The dipole moment of \(\mathrm{LiH}\) is \(1.964 \times 10^{-27} \mathrm{Cm}\) and the intermolecular distance between \(\mathrm{Li}\) and \(\mathrm{H}\) in this molecule is \(1.596 \AA\). What is per cent ionic character in molecule?

Short Answer

Expert verified
The percent ionic character of LiH is approximately 11.17%.

Step by step solution

01

Determine the dipole moment in Debye units

The dipole moment given is in Coulomb-meters (Cm). To find the percent ionic character, it is helpful to first convert this value to Debye units, since Debye is a more common unit for molecular dipole moments. The conversion factor is: 1 Debye = 3.33564 × 10^-30 Cm. Therefore, divide the given dipole moment by this conversion factor to get the value in Debye.
02

Calculate the theoretical dipole moment for 100% ionic character

The theoretical dipole moment (μ_theoretical) assuming 100% ionic character can be calculated by multiplying the elementary charge (e = 1.602 × 10^-19 C) by the distance between the ions (d, in meters). The distance must first be converted from Angstroms to meters (1 Angstrom = 10^-10 meters).
03

Calculate the percent ionic character

The percent ionic character can be found by dividing the measured dipole moment in Debye (μ_measured) by the theoretical dipole moment for a 100% ionic bond (μ_theoretical), and then multiplying by 100 to get the percentage. This gives the proportion of the bond's ionic character relative to a purely ionic bond.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with Vaia!

Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Dipole Moment
A dipole moment is a vector quantity that represents the separation of positive and negative electrical charges within a molecule. It is a crucial concept in chemistry for understanding the polarity of bonds and molecules. The larger the dipole moment, the more polar the bond. To quantify dipole moments, we utilize units called Debye (D).

Starting with a value given in Coulomb-meters, one must convert to Debye to align with conventional chemistry practice. This conversion not only aids in easier comparison with standard tables and literature but also simplifies calculations. It is a fundamental example of the importance of unit conversion in chemistry.
Ionic Bonding
Ionic bonding is characterized by the electrostatic attraction between oppositely charged ions. Typically, it occurs between a metal that loses electrons to become a cation, and a nonmetal that gains electrons to become an anion. These compounds usually have high melting and boiling points due to the strong forces holding the ions together.

In the context of dipole moments and percent ionic character, a purely ionic bond would hypothetically exhibit a dipole moment equal to the product of the charge of an electron and the separation between the ions. However, most ionic compounds have some degree of covalent character – they're not 100% ionic – leading us to calculate their percent ionic character.
Conversion of Units in Chemistry
Unit conversion is a fundamental skill in chemistry, allowing scientists to compare and apply measurements in different units appropriately. For dipole moment calculations, a conversion from Coulomb-meters to Debye units is essential. Knowing the correct conversion factors and how to apply them ensures accurate and consistent results across scientific disciplines.

For instance, the conversion from Angstrom units to meters is frequently necessary since measurements at the atomic scale are often provided in Angstroms. Mastering these conversions is vital for precise scientific communication and analyses.
Electronegativity and Bond Character
Electronegativity is a measure of an atom's tendency to attract electrons in a chemical bond. The difference in electronegativity between bonding atoms can indicate the bond's character, whether it is more ionic or covalent. Large differences usually signify ionic bonds, while small differences point towards covalent bonds.

The percent ionic character of a bond can be thought of as a way to quantify the effect of electronegativity differences on bond polarity. A greater percent ionic character indicates a bond is closer to the ionic end of the bonding spectrum, with a corresponding high dipole moment, and vice versa for covalent bonds which have lower dipole moments.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

Distribution coefficient of an organic acid between water and benzene is \(4.1\) is favour of \(\mathrm{C}_{6} \mathrm{H}_{6}\). If \(5 \mathrm{~g}\) of the acid is distributed in between 50 \(\mathrm{mL}\) of benzene and \(100 \mathrm{~mL}\) of water, calculate the concentration of the acid in two solvents.

\(100 \mathrm{~mL}\) of \(1 \times 10^{-2} M\) aqueous solution of an organic compound were shaken with \(50 \mathrm{~mL}\) of an organic solvent till equilibrium is attained. Calculate the concentration of organic compound in organic solvent. Given that distribution coefficient of organic compound for the given solvent is 50 in favour of organic solvent.

Calculate the \(\%\) ionic character in \(\mathrm{HCl}\) molecule. Given bond length of \(\mathrm{HCl}\) is \(1.275 \AA\) and \(\mu_{\mathrm{HCi}}=1.03\) debye.

Calculate the enthalpy change for the reaction, $$ \mathrm{XeF}_{4} \longrightarrow \mathrm{Xe}^{+}+\mathrm{F}^{-}+\mathrm{F}_{2}+\mathrm{F} $$ The average \(\mathrm{Xe}-\mathrm{F}\) bond energy is \(34 \mathrm{kcal} / \mathrm{mol}\), first I.E. of \(\mathrm{Xe}\) is 279 kcal/mol, electron affinity of \(\mathrm{F}\) is 85 kcal/ mol and bond dissociation energy of \(\mathrm{F}_{2}\) is \(38 \mathrm{kcal} / \mathrm{mol}\).

\(1 \mathrm{~g}\) of \(\mathrm{Mg}\) atoms in the vapour phase absorbs \(50.0 \mathrm{~kJ}\) of energy. Find the composition of \(\mathrm{Mg}^{+}\) and \(\mathrm{Mg}^{2+}\) formed as a result of absorption of energy. \(\mathrm{IE}_{1}\) and \(\mathrm{IE}_{2}\) for \(\mathrm{Mg}\) are 740 and \(1450 \mathrm{~kJ} \mathrm{~mol}^{-1}\) respectively.

See all solutions

Recommended explanations on Chemistry Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.

Sign-up for free