91Ó°ÊÓ

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.

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 \(1.56 \mathrm{D}\).

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.