JEE MAIN - Chemistry (2021 - 26th February Evening Shift - No. 1)
Match List - I with List - II.
Choose the correct answer from the options given below :
| List - I (Molecule) | List - II (Bond order) | ||
|---|---|---|---|
| (a) | $$N{e_2}$$ | (i) | 1 |
| (b) | $${N_2}$$ | (ii) | 2 |
| (c) | $${F_2}$$ | (iii) | 0 |
| (d) | $${O_2}$$ | (iv) | 3 |
Choose the correct answer from the options given below :
(a) $$ \to $$ (i), (b) $$ \to $$ (ii), (c) $$ \to $$ (iii), (d) $$ \to $$ (iv)
(a) $$ \to $$ (iv), (b) $$ \to $$ (iii), (c) $$ \to $$ (ii), (d) $$ \to $$ (i)
(a) $$ \to $$ (iii), (b) $$ \to $$ (iv), (c) $$ \to $$ (i), (d) $$ \to $$ (ii)
(a) $$ \to $$ (ii), (b) $$ \to $$ (i), (c) $$ \to $$ (iv), (d) $$ \to $$ (iii)
Explanation
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
_26th_February_Evening_Shift_en_1_1.png)
Here Na = Anti bonding electron $$=$$ 4 and Nb = 10
(5) $$\,\,\,\,$$ After 14 electrons to 20 electrons molecular orbital configuration is - - -
_26th_February_Evening_Shift_en_1_2.png)
Here Na = 10
and Nb = 10
(a) Molecular orbital configuration of Ne2 (20 electrons) is
$${\sigma _{1{s^2}}}\,\,\sigma _{1{s^2}}^ * \,$$ $${\sigma _{2{s^2}}}\,\,\sigma _{2{s^2}}^ * \,$$ $${\sigma _{2p_z^2}}\,\,{\pi _{2p_x^2}} = {\pi _{2p_y^2}}\,\,\pi _{2p_x^2}^ * \,\, = \pi _{2p_y^2}^ * $$$$\sigma _{2p_z^2}^*$$
$$\therefore\,\,\,\,$$Na = 10
Nb = 10
$$\therefore\,\,\,\,$$ BO = $${1 \over 2}\left[ {10 - 10} \right] = 0$$
(Note : All inert gases has BO = 0 and it does not exist as molecule. Here Ne is also an inert gas.)
(b) Moleculer orbital configuration of $$N_2$$ (14 electrons) 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 = $${1 \over 2}\left[ {10 - 4} \right] = 3$$
(d) Molecular orbital configuration of F2 (18 electrons) is
$${\sigma _{1{s^2}}}\,\,\sigma _{1{s^2}}^ * \,$$ $${\sigma _{2{s^2}}}\,\,\sigma _{2{s^2}}^ * \,$$ $${\sigma _{2p_z^2}}\,\,{\pi _{2p_x^2}} = {\pi _{2p_y^2}}\,\,\pi _{2p_x^2}^ * \,\, = \pi _{2p_y^2}^ * $$
$$\therefore\,\,\,\,$$Na = 8
Nb = 10
$$\therefore\,\,\,\,$$ BO = $${1 \over 2}\left[ {10 - 8} \right] = 1$$
(d) 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 = $${1 \over 2}\left[ {10 - 6} \right] = 2$$
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
(a) Molecular orbital configuration of Ne2 (20 electrons) is
$${\sigma _{1{s^2}}}\,\,\sigma _{1{s^2}}^ * \,$$ $${\sigma _{2{s^2}}}\,\,\sigma _{2{s^2}}^ * \,$$ $${\sigma _{2p_z^2}}\,\,{\pi _{2p_x^2}} = {\pi _{2p_y^2}}\,\,\pi _{2p_x^2}^ * \,\, = \pi _{2p_y^2}^ * $$$$\sigma _{2p_z^2}^*$$
$$\therefore\,\,\,\,$$Na = 10
Nb = 10
$$\therefore\,\,\,\,$$ BO = $${1 \over 2}\left[ {10 - 10} \right] = 0$$
(Note : All inert gases has BO = 0 and it does not exist as molecule. Here Ne is also an inert gas.)
(b) Moleculer orbital configuration of $$N_2$$ (14 electrons) 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 = $${1 \over 2}\left[ {10 - 4} \right] = 3$$
(d) Molecular orbital configuration of F2 (18 electrons) is
$${\sigma _{1{s^2}}}\,\,\sigma _{1{s^2}}^ * \,$$ $${\sigma _{2{s^2}}}\,\,\sigma _{2{s^2}}^ * \,$$ $${\sigma _{2p_z^2}}\,\,{\pi _{2p_x^2}} = {\pi _{2p_y^2}}\,\,\pi _{2p_x^2}^ * \,\, = \pi _{2p_y^2}^ * $$
$$\therefore\,\,\,\,$$Na = 8
Nb = 10
$$\therefore\,\,\,\,$$ BO = $${1 \over 2}\left[ {10 - 8} \right] = 1$$
(d) 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 = $${1 \over 2}\left[ {10 - 6} \right] = 2$$
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