(a) State Heisenberg's uncertainty principle. Show that for a particle in a one-dimensional box having length from 0 to L, the two normalised eigen functions corresponding to the eigenvalues E₁ and E₂ (characterised by quantum numbers 1 and 2 respect

Question

(a) State Heisenberg's uncertainty principle. Show that for a particle in a one-dimensional box having length from 0 to L, the two normalised eigen functions corresponding to the eigenvalues E₁ and E₂ (characterised by quantum numbers 1 and 2 respectively) are orthogonal to each other. (10 marks)
(b) (i) Calculate the bond order for the following : (I) Oxygen, (II) Superoxide, (III) Peroxide, (IV) Dioxygenyl ion. Which will have the highest stability ? (10 marks)
(ii) Draw the molecular orbital diagram for CO. (10 marks)
(c) (i) Calculate the limiting radius ratio for the crystals with coordination number 3 and 6. (10 marks)
(ii) Explain stoichiometric defects with an example. (10 marks)

UPSC Answer Check · Accepted Answer

(a) Heisenberg’s uncertainty principle states that for a particle it is impossible to determine simultaneously its exact position and exact momentum. Mathematically, Δx Δpₓ ≥ ħ/2, Δy Δp_y ≥ ħ/2, Δz Δp_z ≥ ħ/2, and for energy and time ΔE Δt ≥ ħ/2, where ħ = h/2π. It is valid in quantum mechanics because position and momentum operators do not commute.

For a particle in a one-dimensional box 0 ≤ x ≤ L, the normalised eigenfunctions are
ψ₁ = √(2/L) sin(πx/L), ψ₂ = √(2/L) sin(2πx/L).
Their orthogonality integral is
I = ∫₀ᴸ ψ₁ψ₂ dx = (2/L) ∫₀ᴸ sin(πx/L) sin(2πx/L) dx.
Using sin A sin B = 1/2[cos(A−B) − cos(A+B)], with A = πx/L and B = 2πx/L,
I = (1/L) ∫₀ᴸ [cos(πx/L) − cos(3πx/L)] dx
= (1/L)[(L/π) sin(πx/L) − (L/3π) sin(3πx/L)]₀ᴸ
= (1/π)[sin π − sin 0] − (1/3π)[sin 3π − sin 0] = 0.
Hence ψ₁ and ψ₂ are orthogonal.

(b)(i) Using molecular orbital theory, bond order = (N_b − N_a)/2, where N_b and N_a are the numbers of electrons in bonding and antibonding molecular orbitals.
O₂: 16 electrons. Configuration: σ2s² σ*2s² σ2p(z)² π2p(x)² π2p(y)² π*2p(x)¹ π*2p(y)¹. N_b = 8, N_a = 4. Bond order = (8 − 4)/2 = 2.
Superoxide O₂⁻: 17 electrons. One extra electron enters π*. N_b = 8, N_a = 5. Bond order = (8 − 5)/2 = 1.5.
Peroxide O₂²⁻: 18 electrons. π* has four electrons. N_b = 8, N_a = 6. Bond order = (8 − 6)/2 = 1.
Dioxygenyl ion O₂⁺: 15 electrons. One electron is removed from π*. N_b = 8, N_a = 3. Bond order = (8 − 3)/2 = 2.5.

Highest stability: O₂⁺, because it has the highest bond order, 2.5, hence the strongest and shortest bond among the given species.

(b)(ii) CO is a heteronuclear diatomic molecule. Carbon contributes 4 valence electrons and oxygen contributes 6 valence electrons, giving 10 valence electrons; total electrons including cores are 14. The occupied molecular orbital configuration is:
(1σ)²(2σ)²(3σ)²(4σ)²(1π)⁴(5σ)².
Valence configuration: (3σ)²(4σ)²(1π)⁴(5σ)². Bond order = (8 − 2)/2 = 3. The HOMO is 5σ, mainly carbon lone-pair character; the LUMO is 2π*.

Schematic MO diagram, energy increasing upward:
2π*: empty (LUMO)
5σ: ↑↓ (HOMO)
1π: ↑↓ ↑↓
4σ: ↑↓
3σ: ↑↓
2σ: ↑↓, mainly C 1s core
1σ: ↑↓, mainly O 1s core

Carbon 2s and 2p orbitals mix with oxygen 2s and 2p orbitals to form these MOs; oxygen atomic levels lie lower because oxygen is more electronegative.

(c)(i) Let r be cation radius and R be anion radius. At the limiting radius ratio, cation touches all anions and anions just touch each other.

For coordination number 3, the arrangement is trigonal planar. Distance from centre to each anion = r + R. For an equilateral triangle, side = √3(r + R). At the limit, anion–anion distance = 2R. Hence
2R = √3(r + R), so r/R = 2/√3 − 1 = (2√3 − 3)/3 ≈ 0.155.

For coordination number 6, the arrangement is octahedral. Distance from centre to each anion = r + R. Edge length between adjacent anions = √2(r + R). At the limit, edge = 2R. Hence
2R = √2(r + R), so r/R = √2 − 1 ≈ 0.414.

These values assume hard-sphere ions and ideal close contact.

(c)(ii) Stoichiometric defects are point defects in ionic crystals that do not change the stoichiometric ratio of cations to anions. They preserve electrical neutrality and the chemical formula of the crystal. The main types are:
Schottky defect: Equal numbers of cation and anion vacancies are created. This occurs in highly ionic crystals with cations and anions of similar size and high coordination number, e.g. NaCl, KCl, CsCl. Density decreases, and ionic conductivity increases.
Frenkel defect: A smaller ion, usually a cation, leaves its normal lattice site and occupies an interstitial site, leaving a vacancy. Stoichiometry is unchanged. It occurs when cations and anions differ greatly in size and coordination number is low, e.g. ZnS, AgCl, AgBr. Density remains almost unchanged.

Both are stoichiometric because an equal number of missing and displaced or vacant sites maintain the original cation–anion ratio.

(a) State Heisenberg's uncertainty principle. Show that for a particle in a one-dimensional box having length from 0 to L, the…

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