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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—like \( sp, sp^2, \) 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

Draw the Lewis structure for each of the following molecules or ions, and predict their electron-domain and molecular geometries: (a) \(\mathrm{AsF}_{3},\) (b) \(\mathrm{CH}_{3}^{+},\) (c) \(\mathrm{BrF}_{3},\) (d) \(\mathrm{ClO}_{3}^{-},\) (e) \(\mathrm{XeF}_{2}\), (f) \(\mathrm{BrO}_{2}^{-}\).

The \(\mathrm{O}-\mathrm{H}\) bond lengths in the water molecule \(\left(\mathrm{H}_{2} \mathrm{O}\right)\) are \(0.96 \AA\), and the \(\mathrm{H}-\mathrm{O}-\mathrm{H}\) angle is \(104.5^{\circ} .\) The dipole moment of the water molecule is \(1.85 \mathrm{D} .\) (a) In what directions do the bond dipoles of the \(\mathrm{O}-\mathrm{H}\) bonds point? In what direction does the dipole moment vector of the water molecule point? (b) Calculate the magnitude of the bond dipole of the \(\mathrm{O}-\mathrm{H}\) bonds. (Note: You will need to use vector addition to do this.) (c) Compare your answer from part (b) to the dipole moments of the hydrogen halides (Table 8.3). Is your answer in accord with the relative electronegativity of oxygen?

(a) If the valence atomic orbitals of an atom are sp hybridized, how many unhybridized \(p\) orbitals remain in the valence shell? How many \(\pi\) bonds can the atom form? (b) Imagine that you could hold two atoms that are bonded together, twist them, and not change the bond length. Would it be easier to twist (rotate) around a single \(\sigma\) bond or around a double \((\sigma\) plus \(\pi)\) bond, or would they be the same? Explain.

The phosphorus trihalides \(\left(\mathrm{PX}_{3}\right)\) show the following variation in the bond angle \(\mathrm{X}-\mathrm{P}-\mathrm{X}: \mathrm{PF}_{3}, 96.3^{\circ} ; \mathrm{PCl}_{3}, 100.3^{\circ} ; \mathrm{PBr}_{3}\), \(101.0^{\circ} ; \mathrm{PI}_{3}, 102.0^{\circ} .\) The trend is generally attributed to the change in the electronegativity of the halogen. (a) Assuming that all electron domains are the same size, what value of the \(\mathrm{X}-\mathrm{P}-\mathrm{X}\) angle is predicted by the VSEPR model? (b) What is the general trend in the \(\mathrm{X}-\mathrm{P}-\mathrm{X}\) angle as the halide electronegativity increases? (c) Using the VSEPR model, explain the observed trend in \(\mathrm{X}-\mathrm{P}-\mathrm{X}\) angle as the electronegativity of \(X\) changes. (d) Based on your answer to part (c), predict the structure of \(\mathrm{PBrCl}_{4}\).

Would you expect the nonbonding electron-pair domain in \(\mathrm{NH}_{3}\) to be greater or less in size than for the corresponding one in \(\mathrm{PH}_{3}\) ? Explain.

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