JEE MAIN - Chemistry (2021 - 25th February Morning Shift - No. 9)
According to molecular orbital theory, the species among the following that does not exist is :
$${O_2}^{2 - }$$
$$B{e_2}$$
$$H{e_2}^ - $$
$$H{e_2}^ + $$
Explanation
Note :
According to molecules orbital theory, when a molecule have bond order = 0 then that molecule does not exist.
We know, Bond order $$ = {1 \over 2}$$ [Nb $$-$$ Na]
Nb = No of electrons in bonding molecular orbital
Na $$=$$ No of electrons in anti bonding molecular orbital
(4) $$\,\,\,\,$$ upto 14 electrons, molecular orbital configuration is
_25th_February_Morning_Shift_en_9_1.png)
Here Na = Anti bonding electron $$=$$ 4 and Nb = 10
(5) $$\,\,\,\,$$ After 14 electrons to 20 electrons molecular orbital configuration is - - -
_25th_February_Morning_Shift_en_9_2.png)
Here Na = 10
and Nb = 10
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
$$\,\,\,$$ Configuration of $$He_2^ + $$ (3 electrons) is = $${\sigma _{1{s^2}}}\,\,\sigma _{1{s^1}}^ * $$
$$\therefore\,\,\,$$ Bond order = $${1 \over 2}$$ (2 $$-$$1) = 0.5
$$\,\,\,$$ Configuration of $$He_2^ - $$ (5 electrons) is = $${\sigma _{1{s^2}}}\,\,\sigma _{1{s^2}}^ * {\sigma _{2{s^1}}}\,$$
$$\therefore\,\,\,$$ Bond order = $${1 \over 2}$$ (3 $$-$$2) = 0.5
$$\,\,\,$$ Configuration of $$Be_2 $$ (4 electrons) is = $${\sigma _{1{s^2}}}$$ $$\sigma _{1{s^2}}^ * $$
$$\therefore$$ Bond order = $${1 \over 2}$$ (2 $$-$$ 2) = 0
According to molecules orbital theory, when a molecule have bond order = 0 then that molecule does not exist.
We know, Bond order $$ = {1 \over 2}$$ [Nb $$-$$ Na]
Nb = No of electrons in bonding 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
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
$$\,\,\,$$ Configuration of $$He_2^ + $$ (3 electrons) is = $${\sigma _{1{s^2}}}\,\,\sigma _{1{s^1}}^ * $$
$$\therefore\,\,\,$$ Bond order = $${1 \over 2}$$ (2 $$-$$1) = 0.5
$$\,\,\,$$ Configuration of $$He_2^ - $$ (5 electrons) is = $${\sigma _{1{s^2}}}\,\,\sigma _{1{s^2}}^ * {\sigma _{2{s^1}}}\,$$
$$\therefore\,\,\,$$ Bond order = $${1 \over 2}$$ (3 $$-$$2) = 0.5
$$\,\,\,$$ Configuration of $$Be_2 $$ (4 electrons) is = $${\sigma _{1{s^2}}}$$ $$\sigma _{1{s^2}}^ * $$
$$\therefore$$ Bond order = $${1 \over 2}$$ (2 $$-$$ 2) = 0
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