The two ions `A^+` and `B^-` have radii 40 pm and 120 pm respectively. In the closed packed crystal of compound AB, the coordination number of `A^+` would be

Options

A. 6
B. 8
C. 4
D. 12

Answer

To determine the coordination number of the cation \( A^+ \) in the closed packed crystal of compound \( AB \), we can follow these steps:

### Step 1: Identify the radii of the ions
We are given:
- Radius of cation \( A^+ = 40 \) pm
- Radius of anion \( B^- = 120 \) pm

### Step 2: Calculate the radius ratio
The radius ratio \( r \) is calculated using the formula:
\[ r = \frac{R_{cation}}{R_{anion}} \]
Substituting the values:
\[ r = \frac{40 \text{ pm}}{120 \text{ pm}} = \frac{1}{3} = 0.333 \]

### Step 3: Determine the coordination number based on the radius ratio
The coordination number can be determined based on the radius ratio \( r \):
- If \( 0.225 < r < 0.414 \), the coordination number is 4 (the cation occupies tetrahedral voids).

Since we calculated \( r = 0.333 \), which lies between 0.225 and 0.414, we can conclude that the coordination number of \( A^+ \) is 4.

### Final Answer
The coordination number of \( A^+ \) in the closed packed crystal of compound \( AB \) is **4**.

---