Question

What is the bond polarity of the water molecule?

Answer 1

`mu=1.84D`

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=Qxxr`

where, `mu` is the dipole moment,
`Q` is the coulomb charge `Q=1.60xx10^(-19)C`,
and `r` is the bond length or the distance between two ions.

For covalent compounds, the expression becomes:

`mu=deltaxxr`

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=1.5D`, where `D` is the Debye unit where, `1D=3.34xx10^(-30)C*m`.

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

`mu_("total")=2xx1.5Dxxcos(104.5/2)=1.84D`

Note that `104.5^@` is the bonds angle in water.