 Solution

ID: past-exam-of-the-chemistry-a-course-of-the-university-of-cambridge/2025/paper-1/a2/b/solution

Let

a = ϕ_{1 s}

and

b = ϕ_{2 s}

be orthonormal atomic orbitals. The normalized spatial factors are

Φ_{\pm} (1, 2) = \frac{a ( 1 ) b ( 2 ) \pm b ( 1 ) a ( 2 )}{2} .

For the singlet state, use the antisymmetric spin factor

χ_{00} = \frac{α ( 1 ) β ( 2 ) - β ( 1 ) α ( 2 )}{2}, Ψ_{S} = Φ_{+} χ_{00} .

For the three components of the triplet state, use

χ_{11} = α (1) α (2), χ_{10} = \frac{α ( 1 ) β ( 2 ) + β ( 1 ) α ( 2 )}{2}, χ_{1, - 1} = β (1) β (2), Ψ_{T, M} = Φ_{-} χ_{1 M} .

Each total wavefunction obeys the Pauli principle. At coincident spatial coordinates,

Φ_{-} (r, r) = \frac{a ( r ) b ( r ) - b ( r ) a ( r )}{2} = 0.

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.

 New to topics? Read the docs here!