Question

The resultant dipole moment of water is 1.85 D, Calculate the bond moment of $OH$ bond (in D), bond angle in water molecule is ${104}^o$ C. Given: cos ${104}^o$ = -0.25 (Ignore the dipole moment due to lone pair)

Answer 1

We know, dipole moment is a vector quantity
Let the dipole moment of $OH$ bond be x D.
Using formula
R = $\sqrt{P^2+Q^2+2PQcos\theta }$
where R is the resultant dipole moment
P = Q is the dipole moment of each $OH$ bond
put up values,
R = $\sqrt{x^2+x^2+2x^{2}cos{104}^o}$
1.85 = $\sqrt{2x^2-2x^2\times 0.25}$
1.85 = $\sqrt{1.5x^2}$
3.4225 = $1.5x^2$
On solving, x = 1.51 D