N_2 and O_2 are converted in to N_2^+ and O_2 | vedclass

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Question

Bond order $=1 / 2 	imes\left(N_B-N_{A B}ight)$

Order of filling $MO$ is:

$1.\,N _2$ :

valence electrons $=5+5=10$

$(\sigma 2 s )^2(\si

Vedclass · Accepted Answer

$N_2^+$ becomes dimagnetic

Vedclass · Answer

Bond order $=1 / 2 	imes\left(N_B-N_{A B}ight)$

Order of filling $MO$ is:

$1.\,N _2$ :

valence electrons $=5+5=10$

$(\sigma 2 s )^2(\sigma * 2 s )^2(\sigma 2 p )^2(\pi 2 p )^4$

$BO =1 / 2 	imes(6-0)=3$

$2.\, N _2^{+}$:

valence electrons $=5+5-1=9$

$(\sigma 2 s )^2(\sigma * 2 s )^2(\sigma 2 p )^2(\pi 2 p )^3$

$BO =1 / 2 	imes(5-0)=2.5$

$3 . \,O _2 	ext { : }$

$	ext { valence electrons }=6+6=12$

$(\sigma 2 s )^2(\sigma * 2 s )^2(\sigma 2 p )^2(\pi 2 p )^4\left(\pi^* 2 p ight)^2$

$BO =1 / 2 	imes(6-2)=2$

$4.\,O _2^{+}$:

valence electrons $=6+6-1=11$

$(\sigma 2 s )^2(\sigma * 2 s )^2(\sigma 2 p )^2(\pi 2 p )^4\left(\pi^* 2 p ight)^1$

$BO =1 / 2 	imes(6-1)=2.5 	ext {; }$

Bond strength $\propto$ bond order

bond strength/order: $N _2\,> \,N _2^{+}$ and $O _2^{+}\,>\, O _2$

$N _2^{+}$has one unpaired electron therefore it is paramagnetic and not diamagnetic. Therefore $D$ is correct.

$N_2$ and $O_2$ are converted in to $N_2^+$ and $O_2^+$ respectively. Which of the following is not correct ?

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