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Chapter 1: Problem 16

All of the bonds in the carbonate ion \(\left(\mathrm{CO}_{3}{ }^{2-}\right)\) are between \(\mathrm{C}\) and \(\mathrm{O}\). Write Lewis formulas for the major resonance contributors, and use curved arrows to show their relationship. Apply the resonance concept to explain why all of the \(\mathrm{C}-\mathrm{O}\) bond distances in carbonate are equal.

Short Answer

Expert verified
The carbonate ion has three equal \(\mathrm{C}-\mathrm{O}\) bonds due to resonance delocalization of electrons.

Step by step solution

01

Identify the Basic Structure

Write the basic Lewis structure for the carbonate ion, \(\mathrm{CO}_{3}^{2-}\). Place the carbon atom in the center and single bond it to the three oxygen atoms. Allocate a total of 24 valence electrons (4 from carbon, 6 from each of the three oxygens, and 2 extra for the 2- charge).
02

Complete Octets

Place lone pairs around each oxygen atom so that each has an octet of electrons. This will initially require 18 electrons for the three oxygens (6 for each), leaving 6 electrons. Place the remaining electrons as lone pairs on the oxygen atoms.
03

Calculate Formal Charges

Calculate the formal charge for each atom: Formal Charge = (Valence electrons) - (Non-bonding electrons) - 0.5*(Bonding electrons). For carbon in the structure, the formal charge is 0; for oxygen atoms bonded by single bonds, the formal charge is -1; for oxygen involved in a double bond, the formal charge is 0.
04

Draw Resonance Structures

Draw three resonance structures for \(\mathrm{CO}_{3}^{2-}\). In each, one oxygen is double-bonded to carbon (neutral formal charge), while the other two are single-bonded (with -1 formal charge). Use curved arrows to show electron movement: start pointing at a lone pair on a single-bonded oxygen moving towards the single bond to form a double bond, and another arrow from the double bond to the adjacent oxygen, turning it into a lone pair.
05

Explain Resonance Averaging

Understand that resonance structures are not real but theoretical representations. The true structure (resonance hybrid) is an average of the resonance forms, meaning that electron density is delocalized across all \(\mathrm{C}-\mathrm{O}\) bonds. This delocalization explains why all \(\mathrm{C}-\mathrm{O}\) bond lengths in the carbonate ion are equal, as no single structure dominates.

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Key Concepts

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

Lewis Structures
Lewis structures are essential tools in understanding the arrangement of atoms within a molecule. For the carbonate ion, \( \mathrm{CO}_{3}^{2-} \), we start by identifying the central atom and then distribute the valence electrons accordingly. In this ion, carbon is the central atom surrounded by the three oxygen atoms.
  • The total valence electrons available are calculated by adding those from carbon, the three oxygen atoms, and considering the extra 2 electrons due to the ion's charge.
  • We begin by forming single bonds between carbon and each oxygen, using six electrons in this process (two per bond).
  • The remaining electrons are distributed as lone pairs around the oxygen atoms to satisfy their octet requirements.
This step helps establish a basic molecular framework, but it鈥檚 only the starting point for identifying the full picture through resonance structures.
Formal Charge
Calculating the formal charge is crucial in identifying the most stable Lewis structures and understanding the distribution of electrons. The formal charge helps us evaluate each atom's participation in the molecule's electron cloud formation. The carbonate ion, \(\mathrm{CO}_{3}^{2-}\), provides an excellent example.

The formula used is:\[\text{Formal Charge} = \text{Valence Electrons} - \text{Non-bonding Electrons} - \frac{1}{2}\times\text{Bonding Electrons}. \]
  • For the central carbon atom, the formal charge is 0 when it is double-bonded to one of the oxygen atoms.
  • The oxygen atom involved in the double bond also has a formal charge of 0.
  • Oxygen atoms forming single bonds with carbon, however, hold a formal charge of -1.
By calculating these charges, we can choose structures where charges are mostly neutral or distributed optimally, balancing stability.
Bond Delocalization
Bond delocalization refers to the phenomenon where electrons are shared among more than two atoms. This is a key concept when looking at resonance structures, especially in the carbonate ion \(\mathrm{CO}_{3}^{2-}\).

Resonance structures illustrate potential electron sharing patterns, but the true electron distribution is a mix of these forms.
  • In carbonate, each resonance structure alternates which oxygen atom forms a double bond with carbon. Even though these structures may appear different, they share a common feature: electrons are not confined to a single bonding pattern.
  • The concept of resonance averaging explains why this electron delocalization exists: the actual hybrid structure is an average of all resonance forms.
  • This averaging results in all \(\mathrm{C}-\mathrm{O}\) bonds in the carbonate ion being equivalent in length, as none of the single or double bonds remain fixed but are part of a continuum of electron sharing.
Understanding bond delocalization helps explain equal bond energies and lengths in seemingly complex ions.

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Most popular questions from this chapter

Write Lewis formulas, including unshared pairs, for each of the following. Carbon has four bonds in each compound. (a) Propane \(\left(\mathrm{C}_{3} \mathrm{H}_{8}\right)\) (c) Methyl fluoride \(\left(\mathrm{CH}_{3} \mathrm{~F}\right)\) (b) Methanol \(\left(\mathrm{CH}_{4} \mathrm{O}\right)\) (d) Ethyl fluoride \(\left(\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{~F}\right)\) Sample Solution (a) The Lewis formula of propane is analogous to that of ethane but the chain has three carbons instead of two. The ten covalent bonds in the Lewis formula shown account for 20 valence electrons, which. is the same as that calculated from the molecular formula \(\left(\mathrm{C}_{3} \mathrm{H}_{8}\right)\). The eight hydrogens of \(\mathrm{C}_{3} \mathrm{H}_{8}\) contribute 1 electron each and the three carbons 4 each, for a total of \(20(8\) from the hydrogens and 12 from the carbons). Therefore, all the valence electrons are in covalent bonds; propane has no unshared pairs.

Potassium hydride \((\mathrm{KH})\) is a source of the strongly basic hydride ion \(\left(: \mathrm{H}^{-}\right)\). Using curved arrows to track electron movement, write an equation for the reaction of hydride ion with water. What is the conjugate acid of hydride ion?

Which is a stronger base, ethoxide \(\left(\mathrm{CH}_{3} \mathrm{CH}_{2} \ddot{\mathrm{O}}:^{-}\right)\) or tert-butoxide \(\left[\mathrm{CH}_{3}\right)_{3} \mathrm{CO}:\) ]?

Salicylic acid, the starting material for the preparation of aspirin, has a \(K_{\mathrm{a}}\) of \(1.06 \times 10^{-3}\). What is its \(\mathrm{pK}_{\mathrm{a}} ?\)

Referring to the periodic table as needed, write electron configurations for all the elements in the third period. Sample Solution The third period begins with sodium and ends with argon. The atomic number \(Z\) of sodium is 11 , and so a sodium atom has 11 electrons. The maximum number of electrons in the 1s, \(2 \mathrm{~s}\), and \(2 \mathrm{p}\) orbitals is ten, and so the eleventh electron of sodium occupies a 3 s orbital. The electron configuration of sodium is \(1 s^{2} 2 s^{2} 2 p_{x}^{2} 2 p_{y}^{2} 2 p_{z}^{2} 3 s^{1}\).
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