In a compound AB, electro negativity difference between A and B is 1.9. Atomic radius A and B are `4Å and 2Å` . The distance between A and atoms means `d_(A-B)`

Options

A. `6.72Å`
B. `5.82Å`
C. `6.9Å`
D. `7.5Å`

Answer

To solve the problem, we will use the given information about the electronegativity difference and the atomic radii of elements A and B. The goal is to find the distance between the atoms A and B, denoted as \( d_{A-B} \).

### Step-by-Step Solution:

1. **Identify Given Values:**
- Electronegativity difference (\( \Delta \chi \)) = 1.9
- Atomic radius of A (\( r_A \)) = 4 Å
- Atomic radius of B (\( r_B \)) = 2 Å

2. **Use the Steveson Equation:**
The distance between atoms A and B can be calculated using the Steveson equation:
\[ d_{A-B} = r_A + r_B - 0.03 \times \Delta \chi \]

3. **Substitute the Known Values:**
Plug in the values into the equation:
\[ d_{A-B} = 4 \, \text{Å} + 2 \, \text{Å} - 0.03 \times 1.9 \]

4. **Calculate the Electronegativity Adjustment:**
Calculate \( 0.03 \times 1.9 \):
\[ 0.03 \times 1.9 = 0.057 \]

5. **Complete the Calculation:**
Now substitute this back into the equation:
\[ d_{A-B} = 4 + 2 - 0.057 = 6 - 0.057 = 5.943 \, \text{Å} \]

6. **Final Result:**
The distance between atoms A and B is approximately:
\[ d_{A-B} \approx 5.94 \, \text{Å} \]

7. **Select the Correct Option:**
Based on the options provided, the closest option to 5.94 Å is approximately 5.82 Å, which is option B.

### Summary:
The calculated distance \( d_{A-B} \) is approximately 5.94 Å, and the correct answer from the options given is option B.