The dipole moment of `HBr` is ` 1.6 xx 10^(-30) cm` and interatomic spacing is `1 Å`. The `%` ionic character of `HBr` is

Answer

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

### Step 1: Understand the given data
We have the following information:
- Dipole moment of HBr (μ) = \(1.6 \times 10^{-30} \, \text{cm}\)
- Interatomic spacing (d) = \(1 \, \text{Å} = 1 \times 10^{-10} \, \text{cm}\)
- Charge of an electron (Q) = \(1.6 \times 10^{-19} \, \text{C}\)

### Step 2: Write the formula for percentage ionic character
The formula for calculating the percentage ionic character of a molecule is given by:
\[ \text{Percentage ionic character} = \left( \frac{\mu}{d \cdot Q} \right) \times 100 \]
Where:
- μ is the dipole moment
- d is the interatomic spacing
- Q is the charge of an electron

### Step 3: Substitute the values into the formula
Now we can substitute the known values into the formula:
\[ \text{Percentage ionic character} = \left( \frac{1.6 \times 10^{-30} \, \text{cm}}{(1 \times 10^{-10} \, \text{cm}) \cdot (1.6 \times 10^{-19} \, \text{C})} \right) \times 100 \]

### Step 4: Simplify the expression
Calculating the denominator:
\[ (1 \times 10^{-10}) \cdot (1.6 \times 10^{-19}) = 1.6 \times 10^{-29} \, \text{cm} \cdot \text{C} \]

Now substituting this back into the equation:
\[ \text{Percentage ionic character} = \left( \frac{1.6 \times 10^{-30}}{1.6 \times 10^{-29}} \right) \times 100 \]

### Step 5: Calculate the result
\[ \text{Percentage ionic character} = \left( \frac{1.6}{1.6} \times 10^{-30 + 29} \right) \times 100 = (1 \times 10^{-1}) \times 100 = 10\% \]

### Final Answer
Thus, the percentage ionic character of HBr is **10%**.

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