A neutral water molecule (H2O) in its vapour state has an electric dipole moment of magnitude 6.4 × 10-30 C-m . How far apart are the molecules centres of positive and negative charges?

Question

A neutral water molecule (H2O) in its vapour state has an electric dipole moment of magnitude 6.4 × 10-30 C-m . How far apart are the molecules centres of positive and negative charges? (A) 4m (B) 4 mm (C) 4 μ m (D) 4 pm

Tardigrade Admin · Accepted Answer

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

2 l = \frac{p}{10 e} = \frac{6.4 \times 1 0 ^{- 30}}{10 \times 1.6 \times 1 0 ^{- 19}}

= 4 \times 1 0^{- 12} m

= 4 p m

Q. A neutral water molecule $(H_{2} O)$ in its vapour state has an electric dipole moment of magnitude $6.4 \times 1 0^{- 30} C - m$ . How far apart are the molecules centres of positive and negative charges?

4m

4 mm

$4 μ m$

4 pm

Q. A neutral water molecule (H2​O) in its vapour state has an electric dipole moment of magnitude 6.4×10−30C−m . How far apart are the molecules centres of positive and negative charges?

4m

4 mm

4μm

4 pm

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 ie, distance between centres of positive and negative charges is 2l=10ep​=10×1.6×10−196.4×10−30​ =4×10−12m =4pm

Q. A neutral water molecule $(H_{2} O)$ in its vapour state has an electric dipole moment of magnitude $6.4 \times 1 0^{- 30} C - m$ . How far apart are the molecules centres of positive and negative charges?

$4 μ m$