JEE MAIN - Chemistry (2020 - 9th January Morning Slot - No. 3)
If the magnetic moment of a dioxygen species
is 1.73 B.M, it may be :
$$O_2^ - $$ or $$O_2^ + $$
O2, $$O_2^ - $$ or $$O_2^ + $$
O2 or $$O_2^ + $$
O2 or $$O_2^ - $$
Explanation
Magnetic moment = 1.73 BM
$$ \therefore $$ Unpaired electron = 1
(1) $$O_2$$ has 16 electrons.
Moleculer orbital configuration of $$O_2$$ is
$${\sigma _{1{s^2}}}\,\sigma _{1{s^2}}^ * \,\,{\sigma _{2{s^2}}}\,\,\sigma _{2{s^2}}^ * \,\,{\sigma _{2p_z^2}}\,\,{\pi _{2p_x^2}}\,= \,{\pi _{2p_y^2}}\,\pi _{2p_x^1}^ * = \,\,\pi _{2p_y^1}^ * $$
Here 2 unpaired electron present.
(2) $$O_2^{+}$$ has 15 electrons.
Moleculer orbital configuration of $$O_2^{+}$$ is
$${\sigma _{1{s^2}}}\,\sigma _{1{s^2}}^ * \,\,{\sigma _{2{s^2}}}\,\,\sigma _{2{s^2}}^ * \,\,{\sigma _{2p_z^2}}\,\,{\pi _{2p_x^2}}\,= \,{\pi _{2p_y^2}}\,\pi _{2p_x^1}^ * = \,\,\pi _{2p_y^0}^ * $$
Here 1 unpaired electron present.
(3) $$O_2^{-}$$ has 17 electrons.
Moleculer orbital configuration of $$O_2^{-}$$ is
$${\sigma _{1{s^2}}}\,\sigma _{1{s^2}}^ * \,\,{\sigma _{2{s^2}}}\,\,\sigma _{2{s^2}}^ * \,\,{\sigma _{2p_z^2}}\,\,{\pi _{2p_x^2}}\,= \,{\pi _{2p_y^2}}\,\pi _{2p_x^2}^ * = \,\,\pi _{2p_y^1}^ * $$
Here 1 unpaired electron present.
Hence $$O_2^{-}$$ & $$O_2^{+}$$ have one unpaired electron.
$$ \therefore $$ Unpaired electron = 1
(1) $$O_2$$ has 16 electrons.
Moleculer orbital configuration of $$O_2$$ is
$${\sigma _{1{s^2}}}\,\sigma _{1{s^2}}^ * \,\,{\sigma _{2{s^2}}}\,\,\sigma _{2{s^2}}^ * \,\,{\sigma _{2p_z^2}}\,\,{\pi _{2p_x^2}}\,= \,{\pi _{2p_y^2}}\,\pi _{2p_x^1}^ * = \,\,\pi _{2p_y^1}^ * $$
Here 2 unpaired electron present.
(2) $$O_2^{+}$$ has 15 electrons.
Moleculer orbital configuration of $$O_2^{+}$$ is
$${\sigma _{1{s^2}}}\,\sigma _{1{s^2}}^ * \,\,{\sigma _{2{s^2}}}\,\,\sigma _{2{s^2}}^ * \,\,{\sigma _{2p_z^2}}\,\,{\pi _{2p_x^2}}\,= \,{\pi _{2p_y^2}}\,\pi _{2p_x^1}^ * = \,\,\pi _{2p_y^0}^ * $$
Here 1 unpaired electron present.
(3) $$O_2^{-}$$ has 17 electrons.
Moleculer orbital configuration of $$O_2^{-}$$ is
$${\sigma _{1{s^2}}}\,\sigma _{1{s^2}}^ * \,\,{\sigma _{2{s^2}}}\,\,\sigma _{2{s^2}}^ * \,\,{\sigma _{2p_z^2}}\,\,{\pi _{2p_x^2}}\,= \,{\pi _{2p_y^2}}\,\pi _{2p_x^2}^ * = \,\,\pi _{2p_y^1}^ * $$
Here 1 unpaired electron present.
Hence $$O_2^{-}$$ & $$O_2^{+}$$ have one unpaired electron.
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