According to MO theory which of the lists ran | vedclass

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Question

$\mathrm{N}_{2}:(\sigma 1 s)^{2}\left(\sigma^{*} 1 s\right)^{2}(\sigma 2 s)^{2}$$\left(\sigma^{*} 2 s\right)^{2}\left(\pi 2 p_{x}\right)^{2}\left(\pi

Vedclass · Accepted Answer

$N_2^{2-} < N_2^- < N_2$

Vedclass · Answer

$\mathrm{N}_{2}:(\sigma 1 s)^{2}\left(\sigma^{*} 1 sight)^{2}(\sigma 2 s)^{2}$$\left(\sigma^{*} 2 sight)^{2}\left(\pi 2 p_{x}ight)^{2}\left(\pi 2 p_{y}ight)^{2}\left(\sigma 2 p_{z}ight)^{2}$

$	herefore \quad \mathrm{B} . \mathrm{O}.=\frac{10-4}{2}=3$

$\mathrm{N}_{2}^{-}:(\sigma 1 s)^{2}\left(\sigma^{*} 1 sight)^{2}(\sigma 2 s)^{2}\left(\sigma^{*} 2 sight)^{2}\left(\pi 2 p_{x}ight)^{2}$

$\left(\pi 2 p_{y}ight)^{2}\left(\sigma 2 p_{z}ight)^{2}\left(\pi^{*} 2 p_{x}ight)^{1}$

$	herefore \quad \mathrm{B}. \mathrm{O}.=\frac{10-5}{2}=2.5$

$\mathrm{N}_{2}^{2-}:(\sigma 1 s)^{2}\left(\sigma^{*} 1 sight)^{2}(\sigma 2 s)^{2}\left(\sigma^{*} 2 sight)^{2}\left(\pi 2 p_{x}ight)^{2}$

$\left(\pi 2 p_{y}ight)^{2}\left(\sigma 2 p_{z}ight)^{2}\left(\pi^{*} 2 p_{x}ight)^{1}\left(\pi^{*} 2 p_{y}ight)^{1}$

$	herefore \quad \mathrm{B} . \mathrm{O} .=\frac{10-6}{2}=2$

Hence the order: $\mathrm{N}_{2}^{2-}<\mathrm{N}_{2}^{-}<\mathrm{N}_{2}$

According to $MO$ theory which of the lists ranks the nitrogen species in terms of increasing bond order?

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