(a) Find the probability of existence of a particle in a one-dimensional box of length 'a' in the region 0 ≤ x ≤ a/4 for the states n = 1, 2 and 3. (15 marks)

(b) The standard reduction potential of oxygen under acidic conditions at 298 K is +1.23 V

Question

(a) Find the probability of existence of a particle in a one-dimensional box of length 'a' in the region 0 ≤ x ≤ a/4 for the states n = 1, 2 and 3. (15 marks)

(b) The standard reduction potential of oxygen under acidic conditions at 298 K is +1.23 V. What is the standard reduction potential for the four-electron reduction of O₂(g) under basic conditions? (15 marks)

(c) The radii of Zn²⁺ and S²⁻ ions are 0·74 Å and 1·84 Å respectively. Determine the most stable form of arrangement of ions in ZnS crystal lattice. Draw the CCP (Cubic Close Packing) structure of ZnS. (15 marks)

(d) In a sample of NaCl, one of every 10,000 sites, normally occupied by Na⁺, is occupied instead by Ca²⁺. Assuming that all of the Cl⁻ sites are fully occupied, what is the stoichiometry of the sample? (5 marks)

UPSC Answer Check · Accepted Answer

(a) 
For a particle in a one-dimensional box of length a, the normalized wavefunction is
ψₙ(x) = √(2/a) sin(nπx/a).
The probability of finding the particle in 0 ≤ x ≤ a/4 is
Pₙ = ∫ from x = 0 to x = a/4 |ψₙ(x)|² dx
= (2/a) ∫ from x = 0 to x = a/4 sin²(nπx/a) dx.

Let u = nπx/a, so dx = (a/nπ) du. When x = 0, u = 0; when x = a/4, u = nπ/4.
Thus
Pₙ = (2/nπ) ∫ from u = 0 to u = nπ/4 sin²u du.

Using sin²u = (1 − cos2u)/2,
∫ sin²u du = u/2 − sin2u/4.
Therefore
Pₙ = (2/nπ)[u/2 − sin2u/4] from 0 to nπ/4
= 1/4 − sin(nπ/2)/(2nπ).

For n = 1:
P₁ = 1/4 − sin(π/2)/(2π) = 1/4 − 1/(2π) = (π − 2)/(4π) ≈ 0.0908.

For n = 2:
P₂ = 1/4 − sin(π)/(4π) = 1/4 = 0.2500.

For n = 3:
P₃ = 1/4 − sin(3π/2)/(6π) = 1/4 + 1/(6π) ≈ 0.3031.

Final probabilities: P₁ = (π − 2)/(4π), P₂ = 1/4, P₃ = 1/4 + 1/(6π).

(b) 
The acidic reduction half-reaction is
O₂(g) + 4H⁺ + 4e⁻ ⇌ 2H₂O(l), E° = +1.23 V.
Under basic conditions, the corresponding four-electron reduction is
O₂(g) + 2H₂O(l) + 4e⁻ ⇌ 4OH⁻, with E° = ?

Use the Nernst equation for the acidic couple:
E = E° − (0.05916/4) log(1/([H⁺]⁴ p_O₂)).
Taking p_O₂ = 1 bar,
E = E° + 0.05916 log[H⁺] = E° − 0.05916 pH.

At standard basic conditions, [OH⁻] = 1 mol L⁻¹, so pOH = 0 and pH = 14 at 298 K.
Hence the potential of the acidic couple at pH 14 is
E = 1.23 − 14 × 0.05916
= 1.23 − 0.82824
= 0.40176 V.

This is exactly the standard reduction potential of the basic couple, because [OH⁻] = 1 mol L⁻¹ is the standard basic state.

Final answer: E°(O₂/OH⁻) = +0.40176 V ≈ +0.40 V at 298 K.

(c) 
 (i) 
The radius ratio is
r(Zn²⁺)/r(S²⁻) = 0.74 Å / 1.84 Å = 0.402.

For ionic crystals:
0.225 < r₊/r₋ < 0.414 favours tetrahedral coordination, CN = 4.
0.414 < r₊/r₋ < 0.732 favours octahedral coordination, CN = 6.

Here r₊/r₋ = 0.402, which lies just below 0.414. Therefore the stable arrangement is tetrahedral 4:4 coordination. ZnS accordingly adopts the zinc blende (sphalerite) structure, in which S²⁻ ions form a cubic close-packed array and Zn²⁺ ions occupy half of the tetrahedral voids. Each Zn²⁺ is tetrahedrally surrounded by four S²⁻ ions, and each S²⁻ is tetrahedrally surrounded by four Zn²⁺ ions.

(ii) 
In the CCP/zinc blende unit cell, S²⁻ occupies the face-centred cubic lattice positions: eight corners and six face centres. Zn²⁺ occupies four alternate tetrahedral voids. Using fractional coordinates in the cubic unit cell:

S²⁻ positions: (0,0,0), (0,½,½), (½,0,½), (½,½,0), plus the equivalent face-centred positions.
Zn²⁺ positions: (¼,¼,¼), (¼,¾,¾), (¾,¼,¾), (¾,¾,¼).

A schematic representation is:

S²⁻ at corners and face centres of the cube;
Zn²⁺ at four alternate tetrahedral holes inside the cube.

S²⁻ -------- S²⁻
 /| /|
 S²⁻ |------- S²⁻ |
 | | | |
 | S²⁻-----|- S²⁻
 | / Zn | /
 S²⁻ -------- S²⁻

The unit cell contains 4 Zn²⁺ and 4 S²⁻, so the formula is ZnS. The structure is the cubic close-packed zinc blende form.

(d) 
Let the total number of cation sites normally occupied by Na⁺ be 10,000. One site is occupied by Ca²⁺.
Ca²⁺ has one extra positive charge compared with Na⁺, so this creates an excess positive charge of +1 per 10,000 cation sites.

To maintain electrical neutrality, one Na⁺ site must be vacant. Therefore:
Number of Ca²⁺ = 1
Number of Na⁺ vacancies = 1
Number of Na⁺ present = 10,000 − 1 − 1 = 9998
Number of Cl⁻ = 10,000, since all Cl⁻ sites are fully occupied.

Thus the stoichiometry is
Na₉₉₉₈CaCl₁₀₀₀₀, or after normalizing to one Cl⁻,
Na₀.₉₉₉₈Ca₀.₀₀₀₁Cl, with one Na⁺ vacancy per 10,000 cation sites.

(a) Find the probability of existence of a particle in a one-dimensional box of length 'a' in the region 0 ≤ x ≤ a/4 for the…

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