JEE Main 2024 — 8 April, Shift 2 — Question 79

Using molecular orbital theory, we have

$$\text{Bond \;order} = \frac{N_b - N_a}{2}$$

For the given molecules:

1. $\mathrm{C_2}$ (12 e$^-$): MO filling: $(\sigma_{2s})^2(\sigma_{2s}^*)^2(\pi_{2p_x})^2(\pi_{2p_y})^2$

$$N_b = 8,\ N_a = 4 \Rightarrow \text{B.O.} = \frac{8-4}{2} = 2$$

2. $\mathrm{O_2}$ (16 e$^-$): MO filling: $(\sigma_{2s})^2(\sigma_{2s}^*)^2(\sigma_{2p_z})^2(\pi_{2p_x})^2(\pi_{2p_y})^2(\pi_{2p_x}^*)^1(\pi_{2p_y}^*)^1$

$$N_b = 10,\ N_a = 6 \Rightarrow \text{B.O.} = \frac{10-6}{2} = 2$$

3. $\mathrm{Be_2}$ (8 e$^-$):

$$N_b = 4,\ N_a = 4 \Rightarrow \text{B.O.} = 0$$

4. $\mathrm{Li_2}$ (6 e$^-$):

$$N_b = 4,\ N_a = 2 \Rightarrow \text{B.O.} = \frac{4-2}{2} = 1$$

5. $\mathrm{Ne_2}$ (20 e$^-$):

$$N_b = 10,\ N_a = 10 \Rightarrow \text{B.O.} = 0$$

6. $\mathrm{N_2}$ (14 e$^-$):

$$N_b = 10,\ N_a = 4 \Rightarrow \text{B.O.} = \frac{10-4}{2} = 3$$

7. $\mathrm{He_2}$ (4 e$^-$):

$$N_b = 2,\ N_a = 2 \Rightarrow \text{B.O.} = 0$$

Thus, the molecules having bond order $2$ are $\mathrm{C_2\ \;and\;\ O_2}$. Number of such molecules = $2$.

Thus, the correct answer is $2$.