Private: Chapter Five
Key Terms, Key Equations, Summaries, and Exercises (Chapter 5)
Key Terms
antibonding orbital molecular orbital located outside of the region between two nuclei; electrons in an antibonding orbital destabilize the molecule
bond order number of pairs of electrons between two atoms; it can be found by the number of bonds in a Lewis structure or by the difference between the number of bonding and antibonding electrons divided by two
bonding orbital molecular orbital located between two nuclei; electrons in a bonding orbital stabilize a molecule
degenerate orbitals orbitals that have the same energy
diamagnetism phenomenon in which a material is not magnetic itself but is repelled by a magnetic field; it occurs when there are only paired electrons present
homonuclear diatomic molecule molecule consisting of two identical atoms
hybrid orbital orbital created by combining atomic orbitals on a central atom
hybridization model that describes the changes in the atomic orbitals of an atom when it forms a covalent compound
linear combination of atomic orbitals technique for combining atomic orbitals to create molecular orbitals
molecular orbital region of space in which an electron has a high probability of being found in a molecule
molecular orbital diagram visual representation of the relative energy levels of molecular orbitals
molecular orbital theory model that describes the behavior of electrons delocalized throughout a molecule in terms of the combination of atomic wave functions
node plane separating different lobes of orbitals, where the probability of finding an electron is zero
overlap coexistence of orbitals from two different atoms sharing the same region of space, leading to the formation of a covalent bond
paramagnetism phenomenon in which a material is not magnetic itself but is attracted to a magnetic field; it occurs when there are unpaired electrons present
pi bond (π bond) covalent bond formed by side-by-side overlap of atomic orbitals; the electron density is found on opposite sides of the internuclear axis
s-p mixing change that causes σp orbitals to be less stable than πp orbitals due to the mixing of s and p-based molecular orbitals of similar energies.
sigma bond (σ bond) covalent bond formed by overlap of atomic orbitals along the internuclear axis
sp hybrid orbital one of a set of two orbitals with a linear arrangement that results from combining one s and one p
orbital
sp2 hybrid orbital one of a set of three orbitals with a trigonal planar arrangement that results from combining one
s and two p orbitals
sp3 hybrid orbital one of a set of four orbitals with a tetrahedral arrangement that results from combining one s and three p orbitals
sp3d hybrid orbital one of a set of five orbitals with a trigonal bipyramidal arrangement that results from
combining one s, three p, and one d orbital
sp3d2 hybrid orbital one of a set of six orbitals with an octahedral arrangement that results from combining one s, three p, and two d orbitals
valence bond theory description of bonding that involves atomic orbitals overlapping to form σ or π bonds, within which pairs of electrons are shared
π bonding orbital molecular orbital formed by side-by-side overlap of atomic orbitals, in which the electron density is found on opposite sides of the internuclear axis
π* bonding orbital antibonding molecular orbital formed by out of phase side-by-side overlap of atomic orbitals, in which the electron density is found on both sides of the internuclear axis, and there is a node between the nuclei
- bonding orbital molecular orbital in which the electron density is found along the axis of the bond
σ* bonding orbital antibonding molecular orbital formed by out-of-phase overlap of atomic orbital along the axis of the bond, generating a node between the nuclei
Key Equations
- bond order = ⎛number of bonding electron⎞ − ⎛number of antibonding electrons⎞
⎝⎠⎝⎠
2
Summary
Valence bond theory describes bonding as a consequence of the overlap of two separate atomic orbitals on different atoms that creates a region with one pair of electrons shared between the two atoms. When the orbitals overlap along an axis containing the nuclei, they form a σ bond. When they overlap in a fashion that creates a node along this axis, they form a π bond. Dipole moments can be used to determine partial separations of charges between atoms.
We can use hybrid orbitals, which are mathematical combinations of some or all of the valence atomic orbitals, to describe the electron density around covalently bonded atoms. These hybrid orbitals either form sigma (σ) bonds directed toward other atoms of the molecule or contain lone pairs of electrons. We can determine the type of hybridization around a central atom from the geometry of the regions of electron density about it. Two such regions imply sp hybridization; three, sp2 hybridization; four, sp3 hybridization; five, sp3d hybridization; and six, sp3d2 hybridization. Pi (π) bonds are formed from unhybridized atomic orbitals (p or d orbitals).
Multiple bonds consist of a σ bond located along the axis between two atoms and one or two π bonds. The σ bonds are usually formed by the overlap of hybridized atomic orbitals, while the π bonds are formed by the side-by-side overlap of unhybridized orbitals. Resonance occurs when there are multiple unhybridized orbitals with the appropriate alignment to overlap, so the placement of π bonds can vary.
Molecular orbital (MO) theory describes the behavior of electrons in a molecule in terms of combinations of the atomic wave functions. The resulting molecular orbitals may extend over all the atoms in the molecule. Bonding molecular orbitals are formed by in-phase combinations of atomic wave functions, and electrons in these orbitals stabilize a molecule. Antibonding molecular orbitals result from out-of-phase combinations of atomic wave functions and electrons in these orbitals make a molecule less stable. Molecular orbitals located along an internuclear axis are called σ MOs. They can be formed from s orbitals or from p orbitals oriented in an end-to-end fashion. Molecular orbitals formed from p orbitals oriented in a side-by-side fashion have electron density on opposite sides of the
internuclear axis and are called π orbitals.
We can describe the electronic structure of diatomic molecules by applying molecular orbital theory to the valence electrons of the atoms. Electrons fill molecular orbitals following the same rules that apply to filling atomic orbitals; Hund’s rule and the Aufbau principle tell us that lower-energy orbitals will fill first, electrons will spread out before they pair up, and each orbital can hold a maximum of two electrons with opposite spins. Materials with unpaired electrons are paramagnetic and attracted to a magnetic field, while those with all-paired electrons are diamagnetic and repelled by a magnetic field. Correctly predicting the magnetic properties of molecules is in advantage of molecular orbital theory over Lewis structures and valence bond theory.
Exercises
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1.
Explain how σ and π bonds are similar and how they are different.
2.
Use valence bond theory to explain the bonding in F2, HF, and ClBr. Sketch the overlap of the atomic orbitals involved in the bonds.
3.
Use valence bond theory to explain the bonding in O2. Sketch the overlap of the atomic orbitals involved in the bonds in O2.
4.
How many σ and π bonds are present in the molecule HCN?
5.
A friend tells you N2 has three π bonds due to overlap of the three p-orbitals on each N atom. Do you agree?
6.
Draw the Lewis structures for CO2 and CO, and predict the number of σ and π bonds for each molecule.
(a) CO2
(b) CO
5.2 Hybrid Atomic Orbitals
7.
Why is the concept of hybridization required in valence bond theory?
8.
Give the shape that describes each hybrid orbital set:
(a) sp2
(b) sp3d
(c) sp
(d) sp3d2
9.
Explain why a carbon atom cannot form five bonds using sp3d hybrid orbitals.
10.
What is the hybridization of the central atom in each of the following?
(a) BeH2
(b) SF6
(c) PO43−
(d) PCl5
11.
A molecule with the formula AB3 could have one of four different shapes. Give the shape and the hybridization of the central A atom for each.
12.
Methionine, CH3SCH2CH2CH(NH2)CO2H, is an amino acid found in proteins. The Lewis structure of this compound is shown below. What is the hybridization type of each carbon, oxygen, the nitrogen, and the sulfur?
13.
Sulfuric acid is manufactured by a series of reactions represented by the following equations:
S8(𝑠)+8O2(𝑔)⟶8SO2(𝑔)
2SO2(𝑔)+O2(𝑔)⟶2SO3(𝑔)
SO3(𝑔)+H2O(𝑙)⟶H2SO4(𝑙)Draw a Lewis structure, predict the molecular geometry by VSEPR, and determine the hybridization of sulfur for the following:
(a) circular S8 molecule
(b) SO2 molecule
(c) SO3 molecule
(d) H2SO4 molecule (the hydrogen atoms are bonded to oxygen atoms)
14.
Two important industrial chemicals, ethene, C2H4, and propene, C3H6, are produced by the steam (or thermal) cracking process:
2C3H8(𝑔)⟶C2H4(𝑔)+C3H6(𝑔)+CH4(𝑔)+H2(𝑔)
For each of the four carbon compounds, do the following:
(a) Draw a Lewis structure.
(b) Predict the geometry about the carbon atom.
(c) Determine the hybridization of each type of carbon atom.
15.
Analysis of a compound indicates that it contains 77.55% Xe and 22.45% F by mass.
(a) What is the empirical formula for this compound? (Assume this is also the molecular formula in responding to the remaining parts of this exercise).
(b) Write a Lewis structure for the compound.
(c) Predict the shape of the molecules of the compound.
(d) What hybridization is consistent with the shape you predicted?
16.
Consider nitrous acid, HNO2 (HONO).
(a) Write a Lewis structure.
(b) What are the electron pair and molecular geometries of the internal oxygen and nitrogen atoms in the HNO2 molecule?
(c) What is the hybridization on the internal oxygen and nitrogen atoms in HNO2?
17.
Strike-anywhere matches contain a layer of KClO3 and a layer of P4S3. The heat produced by the friction of striking the match causes these two compounds to react vigorously, which sets fire to the wooden stem of the match. KClO3 contains the ClO3−
ion. P4S3 is an unusual molecule with the skeletal structure.
(a) Write Lewis structures for P4S3 and the ClO3–
ion.
(b) Describe the geometry about the P atoms, the S atom, and the Cl atom in these species.
(c) Assign a hybridization to the P atoms, the S atom, and the Cl atom in these species.
(d) Determine the oxidation states and formal charge of the atoms in P4S3 and the ClO3–
ion.
18.
Identify the hybridization of each carbon atom in the following molecule. (The arrangement of atoms is given; you need to determine how many bonds connect each pair of atoms.)
19.
Write Lewis structures for NF3 and PF5. On the basis of hybrid orbitals, explain the fact that NF3, PF3, and PF5 are stable molecules, but NF5 does not exist.
20.
In addition to NF3, two other fluoro derivatives of nitrogen are known: N2F4 and N2F2. What shapes do you predict for these two molecules? What is the hybridization for the nitrogen in each molecule?
5.3 Multiple Bonds
21.
The bond energy of a C–C single bond averages 347 kJ mol−1; that of a C≡C
triple bond averages 839 kJ mol−1. Explain why the triple bond is not three times as strong as a single bond.
22.
For the carbonate ion, CO32−,
draw all of the resonance structures. Identify which orbitals overlap to create each bond.
23.
A useful solvent that will dissolve salts as well as organic compounds is the compound acetonitrile, H3CCN. It is present in paint strippers.
(a) Write the Lewis structure for acetonitrile, and indicate the direction of the dipole moment in the molecule.
(b) Identify the hybrid orbitals used by the carbon atoms in the molecule to form σ bonds.
(c) Describe the atomic orbitals that form the π bonds in the molecule. Note that it is not necessary to hybridize the nitrogen atom.
24.
For the molecule allene, H2C=C=CH2,
give the hybridization of each carbon atom. Will the hydrogen atoms be in the same plane or perpendicular planes?
25.
Identify the hybridization of the central atom in each of the following molecules and ions that contain multiple bonds:
(a) ClNO (N is the central atom)
(b) CS2
(c) Cl2CO (C is the central atom)
(d) Cl2SO (S is the central atom)
(e) SO2F2 (S is the central atom)
(f) XeO2F2 (Xe is the central atom)
(g) ClOF2+
(Cl is the central atom)
26.
Describe the molecular geometry and hybridization of the N, P, or S atoms in each of the following compounds.
(a) H3PO4, phosphoric acid, used in cola soft drinks
(b) NH4NO3, ammonium nitrate, a fertilizer and explosive
(c) S2Cl2, disulfur dichloride, used in vulcanizing rubber
(d) K4[O3POPO3], potassium pyrophosphate, an ingredient in some toothpastes
27.
For each of the following molecules, indicate the hybridization requested and whether or not the electrons will be delocalized:
(a) ozone (O3) central O hybridization
(b) carbon dioxide (CO2) central C hybridization
(c) nitrogen dioxide (NO2) central N hybridization
(d) phosphate ion (PO43−)
central P hybridization
28.
For each of the following structures, determine the hybridization requested and whether the electrons will be delocalized:
(a) Hybridization of each carbon
(b) Hybridization of sulfur
(c) All atoms
29.
Draw the orbital diagram for carbon in CO2 showing how many carbon atom electrons are in each orbital.
5.4 Molecular Orbital Theory
30.
Sketch the distribution of electron density in the bonding and antibonding molecular orbitals formed from two s orbitals and from two p orbitals.
31.
How are the following similar, and how do they differ?
(a) σ molecular orbitals and π molecular orbitals
(b) ψ for an atomic orbital and ψ for a molecular orbital
(c) bonding orbitals and antibonding orbitals
32.
If molecular orbitals are created by combining five atomic orbitals from atom A and five atomic orbitals from atom B combine, how many molecular orbitals will result?
33.
Can a molecule with an odd number of electrons ever be diamagnetic? Explain why or why not.
34.
Can a molecule with an even number of electrons ever be paramagnetic? Explain why or why not.
35.
Why are bonding molecular orbitals lower in energy than the parent atomic orbitals?
36.
Calculate the bond order for an ion with this configuration:
(σ2𝑠)2(σ∗2𝑠)2(σ2𝑝𝑥)2(π2𝑝𝑦,π2𝑝𝑧)4(π∗2𝑝𝑦,π∗2𝑝𝑧)3
37.
Explain why an electron in the bonding molecular orbital in the H2 molecule has a lower energy than an electron in the 1s atomic orbital of either of the separated hydrogen atoms.
38.Predict the valence electron molecular orbital configurations for the following, and state whether they will be stable or unstable ions.
(a) Na22+
(b) Mg22+
(c) Al22+
(d) Si22+
(e) P22+
(f) S22+
(g) F22+
(h) Ar22+
39.
Determine the bond order of each member of the following groups, and determine which member of each group is predicted by the molecular orbital model to have the strongest bond.
(a) H2, H2+,
H2−
(b) O2, O22+,
O22−
(c) Li2, Be2+,
Be2
(d) F2, F2+,
F2−
(e) N2, N2+,
N2−
40.
For the first ionization energy for an N2 molecule, what molecular orbital is the electron removed from?
41.
Compare the atomic and molecular orbital diagrams to identify the member of each of the following pairs that has the highest first ionization energy (the most tightly bound electron) in the gas phase:
(a) H and H2
(b) N and N2
(c) O and O2
(d) C and C2
(e) B and B2
42.
Which of the period 2 homonuclear diatomic molecules are predicted to be paramagnetic?
43.
A friend tells you that the 2s orbital for fluorine starts off at a much lower energy than the 2s orbital for lithium, so the resulting σ2s molecular orbital in F2 is more stable than in Li2. Do you agree?
44.
True or false: Boron contains 2s22p1 valence electrons, so only one p orbital is needed to form molecular orbitals.
45.
What charge would be needed on F2 to generate an ion with a bond order of 2?
46.
Predict whether the MO diagram for S2 would show s-p mixing or not.
47.
Explain why N22+
is diamagnetic, while O24+,
which has the same number of valence electrons, is paramagnetic.
48.
Using the MO diagrams, predict the bond order for the stronger bond in each pair:
(a) B2 or B2+
(b) F2 or F2+
(c) O2 or O22+
(d) C2+
or C2−