The percentage ionic character of a bond having `1.275Å` its length hand 1.03 D its dipole moment is :

Options

A. `10%`
B. `15%`
C. `16.83%`
D. `18.8%`

Explanation

To calculate the percentage ionic character of a bond given its bond length and dipole moment, we can follow these steps:

### Step 1: Understand the Formula
The percentage ionic character can be calculated using the formula:

\[ \text{Percentage Ionic Character} = \left( \frac{\text{Observed Dipole Moment}}{\text{Calculated Dipole Moment}} \right) \times 100 \]

### Step 2: Identify Given Values
From the question, we have:
- Bond length = \(1.275 \, \text{Å}\)
- Observed dipole moment = \(1.03 \, \text{D}\)

### Step 3: Convert Units
1. **Convert Bond Length to Centimeters**:
\[ 1 \, \text{Å} = 10^{-8} \, \text{cm} \]
Therefore,
\[ 1.275 \, \text{Å} = 1.275 \times 10^{-8} \, \text{cm} \]

2. **Convert Dipole Moment from Debye to ESU cm**:
\[ 1 \, \text{D} = 3.33 \times 10^{-18} \, \text{ESU cm} \]
Thus,
\[ 1.03 \, \text{D} = 1.03 \times 3.33 \times 10^{-18} \, \text{ESU cm} \approx 3.43 \times 10^{-18} \, \text{ESU cm} \]

### Step 4: Calculate the Charge
The charge of an electron is:
\[ e = 1.6 \times 10^{-19} \, \text{C} \]

### Step 5: Calculate the Calculated Dipole Moment
The calculated dipole moment (\(d\)) can be found using the formula:
\[ d = \text{Charge} \times \text{Bond Length} \]
Substituting the values:
\[ d = (1.6 \times 10^{-19} \, \text{C}) \times (1.275 \times 10^{-8} \, \text{cm}) \approx 2.04 \times 10^{-27} \, \text{C cm} \]

### Step 6: Convert Calculated Dipole Moment to Debye
To convert this calculated dipole moment to Debye:
\[ d \text{ (in Debye)} = \frac{d \text{ (in ESU cm)}}{3.33 \times 10^{-18}} \approx \frac{2.04 \times 10^{-27}}{3.33 \times 10^{-18}} \approx 6.12 \, \text{D} \]

### Step 7: Calculate Percentage Ionic Character
Now, substituting the observed and calculated dipole moments into the percentage ionic character formula:
\[ \text{Percentage Ionic Character} = \left( \frac{1.03}{6.12} \right) \times 100 \approx 16.83\% \]

### Final Answer
The percentage ionic character of the bond is approximately **16.83%**.

---