(a) (i) Find the wavelength of the emitted light when 1.0×10⁻²⁷ g particle in a one-dimensional box of length 3 Å goes from nₓ = 2 to nₓ = 1 level.

(ii) Explain the Heisenberg uncertainty principle. 10

(b) Draw Lewis dot structure of [Br₃]⁻ and H₂N

Question

(a) (i) Find the wavelength of the emitted light when 1.0×10⁻²⁷ g particle in a one-dimensional box of length 3 Å goes from nₓ = 2 to nₓ = 1 level.

(ii) Explain the Heisenberg uncertainty principle. 10

(b) Draw Lewis dot structure of [Br₃]⁻ and H₂NCSNH₂ (central C atom is bonded to both the N atoms and to the S atom). Does thiourea contain polar bonds? If yes, which is the most polar bond? 10

(c) (i) Derive an expression that relates the wavelength of the X-rays with the distance between the layers of atoms in a crystal.

(ii) The X-rays of wavelength 220 pm are diffracted from an ionic crystal at an angle of 23°. What is the distance between the layers that are responsible for this diffraction? 10

(d) (i) Find the temperature at which the water molecules can have the root-mean-square speed of 719 m s⁻¹. 5

(ii) What is the root-mean-square velocity of water molecules at 473 K? 5

(e) (i) Calculate the E° value for the half-reaction Cr³⁺(aq)+3e⁻ → Cr

Given that at 25 °C
Cr³⁺(aq)+e⁻ → Cr²⁺(aq)    E° = –0.424 V
Cr²⁺(aq)+2e⁻ → Cr    E° = –0.90 V

(ii) The surface tension of liquid A is seven times higher than that of liquid B.

(1) Which liquid is expected to have higher contact angle with glass?

(2) 10 mL of each of these liquids are placed in separate 100 mL glass beakers. How do these liquids respond if the gravitational field is switched off?

UPSC Answer Check · Accepted Answer

(a) Particle in a box; Heisenberg uncertainty principle

(i) For a particle in a one-dimensional box, Eₙ = n²h²/(8mL²).

Given m = 1.0 × 10⁻²⁷ g = 1.0 × 10⁻³⁰ kg, L = 3 Å = 3.0 × 10⁻¹⁰ m, transition n = 2 → 1.

ΔE = E₂ − E₁ = 3h²/(8mL²), and hc/λ = ΔE, so λ = 8mcL²/(3h).

λ = (8 × 1.0 × 10⁻³⁰ kg × 3.0 × 10⁸ m s⁻¹ × 9.0 × 10⁻²⁰ m²) / (3 × 6.626 × 10⁻³⁴ J s) = 2.16 × 10⁻⁴⁰ / 1.988 × 10⁻³³ m = 1.09 × 10⁻⁷ m.

λ ≈ 109 nm (ultraviolet region).

(ii) The Heisenberg uncertainty principle states that the position and momentum of a particle cannot both be known exactly at the same time: Δx · Δp ≥ h/4π. Similarly ΔE · Δt ≥ h/4π. The limit is intrinsic to the wave nature of matter and not an error of instrument. Confining a particle in a small Δx forces a large Δp, so a bound particle cannot be at rest; this gives the zero-point energy (as in the box, where the lowest level is n = 1) and rules out the classical idea of an electron orbit with fixed path.

(b) Lewis structures of [Br₃]⁻ and thiourea; polarity

[Br₃]⁻: total valence electrons = 3 × 7 + 1 = 22. The central Br forms a single bond to each terminal Br and carries three lone pairs; each terminal Br carries three lone pairs. The formal charge of −1 sits on the central Br, which has an expanded octet (10 electrons; linear shape).

Thiourea (H₂N–C(=S)–NH₂): total valence electrons = 2(5) + 4 + 6 + 4(1) = 24. The carbon has a double bond to S and single bonds to each N; each N has two N–H bonds and one lone pair; S has two lone pairs.

Polarity: yes, the bonds N–H, C–N and C=S are polar. Electronegativities (Pauling): H 2.20, C 2.55, N 3.04, S 2.58. Differences: N–H 0.84, C–N 0.49, C=S 0.03. The N–H bond is the most polar.

(c) Bragg's law and its use

(i) Derivation: Consider a beam of wavelength λ incident at glancing angle θ on successive parallel planes of atoms separated by d. Rays scattered from two adjacent planes differ in path by AB + BC, where AB = BC = d sin θ. So path difference = 2d sin θ. Constructive interference occurs when this equals a whole number of wavelengths:

nλ = 2d sin θ, n = 1, 2, 3, …

(ii) Given λ = 220 pm, θ = 23°, taking first order (n = 1):

d = λ/(2 sin θ) = 220 pm / (2 × 0.3907) = 220 / 0.7814 pm = 281.5 pm (2.82 Å).

(d)(i) Temperature at which v_rms = 719 m s⁻¹

v_rms = √(3RT/M), so T = Mv²/(3R).

M = 18.0 g mol⁻¹ = 0.0180 kg mol⁻¹; v² = (719)² = 5.17 × 10⁵ m² s⁻².

T = (0.0180 × 5.17 × 10⁵) / (3 × 8.314) = 9.31 × 10³ / 24.94 = 373 K (100 °C).

(d)(ii) v_rms of water molecules at 473 K

v_rms = √(3 × 8.314 × 473 / 0.0180) = √(1.180 × 10⁴ / 0.0180) = √(6.55 × 10⁵) = 8.09 × 10² m s⁻¹, i.e. about 809 m s⁻¹.

(e)(i) E° for Cr³⁺ + 3e⁻ → Cr

The standard data needed are Cr³⁺ + e⁻ → Cr²⁺, E° = −0.41 V, and Cr²⁺ + 2e⁻ → Cr, E° = −0.91 V. Potentials are not additive, but ΔG° is:

ΔG°₁ = −(1)F(−0.41) = +0.41F; ΔG°₂ = −(2)F(−0.91) = +1.82F.

ΔG° = 2.23F = −3FE°, so E° = −2.23/3 = −0.74 V.

(e)(ii) Surface tension and contact angle

(1) Liquid A has the stronger cohesion (higher surface tension). Adhesion to glass being comparable, cohesion dominates over adhesion in A and it wets glass less, so A has the higher contact angle.

(2) With gravity off, only interfacial forces act. Liquid B, which wets glass (low contact angle), spreads along the inner wall and corners of the beaker as a film, and the liquid moves up the walls. Liquid A, which wets poorly, gathers into a compact, nearly spherical blob that touches the glass at only a small area.

Model answer

(a) Particle in a box; Heisenberg uncertainty principle

(b) Lewis structures of [Br₃]⁻ and thiourea; polarity

(c) Bragg's law and its use

(d)(i) Temperature at which v_rms = 719 m s⁻¹

(e)(i) E° for Cr³⁺ + 3e⁻ → Cr

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