.
2. h/4π appears in the uncertainty principle, not in Bohr's quantisation condition.

Answer: (c) He²⁺

Explanation:
1. Bohr's atomic model is applicable only to the hydrogen atom and H-like species, i.e. one-electron systems.
2. Be³⁺, Li²⁺ and H each have one electron.
3. He²⁺ does not contain any electron, so Bohr's model is not applicable to it.

Teacher's Note:
1. Count electrons: Z minus the charge. He²⁺ has 2 − 2 = 0 electrons.
2. A bare nucleus such as He²⁺ (an alpha particle) has no orbits to describe.

Answer: (d) 2.12

Explanation:
1. r ∝ n²/Z; for the H atom Z = 1.
2. \( \frac{r_1}{r_2} = \frac{n_1^2}{n_2^2} \Rightarrow \frac{0.530}{r_2} = \frac{1^2}{2^2} \)
3. r₂ = 0.530 × 4 = 2.12 Å.

Teacher's Note:
1. The first excited state is n = 2, not n = 1.
2. Radius grows as n², so doubling n makes the orbit four times larger.

Answer: (a) 91 nm

Explanation:
1. \( \frac{1}{\lambda} = R_H \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right) = 1.097 \times 10^7 \left( \frac{1}{1^2} - \frac{1}{\infty^2} \right) \) m⁻¹
2. λ = 1/(1.097 × 10⁷) m = 9.11 × 10⁻⁸ m
3. λ = 91.1 nm ≈ 91 nm.

Teacher's Note:
1. The jump from infinity to n = 1 gives the series limit of the Lyman series, which lies in the UV region.
2. 1/∞² = 0, so λ is simply the reciprocal of R_H.

Answer: (d) 10.2

Explanation:
1. \( \Delta E = 13.6 Z^2 \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right) \) eV
2. \( \Delta E = 13.6 \times 1^2 \left( \frac{1}{1^2} - \frac{1}{2^2} \right) = 13.6 \times \frac{3}{4} \)
3. ΔE = 10.2 eV.

Teacher's Note:
1. E₁ = −13.6 eV and E₂ = −3.4 eV, so the gap is 10.2 eV.
2. 13.6 eV is the ionisation energy (n = 1 to ∞), and 3.4 eV is just the magnitude of E₂; both are traps.

Answer: (d) −82 kJ mol⁻¹

Explanation:
1. E_n = −1312/n² kJ mol⁻¹, so E₂ = −1312/2² = −328 kJ mol⁻¹.
2. E₄ = −1312/4² = −82 kJ mol⁻¹.
3. Equivalently, E₄ = E₂ × (2/4)² = −328/4 = −82 kJ mol⁻¹.

Teacher's Note:
1. Energy varies as 1/n², so going from n = 2 to n = 4 divides it by 4, not by 2.
2. Bohr orbit energies are negative, and they become less negative as n increases.

Answer: (d) All of the above

Explanation:
1. Limitations of Bohr's model are:
2. (i) It fails to account for the finer details (doublets) of the hydrogen atom spectrum observed by using sophisticated spectroscopic techniques.
3. (ii) It is unable to explain the spectrum of atoms other than hydrogen (multi-electron atoms).
4. (iii) It is unable to explain the ability of atoms to form molecules by chemical bonds.
5. Hence all the given statements are limitations.

Teacher's Note:
1. Bohr's model also ignores the dual nature of the electron and the uncertainty principle.
2. It cannot explain the Zeeman (magnetic) and Stark (electric) effects either.

Answer: (b) λ = h/p

Explanation:
1. The relation between wavelength (λ) and momentum (p) of a material particle is λ = h/p = h/mv.
2. This is popularly called de Broglie's equation.

Teacher's Note:
1. Wavelength is inversely proportional to momentum: heavier or faster particles have shorter wavelengths.
2. Only λ = h/p is dimensionally correct, since h has units of momentum × length.

Answer: (b) Alpha particle (He²⁺)

Explanation:
1. From de Broglie's equation, λ = h/mv; at the same speed, λ ∝ 1/m.
2. The greater the mass of the particle, the smaller is its wavelength.
3. Among these, the alpha particle (He²⁺) has the highest mass, therefore it has the shortest wavelength.

Teacher's Note:
1. Mass order: electron ≪ proton ≈ neutron < alpha particle (about 4 u).
2. At equal speeds, de Broglie wavelength simply follows the reverse order of mass.

Answer: (c) 6.625 × 10⁻¹⁷ kg m s⁻¹

Explanation:
1. According to de Broglie's relation, λ = h/mv = h/p, so p = h/λ.
2. p = 6.625 × 10⁻³⁴ J s ÷ 10⁻¹⁷ m
3. p = 6.625 × 10⁻³⁴ × 10¹⁷ = 6.625 × 10⁻¹⁷ kg m s⁻¹.

Teacher's Note:
1. Rearrange first (p = h/λ), then substitute; it avoids power-of-ten slips.
2. 1 J s m⁻¹ = 1 kg m s⁻¹, the unit of momentum.

Answer: (d) E_e > E_p > E_α

Explanation:
1. de Broglie wavelength: λ = h/mv ⇒ v = h/mλ.
2. KE = ½mv² = ½m (h/mλ)² = \( \frac{h^2}{2m\lambda^2} \)
3. Hence KE ∝ 1/m when λ is the same.
4. Order of mass: m_e < m_p < m_α.
5. Thus the order of KE is E_e > E_p > E_α.

Teacher's Note:
1. For the same wavelength the momenta are equal, and KE = p²/2m, so the lightest particle has the most KE.
2. Do not confuse this with equal speeds, where the heaviest particle has the most KE.

Answer: (c) 6.6 × 10⁻³¹ cm

Explanation:
1. According to de Broglie, λ = h/mv.
2. Given m = 100 g, v = 100 cm s⁻¹ and h = 6.6 × 10⁻³⁴ J s = 6.6 × 10⁻²⁷ erg s.
3. λ = 6.6 × 10⁻²⁷ ÷ (100 × 100) = 6.6 × 10⁻³¹ cm.

Teacher's Note:
1. Work entirely in CGS units (g, cm, erg) so the answer comes out directly in cm.
2. Such a tiny wavelength shows why wave nature is never noticed for everyday objects.

Answer: (d) 0.1 kg

Explanation:
1. According to de Broglie's equation, λ = h/mv.
2. 6.62 × 10⁻³⁵ = 6.62 × 10⁻³⁴ ÷ (x × 100)
3. x = 6.62 × 10⁻³⁴ ÷ (6.62 × 10⁻³⁵ × 100) = 0.1 kg.

Teacher's Note:
1. Rearranging to m = h/λv first keeps the arithmetic simple.
2. Here the matching mantissas (6.62) cancel, leaving only powers of 10.

Answer: (a) 25 : 9

Explanation:
1. de Broglie equation: λ = h/mv; KE = ½mv² ⇒ v = √(2KE/m).
2. \( \lambda = \frac{h}{\sqrt{2m\,KE}} \) ⇒ KE ∝ 1/λ².
3. \( \frac{(KE)_1}{(KE)_2} = \frac{\lambda_2^2}{\lambda_1^2} = \left( \frac{5}{3} \right)^2 = \frac{25}{9} \)
4. Hence the ratio of the kinetic energies is 25 : 9.

Teacher's Note:
1. KE varies inversely as the square of the wavelength, so invert the ratio and then square it.
2. 9 : 25 is the trap answer from squaring without inverting.

Answer: (d) The position and velocity of the electrons in the orbit cannot be determined simultaneously

Explanation:
1. Bohr's model assumes electrons revolve in definite circular orbits with quantised energy; the orbit nearest the nucleus has the lowest energy.
2. The statement that the position and velocity of the electrons cannot be determined simultaneously is not part of Bohr's model.
3. It is in fact given by Heisenberg in his uncertainty principle, which actually contradicts Bohr's idea of definite orbits.

Teacher's Note:
1. Bohr's model gives the electron a fixed path, i.e. exact position and velocity, which the uncertainty principle later ruled out.
2. In 'not correct' questions, look for the statement that belongs to a later theory.

Answer: (d) ΔE·Δx ≥ h/(4π)

Explanation:
1. From the Heisenberg uncertainty principle, Δx·Δp ≥ h/4π … (i)
2. Since p = mv, Δx·mΔv ≥ h/4π, i.e. Δx·Δv ≥ h/4πm … (ii)
3. The principle also applies to the pairs energy–time (ΔE·Δt) and angular momentum–angle, along with position–momentum: ΔE·Δt ≥ h/4π … (iii)
4. ΔE·Δx is not a conjugate pair (its product does not have the units of h), so ΔE·Δx ≥ h/4π is not a valid expression.

Teacher's Note:
1. Valid pairs are conjugate variables whose product has units of J s: position–momentum and energy–time.
2. Energy × length has units of J m, which cannot be compared with h.

Answer: (c) 1.92 × 10⁻³ m

Explanation:
1. By Heisenberg's uncertainty principle, Δx·mΔv = h/4π.
2. Δv = 0.005% of 600 m/s = (600 × 0.005)/100 = 0.03 m s⁻¹.
3. Δx = h/(4πmΔv) = 6.6 × 10⁻³⁴ ÷ (4 × 3.14 × 9.1 × 10⁻³¹ × 0.03)
4. Δx = 1.92 × 10⁻³ m.

Teacher's Note:
1. Convert the percentage accuracy into an actual uncertainty in velocity before substituting.
2. Use the electron's real mass, 9.1 × 10⁻³¹ kg.

Answer: (c) 1.93 × 10⁻² m

Explanation:
1. Δx·Δv ≥ h/4πm
2. Δv = 0.001% of 300 m s⁻¹ = 300 × 0.001 × 10⁻² = 3 × 10⁻³ m s⁻¹.
3. Δx = 6.63 × 10⁻³⁴ ÷ (4 × 3.14 × 9.1 × 10⁻³¹ × 300 × 0.001 × 10⁻²)
4. Δx = 0.01933 m = 1.93 × 10⁻² m.

Teacher's Note:
1. 0.001% means multiply by 0.001/100 = 10⁻⁵.
2. For an electron, h/4πm ≈ 5.8 × 10⁻⁵ m² s⁻¹; dividing this by Δv gives Δx quickly.

Answer: (a) 2

Explanation:
1. According to Heisenberg's uncertainty principle, Δx × m × Δv = h/4π.
2. For A: Δx_A × m × 0.05 = h/4π … (i)
3. For B: Δx_B × 5m × 0.02 = h/4π … (ii)
4. Dividing (i) by (ii): \( \frac{\Delta x_A \times m \times 0.05}{\Delta x_B \times 5m \times 0.02} = 1 \)
5. \( \frac{\Delta x_A}{\Delta x_B} = \frac{5m \times 0.02}{m \times 0.05} = 2 \)

Teacher's Note:
1. Δx is inversely proportional to the product mΔv, so compare the products: 0.05m for A and 0.10m for B.
2. The particle with the smaller mΔv has the larger uncertainty in position.

Answer: (c) Both (a) and (b)

Explanation:
1. In the Schrödinger equation Ĥψ = Eψ, Ĥ is a mathematical operator called the Hamiltonian.
2. It was introduced by Schrödinger from the expression for the total energy of the system.
3. Hence both statements (a) and (b) are correct.

Teacher's Note:
1. The Hamiltonian represents the total energy (kinetic + potential) of the system.
2. Solving Ĥψ = Eψ for the hydrogen atom gives the allowed energies and the orbitals (ψ).

Answer: (d) 2s-orbital

Explanation:
1. The graphs between |ψ|² and r are radial density plots having (n − l − 1) radial nodes.
2. 1s: 1 − 0 − 1 = 0 nodes; 2s: 2 − 0 − 1 = 1 node; 3s: 3 − 0 − 1 = 2 nodes; 2p: 2 − 1 − 1 = 0 nodes.
3. The given graph starts at a maximum at the nucleus (an s-orbital) and touches zero once, i.e. it has one radial node.
4. Thus the graph represents the 2s-orbital.

Teacher's Note:
1. Count the points (other than r = 0) where the curve touches zero to get the number of radial nodes.
2. For s orbitals |ψ|² is maximum at the nucleus; for p orbitals it is zero at the nucleus.

Answer: (b) size, shape and orientation

Explanation:
1. The principal quantum number (n) is related to the size of the shell.
2. The azimuthal quantum number (l) is related to the shape of the subshell.
3. The magnetic quantum number (m_l) is related to the orientation of the orbitals.

Teacher's Note:
1. Memory aid: n–l–m ↔ size–shape–orientation.
2. The spin quantum number gives the spin orientation of the electron, not of the orbital.

Answer: (a) \( \sqrt{6}\,\hbar \)

Explanation:
1. Orbital angular momentum = \( \sqrt{l(l+1)}\,\frac{h}{2\pi} = \sqrt{l(l+1)}\,\hbar \)
2. For a d orbital, l = 2.
3. Angular momentum = \( \sqrt{2(2+1)}\,\hbar = \sqrt{6}\,\hbar \)

Teacher's Note:
1. Use l, not n: s = 0, p = 1, d = 2, f = 3.
2. √2 ħ is for a p orbital and 2√3 ħ for an f orbital; an s electron has zero orbital angular momentum.

Answer: (b) 3

Explanation:
1. l can take values 0 to (n − 1), so n must be at least l + 1.
2. Here l = 2, thus n = 2 + 1 = 3.
3. Hence the minimum value of the principal quantum number is 3 (the 3d subshell is the first d subshell).

Teacher's Note:
1. Since l ≤ n − 1, the first d subshell appears only in the third shell.
2. The p subshell starts at n = 2 and the f subshell at n = 4 for the same reason.

Answer: (a) spherical

Explanation:
1. For the s subshell, l = 0.
2. The s orbital is spherical in shape.

Teacher's Note:
1. l = 0 means zero orbital angular momentum and no angular nodes, so the orbital looks the same in all directions.
2. Dumb-bell is the p orbital (l = 1) and double dumb-bell (cloverleaf) is typical of d orbitals (l = 2).

Answer: (d) l = 1; m = −1, 0, +1

Explanation:
1. For n = 2, l can be 0 and 1.
2. For l = 0, m = 0; for l = 1, m = −1, 0, +1.
3. Hence the correct set is l = 1; m = −1, 0, +1.

Teacher's Note:
1. m_l runs from −l to +l, so it can never be numerically larger than l.
2. l = 2 needs n ≥ 3, so option (a) is ruled out immediately.

Answer: (a) l

Explanation:
1. Orbital angular momentum = \( \frac{h}{2\pi} \sqrt{l(l+1)} \).
2. Hence it depends only on l. l can have values ranging from 0 to (n − 1).

Teacher's Note:
1. The formula contains only l, so all orbitals of the same subshell type have the same orbital angular momentum.
2. Contrast this with Bohr's model, where angular momentum depends on n (mvr = nh/2π).

Answer: (c) spin quantum number

Explanation:
1. Two electrons occupying the same orbital have the same n, l and m_l.
2. Their spins are opposite, so the spin quantum number (+½ and −½) distinguishes them.

Teacher's Note:
1. This is a direct consequence of Pauli's exclusion principle.
2. An orbital is defined by n, l and m_l; an electron needs the fourth number, m_s.

Answer: (a) A-2, B-4, C-1, D-3

Explanation:
1. The correct match is A → 2, B → 4, C → 1, D → 3.
2. (A) The principal quantum number determines the size and, to a large extent, the energy of the orbital.
3. (B) The azimuthal quantum number determines the angular momentum of the electron and defines the three-dimensional shape of the orbital.
4. (C) The magnetic quantum number gives information about the spatial orientation of the orbital with respect to a standard set of coordinate axes.
5. (D) The spin quantum number arises from the spectral evidence that an electron, in its motion around the nucleus in an orbit, also rotates or spins about its own axis.

Teacher's Note:
1. Fix the two easiest pairs first (n → size/energy, s → spin); only one code will then fit.
2. Orientation belongs to m_l, and shape belongs to l.

Answer: (a) 25

Explanation:
1. For n = 5, l = 0, 1, 2, 3, 4 and each l has (2l + 1) values of m_l.
2. Total number of orbitals = n² = 5² = 25.
3. Each orbital can hold one electron with m_s = +½ (and one with −½).
4. Hence 25 orbitals can be associated with m_s = +½.

Teacher's Note:
1. m_s describes the electron, not the orbital, so specifying it does not reduce the number of orbitals.
2. If the question asked for electrons with m_s = +½ in n = 5, the answer would also be 25 (half of 2n² = 50).

Answer: (d) 1

Explanation:
1. For a hydrogen-like wave function there are (n − l − 1) radial nodes and (n − 1) total nodes.
2. For 3p: n = 3, l = 1.
3. Number of radial nodes = 3 − 1 − 1 = 1.

Teacher's Note:
1. Total nodes (n − 1) = radial (n − l − 1) + angular (l); for 3p: 2 = 1 + 1.
2. Do not confuse radial nodes with the angular (nodal plane) count, which equals l.

Answer: (c) 9

Explanation:
1. Total number of orbitals associated with the n^th shell = n².
2. For the third shell, n = 3: 3² = 9 orbitals (one 3s, three 3p and five 3d).

Teacher's Note:
1. Orbitals per shell = n², electrons per shell = 2n².
2. 3 is the number of subshells in the third shell, a common distractor.

Answer: (d) A-5, B-4, C-1, D-2, E-3

Explanation:
1. The correct match is A → 5, B → 4, C → 1, D → 2, E → 3.
2. A: lobes lie between the y- and z-axes, so it is d_yz.
3. B: lobes lie along the x- and y-axes, so it is d_x²−y².
4. C: two lobes along the z-axis with a ring (doughnut) in the xy-plane, so it is d_z².
5. D: lobes lie between the x- and y-axes, so it is d_xy.
6. E: lobes lie between the x- and z-axes, so it is d_xz.

Teacher's Note:
1. d_xy, d_yz and d_xz have lobes between the named axes; d_x²−y² has lobes on the x- and y-axes.
2. d_z² is the easiest to spot because of its ring, which fixes C-1 straight away.

Answer: (b) 5p < 6s < 4f < 5d

Explanation:
1. According to the Aufbau principle, the order in which the energies of the orbitals increase is:
2. 1s, 2s, 2p, 3s, 3p, 4s, 3d, 4p, 5s, 4d, 5p, 6s, 4f, 5d, 6p, 7s, 5f, 6d, 7p, 8s
3. Using the (n + l) rule: 5p = 6, 6s = 6, 4f = 7, 5d = 7; for equal (n + l), lower n is lower in energy.
4. Hence the correct order of increasing energy is 5p < 6s < 4f < 5d.

Teacher's Note:
1. Apply the (n + l) rule first; break ties by choosing the smaller n.
2. 4f comes before 5d even though both have n + l = 7, because 4 < 5.

Answer: (d) Cu and Cr

Explanation:
1. The Aufbau principle does not give the correct arrangement of filling of atomic orbitals in copper and chromium.
2. Half-filled and completely filled configurations of Cr and Cu have lower energy and are therefore more stable.
3. Cr (Z = 24) = 1s² 2s² 2p⁶ 3s² 3p⁶ 3d⁵ 4s¹
4. Cu (Z = 29) = 1s² 2s² 2p⁶ 3s² 3p⁶ 3d¹⁰ 4s¹

Teacher's Note:
1. The expected configurations 3d⁴4s² (Cr) and 3d⁹4s² (Cu) are wrong; one 4s electron shifts to 3d.
2. Mn (3d⁵4s²) and Zn (3d¹⁰4s²) follow the Aufbau order normally.

Answer: (c) Pauli exclusion principle

Explanation:
1. The box diagram shows two electrons in the 1s orbital with the same (parallel) spin.
2. According to the Pauli exclusion principle, an orbital can hold a maximum of two electrons and they must have opposite spins.
3. Hence the representation violates the Pauli exclusion principle.

Teacher's Note:
1. Two electrons in one orbital with parallel spins would have all four quantum numbers identical.
2. Hund's rule is about distributing electrons among degenerate orbitals; with only one 1s orbital it does not apply here.

Answer: (c) Hund's rule

Explanation:
1. According to Hund's rule, electron pairing in p, d and f orbitals cannot occur until each orbital of a given subshell is singly occupied.
2. Hence 1s² 2s² 2p_x² 2p_y¹ 2p_z⁰ is not possible for nitrogen.

Teacher's Note:
1. Hund's rule decides how electrons are spread within a subshell; Aufbau decides which subshell fills first.
2. Nitrogen's three unpaired 2p electrons with parallel spins make its half-filled 2p³ configuration extra stable.

Answer: (a) The electronic configuration of N-atom is

Explanation:
1. According to Hund's rule, pairing of electrons in the orbitals of a particular subshell does not take place until all the orbitals of the subshell are singly occupied.
2. Moreover, the singly occupied orbitals must have electrons with parallel spins, i.e. 2p_x¹ 2p_y¹ 2p_z¹ as ↑ ↑ ↑ (or ↓ ↓ ↓).
3. In option (a) the third 2p electron has opposite spin, so the configuration shown is wrong.
4. Statements (b), (c) and (d) are correct: an orbital needs n, l, m_l while an electron needs n, l, m_l, m_s; for s, l = 0 so √(l(l+1)) = 0; and d_z² has m = 0.

Teacher's Note:
1. In a correct ground-state box diagram the singly filled orbitals of a subshell all point the same way.
2. Read every arrow carefully in box-diagram questions; one flipped arrow is the usual trick.

Answer: (d)

Explanation:
1. According to Hund's rule, the three p-orbitals are first singly occupied with parallel spins; the fourth electron then pairs up in one of them.
2. So four p-electrons are arranged as one paired orbital and two singly occupied orbitals with parallel spins: ↑ ↑ ⇅.
3. (a) pairs electrons while an orbital is still empty (violates Hund's rule); (b) has the two unpaired electrons with opposite spins (violates Hund's rule of maximum multiplicity); (c) puts two electrons of the same spin in one orbital (violates Pauli's principle).

Teacher's Note:
1. p⁴ always has exactly two unpaired electrons, and they must have parallel spins.
2. The order of the boxes does not matter because the three p orbitals are degenerate.

Answer: (a) A-3, B-5, C-1, D-4

Explanation:
1. The correct match is A → 3, B → 5, C → 1, D → 4.
2. (A) Hund's rule: pairing of electrons in the orbitals of the same subshell (p, d or f) does not take place until each orbital has got one electron, i.e. is singly occupied.
3. (B) Aufbau principle: in the ground state of atoms, the orbitals are filled in order of their increasing energies.
4. (C) Pauli exclusion principle: no two electrons in an atom can have the same set of four quantum numbers.
5. (D) Heisenberg's uncertainty principle: it is impossible to determine the exact position and exact momentum of a subatomic particle simultaneously.

Teacher's Note:
1. Statement 2 (extra stability of half-filled/full orbitals) is the extra item in Column II and matches none of the rules.
2. Aufbau literally means 'building up', i.e. filling in order of increasing energy.

Answer: (d) The energy of 2s-orbital is less than the energy of 2p-orbital in case of hydrogen-like atoms

Explanation:
1. (a) According to de Broglie's equation, λ = h/mv. Statement (a) is correct.
2. (b) From Δx·Δp ≥ h/4π, writing Δp = mΔv = m·aΔt and F = ma gives Δx·F·Δt ≥ h/4π; since F·Δx is energy, ΔE·Δt ≥ h/4π. Statement (b) is correct.
3. (c) Half-filled and fully filled orbitals have greater stability due to greater exchange energy, greater symmetry and a more balanced arrangement. Statement (c) is correct.
4. (d) For a single-electron species like H, the energy depends only on n and not on l. Hence the energies of 2s and 2p orbitals are equal in hydrogen-like species. Statement (d) is wrong.

Teacher's Note:
1. In hydrogen and hydrogen-like ions, all orbitals of the same shell are degenerate (2s = 2p, 3s = 3p = 3d).
2. The order 2s < 2p applies only to multi-electron atoms, because of shielding.

Answer: (a) I

Explanation:
1. The most stable arrangement is I: 4s ⇅ and 3d ↑ ↑ ↑ ↑ ↑.
2. (i) According to the Aufbau principle, electrons enter the subshells of an atom in increasing order of energy, so 4s is filled before 3d.
3. (ii) According to Hund's rule of maximum multiplicity, pairing of electrons in the orbitals of a subshell does not take place until all orbitals of the subshell are singly occupied.
4. Arrangement I satisfies both rules and has a half-filled 3d subshell with five parallel spins (maximum exchange energy).

Teacher's Note:
1. II and III pair a 3d electron while leaving 4s half-filled, which gives up both a filled 4s and the half-filled 3d stability.
2. IV pairs electrons in 3d while two 3d orbitals are still empty, a clear violation of Hund's rule.

Answer: (c) 5, 0, 0, +½

Explanation:
1. ₃₇Rb = [Kr] 5s¹. Its valence electronic configuration is 5s¹.
2. So the quantum numbers of the 5s¹ electron are n = 5, l = 0 (s-orbital), m = 0 (as m = −l to +l) and s = +½.

Teacher's Note:
1. Rb is in period 5, group 1, so its valence electron is in 5s.
2. For any s electron, l = 0 forces m = 0, which rules out (a) and (d).

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

Explanation:
1. Electronic configuration of Fe²⁺ is [Ar] 3d⁶ 4s⁰; number of d-electrons = 6.
2. Mg: 1s² 2s² 2p⁶ 3s² (s-electrons = 6) — matches.
3. Cl: 1s² 2s² 2p⁶ 3s² 3p⁵ (p-electrons = 11) — does not match.
4. Fe: [Ar] 3d⁶ 4s² (d-electrons = 6) — matches.
5. Ne: 1s² 2s² 2p⁶ (p-electrons = 6) — matches.
6. Hence Cl, with 11 p-electrons, does not match the 6 d-electrons of Fe²⁺.

Teacher's Note:
1. Transition metals lose their 4s electrons first, so Fe²⁺ keeps all six 3d electrons.
2. Count electrons of a type across all shells: Cl has 2p⁶ + 3p⁵ = 11 p-electrons.

Answer: (a) Fe³⁺

Explanation:
1. Fe³⁺ = [Ar] 3d⁵: 5 unpaired electrons.
2. Co³⁺ = [Ar] 3d⁶: 4 unpaired electrons.
3. Ni²⁺ = [Ar] 3d⁸: 2 unpaired electrons.
4. Cu²⁺ = [Ar] 3d⁹: 1 unpaired electron.
5. Hence the maximum number of unpaired electrons (5) is present in Fe³⁺.

Teacher's Note:
1. A d⁵ configuration gives the maximum possible five unpaired d electrons.
2. For dⁿ with n > 5, unpaired electrons = 10 − n.

Answer: (a) A-4, B-3, C-1, D-2

Explanation:
1. The correct match is A → 4, B → 3, C → 1, D → 2.
2. (A) Cr (Z = 24) = 1s² 2s² 2p⁶ 3s² 3p⁶ 3d⁵ 4s¹ = [Ar] 3d⁵ 4s¹
3. (B) Fe²⁺ (Z = 26) = [Ar] 3d⁶ 4s⁰
4. (C) Ni²⁺ (Z = 28) = [Ar] 3d⁸ 4s⁰
5. (D) Cu (Z = 29) = [Ar] 3d¹⁰ 4s¹

Teacher's Note:
1. [Ar] 3d⁶ 4s² (item 5) is neutral Fe, the extra item in Column II.
2. Cr and Cu are the two exceptions in the first transition series with a single 4s electron.

Answer: (b) A-3, B-4, C-1, D-5

Explanation:
1. The correct match is A → 3, B → 4, C → 1, D → 5.
2. (A) Cu (Z = 29): 1s² 2s² 2p⁶ 3s² 3p⁶ 3d¹⁰ 4s¹
3. (B) Cu²⁺ (Z = 29): 1s² 2s² 2p⁶ 3s² 3p⁶ 3d⁹
4. (C) Zn²⁺ (Z = 30): 1s² 2s² 2p⁶ 3s² 3p⁶ 3d¹⁰
5. (D) Cr³⁺ (Z = 24): 1s² 2s² 2p⁶ 3s² 3p⁶ 3d³

Teacher's Note:
1. For cations of transition metals, remove the 4s electrons before the 3d electrons.
2. Cu²⁺ loses its 4s¹ and one 3d electron, leaving 3d⁹.