If the dipole moment of AB molecule is given by 1.6 D and A - B the bond length is `1Å` then % covalent character of the bond is

Options

A. 25
B. 33.33
C. 66.66
D. 75

Answer

To find the percentage covalent character of the bond in the AB molecule given the dipole moment and bond length, we can follow these steps:

### Step 1: Convert the dipole moment from Debye to electrostatic units (esu)
The dipole moment (μ) is given as 1.6 D. We know that:
1 D = \( 3.336 \times 10^{-29} \) C·m

So, we convert 1.6 D to esu:
\[ \mu_{\text{observed}} = 1.6 \, \text{D} = 1.6 \times 3.336 \times 10^{-29} \, \text{C·m} \]
\[ \mu_{\text{observed}} = 1.6 \times 3.336 \times 10^{-29} \approx 5.3376 \times 10^{-29} \, \text{C·m} \]

### Step 2: Calculate the ionic dipole moment (μ_ionic)
The ionic dipole moment can be calculated using the formula:
\[ \mu_{\text{ionic}} = Q \times D \]
Where:
- \( Q \) is the charge at each end of the dipole, which is approximately \( 4.8 \times 10^{-10} \) esu (the charge of an electron).
- \( D \) is the bond length, which is given as \( 1 \, \text{Å} = 1 \times 10^{-8} \, \text{cm} \).

Now, substituting the values:
\[ \mu_{\text{ionic}} = (4.8 \times 10^{-10} \, \text{esu}) \times (1 \times 10^{-8} \, \text{cm}) \]
\[ \mu_{\text{ionic}} = 4.8 \times 10^{-18} \, \text{C·m} \]

### Step 3: Calculate the percentage ionic character
The percentage ionic character can be calculated using the formula:
\[ \text{Percentage Ionic Character} = \left( \frac{\mu_{\text{observed}}}{\mu_{\text{ionic}}} \right) \times 100 \]
Substituting the values we calculated:
\[ \text{Percentage Ionic Character} = \left( \frac{5.3376 \times 10^{-29}}{4.8 \times 10^{-18}} \right) \times 100 \]
Calculating this gives:
\[ \text{Percentage Ionic Character} \approx 11.1\% \]

### Step 4: Calculate the percentage covalent character
The percentage covalent character can be found by subtracting the percentage ionic character from 100%:
\[ \text{Percentage Covalent Character} = 100\% - \text{Percentage Ionic Character} \]
\[ \text{Percentage Covalent Character} = 100\% - 11.1\% = 88.9\% \]

### Final Answer
The percentage covalent character of the bond is approximately **88.9%**.