JEE MAIN - Chemistry (2008 - No. 28)
Which of the following pair of species have the same bond order?
CN- and NO+
CN- and CN+
$$O_2^-$$ and CN-
NO+ and CN+
Explanation
Number of electron in
NO+
= number of electron in CNā
= 14 electrons.
As both have same number of electrons so their bond order is equal.
Moleculer orbital configuration of NO+ (14 electrons) is
= $${\sigma _{1{s^2}}}\,\sigma _{1{s^2}}^ * \,\,{\sigma _{2{s^2}}}\,\,\sigma _{2{s^2}}^ * \,\,{\sigma _{2p_z^2}}\,\,{\pi _{2p_x^2}}\,= \,{\pi _{2p_y^2}}\,\pi _{2p_x^0}^ * = \,\,\pi _{2p_y^0}^ * $$
$$ \therefore $$ B.O = $${1 \over 2}\left[ {10 - 4} \right]$$ = 3
Moleculer orbital configuration of CN- (14 electrons) is
= $${\sigma _{1{s^2}}}\,\sigma _{1{s^2}}^ * \,{\sigma _{2{s^2}}}\,\sigma _{2{s^2}}^ * \,{\pi _{2p_x^2}}\, = \,{\pi _{2p_y^2}}\,{\sigma _{2p_z^2}}$$
$$ \therefore $$ B.O = $${1 \over 2}\left[ {10 - 4} \right]$$ = 3
As both have same number of electrons so their bond order is equal.
Moleculer orbital configuration of NO+ (14 electrons) is
= $${\sigma _{1{s^2}}}\,\sigma _{1{s^2}}^ * \,\,{\sigma _{2{s^2}}}\,\,\sigma _{2{s^2}}^ * \,\,{\sigma _{2p_z^2}}\,\,{\pi _{2p_x^2}}\,= \,{\pi _{2p_y^2}}\,\pi _{2p_x^0}^ * = \,\,\pi _{2p_y^0}^ * $$
$$ \therefore $$ B.O = $${1 \over 2}\left[ {10 - 4} \right]$$ = 3
Moleculer orbital configuration of CN- (14 electrons) is
= $${\sigma _{1{s^2}}}\,\sigma _{1{s^2}}^ * \,{\sigma _{2{s^2}}}\,\sigma _{2{s^2}}^ * \,{\pi _{2p_x^2}}\, = \,{\pi _{2p_y^2}}\,{\sigma _{2p_z^2}}$$
$$ \therefore $$ B.O = $${1 \over 2}\left[ {10 - 4} \right]$$ = 3
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