JEE Main 2007 — Chemical Bonding And Molecular Structure Question with Solution

(A) Moleculer orbital configuration of $$C_2$$ (12 electrons)

= $${\sigma }_{1s^2}{\sigma }_{1s^2}^{\ast }{\sigma }_{2s^2}{\sigma }_{2s^2}^{\ast }{\pi }_{2p_x^2}={\pi }_{2p_y^2}$$

$$\therefore$$N_a = 4

N_b = 8

$$\therefore$$ BO = $$\frac{1}{2}\left[8-4\right]=2$$

Here no unpaired electron present, so it is diamagnetic.

Moleculer orbital configuration of $$C_2^{+}$$ (11 electrons)

= $${\sigma }_{1s^2}{\sigma }_{1s^2}^{\ast }{\sigma }_{2s^2}{\sigma }_{2s^2}^{\ast }{\pi }_{2p_x^2}={\pi }_{2p_y^1}$$

$$\therefore$$N_a = 4

N_b = 7

$$\therefore$$ BO = $$\frac{1}{2}\left[7-4\right]=1.5$$

Here 1 unpaired electron present, so it is paramagnetic.

(B) Moleculer orbital configuration of $$N_2$$ (14 electrons)

= $${\sigma }_{1s^2}{\sigma }_{1s^2}^{\ast }{\sigma }_{2s^2}{\sigma }_{2s^2}^{\ast }{\pi }_{2p_x^2}={\pi }_{2p_y^2}{\sigma }_{2p_z^2}$$

$$\therefore$$N_a = 4

N_b = 10

$$\therefore$$ BO = $$\frac{1}{2}\left[10-4\right]=3$$

Here no unpaired electron present, so it is diamagnetic.

Moleculer orbital configuration of $$N_2^{+}$$ (13 electrons)

= $${\sigma }_{1s^2}{\sigma }_{1s^2}^{\ast }{\sigma }_{2s^2}{\sigma }_{2s^2}^{\ast }{\pi }_{2p_x^2}={\pi }_{2p_y^2}{\sigma }_{2p_z^1}$$

$$\therefore$$N_a = 4

N_b = 9

$$\therefore$$ BO = $$\frac{1}{2}\left[9-4\right]=2.5$$

Here 1 unpaired electron present, so it is paramagnetic.

(C) Moleculer orbital configuration of NO (15 electrons) is

$${\sigma }_{1s^2}{\sigma }_{1s^2}^{\ast }{\sigma }_{2s^2}{\sigma }_{2s^2}^{\ast }{\sigma }_{2p_z^2}{\pi }_{2p_x^2}={\pi }_{2p_y^2}{\pi }_{2p_x^1}^{\ast }={\pi }_{2p_y^0}^{\ast }$$

$$\therefore$$N_a = 5

N_b = 10

$$\therefore$$ BO = $$\frac{1}{2}\left[10-5\right]=2.5$$

Here is 1 unpaired electron, So it is paramagnetic.

Moleculer orbital configuration of NO⁺ (14 electrons) is

$${\sigma }_{1s^2}{\sigma }_{1s^2}^{\ast }{\sigma }_{2s^2}{\sigma }_{2s^2}^{\ast }{\sigma }_{2p_z^2}{\pi }_{2p_x^2}={\pi }_{2p_y^2}{\pi }_{2p_x^0}^{\ast }={\pi }_{2p_y^0}^{\ast }$$

$$\therefore$$N_a = 4

N_b = 10

$$\therefore$$ BO = $$\frac{1}{2}\left[10-4\right]=3$$

Here is no unpaired electron, So it is diamagnetic.

(D) Molecular orbital configuration of O₂ (16 electrons) is

$${\sigma }_{1s^2}{\sigma }_{1s^2}^{\ast }$$ $${\sigma }_{2s^2}{\sigma }_{2s^2}^{\ast }$$ $${\sigma }_{2p_z^2}{\pi }_{2p_x^2}={\pi }_{2p_y^2}{\pi }_{2p_x^1}^{\ast }={\pi }_{2p_y^1}^{\ast }$$

$$\therefore$$N_a = 6

N_b = 10

$$\therefore$$ BO = $$\frac{1}{2}\left[10-6\right]=2$$

Here 2 unpaired electrons present, so it is paramagnetic.

Molecular orbital configuration of O$${}_2^{+}$$ (15 electrons) is

$${\sigma }_{1s^2}{\sigma }_{1s^2}^{\ast }{\sigma }_{2s^2}{\sigma }_{2s^2}^{\ast }{\sigma }_{2p_z^2}{\pi }_{2p_x^2}={\pi }_{2p_y^2}{\pi }_{2p_x^1}^{\ast }={\pi }_{2p_y^o}^{\ast }$$

$$\therefore$$ N_b = 10

N_a = 5

$$\therefore$$ BO = $$\frac{1}{2}\left[10-5\right]$$ = 2.5

Here 1 unpaired electrons present, so it is also paramagnetic.