JEE Advance - Chemistry (2016 - Paper 2 Offline - No. 7)
According to Molecular Orbital Theory, which of the following statements is(are) correct?
$$C_2^{2-}$$ is expected to be diamagnetic
$$O_2^{2+}$$ expected to have a longer bond length than O2
$$N_2^+$$ and $$N_2^-$$ have the same bond order
$$He_2^+$$ has the same energy as two isolated He atoms
Explanation
Option (A) : According to molecular orbital theory, the arrangement of the electrons in the molecular orbitals is as follows :
For $$C_2^{2 - }$$, the total number of electrons is 14. The molecular orbital configuration is
$${\sigma _{1{s^2}}}\,\sigma _{1{s^2}}^ * \,{\sigma _{2{s^2}}}\,\sigma _{2{s^2}}^ * \,{\pi _{2p_x^2}}\, = \,{\pi _{2p_y^2}}\,{\sigma _{2p_z^2}}$$
There is no unpaired electron and thus it is diamagnetic.
Option (B) : For $$O_2^{2 + }$$, the total number of electrons is 14. The molecular orbital configuration is
$${\sigma _{1{s^2}}}\,\,\sigma _{1{s^2}}^ * \,$$ $${\sigma _{2{s^2}}}\,\,\sigma _{2{s^2}}^ * \,$$ $${\sigma _{2p_z^2}}\,\,{\pi _{2p_x^2}} = {\pi _{2p_y^2}}\,\,\pi _{2p_x^0}^ * \,\, = \pi _{2p_y^0}^ * $$
Thus the bond order = (10 $$-$$ 4)/2 = 3;
Molecular orbital configuration of O2 (16 electrons) is
$${\sigma _{1{s^2}}}\,\,\sigma _{1{s^2}}^ * \,$$ $${\sigma _{2{s^2}}}\,\,\sigma _{2{s^2}}^ * \,$$ $${\sigma _{2p_z^2}}\,\,{\pi _{2p_x^2}} = {\pi _{2p_y^2}}\,\,\pi _{2p_x^1}^ * \,\, = \pi _{2p_y^1}^ * $$
So O$$_2$$ has 2 unpaired electrons. for O2, bond order = 2.
As bond order is inversely proportional to bond length, thus the bond length of $$O_2^{2 + }$$ is less than the bond length of O2.
Option (C) : For $$N_2^ + $$, the total number of electrons is 13. The molecular orbital configuration is
$${\sigma _{1{s^2}}}\,\sigma _{1{s^2}}^ * \,{\sigma _{2{s^2}}}\,\sigma _{2{s^2}}^ * \,{\pi _{2p_x^2}}\, = \,{\pi _{2p_y^2}}\,{\sigma _{2p_z^1}}$$
Bond order of $$N_2^ + $$ = (9 $$-$$ 4)/2 = 2.5
$$N_2^{ - }$$ has 15 electrons.
Moleculer orbital configuration of $$N_2^{ - }$$ is
$${\sigma _{1{s^2}}}\,\sigma _{1{s^2}}^ * \,{\sigma _{2{s^2}}}\,\sigma _{2{s^2}}^ * \,{\pi _{2p_x^2}}\, = \,{\pi _{2p_y^2}}\,{\sigma _{2p_z^2}}\,\pi _{2p_x^1}^ * \, = \,\pi _{2p_y^0}^ *$$
$$\therefore\,\,\,\,$$Na = 5
Nb = 10
$$\therefore\,\,\,\,$$ BO = $${1 \over 2}\left[ {10 - 5} \right] = 2.5$$
Thus, the bond orders are the same.
Option (D) : As some energy is released during the formation of $$He_2^ + $$ from two isolated He atoms, thus it has lesser energy as compared to two isolated He atoms.
For $$C_2^{2 - }$$, the total number of electrons is 14. The molecular orbital configuration is
$${\sigma _{1{s^2}}}\,\sigma _{1{s^2}}^ * \,{\sigma _{2{s^2}}}\,\sigma _{2{s^2}}^ * \,{\pi _{2p_x^2}}\, = \,{\pi _{2p_y^2}}\,{\sigma _{2p_z^2}}$$
There is no unpaired electron and thus it is diamagnetic.
Option (B) : For $$O_2^{2 + }$$, the total number of electrons is 14. The molecular orbital configuration is
$${\sigma _{1{s^2}}}\,\,\sigma _{1{s^2}}^ * \,$$ $${\sigma _{2{s^2}}}\,\,\sigma _{2{s^2}}^ * \,$$ $${\sigma _{2p_z^2}}\,\,{\pi _{2p_x^2}} = {\pi _{2p_y^2}}\,\,\pi _{2p_x^0}^ * \,\, = \pi _{2p_y^0}^ * $$
Thus the bond order = (10 $$-$$ 4)/2 = 3;
Molecular orbital configuration of O2 (16 electrons) is
$${\sigma _{1{s^2}}}\,\,\sigma _{1{s^2}}^ * \,$$ $${\sigma _{2{s^2}}}\,\,\sigma _{2{s^2}}^ * \,$$ $${\sigma _{2p_z^2}}\,\,{\pi _{2p_x^2}} = {\pi _{2p_y^2}}\,\,\pi _{2p_x^1}^ * \,\, = \pi _{2p_y^1}^ * $$
So O$$_2$$ has 2 unpaired electrons. for O2, bond order = 2.
As bond order is inversely proportional to bond length, thus the bond length of $$O_2^{2 + }$$ is less than the bond length of O2.
Option (C) : For $$N_2^ + $$, the total number of electrons is 13. The molecular orbital configuration is
$${\sigma _{1{s^2}}}\,\sigma _{1{s^2}}^ * \,{\sigma _{2{s^2}}}\,\sigma _{2{s^2}}^ * \,{\pi _{2p_x^2}}\, = \,{\pi _{2p_y^2}}\,{\sigma _{2p_z^1}}$$
Bond order of $$N_2^ + $$ = (9 $$-$$ 4)/2 = 2.5
$$N_2^{ - }$$ has 15 electrons.
Moleculer orbital configuration of $$N_2^{ - }$$ is
$${\sigma _{1{s^2}}}\,\sigma _{1{s^2}}^ * \,{\sigma _{2{s^2}}}\,\sigma _{2{s^2}}^ * \,{\pi _{2p_x^2}}\, = \,{\pi _{2p_y^2}}\,{\sigma _{2p_z^2}}\,\pi _{2p_x^1}^ * \, = \,\pi _{2p_y^0}^ *$$
$$\therefore\,\,\,\,$$Na = 5
Nb = 10
$$\therefore\,\,\,\,$$ BO = $${1 \over 2}\left[ {10 - 5} \right] = 2.5$$
Thus, the bond orders are the same.
Option (D) : As some energy is released during the formation of $$He_2^ + $$ from two isolated He atoms, thus it has lesser energy as compared to two isolated He atoms.
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