Question

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

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).

Key concepts

O-H Bond

In the water molecule, the \(\text{O-H}\) bond is a fundamental part of its structure. These bonds form between the oxygen ( ext{O}) and hydrogen ( ext{H}) atoms. The length of each \(\text{O-H}\) bond in water is approximately 0.96 Å (angstroms). This distance indicates how far apart the oxygen and hydrogen atoms are bonded together and plays a crucial role in the properties of water.
The geometry of the water molecule itself is not linear, which significantly influences the direction of bond dipoles. Instead, there is an angle of 104.5° between the two hydrogen atoms connected to the oxygen atom. This angle affects how the bond dipoles are directed within the molecule. Importantly, these dipoles point towards the oxygen atom since oxygen is more electronegative than hydrogen. This intrinsic structure contributes to water's bent shape and results in net dipole moments owing to the asymmetrical position of atoms.

Electronegativity

Electronegativity is a measure of an atom's ability to attract electrons in a bond. In the context of the \(\text{O-H}\) bond, oxygen has a higher electronegativity than hydrogen. Oxygen's electronegativity allows it to draw the shared electrons towards itself more strongly, creating a polar bond.
In water, this results in the oxygen exhibiting a partial negative charge ( ext{δ}-), and the hydrogen atoms displaying a partial positive charge ( ext{δ}+). This difference in electronegativity between the two atoms is what gives rise to the bond dipoles that collectively contribute to the overall dipole moment of the water molecule.
Moreover, the greater the difference in electronegativity between two bonded atoms, the more polar the bond tends to be. This polarity is significant in water's ability to dissolve various substances, making it a universal solvent. It also affects the water molecule's interactions with other molecules and can influence physical properties like boiling point and surface tension.

Vector Addition

To determine the compound dipole moment of a water molecule, the vector addition concept is vital. Each \(\text{O-H}\) bond in the water molecule contributes to the molecule’s net dipole moment. These bond dipoles are vector quantities. This means they have both magnitude and direction, and their combined effect must be considered using vector addition.
The dipole moments from the two \(\text{O-H}\) bonds make an angle of 104.5° with each other. To compute their resultant vector, the vector addition equation is used:

• Recognize each bond dipole as separate vectors originating from the oxygen atom.
• Add these vectors using the law of cosines or similar vector addition techniques to find the resultant dipole moment of the molecule.
• This resultant vector represents the overall dipole moment of the water molecule and points perpendicular to the plane of the \(\text{O-H}\) bonds.

The calculated overall dipole moment (1.85 D) highlights the water molecule's polarity, showcasing how vector addition reveals the direction and strength of this net dipole effect.