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 $1.5\;D$

Question

$(A)\;141^{\circ}21' \\ (B)\;145^{\circ} \\(C)\;140^{\circ}20' \\(D)\;139^{\circ}30' $

Answer 1

$\mu =\sqrt {\mu_1^2 +\mu_1^2 +2 \mu_1 \mu_1 \cos \alpha}$

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.