CHEM_REMOVED - CHEMICAL BONDING Question with Solution | TestHub

Number of${\mathrm{e}}^{-}$

$\mathrm{CO}=14, {\mathrm{NO}}^{+}=14$

${\mathrm{O}}_2^{2-}=18, \mathrm{ }{\mathrm{B}}_2=10$.

According to$\mathrm{MOT}$,

$\mathrm{CO}:-\mathrm{\sigma }1{\mathrm{s}}^2<\stackrel{*}{\mathrm{\sigma }}1{\mathrm{s}}^2<\mathrm{\sigma }2{\mathrm{s}}^2<\stackrel{*}{\mathrm{\sigma }}2{\mathrm{s}}^2<\mathrm{\pi }2{{\mathrm{p}}_{\mathrm{x}}}^2=\mathrm{\pi }2{{\mathrm{p}}_{\mathrm{y}}}^2<\mathrm{\sigma }2{{\mathrm{p}}_{\mathrm{z}}}^2$

no unpaired e^-so diamagnetic

$B_2:-\sigma 1s^2<\stackrel{*}{\sigma }1s^2<\sigma 2s^2<\stackrel{*}{\sigma }2s^2<\pi 2p_x{}^1=\pi 2p_y{}^1$

$2$unpaired${\mathrm{e}}^{-}$,so paramagnetic

${\mathrm{O}}_2^{-2}:-\mathrm{\sigma }1{\mathrm{s}}^2<\stackrel{*}{\mathrm{\sigma }}1{\mathrm{s}}^2<\mathrm{\sigma }2{\mathrm{s}}^2<\stackrel{*}{\mathrm{\sigma }}2{\mathrm{s}}^2<\mathrm{\sigma }2{{\mathrm{p}}_{\mathrm{z}}}^2<\mathrm{\pi }2{{\mathrm{p}}_{\mathrm{x}}}^2=\mathrm{\pi }2{{\mathrm{p}}_{\mathrm{y}}}^2<\stackrel{*}{\mathrm{\pi }}2{{\mathrm{p}}_{\mathrm{x}}}^2=\stackrel{*}{\mathrm{\pi }}2{{\mathrm{p}}_{\mathrm{y}}}^2$

no unpaired${\mathrm{e}}^{-}$, so diamagnetic

${\mathrm{NO}}^{+}:-\mathrm{\sigma }1{\mathrm{s}}^2<\stackrel{*}{\mathrm{\sigma }}1{\mathrm{s}}^2<\mathrm{\sigma }2{\mathrm{s}}^2<\stackrel{*}{\mathrm{\sigma }}2{\mathrm{s}}^2<\mathrm{\pi }2{{\mathrm{p}}_{\mathrm{x}}}^2=\mathrm{\pi }2{{\mathrm{p}}_{\mathrm{y}}}^2<\mathrm{\sigma }2{{\mathrm{p}}_{\mathrm{z}}}^2$

no unpaired${\mathrm{e}}^{-}$, so diamagnetic.