A compound AB crystallises in bcc lattice with the unit cell edge length of 380 pm. Calculate (i) the distance between oppositely charged ions in the lattice ,(ii) radius of `B^-` if the radius of `A^+` is 190 pm

Explanation

To solve the problem step by step, we will follow these calculations:

### Step 1: Understanding the BCC Lattice
In a body-centered cubic (BCC) lattice, there are atoms at each corner of the cube and one atom at the center of the cube. For a compound AB, A is the cation (A^+) and B is the anion (B^-).

### Step 2: Calculate the Distance Between Oppositely Charged Ions
The distance between the cation (A^+) and the anion (B^-) can be calculated using the formula for the body diagonal of the cube. The formula for the body diagonal (d) in terms of the edge length (a) is given by:

\[ d = \sqrt{3} \cdot a \]

Since we are interested in the distance between the cation and anion, we take half of the body diagonal:

\[ \text{Distance} = \frac{d}{2} = \frac{\sqrt{3} \cdot a}{2} \]

Given that the edge length \( a = 380 \, \text{pm} \):

\[ \text{Distance} = \frac{\sqrt{3} \cdot 380 \, \text{pm}}{2} \]

Calculating this gives:

\[ \text{Distance} = \frac{1.732 \cdot 380}{2} \]
\[ \text{Distance} \approx \frac{657.36}{2} \]
\[ \text{Distance} \approx 328.68 \, \text{pm} \]

### Step 3: Calculate the Radius of B^-
We know the distance between the cation and anion is equal to the sum of their radii:

\[ \text{Distance} = r_A + r_B \]

Where:
- \( r_A \) is the radius of A^+ (given as 190 pm)
- \( r_B \) is the radius of B^-

Substituting the known values:

\[ 328.68 \, \text{pm} = 190 \, \text{pm} + r_B \]

Now, solving for \( r_B \):

\[ r_B = 328.68 \, \text{pm} - 190 \, \text{pm} \]
\[ r_B \approx 138.68 \, \text{pm} \]

### Final Answers
(i) The distance between oppositely charged ions in the lattice is approximately **328.68 pm**.

(ii) The radius of B^- is approximately **138.68 pm**.

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