Question

the observed value of dipole moment of water molecule is found to be 1.84 D .calculate the H-O-H bond angle in H₂O molecule ,given that the bond moment of O-H BOND is 1.5 D .

Answer 1

Dear​ Mukesh Verma,

$$\begin{gathered} Dipolemomentandbondanglecanberelatedas \\ {\mu }_R=\sqrt{{{\mu }_1}^2+{{\mu }_2}^2+2{\mu }_1{\mu }_2\cos \theta } \\ {\mu }_R=1.84D;{\mu }_1={\mu }_2=1.5D;So\theta canbecalculated \\ 1.84=\sqrt{1.5^2+1.5^2+2\times 1.5\times 1.5\cos \theta } \\ 3.3856=4.5+4.5cos\theta \\ \cos \theta =\frac{3.3856-4.5}{4.5}=-0.2476 \\ \theta =co{s}^{-1}(-0.2476) \\ \theta =104.{34}^{o}approx \end{gathered}$$

Bond angle of H₂O molecule is 104.34 degree approx.

Regards