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

Options

A. `4m`
B. `4 mm`
C. `4 mu m`
D. `4 p m`

Answer

D

Explanation

There are `10` electrons and `10` protons in a neutral water molecule.
So its dipole moment is `p=q (2l)= 10e(2l)`
Hence length of the dipole i.e., distance between centres of positive and negative charge is
`2l=(p)/(10e)=(6.4xx10^(-20))/(10xx1.6xx10^(-19))= 4xx10^(-12)m= 4p m`