/*! 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 73 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
When the two sets of 2\(p\) orbitals come close together to bond, they can form one \(\sigma\) bond (through the overlap of 2\(p_z\) orbitals), two \(\pi\) bonds (formed from the overlap of 2\(p_x\) and 2\(p_y\) orbitals), and three antibonding orbitals (one \(\sigma^*\) and two \(\pi^*\) orbitals formed from the out-of-phase overlap of respective orbitals on both atoms).

Step by step solution

01

Draw 2\(p\) orbitals

First, we need to draw all three 2\(p\) orbitals on one atom and all three 2\(p\) orbitals on another atom. The three 2\(p\) orbitals are 2\(p_x\), 2\(p_y\), and 2\(p_z\). Each orbital is at right angles to each other, representing the x, y, and z axes. Draw two sets of these orbitals, one for each atom.
02

Answer (a) Number of \(\sigma\) bonds

When the two atoms come close together to bond, we need to determine how many \(\sigma\) bonds can be made between the two sets of 2\(p\) orbitals. A \(\sigma\) bond is formed when the orbitals overlap head-on. In this case, only one \(\sigma\) bond can be made: the 2\(p_z\) orbital on the first atom can overlap with the 2\(p_z\) orbital on the second atom.
03

Answer (b) Number of \(\pi\) bonds

Now, let's determine how many \(\pi\) bonds can be made between the two sets of 2\(p\) orbitals. A \(\pi\) bond is formed when the orbitals overlap side-by-side. In this case, two \(\pi\) bonds can be made: 1. The 2\(p_x\) orbital on the first atom can overlap with the 2\(p_x\) orbital on the second atom. 2. The 2\(p_y\) orbital on the first atom can overlap with the 2\(p_y\) orbital on the second atom. Therefore, two \(\pi\) bonds can be formed.
04

Answer (c) Number of antibonding orbitals

Finally, let's find the number and type of antibonding orbitals that can be made from the two sets of 2\(p\) orbitals. Antibonding orbitals are formed when the wave functions of atomic orbitals subtract from each other, creating a node between the atoms and resulting in a higher energy state. For the 2\(p\) orbitals, we can form a total of three antibonding orbitals: 1. One \(\sigma^*\) antibonding orbital can be formed from the out-of-phase overlap of the 2\(p_z\) orbitals on both atoms. 2. Two \(\pi^*\) antibonding orbitals can be formed from the out-of-phase overlap of 2\(p_x\) orbitals and 2\(p_y\) orbitals on both atoms. In summary, the two sets of 2\(p\) orbitals can make one \(\sigma\) bond, two \(\pi\) bonds, and three antibonding orbitals (one \(\sigma^*\) and two \(\pi^*\) orbitals).

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Key Concepts

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

Sigma Bonds
Imagine two puzzle pieces fitting together perfectly. This is similar to how a sigma bond forms when two atomic orbitals join head-on. It's the simplest type of covalent bond and represents a strong interaction between atoms that forms a molecular backbone. In the case of 2p orbitals, only one sigma bond can occur, that between the two 2pz orbitals. This is because the pz orbitals are aligned along the axis connecting the two atoms, allowing for direct overlap.

Why is this important? Well, sigma bonds give molecules their basic shapes. Because they are formed from head-on overlap, these bonds are symmetrical around the bond axis and tend to be stronger and more stable than other types of covalent bonds. Understanding sigma bonds helps us predict how atoms will come together to form molecules and what shape those molecules will take.
Pi Bonds
Now let's move to the atoms holding hands side by side, which is how pi bonds are formed. Unlike sigma bonds, pi bonds result from the side-to-side overlap of atomic orbitals. In the world of 2p orbitals, each atom can form two pi bonds, one each from the px and py orbitals. These are the stages for electron clouds to dance above and below the plane of the atoms in a molecule.

While pi bonds alone are weaker than sigma bonds, they are essential for the stability of double and triple bonds in chemistry. Pi bonds restrict the rotation of bonded atoms, giving rise to interesting phenomena like the rigidity seen in double-bonded compounds. They also have implications in the world of chemistry and materials science, playing a critical role in the properties of molecules like the toughness of synthetic plastics or the conductivity of organic semiconductors.
Antibonding Orbitals
In every relationship, there's a possibility for harmony (bonding) or conflict (antibonding). The latter is what we observe with antibonding orbitals. These are odd ones out, resulting from the out-of-phase overlap of atomic orbitals that create a node where electron density is essentially cancelled out. Think of this as adding two waves that are out of sync, negating each other at certain points. For 2p orbitals, three antibonding orbitals are possible: one sigma-star (\b\(\b\sigma^\*\)\b)) from the out-of-phase overlap of 2pz orbitals, and two pi-star (\b\(\b\pi^\*\)\b)) orbitals from similar interactions between 2px and 2py orbitals.

An understanding of antibonding orbitals is instrumental in grasping molecular orbital theory, which explains the behavior of electrons in a molecule in terms of quantum mechanics. These orbitals are of higher energy and can destabilize a molecule if electrons occupy them. Knowing this helps explain why some reactions occur and others do not, providing insights into the world of reaction mechanisms and molecular dynamics.
Atomic Orbital Overlap
The heart of bonding lies in how well the atomic orbitals can blend their electron clouds together, a concept known as atomic orbital overlap. Good overlap equals a strong bond, much like a firm handshake. This overlap can occur either head-on (forming sigma bonds) or side-by-side (forming pi bonds). The extent of this overlap is a quintessential concept in understanding molecular structure and bonding.

In the interaction of 2p orbitals, proper alignment is crucial. The pz orbitals overlap along the axis connecting two atoms, while the px and py orbitals need to be parallel for pi bonding. The degree of overlap will influence bond strength, bond length, and ultimately, how the molecule behaves in various conditions. It's a simple and elegant way to predict chemical reactivity and physical properties, which are influenced by the nature of these overlaps, their strength, and the type of bonds they create.

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Most popular questions from this chapter

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?

(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?

The structure of borazine, \(\mathrm{B}_{3} \mathrm{N}_{3} \mathrm{H}_{6},\) is a six-membered ring of alternating \(\mathrm{B}\) and \(\mathrm{N}\) atoms. There is one \(\mathrm{H}\) atom bonded to each \(\mathrm{B}\) and to each \(\mathrm{N}\) atom. The molecule is planar. (a) Write a Lewis structure for borazine in which the formal charge on every atom is zero. (b) Write a Lewis structure for borazine in which the octet rule is satisfied for every atom. (c) What are the formal charges on the atoms in the Lewis structure from part (b)? Given the electronegativities of \(\mathrm{B}\) and \(\mathrm{N},\) do the formal charges seem favorable or unfavorable? (d)Do either of the Lewis structures in parts (a) and (b) have multiple resonance structures? (e) What are the hybridizations at the B and N atoms in the Lewis structures from parts (a) and (b)? Would you expect the molecule to be planar for both Lewis structures? (f) The six \(B-N\) bonds in the borazine molecule are all identical in length at 1.44 A. Typical values for the bond lengths of \(\mathrm{B}-\mathrm{N}\) single and double bonds are 1.51 \(\mathrm{A}\) and \(1.31 \mathrm{A},\) respectively. Does the value of the \(\mathrm{B}-\mathrm{N}\) bond length seem to favor one Lewis structure over the other? (g) How many electrons are in the \(\pi\) system of borazine?

What are the electron-domain and molecular geometries of a molecule that has the following electron domains on its central atom? (a) Three bonding domains and no nonbonding domains, (b) three bonding domains and one nonbonding domain, (c) two bonding domains and two nonbonding domains.

Give the electron-domain and molecular geometries for the following molecules and ions: (a) \(\mathrm{HCN},(\mathbf{b}) \mathrm{SO}_{3}^{2-},(\mathbf{c}) \mathrm{SF}_{4}\) \((\mathbf{d}) \mathrm{PF}_{6},(\mathbf{e}) \mathrm{NH}_{3} \mathrm{Cl}^{+},(\mathbf{f}) \mathrm{N}_{3}^{-}\)

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