Class 11 Chemistry Chapter 2: Structure of Atom worksheet with answer key

Answer: (d) Li, Na and K only

An electron is ejected only when hν exceeds the work function W₀. 2.59 eV is more than 2.42, 2.3 and 2.25 eV but less than 3.7, 4.8 and 4.3 eV, so Li, Na and K emit and the other three do not.

Answer: (a) 199F⁻

Na⁺ has 11 − 1 = 10 electrons. Neutrons = A − Z: F⁻ 19 − 9 = 10; O 8; Mg²⁺ 12 (the charge does not change the neutrons); N 7.

Answer: (d) 4018Ar and 4020Ca

Isobars have the same mass number and different atomic numbers: Ar and Ca both have A = 40. The chlorine and carbon pairs are isotopes; ¹⁴C and ¹⁶O have the same number of neutrons (8) but different A.

Answer: (b) the charge to mass ratio of the canal rays

Cathode rays (electrons) do not depend on the gas or the electrodes. Canal rays are positive ions of the gas itself, so their mass, and hence their charge to mass ratio, depends on the gas.

Answer: (c) 5.0 µm

λ = 1/ν̄ = 1/2000 cm = 5.0 × 10⁻⁴ cm = 5.0 × 10⁻⁶ m = 5.0 µm. 50 µm and 0.50 µm slip the cm to µm conversion by ten; 20 µm inverts the wrong quantity.

Hund's rule: in orbitals of the same subshell, electrons do not pair until each orbital holds one electron. N, 2p³: [↑][↑][↑], three unpaired. O, 2p⁴: [↑↓][↑][↑], the fourth electron pairs only after all three are singly filled, two unpaired.

• statement of Hund's rule (1)
• N 2p³ ↑ ↑ ↑ (3 unpaired) and O 2p⁴ ↑↓ ↑ ↑ (2 unpaired) (1)

(i) The Bohr model treats the electron only as a charged particle and ignores its wave character (dual behaviour of matter). (ii) It places the electron in a definite circular orbit, which would need its exact position and velocity at the same time; the uncertainty principle forbids this.

• ignores the wave nature (dual behaviour) of the electron (1)
• definite orbit contradicts the uncertainty principle (1)

The intensities of the wavelengths a hot body emits depend on its temperature: at each temperature intensity rises with wavelength to a maximum and then falls. As temperature rises the maximum shifts to shorter wavelength, so a larger share of the visible light is of short wavelength; the glow moves from red (long λ) towards white and then blue (short λ).

• emission spectrum depends only on temperature, with an intensity maximum (1)
• maximum shifts to shorter wavelength as T rises, hence red → white → blue (1)

(i) Symmetry: a half-filled 3d⁵ subshell has a symmetrical distribution of electrons; electrons in the same subshell shield one another little and are held more strongly, so it is more stable. (ii) Exchange energy: electrons of parallel spin in degenerate orbitals can exchange positions, releasing exchange energy; the number of exchanges is largest for a half-filled set (10 for d⁵, against 6 for d⁴), so the 4s to 3d shift lowers the total energy. Unpaired electrons: five in 3d and one in 4s, six in all.

• symmetrical distribution of a half-filled subshell (1)
• maximum exchange energy for d⁵ (1)
• 6 unpaired electrons (1)

Expected: with positive charge and mass spread evenly through each atom, the α-particles should pass through, slowing and turning only by small angles. Observed and concluded: (i) most passed undeflected, so most of the atom is empty space; (ii) a small fraction was deflected by small angles and (iii) a very few (about 1 in 20,000) bounced back by nearly 180°, so the positive charge and most of the mass are concentrated in a very small region, the nucleus, whose volume is negligible compared with the atom's.

• expected small deflections from uniform charge and mass (Thomson) (1)
• most undeflected: atom mostly empty (1)
• few deflected/rebounded: positive charge concentrated in a tiny nucleus (1)

(i) E (one photon) = hc/λ = (6.626 × 10⁻³⁴ × 3.0 × 10⁸)/(400 × 10⁻⁹) = 4.97 × 10⁻¹⁹ J; one mole: 4.97 × 10⁻¹⁹ × 6.022 × 10²³ = 2.99 × 10⁵ J = 299 kJ mol⁻¹. (ii) 299 kJ is more than 242 kJ, so each photon can break one bond; 299 − 242 = 57 kJ per mole is left over (as kinetic energy of the atoms). (iii) E ∝ 1/λ: one mole of 589 nm photons carries 299 × 400/589 = 203 kJ, less than 242 kJ; each photon is absorbed singly, so the energy of two photons cannot be pooled to break one bond: no.

• 299 kJ mol⁻¹ for 400 nm photons (1)
• yes; 57 kJ mol⁻¹ left over (1)
• 589 nm: 203 kJ mol⁻¹ < 242, cannot break the bond (1)