JEE Chemistry Structure Of Atom MCQs Set 02

Answer: (a) 492 nm

Explanation:
1. First line: n = 4 → 3, \( \dfrac{1}{\lambda_1}=R\left(\dfrac19-\dfrac1{16}\right)=\dfrac{7R}{144} \).
2. Second line: n = 5 → 3, \( \dfrac{1}{\lambda_2}=R\left(\dfrac19-\dfrac1{25}\right)=\dfrac{16R}{225} \).
3. \( \lambda_2=720\times\dfrac{7/144}{16/225}\approx 492 \) nm.

Teacher's Note:
1. Paschen lines end on n = 3; the first line starts from n = 4 and the second from n = 5.
2. Take the ratio of the two Rydberg expressions to avoid needing R.

Answer: (c) \( \dfrac{27}{16} x \)

Explanation:
1. Bohr radius \( r \propto n^2/Z \).
2. Li²⁺ (Z = 3, n = 2): \( r=\dfrac{4}{3}\,a_0 = x \). Be³⁺ (Z = 4, n = 3): \( r'=\dfrac{9}{4}\,a_0 \).
3. \( r'/x = \dfrac{9/4}{4/3} = \dfrac{27}{16} \), so \( r'=\dfrac{27}{16}x \).

Teacher's Note:
1. Radius depends on n² and inversely on the nuclear charge Z.
2. Express both radii in terms of a₀ and take the ratio.

Answer: (d) 5

Explanation:
1. The three p orbitals (px, py, pz) have their lobes along the axes.
2. Among d orbitals, only dz² and dx²−y² lie along the axes; dxy, dyz and dxz lie between the axes.
3. Total = 3 + 2 = 5.

Teacher's Note:
1. Orbitals whose names contain two axis letters (xy, yz, xz) point between the axes.
2. The t2g set (dxy, dyz, dxz) lies between the axes; the eg set (dz², dx²−y²) lies along the axes.

Answer: (b) \( \dfrac{9\lambda}{5} \)

Explanation:
1. Shortest Lyman line (n = ∞ → 1): \( \dfrac1\lambda=R \).
2. Longest Balmer line of He⁺ (Z = 2, n = 3 → 2): \( \dfrac1{\lambda'}=4R\left(\dfrac14-\dfrac19\right)=\dfrac{5R}{9} \).
3. \( \lambda'=\dfrac{9}{5R}=\dfrac{9\lambda}{5} \).

Teacher's Note:
1. For hydrogen-like ions multiply the Rydberg expression by Z².
2. Shortest wavelength in a series corresponds to n₂ = ∞; longest to n₂ = n₁ + 1.

Answer: (c) 270 pm

Explanation:
1. \( r_n \propto n^2/Z \), so \( r_3(\mathrm{He^+})=0.6\times\dfrac{9}{2}=2.7 \) Å.
2. 2.7 Å = 270 pm.

Teacher's Note:
1. 1 Å = 100 pm.
2. Scale the hydrogen radius by n²/Z.

Answer: (b) 798

Explanation:
1. Energy of one photon = hν = 6.626 × 10⁻³⁴ × 2 × 10¹² = 1.325 × 10⁻²² J.
2. For one mole: 1.325 × 10⁻²² × 6.022 × 10²³ ≈ 798 J mol⁻¹.

Teacher's Note:
1. Multiply the energy of one photon by Avogadro's number to get the molar energy.
2. hν is the energy of a single photon.

Answer: (b) 32

Explanation:
1. The capacity of a shell is 2n².
2. For n = 4: 2 × 16 = 32 electrons.

Teacher's Note:
1. Shell capacity = 2n²: 2, 8, 18, 32 for n = 1 to 4.
2. Subshells 4s, 4p, 4d, 4f hold 2 + 6 + 10 + 14 = 32.

Answer: (d) r₀ = 2a₀

Explanation:
1. A radial node occurs where ψ = 0.
2. The exponential never vanishes, so \( 2-\dfrac{r_0}{a_0}=0 \).
3. Hence r₀ = 2a₀.

Teacher's Note:
1. The number of radial nodes is n − l − 1; 2s has one.
2. Set the polynomial factor of the wave function to zero.

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

Explanation:
1. He⁺, 4 → 2: \( \dfrac1\lambda=4R\left(\dfrac14-\dfrac1{16}\right)=\dfrac{3R}{4} \).
2. Hydrogen 2 → 1: \( R\left(1-\dfrac14\right)=\dfrac{3R}{4} \), which matches.
3. So the hydrogen transition is n = 2 to n = 1.

Teacher's Note:
1. Z² scaling of He⁺ lines makes them coincide with some hydrogen lines.
2. Compare the values of 1/λ directly.

Answer: (b) D > B > C > A

Explanation:
1. The orbitals are A = 3s, B = 4s, C = 3p, D = 3d.
2. (n + l) values: 3s = 3, 4s = 4, 3p = 4, 3d = 5.
3. Higher (n + l) means higher energy; for equal (n + l) (4s and 3p) the larger n is higher. Order: 3d > 4s > 3p > 3s, i.e. D > B > C > A.

Teacher's Note:
1. Use the (n + l) rule for multi-electron atoms.
2. When (n + l) is equal, the orbital with the higher n has higher energy.

Answer: (a) 2

Explanation:
1. (A) λ = h/mv = 6 × 10⁻³⁴/(9 × 10⁻³¹ × 1000) = 6.667 × 10⁻⁷ m = 666.67 nm. True.
2. (B) and (D) are false: cathode rays are the same whatever the electrode material or gas.
3. (C) True: the rays travel from cathode to anode. Hence 2 statements are true.

Teacher's Note:
1. Cathode ray particles (electrons) are identical for every gas and every electrode material.
2. de Broglie wavelength λ = h/(mv).