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Represent bonding in the carbon dioxide molecule, \(\mathrm{CO}_{2},\) by \((\mathrm{a})\) a Lewis structure and \((\mathrm{b})\) the valencebond method. Identify \(\sigma\) and \(\pi\) bonds, the necessary hybridization scheme, and orbital overlap.

Short Answer

Expert verified
The Lewis structure for CO2 is \(O = C = O\) showing carbon atom in the center, double bonded to two oxygen atoms. In terms of Valence Bond theory, each double bond contains one \(\sigma\) bond and one \(\pi\) bond. The \(\sigma\) bond is formed by the overlap of an sp hybrid orbital from carbon and a p orbital from oxygen, while the \(\pi\) bond is formed through the sideways overlapping of the p orbitals of both carbon and oxygen atoms. Carbon undergoes sp hybridization in CO2.

Step by step solution

01

Lewis Structure

A Lewis structure shows all the valence electrons in a molecule. Since oxygen has 6 valence electrons and carbon has 4, we total these together and get \(6*2 + 4 = 16\) electrons. These 16 electrons form the structure: \(O = C = O\). That structure indicates carbon atom in the center, double bonded to two oxygen atoms on either side.
02

The Valence Bond Method and Identifying \(\sigma\) and \(\pi\) Bonds

The valence bond method lays the foundation to identify \(\sigma\) and \(\pi\) bonds. Each double bond consists of a \(\sigma\) bond and a \(\pi\) bond. In this molecule, one \(\sigma\) bond is formed by the overlap of an sp hybrid orbital from carbon and a p orbital from oxygen. The \(\pi\) bond is formed by the sideways overlapping of p orbitals.
03

Identifying Hybridization Scheme

In CO2, the carbon atom is the center atom linked to two oxygen atoms through double bonds. Hence, carbon forms 2 sigma bonds with the two oxygen atoms, which requires two hybrid orbitals. Therefore, the hybridization of carbon in CO2 is sp (two hybrid orbitals: one s and one p orbital).

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Key Concepts

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

Lewis Structure
The Lewis structure is a simple way to represent the valence electrons in a molecule. In the case of carbon dioxide (\(\mathrm{CO}_{2}\)), we start by counting the total valence electrons. Each oxygen atom has six valence electrons, and the carbon atom has four. Hence, the total comes to 16 valence electrons (\(6 \times 2 + 4 = 16\)). This electron count helps in constructing the structure.

The carbon atom, which usually acts as the central atom due to its ability to form multiple bonds, is bonded to two oxygen atoms. Each bond consists of sharing electrons to fill the outer shell of each atom. In \(\mathrm{CO}_2\), the electron sharing forms a double bond between carbon and each oxygen atom. Representing this structure, we write \(O = C = O\), showing two double bonds. This structure satisfies the octet rule, where each atom has eight electrons in its valence shell.
Valence Bond Theory
Valence bond theory provides insight into how atoms bond by overlapping their orbitals to form a molecule. It considers atomic orbitals overlapping each other to form those bonds.

In \(\mathrm{CO}_{2}\), the valence bond method allows us to identify different kinds of bonds formed. Each double bond in \(\mathrm{CO}_2\) includes a \(\sigma\) bond and a \(\pi\) bond:
  • \(\sigma\) bond: Formed by the head-on overlap of orbitals. In \(\mathrm{CO}_2\), the \(\sigma\) bond between the carbon and each oxygen atom is formed by the overlap of an sp hybrid orbital from carbon and a p orbital from oxygen.
  • \(\pi\) bond: Arises from the sideways overlap of p orbitals. Once the \(\sigma\) bonds form, p orbitals that didn’t participate in forming \(\sigma\) bonds overlap sideways, creating \(\pi\) bonds.
Hybridization
Hybridization is a concept used to describe how atomic orbitals mix to form new hybrid orbitals suitable for pairing electrons. This mixing explains bond formation's geometry and bond angles in molecules.

For carbon dioxide (\(\mathrm{CO}_2\)), the hybridization of the carbon atom is noted as sp hybridization:
  • Carbon forms two \(\sigma\) bonds with each oxygen atom, requiring two hybrid orbitals.
  • These two hybrid orbitals are formed by mixing one s orbital and one p orbital from carbon, leading to sp hybridization.
  • This results in a linear molecular shape with a bond angle of 180°.
The two remaining p orbitals are used for \(\pi\) bonding, allowing the formation of strong, stable double bonds with oxygen.
Sigma and Pi Bonds
Understanding \(\sigma\) and \(\pi\) bonds is crucial in breaking down the type of bond that forms between atoms in molecules like \(\mathrm{CO}_2\).

  • \(\sigma\) Bonds: These are the strong types of bonds formed by the end-to-end overlap of orbitals along the axis connecting two nuclei. In \(\mathrm{CO}_2\), each carbon-oxygen double bond has one \(\sigma\) bond.
  • \(\pi\) Bonds: Weaker than \(\sigma\) bonds, \(\pi\) bonds are formed when parallel orbitals overlap sideways. In each \(\mathrm{CO}_2\) double bond, one \(\pi\) bond complements the \(\sigma\) bond, adding strength and stability to the bond.
The understanding of these bonds helps in visualizing the structure of molecules and predicting their behavior in various chemical reactions.

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Most popular questions from this chapter

In what type of material is the energy gap between the valence band and the conduction band greatest: metal, semiconductor, or insulator? Explain.

The 60 -cycle alternating electric current (AC) commonly used in households changes direction 120 times per second. That is, in a one-second time period a terminal at an electric outlet is positive 60 times and negative 60 times. In direct electric current (DC), the flow between terminals is in one direction only. A rectifer is a device that converts alternating to direct current. One type of rectifier is the \(p-n\) junction rectifier. It is commonly incorporated in adapters required to operate electronic devices from ordinary house current. In the operation of this rectifier, a \(p\) -type semiconductor and an \(n\) -type semiconductor are in contact along a boundary, or junction. Each semiconductor is connected to one of the terminals in an AC electrical outlet. Describe how this rectifier works. That is, show that when the semiconductors are connected to the terminals in an AC outlet, half the time a large flow of charge occurs and half the time essentially no charge flows across the \(p-n\) junction.

\(\mathrm{He}_{2}\) does not exist as a stable molecule, but there is evidence that such a molecule can be formed between electronically excited He atoms. Write a molecular orbital diagram to account for this.

Match each of the following species with one of these hybridization schemes: \(s p, s p^{2}, s p^{3}, s p^{3} d, s p^{3} d^{2} .\) (a) \(\mathrm{PF}_{6}^{-}\) (b) \(\operatorname{COS} ;\) (c) \(\operatorname{SiCl}_{4} ;\) (d) \(\mathrm{NO}_{3}^{-}\);(e) AsF \(_{5}\)

Fluorine nitrate, \(\mathrm{FONO}_{2}\), is an oxidizing agent used as a rocket propellant. A reference source lists the following data for \(\mathrm{FO}_{\mathrm{a}} \mathrm{NO}_{2}\). (The subscript "a" shows that this O atom is different from the other two.) Bond lengths: \(\mathrm{N}-\mathrm{O}=129 \mathrm{pm}\) $$ \mathrm{N}-\mathrm{O}_{\mathrm{a}}=139 \mathrm{pm} ; \mathrm{O}_{\mathrm{a}}-\mathrm{F}=142 \mathrm{pm} $$ Bond angles: \(\mathrm{O}-\mathrm{N}-\mathrm{O}=125^{\circ}\) $$ \mathrm{F}-\mathrm{O}_{\mathrm{a}}-\mathrm{N}=105^{\circ} $$ \(\mathrm{NO}_{\mathrm{a}} \mathrm{F}\) plane is perpendicular to the \(\mathrm{O}_{2} \mathrm{NO}_{\mathrm{a}}\) plane Use these data to construct a Lewis structure(s), a three-dimensional sketch of the molecule, and a plausible bonding scheme showing hybridization and orbital overlaps.

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