/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 63 Draw a picture that shows all th... [FREE SOLUTION] | 91Ó°ÊÓ

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Draw a picture that shows all three \(2 p\) orbitals on one atom and all three \(2 p\) orbitals on another atom. (a) Imagine the atoms coming close together to bond. How many \(\sigma\) bonds can the two sets of \(2 p\) orbitals make with each other? (b) How many \(\pi\) bonds can the two sets of \(2 p\) orbitals make with each other? (c) How many antibonding orbitals, and of what type, can be made from the two sets of \(2 p\) orbitals?

Short Answer

Expert verified
(a) The two sets of \(2p\) orbitals can form 1 \(\sigma\) bond. (b) The two sets of \(2p\) orbitals can form 2 \(\pi\) bonds. (c) The two sets of \(2p\) orbitals can form 3 antibonding orbitals: 1 \(\sigma^*\) and 2 \(\pi^*\).

Step by step solution

01

Draw the 2p orbitals for each atom

First, let's draw each \(2p\) orbital for both atoms. There are three \(2p\) orbitals for each atom: \(2p_x\), \(2p_y\), and \(2p_z\). Each of them has two lobes with opposite phases and they are oriented along the respective x, y, and z axes.
02

Understand the types of bonds

We have two types of bonds to consider: sigma and pi bonds. When orbitals overlap end-to-end along their axis, a \(\sigma\) bond is formed. When orbitals overlap side-to-side, a \(\pi\) bond is formed.
03

Determine the number of sigma bonds

Since a \(\sigma\) bond is formed by end-to-end overlap of orbitals, only \(2p_z\) orbitals can form a \(\sigma\) bond because they are oriented along the z-axis. So, one \(\sigma\) bond can be formed between the two \(2p_z\) orbitals of the two atoms.
04

Determine the number of pi bonds

A \(\pi\) bond is formed by side-to-side overlap of orbitals. Since \(2p_x\) and \(2p_y\) orbitals are oriented along the x and y axis respectively, they can form \(\pi\) bonds through side-to-side overlap. Hence, two \(\pi\) bonds can be formed between the sets of \(2p\) orbitals on each atom.
05

Determine the number of antibonding orbitals

Antibonding orbitals are formed when the overlap of orbitals results in a decrease in electron density between the two atomic nuclei. For each type of bond (\(\sigma\) bond, \(\pi_x\) bond and \(\pi_y\) bond formed between \(2p\) orbitals), there will also be an antibonding orbital with a higher energy level. So, in total, we have the following antibonding orbitals: 1. 1 \(\sigma^*\) (from \(2p_z\) orbitals) 2. 2 \(\pi^*\) (from \(2p_x\) and \(2p_y\) orbitals) To answer the initial questions: (a) The two sets of \(2p\) orbitals can form 1 \(\sigma\) bond. (b) The two sets of \(2p\) orbitals can form 2 \(\pi\) bonds. (c) The two sets of \(2p\) orbitals can form 3 antibonding orbitals: 1 \(\sigma^*\) and 2 \(\pi^*\).

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Key Concepts

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

Sigma Bonds
Sigma bonds are the strongest type of covalent chemical bonds. They are characterized by the head-on overlap of atomic orbitals, which allows electrons to be shared directly between the nuclei of the bonding atoms. This direct overlap creates a single bond that can rotate freely around its axis.

In the context of the exercise, sigma bonds are formed when two atoms' orbitals overlap end-to-end. Specifically, when considering the overlap of the 2p orbitals, the two 2p_z orbitals align along the same axis, allowing them to form a single sigma bond. This is because they are oriented along the z-axis, enabling the maximum amount of overlap.

Here are some key points about sigma bonds:
  • They are single bonds represented by a single line between two atoms in a structural formula.
  • Their formation allows for better electron sharing, leading to stronger interactions.
  • These bonds can involve orbitals such as s, p, and hybrid orbitals, depending on the participating atoms.
This strong overlap of orbitals provides the basis for molecule stability and is crucial for the configuration and shape of molecules.
Pi Bonds
Pi bonds are typically weaker than sigma bonds and arise from the side-to-side overlap of orbitals. They can't exist alone and are often found in conjunction with a sigma bond, forming what we recognize as double or triple bonds in molecules.

In pi bonds, the electron density is concentrated above and below the axis of the bonding atoms, rather than directly between the nuclei as in sigma bonds. For the 2p orbitals in the exercise, pi bonds are established when the 2p_x and 2p_y orbitals overlap side-to-side.

Several important characteristics of pi bonds include:
  • They restrict rotation around the bond axis, which influences the shape and configuration of the molecules they join, contributing to the rigidity of structures like alkenes and alkynes.
  • A molecule with one sigma and one pi bond is said to have a double bond, while one with one sigma and two pi bonds has a triple bond.
  • Pi bonds are often responsible for the unique reactivity patterns seen in unsaturated hydrocarbons, due to the electron density present in their free orbitals.
Understanding pi bonds allows you to grasp how multiple bonding interactions can affect molecular geometry and chemical characteristics.
Antibonding Orbitals
Antibonding orbitals are formed when atomic orbitals overlap in such a manner that the electron density is concentrated outside the bonding region. This results in a net decrease in the bond's stability due to reduced electron sharing between the atomic nuclei. These orbitals are typically higher in energy compared to their bonding counterparts.

In the case of the 2p orbitals discussed in the exercise, antibonding orbitals are created when overlap, particularly detrimental or ineffective, reduces the constructive interference seen in bonding situations. Specifically, for each type of bond - sigma and pi - there is a corresponding antibonding orbital, labeled with an asterisk (*).

Key aspects of antibonding orbitals include:
  • They are denoted as σ* or Ï€*, indicating their relationship to their respective bonding orbitals (e.g., σ* for sigma bonds).
  • These orbitals can contribute to the overall energy and instability in a molecule, sometimes even leading to breaking bonds if occupied by electrons.
  • During molecular orbital formation, the presence of antibonding orbitals explains why not all available electrons contribute to bond formation.
The concept of antibonding orbitals is essential for understanding the balance of energy in compounds and plays a pivotal role in reactions and molecular interactions.

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Most popular questions from this chapter

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), (c) the \(2 s\) orbital on one atom and the \(2 p_{z}\) orbital on the other atom.

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

Sulfur tetrafluoride (SF \(_{4}\) ) reacts slowly with \(\mathrm{O}_{2}\) to form sulfur tetrafluoride monoxide \(\left(\mathrm{OSF}_{4}\right)\) according to the following unbalanced reaction: $$ \mathrm{SF}_{4}(g)+\mathrm{O}_{2}(g) \longrightarrow \operatorname{OSF}_{4}(g) $$ The \(\mathrm{O}\) atom and the four \(\mathrm{F}\) atoms in \(\mathrm{OSF}_{4}\) are bonded to a central \(S\) atom. (a) Balance the equation. (b) Write a Lewis structure of \(\mathrm{OSF}_{4}\) in which the formal charges of all atoms are zero. (c) Use average bond enthalpies (Table 8.4) to estimate the enthalpy of the reaction. Is it endothermic or exothermic? (d) Determine the electrondomain geometry of \(\mathrm{OSF}_{4}\), and write two possible molecular geometries for the molecule based on this electron-domain geometry. (e) Which of the molecular geometries in part (d) is more likely to be observed for the molecule? Explain.

Determine the electron configurations for \(\mathrm{CN}^{+}, \mathrm{CN}\), and \(\mathrm{CN}^{-}\). (a) Which species has the strongest \(\mathrm{C}-\mathrm{N}\) bond? (b) Which species, if any, has unpaired electrons?

(a) Methane \(\left(\mathrm{CH}_{4}\right)\) and the perchlorate ion \(\left(\mathrm{ClO}_{4}-\right)\) are both described as tetrahedral. What does this indicate about their bond angles? (b) The \(\mathrm{NH}_{3}\) molecule is trigonal pyramidal, while \(\mathrm{BF}_{3}\) is trigonal planar. Which of these molecules is flat?

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