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, we can follow these steps:

### Step 1: Understand the Formula
The percentage covalent character can be calculated using the formula:
\[ \text{Percentage Covalent Character} = \left( \frac{\mu_{\text{observed}}}{\mu_{\text{calculated}}} \right) \times 100 \]
Where:
- \(\mu_{\text{observed}}\) is the dipole moment given in the problem.
- \(\mu_{\text{calculated}}\) is the dipole moment calculated based on the charge and bond length.

### Step 2: Given Values
From the problem, we have:
- \(\mu_{\text{observed}} = 1.6 \, \text{D}\) (Debye)
- Bond length \(d = 1 \, \text{Å} = 1 \times 10^{-8} \, \text{cm}\)

### Step 3: Calculate \(\mu_{\text{calculated}}\)
To calculate \(\mu_{\text{calculated}}\), we need to know the charge \(q\). The dipole moment is given by:
\[ \mu = q \times d \]
We can assume that the charge \(q\) is approximately \(1.6 \times 10^{-19} \, \text{C}\) (the charge of an electron).

Now substituting the values:
\[ \mu_{\text{calculated}} = q \times d = (1.6 \times 10^{-19} \, \text{C}) \times (1 \times 10^{-8} \, \text{cm}) = 1.6 \times 10^{-27} \, \text{C cm} \]

### Step 4: Convert \(\mu_{\text{calculated}}\) to Debye
1 Debye is equal to \(3.336 \times 10^{-29} \, \text{C cm}\). To convert \(\mu_{\text{calculated}}\) to Debye:
\[ \mu_{\text{calculated}} = \frac{1.6 \times 10^{-27} \, \text{C cm}}{3.336 \times 10^{-29} \, \text{C cm/D}} \approx 4.8 \, \text{D} \]

### Step 5: Calculate the Percentage Covalent Character
Now substituting the values into the percentage covalent character formula:
\[ \text{Percentage Covalent Character} = \left( \frac{1.6 \, \text{D}}{4.8 \, \text{D}} \right) \times 100 \]
\[ \text{Percentage Covalent Character} = \left( \frac{1.6}{4.8} \right) \times 100 = 33.33\% \]

### Conclusion
The percentage covalent character of the bond in the AB molecule is approximately **33.33%**.

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