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Chapter 9: Problem 66

Describe the bonding in the \(\mathrm{CO}_{3}^{2-}\) ion using the localized electron model. How would the molecular orbital model describe the \(\pi\) bonding in this species?

Short Answer

Expert verified
The bonding in the carbonate ion (CO₃²⁻) can be described using the localized electron model as a resonance hybrid of three equivalent structures: each oxygen atom can form a double bond with the central carbon atom while the other two oxygen atoms form single bonds. The actual structure results in a 1.33 bond order between carbon and each oxygen atom due to resonance effects. In the molecular orbital model, the π bonds are formed by the lateral overlap of p orbitals on the carbon and oxygen atoms, with π electrons distributed among the three oxygen atoms. The π bonding in the carbonate ion is delocalized rather than localized to a single carbon-oxygen double bond, leading to resonance stabilization.

Step by step solution

01

Identify the Central Atom and Determine Lewis Structure

First, we need to determine the Lewis structure for the carbonate ion (CO₃²⁻). In the carbonate ion, carbon (C) is the central atom, surrounded by three oxygen (O) atoms. In total, we have 24 valence electrons (4 from C, 6 from each O, and 2 from the charge) to be distributed in the ion.
02

Draw Single and Double Bonds between Atoms

Now, we will form bonds between the carbon and each oxygen atom. We can start by creating a single bond between the central carbon atom and each oxygen atom. This uses up 6 of the 24 valence electrons, leaving 18 electrons to be distributed. The remaining electrons will be placed as lone pairs around the oxygen atoms.
03

Distribute Lone Pairs around the Oxygen Atoms

Since each oxygen atom needs 6 more electrons to achieve an octet, we will distribute 6 electrons (3 lone pairs) around each oxygen atom. However, by doing this, we have used up all the 24 valance electrons and the central carbon still lacks an octet. To mitigate this, we move one lone pair from an oxygen atom to form a double bond with the central carbon atom.
04

Consider Resonance Structures

There are three equivalent possible structures for the carbonate ion: each oxygen atom can form a double bond with the central carbon atom, while the other two oxygen atoms form single bonds. Since all structures are equivalent, the actual structure is a combination of all three resonance structures with the double bond delocalized over all three oxygen atoms. This results in 1.33 bond order between carbon and each oxygen atom. Now, we will describe the bonding in the carbonate ion using the molecular orbital model, focusing on the π bonding.
05

Formation of π Bonds in Molecular Orbital Model

In the molecular orbital model, the π bonds are formed by the lateral overlap of the p orbitals on the carbon and oxygen atoms. The central carbon has three p orbitals, one on each oxygen atom.
06

Distribute π Electrons in Molecular Orbitals

The 24 valence electrons of the carbonate ion include six π electrons, four from carbon and two from the two double bonds. In the molecular orbital model, these six π electrons are distributed among the three oxygen atoms.
07

Describe the Delocalization of π Bonds

Due to the continuous overlapping of p orbitals from the carbon and three oxygen atoms, the π bonding in carbonate ion is delocalized rather than being localized to a single carbon-oxygen double bond. This delocalized π bonding lowers the energy of the carbonate ion and provides additional stability, resulting in resonance stabilization. In conclusion, the localized electron model describes the bonding in the carbonate ion as a resonance hybrid of three structures, with each oxygen atom participating in bond formation with the central carbon atom. In the molecular orbital model, the π bonding is delocalized over all three oxygen atoms, leading to resonance stabilization.

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Most popular questions from this chapter

The molecules \(\mathrm{N}_{2}\) and \(\mathrm{CO}\) are isoelectronic but their properties are quite different. Although as a first approximation we often use the same MO diagram for both, suggest how the \(\mathrm{MOs}\) in \(\mathrm{N}_{2}\) and \(\mathrm{CO}\) might be different.

An unusual category of acids known as superacids, which are defined as any acid stronger than 100\(\%\) sulfuric acid, can be prepared by seemingly simple reactions similar to the one below. In this example, the reaction of anhydrous HF with SbF produces the superacid $\left[\mathrm{H}_{2} \mathrm{F}\right]^{+}\left[\mathrm{SbF}_{6}\right]^{-} :$ $$ 2 \mathrm{HF}(l)+\mathrm{SbF}_{5}(l) \longrightarrow\left[\mathrm{H}_{2} \mathrm{F}\right]^{+}\left[\mathrm{SbF}_{6}\right]^{-}(l) $$ a. What are the molecular structures of all species in this reaction? What are the hybridizations of the central atoms in each species? b. What mass of $\left[\mathrm{H}_{2} \mathrm{F}\right]^{+}\left[\mathrm{SbF}_{6}\right]^{-}$ can be prepared when 2.93 \(\mathrm{mL}\) anhydrous \(\mathrm{HF}\) (density $=0.975 \mathrm{g} / \mathrm{mL} )\( and 10.0 \)\mathrm{mL}\( SbFs (density \)=3.10 \mathrm{g} / \mathrm{mL}$ ) are allowed to react?

As compared with \(\mathrm{CO}\) and \(\mathrm{O}_{2}, \mathrm{CS}\) and \(\mathrm{S}_{2}\) are very unstable molecules. Give an explanation based on the relative abilities of the sulfur and oxygen atoms to form \(\pi\) bonds.

Which is the more correct statement: "The methane molecule \(\left(\mathrm{CH}_{4}\right)\) is a tetrahedral molecule because it is $s p^{3}\( hybridized" or "The methane molecule (CH_ \)_{4} )\( is \)s p^{3}$ hybridized because it is a tetrahedral molecule"? What, if anything, is the difference between these two statements?

Describe the bonding in the first excited state of \(\mathrm{N}_{2}\) (the one closest in energy to the ground state) using the molecular orbital model. What differences do you expect in the properties of the molecule in the ground state as compared to the first excited state? (An excited state of a molecule corresponds to an electron arrangement other than that giving the lowest possible energy.)
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