JEE MAIN - Chemistry (2007 - No. 33)
In which of the following ionization processes, the bond order has increased and the magnetic
behaviour has changed?
$$C_2 \to C_2^+$$
$$N_2 \to N_2^+$$
$$NO \to NO^+$$
$$O_2 \to O_2^+$$
Explanation
(A) Moleculer orbital configuration of $$C_2$$ (12 electrons)
= $${\sigma _{1{s^2}}}\,\sigma _{1{s^2}}^ * \,{\sigma _{2{s^2}}}\,\sigma _{2{s^2}}^ * \,{\pi _{2p_x^2}}\, = \,{\pi _{2p_y^2}}$$
$$\therefore\,\,\,\,$$Na = 4
Nb = 8
$$\therefore\,\,\,\,$$ BO = $${1 \over 2}\left[ {8 - 4} \right] = 2$$
Here no unpaired electron present, so it is diamagnetic.
Moleculer orbital configuration of $$C_2^+$$ (11 electrons)
= $${\sigma _{1{s^2}}}\,\sigma _{1{s^2}}^ * \,{\sigma _{2{s^2}}}\,\sigma _{2{s^2}}^ * \,{\pi _{2p_x^2}}\, = \,{\pi _{2p_y^1}}$$
$$\therefore\,\,\,\,$$Na = 4
Nb = 7
$$\therefore\,\,\,\,$$ BO = $${1 \over 2}\left[ {7 - 4} \right] = 1.5$$
Here 1 unpaired electron present, so it is paramagnetic.
(B) Moleculer orbital configuration of $$N_2$$ (14 electrons)
= $${\sigma _{1{s^2}}}\,\sigma _{1{s^2}}^ * \,{\sigma _{2{s^2}}}\,\sigma _{2{s^2}}^ * \,{\pi _{2p_x^2}}\, = \,{\pi _{2p_y^2}}\,{\sigma _{2p_z^2}}$$
$$\therefore\,\,\,\,$$Na = 4
Nb = 10
$$\therefore\,\,\,\,$$ BO = $${1 \over 2}\left[ {10 - 4} \right] = 3$$
Here no unpaired electron present, so it is diamagnetic.
Moleculer orbital configuration of $$N_2^+$$ (13 electrons)
= $${\sigma _{1{s^2}}}\,\sigma _{1{s^2}}^ * \,{\sigma _{2{s^2}}}\,\sigma _{2{s^2}}^ * \,{\pi _{2p_x^2}}\, = \,{\pi _{2p_y^2}}\,{\sigma _{2p_z^1}}$$
$$\therefore\,\,\,\,$$Na = 4
Nb = 9
$$\therefore\,\,\,\,$$ BO = $${1 \over 2}\left[ {9 - 4} \right] = 2.5$$
Here 1 unpaired electron present, so it is paramagnetic.
(C) Moleculer orbital configuration of NO (15 electrons) is
$${\sigma _{1{s^2}}}\,\sigma _{1{s^2}}^ * \,\,{\sigma _{2{s^2}}}\,\,\sigma _{2{s^2}}^ * \,\,{\sigma _{2p_z^2}}\,\,{\pi _{2p_x^2}}\,= \,{\pi _{2p_y^2}}\,\pi _{2p_x^1}^ * = \,\,\pi _{2p_y^0}^ * $$
$$\therefore\,\,\,\,$$Na = 5
Nb = 10
$$\therefore\,\,\,\,$$ BO = $${1 \over 2}\left[ {10 - 5} \right] = 2.5$$
Here is 1 unpaired electron, So it is paramagnetic.
Moleculer orbital configuration of NO+ (14 electrons) is
$${\sigma _{1{s^2}}}\,\sigma _{1{s^2}}^ * \,\,{\sigma _{2{s^2}}}\,\,\sigma _{2{s^2}}^ * \,\,{\sigma _{2p_z^2}}\,\,{\pi _{2p_x^2}}\,= \,{\pi _{2p_y^2}}\,\pi _{2p_x^0}^ * = \,\,\pi _{2p_y^0}^ * $$
$$\therefore\,\,\,\,$$Na = 4
Nb = 10
$$\therefore\,\,\,\,$$ BO = $${1 \over 2}\left[ {10 - 4} \right] = 3$$
Here is no unpaired electron, So it is diamagnetic.
(D) Molecular orbital configuration of O2 (16 electrons) is
$${\sigma _{1{s^2}}}\,\,\sigma _{1{s^2}}^ * \,$$ $${\sigma _{2{s^2}}}\,\,\sigma _{2{s^2}}^ * \,$$ $${\sigma _{2p_z^2}}\,\,{\pi _{2p_x^2}} = {\pi _{2p_y^2}}\,\,\pi _{2p_x^1}^ * \,\, = \pi _{2p_y^1}^ * $$
$$\therefore\,\,\,\,$$Na = 6
Nb = 10
$$\therefore\,\,\,\,$$ BO = $${1 \over 2}\left[ {10 - 6} \right] = 2$$
Here 2 unpaired electrons present, so it is paramagnetic.
Molecular orbital configuration of O$$_2^ + $$ (15 electrons) is
$${\sigma _{1{s^2}}}\,\sigma _{1{s^2}}^ * \,{\sigma _{2{s^2}}}\,\sigma _{2{s^2}}^ * \,{\sigma _{2p_z^2}}\,{\pi _{2p_x^2}}\, = \,{\pi _{2p_y^2}}\,\pi _{2p_x^1}^ * \, = \,\pi _{2p_y^o}^ * $$
$$\therefore\,\,\,\,$$ Nb = 10
Na = 5
$$\therefore\,\,\,\,$$ BO = $${1 \over 2}\left[ {10 - 5} \right]$$ = 2.5
Here 1 unpaired electrons present, so it is also paramagnetic.
= $${\sigma _{1{s^2}}}\,\sigma _{1{s^2}}^ * \,{\sigma _{2{s^2}}}\,\sigma _{2{s^2}}^ * \,{\pi _{2p_x^2}}\, = \,{\pi _{2p_y^2}}$$
$$\therefore\,\,\,\,$$Na = 4
Nb = 8
$$\therefore\,\,\,\,$$ BO = $${1 \over 2}\left[ {8 - 4} \right] = 2$$
Here no unpaired electron present, so it is diamagnetic.
Moleculer orbital configuration of $$C_2^+$$ (11 electrons)
= $${\sigma _{1{s^2}}}\,\sigma _{1{s^2}}^ * \,{\sigma _{2{s^2}}}\,\sigma _{2{s^2}}^ * \,{\pi _{2p_x^2}}\, = \,{\pi _{2p_y^1}}$$
$$\therefore\,\,\,\,$$Na = 4
Nb = 7
$$\therefore\,\,\,\,$$ BO = $${1 \over 2}\left[ {7 - 4} \right] = 1.5$$
Here 1 unpaired electron present, so it is paramagnetic.
(B) Moleculer orbital configuration of $$N_2$$ (14 electrons)
= $${\sigma _{1{s^2}}}\,\sigma _{1{s^2}}^ * \,{\sigma _{2{s^2}}}\,\sigma _{2{s^2}}^ * \,{\pi _{2p_x^2}}\, = \,{\pi _{2p_y^2}}\,{\sigma _{2p_z^2}}$$
$$\therefore\,\,\,\,$$Na = 4
Nb = 10
$$\therefore\,\,\,\,$$ BO = $${1 \over 2}\left[ {10 - 4} \right] = 3$$
Here no unpaired electron present, so it is diamagnetic.
Moleculer orbital configuration of $$N_2^+$$ (13 electrons)
= $${\sigma _{1{s^2}}}\,\sigma _{1{s^2}}^ * \,{\sigma _{2{s^2}}}\,\sigma _{2{s^2}}^ * \,{\pi _{2p_x^2}}\, = \,{\pi _{2p_y^2}}\,{\sigma _{2p_z^1}}$$
$$\therefore\,\,\,\,$$Na = 4
Nb = 9
$$\therefore\,\,\,\,$$ BO = $${1 \over 2}\left[ {9 - 4} \right] = 2.5$$
Here 1 unpaired electron present, so it is paramagnetic.
(C) Moleculer orbital configuration of NO (15 electrons) is
$${\sigma _{1{s^2}}}\,\sigma _{1{s^2}}^ * \,\,{\sigma _{2{s^2}}}\,\,\sigma _{2{s^2}}^ * \,\,{\sigma _{2p_z^2}}\,\,{\pi _{2p_x^2}}\,= \,{\pi _{2p_y^2}}\,\pi _{2p_x^1}^ * = \,\,\pi _{2p_y^0}^ * $$
$$\therefore\,\,\,\,$$Na = 5
Nb = 10
$$\therefore\,\,\,\,$$ BO = $${1 \over 2}\left[ {10 - 5} \right] = 2.5$$
Here is 1 unpaired electron, So it is paramagnetic.
Moleculer orbital configuration of NO+ (14 electrons) is
$${\sigma _{1{s^2}}}\,\sigma _{1{s^2}}^ * \,\,{\sigma _{2{s^2}}}\,\,\sigma _{2{s^2}}^ * \,\,{\sigma _{2p_z^2}}\,\,{\pi _{2p_x^2}}\,= \,{\pi _{2p_y^2}}\,\pi _{2p_x^0}^ * = \,\,\pi _{2p_y^0}^ * $$
$$\therefore\,\,\,\,$$Na = 4
Nb = 10
$$\therefore\,\,\,\,$$ BO = $${1 \over 2}\left[ {10 - 4} \right] = 3$$
Here is no unpaired electron, So it is diamagnetic.
(D) Molecular orbital configuration of O2 (16 electrons) is
$${\sigma _{1{s^2}}}\,\,\sigma _{1{s^2}}^ * \,$$ $${\sigma _{2{s^2}}}\,\,\sigma _{2{s^2}}^ * \,$$ $${\sigma _{2p_z^2}}\,\,{\pi _{2p_x^2}} = {\pi _{2p_y^2}}\,\,\pi _{2p_x^1}^ * \,\, = \pi _{2p_y^1}^ * $$
$$\therefore\,\,\,\,$$Na = 6
Nb = 10
$$\therefore\,\,\,\,$$ BO = $${1 \over 2}\left[ {10 - 6} \right] = 2$$
Here 2 unpaired electrons present, so it is paramagnetic.
Molecular orbital configuration of O$$_2^ + $$ (15 electrons) is
$${\sigma _{1{s^2}}}\,\sigma _{1{s^2}}^ * \,{\sigma _{2{s^2}}}\,\sigma _{2{s^2}}^ * \,{\sigma _{2p_z^2}}\,{\pi _{2p_x^2}}\, = \,{\pi _{2p_y^2}}\,\pi _{2p_x^1}^ * \, = \,\pi _{2p_y^o}^ * $$
$$\therefore\,\,\,\,$$ Nb = 10
Na = 5
$$\therefore\,\,\,\,$$ BO = $${1 \over 2}\left[ {10 - 5} \right]$$ = 2.5
Here 1 unpaired electrons present, so it is also paramagnetic.
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