Question

Which statement is correct about $O_2^{+}$ ?

• A. Paramagnetic and bond order < $O_2$
• B. Paramagnetic and bond order > $O_2$
• C. Diamagnetic and bond order < $O_2$
• D. Diamagnetic and bond order > $O_2$

Answer 1

The correct option is B Paramagnetic and bond order > $O_2$

From MOT Electronic configuration is:

$O_2^{+}$:${\left(\sigma 1s\right)}^2{(\sigma \ast 1s)}^2{\left(\sigma 2s\right)}^2{(\sigma 2\ast s)}^2{\left(\sigma 2p\right)}^2{\left(\pi 2p\right)}^4{(\pi \ast 2p)}^1$

Bond order = $1/2\times (N_B-N_{AB})=1/2\times (6-1)=2.5$

$O_2$ has bond order =$2$ and two unpaired electrons.

Therefore $O_2^{+}$ is paramagnetic due to unpaired electron and has higher bond order than $O_2$

Hence, $B$ is correct.