JEE MAIN - Chemistry (2019 - 9th January Evening Slot - No. 16)
In which of the following process, the bond order has increased and paramagnetic character has charged to diamagnetic ?
NO $$ \to $$ NO+
N2 $$ \to $$ N2+
O2 $$ \to $$ O2+
O2 $$ \to $$ O22$$-$$
Explanation
Molecular orbital configuration of NO (15 electrons) is
$${\sigma _{1{s^2}}}\,\sigma _{1{s^2}}^ * \,{\sigma _{2{s^2}}}\,\sigma _{2{s^2}}^ * \,{\sigma _{2p_z^2}}\,{\pi _{2p_x^2}}\, = \,{\pi _{2p_y^2}}\,\pi _{2p_x^1}^ * \, = \,\pi _{2p_y^o}^ * $$
$$\therefore\,\,\,\,$$ Nb = 10
Na = 5
$$\therefore\,\,\,\,$$ BO = $${1 \over 2}\left[ {10 - 5} \right]$$ = 2.5
And in NO one unpaired electron is present , so it is paramagnetic.
Similarly Molecular orbital configuration of NO+ (14 electrons) is
$${\sigma _{1{s^2}}}\,\sigma _{1{s^2}}^ * \,{\sigma _{2{s^2}}}\,\sigma _{2{s^2}}^ * \,{\sigma _{2p_z^2}}\,{\pi _{2p_x^2}}\, = \,{\pi _{2p_y^2}}\,$$
$$\therefore\,\,\,\,$$ Nb = 10
Na = 4
$$\therefore\,\,\,\,$$ BO = $${1 \over 2}\left[ {10 - 4} \right]$$ = 3
And in NO+ no unpaired electron is present , so it is diamagnetic.
$${\sigma _{1{s^2}}}\,\sigma _{1{s^2}}^ * \,{\sigma _{2{s^2}}}\,\sigma _{2{s^2}}^ * \,{\sigma _{2p_z^2}}\,{\pi _{2p_x^2}}\, = \,{\pi _{2p_y^2}}\,\pi _{2p_x^1}^ * \, = \,\pi _{2p_y^o}^ * $$
$$\therefore\,\,\,\,$$ Nb = 10
Na = 5
$$\therefore\,\,\,\,$$ BO = $${1 \over 2}\left[ {10 - 5} \right]$$ = 2.5
And in NO one unpaired electron is present , so it is paramagnetic.
Similarly Molecular orbital configuration of NO+ (14 electrons) is
$${\sigma _{1{s^2}}}\,\sigma _{1{s^2}}^ * \,{\sigma _{2{s^2}}}\,\sigma _{2{s^2}}^ * \,{\sigma _{2p_z^2}}\,{\pi _{2p_x^2}}\, = \,{\pi _{2p_y^2}}\,$$
$$\therefore\,\,\,\,$$ Nb = 10
Na = 4
$$\therefore\,\,\,\,$$ BO = $${1 \over 2}\left[ {10 - 4} \right]$$ = 3
And in NO+ no unpaired electron is present , so it is diamagnetic.
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