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How do you calculate the dipole moment of water?

seo-qna
Last updated date: 13th Jul 2024
Total views: 391.2k
Views today: 9.91k
Answer
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Hint: Dipole moment is a vector quantity means it has magnitude and direction. Dipole moment is a measure of polarity of chemical bonds present in between two atoms in a particular molecule. In dipole moment the separation of negative and positive charge takes place.

Complete answer:
- In the question they asked how we can calculate the dipole moment of water.
- There is a formula to calculate the dipole moment and it is as follows.
\[\text{Dipole moment (}\mu \text{) = charge (Q) }\times \text{ distance of separation (r)}\]
- Now coming to the calculation of the dipole moment of the water molecule.
- We know that the electrons are localized around the oxygen atom in the water molecule.
- The localization of electrons is due to the high electronegativity of the central oxygen atom in water molecules.
- Due to the presence of lone pairs of electrons on the oxygen molecule the shape of the molecule is going to be a bent shape.
- The shape and charge separation in the water can be seen in the following picture.
seo images

- The bond angle which is present in water is 104.5$^{o}$ .
- The individual bond moment of each hydrogen-oxygen bond is 1.5 D.
- There are two hydrogen in water molecules, the two hydrogen creates their individual dipole moments.
- We have to calculate the individual dipole moments and later we have to do the sum to get the dipole moment of the water molecule.
- The dipole moment caused by the left hydrogen in water molecule = $1.5D\times \cos ({{52.2388}^{o}})=15D\times 0.612 = 0.9187$
(Here, ${{52.2388}^{o}}=\dfrac{{{104.5}^{o}}}{2}$ )
- The dipole moment caused by the right hydrogen in water molecule = $1.5D\times \cos ({{52.2388}^{o}})=15D\times 0.612 = 0.9187$
(Here, ${{52.2388}^{o}}=\dfrac{{{104.5}^{o}}}{2}$ )

Therefore the net dipole moment of the water molecule = $0.9187 + 0.9187 = 1.837 D$.

Note: The unit to measure dipole moment of the molecules is D (Debye). The dipole moment for linear molecules like beryllium difluoride are 0. Because the dipole moment caused by individual fluorine atoms is zero.