Fill in the blanks with appropriate choice.
Bond order of $$N_2^ + $$  is $$\underline P $$  while that of $${N_2}$$  is $$\underline Q .$$  Bond order of $$O_2^ + $$  is $$\underline R $$  while that of $${O_2}$$  is $$\underline S .$$  $$N - N$$  bond distance $$\underline T ,$$  when $${N_2}$$  changes to $$N_2^ + $$  and when $${O_2}$$  changes to $$O_2^ + ,$$  the $$O - O$$  bond distance is $$\underline U .$$

	$$P$$	$$Q$$	$$R$$	$$S$$	$$T$$	$$U$$
(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

Solution :
$$N_2^ + :\left( {\sigma 1{s^2}} \right)\left( {{\sigma ^ * }1{s^2}} \right)\left( {\sigma 2{s^2}} \right)$$     $$\left( {{\sigma ^ * }2{s^2}} \right)\left( {\pi 2p_x^2 = \pi 2p_y^2} \right)\left( {\sigma 2p_z^1} \right)$$
$$\eqalign{ & B.O. = \frac{1}{2} \times \left( {9 - 4} \right) = 2.5 \cr & {N_2}:B.O. = \frac{1}{2} \times \left( {10 - 4} \right) = 3 \cr} $$
$$O_2^ + :\left( {\sigma 1{s^2}} \right)\left( {{\sigma ^ * }1{s^2}} \right)\left( {\sigma 2{s^2}} \right)$$     $$\left( {{\sigma ^ * }2{s^2}} \right)\left( {\sigma 2p_z^2} \right)\left( {\pi 2p_x^2 = \pi 2p_y^2} \right)$$      $$\left( {{\pi ^ * }2p_x^1} \right)$$
$$B.O. = \frac{1}{2} \times \left( {10 - 5} \right) = 2.5;$$      $${O_2}:B.O. = \frac{1}{2} \times \left( {10 - 6} \right) = 2$$
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.