Among `NO_2^(+),KO_2and Na_2O_2 and NaAlO_2` the paramagnetism exist in -

Options

A. `Na_2O_2` only
B. `KO_2 and NO_2^+`
C. `Na_2O_2 and NaAlO_2`
D. `KO_2` only

Answer

To determine which of the compounds \( NO_2^+, KO_2, Na_2O_2, \) and \( NaAlO_2 \) exhibits paramagnetism, we need to analyze the total number of electrons in each compound. A compound is paramagnetic if it has unpaired electrons, which typically occurs when the total number of electrons is odd.

### Step 1: Calculate the total number of electrons in each compound.

1. **For \( Na_2O_2 \)**:
- Sodium (\( Na \)): Atomic number = 11 (2 atoms contribute \( 11 \times 2 = 22 \) electrons)
- Oxygen (\( O \)): Atomic number = 8 (2 atoms contribute \( 8 \times 2 = 16 \) electrons)
- Total electrons = \( 22 + 16 = 38 \) (even)

2. **For \( KO_2 \)**:
- Potassium (\( K \)): Atomic number = 19 (1 atom contributes \( 19 \) electrons)
- Oxygen (\( O \)): Atomic number = 8 (2 atoms contribute \( 8 \times 2 = 16 \) electrons)
- Total electrons = \( 19 + 16 = 35 \) (odd)

3. **For \( NO_2^+ \)**:
- Nitrogen (\( N \)): Atomic number = 7 (1 atom contributes \( 7 \) electrons)
- Oxygen (\( O \)): Atomic number = 8 (2 atoms contribute \( 8 \times 2 = 16 \) electrons)
- Since it is \( NO_2^+ \), we subtract 1 electron due to the positive charge.
- Total electrons = \( 7 + 16 - 1 = 22 \) (even)

4. **For \( NaAlO_2 \)**:
- Sodium (\( Na \)): Atomic number = 11 (1 atom contributes \( 11 \) electrons)
- Aluminum (\( Al \)): Atomic number = 13 (1 atom contributes \( 13 \) electrons)
- Oxygen (\( O \)): Atomic number = 8 (2 atoms contribute \( 8 \times 2 = 16 \) electrons)
- Total electrons = \( 11 + 13 + 16 = 40 \) (even)

### Step 2: Determine which compounds are paramagnetic.

- \( Na_2O_2 \): Total electrons = 38 (even) → Not paramagnetic
- \( KO_2 \): Total electrons = 35 (odd) → Paramagnetic
- \( NO_2^+ \): Total electrons = 22 (even) → Not paramagnetic
- \( NaAlO_2 \): Total electrons = 40 (even) → Not paramagnetic

### Conclusion:
The only compound that exhibits paramagnetism among the given options is \( KO_2 \).

### Final Answer:
**The compound that exhibits paramagnetism is \( KO_2 \).**