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Chapter 4: Problem 103

Select the more polar bond in each of the following pairs: a. \(\mathrm{C}-\mathrm{N}\) or \(\mathrm{C}-\mathrm{O}\) b. \(\mathrm{N}-\mathrm{F}\) or \(\mathrm{N}-\mathrm{Br}\) c. \(\mathrm{Br}-\mathrm{Cl}\) or \(\mathrm{S}-\mathrm{Cl}\) d. \(\mathrm{Br}-\mathrm{Cl}\) or \(\mathrm{Br}-\mathrm{I}\) e. \(\mathrm{N}-\mathrm{F}\) or \(\mathrm{N}-\mathrm{O}\)

Short Answer

Expert verified
a) \(\mathrm{C}-\mathrm{O}\) b) \(\mathrm{N}-\mathrm{F}\) c) \(\mathrm{S}-\mathrm{Cl}\) d) \(\mathrm{Br}-\mathrm{I}\) e) \(\mathrm{N}-\mathrm{F}\)

Step by step solution

01

- Understanding Polarity

To determine the more polar bond, compare the electronegativity values of the atoms in each bond. Polarity increases with the difference in electronegativity between the two atoms.
02

- Compare \(\mathrm{C}-\mathrm{N}\) and \(\mathrm{C}-\mathrm{O}\) Bonds

Carbon (C) has an electronegativity of 2.55, nitrogen (N) has 3.04, and oxygen (O) has 3.44. Calculate the differences: \(\mathrm{C}-\mathrm{N}\): 3.04 - 2.55 = 0.49, \(\mathrm{C}-\mathrm{O}\): 3.44 - 2.55 = 0.89. The \(\mathrm{C}-\mathrm{O}\) bond has a greater difference, indicating it is more polar.
03

- Compare \(\mathrm{N}-\mathrm{F}\) and \(\mathrm{N}-\mathrm{Br}\) Bonds

Fluorine (F) has an electronegativity of 3.98, whereas bromine (Br) has 2.96. Calculate the differences: \(\mathrm{N}-\mathrm{F}\): 3.98 - 3.04 = 0.94, \(\mathrm{N}-\mathrm{Br}\): 3.04 - 2.96 = 0.08. The \(\mathrm{N}-\mathrm{F}\) bond is more polar.
04

- Compare \(\mathrm{Br}-\mathrm{Cl}\) and \(\mathrm{S}-\mathrm{Cl}\) Bonds

Bromine (Br) has an electronegativity of 2.96, chlorine (Cl) has 3.16, and sulfur (S) has 2.58. Calculate the differences: \(\mathrm{Br}-\mathrm{Cl}\): 3.16 - 2.96 = 0.20, \(\mathrm{S}-\mathrm{Cl}\): 3.16 - 2.58 = 0.58. The \(\mathrm{S}-\mathrm{Cl}\) bond is more polar.
05

- Compare \(\mathrm{Br}-\mathrm{Cl}\) and \(\mathrm{Br}-\mathrm{I}\) Bonds

Iodine (I) has an electronegativity of 2.66. Calculate the differences: \(\mathrm{Br}-\mathrm{Cl}\): 3.16 - 2.96 = 0.20, \(\mathrm{Br}-\mathrm{I}\): 2.96 - 2.66 = 0.30. The \(\mathrm{Br}-\mathrm{I}\) bond is more polar.
06

- Compare \(\mathrm{N}-\mathrm{F}\) and \(\mathrm{N}-\mathrm{O}\) Bonds

Calculate the differences: \(\mathrm{N}-\mathrm{F}\): 3.98 - 3.04 = 0.94, \(\mathrm{N}-\mathrm{O}\): 3.44 - 3.04 = 0.40. The \(\mathrm{N}-\mathrm{F}\) bond is more polar.

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Key Concepts

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

Electronegativity
Electronegativity is the ability of an atom to attract electrons towards itself in a chemical bond. This property varies across the periodic table; generally, it increases from left to right across a period and decreases from top to bottom within a group.
  • For example, Fluorine (F) has the highest electronegativity of all elements, making it very effective at attracting electrons.
  • Electronegativity values are helpful when comparing elements to predict bond behavior and polarity.
Understanding electronegativity differences allows us to determine how electrons will be shared between atoms, which is crucial for predicting bond types and the resulting molecular properties.
Polar Bonds
Polar bonds occur when there is a significant difference in electronegativity between the two atoms involved in the bond. This difference results in an unequal sharing of electrons, causing a dipole moment.
  • For instance, in a \( \text{C-O} \) bond, oxygen pulls electron density towards itself more than carbon due to its higher electronegativity, creating a polar bond.
  • On the other hand, a bond with minimal or no electronegativity difference, like \( \text{Cl-Cl} \), is nonpolar, as electrons are shared equally.
In summary, polar bonds are essential in determining a molecule's overall polarity and reactivity.
Chemical Bonding
Chemical bonding involves the interaction of atoms to form molecules by sharing or transferring electrons. There are primarily three types of bonds:
  • Ionic Bonds: Form when electrons are transferred from one atom to another, such as in \( \text{NaCl} \).
  • Covalent Bonds: Happen when atoms share electrons, seen in \( \text{H}_2 \).
  • Metallic Bonds: Involve a 'sea' of shared electrons among metal atoms, allowing for conductivity and malleability.
Understanding the type of bond helps predict physical properties, reactivity, and the stability of compounds.
Molecular Polarity
Molecular polarity arises from the sum of the dipole moments of all the bonds in a molecule. It determines the molecule's behavior in different environments.
  • A molecule like \( \text{CO}_2 \) is nonpolar because its linear shape allows the dipole moments of the \( \text{C=O} \) bonds to cancel out.
  • However, \( \text{H}_2\text{O} \) is polar due to its bent structure causing the dipole moments to add up, creating an overall polar molecule.
In essence, considering molecular geometry in addition to bond polarity is important in determining the overall polarity of the molecule.
Bond Comparison
Comparing bonds involves evaluating their polarity based on the difference in electronegativity between the atoms. Here's a breakdown from the exercise:
  • \( \text{C-N} \) vs. \( \text{C-O} \): The \( \text{C-O} \) bond is more polar due to a larger electronegativity difference.
  • \( \text{N-F} \) vs. \( \text{N-Br} \): \( \text{N-F} \) bond is more polar.
  • \( \text{Br-Cl} \) vs. \( \text{S-Cl} \): \( \text{S-Cl} \) bond is more polar.
  • \( \text{Br-Cl} \) vs. \( \text{Br-I} \): \( \text{Br-I} \) bond is more polar.
  • \( \text{N-F} \) vs. \( \text{N-O} \): \( \text{N-F} \) bond is more polar.
Comparing bonds helps predict which bonds will contribute more to the overall polarity and reactivity of molecules.

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Most popular questions from this chapter

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