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 need to find the distance between atoms A and B in the compound AB, using the given information about their atomic radii and the electronegativity difference. We will apply the Stevenson equation to calculate the distance.

### Step-by-Step Solution:

1. **Identify Given Values**:
- Electronegativity difference (ΔEN) = 1.9
- Atomic radius of A (rA) = 4 Å
- Atomic radius of B (rB) = 2 Å

2. **Use the Stevenson Equation**:
The equation to find the distance (dA-B) between atoms A and B is:
\[ d_{A-B} = r_A + r_B - 0.03 \times \Delta EN \]

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

4. **Calculate the Electronegativity Contribution**:
First, calculate \(0.03 \times 1.9\):
\[ 0.03 \times 1.9 = 0.057 \]

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

6. **Final Result**:
The calculated distance \(d_{A-B}\) is approximately \(5.94 \, \text{Å}\).

7. **Choose the Nearest Option**:
If the options provided include \(5.82 \, \text{Å}\), it is the closest match to our calculated value.

### Conclusion:
The distance between atoms A and B in the compound AB is approximately \(5.94 \, \text{Å}\), and the nearest option is \(5.82 \, \text{Å}\).

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