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Chapter 42: Problem 37

CP (a) The equilibrium separation of the two nuclei in an NaCl molecule is 0.24 \(\mathrm{nm}\) . If the molecule is modeled as charges \(+e\) and \(-e\) separated by 0.24 \(\mathrm{nm}\) , what is the electric dipole moment of the molecule (see Section 21.7\() ?\) (b) The measured electric dipole moment arises from point charges \(+q\) and \(-q\) separated by 0.24 \(\mathrm{nm}\) , what is \(q ?\) (c) A definition of the fractional ionic character of the bond is \(q / e\) . If the sodium atom has charge \(+e\) and the chlorine atom has charge \(-e\) the fractional ionic character would be equal to 1. What is the actual fractional ionic character for the bond in NaCl? (d) The equilibrium distance between nuclei in the hydrogen iodide (HI) molecule is \(0.16 \mathrm{nm},\) and the measured electric dipole moment of the molecule is \(1.5 \times 10^{-30} \mathrm{C} \cdot \mathrm{m}\) . What is the fractional ionic character for the bond in \(\mathrm{HI}\) ? How does your answer compare to that for NaCl calculated in part (c)? Discuss reasons for the difference in these results.

Short Answer

Expert verified
(a) \(3.84 \times 10^{-29} \mathrm{C} \cdot \mathrm{m}\); (b) \(1.6 \times 10^{-19} \mathrm{C}\); (c) 1; (d) 0.0586; HI has more covalent character.

Step by step solution

01

Calculate Dipole Moment for NaCl

The electric dipole moment (p) is given by the formula \( p = q \cdot d \), where \( q \) is the charge and \( d \) is the separation distance. For NaCl, the charges are \(+e\) and \(-e\), and the separation is 0.24 nm or 0.24 x 10^{-9} meters. The elementary charge \( e \) is approximately \( 1.6 \times 10^{-19} \mathrm{C} \). Thus, the dipole moment is \( p = 1.6 \times 10^{-19} \mathrm{C} \times 0.24 \times 10^{-9} \mathrm{m} \approx 3.84 \times 10^{-29} \mathrm{C} \cdot \mathrm{m} \).
02

Calculate Charge for Measured Dipole Moment

Given the NaCl dipole moment calculated in Step 1 as \( 3.84 \times 10^{-29} \mathrm{C} \cdot \mathrm{m} \), if this arises from charges \(+q\) and \(-q\) with separation of 0.24 nm, then \( p = q \cdot d \) gives us \( q = \frac{p}{d} = \frac{3.84 \times 10^{-29}}{0.24 \times 10^{-9}} \approx 1.6 \times 10^{-19} \mathrm{C} \). Thus, \( q \) equals the elementary charge \( e \).
03

Calculate Fractional Ionic Character for NaCl

The fractional ionic character is defined as \( \frac{q}{e} \). For NaCl, we found \( q = e \), so \( \frac{q}{e} = 1 \), meaning the bond in NaCl has a full ionic character.
04

Calculate Fractional Ionic Character for HI

The measured dipole moment for HI is given as \( 1.5 \times 10^{-30} \mathrm{C} \cdot \mathrm{m} \), and the bond separation is 0.16 nm or 0.16 x 10^{-9} meters. Using \( q = \frac{p}{d} \), we find \( q = \frac{1.5 \times 10^{-30}}{0.16 \times 10^{-9}} \approx 9.375 \times 10^{-21} \mathrm{C} \). The fractional ionic character is \( \frac{q}{e} = \frac{9.375 \times 10^{-21}}{1.6 \times 10^{-19}} \approx 0.0586 \), indicating a lower ionic character than NaCl.
05

Compare and Discuss Ionic Characters of NaCl and HI

The ionic character of NaCl is 1, implying a fully ionic bond, while HI has a much lower ionic character of approximately 0.0586, indicating a significant covalent nature. The difference arises because NaCl involves a metal and a non-metal leading to a high electron affinity, while HI is a bond between two non-metals with more sharing of electrons.

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Key Concepts

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

Fractional Ionic Character
Fractional ionic character is an important concept when understanding the nature of chemical bonds. It represents how much a bond between two atoms resembles an ionic bond rather than a covalent one. This is quantified as the ratio of the actual charge separation in the molecule to the elementary charge, expressed as \( \frac{q}{e} \).
In a molecule like sodium chloride (NaCl), the fractional ionic character is very close to 1. This indicates that the bond is predominantly ionic, as electrons are transferred from one atom to another. In contrast, in hydrogen iodide (HI), the fractional ionic character is much lower (approximately 0.0586), indicating that the bond has more covalent features where electrons are more equally shared between atoms.
The fractional ionic character is influenced by:
  • Electronegativity difference between the atoms
  • Distance between the charges
  • Intrinsic properties of the atoms involved (like metals vs non-metals)
Understanding these parameters helps predict the behavior and reactivity of substances in different chemical reactions.
Covalent and Ionic Bonds
Covalent and ionic bonds represent two fundamental types of chemical bonding. Ionic bonds are formed when one atom donates an electron to another, resulting in positively and negatively charged ions. This typically occurs between metals and non-metals, such as NaCl, where sodium (a metal) loses an electron, and chlorine (a non-metal) gains it.
However, when two atoms share electrons, the bond is called covalent. This usually happens between non-metals, such as in HI. The shared electrons "glue" the atoms together, providing stability.
Some key contrasts between ionic and covalent bonds are:
  • Ionic Bonds: High melting and boiling points, conduct electricity when dissolved in water, form crystal lattices.
  • Covalent Bonds: Lower melting and boiling points, do not conduct electricity, often form molecules with discrete units.
Depending on the difference in electronegativity and other factors like atomic size, bonds can exhibit characteristics of both ionic and covalent bonds.
Elementary Charge
The concept of the elementary charge is fundamental in physics and chemistry. It is the smallest unit of electric charge that is considered indivisible. Denoted as \( e \), its value is approximately \( 1.6 \times 10^{-19} \mathrm{C} \). Understanding this concept helps elucidate the charge interactions in bonding, especially in ionic and polar structures.
In the context of molecules, knowing the elementary charge helps determine the dipole moment. For example, in NaCl, calculating the dipole moment involves using the separation between charges (\( +e \) for Na and \( -e \) for Cl) due to their high ionic character. This involves determining the product of the charge value and the distance between the ions.
Some key aspects of the elementary charge include:
  • It defines the charge of protons and electrons, with electrons having \(-e\) and protons having \(+e\).
  • It is a key foundation for understanding electrostatic interactions in chemistry and physics.
  • Aids in calculating the properties of molecules, such as the electric dipole moment.
Mastering this concept involves recognizing its pivotal role in determining the behavior of charged particles and their interactions in various chemical structures.

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Most popular questions from this chapter

A hypothetical \(\mathrm{NH}\) molecule makes a rotational-level transition from \(l=3\) to \(I=1\) and gives off a photon of wavelength 1.780 \(\mathrm{nm}\) in doing so. What is the separation between the two atoms in this molecule if we model them as point masses? The mass of hydrogen is \(1.67 \times 10^{-27} \mathrm{kg},\) and the mass of nitrogen is \(2.33 \times 10^{-26} \mathrm{kg} .\)

Metallic lithium has a bce crystal structure. Each unit cell is a cube of side length \(a=0.35 \mathrm{nm}\) . (a) For a bce lattice, what is the number of atoms per unit volume? Give your answer in terms of \(a\) . (Hint: How many are there per unit cell?) (b) Use the result of part (a) to calculate the zero- temperature Fermi energy \(E_{\mathrm{PO}}\) for metallic lithium. Assume there is one free electron per atom.

CALC The one-dimensional calculation of Example 42.4 (Section 42.3\()\) can be extended to three dimensions. For the three-dimensional fce NaCl lattice, the result for the potential energy of a pair of \(\mathrm{Na}^{+}\) and \(C^{-}\) ions due to the electrostatic interaction with all of the ions in the crystal is \(U=-\alpha e^{2} / 4 \pi \epsilon_{0} r,\) where \(\alpha=1.75\) is the Madelung constant. Another contribution to the potential energy is a repulsive interaction at small ionic separation \(r\) due to overlap of the electron clouds. This contribution can be represented by \(A / r^{8},\) where \(A\) is a positive constant, so the expression for the total potential energy is $$U_{\mathrm{tot}}=-\frac{\alpha e^{2}}{4 \pi \epsilon_{0} r}+\frac{A}{r^{8}}$$ (a) Let \(r_{0}\) be the value of the ionic separation \(r\) for which \(U_{\text { ot }}\) is a minimum. Use this definition to find an equation that relates \(r_{0}\) and \(A,\) and use this to write \(U_{\text { ot }}\) in terms of \(r_{0 .}\) For \(\mathrm{NaCl}\) , \(r_{0}=0.281 \mathrm{nm} .\) Obtain a numerical value (in electron volts) of \(U_{\mathrm{tot}}\) for NaCl. (b) The quantity \(-U_{\text { tot }}\) is the energy required to remove a \(\mathrm{Na}^{+}\) ion and a \(\mathrm{Cl}^{-}\) ion from the crystal. Forming a pair of neutral atoms from this pair of ions involves the release of 5.14 eV (the ionization energy of \(\mathrm{Na}\) ) and the expenditure of 3.61 \(\mathrm{eV}\) (the electron affinity of Cl). Use the result of part (a) to calculate the energy required to remove a pair of neutral Na and Cl atoms from the crystal. The experimental value for this quantity is \(6.39 \mathrm{eV} ;\) how well does your calculation agree?

Hydrogen is found in two naturally occurring isotopes; normal hydrogen (containing a single proton in its nucleus) and deuterium (having a proton and a neurron). Assuming that both molecules are the same size and that the proton and neutron have the same mass (which is almost the case), find the ratio of (a) the energy of any given rotational state in a diatomic hydrogen molecule to the energy of the same state in a diatomic deuterium molecule and (b) the energy of any given vibrational state in hydrogen to the same state in deuterium (assuming that the force constant is the same for both molecules) Why is it physically reasonable that the force constant would be the same for hydrogen and deuterium molecules?

(a) A molecule decreases its vibrational energy by 0.250 \(\mathrm{eV}\) by giving up a photon of light. What wavelength of light does it give up during this process, and in what part of the electromagnetic spectrum does that wavelength of light lie? (b) An atom decreases its energy by 8.50 eV by giving up a photon of light. What wavelength of light does it give up during this process, and in what part of the electromagnetic spectrum does that wavelength of light lie? (c) A molecule decreases its rotational energy by \(3.20 \times 10^{-3} \mathrm{eV}\) by giving up a photon of light. What wavelength of light does it give up during this process, and in what part of the electromagnetic spectrum does that wavelength of light lie?
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