The correct stability order for N_2 and its g | vedclass

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Question

Bond stability $\propto$ bond order

Bond order $=1 / 2 	imes\left(\mathrm{N}_{\mathrm{B}}-\mathrm{N}_{\mathrm{AB}}ight)$

Order of filling MO

Vedclass · Accepted Answer

$N_2 > N_2^+ >  N_2^- >  N_2^{-2}$

Vedclass · Answer

Bond stability $\propto$ bond order

Bond order $=1 / 2 	imes\left(\mathrm{N}_{\mathrm{B}}-\mathrm{N}_{\mathrm{AB}}ight)$

Order of filling MO is:

1. $\mathrm{N}_{2}$ :

valence electrons $=5+5=10$

$(\sigma 2 \mathrm{s})^{2}(\sigma * 2 \mathrm{s})^{2}(\sigma 2 \mathrm{p})^{2}(\pi 2 \mathrm{p})^{4}$

$\mathrm{BO}=1 / 2 	imes(6-0)=3$

2. $\mathrm{N}_{2}^{+}$

valence electrons $=5+5-1=9$

$(\sigma 2 \mathrm{s})^{2}(\sigma * 2 \mathrm{s})^{2}(\sigma 2 \mathrm{p})^{2}(\pi 2 \mathrm{p})^{3}$

$\mathrm{BO}=1 / 2 	imes(5-0)=2.5$

$3 . \mathrm{N}_{2}^{-}$

valence electrons $=5+5+1=11$

$(\sigma 2 s)^{2}(\sigma * 2 s)^{2}(\sigma 2 p)^{2}(\pi 2 p)^{4}\left(\pi^{*} 2 pight)^{1}$

$\mathrm{BO}=1 / 2 	imes(6-1)=2.5$

4. $\mathrm{N}_{2}^{2-}$

valence electrons $=5+5+2=12$

$(\sigma 2 \mathrm{s})^{2}(\sigma * 2 \mathrm{s})^{2}(\sigma 2 \mathrm{p})^{2}(\pi 2 \mathrm{p})^{4}\left(\pi^{*} 2 \mathrm{p}ight)^{2}$

$\mathrm{BO}=1 / 2 	imes(6-2)=2$

bond strength or stability: $\mathrm{N}_{2}>\mathrm{N}_{2}^{+}>\mathrm{N}_{2}^{-}>\mathrm{N}_{2}^{2-}$

$\mathrm{N}_{2}^{+}$

here $\mathrm{N}_{2}^{+}$ and $\mathrm{N}_{2}^{-}$ have same bond order but later has one electron in antibonding orbital which reduces its stability as compared to $\mathrm{N}_{2}^{+}$

Therefore, A is correct.

The correct stability order for $N_2$ and its given ions is :-

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