Paramagnetic pairs (s) among the following is (are) .

Options

A. `[BaO_(2),NO_(2)]`
B. `[KO_(2),NO]`
C. `[H_(2)O_(2),NO]`
D. `K_(3)[Fe(CN)_(6)],CuCI_(2)`

Explanation

To determine the paramagnetic pairs among the given options, we need to analyze the electronic configurations of the molecules or ions involved. Paramagnetic substances are characterized by the presence of unpaired electrons. Let's go through each option step by step.

### Step 1: Analyze Option A - BaO2 and NO2
1. **Barium Peroxide (BaO2)**:
- Barium (Ba) has an oxidation state of +2, and the peroxide ion (O2^2-) has a specific electronic configuration.
- The electronic configuration of the peroxide ion results in all electrons being paired.
- Therefore, BaO2 is **diamagnetic**.

2. **Nitrogen Dioxide (NO2)**:
- NO2 has an odd number of electrons (15 electrons).
- The electronic configuration shows that there is one unpaired electron.
- Therefore, NO2 is **paramagnetic**.

**Conclusion for Option A**: Since BaO2 is diamagnetic and NO2 is paramagnetic, Option A is **false**.

### Step 2: Analyze Option B - KO2 and NO
1. **Potassium Superoxide (KO2)**:
- KO2 contains the superoxide ion (O2^-).
- The electronic configuration of the superoxide ion shows one unpaired electron.
- Therefore, KO2 is **paramagnetic**.

2. **Nitric Oxide (NO)**:
- NO has 15 electrons as well.
- Its electronic configuration reveals that there is one unpaired electron.
- Therefore, NO is **paramagnetic**.

**Conclusion for Option B**: Both KO2 and NO are paramagnetic, so Option B is **correct**.

### Step 3: Analyze Option C - H2O2
- **Hydrogen Peroxide (H2O2)**:
- H2O2 contains the peroxide ion (O2^2-).
- As discussed earlier, all electrons in the peroxide ion are paired.
- Therefore, H2O2 is **diamagnetic**.

**Conclusion for Option C**: H2O2 is diamagnetic, so Option C is **false**.

### Step 4: Analyze Option D - K3Fe(CN)6 and CuCl2
1. **Potassium Ferricyanide (K3Fe(CN)6)**:
- Iron in this compound has an oxidation state of +3.
- The electronic configuration of Fe^3+ shows that there is one unpaired electron after pairing occurs due to the strong field ligand CN^-.
- Therefore, K3Fe(CN)6 is **paramagnetic**.

2. **Copper(II) Chloride (CuCl2)**:
- Copper in this compound has an oxidation state of +2.
- The electronic configuration of Cu^2+ shows one unpaired electron.
- Therefore, CuCl2 is **paramagnetic**.

**Conclusion for Option D**: Both K3Fe(CN)6 and CuCl2 are paramagnetic, so Option D is **correct**.

### Final Conclusion:
The paramagnetic pairs among the options are found in **Option B** (KO2 and NO) and **Option D** (K3Fe(CN)6 and CuCl2).