Bond Order S2 2 38B4A3

1. **State the problem:** Calculate the molecular bond order for the $\mathrm{S}_2^{2-}$ anion. 2. **Recall the formula for bond order:** $$\text{Bond order} = \frac{\text{Number of bonding electrons} - \text{Number of antibonding electrons}}{2}$$ 3. **Determine the total number of valence electrons:** - Each sulfur atom has 6 valence electrons. - For $\mathrm{S}_2$, total valence electrons = $6 \times 2 = 12$. - The $2-$ charge adds 2 more electrons. - Total electrons = $12 + 2 = 14$. 4. **Write the molecular orbital electron configuration for $\mathrm{S}_2^{2-}$:** - Sulfur is in period 3, so the molecular orbitals resemble those of $\mathrm{O}_2$ but with reversed order of $\pi$ and $\sigma$ orbitals in the $2p$ level. - The order is: $\sigma_{2s}^2, \sigma_{2s}^*{}^2, \sigma_{2p_z}^2, \pi_{2p_x}^4, \pi_{2p_y}^4, \pi_{2p_x}^*{}^0, \pi_{2p_y}^*{}^0, \sigma_{2p_z}^*{}^0$. - Total electrons fill up to $\pi_{2p_x}$ and $\pi_{2p_y}$ orbitals. 5. **Count bonding and antibonding electrons:** - Bonding electrons: $\sigma_{2s}^2 (2) + \sigma_{2p_z}^2 (2) + \pi_{2p_x}^4 (4) + \pi_{2p_y}^4 (4) = 12$ - Antibonding electrons: $\sigma_{2s}^*{}^2 (2)$ 6. **Calculate bond order:** $$\text{Bond order} = \frac{12 - 2}{2} = \frac{10}{2} = 5$$ 7. **Interpretation:** - The bond order of $\mathrm{S}_2^{2-}$ is $5$, indicating a very strong bond between the two sulfur atoms.