Question
Which of the following pairs will have same bond order?
A.
$${F_2}\,{\text{and}}\,O_2^{2 - }$$
B.
$${N_2}\,{\text{and}}\,C{O_2}$$
C.
$${O_2}\,{\text{and}}\,O_2^ - $$
D.
$${N_2}\,{\text{and}}\,N_2^ + $$
Answer :
$${F_2}\,{\text{and}}\,O_2^{2 - }$$
Solution :
$${F_2}\,{\text{and}}\,O_2^{2 - }\,{\text{are}}\,\,{\text{isoelectronic}}{\text{.}}$$
$${F_2}:\left( {\sigma 1{s^2}} \right)\left( {{\sigma ^ * }1{s^2}} \right)\left( {\sigma 2{s^2}} \right)$$ $$\left( {{\sigma ^ * }2{s^2}} \right)\left( {\sigma 2p_z^2} \right)\left( {\pi 2p_x^2 = \pi 2p_y^2} \right)$$ $$\left( {{\pi ^ * }2p_x^2 = {\pi ^ * }2p_y^2} \right)$$
$$B.O. = \frac{1}{2} \times \left( {10 - 8} \right) = 1$$
$$O_2^{2 - }:\left( {\sigma 1{s^2}} \right)\left( {{\sigma ^ * }1{s^2}} \right)\left( {\sigma 2{s^2}} \right)$$ $$\left( {{\sigma ^ * }2{s^2}} \right)\left( {\sigma 2p_z^2} \right)\left( {\pi 2p_x^2 = \pi 2p_y^2} \right)$$ $$\left( {{\pi ^ * }2p_x^2 = {\pi ^ * }2p_y^2} \right)$$
$$B.O. = \frac{{10 - 8}}{2} = 1$$