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. `1.0`
C. `0.5`
D. `0.05`

Explanation

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

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

\[ 1 \, \text{Å} = 10^{-8} \, \text{cm} \]
\[ \text{Bond Length (L)} = 1.617 \, \text{Å} = 1.617 \times 10^{-8} \, \text{cm} \]

### Step 2: Convert Dipole Moment to Centimeters
The dipole moment is given as \(0.38 \, \text{D}\). We convert this to the appropriate units.

\[ 1 \, \text{D} = 3.335 \times 10^{-30} \, \text{C m} = 1 \times 10^{-18} \, \text{esu cm} \]
\[ \text{Dipole Moment} (\mu) = 0.38 \, \text{D} = 0.38 \times 10^{-18} \, \text{esu cm} \]

### Step 3: Use the Dipole Moment Formula
The dipole moment (\(\mu\)) is related to the charge (\(Q\)) and the bond length (\(L\)) by the formula:

\[ \mu = Q \times L \]

Rearranging this gives us:

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

### Step 4: Substitute Values to Find Q
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 Q
Calculating the above expression:

\[ Q = \frac{0.38 \times 10^{-18}}{1.617 \times 10^{-8}} = 0.2347 \times 10^{-10} \, \text{esu} \]

### Step 6: Calculate the Fractional Charge
The fractional charge on each atom can be found by dividing the charge \(Q\) by the elementary charge \(e_0\):

\[ e_0 = 4.802 \times 10^{-10} \, \text{esu} \]

Thus, the fractional charge \(q\) is:

\[ q = \frac{Q}{e_0} = \frac{0.2347 \times 10^{-10} \, \text{esu}}{4.802 \times 10^{-10} \, \text{esu}} \]

### Step 7: Calculate the Fractional Charge
Calculating the above expression gives:

\[ q \approx 0.0488 \]

### Conclusion
The absolute magnitude of the fractional charge on each atom is approximately \(0.05\).