The experimental dipole moment of water molecule is ` 1.84D` . Calculater the bond angle `H-O-H` in water molecule if dispole moment of `OH` bond is ` 1.5D`.

Explanation

` mu = sqrt ( mu_1^2 + mu_1^2 + 2 mu_1mu_1 cos alpha)`

In ` H_2 O` only , tow dipoles equal to ` nu_12` are operating due to two ` O-H` bonds .
`Thus , `1.84 = sqrt (( 1.5)^2 + (1 .5)^2 +2 xx (1.5) xx (1.5) cso alpha)`
`:. cos alpha =- 0.2476`
` :. alpha = 104^@ 20`.