What is the bond polarity of the water molecule?

1 Answer

Answer:

#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

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

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