Solid AB has ZnS type structure. If the radius of `A^(+)` is 22.5 pm then the radius of B will be ……..

Answer

To find the radius of ion B in the ZnS type structure where the radius of ion A is given, we can follow these steps:

### Step 1: Understand the structure
ZnS has a structure where the cation (A^+) occupies tetrahedral voids and the anion (B^-) occupies the face-centered cubic lattice points. The radius ratio of cation to anion is crucial for determining the size of the anion.

### Step 2: Use the radius ratio rule
The radius ratio rule states that the ratio of the radius of the cation (r+) to the radius of the anion (r-) can be expressed as:
\[ \frac{r^+}{r^-} = 0.225 \]
where r^+ is the radius of the cation (A^+) and r^- is the radius of the anion (B^-).

### Step 3: Substitute the known values
We know the radius of A^+ (r^+) is 22.5 pm. We can substitute this value into the equation:
\[ \frac{22.5 \text{ pm}}{r^-} = 0.225 \]

### Step 4: Solve for r^-
To find r^-, we can rearrange the equation:
\[ r^- = \frac{22.5 \text{ pm}}{0.225} \]

### Step 5: Perform the calculation
Now, we can calculate r^-:
\[ r^- = \frac{22.5}{0.225} = 100 \text{ pm} \]

### Conclusion
The radius of ion B (r^-) is 100 pm.

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