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The experimental dipole moment of water molecule is $1.84\;D$ . Calculate the bond angle $H-O-H$ in water molecule , if dipole moment of OH bond is $1.5\;D$

$(A)\;141^{\circ}21' \\ (B)\;145^{\circ} \\(C)\;140^{\circ}20' \\(D)\;139^{\circ}30' $
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$\mu =\sqrt {\mu_1^2 +\mu_1^2 +2 \mu_1 \mu_1 \cos \alpha}$
In $H_2O$ only two dipoles equal to $\mu_1$ are operating due to two $O-H$ bonds.
Thus, $1.84 =\sqrt {(1.5)^2+(1.5)^2+2 (1.5 )\times (1.5) \cos \alpha}$
$\cos \alpha=-0.2476$
$\alpha= 140^{\circ}20'$
Hence C is the correct answer.
answered Apr 4, 2014 by meena.p

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