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 find the percentage ionic character of the bond in the AB molecule, we can follow these steps:

### Step 1: Convert the dipole moment from Debye to esu cm
The dipole moment (μ) is given as 1.2 D. We know that:
1 Debye = \(10^{-18}\) esu cm

So, we convert:
\[ \mu = 1.2 \, \text{D} = 1.2 \times 10^{-18} \, \text{esu cm} \]

### Step 2: Calculate the charge (q) of the electron
The charge of an electron is:
\[ q = 4.8 \times 10^{-10} \, \text{esu} \]

### Step 3: Convert bond length from Ångström to cm
The bond length is given as 1 Å. We convert this to cm:
\[ 1 \, \text{Å} = 10^{-8} \, \text{cm} \]

### Step 4: Calculate the theoretical dipole moment (μ_calculated)
The dipole moment can be calculated using the formula:
\[ \mu = q \times r \]
where \(r\) is the bond length. Substituting the values:
\[ \mu_{\text{calculated}} = (4.8 \times 10^{-10} \, \text{esu}) \times (1 \times 10^{-8} \, \text{cm}) = 4.8 \times 10^{-18} \, \text{esu cm} \]

### Step 5: Calculate the percentage ionic character
The percentage ionic character can be calculated using the formula:
\[ \text{Percentage ionic character} = \left( \frac{\mu_{\text{given}}}{\mu_{\text{calculated}}} \right) \times 100 \]
Substituting the values:
\[ \text{Percentage ionic character} = \left( \frac{1.2 \times 10^{-18}}{4.8 \times 10^{-18}} \right) \times 100 \]

### Step 6: Simplify the calculation
\[ \text{Percentage ionic character} = \left( \frac{1.2}{4.8} \right) \times 100 = \left( \frac{1}{4} \right) \times 100 = 25\% \]

### Final Answer
The percentage ionic character of the bond is **25%**.

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