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(a) Draw a picture showing how two \(p\) orbitals on two different atoms can be combined to make a \(\sigma\) bond. (b) Sketch a \(\pi\) bond that is constructed from \(p\) orbitals. (c) Which is generally stronger, a \(\sigma\) bond or a \(\pi\) bond? Explain. (d) Can two s orbitals combine to form a \(\pi\) bond? Explain.

Short Answer

Expert verified
(a) A \(\sigma\) bond formed by two \(p\) orbitals is created through the head-on overlap of their dumbbell-shaped lobes. (b) A \(\pi\) bond formed by two \(p\) orbitals occurs when the orbitals overlap side-by-side, with electron density spread out above and below the plane of the atoms. (c) A \(\sigma\) bond is generally stronger than a \(\pi\) bond due to the greater extent of overlap and concentrated electron density. (d) Two \(s\) orbitals cannot form a \(\pi\) bond as they have a spherical shape, lacking distinct lobes for side-by-side overlap; they can only form \(\sigma\) bonds.

Step by step solution

01

(a) Drawing a \(\sigma\) bond formed by \(p\) orbitals

To draw a \(\sigma\) bond formed by two \(p\) orbitals, we must first understand that \(p\) orbitals have a dumbbell shape with two lobes. These orbitals can overlap end-to-end, with their lobes pointing towards each other. This is an example of a head-on overlap, which forms a \(\sigma\) bond. Step 1: Draw two atoms with their \(p\) orbitals, both having dumbbell shapes and pointing towards each other. Step 2: Show the overlapping region of two \(p\) orbitals in a different shade or color, if possible. Step 3: Indicate that this overlap is forming a \(\sigma\) bond in the molecule.
02

(b) Sketching a \(\pi\) bond formed by \(p\) orbitals

To sketch a \(\pi\) bond formed by two \(p\) orbitals, keep in mind that \(p\) orbitals have a dumbbell shape with two lobes. These orbitals can overlap side-by-side, with their lobes parallel to each other. This is a parallel overlap, which forms a \(\pi\) bond. Step 1: Draw two atoms with their \(p\) orbitals, both having dumbbell shapes and oriented parallel to each other. Step 2: Show the overlapping region of two \(p\) orbitals on the top and bottom, in a different shade or color, if possible. Step 3: Indicate that this side-by-side overlap is forming a \(\pi\) bond in the molecule.
03

(c) Comparing \(\sigma\) and \(\pi\) bond strengths

Generally, a \(\sigma\) bond is stronger than a \(\pi\) bond. The reason for this difference in strength can be explained by the extent of the overlap between orbitals. In a \(\sigma\) bond, the orbitals overlap end-to-end, which means that the electron density is concentrated between the two atoms. This leads to a greater extent of overlap and stronger bond formation. On the other hand, in a \(\pi\) bond, the orbitals overlap side-by-side, and the electron density is spread out above and below the plane of the atoms, resulting in a lesser extent of overlap and a weaker bond.
04

(d) Forming a \(\pi\) bond with two \(s\) orbitals

No, two \(s\) orbitals cannot combine to form a \(\pi\) bond. The reason for this is that \(s\) orbitals have a spherical shape, while \(\pi\) bonds are formed by the side-by-side overlap of orbitals with distinct lobes, such as \(p\) orbitals. Since \(s\) orbitals do not have distinct lobes and their electron density is spread uniformly around the nucleus, they cannot participate in the formation of a \(\pi\) bond. Instead, when two \(s\) orbitals overlap, they form a \(\sigma\) bond.

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Key Concepts

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

Sigma Bond
In the world of chemical bonding, the sigma bond (\(\sigma\) bond) is a fundamental concept. It represents the strongest type of covalent bond due to the manner in which atomic orbitals overlap. Sigma bonds are formed when the orbitals overlap end-to-end or axially. This kind of overlap is often described as head-on collision.

The most common scenario involves the overlap of two \(p\) orbitals, which have a dumbbell shape. These orbitals align along the internuclear axis, which is the straight line that connects the nuclei of the bonding atoms. This alignment allows for a maximum overlap of the lobes of \(p\) orbitals, resulting in greater electron density concentrated directly between the atom's nuclei.

Sigma bonds are not limited to \(p\) orbitals. Bonds between \(s\) orbitals or even \(s\) and \(p\) orbitals can also form sigma bonds due to their ability to overlap directly in a similar manner.
Pi Bond
Pi bonds (\(\pi\) bonds) are another type of covalent bond that arises from the overlap of orbitals; however, they differ significantly from sigma bonds. Pi bonds result from side-to-side overlap of orbitals, such as \(p\) orbitals, that are oriented parallel to one another.

This kind of overlap occurs above and below the plane of the atomic nuclei, rather than directly between them. Because the electron density in a \(\pi\) bond is located in this manner, it allows for the formation of multiple bonds between two atoms – one being a sigma bond, ensuring stability, and the other a pi bond(s), providing additional connection without altering the sigma bond.

A key feature of \(\pi\) bonds is that they are generally weaker than \(\sigma\) bonds. This is due to the fact that their sideways overlap does not allow as much electron density as the direct overlap of a \(\sigma\) bond.
Orbital Overlap
The concept of orbital overlap is foundational to understanding how covalent bonds are formed. At its simplest, orbital overlap refers to the interaction between the electron clouds of two atoms that occur when they come close together.

There are primarily two types of orbital overlap that lead to bonding:
  • End-to-End Overlap: This occurs in sigma bonds, where orbitals engage directly along their axes. This alignment maximizes contact and electron density between the two atomic nuclei.
  • Side-by-Side Overlap: Found in pi bonds, this interaction places electron density above and below the plane of the nuclei, resulting in less direct interaction.
Both types of overlap have their own implications for bond properties, such as strength and length, influencing the overall molecular stability and geometry.
Molecular Orbitals
The formation of molecular orbitals is a central theme in the bonding theory connecting the overlap of atomic orbitals to the creation of covalent bonds. When atomic orbitals combine, they form new orbitals known as molecular orbitals that are spread out over a molecule rather than being localized around a single atom.

Molecular orbitals can be categorized into bonding and antibonding orbitals, playing a role in determining the stability of the bond:
  • Bonding Molecular Orbitals: These result from constructive interference of wave functions, leading to increased electron density between the nuclei, thus stabilizing the bond.
  • Antibonding Molecular Orbitals: Formed from destructive interference, these have a nodal plane between the nuclei, which can destabilize the molecule if occupied.
Understanding molecular orbitals provides valuable insight into the electronic arrangement within molecules and helps to predict and rationalize molecular properties such as bond order, magnetism, and electronic transitions.

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Most popular questions from this chapter

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. \((\mathbf{b})\) 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 thev be expected to have dipole moments?

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) \(\mathrm{CH}_{3}^{+},(\mathbf{c}) \mathrm{Br} \mathrm{F}_{3}\) (d) \(\mathrm{ClO}_{3}^{-}\) (e) \(\mathrm{XeF}_{2}\) (f) \(\mathrm{BrO}_{2}^{-}\).

(a) What are the relationships among bond order, bond length, and bond energy? (b) According to molecular orbital theory, would either \(\mathrm{Be}\), or \(\mathrm{Be}_{2}^{+}\) be expected to exist? Explain.

Draw sketches illustrating the overlap between the following orbitals on two atoms: (a) the \(2 s\) orbital on each atom, (b) the \(2 p_{z}\) orbital on each atom (assume both atoms are on the \(z\) -axis), \((\mathbf{c})\) the 2 s orbital on one atom and the \(2 p_{2}\) orbital on the other atom.

Consider the \(\mathrm{H}_{2}^{+}\) ion. (a) Sketch the molecular orbitals of the ion and draw its energy-level diagram. (b) How many electrons are there in the \(\mathrm{H}_{2}^{+}\) ion? (c) Write the electron configuration of the ion in terms of its MOs. (d) What is the bond order in \(\mathrm{H}_{2}^{+}\) ? (e) Suppose that the ion is excited by light so that an electron moves from a lower-energy to a higher- energy MO. Would you expect the excited-state \(\mathrm{H}_{2}^{+}\) ion to be stable or to fall apart? (f) Which of the following statements about part (e) is correct: (i) The light excites an electron from a bonding orbital to an antibonding orbital, (ii) The light excites an electron from an antibonding orbital to a bonding orbital, or (iii) In the excited state there are more bonding electrons than antibonding electrons?

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