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Chapter 9: Problem 70

What hybridization do you expect for the atom indicated in red in each of the following species? (a) \(\mathrm{CH}_{3} \mathrm{CO}_{2}^{-} ;\) (b) \(\mathrm{PH}_{4}^{+}\) (c) \(\mathrm{AlF}_{3}\) (d) \(\mathrm{H}_{2} \mathrm{C}=\mathrm{CH}-\mathrm{CH}_{2}^{+}\)

Short Answer

Expert verified
The hybridizations for the atoms indicated in red in each species are: (a) sp2 (b) sp3 (c) sp2 (d) sp2

Step by step solution

01

(a) CH3CO2-

For this species, we need to determine the hybridization of the carbon atom in red. To do this, we need to consider the electron domains around the carbon atom. The central carbon atom is bonded to 3 atoms (one carbon and two oxygen atoms) and has no lone pairs of electrons. This results in 3 electron domains. Thus, the hybridization of this carbon atom is sp2.
02

(b) PH4+

For this species, we need to determine the hybridization of the phosphorus atom. To do this, we need to consider the electron domains around the phosphorus atom. The central phosphorus atom is bonded to 4 atoms (four hydrogen atoms) and has no lone pairs of electrons. This results in 4 electron domains. Therefore, the hybridization of the phosphorus atom is sp3.
03

(c) AlF3

For this species, we need to determine the hybridization of the aluminum atom. To do this, we need to consider the electron domains around the aluminum atom. The central aluminum atom is bonded to 3 atoms (three fluorine atoms) and has no lone pairs of electrons. This leads to 3 electron domains. Consequently, the hybridization of the aluminum atom is sp2.
04

(d) H2C=CH-CH2+

For this species, we need to determine the hybridization of the carbon atom in red. To do this, we need to consider the electron domains around the carbon atom. The carbon atom in red is bonded to 3 atoms (two carbon atoms and one hydrogen atom) and has no lone pairs of electrons. There are 3 electron domains around it. Therefore, the hybridization of this carbon atom is sp2.

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Key Concepts

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

Electron Domains
Electron domains play a crucial role in determining the hybridization of an atom. The term 'electron domain' refers to the regions around an atom where electrons are most likely to be found. These regions can include:
  • Bonds with other atoms
  • Lone pairs of electrons that reside on the atom itself
For instance, a central atom surrounded by three bonded atoms (with no lone pairs) has three electron domains. Understanding the concept of electron domains allows us to predict the geometry and hybridization of an atom in a molecule.
By counting these domains, we can infer the hybridization鈥攍ike \( sp, sp^2, \) or \( sp^3 \). This foundational concept is vital for comprehending molecular structure and properties.
sp2 Hybridization
In \( sp^2 \) hybridization, one s orbital mixes with two p orbitals to form three equivalent \( sp^2 \) hybrid orbitals. This kind of hybridization arises when there are three electron domains surrounding a central atom. Each of these hybrid orbitals has one-third s character and two-thirds p character.
A common example of \( sp^2 \) hybridization is seen in molecules like ethene (C=C), where each carbon atom is bonded to two other atoms and has one \( \pi \) bond (pi bond).
  • The result is a trigonal planar geometry
  • Angles of approximately \( 120^\circ \)
This configuration also affects the molecule's reactivity and physical properties due to the presence of the \( \pi \) bond, which allows for rotations and has a significant impact on molecular interactions.
sp3 Hybridization
\( sp^3 \) hybridization occurs when one s orbital combines with three p orbitals to form four equivalent \( sp^3 \) hybrid orbitals. This typically happens when a central atom has four electron domains, which can consist of bonds or lone pairs.
The classic example of \( sp^3 \) hybridization is seen in methane (\( \text{CH}_4 \)), where the central carbon atom is surrounded by four hydrogen atoms.
  • The resulting shape is tetrahedral
  • Angle of \( 109.5^\circ \)
Each \( sp^3 \) orbital has a character of one-fourth s and three-fourths p, allowing for a uniform distribution of electrons in space. This leads to a stable, tetrahedral geometry making the molecules nonplanar and enhancing their three-dimensional nature.
Molecular Geometry
Molecular geometry refers to the three-dimensional arrangement of atoms within a molecule. This geometric shape directly influences the molecule's physical and chemical properties.
  • The determination of molecular geometry primarily relies on the electron domains around the central atom.
  • Each molecule adopts a shape that minimizes repulsions between electron domains.
Typical shapes include:
  • Linear
  • Trigonal planar
  • Tetrahedral

For example, in a molecule like \( \text{CH}_4 \) with \( sp^3 \) hybridization, a tetrahedral shape with four equivalent bonds is observed. This geometry ensures that electron repulsions are minimized to create a stable molecular structure. Understanding the molecular geometry gives insights into the synthesize and functional interactions of the species.

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Most popular questions from this chapter

(a) Starting with the orbital diagram of a boron atom, describe the steps needed to construct hybrid orbitals appropriate to describe the bonding in \(\mathrm{BF}_{3}\). (b) What is the name given to the hybrid orbitals constructed in (a)? (c) Sketch the large lobes of the hybrid orbitals constructed in part (a). (d) Are any valence atomic orbitals of \(\mathrm{B}\) left unhybridized? If so, how are they oriented relative to the hybrid orbitals?

The nitrogen atoms in \(\mathrm{N}_{2}\) participate in multiple bonding, whereas those in hydrazine, \(\mathrm{N}_{2} \mathrm{H}_{4},\) do not. (a) Draw Lewis structures for both molecules. (b) What is the hybridization of the nitrogen atoms in each molecule? (c) Which molecule has the stronger \(\mathrm{N}-\mathrm{N}\) bond?

Dichloroethylene \(\left(\mathrm{C}_{2} \mathrm{H}_{2} \mathrm{Cl}_{2}\right)\) has three forms (isomers), each of which is a different substance. (a) Draw Lewis structures of the three isomers, all of which have a carbon-carbon double bond. (b) Which of these isomers has a zero dipole moment? (c) How many isomeric forms can chloroethylene, \(\mathrm{C}_{2} \mathrm{H}_{3} \mathrm{Cl},\) have? Would they be expected to have dipole moments?

Give the electron-domain and molecular geometries of a molecule that has the following electron domains on its central atom: (a) four bonding domains and no nonbonding domains, (b) three bonding domains and two nonbonding domains, (c) five bonding domains and one nonbonding domain, (d) four bonding domains and two nonbonding domains.

Azo dyes are organic dyes that are used for many applications, such as the coloring of fabrics. Many azo dyes are derivatives of the organic substance azobenzene, \(\mathrm{C}_{12} \mathrm{H}_{10} \mathrm{~N}_{2}\). A closely related substance is hydrazobenzene, \(\mathrm{C}_{12} \mathrm{H}_{12} \mathrm{~N}_{2} .\) The Lewis structures of these two substances are (Recall the shorthand notation used for benzene.) (a) What is the hybridization at the \(\mathrm{N}\) atom in each of the substances? (b) How many unhybridized atomic orbitals are there on the \(\mathrm{N}\) and the \(C\) atoms in each of the substances? (c) Predict the \(\mathrm{N}-\mathrm{N}-\mathrm{C}\) angles in each of the substances. (d) Azobenzene is said to have greater delocalization of its \(\pi\) electrons than hydrazobenzene. Discuss this statement in light of your answers to (a) and (b). (e) All the atoms of azobenzene lie in one plane, whereas those of hydrazobenzene do not. Is this observation consistent with the statement in part (d)? (f) Azobenzene is an intense red-orange color, whereas hydrazobenzene is nearly colorless. Which molecule would be a better one to use in a solar energy conversion device? (See the "Chemistry Put to Work" box for more information about solar cells.)
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