In the `MO` diagram for `O_(2)^(Theta)` ion the highest occupied orbital is .

Options

A. `pi MO` orbital
B. `sigma MO` orbital
C. `pi^(**) MO` orbital
D. `sigma MO` orbital

Answer

To determine the highest occupied molecular orbital (HOMO) in the molecular orbital (MO) diagram for the \( O_2^- \) ion, we can follow these steps:

### Step 1: Determine the Electronic Configuration of Oxygen
Oxygen has an atomic number of 8. The electronic configuration can be written as:
- \( 1s^2 \)
- \( 2s^2 \)
- \( 2p^4 \)

### Step 2: Account for the Extra Electron in the Ion
Since we are dealing with the \( O_2^- \) ion, we need to add one extra electron to the configuration of \( O_2 \). The neutral \( O_2 \) molecule has 16 electrons (8 from each oxygen atom), so the \( O_2^- \) ion will have:
- Total electrons = 16 + 1 = 17 electrons.

### Step 3: Fill the Molecular Orbitals
The molecular orbitals for \( O_2 \) and \( O_2^- \) can be filled as follows:
- The order of filling for \( O_2 \) is:
- \( \sigma_{2s}^2 \)
- \( \sigma_{2s}^*^2 \)
- \( \sigma_{2p_z}^2 \)
- \( \pi_{2p_x}^2 \)
- \( \pi_{2p_y}^2 \)
- \( \pi_{2p_x}^* \) and \( \pi_{2p_y}^* \) (for \( O_2 \), both are empty)

For \( O_2^- \):
- Fill the extra electron into one of the \( \pi^* \) orbitals:
- \( \sigma_{2s}^2 \)
- \( \sigma_{2s}^*^2 \)
- \( \sigma_{2p_z}^2 \)
- \( \pi_{2p_x}^2 \)
- \( \pi_{2p_y}^2 \)
- \( \pi_{2p_x}^*^1 \) (the extra electron goes here)

### Step 4: Identify the Highest Occupied Orbital
The highest occupied molecular orbital (HOMO) in the \( O_2^- \) ion is the \( \pi_{2p_x}^* \) orbital, as it is the last orbital that gets filled with electrons.

### Conclusion
Thus, the highest occupied orbital in the \( O_2^- \) ion is \( \pi^* \).