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Arrange the bonds in each of the following sets in order of increasing polarity: (a) \(\mathrm{C}-\mathrm{F}, \mathrm{O}-\mathrm{F}, \mathrm{Be}-\mathrm{F}\) ; (b) \(\mathrm{O}-\mathrm{Cl}, \mathrm{S}-\mathrm{Br}, \mathrm{C}-\mathrm{P} ;(\mathbf{c}) \mathrm{C}-\mathrm{S}, \mathrm{B}-\mathrm{F}, \mathrm{N}-\mathrm{O}\)

Short Answer

Expert verified
The bonds arranged in order of increasing polarity are: (a) O-F, C-F, Be-F (b) S-Br, C-P, O-Cl (c) C-S, N-O, B-F

Step by step solution

01

Know the electronegativity trend in the periodic table.

Electronegativity generally increases across a period from left to right and decreases down a group. The most electronegative element is Fluorine (F), while the least electronegative element is Francium (Fr). Memorize or refer to the electronegativity values of the relevant elements to compare the electronegativity differences.
02

Calculate the electronegativity difference for each bond in group (a).

- For C-F, the electronegativity difference is \(|2.5 - 4.0| = 1.5\) - For O-F, the difference is \(|3.5 - 4.0| = 0.5\) - For Be-F, the difference is \(|1.5 - 4.0| = 2.5\)
03

Arrange group (a) bonds in order of increasing polarity.

Since the order of electronegativity differences is O-F < C-F < Be-F, the order of increasing polarity is: O-F, C-F, Be-F.
04

Calculate the electronegativity difference for each bond in group (b).

- For O-Cl, the difference is \(|3.5 - 3.0| = 0.5\) - For S-Br, the difference is \(|2.5 - 2.8| = 0.3\) - For C-P, the difference is \(|2.5 - 2.1| = 0.4\)
05

Arrange group (b) bonds in order of increasing polarity.

Since the order of electronegativity differences is S-Br < C-P < O-Cl, the order of increasing polarity is: S-Br, C-P, O-Cl.
06

Calculate the electronegativity difference for each bond in group (c).

- For C-S, the difference is \(|2.5 - 2.5| = 0.0\) - For B-F, the difference is \(|2.0 - 4.0| = 2.0\) - For N-O, the difference is \(|3.0 - 3.5| = 0.5\)
07

Arrange group (c) bonds in order of increasing polarity.

Since the order of electronegativity differences is C-S < N-O < B-F, the order of increasing polarity is: C-S, N-O, B-F. So, the final arrangement of all bonds in order of increasing polarity are: (a) O-F, C-F, Be-F (b) S-Br, C-P, O-Cl (c) C-S, N-O, B-F

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Key Concepts

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

Bond Polarity
Bond polarity is a critical concept in chemistry that helps us understand how electrons are distributed between atoms in a bond. A bond becomes polar when there is a difference in electronegativity between the two atoms participating in the bond. This difference creates an unequal sharing of electrons, with the more electronegative atom pulling electron density towards itself, forming a dipole moment. Consider a bond between carbon (C) and fluorine (F). Fluorine, being more electronegative, attracts the shared electrons more strongly than carbon, resulting in a polar bond. Conversely, when two atoms share electrons equally, such as in the case of an oxygen (O) bond to fluorine (F), the bond is considered nonpolar.

  • A bond is considered **polar** if there is a significant difference in electronegativity between the two bonded atoms.
  • A bond is **nonpolar** if the electronegativity difference is negligible or zero.
By identifying the electron pulling capability, or electronegativity, of each atom in a bond, you can determine its polarity and predict the bond's behavior in compounds.
Periodic Table Trends
Understanding periodic table trends is essential to mastering the concept of electronegativity and, consequently, bond polarity. Electronegativity often follows predictable trends across the periodic table, which can help in evaluating how atoms interact in compounds. As you move from left to right across a period, electronegativity generally increases. Conversely, moving down a group, electronegativity decreases. This is due to the increasing number of electron shells which cause a greater shielding effect, reducing the nucleus's pull on valent electrons. For example, in the periodic table, fluorine sits in the top right corner, while oxygen is just left of it. This placement indicates that fluorine has a higher electronegativity than oxygen.

To understand these trends in-depth, consider the following key points:
  • **Across a period:** Electronegativity increases due to the increase in nuclear charge without a significant increase in shielding.
  • **Down a group:** Electronegativity decreases because of increased shielding from the inner electron shells.
  • **Fluorine:** As the most electronegative element, it sets the benchmark for electronegativity.
By analyzing these trends, you quickly establish which bonds will exhibit greater polarity, thus helping predict molecular behavior.
Chemical Bonds
Chemical bonds are the forces that hold atoms together in molecules and compounds. They are essential to forming the variety of substances around us. The nature of a chemical bond—whether it is ionic, covalent, or metallic—depends on how electrons are shared or transferred between atoms. Two major types of bonds related to polarity are ionic and covalent bonds:

  • **Ionic Bonds:** These occur when electrons are transferred from one atom to another, forming ions. This usually happens between atoms with a large difference in electronegativity, such as sodium and chlorine in table salt (NaCl).
  • **Covalent Bonds:** These involve the sharing of electron pairs between atoms. Covalent bonds can be polar or nonpolar. Polar covalent bonds form when there is an unequal sharing due to differences in electronegativity, like the hydrogen and oxygen in water (H₂O).
Understanding the type and polarity of a bond helps predict a molecule's properties such as solubility, melting point, and electrical conductivity. It also provides insights into how molecules will interact with each other, vital for chemical reactions and biological processes. For example, the polar bonds in water give it unique properties like its high boiling point and excellent solvent abilities.

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Most popular questions from this chapter

An ionic substance of formula MX has a lattice energy of \(6 \times 10^{3} \mathrm{k} / \mathrm{mol} .\) Is the charge on the ion M likely to be \(1+, 2+,\) or \(3+?\) Explain.

(a) Triazine, \(\mathrm{C}_{3} \mathrm{H}_{3} \mathrm{N}_{3},\) is like benzene except that in triazine every other \(\mathrm{C}-\mathrm{H}\) group is replaced by a nitrogen atom. Draw the Lewis structure(s) for the triazine molecule. (b) Estimate the carbon-nitrogen bond distances in the ring.

Formic acid has the chemical formula HCOOH. It is a colorless liquid that has a density of 1.220 \(\mathrm{g} / \mathrm{mL}\) . (a) The carbon atom in formic acid is bound to one \(\mathrm{H}\) and both \(\mathrm{O}^{\prime}\) 's. Draw the Lewis structure for formic acid, showing resonance if present. (b) Formic acid can react with NaOH in aqueous solution to produce the formate ion, HCOO- . Write the balanced chemical equation for this reaction. (c) Draw the Lewis structure of the formate ion, showing resonance if present. (d) How many milliliters of a 0.100 M solution of NaOH would it take to completely react with 0.785 \(\mathrm{mL}\) of formic acid?

Write electron configurations for the following ions, and determine which have noble-gas configurations: (a) \(\mathrm{Cd}^{2+}\) , \((\mathbf{b}) \mathrm{P}^{3-},(\mathbf{c}) \mathrm{Zr}^{4+},(\mathbf{d}) \mathrm{Ru}^{3+},(\mathbf{e}) \mathrm{As}^{3-},(\mathbf{f}) \mathrm{Ag}^{+}\)

True or false: (a) The \(\mathrm{C}-\) Cbonds in benzene are all the same length and correspond to typical single \(\mathrm{C}-\mathrm{Cbond}\) lengths. (b) The \(\mathrm{C}-\mathrm{C}\) bond in acetylene, HCCH, is longer than the average \(\mathrm{C}-\mathrm{C}\) bond length in benzene.

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