Draw the energy level diagram of `H_(2)^(+)` ion and discuss its stability in terms of bond order.

Answer

### Step-by-Step Solution:

1. **Identify the Ion and its Electrons:**
- The ion in question is \( H_2^+ \). This ion is derived from the hydrogen molecule \( H_2 \) by removing one electron.
- The total number of electrons in \( H_2^+ \) is 1 (since \( H_2 \) has 2 electrons and we remove 1).

2. **Draw the Molecular Orbital Diagram:**
- The molecular orbitals for \( H_2 \) are formed from the combination of the atomic orbitals of the two hydrogen atoms.
- The bonding molecular orbital is denoted as \( \sigma_{1s} \) and the antibonding molecular orbital is denoted as \( \sigma_{1s}^* \).
- In \( H_2^+ \), we fill the available molecular orbitals with the single electron:
- \( \sigma_{1s}^1 \) (1 electron in the bonding orbital)
- \( \sigma_{1s}^* \) (0 electrons in the antibonding orbital)

3. **Write the Molecular Orbital Configuration:**
- The molecular orbital configuration for \( H_2^+ \) is:
\[ \sigma_{1s}^1 \]

4. **Calculate the Bond Order:**
- The bond order can be calculated using the formula:
\[ \text{Bond Order} = \frac{N_b - N_a}{2} \]
where \( N_b \) is the number of bonding electrons and \( N_a \) is the number of antibonding electrons.
- For \( H_2^+ \):
- \( N_b = 1 \) (1 electron in the bonding orbital)
- \( N_a = 0 \) (0 electrons in the antibonding orbital)
- Substituting the values:
\[ \text{Bond Order} = \frac{1 - 0}{2} = 0.5 \]

5. **Discuss the Stability:**
- A bond order of 0.5 indicates that the bond between the two hydrogen atoms in \( H_2^+ \) is relatively weak.
- Generally, a bond order of 1 or greater indicates a stable bond, while a bond order less than 1 suggests that the bond is less stable or partially formed.
- Therefore, \( H_2^+ \) is less stable compared to \( H_2 \) (which has a bond order of 1).

### Summary:
- The energy level diagram for \( H_2^+ \) shows one electron in the bonding molecular orbital \( \sigma_{1s} \) and none in the antibonding orbital \( \sigma_{1s}^* \).
- The bond order of \( H_2^+ \) is 0.5, indicating that it is less stable than the neutral \( H_2 \) molecule.