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(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 \((\sigma\) plust (rotate) around a single \(\sigma\) bond same?

Short Answer

Expert verified
(a) In sp hybridization, there is one unhybridized p orbital remaining, and the atom can form one π bond. (b) It would be easier to twist (rotate) around a single σ bond than around a double bond (σ plus π) due to the π bond overlap restriction in the double bond.

Step by step solution

01

Understanding sp hybridization and counting unhybridized p orbitals

An atom with sp hybridized orbitals has one s orbital and one p orbital combining to form two equivalent sp hybrid orbitals. Therefore, the atom has one remaining unhybridized p orbital, as there are initially three p orbitals in the valence shell of most elements (p_x, p_y, p_z).
02

Calculating the number of π bonds possible

The unhybridized p orbital can overlap with another unhybridized p orbital from another atom to form a π bond. Since there is only one unhybridized p orbital, the atom can form a maximum of one π bond. Answer for part (a): There is one unhybridized p orbital remaining and the atom can form one π bond.
03

Comparing the ease of rotation around single σ and double bonds

A single bond, which is a σ bond, allows for free rotation around the bond axis because its electron density is distributed symmetrically between the two bonded atoms. On the other hand, a double bond consists of one σ bond and one π bond. The π bond electron density is distributed above and below the plane formed by the two atoms and the σ bond. This distribution restricts the rotation around the bond axis because rotating the bond breaks the π bond overlap. Answer for part (b): It would be easier to twist (rotate) around a single σ bond than around a double bond (σ plus π) because of the π bond overlap restriction in the double bond.

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Key Concepts

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

sp Hybridization
When atoms undergo hybridization, their atomic orbitals mix to create new hybrid orbitals. In the case of sp hybridization, one s orbital combines with one p orbital. This results in the formation of two equivalent sp hybrid orbitals.
These sp hybrid orbitals are oriented linearly at an angle of 180°, making them ideal for linear geometry in molecules, such as acetylene (C₂H₂).
It's important to note that during sp hybridization, not all p orbitals are used in the hybridization process, leading to the presence of unhybridized p orbitals.
Unhybridized Orbitals
Within the valence shell of most atoms, there are three p orbitals: p_x, p_y, and p_z. When sp hybridization occurs, only one p orbital is used along with one s orbital, resulting in sp hybrids.
This means that out of the three p orbitals, two remain untouched, but only one stays completely unhybridized as usable in the electron configuration. This unhybridized p orbital plays a crucial role in forming additional bonds, such as π bonds when interacting with other atomic orbitals.
Pi Bonds
A π bond is formed from the sideways overlap of two unhybridized p orbitals from adjacent atoms. Unlike sigma (σ) bonds which are formed by head-on overlapping, π bonds involve the overlap above and below the bonding axis.
The presence of π bonds gives rise to significant bonding implications:
  • They provide additional bond strength and stability.
  • Restricting rotational freedom around the bond axis, as in double and triple bonds.
An atom with a single unhybridized p orbital, as in the case of sp hybridization, can form only one π bond.
Sigma Bonds
Sigma bonds (σ bonds) arise from the head-on or axial overlap of orbitals. These bonds form the framework for most single bonds in covalent compounds.
Due to this direct overlap:
  • σ bonds allow free rotation around their bond axis without disrupting the bond.
  • They are stronger and more stabilized compared to π bonds given their direct line of overlap.
This free rotation is intrinsic to single bonds, differentiating them from double bonds that restrict such movement due to accompanying π bonds.
Valence Shell
The valence shell is the outermost electron shell of an atom, crucial for chemical bond formation. It contains the electrons that participate in hybridization and bond formation.
Typically, the valence shell consists of s and p orbitals, which combined through hybridization, can form sp, sp², or sp³ hybrids depending on the number of orbitals mixing.
In sp hybridization, the valence shell is adapted to form specific linear structures, and only part of its p orbitals participate, leading to significant changes in the atom’s bonding capabilities.

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Most popular questions from this chapter

Predict whether each of the following molecules is polar or nonpolar: (a) IF, (b) \(\mathrm{CS}_{2},(\mathbf{c}) \mathrm{SO}_{3},(\mathbf{d}) \mathrm{PCl}_{3},(\mathbf{e}) \mathrm{SF}_{6},(\mathbf{f}) \mathrm{IF}_{5}\)

From their Lewis structures, determine the number of \(\sigma\) and \(\pi\) bonds in each of the following molecules or ions: (a) \(\mathrm{CO}_{2} ;\) (b) cyanogen,\((\mathrm{CN})_{2} ;(\mathbf{c})\) formaldehyde, \(\mathrm{H}_{2} \mathrm{CO}\) (d) formic acid, HCOOH, which has one H and two O atoms attached to \(\mathrm{C}\) .

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.

Place the following molecules and ions in order from smallest to largest bond order: \(\mathrm{H}_{2}^{+}, \mathrm{B}_{2}, \mathrm{N}_{2}^{+}, \mathrm{F}_{2}^{+},\) and \(\mathrm{Ne}_{2}\) .

Consider the molecule \(\mathrm{PF}_{4} \mathrm{Cl}\) (a) Draw a Lewis structure for the molecule, and predict its electron-domain geometry. (b) Which would you expect to take up more space, a \(\mathrm{P}-\mathrm{F}\) bond or a \(\mathrm{P}-\mathrm{Cl}\) bond? Explain. (c) Predict the molecular geometry of \(\mathrm{PF}_{4} \mathrm{Cl} .\) How did your answer for part (b) influence your answer here in part \((\mathrm{c}) ?(\mathbf{d})\) Would you expect the molecule to distort from its ideal electron-domain geometry? If so, how would it distort?

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