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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

01

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掳, we can visualize this as two vectors originating from the oxygen atom, each making an angle of 104.5掳 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#
02

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掳), we need to use vector addition to calculate their sum. We can use the following equation to determine the bond dipole: \[\mu_{H_2O}^2 = 2\mu_{OH}^2 + 2\mu_{OH}^2 \cos{(104.5掳)}\] We are given \(\mu_{H_2O} = 1.85\) D, and we need to find \(\mu_{OH}\). Solving for \(\mu_{OH}\), we get: \[\mu_{OH} = \sqrt{\frac{\mu_{H_2O}^2}{2(1+\cos{(104.5掳)})}} \] \[\mu_{OH} \approx 1.73 \text{ D}\] #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).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

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 脜 (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掳 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{未}-), and the hydrogen atoms displaying a partial positive charge ( ext{未}+). 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鈥檚 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掳 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.

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Most popular questions from this chapter

The lactic acid molecule, \(\mathrm{CH}_{3} \mathrm{CH}(\mathrm{OH}) \mathrm{COH},\) gives sour milk its unpleasant, sour taste. (a) Draw the Lewis structure for the molecule, assuming that carbon always forms four bonds in its stable compounds. (b) How many \(\pi\) and how many \(\sigma\) bonds are in the molecule? (c) Which CO bond is shortest in the molecule? (d) What is the hybridization of atomic orbitals around the carbon atom associated with that short bond? (e) What are the approximate bond angles around each carbon atom in the molecule?

Sulfur tetrafluoride \(\left(\mathrm{SF}_{4}\right)\) reacts slowly with \(\mathrm{O}_{2}\) to form sulfur tetrafluoride monoxide (OSF_ \(_{4} )\) according to the following unbalanced reaction: \begin{equation}\mathrm{SF}_{4}(g)+\mathrm{O}_{2}(g) \longrightarrow \mathrm{OSF}_{4}(g) \end{equation} The O atom and the four \(\mathrm{F}\) atoms in OSF \(_{4}\) are bonded to a central \(\mathrm{S}\) atom. (a) Balance the equation. (b) Write a Lewis structure of OSF_ in which the formal charges of all atoms are zero.(c) Use average bond enthalpies (Table 8.3 ) to estimate the enthalpy of the reaction. Is it endothermic or exothermic? (d) Determine the electron-domain geometry of \(\mathrm{OSF}_{4}\), and write two possible molecular geometries for the molecule based on this electron-domain geometry. (e) For each of the molecules you drew in part (d), state how many fluorines are equatorial and how many are axial.

(a) An AB \(_{6}\) molecule has no lone pairs of electrons on the A atom. What is its molecular geometry? (b) An AB \(_{4}\) molecule has two lone pairs of electrons on the A atom (in addition to the four B atoms). What is the electron-domain geometry around the A atom? (c) For the AB \(_{4}\) molecule in part (b), predict the molecular geometry.

Consider the molecule \(\mathrm{BF}_{3}\). (a) What is the electron configuration of an isolated B atom? (b) What is the electron configuration of an isolated F atom? (c) What hybrid orbitals should be constructed on the B atom to make the B鈥揊 bonds in \(\mathrm{B} \mathrm{F}_{3}\)?(d) What valence orbitals, if any, remain unhybridized on the B atom in \(\mathrm{BF}_{3} ?\)

Molecules that are brightly colored have a small energy gap between filled and empty electronic states (the HOMO-LUMO gap; see Exercise 9.104 ). Sometimes you can visually tell which HOMO-LUMO gap is larger for one molecule than another. Suppose you have samples of two crystalline powders-one is white, and one is green. Which one has the larger HOMO-LUMO gap?
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