Question

The experimental dipole moment of water molecule is 1.84D. Calculate the bond angle H-O-H in water molecule if the dipole moment of OH bond is 1.5D. [Take nearest integer]

Answer 1

HINT: Dipole moment is the product of the charge and the nuclear separation of the two charges. To solve this question you can use the relation ${{\mu }_{\exp }}=\sqrt{{{\mu }_{1}}^{2}+{{\mu }_{2}}^{2}+2{{\mu }_{1}}{{\mu }_{2}}\cos \theta }$. Use the experimental value of the dipole moment which is given in the question.

COMPLETE STEP BY STEP SOLUTION: We know that dipole moment is a quantity that we use to describe that two opposite charges are separated by a distance.
In the question the experimental dipole moment is given to us and the dipole moment of the OH bond is given to us.
There is a relation between the experimental dipole moment and the dipole moment of each bond along with bond angles that we can use to find out the bond angle between the atoms involved.
The relation is-
${{\mu }_{\exp }}=\sqrt{{{\mu }_{1}}^{2}+{{\mu }_{2}}^{2}+2{{\mu }_{1}}{{\mu }_{2}}\cos \theta }$
Where, $\mu$ is the experimental dipole moment of a given molecule.
${{\mu }_{1}}$ and ${{\mu }_{2}}$ is the dipole moment of the bonds operating in the molecule.
$\operatorname{Cos}\theta$ is the angle between the two bonds.
Now, let us use this relation to find out the bond angle of H-O-H bond in a water molecule.
Here, we can see that there are two equivalent O-H bonds present. Therefore, they have the same dipole moment which is given to us as 1.5D.
The experimental value of the dipole moment is also given to us which is 1.84D.
Therefore, putting these values in the above equation, we will get-
$\begin{aligned} & 1.84=\sqrt{{{\left( 1.5 \right)}^{2}}+{{\left( 1.5 \right)}^{2}}+2\times 1.5\times 1.5\cos \theta } \\\ & or,{{\left( 1.84 \right)}^{2}}=4.5+4.5\cos \theta \\\ & or,3.3856-4.5=4.5\cos \theta \\\ & or,-1.1=4.5\cos \theta \text{ (}3.3856\simeq 3.4) \\\ & or,\cos \theta =-0.244 \\\ & or,\theta ={{\cos }^{-1}}\left( -0.244 \right)=104.123 \\\ \end{aligned}$
We can see from the above calculation that the bond angle is 104.123 which can be written as ${{104}^{\circ }}{{12}^{'}}$.

Therefore, the answer is ${{104}^{\circ }}{{12}^{'}}$.

NOTE: Dipole moment is the quantity that we can measure for any molecule. From this dipole moment we can determine the size of the partial charge on the molecule if we know the bond length of the molecule.
The dipole moment is given by- $\mu =q\times r$, where q is the charge and r is the nuclear separation.