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Predict the ordering of the \(\mathrm{C}-\mathrm{O}\) bond lengths in \(\mathrm{CO}\), \(\mathrm{CO}_{2}\), and \(\mathrm{CO}_{3}^{2-}\)

Short Answer

Expert verified
The ordering of C-O bond lengths in CO, CO2, and CO3^2- is: CO (shortest) < CO2 (intermediate) < CO3^2- (longest) This order is based on the fact that CO has a triple bond, CO2 has a double bond, and CO3^2- has resonance structures that give partial double bond character. Bond lengths decrease as the bond order increases.

Step by step solution

01

Draw the Lewis structures

Draw the Lewis structures for CO, CO2, and CO3^2-. CO: The molecule is neutral. Carbon has 4 valence electrons and Oxygen has 6. They share 3 pairs of electrons to form a triple bond, and Oxygen still has a lone pair left on it. CO2: Carbon (4 valence electrons) forms a double bond with each Oxygen atom (6 valence electrons each). No formal charges are present. CO3^2-: Carbon (4 valence electrons) forms a single bond with each Oxygen atom (6 valence electrons each), and one of the Oxygen atoms forms an additional bond with another lone pair. There is a -1 formal charge on 2 of the Oxygen atoms, resulting in a total charge of -2 for the species.
02

Determine resonance structures

For each species, see if there are any resonance structures and consider the effect of resonance on the bond lengths. CO: There are no resonance structures for CO, so the bond length will be based on the triple bond C≔O. CO2: No resonance structures can be drawn for CO2, as each oxygen atom is already double bonded to the central carbon atom. CO3^2-: In the carbonate ion (CO3^2-), we can draw 3 resonance structures where each of C-O bond has a double bond in one and single bond in the other two. Due to these resonance structures, the C-O bonds will have bond characteristics that are intermediate between single and double bonds.
03

Predict bond lengths

CO has a triple bond, CO2 has a double bond, and CO3^2- has resonance structures that give partial double bond character. As a general rule, single bonds are longer than double bonds, which in turn are longer than triple bonds. Since CO has the shortest bond (triple bond), its C-O bond will be the shortest. CO2 has a double bond, so its bond length will be intermediate. Lastly, CO3^2- has resonance structures, giving it a bond characteristic between single and double bonds, resulting in the longest bond length among the three species. Hence, the ordering of C-O bond lengths is: CO (shortest) < CO2 (intermediate) < CO3^2- (longest)

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Key Concepts

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

Lewis Structures
Understanding Lewis structures is fundamental when comparing bond lengths in molecules. Lewis structures are diagrams that show the bonding between atoms and the presence of lone pairs of electrons in a molecule. For instance, when predicting the C-O bond length in molecules like CO, CO2, and CO3^2-, we begin by drawing their Lewis structures.

In CO, a triple bond is formed as Carbon shares three of its valence electrons with those of Oxygen, resulting in an arrangement where the electron pairs are shared unequally. The Lewis structure directly impacts the bond length as it provides a way to visualize the sharing of electron pairs between atoms, which in turn relates to bond strength and length. Learning how to accurately draw Lewis structures is critical and students are often advised to practice by identifying the central atom, counting valence electrons, and arranging them to satisfy the octet rule (or duet for hydrogen) while minimizing formal charges.
Resonance Structures
Another important concept is resonance structures. These are alternative Lewis structures for a molecule where the chemical connectivity is the same, but the distribution of electrons is different. In molecules like the carbonate ion (CO3^2-), resonance structures significantly influence the observed C-O bond lengths. By drawing the different resonance structures, it becomes evident that each oxygen can be double-bonded to the central carbon in one structure and single-bonded in others.

This equal distribution of multiple bonding scenarios across the same set of atoms leads to an intermediate bond length and strength. The resonance in CO3^2- causes its C-O bonds to have characteristics between a single and double bond, explaining why, in these ions, the C-O bonds are equivalent and intermediate in length. Students should note that resonance structures are a way to represent the delocalization of electrons and should not be confused with a molecule flipping between forms; it is a hybrid of the depicted structures.
Triple, Double, and Single Bonds

Bond Lengths and Strengths

When it comes to bond lengths of triple, double, and single bonds, there's a simple pattern: triple bonds are the shortest and strongest, followed by double bonds, with single bonds being the longest and weakest. This is due to the increased number of shared electron pairs in triple and double bonds, which pull the atoms closer together, resulting in shorter bond distances.

For example, in CO, the carbon and oxygen are held together by a triple bond, which is significantly shorter and stronger than the double bonds found in CO2. CO2's C-O bonds are intermediate in length and strength, while the single C-O bonds in CO3^2- are technically the longest due to the resonance stabilization. However, due to resonance, the effective bond length in CO3^2- is actually shorter than a typical single bond. Students should associate the 'triple < double < single' bond hierarchy with 'shortest < intermediate < longest' in bond lengths while recognizing the influence of resonance in some cases like CO3^2-.

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Most popular questions from this chapter

(a) Does the lattice energy of an ionic solid increase or decrease (i) as the charges of the ions increase, (ii) as the sizes of the ions increase? (b) Using a periodic table, arrange the following substances according to their expected lattice energies, listing them from lowest lattice energy to the highest: \(\mathrm{ScN}, \mathrm{KBr}, \mathrm{MgO}, \mathrm{NaF}\). Compare your list with the data in Table \(8.2\).

Using Lewis symbols, diagram the reaction between magnesium and oxygen atoms to give the ionic substance \(\mathrm{MgO}\).

Write Lewis structures for the following: (a) \(\mathrm{H}_{2} \mathrm{CO}\) (both \(\mathrm{H}\) atoms are bonded to \(\mathrm{C}\) ), (b) \(\mathrm{H}_{2} \mathrm{O}_{2}\), (c) \(\mathrm{C}_{2} \mathrm{~F}_{6}\) (contains a \(\mathrm{C}-\mathrm{C}\) bond \(),\) (d) \(\mathrm{AsO}_{3}{ }^{3-}\), (e) \(\mathrm{H}_{2} \mathrm{SO}_{3}(\mathrm{H}\) is bonded to \(\mathrm{O})\), (f) \(\mathrm{C}_{2} \mathrm{H}_{2}\)

You and a partner are asked to complete a lab entitled "Oxides of Ruthenium" that is scheduled to extend over two lab periods. The first lab, which is to be completed by your partner, is devoted to carrying out compositional analysis. In the second lab, you are to determine melting points. Upon going to lab you find two unlabeled vials, one containing a soft yellow substance and the other a black powder. You also find the following notes in your partner's notebook-Compound 1: \(76.0 \%\) Ru and \(24.0 \%\) O (by mass), Compound 2: \(61.2 \%\) Ru and \(38.8 \% \mathrm{O}\) (by mass). (a) What is the empirical formula for Compound \(1 ?\) (b) What is the empirical formula for Compound 2? (c) Upon determining the melting points of these two compounds, you find that the yellow compound melts at \(25^{\circ} \mathrm{C}\), while the black powder does not melt up to the maximum temperature of your apparatus, \(1200^{\circ} \mathrm{C}\). What is the identity of the yellow compound? What is the identity of the black compound? Be sure to use the appropriate naming convention depending upon whether the compound is better described as a molecular or ionic compound.

Use Lewis symbols to represent the reaction that occurs between \(\mathrm{Ca}\) and \(\mathrm{F}\) atoms.

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