JEE MAIN - Chemistry (2019 - 10th January Morning Slot - No. 17)
Two pi and half sigma bonds are present in :
O2
N$$_2^ + $$
O$$_2^ + $$
N2
Explanation
Two pi and half sigma bonds are presents in molecule with bond order 2.5.
Moleculer orbital configuration of $$N_2^+$$ (13 electrons)
= $${\sigma _{1{s^2}}}\,\sigma _{1{s^2}}^ * \,{\sigma _{2{s^2}}}\,\sigma _{2{s^2}}^ * \,{\pi _{2p_x^2}}\, = \,{\pi _{2p_y^2}}\,{\sigma _{2p_z^1}}$$
Bond order = $${1 \over 2}\left( {9 - 4} \right)$$ = 2.5
Moleculer orbital configuration of $$N_2^+$$ (13 electrons)
= $${\sigma _{1{s^2}}}\,\sigma _{1{s^2}}^ * \,{\sigma _{2{s^2}}}\,\sigma _{2{s^2}}^ * \,{\pi _{2p_x^2}}\, = \,{\pi _{2p_y^2}}\,{\sigma _{2p_z^1}}$$
Bond order = $${1 \over 2}\left( {9 - 4} \right)$$ = 2.5
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