Question

Which of the following facts regarding change in bond length is correct?

• A. Increases in going from $N_2$ to $N_2^{+}$, decreases in going from $O_2$ to $O_2^{+}$
• B. Decreases in going from $N_2$ to $N_2^{+}$, increases in going from $O_2$ to $O_2^{+}$
• C. Increases in going from $N_2$ to $N_2^{+}$ and $O_2$ to $O_2^{+}$
• D. Decrease in going from $N_2$ to $N_2^{+}$ and $O_2$ to $O_2^{+}$

Answer 1

The correct option is A Increases in going from $N_2$ to $N_2^{+}$, decreases in going from $O_2$ to $O_2^{+}$

Bond order = $1/2\times$[(Number of electrons in bonding molecules) - (Number of electrons in antibonding molecules)]

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

1. $N_2$:

valence electrons=$5+5=10$

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

BO=$1/2\times (6-0)=3$;

2. $N_2^{+}$:

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

3.$O_2$:

valence electrons=$6+6=12$

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

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

3.$O_2^{+}$:

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

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

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

Since, bond length $\propto \frac{1}{bondorder}$

Order of bond length is: $N_2$<$N_2^{+}$ and $O_2$>$O_2^{+}$

Hence, option A is correct.