If the dipole moment of HCl is 1.08 D and the bond distance is `1.27Å`, the partial charge on hydrogen and chlorine , respectively are

Options

A. `+1.0 and -1.0`
B. `+.85 and -.85`
C. `+0.356 and -0.356`
D. `+0.177 and -0.177`

Answer

To find the partial charges on hydrogen (H) and chlorine (Cl) in HCl, we can use the formula for dipole moment (\( \mu \)):

\[ \mu = q \times d \]

Where:
- \( \mu \) is the dipole moment in Debye (D),
- \( q \) is the magnitude of the partial charge,
- \( d \) is the bond distance in meters.

### Step 1: Convert the bond distance from Ångstroms to meters
Given:
- Bond distance \( d = 1.27 \, \text{Å} \)

To convert Ångstroms to meters:
\[ 1 \, \text{Å} = 1 \times 10^{-10} \, \text{m} \]
Thus,
\[ d = 1.27 \, \text{Å} = 1.27 \times 10^{-10} \, \text{m} \]

### Step 2: Convert the dipole moment from Debye to Coulomb-meters
Given:
- Dipole moment \( \mu = 1.08 \, \text{D} \)

To convert Debye to Coulomb-meters:
\[ 1 \, \text{D} = 3.336 \times 10^{-29} \, \text{C m} \]
Thus,
\[ \mu = 1.08 \, \text{D} = 1.08 \times 3.336 \times 10^{-29} \, \text{C m} \]
Calculating this gives:
\[ \mu \approx 3.60768 \times 10^{-29} \, \text{C m} \]

### Step 3: Rearrange the dipole moment formula to find the partial charge \( q \)
Using the formula:
\[ q = \frac{\mu}{d} \]
Substituting the values we have:
\[ q = \frac{3.60768 \times 10^{-29} \, \text{C m}}{1.27 \times 10^{-10} \, \text{m}} \]

### Step 4: Calculate the partial charge \( q \)
Calculating the above expression:
\[ q \approx \frac{3.60768 \times 10^{-29}}{1.27 \times 10^{-10}} \approx 2.84 \times 10^{-19} \, \text{C} \]

### Step 5: Determine the partial charges on H and Cl
In HCl, the partial charge on hydrogen (\( q_H \)) and chlorine (\( q_{Cl} \)) can be considered equal in magnitude but opposite in sign:
\[ q_H \approx +2.84 \times 10^{-19} \, \text{C} \]
\[ q_{Cl} \approx -2.84 \times 10^{-19} \, \text{C} \]

### Final Result
The partial charges on hydrogen and chlorine are approximately:
- \( q_H \approx +2.84 \times 10^{-19} \, \text{C} \)
- \( q_{Cl} \approx -2.84 \times 10^{-19} \, \text{C} \)