Question Statement
In the vibrational-rotational infrared spectrum of a diatomic molecule, why is the pure vibrational band origin ($Q$-branch, $\Delta J = 0$) absent in a $^1\Sigma \to ^1\Sigma$ electronic transition?
(A)
Photons carry spin angular momentum $1\hbar$ and orbital angular momentum cannot change without photon emission
(B)
The selection rule for a diatomic state with zero electronic angular momentum along the internuclear axis ($\Lambda = 0$) strictly forbids $\Delta J = 0$
(C)
Vibrational transitions require nuclear dipole moments to be zero
(D)
The rotational constant $B$ is identically zero in the upper vibrational state
Step-by-Step KaTeX Derivation & Detailed Explanation
### Step-by-Step Derivation & Verification: 1. **Angular Momentum Projections**: In diatomic molecules, $\Lambda$ is the projection of electronic orbital angular momentum on the internuclear axis. 2. **Selection Rules**: - If either state has $\Lambda > 0$ (e.g., $\Pi \to \Sigma$): $\Delta J = 0, \pm 1$ ($P, Q, R$ branches exist). - If both states have $\Lambda = 0$ (i.e., $\Sigma \to \Sigma$ transitions, including ordinary infrared fundamental vibrational bands of closed-shell diatomics like $\text{HCl}, \text{CO}$): $\Delta J = \pm 1 \quad (\Delta J = 0 \text{ is strictly forbidden})$ 3. **Resulting Spectrum**: The spectrum consists solely of the $P$-branch ($\Delta J = -1$) and $R$-branch ($\Delta J = +1$) with a missing central $Q$-branch at $\nu_0$. 4. **Conclusion**: Option (B) is the exact physical rationale.