The pair(s) of reagents that yield paramagnetic species is/are

Options

A. Na and excess of `NH_(3)`
B. K and excess of `O_(2)`
C. Cu and dilute `HNO_(3)`
D. `O_(2)` and 2-ethylanthraquinol

Explanation

To determine which pairs of reagents yield paramagnetic species, we need to analyze each option based on the presence of unpaired electrons in the resulting compounds. Here’s a step-by-step breakdown of the solution:

### Step 1: Analyze the first option - Na and excess NH3
- Sodium (Na) reacts with ammonia (NH3) to form a solvated electron.
- The reaction can be represented as:
\[ \text{Na} + \text{NH}_3 \rightarrow \text{Na(NH}_3\text{)}_x + \text{e}^- \]
- The presence of the solvated electron (e-) indicates that there is an unpaired electron.
- **Conclusion**: This results in a paramagnetic species.

### Step 2: Analyze the second option - K and excess O2
- Potassium (K) reacts with oxygen (O2) to form potassium superoxide (KO2).
- The reaction can be represented as:
\[ \text{K} + \text{O}_2 \rightarrow \text{KO}_2 \]
- In KO2, the O2- ion has unpaired electrons, making it paramagnetic.
- **Conclusion**: This results in a paramagnetic species.

### Step 3: Analyze the third option - Cu and dilute HNO3
- Copper (Cu) reacts with dilute nitric acid (HNO3) to produce copper(II) nitrate (Cu(NO3)2), nitrogen monoxide (NO), and water (H2O).
- The reaction can be represented as:
\[ 3\text{Cu} + 8\text{HNO}_3 \rightarrow 2\text{Cu(NO}_3\text{)}_2 + 2\text{NO} + 4\text{H}_2\text{O} \]
- Nitric oxide (NO) is known to have unpaired electrons, making it paramagnetic.
- **Conclusion**: This results in a paramagnetic species.

### Step 4: Analyze the fourth option - O2 and 2 ethyl anthraquinol
- Oxygen (O2) reacts with ethyl anthraquinol to form hydrogen peroxide (H2O2).
- The reaction can be represented as:
\[ \text{O}_2 + 2 \text{C}_{14}\text{H}_{10}\text{O} \rightarrow \text{H}_2\text{O}_2 + \text{other products} \]
- Hydrogen peroxide (H2O2) has no unpaired electrons, making it diamagnetic.
- **Conclusion**: This does not yield a paramagnetic species.

### Final Conclusion
- The pairs of reagents that yield paramagnetic species are:
- Option A: Na and excess NH3
- Option B: K and excess O2
- Option C: Cu and dilute HNO3
- Option D does not yield a paramagnetic species.

### Correct Answers:
- **Option A, B, and C yield paramagnetic species.**