The `H-O-H` bond angle in the water molecule is `105^@` , the`H -O` bond distance being `0.94 Å`, The dipole moment for the moelcule is `1.85D`. Calculate the charge on the oxygen atom .

Explanation

`mu_(H_(2)O) = sqrt(mu_(O-H)^(2) + mu_(O-H)^(2)+ 2 mu_(OH)^(2)cos 105^(@))`
`(1.84)^(2) = 2mu_(O-H)^(2) + 2mu_(O-H)^(2)xx(-0.2588)(because cos 105^(@) = cos (180^(@) - 75^(@)) = - cos 75^(@))`
` 3.3856 = mu_(O-H)^(2) (1 - 0.2588)`

or ` 1.4824 mu_(O-H)^(2)= 3.3856 or mu_(O-H) = 1.51 D`
but ` mu _(O-H) ` = charge `(delta) xx ` bond distance (d)
` therefore 1.51 xx10^(-18)` esu cm = ` delta xx (-0.94 xx10^(-10) cm)`
or ` delta = 1. 606 xx10^(-10)` esu
As O-atom acquires charge = ` 2delta ` (one from each O-H bond) , therefore , charge on O-atom
` = 2xx 1.606 xx10^(-10) equ = 3.212xx10^(-10) ` esu .