Question

The dipole moment of HF is \(1.83 \mathrm{D}\) and the bond length is \(91.7 \mathrm{pm} .\) Calculate the amount of charge (in electronic charge units) on the hydrogen and the fluorine atoms in the HF molecule.

Short answer

The amount of charge on the hydrogen and fluorine atoms in the HF molecule is approximately \textpm 4.15 e.

Step-by-step solution

Convert dipole moment to Coulomb-meters

First, convert the dipole moment from Debye to Coulomb-meters using the conversion factor: 1 Debye = 3.33564 x 10^-30 Coulomb-meters. So the dipole moment is: 1.83 D * 3.33564 x 10^-30 Cm/D = 6.10422 x 10^-30 Cm.

Convert bond length to meters

Next, convert the bond length from picometers to meters: 91.7 pm = 91.7 x 10^-12 m.

Calculate the charge

The amount of charge (q) on the atoms can be calculated by rearranging the formula for the dipole moment: p = q * d, where p is the dipole moment and d is the bond length. Solving for q gives: q = p / d = (6.10422 x 10^-30 Cm) / (91.7 x 10^-12 m) = 6.6528 x 10^-19 Coulombs.

Convert charge to electronic charge units

Lastly, convert the computed charge into electronic charge units (e) where 1 e = 1.602 x 10^-19 Coulombs. This is done by dividing the charge by the electronic charge value: (6.6528 x 10^-19 C) / (1.602 x 10^-19 C/e) = 4.15 e.

Key concepts

Electrostatics in Molecules

Understanding electrostatics in molecules is crucial when studying how atoms bond and interact within a compound. Molecules often have regions of partial positive and negative charge that arise due to the uneven distribution of electrons between the atoms. This separation of charges within a molecule is termed as molecular polarity, which results in a characteristic called the dipole moment. \( \text{The dipole moment is a vector quantity that conveys both the magnitude and direction of the molecular polarity.} \) To quantify this, the dipole moment is represented by the product of the amount of charge separated and the distance between the charges. If we imagine the molecule as a tiny bar magnet, the dipole moment points from the positive to the negative end, ideally providing insights into the molecule's interaction with electric fields and other charged entities. An important note for students is to recognize the difference in electronegativity of the atoms involved, as this affects the degree of charge separation and hence the size of the dipole moment. For example, in a hydrogen fluoride (HF) molecule, the high electronegativity of fluorine pulls more electron density towards itself, creating a more significant dipole moment than molecules with less electronegative partners.

Molecular Bond Length

The molecular bond length is the average distance between the nuclei of two bonded atoms in a molecule. This measurement is pivotal in understanding chemical properties and reactions, as it affects the energy and the stability of the bond. Bond lengths vary widely among different types of atoms and molecules, influenced by factors such as atomic sizes, hybridization states, and the presence of multiple bonds (single, double, or triple).In electrostatic context, bond length plays a role in the calculation of the dipole moment. When considering a diatomic molecule like HF, its dipole moment is the product of the charge separated and the bond length. We use the bond length to relate the calculated physical dipole moment to the charges on the individual atoms, \( p = q \times d \), where \( p \) is the dipole moment, \( q \) represents the charge on atoms, and \( d \) symbolizes the bond length. In the context of the given exercise, converting the bond length from picometers to meters was essential for calculating the dipole moment in the correct units, which is directly related to the next concept of electronic charge units.

Electronic Charge Units

Electronic charge units, often denoted by the symbol \( e \), represent the natural unit of electrical charge. One electronic charge is approximately equated to the charge of a single proton (positive) or electron (negative), which is a constant, \( 1.602 \times 10^{-19} \) Couombs. In the realm of atomic and molecular physics, these units simplify calculations as they convert large or small decimal numbers into comprehensible integers or fractions of the fundamental charge.When students calculate the charge separation in a molecule, determining the charge in electronic charge units gives a more tangible illustration of how many 'protons or electrons worth' of charge is involved. This concept ties directly to the textbook exercise, where the charges on hydrogen and fluorine in an HF molecule were calculated. Upon computing the charge in Coulombs, converting to electronic charge units made the answer more relatable, ending up as a number a little above 4, indicating a charge distribution relatively close to the charge of four electrons. This serves as a powerful visual for students to better grasp molecule polarity and electric characteristics.