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(a) Is \(\mathrm{CO}_{2}\) linear or bent? (b) What hybridization is appropriate for the \(C\) atom? (c) Outline a bonding scheme for \(\mathrm{CO}_{2}\) using the hybridization scheme you have suggested. (d) What \(C-O\) bond order does your scheme imply? (e) Draw a Lewis structure for \(\mathrm{CO}_{2}\). Is this structure consistent with the results you obtained in parts (c) and (d)?

Short Answer

Expert verified
(a) Linear; (b) \(sp\); (c) \(sp\) for \(C\), \(sp^2\) for \(O\); (d) 2; (e) Yes.

Step by step solution

01

Determining Geometry of CO2

To determine if \(\mathrm{CO}_{2}\) is linear or bent, we consider the number of regions of electron density around the central carbon atom. \(\mathrm{CO}_{2}\) has two double bonds and no lone pairs around its central carbon atom, which leads to a linear geometry as predicted by VSEPR theory.
02

Identifying Carbon's Hybridization

The carbon atom in \(\mathrm{CO}_{2}\) forms two double bonds, with two regions of electron density. This corresponds to \(sp\) hybridization, as \(sp\)-hybridized orbitals form a linear arrangement, appropriate for the linear geometry of \(\mathrm{CO}_{2}\).
03

Outlining the Bonding Scheme

In the \(\mathrm{CO}_{2}\) molecule, carbon forms bonds using \(sp\) hybrid orbitals for the sigma bonds with oxygen. Each carbon-oxygen double bond consists of one \(\sigma\) bond (from the overlap of an \(sp\) hybrid orbital on carbon with an \(sp^{2}\) hybrid orbital on oxygen) and one \(\pi\) bond (from the side-to-side overlap of unhybridized \( p \) orbitals).
04

Determining the C-O Bond Order

Each \(C-O\) double bond in \(\mathrm{CO}_{2}\) consists of one \(\sigma\) and one \(\pi\) bond. The bond order is calculated as 2 (one \(\sigma\) plus one \(\pi\) bond), which represents a double bond.
05

Drawing the Lewis Structure

The Lewis structure for \(\mathrm{CO}_{2}\) shows carbon (\(C\)) double-bonded to each oxygen (\(O\)) atom, represented as \(:O=C=O:\). This structure matches the linear geometry and bond order of 2 indicated in previous steps, and contains no lone pairs on the central carbon, consistent with \(sp\) hybridization.

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Key Concepts

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

VSEPR theory
Valence Shell Electron Pair Repulsion (VSEPR) theory is essential in determining the molecular shape of compounds like CO2. According to this theory, electron pairs around a central atom will arrange themselves to minimize repulsion and hence assume a shape that keeps them as far apart as possible.

For carbon dioxide (CO2), the central carbon atom has two regions of electron density corresponding to the double bonds it forms with oxygen. Since CO2 has no lone pairs, the two double bonds dictate its geometry. These double bonds are positioned directly opposite each other around the central carbon, resulting in a linear configuration. This linear shape aligns perfectly with the predictions of VSEPR theory and explains why CO2 is not bent but linear.
sp hybridization
Hybridization is an important concept that helps explain the bonding and shape of molecules in chemistry. In the case of CO2, the central carbon atom undergoes a specific type of hybridization known as sp hybridization.

In sp hybridization, one s orbital and one p orbital in the carbon atom mix together, forming two equivalent sp hybrid orbitals. This combination best fits the two double bonds in CO2 because sp hybridized orbitals arrange in a linear geometry, 180 degrees apart.

This orbital arrangement is what allows CO2 to maintain its linear form, providing strong directional bonds with the two oxygen atoms. It is this simple yet insightful principle that demonstrates how molecular shapes arise from the hybridization of atomic orbitals.
bonding scheme in CO2
The bonding scheme in CO2 uses the concept of hybridization to detail the way atoms connect. In CO2, the carbon atom uses its sp hybrid orbitals to form sigma (σ) bonds with each oxygen atom.

In addition to these sigma bonds, which are end-to-end overlaps of orbitals, carbon and oxygen form pi (π) bonds. These π bonds result from the lateral or side-by-side overlap of unhybridized p orbitals participating in the double bonds. Each carbon-oxygen double bond thus consists of one σ bond and one π bond, confirming the strong and stable double bond nature. This arrangement explains both the strength of the bonds and their contribution to the linear shape of CO2.

By understanding the bonding scheme, we get comprehensive insight into why CO2 behaves as it does and remains a stable, linear molecule.
Lewis structure of CO2
The Lewis structure provides a useful visual representation of the electron arrangement around atoms within a molecule. For CO2, the Lewis structure is drawn as ":O=C=O:", reflecting carbon's double bonds with each oxygen atom.

This straightforward depiction shows no lone pairs on the central carbon, aligning with its sp hybridization state. Each oxygen atom is double-bonded to the central carbon, displaying a bond order of 2, consistent with the double bond nature observed in CO2's bonding scheme.

Lewis structures effectively emphasize how valence electrons are shared among atoms, giving insight into the chemical properties and molecular structure. Thus, this visual representation of CO2 not only reveals its linear shape but also maintains a coherent bond order and hybridization model.

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Most popular questions from this chapter

(a) Draw the structure of \(\left[\mathrm{CO}_{3}\right]^{2-}\). (b) If all the \(\mathrm{C}-\mathrm{O}\) bond distances are equal, write a set of resonance structures to describe the bonding in \(\left[\mathrm{CO}_{3}\right]^{2-}\). (c) Describe the bonding in \(\left[\mathrm{CO}_{3}\right]^{2-}\) in terms of a hybridization scheme and compare the result with that obtained in part (b).

What is meant by a ligand group orbital?

(a) \(\mathrm{BCl}_{3}\) has \(D_{3 \mathrm{h}}\) symmetry. Draw the structure of \(\mathrm{BCl}_{3}\) and give values for the bond angles. \(\mathrm{NCl}_{3}\) has \(C_{3 \mathrm{v}}\) symmetry. Is it possible to state the bond angles from this information? (b) Derive the symmetry labels for the atomic orbitals on \(\mathrm{B}\) in \(\mathrm{BCl}_{3}\) and on \(\mathrm{N}\) in \(\mathrm{NCl}_{3}\).

\(\operatorname{In}\left[\mathrm{B}_{2} \mathrm{H}_{7}\right]^{-}(5.11),\) each \(\mathrm{B}\) atom is approximately tetrahedral. (a) How many valence electrons are present in the anion? (b) Assume that each \(\mathrm{B}\) atom is \(s p^{3}\) hybridized. After localization of the three terminal \(\mathrm{B}-\mathrm{H}\) bonds per \(\mathrm{B}\), what B-centred orbital remains for use in the bridging interaction? (c) Following from your answer to part (b), construct an approximate orbital diagram to show the formation of \(\left[\mathrm{B}_{2} \mathrm{H}_{7}\right]^{-}\) from two \(\mathrm{BH}_{3}\) units and \(\mathrm{H}^{-}\). What does this approach tell you about the nature of the \(\mathrm{B}-\mathrm{H}-\mathrm{B}\) bridge?

Suggest an appropriate hybridization scheme for the central atom in each of the following species: (a) \(\mathrm{SiF}_{4}\); (b) \(\left[\mathrm{PdCl}_{4}\right]^{2-} ;(\mathrm{c}) \mathrm{NF}_{3} ;(\mathrm{d}) \mathrm{F}_{2} \mathrm{O} ;(\mathrm{e})\left[\mathrm{CoH}_{5}\right]^{4-}\); \((\mathrm{f})\left[\mathrm{FeH}_{6}\right]^{4-} ;(\mathrm{g}) \mathrm{CS}_{2} ;(\mathrm{h}) \mathrm{BF}_{3}\).

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