Which of the following orders is correct for the bond dissociation energy of $\mathrm{O}_{2}, \mathrm{O}_{2}^{+}, \mathrm{O}_{2}^{-}$and $\mathrm{O}_{2}^{2-} ?$

Question

Aicharya · Accepted Answer

Bond Dissociation Energy $\uparrow ightarrow$ Bond Order $\uparrow$
Bond Order $(\mathrm{B} . \mathrm{O})=\frac{1}{2} x\left\{\left[\mathrm{No} .ight.ight.$ of $e^{-1}$ in Antibonding M.O.]- $\left(\mathrm{No} {\text { of }} e^{-}ight.$in bonding M.O. $\left.]ight\}$
For $\mathrm{O}_{2}$ molecule, electronic configuration is
$\mathrm{O}_{2}=\sigma 1 s^{2}, \sigma^{*} 1 s^{2}, \sigma 2 s^{2}, \sigma^{*} 2 s^{2}, \sigma 2 p_{z}^{2}, \Pi 2 p_{x}^{2}=\Pi 2 p_{y}^{2}, \Pi^{* 2} p_{x}^{1}=\Pi^{*} 2 p^{1} y$
B.O. $\cdot\left(\mathrm{O}_{2}ight)=\frac{10-6}{2}=2$
(a) For $\mathrm{O}_{2}^{+}$molecule, an electron is removed from $\pi^{*} 2 p y$ orbital
B.O. $\left(\mathrm{O}_{2}^{+}ight)=\frac{10-5}{2}=2.5$
(b) B.O. $\left(\mathrm{O}_{2}ight)=2$
(c) B.O. $\left(\mathrm{O}_{2}^{-}ight):$electron is added to $\pi^{*} 2 p x^{1} y$
$$\mathrm{B} \cdot \mathrm{O} \cdot\left(\mathrm{O}_{2}^{2}ight)=\frac{10-7}{2}=1.5$$
(d) For $\mathrm{O}_{2}^{2-}$ molecule, two electrons are added to $\pi^{*} 2 p^{1} x$ ant $\pi^{*} 2 p^{1} y$
B.O. $\left(\mathrm{O}_{2}^{2-}ight)=\frac{10-8}{2}=1$
$\therefore$ The order is:
$$\mathrm{O}_{2}^{+}>\mathrm{O}_{2}>\mathrm{O}_{2}^{-}>\mathrm{O}_{2}^{2-}$$
Higher B.O, Higher Dissociation Energy ^

Which of the following orders is correct for the bond dissociation energy of $\mathrm{O}_{2},…

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