Which of the following is paramagnetic?

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(C) To determine the magnetic behavior,we use Molecular Orbital Theory $(MOT)$: $N_2$ ($14$ electrons): $KK \sigma(2s)^2 \sigma^*(2s)^2 \pi(2p_x)^2 \p

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$N_2^+$

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(C) To determine the magnetic behavior,we use Molecular Orbital Theory $(MOT)$: $N_2$ ($14$ electrons): $KK \sigma(2s)^2 \sigma^*(2s)^2 \pi(2p_x)^2 \pi(2p_y)^2 \sigma(2p_z)^2$. All electrons are paired,so it is diamagnetic. $C_2$ ($12$ electrons): $KK \sigma(2s)^2 \sigma^*(2s)^2 \pi(2p_x)^2 \pi(2p_y)^2$. All electrons are paired,so it is diamagnetic. $N_2^+$ ($13$ electrons): $KK \sigma(2s)^2 \sigma^*(2s)^2 \pi(2p_x)^2 \pi(2p_y)^2 \sigma(2p_z)^1$. There is one unpaired electron in the $\sigma(2p_z)$ orbital,so it is paramagnetic. $O_2^{2-}$ ($18$ electrons): $KK \sigma(2s)^2 \sigma^*(2s)^2 \sigma(2p_z)^2 \pi(2p_x)^2 \pi(2p_y)^2 \pi^*(2p_x)^2 \pi^*(2p_y)^2$. All electrons are paired,so it is diamagnetic. Therefore,the correct option is $C$.

List-$I$	List-$II$
$A. Li_2$	$i. 3$
$B. N_2$	$ii. 1.5$
$C. Be_2$	$iii. 1.0$
$D. O_2$	$iv. 0$
	$v. 2$

Similar Questions

Match the molecules in List-$I$ with their respective bond orders in List-$II$.
List-$I$ List-$II$
$A. Li_2$ $i. 3$
$B. N_2$ $ii. 1.5$
$C. Be_2$ $iii. 1.0$
$D. O_2$ $iv. 0$
$v. 2$

The bond order is not $3$ for

The bond order in $O_2^+$ is

Number of molecules having bond order $2$ from the following molecules is: $C_2, O_2, Be_2, Li_2, Ne_2, N_2, He_2$

Molecular orbital theory was developed mainly by

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