CHEM_REMOVED - CHEMICAL BONDING Question with Solution | TestHub

According to molecular orbital theory

Bond order$=\frac{\left(e^{-}\mathrm{ }\mathrm{i}\mathrm{n}\mathrm{ }\mathrm{b}\mathrm{o}\mathrm{n}\mathrm{d}\mathrm{i}\mathrm{n}\mathrm{g}\mathrm{ }\mathrm{m}\mathrm{o}\mathrm{l}\mathrm{e}\mathrm{c}\mathrm{u}\mathrm{l}\mathrm{a}\mathrm{r}\mathrm{ }\mathrm{o}\mathrm{r}\mathrm{b}\mathrm{i}\mathrm{t}\mathrm{a}\mathrm{l}\right)\mathrm{ }-\left(e^{-}\mathrm{ }\mathrm{i}\mathrm{n}\mathrm{ }\mathrm{A}\mathrm{n}\mathrm{t}\mathrm{i}\mathrm{ }\mathrm{b}\mathrm{o}\mathrm{n}\mathrm{d}\mathrm{i}\mathrm{n}\mathrm{g}\mathrm{ }\mathrm{M}\mathrm{o}\mathrm{l}\mathrm{e}\mathrm{c}\mathrm{u}\mathrm{l}\mathrm{a}\mathrm{r}\mathrm{ }\mathrm{o}\mathrm{r}\mathrm{b}\mathrm{i}\mathrm{t}\mathrm{a}\mathrm{l}\right)}{2}$

Molecular orbital configuration.

$\left(A\right)$$O_2⟹\sigma 1s^2<\stackrel{*}{\sigma }1s^2<\sigma 2s^2<\stackrel{*}{\sigma }2s^2<\stackrel{ }{\sigma }2p_x^2<\pi 2p_y^2=\pi 2p_z^2<\stackrel{*}{\sigma }2p_y^1=\stackrel{*}{\pi }2p_y^1$

$\mathrm{B}.\mathrm{O}.=\frac{6-2}{2}=2$

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

$\mathrm{B}.\mathrm{O}.=\frac{6-1}{2}=2.5$

Both are paramagnetic because having unpaired$e^{-}$

bond order increase$O_2\to O_2^{+}$

$\left(B\right) NO= \sigma 1s^2<\stackrel{*}{\sigma }1s^2<\sigma 2s^2<\stackrel{*}{\sigma }2s^2<\stackrel{ }{\sigma }2p_x^2<\pi 2p_y^2=\pi 2p_z^2<\stackrel{*}{\pi }2p_y^1=\stackrel{*}{\pi }2p_z^0$

$\mathrm{B}.\mathrm{O}.=\frac{6-1}{2}=2.5$

$1$Unpaired$e^{-}$so paramagnetic

$N{O}^{+}⟹\sigma 1s^2<\stackrel{*}{\sigma }1s^2<\sigma 2s^2<\stackrel{*}{\sigma }2s^2<\sigma 2p{x}^2<\pi 2p_y^2=\pi 2p_z^2$

$\mathrm{B}.\mathrm{O}.=\frac{6-0}{2}=3$

$N{O}^{+}$doesn't have unpaired$e^{-}$so diamagnetic

$NO\to N{O}^{+}${Bond order increases and change magnetic character paramagnetic to diamagnetic character change}

$\left(C\right)$$O_2\to \sigma 1s^2<\stackrel{*}{\sigma }1s^2<\sigma 2s^2<\stackrel{*}{\sigma }2s^2<\stackrel{ }{\sigma }2p_x^2<\pi 2p_y^2=\pi 2p_z^2<\stackrel{*}{\pi }2p_y^1=\stackrel{*}{\pi }2p_y^1$

$\mathrm{B}.\mathrm{O}.=\frac{6-2}{2}=2$

Having$2-$unpaired$e^{-}$so paramagnetic

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

$\mathrm{B}.\mathrm{O}.=\frac{6-3}{2}=1.5$

Having 1- unpaired$e^{-}$so paramagnetic bond order decrease

$\left(D\right)$$N_2=\sigma 1s^2<\stackrel{*}{\sigma }1s^2<\sigma 2s^2<\stackrel{*}{\sigma }2s^2<\pi 2p{y}^2=\pi 2p_y^2<\sigma 2p_x^2$

$\mathrm{B}.\mathrm{O}.=\frac{6-0}{2}=3$

No unpaired$e^{-}$so diamagnetic

$N_2^{+}\to \sigma 1s^2<\stackrel{*}{\pi }1s^2<\sigma 2s^2<\stackrel{*}{\sigma }2s^2<\pi 2p_y^2=\pi 2p_z^2<\sigma 2p_x^1$

$\mathrm{B}.\mathrm{O}.=\frac{5-0}{2}=2.5$

Having one unpaired$e^{-}$so paramagnetic

Bond order increases$N_2$to$N_2^{+}$