Question

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 \(1.85\) 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

The bond dipoles of the O-H bonds in a water molecule point from the partial positive charge of H to the partial negative charge on O, along the H-O bonds. The total dipole moment vector of the water molecule points approximately towards the centroid of the imaginary equilateral triangle formed by the hydrogen and oxygen atoms if all the atoms were in the same plane. The magnitude of the bond dipole of the O-H bonds, calculated using vector addition, is found to be 1.65 D. Comparing this value with the dipole moments of the hydrogen halides, the O-H bond dipole is greater, suggesting that the electronegativities of oxygen and hydrogen are more different than those of hydrogen and the halogens. This is consistent with the known fact that oxygen is more electronegative than the halogens.

Step-by-step solution

1a. Bond dipole direction of O-H bonds

In a water molecule, the O-H bond dipoles point in the direction from the partial positive charge of H to the partial negative charge on O. This direction is along the H-O bond in both O-H bonds.

1b. Direction of the dipole moment vector of the water molecule

The total dipole moment vector of the water molecule is the vector sum of the individual bond dipoles. As both bond dipoles are not collinear, their vector sum points along a direction in between the two bond dipoles. It would point approximately towards the centroid of the imaginary equilateral triangle formed by the hydrogen and oxygen atoms if all the atoms were in the same plane. ###Step 2: Calculate the magnitude of the O-H bond dipole###

2a. Calculate the vector components of individual bond dipoles

Each bond dipole can be described by two vector components. The component along the x-axis is given by \(\mu_{H-O} \times \cos(\theta / 2)\) and the component along the y-axis is given by \(\mu_{H-O} \times \sin(\theta / 2)\), where \(\theta\) is the H-O-H angle.

2b. Calculate the vector components of the water molecule dipole moment

Let \(D_x\) and \(D_y\) be the x and y components of the water molecule dipole moment. Then, \(D_x = 2 \times \mu_{H-O} \times \cos(\theta / 2) = 1.85\) D since there are two O-H bonds. And \(D_y = 2 \times \mu_{H-O} \times \sin(\theta / 2) = 0\) since the y-components of the bond dipoles cancel each other out.

2c. Calculate the magnitude of the bond dipole of the O-H bonds

Rearranging the equations in 2b, we get \(\mu_{H-O} = \frac{D_x}{2 \times \cos(\theta / 2)}\). Substituting the given values, we get \(\mu_{H-O} = \frac{1.85}{2 \times \cos(104.5^\circ / 2)} = 1.65\) D (rounded to two decimal places). ###Step 3: Compare O-H bond dipole with hydrogen halides and relate to electronegativity###

3a. Compare O-H bond dipole with hydrogen halides

The dipole moments of hydrogen halides (Table 8.3) are lower than the calculated bond dipole for the O-H bonds (1.65 D vs. 1.04 D to 1.22 D).

3b. Relate the findings to the electronegativity of oxygen

The larger dipole moment for the O-H bond suggests that the electronegativities of oxygen and hydrogen are more different than those of hydrogen and the halogens in comparison. This is in accord with the known electronegativity values that oxygen is more electronegative than the halogens.

Key concepts

Bond Dipole

In the water molecule (H₂O), bond dipoles arise due to the difference in electronegativity between the oxygen and hydrogen atoms. Oxygen is more electronegative and therefore attracts the shared electrons in the O-H bond more strongly than hydrogen. This creates a partial negative charge on oxygen and a partial positive charge on each hydrogen atom.

As a result, each O-H bond becomes polar, forming a bond dipole. The direction of the bond dipole is from the hydrogen (positive end) towards the oxygen (negative end), along the O-H bond axis. Since the molecule is V-shaped with an angle of 104.5°, these two bond dipoles are not in a straight line, which gives the water molecule its overall dipole moment.

Vector Addition

The total dipole moment of the water molecule is found using vector addition. Here, the vectors are the bond dipoles, and the angle between these dipoles is a crucial factor. The dipole moment is not simply the sum of magnitudes but depends on the angle between vectors and their direction.

The two O-H bond dipoles in a water molecule form an angle of 104.5°. By breaking down each bond dipole into x and y components, we can sum them to find the molecule's total dipole moment. The x-components add up, while the y-components cancel each other due to the symmetry of the structure. This results in a net dipole moment that roughly bisects the H-O-H angle, pointing between the hydrogen atoms.

Electronegativity

Electronegativity refers to an atom's ability to attract and hold onto electrons. In the context of a water molecule, oxygen is highly electronegative compared to hydrogen. This electronegativity difference is why oxygen pulls the shared electrons in the O-H bonds closer, causing polarity in the molecule.

Comparatively, in hydrogen halides, the halogen atoms like chlorine or bromine are less electronegative than oxygen. Consequently, the dipole moments of hydrogen halides are lower than that of the O-H bonds. The electronegativity scales agree with this, showing oxygen's stronger ability to attract electrons than halogens such as chlorine and bromine.

O-H Bond Length

The O-H bond length in water is about 0.96 Å, a typical distance for the bond between hydrogen and a more electronegative atom like oxygen. This bond length affects the strength and polarity of the bond, as the electrons shared in the bond are held closer to the more electronegative oxygen atom.

Bond lengths are crucial in determining the molecular geometry and overall dipole moments. The short bond length in water allows the strong interaction between oxygen and hydrogen, but the angle and other factors like electron distribution also play significant roles in shaping the molecule's properties as a solvent and its high polarity.