Vsepr & Hybridisation

Question

Bond order of O2 according to MO theory is:

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1. **Total Electrons**: $16$ ($8 	imes 2$).
2. **MO Configuration**: $\sigma 1s^2, \sigma^* 1s^2, \sigma 2s^2, \sigma^* 2s^2, \sigma 2p_z^2, (\pi 2p_x^2, \pi 2p_y^2), (\pi^* 2p_x^1, \pi^* 2p_y^1)$.
3. **Identify Electrons**:
   - Bonding electrons ($N_b$): 10 (2 from $\sigma 1s$, 2 from $\sigma 2s$, 6 from $2p$ bonding MOs).
   - Anti-bonding electrons ($N_a$): 6 (2 from $\sigma^* 1s$, 2 from $\sigma^* 2s$, 2 from $\pi^* 2p$).
4. **Bond Order Formula**: $BO = (N_b - N_a) / 2$.
5. **Calculation**: $BO = (10 - 6) / 2 = 2$.
6. **Note**: The presence of two unpaired electrons in $\pi^*$ orbitals correctly explains $O_2$'s paramagnetism.

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