A crystalline solid AB has NaCl type structure with radius of `B^(-)` ion is 250 pm. Which of the following cation can be made to slip into tetrahedral site of crystals of `A^(+)B^(-)` ?

Options

A. `P^(+)` (radius = 180 pm)
B. `Q^(+)=` (radius = 56 pm)
C. `R^(+)=` (radius = 200 pm)
D. `S^(+)=` (radius = 150 pm)

Answer

To solve the problem of determining which cation can slip into the tetrahedral site of a crystalline solid AB with NaCl type structure, we will follow these steps:

### Step 1: Understand the Structure
The crystalline solid AB has a NaCl type structure, which means it has a face-centered cubic (FCC) arrangement of anions (B^(-)) with cations (A^(+)) occupying the octahedral sites and tetrahedral sites.

### Step 2: Identify the Radius of the Anion
The radius of the B^(-) ion is given as 250 pm (picometers).

### Step 3: Use the Radius Ratio for Tetrahedral Sites
For cations to occupy tetrahedral sites, the radius ratio (r^+ / r^-) must fall within the range of 0.225 to 0.414. Here, r^+ is the radius of the cation and r^- is the radius of the anion.

### Step 4: Calculate the Minimum Radius of the Cation
Using the lower limit of the radius ratio:
\[ \frac{r^+}{r^-} = 0.225 \]
Substituting the radius of the anion:
\[ r^+ = 0.225 \times 250 \text{ pm} = 56.25 \text{ pm} \]

### Step 5: Calculate the Maximum Radius of the Cation
Using the upper limit of the radius ratio:
\[ \frac{r^+}{r^-} = 0.414 \]
Substituting the radius of the anion:
\[ r^+ = 0.414 \times 250 \text{ pm} = 103.5 \text{ pm} \]

### Step 6: Determine the Range for the Cation Radius
From the calculations, the radius of the cation (r^+) must be between 56.25 pm and 103.5 pm to fit into the tetrahedral site.

### Step 7: Identify the Suitable Cation
Now, we need to check the given options for cations and see which one has a radius that falls within the calculated range (56.25 pm to 103.5 pm).

### Conclusion
The cation that lies within this range can slip into the tetrahedral site of the AB crystal structure.