In two different polar molecules, the ionic charge is `9.6` x `10^-10 e.s.u` and `3.2` x `10^-19` coulombs respectively. If inter ionic distance in both the molecules is` 1 A` unit, then the dipole moment are respectively

Options

A. `4.8D, 4.8D`
B. `1D, 4.8D`
C. `9.6D`, `9.6D`
D. `3.33D`, `4.8D`

Answer

To find the dipole moments of the two polar molecules given their ionic charges and inter-ionic distance, we will use the formula for dipole moment (μ):

\[ \mu = q \times d \]

where:
- \( \mu \) is the dipole moment,
- \( q \) is the charge, and
- \( d \) is the distance between the charges.

### Step 1: Calculate the dipole moment for the first molecule

1. **Identify the charge and distance**:
- Charge \( q_1 = 9.6 \times 10^{-10} \) e.s.u.
- Distance \( d = 1 \) Å = \( 1 \times 10^{-8} \) cm.

2. **Apply the formula**:
\[ \mu_1 = q_1 \times d = (9.6 \times 10^{-10} \, \text{e.s.u.}) \times (1 \times 10^{-8} \, \text{cm}) \]

3. **Calculate**:
\[ \mu_1 = 9.6 \times 10^{-18} \, \text{e.s.u. cm} \]

### Step 2: Calculate the dipole moment for the second molecule

1. **Identify the charge and distance**:
- Charge \( q_2 = 3.2 \times 10^{-19} \) coulombs.
- Distance \( d = 1 \) Å = \( 1 \times 10^{-8} \) cm.

2. **Convert charge from coulombs to e.s.u.**:
- The conversion factor is \( 1 \, \text{C} = 3 \times 10^9 \, \text{e.s.u.} \).
\[ q_2 \text{ in e.s.u.} = 3.2 \times 10^{-19} \, \text{C} \times 3 \times 10^9 \, \text{e.s.u./C} = 9.6 \times 10^{-10} \, \text{e.s.u.} \]

3. **Apply the formula**:
\[ \mu_2 = q_2 \times d = (9.6 \times 10^{-10} \, \text{e.s.u.}) \times (1 \times 10^{-8} \, \text{cm}) \]

4. **Calculate**:
\[ \mu_2 = 9.6 \times 10^{-18} \, \text{e.s.u. cm} \]

### Final Results

The dipole moments for the two polar molecules are:
- For the first molecule: \( \mu_1 = 9.6 \times 10^{-18} \, \text{e.s.u. cm} \)
- For the second molecule: \( \mu_2 = 9.6 \times 10^{-18} \, \text{e.s.u. cm} \)