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

(a) Heisenberg's uncertainty principle; orthogonality of particle-in-a-box eigenfunctions

Statement. It is impossible to know simultaneously and exactly both the position and the momentum of a particle along the same coordinate. The product of their uncertainties has a lower limit:

Δx · Δpₓ ≥ ħ/2 = h/4π

• The principle follows from the fact that the position and momentum operators do not commute: [x̂, p̂ₓ] = iħ.
• It is not a limitation of instruments; it is a property of matter waves.
• Consequence: an electron cannot move in a sharply defined Bohr orbit, so we can speak only of probabilities (orbitals).

Particle in a box (0 ≤ x ≤ L).

• The potential is V = 0 inside the box and ∞ outside. The boundary conditions ψ(0) = ψ(L) = 0 give ψₙ = A sin(nπx/L) and Eₙ = n²h²/8mL², with n = 1, 2, 3, …
• Normalisation, ∫₀ᴸ A² sin²(nπx/L) dx = A²L/2 = 1, gives A = √(2/L).

So:

• ψ₁ = √(2/L) sin(πx/L), with E₁ = h²/8mL²;
• ψ₂ = √(2/L) sin(2πx/L), with E₂ = 4h²/8mL².

Orthogonality. We must show I = ∫₀ᴸ ψ₁ψ₂ dx = 0.

I = (2/L) ∫₀ᴸ sin(πx/L) sin(2πx/L) dx

Using sin A sin B = ½[cos(A − B) − cos(A + B)]:

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.

General result (n ≠ m).

∫₀ᴸ ψₙψₘ dx = (1/L) ∫₀ᴸ [cos((n − m)πx/L) − cos((n + m)πx/L)] dx = (1/π) [ sin((n − m)πx/L)/(n − m) − sin((n + m)πx/L)/(n + m) ]₀ᴸ = 0,

because sin(kπ) = 0 for any integer k. For n = m, the same integral equals 1. So ∫ψₙψₘ dx = δₙₘ: the set is orthonormal. This is expected, since eigenfunctions of a Hermitian operator (Ĥ) that belong to different eigenvalues (E₁ ≠ E₂) are always orthogonal.

(b)(i) Bond orders of O₂, O₂⁻, O₂²⁻ and O₂⁺

For O₂ and its ions, the 2s–2p mixing is small, so σ2p_z lies below π2p. Counting valence electrons only (the 1s cores, KK, cancel out), BO = (N_b − N_a)/2.

O₂ (12 valence e⁻): KK σ2s² σ*2s² σ2p_z² π2p_x² π2p_y² π*2p_x¹ π*2p_y¹.

• N_b = 8, N_a = 4, so BO = 2.
• Two unpaired electrons: paramagnetic.

Superoxide O₂⁻ (13 e⁻): π*³.

• N_b = 8, N_a = 5, so BO = 1.5.
• One unpaired electron: paramagnetic (e.g. KO₂).

Peroxide O₂²⁻ (14 e⁻): π*⁴.

• N_b = 8, N_a = 6, so BO = 1.
• All electrons paired: diamagnetic (e.g. Na₂O₂, H₂O₂).

Dioxygenyl O₂⁺ (11 e⁻): π*¹.

• N_b = 8, N_a = 3, so BO = 2.5.
• One unpaired electron: paramagnetic (e.g. O₂⁺[PtF₆]⁻).

Stability order: O₂⁺ > O₂ > O₂⁻ > O₂²⁻. Stability and bond strength rise with bond order, so O₂⁺ is the most stable and O₂²⁻ the least. Bond lengths follow the reverse order: O₂⁺ ≈ 112 pm < O₂ 121 pm < O₂⁻ ≈ 133 pm < O₂²⁻ ≈ 149 pm.

(b)(ii) Molecular orbital diagram of CO

• Electron count. CO has 14 electrons in all, 10 of them valence electrons (C 2s²2p², O 2s²2p⁴). It is isoelectronic with N₂.
• Why the atomic levels differ. Oxygen is more electronegative, so its 2s and 2p atomic orbitals lie well below those of carbon. As a result, each bonding MO is weighted towards O and each antibonding MO towards C.
• Why the MO order is N₂-like. The C 2s–2p energy gap is small, so s–p mixing is strong. This pushes the σ orbital from 2p (5σ) above the π set (1π).

Configuration. KK (3σ)² (4σ)² (1π)⁴ (5σ)². In the older notation this is σ2s² σ*2s² π2p⁴ σ2p².

• Bond order: bonding electrons 3σ² + 1π⁴ + 5σ² = 8, antibonding 4σ² = 2, so BO = (8 − 2)/2 = 3. This is consistent with the short, strong bond (112.8 pm, bond energy ≈ 1072 kJ mol⁻¹).
• Magnetism: CO is diamagnetic, since no electrons are unpaired.
• HOMO = 5σ: a lone pair concentrated on carbon. CO therefore acts as a σ-donor through carbon in metal carbonyls.
• LUMO: 2π*, an empty antibonding π orbital, also larger on carbon. It accepts electron density from filled metal d-orbitals (π back-bonding), which is why the C–O stretching frequency falls from 2143 cm⁻¹ in free CO in carbonyls.
• Dipole: despite oxygen's higher electronegativity, the dipole moment is tiny (≈ 0.11 D) and its negative end lies on carbon. The carbon lone pair (5σ) offsets the polarity of the bonding orbitals.

(c)(i) Limiting radius ratios for coordination numbers 3 and 6

Let r be the cation radius and R the anion radius. At the limiting ratio, the cation touches every surrounding anion, and those anions just touch one another. If r/R falls below this value, the cation "rattles" in its hole, anion–anion repulsion dominates, and a lower coordination number becomes favoured.

CN = 3 (trigonal planar).

• The three anion centres form an equilateral triangle of side 2R.
• The cation sits at the centroid, at a distance (r + R) from each anion centre.
• In the right triangle formed by the centroid, one vertex and the mid-point of a side, the angle at the centroid is 60°. So sin 60° = R/(r + R).
• r + R = R/sin 60° = 2R/√3, which gives r/R = 2/√3 − 1 = 0.155.

CN = 6 (octahedral).

• The equatorial plane of the octahedron contains four anions at the corners of a square of side 2R, with the cation at the centre. The other two anions lie above and below this plane.
• The half-diagonal equals r + R, so (r + R) = √2 · R (equivalently, cos 45° = R/(r + R)).
• This gives r/R = √2 − 1 = 0.414.

Stable ranges:

• CN 3 is stable for 0.155–0.225 (e.g. B₂O₃).
• CN 4 (tetrahedral) for 0.225–0.414 (ZnS).
• CN 6 for 0.414–0.732 (NaCl, with r/R ≈ 0.52).
• CN 8 (cubic) for 0.732 or more (CsCl, with r/R ≈ 0.93).

These values rest on a hard-sphere ionic model, so they are guides rather than rules; covalent crystals often violate them.

(c)(ii) Stoichiometric defects

Stoichiometric (intrinsic) defects are point defects in an ionic solid that leave the cation : anion ratio exactly as given by the formula, and keep the crystal electrically neutral. They form spontaneously because their small enthalpy cost is repaid by a gain in configurational entropy. For this reason every crystal above 0 K contains some, and their number rises with temperature.

1. Schottky defect.

• An equal number of cation and anion vacancies (one pair for a 1:1 salt), so neutrality and stoichiometry are both preserved.
• It is favoured in highly ionic solids with cations and anions of similar size and a high coordination number.
• Examples: NaCl, KCl, KBr, CsCl. NaCl contains roughly 10⁶ Schottky pairs per cm³ at room temperature.
• Consequences: density falls, because mass is lost while the volume is unchanged; ionic conductivity rises, because ions hop into vacancies; and lattice energy falls slightly.

2. Frenkel (dislocation) defect.

• An ion, usually the smaller cation, leaves its lattice site for an interstitial position, creating a vacancy–interstitial pair.
• It is favoured by a large size difference between the ions and a low coordination number.
• Examples: AgCl, AgBr, AgI, ZnS.
• Consequences: density is essentially unchanged, because no ion leaves the crystal. Conductivity and the dielectric constant increase.
• AgBr shows both Schottky and Frenkel defects.

Contrast. In non-stoichiometric defects the ratio does change. In metal-excess NaCl, for example, heating in Na vapour creates anion vacancies occupied by trapped electrons (F-centres), which make the crystal yellow. Such defects are not stoichiometric.