Molecule AB has a bond length of `1.617Å` and a dipole moment of 0.38 D. The fractional charge on each atom (absolute magnitude) is :
`(e_(0)=4.802xx10^(-10)"esu")`

Options

A. 0
B. 0.05
C. 0.5
D. `1.0`

Explanation

To find the fractional charge on each atom in the molecule AB, we will follow these steps:

### Step 1: Convert the bond length to centimeters
The given bond length is \(1.617 \, \text{Å}\). We need to convert this to centimeters.

\[ 1 \, \text{Å} = 10^{-10} \, \text{m} \quad \text{and} \quad 1 \, \text{m} = 100 \, \text{cm} \]
Thus,
\[ 1.617 \, \text{Å} = 1.617 \times 10^{-10} \, \text{m} = 1.617 \times 10^{-8} \, \text{cm} \]

### Step 2: Convert the dipole moment to esu cm
The given dipole moment is \(0.38 \, \text{D}\). We need to convert this to esu cm.

\[ 1 \, \text{D} = 1 \times 10^{-18} \, \text{esu cm} \]
Thus,
\[ 0.38 \, \text{D} = 0.38 \times 10^{-18} \, \text{esu cm} \]

### Step 3: Use the formula for dipole moment
The dipole moment (\(\mu\)) is related to the charge (\(q\)) and the distance (\(L\)) by the formula:

\[ \mu = q \times L \]
We can rearrange this to find the charge:

\[ q = \frac{\mu}{L} \]

### Step 4: Substitute the values
Substituting the values we have:

\[ q = \frac{0.38 \times 10^{-18} \, \text{esu cm}}{1.617 \times 10^{-8} \, \text{cm}} \]

### Step 5: Calculate the charge
Calculating the charge:

\[ q = \frac{0.38 \times 10^{-18}}{1.617 \times 10^{-8}} = 0.235 \times 10^{-10} \, \text{esu} \]

### Step 6: Calculate the fractional charge
The absolute value of the charge on each atom (\(e_0\)) is given as \(4.802 \times 10^{-10} \, \text{esu}\). The fractional charge (\(f\)) is given by:

\[ f = \frac{q}{e_0} \]

Substituting the values:

\[ f = \frac{0.235 \times 10^{-10}}{4.802 \times 10^{-10}} \]

### Step 7: Perform the division
Calculating the fractional charge:

\[ f = 0.05 \]

### Final Answer
The fractional charge on each atom (absolute magnitude) is \(0.05\).

---