Choose the paramagnetic oxide in the following

Options

A. `Na_(2) O `
B. `MgO `
C. `BeO `
D. `KO_(2)`

Explanation

To determine which oxide is paramagnetic among the given options, we need to analyze the electronic configurations and the presence of unpaired electrons in each oxide.

### Step-by-Step Solution:

1. **Understanding Paramagnetism**:
- Paramagnetic substances have unpaired electrons in their molecular or atomic orbitals. These unpaired electrons contribute to the magnetic properties of the substance.

2. **Analyzing the Given Oxides**:
- We have the following oxides to consider: A2O, MgO, BeO, and KO2.

3. **Examining A2O**:
- Let's assume A is sodium (Na). The oxide would be Na2O.
- Sodium (Na) has an electronic configuration of [Ne]3s¹. In Na2O, Na loses one electron to become Na⁺, and O gains two electrons to become O²⁻.
- In Na2O, all electrons are paired (Na⁺ has no unpaired electrons and O²⁻ has paired electrons).
- **Conclusion**: Na2O is not paramagnetic.

4. **Examining MgO**:
- Magnesium (Mg) has an electronic configuration of [Ne]3s². In MgO, Mg loses two electrons to become Mg²⁺, and O gains two electrons to become O²⁻.
- Both Mg²⁺ and O²⁻ have all paired electrons.
- **Conclusion**: MgO is not paramagnetic.

5. **Examining BeO**:
- Beryllium (Be) has an electronic configuration of [He]2s². In BeO, Be loses two electrons to become Be²⁺, and O gains two electrons to become O²⁻.
- Again, both Be²⁺ and O²⁻ have all paired electrons.
- **Conclusion**: BeO is not paramagnetic.

6. **Examining KO2**:
- Potassium (K) has an electronic configuration of [Ar]4s¹. In KO2, K loses one electron to become K⁺, and O2 forms a superoxide ion (O2²⁻).
- The superoxide ion (O2²⁻) has an unpaired electron. It can be represented as O2²⁻ = O-O with a bond order of 1.5, which means it has unpaired electrons.
- **Conclusion**: KO2 is paramagnetic due to the presence of unpaired electrons in the superoxide ion.

7. **Final Answer**:
- The paramagnetic oxide among the given options is KO2.