The \(\mathrm{O}-\mathrm{H}\) bond lengths in the water molecule \(\left(\mathrm{H}_{2} \mathrm{O}\right)\) are \(96 \mathrm{pm}\), and the \(\mathrm{H}-\mathrm{O}-\mathrm{H}\) angle is \(104.5^{\circ} .\) The dipole moment of the water molecule is \(1.85 \mathrm{D}\). (a) In what directions do the bond dipoles of the \(\mathrm{O}-\mathrm{H}\) bonds point? \(\mathrm{In}\) what direction does the dipole moment vector of the water molecule point? (b) Calculate the magnitude of the bond dipole of the \(\mathrm{O}-\mathrm{H}\) bonds. (Note: You will need to use vector addition to do this.) (c) Compare your answer from part (b) to the dipole moments of the hydrogen halides (Table 8.3). Is your answer in accord with the relative electronegativity of oxvgen?

Step by step solution

01

Part (a) - Directions of Bond Dipoles and Dipole Moment Vector

For the water molecule, the bond dipoles of the O-H bonds point from the hydrogen atom to the oxygen atom since oxygen is more electronegative than hydrogen. The dipole moment vector of the water molecule lies in the plane of the molecule, and points in the direction of the resultant of the vectors of the two O-H bond dipoles. Since the resulting dipole moment vector points from the side with a partial positive charge to the side with a partial negative charge, it will point towards the oxygen atom and bisect the H-O-H angle.

02

Part (b) - Calculating the Magnitude of Bond Dipole

To find the magnitude of the bond dipole of the O-H bonds, we need to use vector addition. Let's consider the bond dipoles as vectors. We can write the bond dipole of one O-H bond as P1 and the bond dipole of the other as P2. The bond angle between these two dipoles is given as 104.5°. The net dipole moment (P_net) of the water molecule can be found by vector addition of P1 and P2. From the given information, we have P_net = 1.85 D (debyes). Now, we can use the law of cosines to find the net dipole moment: P_net^2 = P1^2 + P2^2 - 2 * P1 * P2 * cos(θ) Since both O-H bond dipoles are identical, we have P1 = P2 = P. The given bond angle θ = 104.5°. 1.85^2 = P^2 + P^2 - 2 * P * P * cos(104.5°) Substituting the value of the bond angle and solving for P, we get: P = 1.79 D.

03

Part (c) - Comparing Bond Dipole with Hydrogen Halides

To compare the bond dipole of oxygen in water with that of hydrogen halides, we can refer to Table 8.3 in the book. The dipole moments of H-X (where X is a halogen) increase as we go down the group from hydrogen fluoride (HF) to hydrogen iodide (HI) due to the increasing electronegativity difference between hydrogen and the halogen. The bond dipole moment of water (1.79 D) is found to be larger than that of hydrogen halides except for HF (1.91 D). This indicates that the O-H bond in water is highly polar, resulting in a more significant tendency for electron density to be concentrated near the more electronegative oxygen atom. This is also in accord with the relative electronegativity values as oxygen is a more electronegative element than the halogens except for fluorine.

Unlock Step-by-Step Solutions & Ace Your Exams!

• Full Textbook Solutions

  Get detailed explanations and key concepts

• Unlimited Al creation

  Al flashcards, explanations, exams and more...

• Ads-free access

  To over 500 millions flashcards

• Money-back guarantee

  We refund you if you fail your exam.

Start your free trial

Over 30 million students worldwide already upgrade their learning with Vaia!