Which of the following molecule is paramagnetic

Options

A. Chlorine
B. Nitrogen
C. Oxygen
D. Hydrogen

Explanation

To determine which of the given molecules is paramagnetic, we need to analyze the electron configurations of the molecules in question. The molecules provided are Cl2, N2, O2, and H2.

### Step-by-Step Solution:

1. **Understanding Paramagnetism and Diamagnetism**:
- A molecule is paramagnetic if it has unpaired electrons in its molecular orbitals.
- A molecule is diamagnetic if all its electrons are paired.

2. **Counting Electrons**:
- We need to determine the total number of valence electrons for each molecule:
- **Cl2**: Each chlorine atom has 7 valence electrons. Thus, Cl2 has \(7 + 7 = 14\) valence electrons.
- **N2**: Each nitrogen atom has 5 valence electrons. Thus, N2 has \(5 + 5 = 10\) valence electrons.
- **O2**: Each oxygen atom has 6 valence electrons. Thus, O2 has \(6 + 6 = 12\) valence electrons.
- **H2**: Each hydrogen atom has 1 valence electron. Thus, H2 has \(1 + 1 = 2\) valence electrons.

3. **Molecular Orbital Theory (MOT) Configuration**:
- We will use the Molecular Orbital Theory to fill the molecular orbitals for O2:
- The order of filling for O2 is:
- \(\sigma_{1s}^2\)
- \(\sigma_{1s}^*^2\)
- \(\sigma_{2s}^2\)
- \(\sigma_{2s}^*^2\)
- \(\sigma_{2p_z}^2\)
- \(\pi_{2p_x}^2\)
- \(\pi_{2p_y}^2\)
- \(\pi_{2p_x}^*^1\)
- \(\pi_{2p_y}^*^1\)

4. **Identifying Unpaired Electrons**:
- From the above filling, we see that O2 has two unpaired electrons in the \(\pi^*\) orbitals (one in each of the \(\pi_{2p_x}^*\) and \(\pi_{2p_y}^*\) orbitals).
- Therefore, O2 is paramagnetic because it has unpaired electrons.

5. **Conclusion**:
- Among the given options (Cl2, N2, O2, H2), the molecule that is paramagnetic is **O2**.

### Final Answer:
The paramagnetic molecule is **O2**.