Among `KO_(2), KAlO_(2), CaO_(2) and NO_(2)^(+)`, unpaired electrons is present in :

Options

A. `NO_(2)^(+) and CaO_(2)`
B. `KO_(2) and KAlO_(2)`
C. `KO_(2)` only
D. `CaO_(2)` only

Answer

To determine which of the given compounds has unpaired electrons, we will analyze each compound one by one and calculate the total number of electrons present in each.

### Step-by-Step Solution:

1. **Identify the Compounds:**
We have four compounds to analyze: KO₂, KAlO₂, CaO₂, and NO₂⁺.

2. **Calculate Total Electrons for Each Compound:**
- **KO₂ (Potassium Superoxide):**
- Potassium (K) has an atomic number of 19, contributing 19 electrons.
- Oxygen (O) has an atomic number of 8, and since there are 2 oxygen atoms, this contributes \(8 \times 2 = 16\) electrons.
- Total electrons in KO₂ = \(19 + 16 = 35\).

- **KAlO₂ (Potassium Aluminum Oxide):**
- Potassium (K) contributes 19 electrons.
- Aluminum (Al) has an atomic number of 13, contributing 13 electrons.
- Oxygen (O) contributes \(8 \times 2 = 16\) electrons.
- Total electrons in KAlO₂ = \(19 + 13 + 16 = 48\).

- **CaO₂ (Calcium Peroxide):**
- Calcium (Ca) has an atomic number of 20, contributing 20 electrons.
- Oxygen (O) contributes \(8 \times 2 = 16\) electrons.
- Total electrons in CaO₂ = \(20 + 16 = 36\).

- **NO₂⁺ (Nitronium Ion):**
- Nitrogen (N) has an atomic number of 7, contributing 7 electrons.
- Oxygen (O) contributes \(8 \times 2 = 16\) electrons.
- Since it is a positive ion (NO₂⁺), we subtract 1 electron.
- Total electrons in NO₂⁺ = \(7 + 16 - 1 = 22\).

3. **Determine the Presence of Unpaired Electrons:**
- KO₂ has 35 electrons (odd number) → **unpaired electrons present**.
- KAlO₂ has 48 electrons (even number) → **no unpaired electrons**.
- CaO₂ has 36 electrons (even number) → **no unpaired electrons**.
- NO₂⁺ has 22 electrons (even number) → **no unpaired electrons**.

4. **Conclusion:**
The only compound among KO₂, KAlO₂, CaO₂, and NO₂⁺ that has unpaired electrons is **KO₂**.

### Final Answer:
**KO₂** has unpaired electrons.

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