CHEMISTRY: ATOMS FIRST VOL 1 W/CONNECT
CHEMISTRY: ATOMS FIRST VOL 1 W/CONNECT
14th Edition
ISBN: 9781259327933
Author: Burdge
Publisher: MCG
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Textbook Question
Chapter 7, Problem 7.129QP

The BO+ ion is paramagnetic. Determine (a) whether the order of molecular-orbital energies is like that in B2 or in O2; (b) the bond order; and (c) the number of unpaired electrons in the ion.

(a)

Expert Solution
Check Mark
Interpretation Introduction

Interpretation: To find whether the order of molecular-orbital energies of BO+ is like that in O2 or B2. Also the bond order and the number of unpaired electrons in BO+ should be found.

Concept Introduction:

  • In molecular orbital theory, when the bonding takes place the atomic orbitals that take part combine to get a new orbital that has the properties of the whole molecule.
  • The orbital with lower energy gets the electron first and each orbital only contain a maximum of two electrons. It obeys Hund's Rule.
  • There are two types of molecular orbitals- bonding molecular orbitals and antibonding molecular orbitals.
  • If molecular species contains at least one unpaired electron, then they are paramagnetic. They tend to get attracted towards the magnetic field. Whereas in diamagnetic species there is no unpaired electrons and are slightly repels the magnetic field.
  • Stability of a molecule can be obtained from the bond order value. As the bond order increases, stability increases. Bond order can be obtained by,

BondOrder=(Numberofelectronsinbondingmolecularorbitals)(Numberofelectronsinantibondingmolecularorbitals)2

Explanation of Solution

In molecular orbital theory, when the bonding takes place the atomic orbitals that take part combine to get a new orbital that has the properties of the whole molecule. The newly formed orbitals are known as molecular orbitals and only contain a maximum of two electrons. The number of newly formed molecular orbital is equal to the number of atomic orbitals involved in the bonding.

There are two types of molecular orbitals,

  1. a) Bonding molecular orbitals: sharing of electron density is between the nuclei and has comparatively lower energy and fills first.
  2. b) Antibonding molecular orbitals: Two nuclei are pulled by the electrons density in opposite direction and have higher energy comparing to bonding molecular orbital.

The orbital with lower energy gets the electron first.

Each orbital only contain a maximum of two electrons.

It obeys Hund's Rule: The orbital having maximum number of electrons with same spin in separate orbitals is the most stable arrangement in an orbital.

Electronic configuration of O is 1s22s22p1 and the electronic configuration of B is 1s22s22p4

The total charge of BO+ is +1 so an electron is reduced from the total valence electrons. Hence the total number of valence electrons in BO+ is 8. By molecular orbital theory, it has unpaired electrons and thus is a paramagnetic species.

To compare the order of molecular orbital energies, the valence electrons in O2 and B2 is taken as eight. BO+ is paramagnetic so the orbitals are arranged as that of O2.

Molecular orbital diagram of B2

σ2pz* CHEMISTRY: ATOMS FIRST VOL 1 W/CONNECT, Chapter 7, Problem 7.129QP , additional homework tip  1

π2px*π2py* CHEMISTRY: ATOMS FIRST VOL 1 W/CONNECT, Chapter 7, Problem 7.129QP , additional homework tip  2 CHEMISTRY: ATOMS FIRST VOL 1 W/CONNECT, Chapter 7, Problem 7.129QP , additional homework tip  3

σ2pz CHEMISTRY: ATOMS FIRST VOL 1 W/CONNECT, Chapter 7, Problem 7.129QP , additional homework tip  4

π2pxπ2py CHEMISTRY: ATOMS FIRST VOL 1 W/CONNECT, Chapter 7, Problem 7.129QP , additional homework tip  5 CHEMISTRY: ATOMS FIRST VOL 1 W/CONNECT, Chapter 7, Problem 7.129QP , additional homework tip  6

σ2s* CHEMISTRY: ATOMS FIRST VOL 1 W/CONNECT, Chapter 7, Problem 7.129QP , additional homework tip  7

σ2s CHEMISTRY: ATOMS FIRST VOL 1 W/CONNECT, Chapter 7, Problem 7.129QP , additional homework tip  8

Molecular orbital diagram of O2

σ2pz* CHEMISTRY: ATOMS FIRST VOL 1 W/CONNECT, Chapter 7, Problem 7.129QP , additional homework tip  9

π2px*π2py* CHEMISTRY: ATOMS FIRST VOL 1 W/CONNECT, Chapter 7, Problem 7.129QP , additional homework tip  10 CHEMISTRY: ATOMS FIRST VOL 1 W/CONNECT, Chapter 7, Problem 7.129QP , additional homework tip  11

π2pxπ2py CHEMISTRY: ATOMS FIRST VOL 1 W/CONNECT, Chapter 7, Problem 7.129QP , additional homework tip  12 CHEMISTRY: ATOMS FIRST VOL 1 W/CONNECT, Chapter 7, Problem 7.129QP , additional homework tip  13

σ2pz CHEMISTRY: ATOMS FIRST VOL 1 W/CONNECT, Chapter 7, Problem 7.129QP , additional homework tip  14

σ2s* CHEMISTRY: ATOMS FIRST VOL 1 W/CONNECT, Chapter 7, Problem 7.129QP , additional homework tip  15

σ2s CHEMISTRY: ATOMS FIRST VOL 1 W/CONNECT, Chapter 7, Problem 7.129QP , additional homework tip  16

The molecular orbital diagram of O2 is similar to BO+.

(b)

Expert Solution
Check Mark
Interpretation Introduction

Interpretation: To find whether the order of molecular-orbital energies of BO+ is like that in O2 or B2. Also the bond order and the number of unpaired electrons in BO+ should be found.

Concept Introduction:

  • In molecular orbital theory, when the bonding takes place the atomic orbitals that take part combine to get a new orbital that has the properties of the whole molecule.
  • The orbital with lower energy gets the electron first and each orbital only contain a maximum of two electrons. It obeys Hund's Rule.
  • There are two types of molecular orbitals- bonding molecular orbitals and antibonding molecular orbitals.
  • If molecular species contains at least one unpaired electron, then they are paramagnetic. They tend to get attracted towards the magnetic field. Whereas in diamagnetic species there is no unpaired electrons and are slightly repels the magnetic field.
  • Stability of a molecule can be obtained from the bond order value. As the bond order increases, stability increases. Bond order can be obtained by,

BondOrder=(Numberofelectronsinbondingmolecularorbitals)(Numberofelectronsinantibondingmolecularorbitals)2

Explanation of Solution

Stability of a molecule can be obtained from the bond order value. As the bond order increases, stability increases. Bond order can be obtained by,

BondOrder=(Numberofelectronsinbondingmolecularorbitals)(Numberofelectronsinantibondingmolecularorbitals)2

Bondorder=(62)2=2

(c)

Expert Solution
Check Mark
Interpretation Introduction

Interpretation: To find whether the order of molecular-orbital energies of BO+ is like that in O2 or B2. Also the bond order and the number of unpaired electrons in BO+ should be found.

Concept Introduction:

  • In molecular orbital theory, when the bonding takes place the atomic orbitals that take part combine to get a new orbital that has the properties of the whole molecule.
  • The orbital with lower energy gets the electron first and each orbital only contain a maximum of two electrons. It obeys Hund's Rule.
  • There are two types of molecular orbitals- bonding molecular orbitals and antibonding molecular orbitals.
  • If molecular species contains at least one unpaired electron, then they are paramagnetic. They tend to get attracted towards the magnetic field. Whereas in diamagnetic species there is no unpaired electrons and are slightly repels the magnetic field.
  • Stability of a molecule can be obtained from the bond order value. As the bond order increases, stability increases. Bond order can be obtained by,

BondOrder=(Numberofelectronsinbondingmolecularorbitals)(Numberofelectronsinantibondingmolecularorbitals)2

Explanation of Solution

The total charge of BO+ is +1 so an electron is reduced from the total valence electrons. Hence the total number of valence electrons in BO+ is 8. The molecular orbital diagram of O2 is similar to BO+ since it is paramagnetic.

Molecular orbital diagram of BO+

σ2pz* CHEMISTRY: ATOMS FIRST VOL 1 W/CONNECT, Chapter 7, Problem 7.129QP , additional homework tip  17

π2px*π2py* CHEMISTRY: ATOMS FIRST VOL 1 W/CONNECT, Chapter 7, Problem 7.129QP , additional homework tip  18 CHEMISTRY: ATOMS FIRST VOL 1 W/CONNECT, Chapter 7, Problem 7.129QP , additional homework tip  19

π2pxπ2py CHEMISTRY: ATOMS FIRST VOL 1 W/CONNECT, Chapter 7, Problem 7.129QP , additional homework tip  20 CHEMISTRY: ATOMS FIRST VOL 1 W/CONNECT, Chapter 7, Problem 7.129QP , additional homework tip  21

σ2pz CHEMISTRY: ATOMS FIRST VOL 1 W/CONNECT, Chapter 7, Problem 7.129QP , additional homework tip  22

σ2s* CHEMISTRY: ATOMS FIRST VOL 1 W/CONNECT, Chapter 7, Problem 7.129QP , additional homework tip  23

σ2s CHEMISTRY: ATOMS FIRST VOL 1 W/CONNECT, Chapter 7, Problem 7.129QP , additional homework tip  24

BO+ has two unpaired electrons.

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Chapter 7 Solutions

CHEMISTRY: ATOMS FIRST VOL 1 W/CONNECT

Ch. 7.1 - Prob. 7.1.3SRCh. 7.1 - Prob. 7.1.4SRCh. 7.2 - Prob. 7.3WECh. 7.2 - Prob. 3PPACh. 7.2 - For each of the following hypothetical molecules,...Ch. 7.2 - Which of these models could represent a polar...Ch. 7.2 - Prob. 7.2.1SRCh. 7.2 - Prob. 7.2.2SRCh. 7.3 - Prob. 7.4WECh. 7.3 - Prob. 4PPACh. 7.3 - Prob. 4PPBCh. 7.3 - Prob. 4PPCCh. 7.3 - Prob. 7.3.1SRCh. 7.3 - Prob. 7.3.2SRCh. 7.4 - Hydrogen selenide (H2Se) is a foul-smelling gas...Ch. 7.4 - Prob. 5PPACh. 7.4 - For which molecule(s) can we not use valence bond...Ch. 7.4 - Which of these models could represent a species...Ch. 7.4 - Prob. 7.4.1SRCh. 7.4 - Prob. 7.4.2SRCh. 7.5 - Prob. 7.6WECh. 7.5 - Use hybrid orbital theory to describe the bonding...Ch. 7.5 - Prob. 6PPBCh. 7.5 - Prob. 6PPCCh. 7.5 - Prob. 7.5.1SRCh. 7.5 - Prob. 7.5.2SRCh. 7.6 - Thalidomide (C13H10N2O4) is a sedative and...Ch. 7.6 - The active ingredient in Tylenol and a host of...Ch. 7.6 - Determine the total number of sigma and pi bonds...Ch. 7.6 - In terms of valence bond theory and hybrid...Ch. 7.6 - In addition to its rise in aqueous solution as a...Ch. 7.6 - Use valence bond theory and hybrid orbitals to...Ch. 7.6 - Use valence bond theory and hybrid orbitals to...Ch. 7.6 - Explain why hybrid orbitals are necessary to...Ch. 7.6 - Prob. 7.6.1SRCh. 7.6 - Prob. 7.6.2SRCh. 7.6 - Prob. 7.6.3SRCh. 7.6 - Prob. 7.6.4SRCh. 7.7 - Prob. 7.9WECh. 7.7 - Use molecular orbital theory to determine whether...Ch. 7.7 - Use molecular orbital theory to determine whether...Ch. 7.7 - For most of the homonuclear diatomic species shown...Ch. 7.7 - Prob. 7.7.1SRCh. 7.7 - Prob. 7.7.2SRCh. 7.7 - Prob. 7.7.3SRCh. 7.7 - Prob. 7.7.4SRCh. 7.8 - It takes three resonance structures to represent...Ch. 7.8 - Use a combination of valence bond theory and...Ch. 7.8 - Use a combination of valence bond theory and...Ch. 7.8 - Prob. 10PPCCh. 7.8 - Prob. 7.8.1SRCh. 7.8 - Prob. 7.8.2SRCh. 7.8 - Prob. 7.8.3SRCh. 7.8 - Prob. 7.8.4SRCh. 7 - Prob. 7.1QPCh. 7 - Sketch the shape of a linear triatomic molecule, a...Ch. 7 - Prob. 7.3QPCh. 7 - Prob. 7.4QPCh. 7 - In the trigonal bipyramidal arrangement, why does...Ch. 7 - Prob. 7.6QPCh. 7 - Predict the geometry of the following molecules...Ch. 7 - Prob. 7.8QPCh. 7 - Predict the geometries of the following species...Ch. 7 - Predict the geometries of the following ions: (a)...Ch. 7 - Prob. 7.11QPCh. 7 - Prob. 7.12QPCh. 7 - Prob. 7.13QPCh. 7 - Describe the geometry about each of the central...Ch. 7 - Prob. 7.15QPCh. 7 - Prob. 7.16QPCh. 7 - Prob. 7.17QPCh. 7 - Prob. 7.18QPCh. 7 - Prob. 7.19QPCh. 7 - Prob. 7.20QPCh. 7 - Prob. 7.21QPCh. 7 - Prob. 7.22QPCh. 7 - Explain the term polarizability. What kind of...Ch. 7 - Prob. 7.24QPCh. 7 - What physical properties are determined by the...Ch. 7 - Prob. 7.26QPCh. 7 - Describe the types of intermolecular forces that...Ch. 7 - The compounds Br2 and ICl are isoelectronic (have...Ch. 7 - If you lived in Alaska, which of the following...Ch. 7 - The binary hydrogen compounds of the Group 4A...Ch. 7 - List the types of intermolecular forces that exist...Ch. 7 - Prob. 7.32QPCh. 7 - Prob. 7.33QPCh. 7 - Prob. 7.34QPCh. 7 - Diethyl ether has a boiling point of 34.5C, and...Ch. 7 - Prob. 7.36QPCh. 7 - Which substance in each of the following pairs...Ch. 7 - Prob. 7.38QPCh. 7 - What kind of attractive forces must be overcome to...Ch. 7 - Prob. 7.40QPCh. 7 - Prob. 7.41QPCh. 7 - The following compounds have the same molecular...Ch. 7 - Prob. 7.43QPCh. 7 - Prob. 7.44QPCh. 7 - Use valence bond theory to explain the bonding in...Ch. 7 - Prob. 7.46QPCh. 7 - Prob. 7.47QPCh. 7 - Prob. 7.48QPCh. 7 - Prob. 7.49QPCh. 7 - What is the hybridization of atomic orbitals? Why...Ch. 7 - Prob. 7.51QPCh. 7 - Prob. 7.52QPCh. 7 - Prob. 7.53QPCh. 7 - Describe the bonding scheme of the AsH3 molecule...Ch. 7 - Prob. 7.55QPCh. 7 - Prob. 7.56QPCh. 7 - Describe the hybridization of phosphorus in PF5.Ch. 7 - Prob. 7.58QPCh. 7 - Prob. 7.59QPCh. 7 - Prob. 7.1VCCh. 7 - Prob. 7.2VCCh. 7 - Prob. 7.3VCCh. 7 - Prob. 7.4VCCh. 7 - Prob. 7.60QPCh. 7 - Which of the following pairs of atomic orbitals of...Ch. 7 - Prob. 7.62QPCh. 7 - Prob. 7.63QPCh. 7 - Prob. 7.64QPCh. 7 - Prob. 7.65QPCh. 7 - Prob. 7.66QPCh. 7 - Prob. 7.67QPCh. 7 - Prob. 7.68QPCh. 7 - Benzo[a]pyrene is a potent carcinogen found in...Ch. 7 - What is molecular orbital theory? How does it...Ch. 7 - Define the following terms: bonding molecular...Ch. 7 - Prob. 7.72QPCh. 7 - Prob. 7.73QPCh. 7 - Prob. 7.74QPCh. 7 - Prob. 7.75QPCh. 7 - Draw a molecular orbital energy level diagram for...Ch. 7 - Prob. 7.77QPCh. 7 - Prob. 7.78QPCh. 7 - Prob. 7.79QPCh. 7 - Acetylene (C2H2) has a tendency to lose two...Ch. 7 - Compare the Lewis and molecular orbital treatments...Ch. 7 - Prob. 7.82QPCh. 7 - Prob. 7.83QPCh. 7 - Prob. 7.84QPCh. 7 - Prob. 7.85QPCh. 7 - Draw the molecular orbital diagram for the cyanide...Ch. 7 - Given that BeO is diamagnetic, use a molecular...Ch. 7 - Prob. 7.88QPCh. 7 - Prob. 7.89QPCh. 7 - Both ethylene (C2H4) and benzene (C6H6) contain...Ch. 7 - Chemists often represent benzene with the...Ch. 7 - Determine which of these molecules has a more...Ch. 7 - Nitryl fluoride (FNO2) is used in rocket...Ch. 7 - Describe the bonding in the nitrate ion NO3 in...Ch. 7 - Prob. 7.95QPCh. 7 - Prob. 7.96QPCh. 7 - Prob. 7.97QPCh. 7 - Prob. 7.98QPCh. 7 - Prob. 7.99QPCh. 7 - Antimony pentafluoride (SbF5) combines with XeF4...Ch. 7 - Prob. 7.101QPCh. 7 - The molecular model of nicotine (a stimulant) is...Ch. 7 - Predict the bond angles for the following...Ch. 7 - The germanium pentafluoride anion (GeF5) has been...Ch. 7 - Draw Lewis structures and give the other...Ch. 7 - Which figure best illustrates the hybridization of...Ch. 7 - Prob. 7.107QPCh. 7 - Prob. 7.108QPCh. 7 - Prob. 7.109QPCh. 7 - Prob. 7.110QPCh. 7 - Prob. 7.111QPCh. 7 - Cyclopropane (C3H6) has the shape of a triangle in...Ch. 7 - The compound 1,2-dichloroethane (C2H4Cl2) is...Ch. 7 - Prob. 7.114QPCh. 7 - Prob. 7.115QPCh. 7 - Prob. 7.116QPCh. 7 - Prob. 7.117QPCh. 7 - Prob. 7.118QPCh. 7 - The amino acid selenocysteine is one of the...Ch. 7 - Prob. 7.120QPCh. 7 - Prob. 7.121QPCh. 7 - Prob. 7.122QPCh. 7 - Gaseous or highly volatile liquid anesthetics are...Ch. 7 - Prob. 7.124QPCh. 7 - Prob. 7.125QPCh. 7 - Two of the drugs that are prescribed for the...Ch. 7 - Prob. 7.127QPCh. 7 - Prob. 7.128QPCh. 7 - The BO+ ion is paramagnetic. Determine (a) whether...Ch. 7 - Use molecular orbital theory to explain the...Ch. 7 - Which best illustrates the change in geometry...Ch. 7 - Prob. 7.132QPCh. 7 - Prob. 7.133QPCh. 7 - Aluminum trichloride (AlCl3) is an...Ch. 7 - Prob. 7.135QPCh. 7 - Prob. 7.136QPCh. 7 - Prob. 7.137QPCh. 7 - Consider an N2 molecule in its first excited...Ch. 7 - The Lewis structure for O2 is Use molecular...Ch. 7 - Draw the Lewis structure of ketene (C2H2O) and...Ch. 7 - The compound TCDD, or...Ch. 7 - Name the kinds of attractive forces that must be...Ch. 7 - Carbon monoxide (CO) is a poisonous compound due...Ch. 7 - Prob. 7.144QPCh. 7 - Prob. 7.145QPCh. 7 - Prob. 7.146QPCh. 7 - Prob. 7.147QPCh. 7 - Prob. 7.148QPCh. 7 - Prob. 7.1KSPCh. 7 - Which of the following species does not have...Ch. 7 - Prob. 7.3KSPCh. 7 - Prob. 7.4KSP
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