Question

derive the term of H2 molecule in ground and excited state

Answer 1

To discuss electronic transitions, we first need an efficient way to describe these states. First, we will use our MO diagrams to construct electron configurations. For $\mathrm{H}_{2}$ , the MO diagram leads to the electron configuration $(1\sigma_{\mathrm{g}})^{2}$ . Now, we can use these configurations to define molecular term symbols which describe the electronic state of the molecule.

Begin by adding the angular momentum and spin for each electron together:

$$M _ {L} = \sum_ {i = 1} ^ {n} m _ {\ell_ {i}}, \quad M _ {S} = \sum_ {i = 1} ^ {n} m _ {s _ {i}}$$

Electron in $\sigma$ orbital: $m = 0$ . Electron in $\pi$ orbital: $m = 1$ .

Then, we calculate the allowed values of the total angular momentum (L) and spin (S) from:

$$- L \leq M _ {L} \leq L, \quad - S \leq M _ {S} \leq S$$

With L and S, molecular term symbols are constructed as follows:

$${ } ^ { 2 S + 1 } \Lambda _ { g / u } , \quad \Lambda = \left| M _ { L } \right|$$

We have a shorthand to keep track of $\wedge$ values:

$$\begin{array}{c c c c c} \Lambda & 0 & 1 & 2 & 3 \end{array}$$ $$\Sigma \quad \Pi \quad \Delta \quad \Phi$$

Finally, $\mathbf{g}$ or $\mathbf{u}$ subscript is determined using the following symmetry relationships:

$$g \times g = u \times u = g, \quad u \times g = g \times u = u$$

Let's apply these rules to $\mathrm{H}_2$ :

$$\begin{array}{l} (1 \sigma_ {0}) ^ {2} \\ M _ {L} = 0 + 0 = 0, \quad M _ {S} = \frac {1}{2} + \left(- \frac {1}{2}\right) = 0 \\ L = 0, \quad S = 0 \\ g \times g = g \\ { } ^ { 2 S + 1 } \Lambda _ { g / u } \Rightarrow { } ^ { 1 } \Sigma _ { g } \\ \end{array}$$

Excited state of H2: $(1\sigma_0)(1\sigma_u^*)$

Now the electrons do not have to be spin paired; therefore, we have the possibility of both singlet and triplet states!

$$\begin{array}{l} M _ {L} = 0 + 0 = 0, \quad M _ {S} = - 1, 0, 1 \\ L = 0, \quad S = 0, 1 \\ g \times u = u \\ { } ^ { 2 S + 1 } \Lambda _ { g / u } \Rightarrow { } ^ { 1 } \Sigma _ { u } , { } ^ { 3 } \Sigma _ { u } \\ \end{array}$$

Hund's Rule: State with greatest spin multiplicity is lowest in energy.