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

`:. muH_(2O) = sqrt ( mu_(OH)^2 + mu_(OH)^2 + 2mu_(OH)^2 cos 105^@)`
Since ` H-2O` has tow vectors of `O-H` bond acting at `105^@` . Let dipole moment of `O-H` bond be `a`
`:. 1.85 = sqrt ( 2a^2 (1+ cos 105^@))`
or a,i.e., `mu O-H = 1.52` debye `= 152 xx10^(-18)` esu cm
Now `mu _(O-H) = delta xx 0.94 xx 10^(-8)`
` :. delta = 1.617 xx 10^(-10)` esu
Since O acquires ` 2 delta` charge one ` delta` charge frome each bod and thus .
Charge on O-atom ` -2 delta`
`2 xx 1.617 xx10^(-10)`
` = 3.23 xx 10^(-10) `esu cm.