The \(\mathrm{O}-\mathrm{H}\) bond lengths in the water molecule \(\left(\mathrm{H}_{2} \mathrm{O}\right)\) are \(0.96 \AA\), 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.) (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
(a) The bond dipoles of the O-H bonds point towards the oxygen atom, and the overall dipole moment vector of the water molecule also points towards the oxygen atom. (b) The magnitude of the O-H bond dipole is approximately \(1.56 \mathrm{D}\). (c) The O-H bond dipole moment aligns with the relative electronegativity of oxygen, as it is larger than the H-Cl bond (\(1.11 \mathrm{D}\)) and smaller than the H-F bond (\(1.91 \mathrm{D}\)).

Step by step solution

01

(a) Bond dipole and dipole moment directions

Since oxygen is more electronegative than hydrogen, the bond dipoles point towards the oxygen atom in both O-H bonds. To find the direction of the overall dipole moment of the water molecule, we have to sum the bond dipoles as vectors. Due to the V-shaped structure and similar bond dipoles, the resultant dipole moment vector also points towards the oxygen atom.
02

(b) Magnitude of O-H bond dipole

Let's denote the O-H bond dipoles as vectors \(\vec{P_1}\) and \(\vec{P_2}\), and the dipole moment of the water molecule as \(\vec{P}\). The magnitude of \(\vec{P}\) is given as \(1.85 \mathrm{D}\). First, we find the angle between the bond dipoles. Since the angle between O-H bonds is \(104.5^{\circ}\), the angle between bond dipoles (\(\theta\)) is also \(104.5^{\circ}\). Using the law of cosines for vector addition, we can express \(\vec{P}\) in terms of \(\vec{P_1}\) and \(\vec{P_2}\). Let \(p\) denote the magnitude of \(\vec{P_1}\), \(\vec{P_2}\), and \(\vec{P}\). \[p^2 = p_1^2 + p_2^2 - 2p_1p_2\cos{\theta}\] Since both O-H bond dipoles are equal in magnitude, we have \(p_1 = p_2\). Therefore, \[p^2 = 2p_1^2 - 2p_1^2\cos{104.5^{\circ}}\] We can now solve for \(p_1\): \[p_1^2 = \frac{p^2}{2 - 2\cos{104.5^{\circ}}}\] \[p_1 = \sqrt{\frac{(1.85 \mathrm{D})^2}{2 - 2\cos{104.5^{\circ}}}}\] \[p_1 \approx 1.56 \mathrm{D}\] Hence, the magnitude of the O-H bond dipole is approximately \(1.56 \mathrm{D}\).
03

(c) Comparison with hydrogen halides and electronegativity

Comparing the dipole moment of the O-H bond with the hydrogen halides (Table 8.3), we see that it lies in the expected range, considering the relative electronegativities of atom pairs. The O-H bond has a larger dipole moment than the H-Cl bond (\(1.11 \mathrm{D}\)), and smaller dipole moments than the H-F bond (\(1.91 \mathrm{D}\)). Furthermore, since oxygen is more electronegative than other elements such as nitrogen and carbon, the O-H bond dipole moment is indeed in accord with the relative electronegativity of oxygen.

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Most popular questions from this chapter

(a) If the valence atomic orbitals of an atom are sp hybridized, how many unhybridized \(p\) orbitals remain in the valence shell? How many \(\pi\) bonds can the atom form? (b) Imagine that you could hold two atoms that are bonded together, twist them, and not change the bond length. Would it be easier to twist (rotate) around a single \(\sigma\) bond or around a double \((\sigma\) plus \(\pi)\) bond, or would they be the same? Explain.

Consider the bonding in an \(\mathrm{MgH}_{2}\) molecule. (a) Draw a Lewis structure for the molecule, and predict its molecular geometry. (b) What hybridization scheme is used in \(\mathrm{MgH}_{2}\) ? (c) Sketch one of the two-electron bonds between an \(\mathrm{Mg}\) hybrid orbital and an \(\mathrm{H} 1 \mathrm{~s}\) atomic orbital.

The three species \(\mathrm{NH}_{2}^{-}, \mathrm{NH}_{3}\), and \(\mathrm{NH}_{4}^{+}\) have \(\mathrm{H}-\mathrm{N}-\mathrm{H}\) bond angles of \(105^{\circ}, 107^{\circ}\), and \(109^{\circ}\), respectively. Explain this variation in bond angles.

Consider the molecule \(\mathrm{PF}_{4}\) Cl. (a) Draw a Lewis structure for the molecule, and predict its electron-domain geometry. (b) Which would you expect to take up more space, a \(\mathrm{P}-\mathrm{F}\) bond or a \(\mathrm{P}-\mathrm{Cl}\) bond? Explain. (c) Predict the molecular geometry of \(\mathrm{PF}_{4} \mathrm{Cl}\). How did your answer for part (b) influence your answer here in part (c)? (d) Would you expect the molecule to distort from its ideal electron-domain geometry? If so, how would it distort?

The azide ion, \(\mathrm{N}_{3}^{-}\), is linear with two \(\mathrm{N}-\mathrm{N}\) bonds of equal length, \(1.16 \AA\). (a) Draw a Lewis structure for the azide ion. (b) With reference to Table \(8.5,\) is the observed \(\mathrm{N}-\mathrm{N}\) bond length consistent with your Lewis structure? (c) What hybridization scheme would you expect at each of the nitrogen atoms in \(\mathrm{N}_{3}^{-}\) ? (d) Show which hybridized and unhybridized orbitals are involved in the formation of \(\sigma\) and \(\pi\) bonds in \(\mathrm{N}_{3}^{-}\). (e) It is often observed that \(\sigma\) bonds that involve an \(s p\) hybrid orbital are shorter than those that involve only \(s p^{2}\) or \(s p^{3}\) hybrid orbitals. Can you propose a reason for this? Is this observation applicable to the observed bond lengths in \(\mathrm{N}_{3}^{-}\) ?

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