JEE MAIN - Chemistry (2018 - 15th April Morning Slot - No. 19)
In the molecular orbital diagram for the molecular ion, N2+, the number of electrons in the $$\sigma $$2pz molecular orbital is :
0
1
2
3
Explanation
Electrons in $$N_2^ + $$ = 7 $$ \times $$ 2 $$-$$ 1 = 13
Moleculer orbital configuration of $$N_2^ + $$ (13)
= $${\sigma _{1{s^2}}}\,\sigma _{1{s^2}}^ * \,{\sigma _{2{s^2}}}\,\sigma _{2{s^2}}^ * \,{\pi _{2p_x^2}}\, = \,{\pi _{2p_y^2}}\,{\sigma _{2p_z^1}}$$
$$\therefore\,\,\,$$ $$\,{\sigma _{2{p_z}}}$$ no. of electrons = 1.
Moleculer orbital configuration of $$N_2^ + $$ (13)
= $${\sigma _{1{s^2}}}\,\sigma _{1{s^2}}^ * \,{\sigma _{2{s^2}}}\,\sigma _{2{s^2}}^ * \,{\pi _{2p_x^2}}\, = \,{\pi _{2p_y^2}}\,{\sigma _{2p_z^1}}$$
$$\therefore\,\,\,$$ $$\,{\sigma _{2{p_z}}}$$ no. of electrons = 1.
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