Chemistry & Chemical Reactivity
Chemistry & Chemical Reactivity
9th Edition
ISBN: 9781133949640
Author: John C. Kotz, Paul M. Treichel, John Townsend, David Treichel
Publisher: Cengage Learning
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Chapter 9, Problem 26PS

Consider the following list of small molecules and ions: C2 O2, CN, O2. CO, NO, NO+, C22, OF. Identify

(a) all species that have a bond order of 3

(b) all species that are paramagnetic

(c) species that have a fractional bond order

(a)

Expert Solution
Check Mark
Interpretation Introduction

Interpretation:

All the species which have a bond order of 3 should be identified from the given list of molecules. 

Concept Introduction:

Molecular orbital (MO) theory:  is a method for determining molecular structure in which electrons are not assigned to individual bonds between atoms, but are treated as moving under the influence of the nuclei in the whole molecule.

Bond order: It is the measure of number of electron pairs shared between two atoms.

Bondorder=12(NumberofelectronsinbondoingMOs-NumberofelectronsinantibondingMOs)

Answer to Problem 26PS

The species which have a bond order of 3 are,

CN-,CO,NO+andC22-

Explanation of Solution

The list of given molecules and their electronic configuration in MO term and bond order are given below,

MoleculeNo.ofelectronsElectronconfigurationBondorder=12(NbNa)C212(σ1s)2(σ*1s)2(σ2s)2(σ*2s)2( π2p)4842=1O217(σ1s)2(σ*1s)2(σ2s)2(σ*2s)2(σ2p)2( π2p)4( π*2p)3 1072=1.5CN14(σ1s)2(σ*1s)2(σ2s)2(σ*2s)2( π2p)4(σ2p)21042=3O216(σ1s)2(σ*1s)2(σ2s)2(σ*2s)2(σ2p)2( π2p)4( π*2p)2 1062=2CO14(σ1s)2(σ*1s)2(σ2s)2(σ*2s)2( π2p)4(σ2p)21042=3NO15(σ1s)2(σ*1s)2(σ2s)2(σ*2s)2(σ2p)2( π2p)4( π*2p)11052=2.5NO+14(σ1s)2(σ*1s)2(σ2s)2(σ*2s)2( π2p)4(σ2p)21042=3C2214(σ1s)2(σ*1s)2(σ2s)2(σ*2s)2( π2p)4(σ2p)21042=3OF18(σ1s)2(σ*1s)2(σ2s)2(σ*2s)2(σ2p)2( π2p)4( π*2p)41082=1NbNumberofbondingelectrons,NaNumberofantibondingelectrons

According to the above table it is clear that there are four molecules have a bond order of 3. And they are mentioned below,

CN-,CO,NO+andC22-

(b)

Expert Solution
Check Mark
Interpretation Introduction

Interpretation:

All the species which are paramagnetic should be identified from the given list of molecules. 

Concept Introduction:

Molecular orbital (MO) theory:  is a method for determining molecular structure in which electrons are not assigned to individual bonds between atoms, but are treated as moving under the influence of the nuclei in the whole molecule.

Bond order: It is the measure of number of electron pairs shared between two atoms.

Bondorder=12(NumberofelectronsinbondoingMOs-NumberofelectronsinantibondingMOs)

Atoms with unpaired electrons are called Paramagnetic. Paramagnetic atoms are attracted to a magnet.

Atoms with paired electrons are called diamagnetic. Diamagnetic atoms are repelled by  a magnet

Answer to Problem 26PS

The species which have paramagnetic nature are,

O2-andNO

Explanation of Solution

The list of given molecules and their electronic configuration in MO term and bond order are given below,

MoleculeNo.ofelectronsElectronconfigurationBondorder=12(NbNa)C212(σ1s)2(σ*1s)2(σ2s)2(σ*2s)2( π2p)4842=1O217(σ1s)2(σ*1s)2(σ2s)2(σ*2s)2(σ2p)2( π2p)4( π*2p)3 1072=1.5CN14(σ1s)2(σ*1s)2(σ2s)2(σ*2s)2( π2p)4(σ2p)21042=3O216(σ1s)2(σ*1s)2(σ2s)2(σ*2s)2(σ2p)2( π2p)4( π*2p)2 1062=2CO14(σ1s)2(σ*1s)2(σ2s)2(σ*2s)2( π2p)4(σ2p)21042=3NO15(σ1s)2(σ*1s)2(σ2s)2(σ*2s)2(σ2p)2( π2p)4( π*2p)11052=2.5NO+14(σ1s)2(σ*1s)2(σ2s)2(σ*2s)2( π2p)4(σ2p)21042=3C2214(σ1s)2(σ*1s)2(σ2s)2(σ*2s)2( π2p)4(σ2p)21042=3OF18(σ1s)2(σ*1s)2(σ2s)2(σ*2s)2(σ2p)2( π2p)4( π*2p)41082=1NbNumberofbondingelectrons,NaNumberofantibondingelectrons

Molecules with fractional bond order value will show paramagnetic character. The fractional values are the result of odd number of electrons and odd number of electrons shows the presence of un-paired electrons.

According to the above table it is clear that there are two molecules have fractional bond order; they will exhibit paramagnetic character.

The species which have paramagnetic nature are,

O2-andNO

(c)

Expert Solution
Check Mark
Interpretation Introduction

Interpretation:

All the species which have fractional bond order should be identified from the given list of molecules. 

Concept Introduction:

Molecular orbital (MO) theory:  is a method for determining molecular structure in which electrons are not assigned to individual bonds between atoms, but are treated as moving under the influence of the nuclei in the whole molecule.

Bond order: It is the measure of number of electron pairs shared between two atoms.

Bondorder=12(NumberofelectronsinbondoingMOs-NumberofelectronsinantibondingMOs)

Answer to Problem 26PS

The species which have fractional bond order are,

O2-andNO

Explanation of Solution

The list of given molecules and their electronic configuration in MO term and bond order are given below,

MoleculeNo.ofelectronsElectronconfigurationBondorder=12(NbNa)C212(σ1s)2(σ*1s)2(σ2s)2(σ*2s)2( π2p)4842=1O217(σ1s)2(σ*1s)2(σ2s)2(σ*2s)2(σ2p)2( π2p)4( π*2p)3 1072=1.5CN14(σ1s)2(σ*1s)2(σ2s)2(σ*2s)2( π2p)4(σ2p)21042=3O216(σ1s)2(σ*1s)2(σ2s)2(σ*2s)2(σ2p)2( π2p)4( π*2p)2 1062=2CO14(σ1s)2(σ*1s)2(σ2s)2(σ*2s)2( π2p)4(σ2p)21042=3NO15(σ1s)2(σ*1s)2(σ2s)2(σ*2s)2(σ2p)2( π2p)4( π*2p)11052=2.5NO+14(σ1s)2(σ*1s)2(σ2s)2(σ*2s)2( π2p)4(σ2p)21042=3C2214(σ1s)2(σ*1s)2(σ2s)2(σ*2s)2( π2p)4(σ2p)21042=3OF18(σ1s)2(σ*1s)2(σ2s)2(σ*2s)2(σ2p)2( π2p)4( π*2p)41082=1NbNumberofbondingelectrons,NaNumberofantibondingelectrons

According to the above table it is clear that there are two molecules have a fractional bond order. And they are mentioned below,

O2-andNO

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