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Q.

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?

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a

3.22×1010 esu 

b

2×1010 esu 

c

1.602×1019C

d

3.28×1010 esu 

answer is C.

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Detailed Solution

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 cm
Now μO-H=δ×0.94×10-8
δ=1.617×10-10 esu
Since O acquires 2δ charge one δ charge from each bond and thus .
Charge on O-atom-2δ
2×1.617×10-10
=3.23×10-10esu cm..

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