Question

A neutral water molecule $(H_2O)$ in its vapour state has an electric dipole moment of magnitude $6.4 \times 10^{-30} C-m$ . How far apart are the molecules centres of positive and negative charges?

• A. 4m
• B. 4 mm
• C. $4 \mu m$
• D. 4 pm

Answer 1

Answer: (D)

There are 10 electrons and 10 protons in a neutral water molecule.
So, its dipole moment is $p=q(2 l)=10 e(2 l)$
Hence length of the dipole ie, distance between centres of positive and negative charges is
$2l=\frac{p}{10 e} =\frac{6.4 \times 10^{-30}}{10 \times 1.6 \times 10^{-19}}$
$=4 \times 10^{-12} m$
$=4\, pm$