Chapter 9: Problem 95

The \(\mathrm{O}-\mathrm{H}\) bond lengths in the water molecule \(\left(\mathrm{H}_{2} \mathrm{O}\right)\) are\(0.96 \mathrm{A},\) 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? 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. \((\mathbf{c})\) 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 oxygen?

Short Answer

Expert verified

In a water molecule, the O-H bond dipoles point towards the oxygen atom, and the dipole moment vector bisects the H-O-H angle. The magnitude of the O-H bond dipole is approximately 1.73 D, larger than the dipole moments of hydrogen halides. This is in accord with the relative electronegativity of oxygen, which is more electronegative than other elements in the hydrogen halides.

Step by step solution

In a water molecule, the O-H bonds are directed along the axes connecting the oxygen atom with the two hydrogen atoms. Since the H-O-H angle is 104.5°, we can visualize this as two vectors originating from the oxygen atom, each making an angle of 104.5° relative to each other. The bond dipoles point towards the more electronegative atom, which in this case is oxygen. #Step 2: Determine the direction of the dipole moment vector of the water molecule#

To determine the direction of the dipole moment vector, we need to consider the O-H bond dipoles. The dipole moment vector of the water molecule is the vector sum of the two O-H bond dipoles. Since these dipoles are not in the same plane, the resultant vector will point in a direction normal to the plane formed by the O-H bond dipoles. In the case of the water molecule, this vector points along an axis perpendicular to the bisector of the H-O-H angle (i.e., the dipole moment vector bisects the H-O-H angle). #Step 3: Calculate the magnitude of the bond dipole of the O-H bonds#

Let's denote the bond dipole of each O-H bond as \(\mu_{OH}\). The total dipole moment of the water molecule is given by \(\mu_{H_2O}\). Since the bond dipoles form a non-zero angle with each other (104.5°), we need to use vector addition to calculate their sum. We can use the following equation to determine the bond dipole: \[\mu_{H_2O}^2 = 2\mu_{OH}^2 + 2\mu_{OH}^2 \cos{(104.5°)}\] We are given \(\mu_{H_2O} = 1.85\) D, and we need to find \(\mu_{OH}\). Solving for \(\mu_{OH}\), we get: \[\mu_{OH} = \sqrt{\frac{\mu_{H_2O}^2}{2(1+\cos{(104.5°)})}} \] \[\mu_{OH} \approx 1.73 \text{ D}\] #Step 4: Compare the calculated bond dipole with the dipole moments of hydrogen halides and its accord with the relative electronegativity of oxygen#

The calculated bond dipole of the O-H bond (1.73 D) is larger than the dipole moments of hydrogen halides, such as the dipole moment of H-Cl (1.08 D). This is in accord with the relative electronegativity of oxygen. Oxygen is more electronegative than other elements in the hydrogen halides, which means that the bond dipoles in O-H bonds will be larger than those in H-X bonds (where X is a halogen).