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

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

$O_2^{2+}\Rightarrow B.O=\frac{10-4}{2}=3.0$

$O_2^{-}<O_2^{+}<O_2^{2+}$