Dipole moment of H₂O is 1.84 D. If the bond angle is 105° and O-H bond length is 0.94 $\stackrel{\mathrm{o}}{\mathrm{A}}$, what is the magnitude of charge on the oxygen atom in water molecule?

${\displaystyle \therefore \mu H_{2O}=\sqrt{{\mu }_{OH}^2+{\mu }_{OH}^2+2{\mu }_{OH}^2{\cos 105}^{\circ }}}$
Since ${\displaystyle H-2O}$ has tow vectors of${\displaystyle O-H}$ bond acting at ${\displaystyle {105}^{\circ }}$ . Let dipole moment of ${\displaystyle O-H}$ bond be a${\displaystyle a}$
${\displaystyle \therefore 1.85=\sqrt{2a^2(1+{\cos 105}^{\circ })}}$
${\displaystyle \mu O-H=1.52}$ debye =${\displaystyle =152\times {10}^{-18}}$ esu cm
Now ${\displaystyle {\mu }_{O-H}=\delta \times 0.94\times {10}^{-8}}$
${\displaystyle \therefore \delta =1.617\times {10}^{-10}}$ esu
Since O acquires ${\displaystyle 2\delta }$ charge one ${\displaystyle \delta }$ charge from each bond and thus .
Charge on O-atom${\displaystyle -2\delta }$
${\displaystyle 2\times 1.617\times {10}^{-10}}$
${\displaystyle =3.23\times {10}^{-10}}$esu cm..