JEE MAIN - Chemistry (2019 - 8th April Evening Slot - No. 13)
Among the following molecules / ions,
$$C_2^{2 - },N_2^{2 - },O_2^{2 - },{O_2}$$
which one is diamagnetic and has the shortest
bond length?
$${O_2}$$
$$C_2^{2 - }$$
$$N_2^{2 - }$$
$$O_2^{2 - }$$
Explanation
Note :
(1) $$\,\,\,\,$$ Bond strength $$ \propto $$ Bond order
(2) $$\,\,\,\,$$ Bond length $$ \propto $$ $${1 \over {Bond\,\,order}}$$
(3) $$\,$$ Bond order $$ = {1 \over 2}$$ [Nb $$-$$ Na]
Nb = No of electrons in bending molecular orbital
Na $$=$$ No of electrons in anti bonding molecular orbital
(4) $$\,\,\,\,$$ upto 14 electrons, molecular orbital configuration is
_8th_April_Evening_Slot_en_13_1.png)
Here Na = Anti bonding electron $$=$$ 4 and Nb = 10
(5) $$\,\,\,\,$$ After 14 electrons to 20 electrons molecular orbital configuration is - - -
_8th_April_Evening_Slot_en_13_2.png)
Here Na = 10
and Nb = 10
(A) In O atom 8 electrons present, so in O2, 8 $$ \times $$ 2 = 16 electrons present.
Then in $$O_2^{2 - }$$ no of electrons = 18
$$\therefore\,\,\,\,$$ 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 are present so it is paramagnetic.
(D) Molecular orbital configuration of O $$_2^{2 - }$$ (18 electrons) is
$${\sigma _{1{s^2}}}\,\sigma _{1{s^2}}^ * \,{\sigma _{2{s^2}}}\,\sigma _{2{s^2}}^ * \,{\sigma _{2p_z^2}}\,{\pi _{2p_x^2}}\, = \,{\pi _{2p_y^2}}\,\pi _{2p_x^2}^ * \, = \,\pi _{2p_y^2}^ * $$
$$\therefore\,\,\,\,$$ Nb = 10
Na = 8
$$\therefore\,\,\,\,$$ BO = $${1 \over 2}$$ [ 10 $$-$$ 8] = 1
Here no unpaired electrons are present so it is diamagnetic.
(B) $$C_2^{2 - }$$ has 14 electrons.
Moleculer orbital configuration of $$C_2^{2 - }$$ is
$${\sigma _{1{s^2}}}\,\sigma _{1{s^2}}^ * \,{\sigma _{2{s^2}}}\,\sigma _{2{s^2}}^ * \,{\pi _{2p_x^2}}\, = \,{\pi _{2p_y^2}}\,{\sigma _{2p_z^2}}$$
$$\therefore\,\,\,\,$$Na = 4
Nb = 10
$$\therefore\,\,\,\,$$ BO = $${1 \over 2}\left[ {10 - 4} \right] = 3$$
Here no unpaired electron present, so it is diamagnetic.
(C) $$N_2^{2 - }$$ has 16 electrons.
Moleculer orbital configuration of $$N_2^{2 - }$$ is
$${\sigma _{1{s^2}}}\,\sigma _{1{s^2}}^ * \,{\sigma _{2{s^2}}}\,\sigma _{2{s^2}}^ * \,{\pi _{2p_x^2}}\, = \,{\pi _{2p_y^2}}\,{\sigma _{2p_z^2}}\,\pi _{2p_x^1}^ * \, = \,\pi _{2p_y^1}^ *$$
$$\therefore\,\,\,\,$$Na = 6
Nb = 10
$$\therefore\,\,\,\,$$ BO = $${1 \over 2}\left[ {10 - 6} \right] = 2$$
Here 2 unpaired electron present, so it is paramagnetic.
As Bond length $$ \propto $$ $${1 \over {Bond\,\,order}}$$
so among two diamagnetic ions $$C_2^{2 - }$$ and $$O_2^{2 - }$$, bond order of $$C_2^{2 - }$$ is more so it will have shorter bond length.
(1) $$\,\,\,\,$$ Bond strength $$ \propto $$ Bond order
(2) $$\,\,\,\,$$ Bond length $$ \propto $$ $${1 \over {Bond\,\,order}}$$
(3) $$\,$$ Bond order $$ = {1 \over 2}$$ [Nb $$-$$ Na]
Nb = No of electrons in bending molecular orbital
Na $$=$$ No of electrons in anti bonding molecular orbital
(4) $$\,\,\,\,$$ upto 14 electrons, molecular orbital configuration is
Here Na = Anti bonding electron $$=$$ 4 and Nb = 10
(5) $$\,\,\,\,$$ After 14 electrons to 20 electrons molecular orbital configuration is - - -
Here Na = 10
and Nb = 10
(A) In O atom 8 electrons present, so in O2, 8 $$ \times $$ 2 = 16 electrons present.
Then in $$O_2^{2 - }$$ no of electrons = 18
$$\therefore\,\,\,\,$$ 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 are present so it is paramagnetic.
(D) Molecular orbital configuration of O $$_2^{2 - }$$ (18 electrons) is
$${\sigma _{1{s^2}}}\,\sigma _{1{s^2}}^ * \,{\sigma _{2{s^2}}}\,\sigma _{2{s^2}}^ * \,{\sigma _{2p_z^2}}\,{\pi _{2p_x^2}}\, = \,{\pi _{2p_y^2}}\,\pi _{2p_x^2}^ * \, = \,\pi _{2p_y^2}^ * $$
$$\therefore\,\,\,\,$$ Nb = 10
Na = 8
$$\therefore\,\,\,\,$$ BO = $${1 \over 2}$$ [ 10 $$-$$ 8] = 1
Here no unpaired electrons are present so it is diamagnetic.
(B) $$C_2^{2 - }$$ has 14 electrons.
Moleculer orbital configuration of $$C_2^{2 - }$$ is
$${\sigma _{1{s^2}}}\,\sigma _{1{s^2}}^ * \,{\sigma _{2{s^2}}}\,\sigma _{2{s^2}}^ * \,{\pi _{2p_x^2}}\, = \,{\pi _{2p_y^2}}\,{\sigma _{2p_z^2}}$$
$$\therefore\,\,\,\,$$Na = 4
Nb = 10
$$\therefore\,\,\,\,$$ BO = $${1 \over 2}\left[ {10 - 4} \right] = 3$$
Here no unpaired electron present, so it is diamagnetic.
(C) $$N_2^{2 - }$$ has 16 electrons.
Moleculer orbital configuration of $$N_2^{2 - }$$ is
$${\sigma _{1{s^2}}}\,\sigma _{1{s^2}}^ * \,{\sigma _{2{s^2}}}\,\sigma _{2{s^2}}^ * \,{\pi _{2p_x^2}}\, = \,{\pi _{2p_y^2}}\,{\sigma _{2p_z^2}}\,\pi _{2p_x^1}^ * \, = \,\pi _{2p_y^1}^ *$$
$$\therefore\,\,\,\,$$Na = 6
Nb = 10
$$\therefore\,\,\,\,$$ BO = $${1 \over 2}\left[ {10 - 6} \right] = 2$$
Here 2 unpaired electron present, so it is paramagnetic.
As Bond length $$ \propto $$ $${1 \over {Bond\,\,order}}$$
so among two diamagnetic ions $$C_2^{2 - }$$ and $$O_2^{2 - }$$, bond order of $$C_2^{2 - }$$ is more so it will have shorter bond length.
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