Question

The species having highest bond order is:

• A. $O_2$
• B. $O_2^{-}$
• C. $O_2^{+}$
• D. $O_2^{2-}$

Answer 1

Answer: (C)

$O_2^{+}$

$O_2$:

valence electrons=$6+6=12$

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

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

$O_2^{+}$:

valence electrons=$6+6-1=11$

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

BO=$1/2\times (6-1)=2.5$;

$O_2^{-}$:

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

$O_2^{2-}$:

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

highest bond order is for $O_2^{+}$