(a) (i) Calculate the number of permitted electrons in a sub-shell and a shell in an atom. (ii) Explain how Pauli's exclusion…

(a) (i) Maximum number of electrons in a sub-shell and a shell By Pauli's exclusion principle no two electrons in an atom have the same set (n, l, mₗ, mₛ). Sub-shell (fixed n, l): mₗ = −l, …, 0, …, +l gives 2l + 1 orbitals; each takes mₛ = +½ and −½. N_sub-shell = 2(2l + 1): s 2, p 6, d 10, f 14. Shell (fixed n): l = 0, 1, …, n − 1, so N_shell = Σₗ₌₀ⁿ⁻¹ 2(2l + 1) = 2[2 · n(n − 1)/2 + n] = 2n²: K 2, L 8, M 18, N 32.

(a) (ii) Role of Pauli's exclusion principle in electronic configuration Statement: no two electrons in an atom can have all four quantum numbers (n, l, mₗ, mₛ) the same; equivalently, the total wavefunction of electrons (fermions) must be antisymmetric under exchange of any two electrons (a Slater determinant vanishes if two electrons share a state). How it fixes the configuration:

• It limits each orbital to two electrons of opposite spin, each sub-shell to 2(2l + 1) and each shell to 2n² electrons. Without it, all electrons would fall into 1s and every atom would behave alike.
• Electrons therefore fill successive orbitals in order of increasing energy (Aufbau, n + l rule): 1s, 2s, 2p, 3s, 3p, 4s, 3d, …
• Within a partly filled sub-shell, Hund's rule applies: electrons occupy different mₗ orbitals with parallel spins before pairing, because the antisymmetry that Pauli's principle imposes keeps parallel-spin electrons apart and lowers their Coulomb repulsion (exchange energy). For example, N (1s² 2s² 2p³) has three unpaired 2p electrons; O (2p⁴) has the fourth electron paired.
• Examples: Na (Z = 11) = 1s² 2s² 2p⁶ 3s¹, one electron outside a closed neon core, hence an alkali metal. Closed shells and sub-shells (He, Ne, Ar) are chemically inert. Thus Pauli's principle produces the shell structure, the periods of lengths 2, 8, 8, 18, …, and the periodicity of chemical properties.

(b) (i) Importance of relaxation in NMR In a field B₀ the nuclear spin levels are populated according to the Boltzmann law, giving a very small population excess in the lower level; the net absorption signal depends on this excess. Relaxation is the return of the spin system to this equilibrium.

• Spin–lattice (longitudinal) relaxation, time T₁: the spins give energy to the surroundings (lattice) and the equilibrium population difference, i.e. M_z, is restored. Without it, continuous absorption would equalise the populations (saturation) and the signal would vanish. T₁ also sets how often pulses can be repeated for signal averaging.
• Spin–spin (transverse) relaxation, time T₂: spins lose phase coherence and M_xy decays. It determines the lifetime of the free induction decay and the natural line width, Δν ≈ 1/(πT₂). Very short T₂ (as in solids) gives broad lines; long T₂ gives the sharp lines needed to resolve chemical shifts and spin–spin coupling.
• T₁ and T₂ depend on molecular motion and environment, so they give information on structure and dynamics, and they provide contrast in MRI. So relaxation makes a steady, sharp and repeatable NMR signal possible and is itself a source of information.

(b) (ii) Why ¹²C and ¹⁶C are unsuitable for NMR NMR needs a nucleus with non-zero spin I, since the magnetic moment μ = γħI and the Zeeman splitting into 2I + 1 levels both vanish for I = 0. Both ¹²C (6 protons, 6 neutrons) and ¹⁶C (6 protons, 10 neutrons) are even–even nuclei. Their protons and neutrons pair off to total angular momentum zero, so I = 0 and μ = 0. There is no Zeeman splitting and no resonance. (¹⁶C is also radioactive, with half-life about 0.75 s.) Carbon NMR therefore uses ¹³C (I = ½, natural abundance about 1.1%).

(c) (i) Quantum numbers of hydrogen and n² degeneracy

• Principal quantum number n = 1, 2, 3, …: fixes the energy Eₙ = −13.6/n² eV and the size of the orbital (number of radial nodes n − l − 1).
• Orbital quantum number l = 0, 1, …, n − 1: fixes the magnitude of orbital angular momentum, L² = l(l + 1)ħ², and the shape (s, p, d, …).
• Magnetic quantum number m = −l, −l + 1, …, +l: fixes L_z = mħ, the orientation of L. Energy of hydrogen depends only on n (Coulomb 1/r potential). For a given n the number of states is Σₗ₌₀ⁿ⁻¹ (2l + 1) = 1 + 3 + 5 + … + (2n − 1) = n² So each level Eₙ is n²-fold degenerate (2n² including spin). Example: n = 2 has 2s (1) + 2p (3) = 4 states.

(c) (ii) z-component of angular momentum L̂_z = −iħ ∂/∂φ L̂_zψ = −iħ ∂/∂φ [A R(r) sin θ cos θ e^−iφ] = −iħ(−i)ψ = −ħψ So ψ is an eigenfunction of L̂_z with L_z = −ħ (m = −1). Since sin θ cos θ e^−iφ ∝ Y₂⁻¹(θ, φ), the state also has l = 2 and L² = 6ħ².