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Consider a series of molecules in which carbon is attached by single covalent bonds to atoms of second-period elements: \(\mathrm{C}-\mathrm{O}, \mathrm{C}-\mathrm{F}, \mathrm{C}-\mathrm{N}, \mathrm{C}-\mathrm{C}\) and \(\mathrm{C}-\mathrm{B}\). Place these bonds in order of increasing bond length.

Short Answer

Expert verified
Bond length order: C-F < C-O < C-N < C-C < C-B.

Step by step solution

01

Understanding Bond Length

Bond length is the average distance between the nuclei of two bonded atoms in a molecule. Generally, bond length decreases as the bond strength increases and as the atomic size of the bonded atoms decreases.
02

Analyzing Electronegativity and Bond Strength

For second-period elements, the electronegativity generally decreases as you move from fluorine to boron. Stronger bonds are usually shorter, and bonds involving highly electronegative atoms tend to be shorter due to stronger attraction.
03

Comparing Atomic Radii

Second-period elements from B to F have decreasing atomic radii. Therefore, bonds with atoms with larger atomic radii tend to be longer because the nuclei are further apart.
04

Evaluating Each Bond

C-O, C-F, C-N, C-C, and C-B bonds differ in length. C-F is the bond between carbon and the most electronegative atom (fluorine) with a smaller atomic radius, resulting in a shorter bond. C-O and C-N are next due to similar trends. C-C bond length is typically longer than these as it involves identical atoms. C-B bond involves larger boron atoms, likely resulting in the longest bond.
05

Arranging Bonds by Length

Based on electronegativity and atomic radii analysis, the order of bond lengths from shortest to longest is C-F, C-O, C-N, C-C, and C-B. This is because fluorine forms the most tightly bound pairs with carbon, resulting in shorter bonds, while larger atoms like boron form longer bonds.

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Key Concepts

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

Covalent Bonds
Covalent bonds are a fundamental type of chemical bond where atoms share electron pairs. These bonds create strong connections between atoms, forming molecules. Covalent bonding occurs when the involved atoms have similar electronegativities, allowing them to share electrons instead of transferring them completely as in ionic bonds.
Covalent bonds are represented by Lewis structures, which show shared pairs of electrons and any lone pairs present. Consider the carbon-fluorine (\(\mathrm{C}-\mathrm{F}\)) bond in the exercise. This bond is covalent because both carbon and fluorine share electrons, creating a strong linkage. As elements form covalent bonds, the number of shared electron pairs can vary, resulting in single, double, or triple bonds. Single covalent bonds, like the carbon bonds in the exercise, involve one pair of shared electrons.
  • Single bonds involve sharing one pair of electrons.
  • Double bonds share two pairs, while triple bonds involve three pairs.
The bond length for covalent bonds will differ based on the nature of the involved atoms, playing a critical role as seen in the exercise's evaluation of bond lengths.
Atomic Radii
Atomic radii refer to the size of an atom and specifically to the distance from the nucleus to the outermost boundary of the surrounding cloud of electrons. It is a key factor in determining the length of covalent bonds. The atomic radius influences how close atoms can get to one another.
As you move across a period in the periodic table from left to right, atomic radii generally decrease. This is because the addition of protons increases the positive charge of the nucleus, pulling the electron cloud closer to the nucleus.
In the case of second-period elements such as those in the exercise (B, C, N, O, F), this trend means that the atomic radius decreases from boron to fluorine. Since shorter atomic radii allow for shorter bond lengths, understanding this trend helps in predicting bond lengths in molecules.
Knowing the relative size of atoms, as discussed for the covalent bonds in the exercise, allows you to determine that bonds between smaller atoms like fluorine and carbon are shorter compared to those involving larger atoms like boron.
Electronegativity
Electronegativity is the tendency of an atom to attract shared electrons in a chemical bond. It plays a vital role in determining the characteristics of a bond, including its length and strength.
Different elements have varying electronegativities, and this property influences the bond lengths in molecules. When you rank the second-period elements by electronegativity, fluorine stands as the most electronegative, followed by oxygen, nitrogen, carbon, and boron.
  • Higher electronegativity means a stronger pull on electrons and usually results in shorter, stronger bonds.
  • Bonds involving highly electronegative atoms, like C-F, are often shorter because the shared electrons are attracted more strongly towards one nucleus.
In the exercise, C-F bonds are identified as the shortest due to fluorine's high electronegativity. Comparing bonds, the higher the electronegativity difference, the shorter and stronger the bond tends to be, impacting the bond order: C-F, C-O, C-N, C-C, C-B.
Second-Period Elements
The second-period elements refer to the set of elements located in the second row of the periodic table, ranging from lithium (Li) to neon (Ne). This period includes elements that are fundamental to chemistry, particularly in forming molecules with carbon, like in the covalent bonds discussed here.
These elements, particularly B, C, N, O, and F, exhibit specific trends as one moves across the period. Some of these trends include decreasing atomic radii and increasing electronegativity. These properties significantly impact their bonding behavior and influence the bond lengths of compounds they form.
  • For example, as atomic radii decrease, bond lengths tend to shorten when comparing similar bonds across the period.
  • Increasing electronegativity means some elements like fluorine will form stronger and shorter bonds with carbon and other elements.
In the exercise, understanding these properties of second-period elements helps determine the bond length order: C-F, C-O, C-N, C-C, C-B. These properties, derived from periodic trends, directly impact the covalent bonding patterns observed with carbon.

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Most popular questions from this chapter

The following molecules or ions all have two oxygen atoms attached to a central atom. Draw a Lewis structure for each one, and then describe the electron-pair geometry and molecular geometry around the central atom. Comment on similarities and differences in the series. (a) \(\mathrm{CO}_{2}\) (b) \(\mathrm{NO}_{2}\) (c) \(\mathrm{O}_{3}\) (d) \(\mathrm{ClO}_{2}\)

Draw a Lewis structure for each of the following molecules or ions. Describe the electron-pair geometry and molecular geometry around the central atom. (a) \(\mathrm{NH}_{2} \mathrm{Cl}\) (b) \(\mathrm{Cl}_{2} \mathrm{O}(\mathrm{O}\) is the central atom) (c) SCN (d) HOF

For each of the bonds listed below, tell which atom is the more negatively charged. (a) \(\mathrm{C}-\mathrm{N}\) (b) \(\mathrm{C}-\mathrm{H}\) (c) \(\mathrm{C}-\mathrm{Br}\) (d) \(\mathrm{S}-\mathrm{O}\)

\- Acrylamide, \(\mathrm{H}_{2} \mathrm{C}=\mathrm{CHCONH}_{2},\) is a known neurotoxin and possible carcinogen. It was a shock to all consumers of potato chips and french fries a few years ago when it was found to occur in those products. (a) Sketch the molecular structure of acrylamide and identify all bond angles. (b) Indicate which carbon-carbon bond is the stronger of the two. (c) Is the molecule polar or nonpolar? (d) The amount of acrylamide found in potato chips is \(1.7 \mathrm{mg} / \mathrm{kg}\). If a serving of potato chips is \(28 \mathrm{g}\), how many moles of acrylamide are you consuming?

In each pair of bonds, predict which is shorter. (a) \(\mathrm{Si}-\mathrm{N}\) or \(\mathrm{Si}-\mathrm{O}\) (b) \(\mathrm{Si}-\mathrm{O}\) or \(\mathrm{C}-\mathrm{O}\) (c) \(\mathrm{C}-\mathrm{F}\) or \(\mathrm{C}-\mathrm{Br}\) (d) The \(\mathrm{C}-\mathrm{N}\) bond or the \(\mathrm{C} \equiv \mathrm{N}\) bond in \(\mathrm{H}_{2} \mathrm{NCH}_{2} \mathrm{C} \equiv \mathrm{N}\)

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