Chemistry - CHEMICAL BONDING Question with Solution | TestHub

$\mathrm{B}.\mathrm{O}=\frac{{\mathrm{N}}_{\mathrm{b}}\mathrm{ }-{\mathrm{N}}_{\mathrm{a}}}{2}$

$O_2^{-}\Rightarrow \sigma 1s^2<\stackrel{*}{\sigma }1s^2<\sigma 2s^2<\stackrel{*}{\sigma }2s^2<\sigma 2p_z^2<\pi 2p_x^2=\pi 2p_y^2<\stackrel{*}{\pi }2p_x^2=\stackrel{*}{\pi }2p_y^1$

B.O for${\mathrm{O}}_2^{-}=1.5\Rightarrow \frac{10\mathrm{ }-\mathrm{ }7}{2}=1.5$

$O_2\Rightarrow \sigma 1s^2<\stackrel{*}{\sigma }1s^2<\sigma 2s^2<\stackrel{*}{\sigma }2s^2<\sigma 2p_z^2<\pi 2p_x^2=\pi 2p_y^2<\stackrel{*}{\pi }2p_x^1=\stackrel{*}{\pi }2p_y^1$

B.O for${\mathrm{O}}_2=2.0\Rightarrow \frac{10\mathrm{ }-\mathrm{ }6}{2}=2$

$O_2^{+}\Rightarrow \sigma 1s^2<\stackrel{*}{\sigma }1s^2<\sigma 2s^2<\stackrel{*}{\sigma }2s^2<\sigma 2p_z^2<\pi 2p_x^2<\pi 2p_y^2<\stackrel{*}{\pi }2p_x^1<\stackrel{*}{\pi }2p_y^0$

B.O for${\mathrm{O}}_2^{+}=2.5\Rightarrow \frac{10\mathrm{ }-\mathrm{ }5}{2}=2.5$

As per M.O.T electronic configurations

${\mathrm{O}}_2^{-}<{\mathrm{O}}_2<{\mathrm{O}}_2^{+}$