The three most stable oxides of carbon are carbon monoxide \((\mathrm{CO}),\) carbon dioxide \(\left(\mathrm{CO}_{2}\right),\) and carbon suboxide \(\left(\mathrm{C}_{3} \mathrm{O}_{2}\right) .\) The space-filling models for these three compounds are For each oxide, draw the Lewis structure, predict the molecular structure, and describe the bonding (in terms of the hybrid orbitals for the carbon atoms).

Short Answer

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The Lewis structures for the three most stable oxides of carbon are: CO: \( \: \: \: \: \: \: \: \: \: C \equiv O : \) CO2: \( \: : O = C = O : \) C3O2: \( \: : O = C = C = C = O : \) The molecular structures of CO, CO2, and C3O2 are all linear. In CO, the carbon atom is sp hybridized, forming a triple bond with an oxygen atom. In CO2, the central carbon atom is sp hybridized, forming double bonds with two oxygen atoms. In C3O2, the terminal carbon atoms are sp2 hybridized, forming double bonds with oxygen atoms and single bonds with the middle carbon atom. The middle carbon atom is sp hybridized, forming double bonds with the two adjacent carbon atoms.

Step by step solution

01

Draw the Lewis Structures

To draw the Lewis structures, we need to follow these steps: 1. Calculate the total number of valence electrons for each compound. 2. Write the symbols for each atom, arranging them to show their connectivity. 3. Distribute the valence electrons as shared (bonding) pairs and lone (non-bonding) pairs. For CO: - Carbon and oxygen have 4 and 6 valence electrons, respectively, so we have a total of 10 valence electrons. - The Lewis structure for CO is: \( \: \: \: \: \: \: \: \: \: \: C \equiv O : \) For CO2: - Carbon and oxygen have 4 and 6 valence electrons, respectively, so for two oxygen atoms and one carbon atom, we have a total of 16 valence electrons. - The Lewis structure for CO2 is: \( \: : O = C = O : \) For C3O2: - Carbon has 4 valence electrons, and oxygen has 6 valence electrons, so for three carbon atoms and two oxygen atoms, we have a total of 22 valence electrons. - The Lewis structure for C3O2 is: \( \: : O = C = C = C = O : \)
02

Predict the Molecular Structure

To predict the molecular structure, we need to determine the electron domain geometry for each central atom. For CO: - The carbon atom is bonded to an oxygen atom through a triple bond, forming a linear electron domain geometry. - Thus, the molecular structure of CO is linear. For CO2: - The carbon atom is doubly bonded to two oxygen atoms, forming a linear electron domain geometry. - Thus, the molecular structure of CO2 is linear. For C3O2: - In C3O2, we have three carbon atoms and two oxygen atoms connected in a linear chain. - The two terminal carbon atoms are doubly bonded to an oxygen atom, forming a linear electron domain geometry. - The middle carbon atom has two double bonds, one to each adjacent carbon atom, also forming a linear electron domain geometry. - Thus, the molecular structure of C3O2 is linear.
03

Determine Carbon Hybridization and Bonding

To determine the hybridization of the carbon atoms and describe the bonding in terms of hybrid orbitals, we will examine each compound individually. For CO: - The carbon atom in CO forms a triple bond with an oxygen atom, which consists of one sigma bond and two pi bonds. - The carbon atom is sp hybridized, forming two sp orbitals. - One of the sp orbitals forms a sigma bond with an oxygen atom's 2p orbital, and the other sp orbital has a lone pair of electrons. - The two pi bonds are formed between the unhybridized 2p orbitals of the carbon and oxygen atoms. For CO2: - The central carbon atom in CO2 forms two double bonds with two oxygen atoms, each double bond consisting of one sigma bond and one pi bond. - The central carbon atom is sp hybridized, forming two sp orbitals. - Each oxygen atom forms a sigma bond with one of the sp orbitals of the central carbon atom. - The pi bonds are formed between the unhybridized 2p orbitals of the carbon and oxygen atoms. For C3O2: - The terminal carbon atoms in C3O2 form a double bond with an oxygen atom, which consists of one sigma bond and one pi bond. They also form a sigma bond with the middle carbon atom. - The terminal carbon atoms are sp2 hybridized, forming three sp2 orbitals. - One of the sp2 orbitals forms a sigma bond with an oxygen atom's 2p orbital, another sp2 orbital forms a sigma bond with the middle carbon atom, and the third sp2 orbital has a lone pair of electrons. - The pi bond with the oxygen atom is formed between the unhybridized 2p orbitals of the carbon and oxygen atoms. - The middle carbon atom in C3O2 forms two double bonds with two adjacent carbon atoms, each double bond consisting of one sigma bond and one pi bond. - The middle carbon atom is sp hybridized, forming two sp orbitals. - Each terminal carbon atom forms a sigma bond with one of the sp orbitals of the middle carbon atom. - The pi bonds are formed between the 2p orbitals of the middle and terminal carbon atoms.

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Most popular questions from this chapter

Predict the molecular structure for each of the following. (See Exercises 25 and \(26 .\) ) a. \(\mathrm{BrFI}_{2}\) b. \(\mathrm{XeO}_{2} \mathrm{F}_{2}\) c. \(\operatorname{TeF}_{2} \mathrm{Cl}_{3}^{-}\) For each formula there are at least two different structures that can be drawn using the same central atom. Draw all possible structures for each formula.

Bond energy has been defined in the text as the amount of energy required to break a chemical bond, so we have come to think of the addition of energy as breaking bonds. However, in some cases the addition of energy can cause the formation of bonds. For example, in a sample of helium gas subjected to a high-energy source, some He_ molecules exist momentarily and then dissociate. Use MO theory (and diagrams) to explain why \(\mathrm{He}_{2}\) molecules can come to exist and why they dissociate.

What are molecular orbitals? How do they compare with atomic orbitals? Can you tell by the shape of the bonding and antibonding orbitals which is lower in energy? Explain.

Draw the Lewis structures, predict the molecular structures, and describe the bonding (in terms of the hybrid orbitals for the central atom) for the following. a. \(\mathrm{XeO}_{3}\) b. \(\mathrm{XeO}_{4}\) c. \(\mathrm{XeOF}_{4}\) d. \(\mathrm{XeOF}_{2}\) e. \(\mathrm{XeO}_{3} \mathrm{F}_{2}\)

Which of the following are predicted by the molecular orbital model to be stable diatomic species? a. \(\mathrm{N}_{2}^{2-}, \mathrm{O}_{2}^{2-}, \mathrm{F}_{2}^{2-}\) b. \(\mathrm{Be}_{2}, \mathrm{B}_{2}, \mathrm{Ne}_{2}\)

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