Question

Among the following, which one is paramagnetic?

• A. $H{e}_2$
• B. $O_2^{-}$
• C. $O_2^{2-}$
• D. $N{e}_2$

Answer 1

The correct option is B $O_2^{-}$

Bond order = $1/2\times (N_B-N_{AB})$

Order of filling MO is:

1. $H{e}_2$:

valence electrons=$2+2=4$

${\left(\sigma 1s\right)}^2{\left({\sigma }^{\ast }1s\right)}^2$

BO=$1/2\times (2-2)=0$;

2. $O_2^{-}$:

valence electrons=$6+6+1=13$

${\left(\sigma 2s\right)}^2{(\sigma \ast 2s)}^2{\left(\sigma 2p\right)}^2{\left(\pi 2p\right)}^4{\left({\pi }^{\ast }2p\right)}^3$

BO=$1/2\times (6-3)=1.5$;

3. $O_2^{-2}$:

valence electrons=$6+6+2=14$

${\left(\sigma 2s\right)}^2{(\sigma \ast 2s)}^2{\left(\sigma 2p\right)}^2{\left(\pi 2p\right)}^4{\left({\pi }^{\ast }2p\right)}^4$

BO=$1/2\times (6-4)=1$;

4. 3. $N{e}_2$:

valence electrons=$8+8=16$

${\left(\sigma 2s\right)}^2{(\sigma \ast 2s)}^2{\left(\sigma 2p\right)}^2{\left(\pi 2p\right)}^4{\left({\pi }^{\ast }2p\right)}^4{\left({\sigma }^{\ast }2p\right)}^2$

BO=$1/2\times (6-6)=0$;

Among the given molecules only $O_2^{-2}$ has one unpaired electron and is therefore paramagnetic.

Therefore B is correct.