Among ` ko_2 , AlO_(2)^(-) Bao_2` and `NO_2^+` unpaired electron is present in :

Options

A. `NO_2^+` and `BaO_2`
B. ` KO_2` and `AlO_2^-`
C. ` KO_2` only
D. `BaO_2` only

Explanation

To determine which of the compounds \( KO_2, AlO_2^-, BaO_2, \) and \( NO_2^+ \) has unpaired electrons, we will analyze each compound step by step.

### Step 1: Analyze \( KO_2 \)
- **Composition**: \( KO_2 \) consists of \( K^+ \) and \( O_2^{2-} \).
- **Oxygen Analysis**: The \( O_2^{2-} \) ion can be analyzed using Molecular Orbital Theory (MOT).
- **Electronic Configuration**: For \( O_2^{2-} \), the electronic configuration is:
- \( 1s^2 \, 2s^2 \, 2p^6 \) (for two oxygen atoms, each contributing \( 2p^5 \) and gaining 2 electrons).
- **Filling Order**: The electrons fill the molecular orbitals as follows:
- \( \sigma_{1s}^2 \, \sigma_{1s}^* \, \sigma_{2s}^2 \, \sigma_{2s}^* \, \sigma_{2p_z}^2 \, \pi_{2p_x}^2 \, \pi_{2p_y}^2 \, \pi_{2p_x}^* \, \pi_{2p_y}^* \)
- **Unpaired Electrons**: There is one unpaired electron in the \( \pi^* \) orbitals.

### Step 2: Analyze \( AlO_2^- \)
- **Composition**: \( AlO_2^- \) consists of \( Al^{3+} \) and \( O_2^{2-} \).
- **Aluminum Analysis**: Aluminum has a valence of 3 and forms bonds with oxygen.
- **Electronic Configuration**: The \( O_2^{2-} \) has already been analyzed above.
- **Bonding**: The aluminum atom forms three bonds with oxygen, leading to a stable configuration with no unpaired electrons.

### Step 3: Analyze \( BaO_2 \)
- **Composition**: \( BaO_2 \) consists of \( Ba^{2+} \) and \( O_2^{2-} \).
- **Barium Analysis**: Barium is a group 2 element and loses 2 electrons to form \( Ba^{2+} \).
- **Electronic Configuration**: The \( O_2^{2-} \) has already been analyzed.
- **Bonding**: The \( O_2^{2-} \) has no unpaired electrons, and \( Ba^{2+} \) also has no unpaired electrons.

### Step 4: Analyze \( NO_2^+ \)
- **Composition**: \( NO_2^+ \) consists of nitrogen and two oxygen atoms.
- **Nitrogen Analysis**: Nitrogen has a valence of 5, and with the positive charge, it effectively has 4 valence electrons.
- **Electronic Configuration**: The structure can be analyzed similarly to \( O_2 \).
- **Bonding**: The nitrogen forms bonds with the two oxygen atoms, leading to a stable configuration with no unpaired electrons.

### Conclusion
After analyzing all four compounds, we find that:
- \( KO_2 \) has one unpaired electron.
- \( AlO_2^- \), \( BaO_2 \), and \( NO_2^+ \) have no unpaired electrons.

Thus, the compound with unpaired electrons is **\( KO_2 \)**.