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Question

The HOH bond angle in water is 105o. The HOH bond distance being 0.94 A. The dipole moment for the molecule is 1.85 D. Calculate the charge on oxygen atom.
Given: cos 105o = 0.25

A
1.617×1010 esu
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B
1.617×1010 C
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C
1.167×1012 esu
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D
1.167×1015 esu
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Solution

The correct option is A 1.617×1010 esu
We know, resulatnt dipole moment can be calculated using
R = P2+Q2+2PQcosθ
where R = μH2O = Resultant dipole moment
P = μOH = dipole moment of OH bond
So, μH2O = μ2OH+μ2OH+2μ2OHcosθ
Put up the values,
1.85 D = 2μ2OH(1+cos105o)
On solving, we get μOH = 1.52 D
Now 1 D = 1018 esu cm
So, μOH = 1.52×1018 esu cm
Also, μOH = δ×d
where δ = partial charge on oxygen atom
d = bond distance of OH bond = 0.94×108 cm
δ = 1.52×10180.94×108 = 1.617×1010 esu

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