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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 (CO鈧兟测伝) 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 蟺 bonds are formed by the lateral overlap of p orbitals on the carbon and oxygen atoms, with 蟺 electrons distributed among the three oxygen atoms. The 蟺 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 (CO鈧兟测伝). 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 蟺 bonding.
05

Formation of 蟺 Bonds in Molecular Orbital Model

In the molecular orbital model, the 蟺 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 蟺 Electrons in Molecular Orbitals

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

Describe the Delocalization of 蟺 Bonds

Due to the continuous overlapping of p orbitals from the carbon and three oxygen atoms, the 蟺 bonding in carbonate ion is delocalized rather than being localized to a single carbon-oxygen double bond. This delocalized 蟺 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 蟺 bonding is delocalized over all three oxygen atoms, leading to resonance stabilization.

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Most popular questions from this chapter

The allene molecule has the following Lewis structure: Must all hydrogen atoms lie the same plane? If not, what is their spatial relationship? Explain.

The diatomic molecule OH exists in the gas phase. OH plays an important part in combustion reactions and is a reactive oxidizing agent in polluted air. The bond length and bond energy have been measured to be 97.06 \(\mathrm{pm}\) and 424.7 \(\mathrm{kJ} / \mathrm{mol}\) respectively. Assume that the OH molecule is analogous to the HF molecule discussed in the chapter and that the MOs result from the overlap of a \(p_{z}\) orbital from oxygen and the 1\(s\) orbital of hydrogen (the O-H bond lies along the z axis). a. Draw pictures of the sigma bonding and antibonding molecular orbitals in OH. b. Which of the two MOs has the greater hydrogen 1\(s\) character? c. Can the 2\(p_{x}\) orbital of oxygen form MOs with the 1\(s\) orbital of hydrogen? Explain. d. Knowing that only the 2\(p\) orbitals of oxygen interact significantly with the 1\(s\) orbital of hydrogen, complete the MO energy-level diagram for OH. Place the correct number of electrons in the energy levels. e. Estimate the bond order for OH. f. Predict whether the bond order of \(\mathrm{OH}^{+}\) is greater than, less than, or the same as that of OH. Explain.

In Exercise 95 in Chapter 8 , the Lewis structures for benzene \(\left(\mathrm{C}_{6} \mathrm{H}_{6}\right)\) were drawn. Using one of the Lewis structures, estimate \(\Delta H_{\mathrm{f}}^{\circ}\) for \(\mathrm{C}_{6} \mathrm{H}_{6}(g)\) using bond energies and given that the standard enthalpy of formation of \(\mathrm{C}(g)\) is 717 \(\mathrm{kJ} / \mathrm{mol}\) . The experimental \(\Delta H_{\mathrm{f}}^{\circ}\) value of \(\mathrm{C}_{6} \mathrm{H}_{6}(g)\) is 83 \(\mathrm{kJ} / \mathrm{mol} .\) Explain the discrepancy between the experimental value and the calculated \(\Delta H_{\mathrm{f}}^{\circ}\) value for \(\mathrm{C}_{6} \mathrm{H}_{6}(g)\)

Show how a \(d_{x z}\) atomic orbital and a \(p_{z}\) atomic orbital combine to form a bonding molecular orbital. Assume the \(x\) -axis is the internuclear axis. Is a \(\sigma\) or a \(\pi\) molecular orbital formed? Explain.

A flask containing gaseous \(\mathrm{N}_{2}\) is irradiated with 25 -nm light. a. Using the following information, indicate what species can form in the flask during irradiation. $$ \begin{array}{ll}{\mathrm{N}_{2}(g) \longrightarrow 2 \mathrm{N}(g)} & {\Delta H=941 \mathrm{kJ} / \mathrm{mol}} \\ {\mathrm{N}_{2}(g) \longrightarrow \mathrm{N}_{2}^{+}(g)+\mathrm{e}^{-}} & {\Delta H=1501 \mathrm{kJ} / \mathrm{mol}} \\ {\mathrm{N}(g) \longrightarrow \mathrm{N}^{+}(g)+\mathrm{e}^{-}} & {\Delta H=1402 \mathrm{kJ} / \mathrm{mol}}\end{array} $$ b. What range of wavelengths will produce atomic nitrogen in the flask but will not produce any ions? c. Explain why the first ionization energy of \(\mathrm{N}_{2}(1501 \mathrm{kJ} /\) mol) is greater than the first ionization energy of atomic nitrogen \((1402 \mathrm{kJ} / \mathrm{mol})\)
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