Largest difference in radii is found in case of the pair

Options

A. Li, Na
B. Na, K
C. K, Rb
D. Rb, Cs

Explanation

To solve the question regarding which pair of alkali metals has the largest difference in radii, we will analyze the atomic radii of the alkali metals and identify the pair with the greatest difference.

### Step-by-Step Solution:

1. **Identify the Alkali Metals**: The alkali metals in question are lithium (Li), sodium (Na), potassium (K), rubidium (Rb), and cesium (Cs).

2. **List the Atomic Radii**: The metallic radii (in picometers) for these alkali metals are as follows:
- Lithium (Li): 152 pm
- Sodium (Na): 186 pm
- Potassium (K): 227 pm
- Rubidium (Rb): 248 pm
- Cesium (Cs): 265 pm

3. **Calculate the Differences in Radii**: We will calculate the differences in radii for each consecutive pair:
- Na and K: 227 pm - 186 pm = 41 pm
- K and Rb: 248 pm - 227 pm = 21 pm
- Rb and Cs: 265 pm - 248 pm = 17 pm
- Li and Na: 186 pm - 152 pm = 34 pm

4. **Identify the Largest Difference**: From the calculated differences:
- Na and K: 41 pm
- Li and Na: 34 pm
- K and Rb: 21 pm
- Rb and Cs: 17 pm

The largest difference in radii is between sodium (Na) and potassium (K), which is 41 pm.

5. **Conclusion**: Therefore, the pair with the largest difference in radii is sodium (Na) and potassium (K).

### Final Answer:
The pair with the largest difference in radii is **sodium (Na) and potassium (K)**.