# What is the bond polarity of the water molecule?

Oct 7, 2015

$\mu = 1.84 D$

#### Explanation:

The polarity of water can be calculated by finding the sum of the two dipole moments of both $O - H$ bonds.

For ionic compounds, the dipole moment could be calculated by:

$\mu = Q \times r$

where, $\mu$ is the dipole moment,
$Q$ is the coulomb charge $Q = 1.60 \times {10}^{- 19} C$,
and $r$ is the bond length or the distance between two ions.

For covalent compounds, the expression becomes:

$\mu = \delta \times r$

where, $\delta$ is the partial charge on atoms.

For water, the partial charges are distributed as follows:

""^(+delta)H-O^(2delta-)-H^(delta+)

It is more complicated to calculate the partial charge on each atom, that is why I will skip this part.

The dipole moment of the $O - H$ bond is $\mu = 1.5 D$, where $D$ is the Debye unit where, $1 D = 3.34 \times {10}^{- 30} C \cdot m$.

So the net dipole moment of water could be calculated by summing the two dipole moments of both $O - H$ bonds

${\mu}_{\text{total}} = 2 \times 1.5 D \times \cos \left(\frac{104.5}{2}\right) = 1.84 D$

Note that ${104.5}^{\circ}$ is the bonds angle in water.