Past exam of the chemistry A course of the University of Cambridge / 2025 / Paper 1 / A2

The Pauli principle requires the complete wavefunction of identical fermions to be antisymmetric under exchanging the full coordinates, including spin:It is the total spatial-and-spin state that changes sign; the spatial factor alone need not be antisymmetric.

Let $a={\varphi }_{1s}$ and $b={\varphi }_{2s}$ be orthonormal atomic orbitals. The normalized spatial factors areFor the singlet state, use the antisymmetric spin factorFor the three components of the triplet state, useEach total wavefunction obeys the Pauli principle. At coincident spatial coordinates,Consequently the triplet pair probability density vanishes at coincidence, even for its $M=0$ component. This zero is stronger than the trivial statement that any exact point has zero integrated probability in a continuum.

In atomic units, define the Coulomb integral and exchange integral byExpanding the four products in the expectation of $r_{12}^{-1}$ yieldsThe one-electron terms have the same value for these two states. Within this common-orbital approximation,so $\boxed{E_S-E_T=2K}$. The exchange integral is nonnegative: it is the positive Coulomb quadratic form of the transition density $a^{\ast }b$, or equivalently has a nonnegative Fourier representation. The triplet is lower. Actual orbital relaxation and correlation make the common-orbital $2K$ an approximate physical splitting.

Keeping the same two spin states but incorrectly imposing boson exchange symmetry would require a symmetric total wavefunction. The spin-antisymmetric singlet state would then accompany ${\Phi }_{-}$, while the spin-symmetric triplet state would accompany ${\Phi }_{+}$. Thusand $\boxed{E_T^{\mathrm{hypothetical}}-E_S^{\mathrm{hypothetical}}=2K}$. The ordering reverses: the triplet would be destabilized by $2K$ relative to the singlet. A genuinely spinless boson would have no singlet/triplet spin distinction; this comparison changes exchange symmetry while retaining the question's spin degrees of freedom.

If two fermions occupied the same complete spin orbital $u$, their antisymmetrized product would beThe zero vector cannot represent a normalized quantum state. This is the Pauli exclusion principle: at most one identical fermion occupies any complete one-particle state. Two electrons may occupy the same spatial orbital with opposite spins, because $a\alpha$ and $a\beta$ are different spin orbitals.

The $3d$ and $4s$ atomic orbitals of chromium are close enough in energy that electron-electron terms can overturn the simple filling order. The configuration $[\mathrm{Ar}]3d^{5}4s^1$ permits five singly occupied $3d$ orbitals and a parallel-spin $4s$ electron. Relative to $3d^{4}4s^2$, it removes the $4s$ pair and permits more parallel-spin exchange interactions, lowering the Coulomb-and-exchange contribution enough to offset the orbital promotion cost.

As qualitative bookkeeping, six parallel electron spins give $\left(\genfrac{}{}{0ex}{}{6}{2}\right)=15$ same-spin pairs, compared with $\left(\genfrac{}{}{0ex}{}{5}{2}\right)=10$ in high-spin $3d^{4}4s^2$. The individual exchange integrals are unequal, so this is an explanation of the tendency, rather than a quantitative atomic energy calculation. The observed ground state configuration is $\boxed{[\mathrm{Ar}]3d^{5}4s^1}$.^[ref]

The closed argon core contributes $L=S=0$. In the half-filled $3d^5$ shell, Hund's rules put one parallel-spin electron in each $m_{\ell }=-2,-1,0,1,2$ orbital. This gives $S_d=5/2$ and the orbital term $L_d=0$. Adding the parallel $4s$ electron, with $\ell =0$ and $s=1/2$, givesThe spin multiplicity is seven. The magnetic projection labels run $M_S,M_J=-3,-2,\dots ,3$, while $M_L=0$. The term symbol agrees with the observed neutral-chromium ground level.^[ref]