JEE MAIN - Chemistry (2022 - 26th June Morning Shift - No. 3)
Consider the ions/molecule
O$$_2^ + $$, O2, O$$_2^ - $$, O$$_2^ {2-} $$
For increasing bond order the correct option is :
O$$_2^ {2-} $$ < O$$_2^ - $$ < O2 < O$$_2^ + $$
O$$_2^ - $$ < O$$_2^ {2-} $$ < O2 < O$$_2^ + $$
O$$_2^ - $$ < O$$_2^ {2-} $$ < O$$_2^ + $$ < O2
O$$_2^ - $$ < O$$_2^ + $$ < O$$_2^ {2-} $$ < O2
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 = Number of electrons in bonding molecular orbital
Na $$=$$ Number of electrons in anti bonding molecular orbital
(4) $$\,\,\,\,$$ upto 14 electrons, molecular orbital configuration is
_26th_June_Morning_Shift_en_3_1.png)
Here Na = Anti bonding electron $$=$$ 4 and Nb = 10
(5) $$\,\,\,\,$$ After 14 electrons to 20 electrons molecular orbital configuration is - - -
_26th_June_Morning_Shift_en_3_2.png)
Here Na = 10
and Nb = 10
In O atom 8 electrons present, so in O2, 8 $$ \times $$ 2 = 16 electrons present.
Then in $$O_2^ + $$ no of electrons = 15
in $$O_2^ - $$ no of electrons = 17
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 = $${1 \over 2}\left[ {10 - 6} \right] = 2$$
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
Molecular orbital configuration of $$O_2^ - $$ (17 electrons) is
$${\sigma _{1{s^2}}}\,\sigma _{1{s^2}}^ * \,{\sigma _{2{s^2}}}\,\sigma _{2{s^2}}^ * \,{\sigma _{2p_z^2}}\,{\pi _{2p_x^2}}\, = \,{\pi _{2p_y^2}}\,\pi _{2p_x^2}^ * \, = \,\pi _{2p_y^1}^ * $$
$$\therefore\,\,\,\,$$ Nb = 10
Na = 7
$$\therefore\,\,\,\,$$ BO = $${1 \over 2}\left[ {10 - 7} \right]$$ = 1.5
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
So, correct order of Bond order is
$$O_2^{2 - } < O_2^ - < {O_2} < O_2^ + $$
(1) $$\,\,\,\,$$ Bond strength $$ \propto $$ Bond order
(2) $$\,\,\,\,$$ Bond length $$ \propto $$ $${1 \over {Bond\,\,order}}$$
(3) $$\,$$ Bond order $$ = {1 \over 2}$$ [Nb $$-$$ Na]
Nb = Number of electrons in bonding molecular orbital
Na $$=$$ Number 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
In O atom 8 electrons present, so in O2, 8 $$ \times $$ 2 = 16 electrons present.
Then in $$O_2^ + $$ no of electrons = 15
in $$O_2^ - $$ no of electrons = 17
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 = $${1 \over 2}\left[ {10 - 6} \right] = 2$$
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
Molecular orbital configuration of $$O_2^ - $$ (17 electrons) is
$${\sigma _{1{s^2}}}\,\sigma _{1{s^2}}^ * \,{\sigma _{2{s^2}}}\,\sigma _{2{s^2}}^ * \,{\sigma _{2p_z^2}}\,{\pi _{2p_x^2}}\, = \,{\pi _{2p_y^2}}\,\pi _{2p_x^2}^ * \, = \,\pi _{2p_y^1}^ * $$
$$\therefore\,\,\,\,$$ Nb = 10
Na = 7
$$\therefore\,\,\,\,$$ BO = $${1 \over 2}\left[ {10 - 7} \right]$$ = 1.5
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
So, correct order of Bond order is
$$O_2^{2 - } < O_2^ - < {O_2} < O_2^ + $$
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