For $\mathrm{H}_{2}$ molecule in the excited state $\sigma_{\mathrm{g}}^{1} \sigma_{\mathrm{u}}^{1}$ spin part of the triplet state with $\mathrm{ms}=0$ is proportional to
(a)$\alpha(1) \beta(2)$
(b)$[\alpha(1) \beta(2)-\beta(1) \alpha(2)]$
(c)$\alpha(1) \alpha(2)$
(d)$[\alpha(1) \beta(2)+\beta(1) \alpha(2)]$
Answer
Answer: D ✓ checked by 4AB · confidence high
The source book printed no answer; this one was worked out and checked — see the explanation.
Explanation
The m_S = 0 component of a triplet is the symmetric combination α(1)β(2) + β(1)α(2); the antisymmetric one is the singlet.