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$N_2$ and $O_2$ are converted in to $N_2^+$ and $O_2^+$ respectively. Which of the following is not correct ?
In $N_2^+$ , the $N-N$ bond weakens
In $O_2^+$ , the $O-O$ bond length decreases
In $O_2^+$ , paramagnetism decreases
$N_2^+$ becomes dimagnetic
Solution
Bond order $=1 / 2 \times\left(N_B-N_{A B}\right)$
Order of filling $MO$ is:
$1.\,N _2$ :
valence electrons $=5+5=10$
$(\sigma 2 s )^2(\sigma * 2 s )^2(\sigma 2 p )^2(\pi 2 p )^4$
$BO =1 / 2 \times(6-0)=3$
$2.\, N _2^{+}$:
valence electrons $=5+5-1=9$
$(\sigma 2 s )^2(\sigma * 2 s )^2(\sigma 2 p )^2(\pi 2 p )^3$
$BO =1 / 2 \times(5-0)=2.5$
$3 . \,O _2 \text { : }$
$\text { valence electrons }=6+6=12$
$(\sigma 2 s )^2(\sigma * 2 s )^2(\sigma 2 p )^2(\pi 2 p )^4\left(\pi^* 2 p \right)^2$
$BO =1 / 2 \times(6-2)=2$
$4.\,O _2^{+}$:
valence electrons $=6+6-1=11$
$(\sigma 2 s )^2(\sigma * 2 s )^2(\sigma 2 p )^2(\pi 2 p )^4\left(\pi^* 2 p \right)^1$
$BO =1 / 2 \times(6-1)=2.5 \text {; }$
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.
