Question

The net dipole moment of water is \(1.84\) debyes and the bond angle is \(104.45^{\circ}\). What moment can be assigned to each \(\mathrm{O}-\mathrm{H}\) bond?

Short answer

The dipole moment assigned to each O-H bond in the water molecule is approximately \(0.846\) Debye.

Step-by-step solution

Understand the geometry of the water molecule

Water molecule (H2O) has a bent geometry with an O-H-O bond angle of 104.45 degrees. The molecule has a net dipole moment due to the electronegativity difference between hydrogen and oxygen atoms.

Create a vector diagram representing the O-H bond dipole moments

Let the dipole moment assigned to each O-H bond be x (in Debye). We can represent the dipole moment of each bond as vectors pointing from the hydrogen atom towards the oxygen atom. Since there's an angle between the vectors, we can use vector addition to find the net dipole moment.

Use vector addition to find the net dipole moment

The net dipole moment vector (µ_net) is the addition of the dipole moment vectors of the two O-H bonds. Let's use the law of cosines to compute this: \( \mu_\text{net}^2 = \mu_\text{OH1}^2 + \mu_\text{OH2}^2 - 2(\mu_\text{OH1})(\mu_\text{OH2})\cos(104.45) \) Since each O-H bond has the same dipole moment (x), we can substitute \(\mu_\text{OH1} = \mu_\text{OH2} = x\): \( \mu_\text{net}^2 = x^2 + x^2 - 2(x)(x)\cos(104.45) \)

Solve for the dipole moment of each O-H bond (x)

We are given that the net dipole moment is 1.84 Debye. So, replace \(\mu_\text{net}\) with 1.84 and solve for x: \( 1.84^2 = x^2 + x^2 - 2(x)(x)\cos(104.45) \) Now, solve for x: \( 3.3856 = 2x^2 - 2x^2\cos(104.45) \) Divide both sides by 2: \( 1.6928 = x^2 - x^2\cos(104.45) \) Rearrange and factor out x^2: \( x^2(1 - \cos(104.45)) = 1.6928\) Now, calculate the value of x: \( x^2 = \frac{1.6928}{1 - \cos(104.45)} \) \( x^2 = 0.7163 \) \( x = \sqrt{0.7163} \) \( x = 0.846 \text{ Debye} \)

Report the result

The dipole moment assigned to each O-H bond in the water molecule is approximately 0.846 Debye.

Key concepts

Water Molecule

The water molecule, known chemically as H₂O, has a simple yet fascinating structure that is crucial for many biological and chemical processes. This molecule consists of two hydrogen atoms covalently bonded to a single oxygen atom. What makes water particularly interesting is its bent shape, forming an angle between the two hydrogen atoms. This geometry significantly affects its chemical properties and interactions. In water, the oxygen atom has a partial negative charge while the hydrogen atoms have partial positive charges. This charge separation occurs because oxygen is more electronegative than hydrogen, meaning it pulls electrons towards itself more effectively. This difference leads to a dipole moment in the molecule, making water a polar molecule. The polarity of water is a primary reason it is an excellent solvent and plays a crucial role in hydrogen bonding, which affects phenomena such as surface tension and capillary action.

O-H Bond

The O-H bond in the water molecule is a type of covalent bond. This bond results from the sharing of electrons between the oxygen and hydrogen atoms. However, due to oxygen's higher electronegativity, it attracts the shared electrons more strongly, causing a partial negative charge on the oxygen and a partial positive charge on the hydrogen. These polar bonds are essential in giving the water molecule its unique properties. Each O-H bond has its own dipole moment. In simple terms, a dipole moment is a measure of the separation of positive and negative charges within the bond. In water, each O-H bond contributes to the overall dipole moment of the molecule. Understanding the dipole moments of these bonds helps chemists predict how water interacts with other substances. The strength and properties of O-H bonds can also affect the behavior of water in different environments, such as its role as a solvent or its ability to form hydrogen bonds with other molecules.

Bond Angle

In a water molecule, the bond angle refers to the angle formed between its two hydrogen atoms and the oxygen atom. This angle is approximately 104.45 degrees, which is less than the ideal tetrahedral angle of 109.5 degrees. The bond angle in water arises from the lone pairs of electrons on the oxygen atom, which repel the hydrogen atoms. These lone pairs push the hydrogen atoms closer together, resulting in a smaller bond angle and giving water its characteristic V-shape. The angle does not just determine the molecule’s shape; it also influences its chemical properties. For example, the bond angle contributes to water's polarity, affecting its boiling and melting points, and its density in different states. Understanding how the bond angle affects water can help explain many of its unique behaviors and properties.

Vector Addition

Vector addition is a mathematical concept used to determine the overall effect of two or more vectors acting at an angle to each other. In the context of the water molecule, vector addition helps calculate the net dipole moment by considering the individual dipole moments of the O-H bonds.Each O-H bond in water has its own dipole vector, pointing from the hydrogen towards the more electronegative oxygen atom. Because of the bond angle, these vectors do not point in the same direction. To find the net dipole moment, we use vector addition, which involves adding the vectors taking their directions and magnitudes into account.The law of cosines is particularly useful in this scenario. By applying \[ \mu_\text{net} = \sqrt{2x^2(1 - \cos(104.45))} \] we consider both the magnitudes of the dipole moments (denoted as x in the equation) and the angle between them. This process results in an accurate calculation of the net dipole moment, a critical factor for understanding water's properties.