AdvancedChemistryHigh School (11–12)Chemical BondingNumericalNEET JEE Main JEE Advanced

The dipole moment of a molecule is the vector sum of its bond dipoles…

Try it on paper first. A hint, the key idea and the worked solution are here when you need them.

The dipole moment of a molecule is the vector sum of its bond dipoles. The figure shows a water molecule and a hydrogen sulphide molecule. Each one is bent, with two lone pairs on the central atom. A crossed arrow beside each bond shows its bond dipole. The vertical arrow shows the measured dipole moment of the whole molecule. The two bond dipoles of one molecule are equal in size.

Two bent molecules drawn side by side. On the left, water: an oxygen atom at the top with two lone pairs, and two hydrogen atoms below it to the left and right. The H-O-H angle is marked 104.5 degrees. Beside each O-H bond is a crossed arrow, the cross at the hydrogen end and the head at the oxygen end, labelled mu O-H. Below the molecule a vertical arrow points up towards the oxygen and is labelled mu = 1.85 D. On the right, hydrogen sulphide drawn the same way: a sulphur atom with two lone pairs, the H-S-H angle marked 92 degrees, a crossed arrow beside each bond pointing towards sulphur labelled mu S-H, and a vertical arrow labelled mu = 0.95 D.

For water, $\mu = 1.85\ \text{D}$ , the H-O-H angle is $104.5^\circ$ and the O-H bond length is $96\ \text{pm}$ . For hydrogen sulphide, $\mu = 0.95\ \text{D}$ , the H-S-H angle is $92^\circ$ and the S-H bond length is $134\ \text{pm}$ . Treat each molecular dipole as the sum of the two bond dipoles only. Take $1\ \text{D} = 3.33564 \times 10^{-30}\ \text{C m}$ and $e = 1.602 \times 10^{-19}\ \text{C}$ .

What is the dipole moment of one O-H bond, in debye?
What partial charge on each hydrogen atom of water does this bond dipole give, in coulombs?
What is the per cent ionic character of the O-H bond?
What is the dipole moment of one S-H bond, in debye?
What is the per cent ionic character of the S-H bond?

The worked solution

Have you tried it?

Work it out on paper first. It is fine to get stuck or make a mistake: that is how we learn. When you are ready, go through the solution one step at a time, and check each step against your own.

Given. The molecular dipole, the bond angle and the bond length of each molecule.

Plan. The two equal bond dipoles make an angle $\theta$ . Their parts across the vertical line cancel. Their parts along it add, so $\mu = 2\mu_b \cos(\theta/2)$ . We want $\mu_b$ , so $\mu_b = \dfrac{\mu}{2\cos(\theta/2)}$ .

Step 1. (a) The O-H bond dipole. Half the angle is $52.25^\circ$ , and $\cos 52.25^\circ = 0.6122$ . So $\mu_b = \dfrac{1.85\ \text{D}}{2 \times 0.6122}$ , which is $1.51\ \text{D}$ .

Step 2. (b) The partial charge. $q = \dfrac{\mu_b}{d}$ . In SI units, $\mu_b = 1.511 \times 3.33564 \times 10^{-30}\ \text{C m}$ . Divide by $96 \times 10^{-12}\ \text{m}$ : $q = 5.25 \times 10^{-20}\ \text{C}$ .

Step 3. (c) Per cent ionic character of O-H. $\dfrac{q}{e} \times 100 = \dfrac{5.250 \times 10^{-20}}{1.602 \times 10^{-19}} \times 100$ . This is $32.8\ \%$ .

Step 4. (d) The S-H bond dipole. Half the angle is $46^\circ$ , and $\cos 46^\circ = 0.6947$ . So $\mu_b = \dfrac{0.95\ \text{D}}{2 \times 0.6947}$ , which is $0.68\ \text{D}$ .

Step 5. (e) Per cent ionic character of S-H. The charge is $\dfrac{0.684 \times 3.33564 \times 10^{-30}}{134 \times 10^{-12}}\ \text{C}$ . As a percentage of $e$ , this is $10.6\ \%$ .

Answer: (a) $1.51\ \text{D}$ . (b) $5.25 \times 10^{-20}\ \text{C}$ . (c) $32.8\ \%$ . (d) $0.68\ \text{D}$ . (e) $10.6\ \%$ .

Check. Add the two O-H bond dipoles back: $2 \times 1.511 \times 0.6122 = 1.85\ \text{D}$ , the measured value. The O-H bond also comes out much more ionic than the S-H bond, as the larger electronegativity of oxygen predicts.

Watch out. Halving the molecular dipole gives $0.925\ \text{D}$ for one bond. That would be right only if the two bonds pointed the same way. In a bent molecule the bond dipoles partly cancel, so each bond dipole is larger than half the total.

← All Problems

← Hydrogen chloride and hydrogen fluoride are both polar molecules…A student studies a diatomic species formed by two oxygen atoms… →

More onChemical Bonding

Discussion (0)

Loading…