Which of the following compounds are paramagnetic in nature ?

Options

A. `KO_(2)`
B. `K_(2)O_(2)`
C. `Na_(2)O_(2)`
D. `RbO_(2)`

Explanation

To determine which of the given compounds are paramagnetic in nature, we need to analyze the total number of electrons in each compound and check for unpaired electrons. Here's a step-by-step solution:

### Step 1: Understand the Concept of Paramagnetism
- Paramagnetic substances have unpaired electrons, which means they will have an odd total number of electrons.
- Diamagnetic substances have all paired electrons, resulting in an even total number of electrons.

### Step 2: Analyze Each Compound
We will analyze each compound provided in the question.

#### Compound 1: KO2 (Potassium Superoxide)
- Potassium (K) has 19 electrons.
- Oxygen (O) has 8 electrons, and since there are 2 oxygen atoms, we have 2 × 8 = 16 electrons.
- Total electrons in KO2 = 19 (from K) + 16 (from O) = 35 electrons.
- Since 35 is odd, KO2 is **paramagnetic**.

#### Compound 2: K2O2 (Potassium Peroxide)
- There are 2 potassium atoms: 2 × 19 = 38 electrons.
- There are 2 oxygen atoms: 2 × 8 = 16 electrons.
- Total electrons in K2O2 = 38 + 16 = 54 electrons.
- Since 54 is even, K2O2 is **diamagnetic**.

#### Compound 3: Na2O2 (Sodium Peroxide)
- There are 2 sodium atoms: 2 × 11 = 22 electrons.
- There are 2 oxygen atoms: 2 × 8 = 16 electrons.
- Total electrons in Na2O2 = 22 + 16 = 38 electrons.
- Since 38 is even, Na2O2 is **diamagnetic**.

#### Compound 4: RbO2 (Rubidium Superoxide)
- Rubidium (Rb) has 37 electrons.
- There are 2 oxygen atoms: 2 × 8 = 16 electrons.
- Total electrons in RbO2 = 37 + 16 = 53 electrons.
- Since 53 is odd, RbO2 is **paramagnetic**.

### Step 3: Conclusion
Based on the analysis:
- **Paramagnetic Compounds**: KO2, RbO2
- **Diamagnetic Compounds**: K2O2, Na2O2

### Final Answer
The compounds that are paramagnetic in nature are **KO2 and RbO2**.