A neutral water molecule `(H_(2)O)` in its vapor state has an electric dipole moment of magnitudes `6.4xx10^(-30)C-m`. How far apart are the molecules centers of positive and negative charge

Options

A. `4 xx 10^(-10)`
B. `4 xx 10^(-11)`
C. `4 xx 10^(-12)`
D. `4 xx 10^(-13)`

Answer

To find the distance between the centers of positive and negative charge in a neutral water molecule (H₂O) given its electric dipole moment, we can follow these steps:

### Step-by-Step Solution:

1. **Understand the Electric Dipole Moment**:
The electric dipole moment (P) is defined as the product of the charge (q) and the distance (L) between the centers of positive and negative charges:
\[ P = q \times L \]
In a water molecule, the dipole moment is given as \(6.4 \times 10^{-30} \, \text{C-m}\).

2. **Identify the Charge**:
In a water molecule, there are 10 electrons and 10 protons, but we consider the effective charge that contributes to the dipole moment. The charge of a single proton (or electron) is approximately:
\[ e = 1.6 \times 10^{-19} \, \text{C} \]
Since the water molecule has a dipole moment due to the separation of charges, we can consider the effective charge contributing to the dipole moment as \(q = 1.6 \times 10^{-19} \, \text{C}\).

3. **Set Up the Equation**:
Substitute the values of \(P\) and \(q\) into the dipole moment formula:
\[ 6.4 \times 10^{-30} = (1.6 \times 10^{-19}) \times L \]

4. **Solve for L**:
Rearranging the equation to solve for \(L\):
\[ L = \frac{6.4 \times 10^{-30}}{1.6 \times 10^{-19}} \]

5. **Calculate L**:
Performing the calculation:
\[ L = \frac{6.4}{1.6} \times 10^{-30 + 19} = 4 \times 10^{-11} \, \text{m} \]

6. **Convert to Picometers**:
Since \(1 \, \text{picometer} = 10^{-12} \, \text{m}\), we can express \(L\) in picometers:
\[ L = 4 \times 10^{-12} \, \text{m} = 4 \, \text{pm} \]

### Final Answer:
The distance between the centers of positive and negative charge in a neutral water molecule is \(4 \, \text{pm}\) (picometers).