Dipole moment of H2O is 1.84 D. If the bond angle is 105° and O-H bond length is 0.94 Ao, what is the magnitude of charge on the oxygen atom in water molecule?

∴μH2O=μOH2+μOH2+2μOH2cos105∘Since H-2O has tow vectors ofO-H bond acting at 105∘ . Let dipole moment of O-H bond be aa∴1.85=2a2(1+cos105∘)μO-H=1.52 debye ==152×10-18 esu cmNow μO-H=δ×0.94×10-8∴δ=1.617×10-10 esuSince O acquires 2δ charge one δ charge from each bond and thus .Charge on O-atom-2δ2×1.617×10-10=3.23×10-10esu cm..