Question

The experimental dipole moment of water molecule is $1.84D$. Calculate the bond angle $H-O-H$ in water molecule, if dipole moment of $OH$ bond is $1.5D$.

[Take nearest integer]

Answer 1

$\mu =\sqrt{{{\mu }_1}^2+{{\mu }_1}^2+2{\mu }_1{\mu }_1\cos \alpha }$
In $H_{2}O$ only two dipoles equal to ${\mu }_1$ are operating due to two $O-H$ bonds
Thus
$1.84=\sqrt{{\left(1.5\right)}^2+{\left(1.5\right)}^2+2\times \left(1.5\right)\times \left(1.5\right)\cos \alpha }$
$\therefore \cos \alpha =-0.2476$
$\alpha ={104}^o{34}^{\prime }$