Question

Fill in the blanks with appropriate choice.

Bond order of $N_{2}^{+}$is P while that of $N_{2}$is Q.

Bond order of $O_{2}^{+}$is R while that of $O_{2}$is S

N – N bond distance T when $N_{2}$changes to $N_{2}^{+}$and when $O_{2}$changes to $O_{2}^{+}$the O – O bond distance U

P Q R S T U

• A. 2 $2.5`$ $2.5`$ 1 increases decreases
• B. $2.5$ 3 2 $1.5$ decreases increases
• C. 3 2 $1.5$ 1 increases decreases
• D. $2.5$ 3 $2.5`$ 2 increases decreases

Answer 1

Answer: (D)

$2.5$ 3 $2.5`$ 2 increases decreases

Answer 2

: $N_{2}^{+}:(\sigma 1s^{2})(\sigma*1s^{2})(\sigma 2s^{2})(\sigma*2s^{2})(\pi 2p_{x}^{2} = \pi 2p_{y}^{2})(\sigma 2p_{z}^{1})$ B.O. $= \frac{1}{2} \times (9 - 4) = 2.5$

$N_{2}:B.O. = \frac{1}{2} \times (10 - 4) = 3$

$O_{2}^{+}(\sigma 1s^{2})(\sigma*1s^{2})(\sigma 2s^{2})(\sigma*2s^{2})(\sigma 2p_{z}^{2})$

$(\pi 2py_{x}^{2} = \pi 2p_{y}^{2})(\pi*2p_{x}^{1})$

B.O. $= \frac{1}{2} \times (10 - 5) = 2.5$

$O_{2}:B.O. = \frac{1}{2} \times (10 - 6) = 2$

Since $N_{2}^{+}$has lower bond order than $N_{2}$bond length of N – N in $N_{2}^{+}$increases. In $O_{2}^{+}$bond order increases from 2 to 2.5 hence, bond length decreases.