QJEEM24Chfebb: JEE Main 2024 Chemistry - Class 11 JEEM24Chfebb Solution: Number of molecules having bond order 2 from the following…

To determine the number of molecules with a bond order of 2 from the given molecules, we need to first calculate the bond order for each molecule. The bond order can be determined using Molecular Orbital Theory (MOT). The bond order is given by the formula:

$\text{Bond Order} = \frac{\text{Number of bonding electrons} - \text{Number of anti-bonding electrons}}{2}$

Let's evaluate each molecule individually:

1. $\mathrm{C}_2$ :

For $\mathrm{C}_2$ , the total number of electrons is 12. The molecular orbital configuration will be $\left( \sigma_{1s} \right)^2 \left( \sigma_{1s}^* \right)^2 \left( \sigma_{2s} \right)^2 \left( \sigma_{2s}^* \right)^2 \left( \pi_{2p_x} \right)^2 \left( \pi_{2p_y} \right)^2$ . There are 8 bonding electrons and 4 anti-bonding electrons:

$\text{Bond Order} = \frac{8 - 4}{2} = 2$

2. $\mathrm{O}_2$ :

For $\mathrm{O}_2$ , the total number of electrons is 16. The molecular orbital configuration will be $\left( \sigma_{1s} \right)^2 \left( \sigma_{1s}^* \right)^2 \left( \sigma_{2s} \right)^2 \left( \sigma_{2s}^* \right)^2 \left( \sigma_{2p_z} \right)^2 \left( \pi_{2p_x} \right)^2 \left( \pi_{2p_y} \right)^2 \left( \pi_{2p_x}^* \right)^1 \left( \pi_{2p_y}^* \right)^1$ . There are 10 bonding electrons and 6 anti-bonding electrons:

$\text{Bond Order} = \frac{10 - 6}{2} = 2$

3. $\mathrm{Be}_2$ :

For $\mathrm{Be}_2$ , the total number of electrons is 8. The molecular orbital configuration will be $\left( \sigma_{1s} \right)^2 \left( \sigma_{1s}^* \right)^2 \left( \sigma_{2s} \right)^2 \left( \sigma_{2s}^* \right)^2$ . There are 4 bonding electrons and 4 anti-bonding electrons:

$\text{Bond Order} = \frac{4 - 4}{2} = 0$

4. $\mathrm{Li}_2$ :

For $\mathrm{Li}_2$ , the total number of electrons is 6. The molecular orbital configuration will be $\left( \sigma_{1s} \right)^2 \left( \sigma_{1s}^* \right)^2 \left( \sigma_{2s} \right)^2$ . There are 4 bonding electrons and 2 anti-bonding electrons:

$\text{Bond Order} = \frac{4 - 2}{2} = 1$

5. $\mathrm{Ne}_2$ :

For $\mathrm{Ne}_2$ , the total number of electrons is 20. The molecular orbital configuration will be $\left( \sigma_{1s} \right)^2 \left( \sigma_{1s}^* \right)^2 \left( \sigma_{2s} \right)^2 \left( \sigma_{2s}^* \right)^2 \left( \sigma_{2p_z} \right)^2 \left( \pi_{2p_x} \right)^2 \left( \pi_{2p_y} \right)^2 \left( \pi_{2p_x}^* \right)^2 \left( \pi_{2p_y}^* \right)^2 \left( \sigma_{2p_z}^* \right)^2$ . There are 10 bonding electrons and 10 anti-bonding electrons:

$\text{Bond Order} = \frac{10 - 10}{2} = 0$

6. $\mathrm{N}_2$ :

For $\mathrm{N}_2$ , the total number of electrons is 14. The molecular orbital configuration will be $\left( \sigma_{1s} \right)^2 \left( \sigma_{1s}^* \right)^2 \left( \sigma_{2s} \right)^2 \left( \sigma_{2s}^* \right)^2 \left( \sigma_{2p_z} \right)^2 \left( \pi_{2p_x} \right)^2 \left( \pi_{2p_y} \right)^2$ . There are 10 bonding electrons and 4 anti-bonding electrons:

$\text{Bond Order} = \frac{10 - 4}{2} = 3$

7. $\mathrm{He}_2$ :

For $\mathrm{He}_2$ , the total number of electrons is 4. The molecular orbital configuration will be $\left( \sigma_{1s} \right)^2 \left( \sigma_{1s}^* \right)^2$ . There are 2 bonding electrons and 2 anti-bonding electrons:

$\text{Bond Order} = \frac{2 - 2}{2} = 0$

Thus, the molecules with a bond order of 2 from the given list are $\mathrm{C}_2$ and $\mathrm{O}_2$ . Therefore, the number of molecules having bond order 2 is 2.