Question

Which charge(s) on the $N_2$ molecule would give a bond order of $2.5$?

• A. $+1$
• B. $+2$
• C. $-1$
• D. $-2$

Answer 1

Answer: (A), (C)

$+1$
$-1$

1. $N_2^{+}$:

valence electrons=$5+5-1=9$

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

BO=$1/2\times (5-0)=2.5$;

1. $N_2^{+2}$:

valence electrons=$5+5-2=8$

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

BO=$1/2\times (4-0)=2$;

3.$N_2^{-1}$:

valence electrons=$5+5+1=11$

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

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

4. $N_2^{-2}$

valence electrons=$5+5+2=12$

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

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

Therefore $+1$ and $-1$ charge on $N_2$ has bond order 0f $2.5$.