/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 54 Consider the formate ion, \(\mat... [FREE SOLUTION] | 91Ó°ÊÓ

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Consider the formate ion, \(\mathrm{HCO}_{2}^{-}\), which is the anion formed when formic acid loses an \(\mathrm{H}^{+}\) ion. The \(\mathrm{H}\) and the two \(\mathrm{O}\) atoms are bonded to the central C atom. (a) Draw the best Lewis structure(s) for this ion. (b) Are resonance structures needed to describe the structure? (c) Would you predict that the \(\mathrm{C}-\mathrm{O}\) bond lengths in the formate ion would be longer or shorter relative to those in \(\mathrm{CO}_{2}\) ?

Short Answer

Expert verified
The best Lewis structure for \(\mathrm{HCO}_2^-\) involves resonance between two forms. The C-O bonds are longer than in \(\mathrm{CO}_2\).

Step by step solution

01

Determine Total Valence Electrons

Calculate the total number of valence electrons available for the \(\mathrm{HCO}_2^-\) ion. Hydrogen contributes 1, carbon 4, each oxygen 6, and the negative charge adds an extra electron, summing up to a total of 18 valence electrons.
02

Create Initial Skeleton Structure

Arrange the atoms with hydrogen and the two oxygen atoms bonded to the central carbon. Place single bonds (2 electrons each) between the carbon atom and each of the three other atoms.
03

Complete the Octets of Outer Atoms

Fill the octets for the two oxygen atoms by adding lone pairs. You will need 6 electrons per oxygen atom (excluding the bond with carbon), so 12 electrons are used to satisfy the octets of both oxygens.
04

Assign Remaining Electrons to Central Atom

Place any remaining electrons on the central carbon atom. After placing lone pairs on oxygen, you will have 2 electrons left, which are initially added to carbon.
05

Form Double Bond to Satisfy Octet Rule

Convert one lone pair from an oxygen into a double bond with carbon to satisfy carbon's octet rule. Choose one oxygen to share its lone pair with carbon, forming a double bond.
06

Consider Resonance Structures

Determine if resonance structures are necessary. Since there are two possible ways to position the double bond (with either oxygen), the formate ion has two resonance structures.
07

Compare C-O Bond Lengths with CO2

In \(\mathrm{CO}_2\), each C-O bond is a double bond, resulting in shorter bond lengths. For \(\mathrm{HCO}_2^-\), each C-O bond exists as a resonance hybrid of a single and a double bond, making the bonds longer than a pure double bond but shorter than a single bond.

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Key Concepts

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

Lewis Structure of the Formate Ion
Understanding the Lewis structure for the formate ion \(\mathrm{HCO}_2^-\) is crucial to visualizing the distribution of electrons around the molecule. Let's explore this step by step.To start, the total number of valence electrons available is 18. Here's how: hydrogen provides 1 electron, carbon contributes 4, each oxygen 6, and the extra negative charge adds 1 more electron.Next, the skeleton structure is organized with carbon at the center, bonded to hydrogen and the two oxygen atoms. Initially, we use single bonds, each accounting for 2 electrons, which leaves us with 12 electrons.The outer atoms, in this case, the oxygens, should have complete octets. To do so, we distribute the remaining electrons, giving each oxygen 6 electrons to complete their octets (excluding the bond). After this process, we are left with 2 electrons, which initially sit on the carbon atom.Since carbon needs an octet too, we convert a lone pair from an oxygen to form a double bond with carbon. This adjustment satisfies carbon's requirement for 8 electrons around it.
Resonance Structures
In some molecules, a single Lewis structure can't fully illustrate electron distribution. The formate ion is a perfect example of resonance, which means multiple structures can describe the same molecule. For the formate ion, once we've established one Lewis structure, it's apparent that there are two places where we can form a double bond. This happens because the double bond can be placed with either of the oxygen atoms. However, this does not mean that the molecule is constantly switching between these forms. Instead, it is often best described as a resonance hybrid, which effectively means that the actual structure is a blend or average of the resonance forms. Due to this hybrid nature, the double-bond character is shared between the two C-O bonds.
Resonance structures often show up in various chemical species and are key to understanding the true nature of atomic interactions and bond lengths, as they affect the molecule's stability and reactivity.
C-O Bond Length in the Formate Ion
Bond lengths are important to understanding molecular shape and function. They vary based on the types of bonds present, like single, double, or even triple bonds.In carbon dioxide, each carbon-oxygen bond is a double bond, resulting in shorter bond lengths due to stronger overlapping between the bonding orbitals. However, in the formate ion, the situation is a little different due to resonance.In \(\mathrm{HCO}_2^-\), each C-O bond is neither purely single nor purely double. Instead, the bond length is an intermediate, due to its nature as a resonance hybrid. The average bond order, which is a measure of bond strength, in a resonance situation may be around 1.5 for formate ion, symbolizing this intermediate length.Overall, the C-O bond lengths within the formate ion are longer than the typical double bond found in \(\mathrm{CO}_2\), but notably shorter than a typical single bond. Understanding this helps predict how the formate ion interacts in different chemical environments.

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Most popular questions from this chapter

We can define average bond enthalpies and bond lengths for ionic bonds, just like we have for covalent bonds. Which ionic bond is predicted to have the smaller bond enthalpy, \(\mathrm{Li}-\mathrm{F}\) or \(\mathrm{Cs}-\mathrm{F} ?\)

1,2-dihydroxybenzene is obtained when two of the adjacent hydrogen atoms in benzene are replaced with an OH group. A skeleton of the molecule is shown here. (a) Complete a Lewis structure for the molecule using bonds and electron pairs as needed. (b) Are there any resonance structures for the molecule? If so, sketch them. (c) Are the resonance structures in (a) and (b) equivalent to one another as they are in benzene?

You and a partner are asked to complete a lab entitled "Carbonates of Group 2 metal" 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 and determine the identity of the Group 2 metal (M). In the second lab, you are to determine the melting point of this compound. Upon going to lab you find two unlabeled vials containing white powder. You also find the following notes in your partner's notebook-Compound \(1: 40.04 \% \mathrm{M}\) and \(12.00 \%\) C, \(47.96 \%\) O (by mass), Compound \(2: 69.59 \%\) M, \(6.09 \% \mathrm{C},\) and \(24.32 \% \mathrm{O}\) (by mass). (a) What is the empirical formula for Compound 1 and the identity of M? (b) What is the empirical formula for Compound 2 and the identity of M? Upon determining the melting points of these two compounds, you find that both compounds do not melt up to the maximum temperature of your apparatus, instead, the compounds decompose and liberate colorless gas. (c) What is the identity of the colorless gas? (d) Write the chemical equation for the decomposition reactions of compound 1 and 2. \((\mathbf{e})\) Are compounds 1 and 2 ionic or molecular?

(a) Using Lewis symbols, make a sketch of the reaction between potassium and bromine atoms to give the ionic substance KBr. (b) How many electrons are transferred? (c) Which atom loses electrons in the reaction?

(a) Which of these compounds is an exception to the octet rule: carbon dioxide, water, ammonia, phosphorus trifluoride, or arsenic pentafluoride? (b) Which of these compounds or ions is an exception to the octet rule: borohydride \(\left(\mathrm{BH}_{4}^{-}\right)\), borazine \(\left(\mathrm{B}_{3} \mathrm{~N}_{3} \mathrm{H}_{6},\right.\) which is analogous to benzene with alternating \(\mathrm{B}\) and \(\mathrm{N}\) in the ring), or boron trichloride?

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