Question

How do you calculate the dipole moment of water?

Answer 1

Step 1: Formula for Dipole moment:

The Dipole moment is known to be a vector quantity having both, magnitude and direction.

The formula for calculating the dipole moment can be given as:

$\mu (\mathrm{Dipole}\mathrm{moment})=\mathrm{q}(\mathrm{magnitude}\mathrm{of}\mathrm{the}\mathrm{charge})\times \mathrm{l}(\mathrm{distance}\mathrm{of}\mathrm{separation})$

Step 2: Calculation of the dipole moment of water:

For the calculation of the dipole moment of the water molecule, it is known that the electrons are localized around the oxygen atom in the water molecule because of the high electronegativity of the central oxygen atom.

Due to the presence of lone pairs of electrons on the oxygen molecule, the shape of the molecule has been known to be a bent shape, which can be shown in the figure below;

Here the bond angle for water is given as $=104.5°$

There are two hydrogens in water molecules that create their individual dipole moments, and to reach them the individual bond moment is $1.5\mathrm{D}.$

The dipole moment caused by the left hydrogen in the water molecule = $1.5\mathrm{D}\times \cos \left({52.2388}^{\mathrm{o}}\right)=15\mathrm{D}\times 0.612=0.9187$
(Here,${52.2388}^o={\displaystyle \frac{{104.5}^o}{2}}$ )

The dipole moment caused by the right hydrogen in the water molecule =$1.5D\times \cos ({52.2388}^o)=15D\times 0.612=0.9187$
(Here, ${52.2388}^o={\displaystyle \frac{{104.5}^o}{2}}$ )

Step 3: Calculating the resultant dipole moment of the water:

The resultant dipole moment of the water molecule can be calculated by summing up the dipole moment of the two Hydrogens, given as:

The net dipole moment of water =$0.9187+0.9187=1.837D$.

Hence, the dipole moment of water is calculated to be as:$1.837\mathrm{D}.$