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Chapter 1: Problem 54

The compounds FCl and ICl have dipole moments \(\mu\) that are similar in magnitude \((0.9\) and \(0.7 \mathrm{D}\), respectively \()\) but opposite in direction. In one compound, chlorine is the positive end of the dipole; in the other it is the negative end. Specify the direction of the dipole moment in each compound, and explain your reasoning.

Short Answer

Expert verified
In FCl, the dipole points towards fluorine; in ICl, the dipole points toward chlorine.

Step by step solution

01

Understanding Dipole Moments

Dipole moments are vector quantities that have both magnitude and direction. They arise due to differences in electronegativity between bonded atoms.
02

Analyzing Electronegativity

The electronegativity of chlorine (Cl) is higher than that of iodine (I) but lower than that of fluorine (F). Therefore, chlorine will be more electronegative in the ICl bond but less electronegative in the FCl bond.
03

Determining Direction in FCl

In FCl, since fluorine is more electronegative than chlorine, fluorine will be the negative end of the dipole. Thus, chlorine will be the positive end.
04

Determining Direction in ICl

In ICl, chlorine is more electronegative than iodine, making chlorine the negative end of the dipole. Iodine will be the positive end.

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Key Concepts

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

Electronegativity
Electronegativity is a measure of an atom's ability to attract and hold onto electrons in a chemical bond. When atoms form bonds, they share electrons, but if one atom is more electronegative than the other, it will pull the electrons closer to itself. This uneven distribution of electrons creates partial charges and influences chemical properties.
Some key points about electronegativity:
  • Atoms with high electronegativity, like fluorine, have a strong tendency to attract electrons.
  • Elements on the top right of the periodic table generally have higher electronegativities.
  • Differences in electronegativity between bonded atoms determine bond polarity and dipole moments.
In the exercise, the differences in electronegativity between chlorine, fluorine, and iodine help us predict the direction of dipole moments.
Molecular Polarity
Molecular polarity refers to the distribution of electrical charge over the atoms in a molecule. If the charge distribution is even, the molecule is nonpolar; if uneven, the molecule is polar. Polarity is crucial because it affects a molecule's interaction with other substances and its physical properties.
To understand molecular polarity, consider:
  • The dipole moment is a vector quantity, meaning direction matters as much as magnitude.
  • Unequal sharing of electrons due to differences in electronegativity results in dipole moments.
  • The geometry of a molecule can cause its polarity; even if individual bonds are polar, the overall molecule can be nonpolar.
In FCl and ICl, their polar nature arises from the imbalance in electronegativity, leading to dipole moments pointing in opposite directions due to the different elements involved.
Covalent Bonds
Covalent bonds are chemical bonds formed when atoms share pairs of electrons. The sharing can be equal or unequal, depending on the atoms' electronegativities. In typical covalent bonds, both atoms have similar electronegativity, leading to the equal sharing of electrons.
Important aspects of covalent bonds include:
  • They are a fundamental type of chemical bond found in many molecules, like water and organic compounds.
  • Polar covalent bonds occur when electronegativity differences lead to unequal sharing, resulting in partial charges.
  • Nonpolar covalent bonds have equal sharing of electrons due to similar electronegativity.
In the FCl and ICl compounds from the exercise, the covalent bonds display polar characteristics, resulting from differences in electronegativity between the chlorine, fluorine, and iodine atoms.

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Most popular questions from this chapter

Nitrosomethane and formaldoxime both have the molecular formula \(\mathrm{CH}_{3} \mathrm{NO}\) and the connectivity CNO. All of the hydrogens are bonded to carbon in nitrosomethane. In formaldoxime, two of the hydrogens are bonded to carbon and one to oxygen. Write Lewis formulas for (a) nitrosomethane and (b) formaldoxime that satisfy the octet rule and are free of charge separation.

Write structural formulas for all the constitutional isomers that have the given molecular formula. (a) \(\mathrm{C}_{2} \mathrm{H}_{7} \mathrm{~N}\) (b) \(\mathrm{C}_{3} \mathrm{H}_{7} \mathrm{Cl}\) (c) \(\mathrm{C}_{3} \mathrm{H}_{8} \mathrm{O}\) Sample Solution (a) The molecular formula \(\mathrm{C}_{2} \mathrm{H}_{7} \mathrm{~N}\) requires 20 valence electrons. Two carbons contribute a total of eight, nitrogen contributes five, and seven hydrogens contribute a total of seven. Nitrogen and two carbons can be connected in the order CCN or CNC. Assuming four bonds to each carbon and three to nitrogen, we write these connectivities as Place a hydrogen on each of the seven available bonds of each framework. The nine bonds in each structural formula account for 18 electrons. Add an unshared pair to each nitrogen to complete its octet and give a total of 20 valence electrons as required by the molecular formula. These two are constitutional isomers.

Write structural formulas for all the constitutional isomers of molecular formula \(\mathrm{C}_{3} \mathrm{H}_{6} \mathrm{O}\) that contain (a) Only single bonds (b) One double bond

Referring to the periodic table as needed, write electron configurations for all the elements in the third period. Sample Solution The third period begins with sodium and ends with argon. The atomic number \(Z\) of sodium is 11 , and so a sodium atom has 11 electrons. The maximum number of electrons in the 1s, \(2 \mathrm{~s}\), and \(2 \mathrm{p}\) orbitals is ten, and so the eleventh electron of sodium occupies a 3 s orbital. The electron configuration of sodium is \(1 s^{2} 2 s^{2} 2 p_{x}^{2} 2 p_{y}^{2} 2 p_{z}^{2} 3 s^{1}\).

How many electrons does carbon have? How many are valence electrons? What third-row element has the same number of valence electrons as carbon?
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