To calculate the fractional charge on each atom in the HBr molecule, we can use the formula for the dipole moment (μ):
\[
\mu = q \times d
\]
where:
- \( \mu \) is the dipole moment,
- \( q \) is the fractional charge on each atom,
- \( d \) is the bond distance.
### Step 1: Convert the dipole moment from Debye to esu·cm
The dipole moment is given as \( \mu = 0.78 \, D \). To convert Debye to esu·cm, we use the conversion factor:
\[
1 \, D = 3.3356 \times 10^{-30} \, C \cdot m
\]
Since \( 1 \, D = 3.3356 \times 10^{-30} \, C \cdot m \) and \( 1 \, m = 100 \, cm \), we can convert it as follows:
\[
\mu = 0.78 \, D \times 3.3356 \times 10^{-30} \, C \cdot m \times 100 \, cm/m
\]
Calculating this gives:
\[
\mu = 0.78 \times 3.3356 \times 10^{-28} \, esu \cdot cm
\]
### Step 2: Calculate the bond distance in cm
The bond distance is given as \( 1.41 \, Å \). We need to convert this to centimeters:
\[
1 \, Å = 10^{-8} \, cm
\]
Thus,
\[
d = 1.41 \, Å = 1.41 \times 10^{-8} \, cm
\]
### Step 3: Rearranging the dipole moment formula to find q
Now we can rearrange the dipole moment formula to solve for the charge \( q \):
\[
q = \frac{\mu}{d}
\]
### Step 4: Substitute the values into the equation
Substituting the values we calculated:
\[
q = \frac{0.78 \times 3.3356 \times 10^{-28} \, esu \cdot cm}{1.41 \times 10^{-8} \, cm}
\]
### Step 5: Calculate q
Now we can calculate \( q \):
\[
q = \frac{0.78 \times 3.3356 \times 10^{-28}}{1.41 \times 10^{-8}} \approx \frac{2.604 \times 10^{-28}}{1.41 \times 10^{-8}} \approx 1.84 \times 10^{-20} \, esu
\]
### Step 6: Determine the fractional charge on each atom
The fractional charge on each atom in HBr can be assumed to be equal and opposite due to the nature of the molecule. Therefore, if we denote the fractional charge on H as \( q_H \) and on Br as \( q_{Br} \):
\[
q_H = +q \quad \text{and} \quad q_{Br} = -q
\]
Thus, the fractional charges are approximately:
\[
q_H \approx +0.92 \times 10^{-20} \, esu
\]
\[
q_{Br} \approx -0.92 \times 10^{-20} \, esu
\]
### Final Answer:
The fractional charge on each atom in HBr is approximately:
- \( q_H \approx +0.92 \times 10^{-20} \, esu \) (for Hydrogen)
- \( q_{Br} \approx -0.92 \times 10^{-20} \, esu \) (for Bromine)
