Question

The bond order of $O_{2}^{+},\,\,{{O}_{2}},\,\,O_{2}^{2-}$ and $O_{2}^{-}$ varies in the order:

• A. $O_{2}^{+}>{{O}_{2}}>O_{2}^{2-}>O_{2}^{-}$
• B. $O_{2}^{2-}>O_{2}^{-}>{{O}_{2}}>O_{2}^{+}$
• C. $O_{2}^{+}>{{O}_{2}}>O_{2}^{-}>O_{2}^{2-}$
• D. ${{O}_{2}}>O_{2}^{+}>O_{2}^{-}>O_{2}^{2-}$

Answer 1

The Correct Option is C

Electronic configuration of $O _{2}^{+}$: $\sigma 1 s^{2}, \sigma^{*} 1 s^{2}, \sigma 2 s^{2}, \sigma^{*} 2 s^{2}, \sigma 2 p_{x}^{2}, \pi 2 p_{y}^{2}$ $=\pi 2 p_{z}^{2}, \pi^{*} 2 p_{y}^{1}=\pi^{*} 2 p_{z}^{0}$ $\therefore$ bond order of $\sigma_{2}^{+}=\frac{1}{2}$ (bonding electron - antibonding electron) $=\frac{1}{2}(10-5)$ $=\frac{1}{2} \times 5=2.5$ Electronic configuration of $O _{2}^{+}$: $\sigma 1 s^{2}, \sigma^{*} 1 s^{2}, \sigma 2 s^{2}, \sigma^{*} 2 s^{2}, \sigma 2 p_{x}^{2}, \pi 2 p_{y}^{2}$ $=\pi 2 p_{x}^{2}, \pi * 2 p_{y}^{1}=\pi^{*} 2 p_{z}^{1}$ $\therefore$ bond order of $O_{2}=\frac{1}{2}(10-6)$ $=\frac{1}{2} \times 4=2$ Electronic configuration of $O _{2}^{2-}:$ $\sigma 1 s^{2}, \sigma^{*} 1 s^{2}, \sigma 2 s^{2}, \sigma^{*} 2 s^{2}, \sigma 2 p_{x}^{2}, \pi 2 p_{y}^{2}$ $\pi 2 p_{z}^{2}, \pi^{*} 2 p_{y}^{2}=\pi^{*} 2 p_{z}^{2}$ $\therefore$ bond order of $O _{2}^{2-}=\frac{1}{2}(10-8)$ $=\frac{1}{2} \times 2=1$ Electronic configuration of $O _{2}^{-}$: $\sigma 1 s^{2}, \sigma^{*} 1 s^{2}, \sigma 2 s^{2}, \sigma^{*} 2 s^{2}, \sigma 2 p_{x}^{2}, \pi 2 p_{y}^{2}$ $=\pi 2 p_{z}^{2}, \pi^{*} 2 p_{y}^{2}=\pi^{*} 2 p_{z}^{1}$ $\therefore$ bond order of $O_{2}^{-}=\frac{1}{2}(10-7)$ $=\frac{1}{2} \times 3=1.5$ Hence, the bond order of $O _{2}^{+}, O _{2}, O _{2}^{2-}$ and $O _{2}^{-}$ varies in the order : $O _{2}^{+}> O _{2}> O _{2}^{-}> O _{2}^{2-}$