Question

Observe the following molecules / ions
$H _{2}, N _{2}, O _{2}, N _{2}^{+}, O _{2}^{+}, O _{2}^{-}, F _{2} .$ Identify correct
statement.

• A. $H _{2}, N _{2}, O _{2}, F _{2}$ show diamagnetic property
• B. $O _{2}, O _{2}^{+}, O _{2}^{-}, N _{2}^{+}$ show paramagnetic property
• C. $N _{2}, F _{2}, O _{2}^{+}, O _{2}^{-}$ show diamagnetic property
• D. $H _{2}, N _{2}^{+}, O _{2}^{+}, O _{2}^{-}$ show paramagnetic property

Answer 1

Answer: (B)

$H _{2}-(\sigma l s)^{2}$

$H _{2}$ does not have any unpaired electron hence diamagnetic in nature.

$N _{2} \rightarrow K K(\sigma 2 s )^{2}\left(\sigma^{*} 2 s \right)^{2}\left(\pi 2 p_{x}\right)^{2}\left(\pi 2 p_{y}\right)^{2}\left(\sigma p_{z}\right)^{2}$

$N _{2}$ does not have any unpaired electron hence diamagnetic in nature.

$O _{2} \rightarrow K K^{1}(\sigma 2 s)^{2}\left(\sigma^{\star} 2 s\right)^{2}\left(\sigma 2 p_{z}\right)^{2}\left(\pi 2 p_{x}\right)^{2}\left(\pi 2 p_{y}\right)^{2}$

$\left(\pi^{*} 2 p_{x}\right)^{1}\left(\pi^{\star} 2 p_{y}\right)^{1}$

$O _{2}$ have two unpaired electrons hence

paramagnetic in nature.

$N _{2}^{+} \rightarrow K K^{1}(\sigma 2 s)^{2}\left(\sigma^{*} 2 s\right)^{2}\left(\pi 2 p_{x}\right)^{2}\left(\pi 2 p_{y}\right)^{2}\left(\pi 2 p_{z}\right)^{1}$

$N _{2}$ have one unpaired electron, so it paramagnetic in nature.

$O _{2}^{+} \rightarrow K K^{\prime}(\sigma 2 s)^{2}\left(\sigma^{*} 2 s\right)^{2}\left(\sigma 2 p_{z}\right)^{2}\left(\pi 2 p_{x}\right)^{2}\left(\pi 2 p_{y}\right)^{2} $

$\left(\pi_{2}^{*} p_{x}\right)^{1}\left(\pi^{*} 2 p_{y}\right)^{0}$

$O _{2}^{+}$ has unpaired electron, so it is paramagnetic in nature.

$O _{2}^{-} \rightarrow K K^{\prime}(\sigma 2 s)^{2}\left(\sigma^{*} 2 s\right)^{2}\left(\sigma 2 p_{z}\right)^{2}\left(\pi 2 p_{x}\right)^{2}\left(\pi 2 p_{y}\right)^{2} $

$\left(\pi^{*} 2 p_{x}\right)^{2}\left(\pi^{*} 2 p_{y}\right)^{1} $

$F _{2} \rightarrow K K^{\prime}(\sigma 2 s)^{2}\left(\sigma^{*} 2 s\right)^{2}\left(\sigma 2 p_{z}\right)^{2}\left(\pi 2 p_{x}\right)^{2}\left(\pi 2 p_{y}\right)^{2} $

$\left(\pi^{*} 2 p_{x}\right)^{2}\left(\pi^{*} 2 p_{y}\right)^{2} $

$F _{2}$ does not have unpaired electron, hence diamagnetic in nature. Hence, the molecules, $N _{2}^{+}, O _{2}, O _{2}^{+}, O _{2}^{-}$ and $H _{2}, N _{2}, F _{2}$

are paramagnetic and diamagnetic respectively due to presence of unpaired and paired electrons.