The dipole moment of HBr is `0.78xx 10^(-18)` esu cm. The bond length of HBr is 1.41 Å. The % ionic character is

Options

A. 7.5
B. 11.52
C. 15
D. 27

Explanation

To find the percentage ionic character of HBr, we can follow these steps:

### Step 1: Understand the formula for percentage ionic character
The formula for calculating the percentage ionic character is given by:

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

### Step 2: Identify the observed dipole moment
From the question, the observed dipole moment of HBr is given as:

\[ \text{Observed Dipole Moment} = 0.78 \times 10^{-18} \text{ esu cm} \]

### Step 3: Calculate the calculated dipole moment
The calculated dipole moment can be found using the formula:

\[ \text{Dipole Moment} = \text{Charge} \times \text{Bond Length} \]

For HBr, we consider the charge of one electron, which is approximately \(1.6 \times 10^{-19}\) C or \(4.8 \times 10^{-10}\) esu. The bond length is given as:

\[ \text{Bond Length} = 1.41 \text{ Å} = 1.41 \times 10^{-8} \text{ cm} \]

Now, we can calculate the calculated dipole moment:

\[ \text{Calculated Dipole Moment} = (4.8 \times 10^{-10} \text{ esu}) \times (1.41 \times 10^{-8} \text{ cm}) \]

Calculating this gives:

\[ \text{Calculated Dipole Moment} = 6.768 \times 10^{-18} \text{ esu cm} \]

### Step 4: Substitute values into the percentage ionic character formula
Now, we can substitute the observed and calculated dipole moments into the formula:

\[ \text{Percentage Ionic Character} = \left( \frac{0.78 \times 10^{-18}}{6.768 \times 10^{-18}} \right) \times 100 \]

### Step 5: Perform the calculation
Calculating the fraction:

\[ \frac{0.78}{6.768} \approx 0.1152 \]

Now multiply by 100 to get the percentage:

\[ \text{Percentage Ionic Character} \approx 11.52\% \]

### Conclusion
Thus, the percentage ionic character of HBr is approximately **11.52%**.

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