If the dipole moment of AB molecule is given by 1.2 D and A - B the bond length is `1­Å` then `%` ionic character of the bond is [Given : 1 debye `=10^(-18)` esu. Cm]

Options

A. 75
B. 50
C. 60
D. 25

Answer

To solve the problem of finding the percentage ionic character of the bond in the AB molecule, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Given Data:**
- Dipole moment (µ) of the AB molecule = 1.2 D
- Bond length (d) = 1 Å = \(1 \times 10^{-8}\) cm (since 1 Å = \(10^{-10}\) m = \(10^{-8}\) cm)

2. **Convert Dipole Moment to ESU cm:**
- Given that \(1 \, \text{D} = 10^{-18} \, \text{esu cm}\), we can convert the dipole moment:
\[ \mu = 1.2 \, \text{D} = 1.2 \times 10^{-18} \, \text{esu cm} \]

3. **Use the Formula for Dipole Moment:**
- The formula for dipole moment is given by:
\[ \mu = Q \times d \]
where \(Q\) is the charge and \(d\) is the bond length.

4. **Rearrange to Find Charge (Q):**
- Rearranging the formula gives:
\[ Q = \frac{\mu}{d} \]
- Substitute the values of µ and d:
\[ Q = \frac{1.2 \times 10^{-18} \, \text{esu cm}}{1 \times 10^{-8} \, \text{cm}} = 1.2 \times 10^{-10} \, \text{esu} \]

5. **Compare with Actual Charge:**
- The actual charge of an electron (or proton) is given as \(Q_e = 4.8 \times 10^{-10} \, \text{esu}\).

6. **Calculate the Percentage Ionic Character:**
- The percentage ionic character can be calculated using the formula:
\[ \text{Percentage Ionic Character} = \left( \frac{Q}{Q_e} \right) \times 100 \]
- Substitute the values:
\[ \text{Percentage Ionic Character} = \left( \frac{1.2 \times 10^{-10}}{4.8 \times 10^{-10}} \right) \times 100 \]
- Simplifying this gives:
\[ \text{Percentage Ionic Character} = \left( \frac{1.2}{4.8} \right) \times 100 = 25\% \]

### Final Answer:
The percentage ionic character of the bond in the AB molecule is **25%**.

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