Atkins' Physical Chemistry
Atkins' Physical Chemistry
11th Edition
ISBN: 9780198769866
Author: ATKINS, P. W. (peter William), De Paula, Julio, Keeler, JAMES
Publisher: Oxford University Press
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Chapter 9, Problem 9E.1P
Interpretation Introduction

Interpretation:

The Huckel secular equations for π electrons of CO32 ion have to be set up and solved.  The energies have to be expressed in the form of Coulomb integrals αO and αC and resonance integral β.  The delocalization energy of the ion has to be estimated.

Concept introduction:

Delocalization of energy refers to the extra stabilization of the molecule which is due to allowing the electrons to spread over the entire molecule. To calculate the delocalization energy, the energy of the molecule in question is compared with a molecule in which electrons cannot spread over the molecule.

Expert Solution & Answer
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Answer to Problem 9E.1P

The Huckel secular equations for π electrons of CO32 ion are set as shown below.

    ((EαO)(EαC)3β2)=0E2E(αO+αC)+αOαCβ2=0

The energies are expressed in the form of Coulomb integrals αO and αC and resonance integral β as shown below.

    E+=12((αO+αC)+(αOαC)1+12β2(αOαC)2)E=12((αO+αC)(αOαC)1+12β2(αOαC)2)E±(local)=12[(αO+αC)±(αOαC)1+4β2(αOαC)2]

The delocalization energy of CO32 is given below.

    Edelocal=4β2(αOαC)

Explanation of Solution

In CO32 ion 4 atoms are involved in delocalization.

Therefore, the secular determinant for the compound is of the order 4×4.

The secular determinant for CO32 ion is written as shown below.

  |αOEβ00βαOEβ00βαOEβ00βαCE|=0

The above determinant is solved as shown below.

    |αOEβ00βαOEβ00βαOEβ00βαC|=0(αOE)|αOEβ0βαOEβ0βαC|β|ββ00αOEβ0βαC|+0+0=0(αOE)[[(αOE)((αOE)×αC(β))2]β(βαC)]β[[(αOE)αCβ2]β(0)]=0(EαO)2×((EαO)(EαC)3β2)=0

The above equation is solved as shown below.

    ((EαO)(EαC)3β2)=0E2E(αO+αC)+αOαC3β2=0E±=(αO+αC)±(αO+αC)2+12β22E±=12((αO+αC)±(αOαC)1+12β2(αOαC)2)

The value of energy obtained from above equation is given below.

    E+=12((αO+αC)+(αOαC)1+12β2(αOαC)2)E=12((αO+αC)(αOαC)1+12β2(αOαC)2)

The localization energy of π bond is calculated using the expression given below.

  |αOEββαCE|=0

The determinant is solved to calculate the localization energy as shown below.

    |αOEββαCE|=0(αOE)(αCE)β2=0(EαO)(EαC)β2=0E2E(αO+αC)+αOαCβ2=0        (1)

The roots of the quadratic equation in equation (1) are calculated as shown below.

    E±(local)=(αO+αC)±(αO+αC)2(4×1×αOαCβ2)2=(αO+αC)±αO2+αC22αOαC(4×1×αOαCβ2)2=(αO+αC)±(αOαC)2+4β22=12[(αO+αC)±(αOαC)1+4β2(αOαC)2]        (2)

The value of localization energy from equation (2) is given below.

    E+local=12((αO+αC)+(αOαC)1+4β2(αOαC)2)Elocal=12((αO+αC)(αOαC)1+4β2(αOαC)2)

The number of π electrons present in the compound is 2.  These electrons are present in the ground state energy, E+.

Therefore, the formula to calculate the delocalization energy is given below.

    Edelocal=2E+2E+local

Substitute the value of E+ and E+local in above equation as shown below.

    Edelocal=[2[12((αO+αC)+(αOαC)1+12β2(αOαC)2)]2[12((αO+αC)+(αOαC)1+4β2(αOαC)2)]]Edelocal=(αOαC)(1+12β2(αOαC)21+4β2(αOαC)2)

The value of 12β2(αOαC)2, and the value of 4β2(αOαC)2.

Therefore, the approximations are applied to the equation as shown below.

    1+12β2(αOαC)2=1+12β22(αOαC)2

    1+4β2(αOαC)2=1+4β22(αOαC)2

Therefore, the equation for delocalization energy can be written as shown below.

    Edelocal=(αOαC)(1+12β22(αOαC)2(1+4β22(αOαC)2))=(αOαC)(6β2(αOαC)22β2(αOαC)2)=(αOαC)(4β2(αOαC)2)=4β2(αOαC)

Therefore, the delocalization energy of CO32 is calculated below.

    Edelocal=4β2(αOαC)

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Chapter 9 Solutions

Atkins' Physical Chemistry

Ch. 9 - Prob. 9A.3AECh. 9 - Prob. 9A.3BECh. 9 - Prob. 9A.4AECh. 9 - Prob. 9A.4BECh. 9 - Prob. 9A.5AECh. 9 - Prob. 9A.5BECh. 9 - Prob. 9A.6AECh. 9 - Prob. 9A.6BECh. 9 - Prob. 9A.7AECh. 9 - Prob. 9A.7BECh. 9 - Prob. 9A.8AECh. 9 - Prob. 9A.8BECh. 9 - Prob. 9A.1PCh. 9 - Prob. 9A.2PCh. 9 - Prob. 9A.3PCh. 9 - Prob. 9B.2DQCh. 9 - Prob. 9B.3DQCh. 9 - Prob. 9B.1AECh. 9 - Prob. 9B.1BECh. 9 - Prob. 9B.2AECh. 9 - Prob. 9B.2BECh. 9 - Prob. 9B.3AECh. 9 - Prob. 9B.3BECh. 9 - Prob. 9B.4AECh. 9 - Prob. 9B.4BECh. 9 - Prob. 9B.1PCh. 9 - Prob. 9B.2PCh. 9 - Prob. 9B.3PCh. 9 - Prob. 9C.1DQCh. 9 - Prob. 9C.2DQCh. 9 - Prob. 9C.3DQCh. 9 - Prob. 9C.4DQCh. 9 - Prob. 9C.1AECh. 9 - Prob. 9C.1BECh. 9 - Prob. 9C.2AECh. 9 - Prob. 9C.2BECh. 9 - Prob. 9C.3AECh. 9 - Prob. 9C.3BECh. 9 - Prob. 9C.4AECh. 9 - Prob. 9C.4BECh. 9 - Prob. 9C.5AECh. 9 - Prob. 9C.5BECh. 9 - Prob. 9C.6AECh. 9 - Prob. 9C.6BECh. 9 - Prob. 9C.2PCh. 9 - Prob. 9C.4PCh. 9 - Prob. 9D.1DQCh. 9 - Prob. 9D.2DQCh. 9 - Prob. 9D.3DQCh. 9 - Prob. 9D.4DQCh. 9 - Prob. 9D.1AECh. 9 - Prob. 9D.1BECh. 9 - Prob. 9D.2AECh. 9 - Prob. 9D.2BECh. 9 - Prob. 9D.3AECh. 9 - Prob. 9D.3BECh. 9 - Prob. 9D.4AECh. 9 - Prob. 9D.4BECh. 9 - Prob. 9D.5AECh. 9 - Prob. 9D.5BECh. 9 - Prob. 9D.6AECh. 9 - Prob. 9D.6BECh. 9 - Prob. 9D.7AECh. 9 - Prob. 9D.7BECh. 9 - Prob. 9D.1PCh. 9 - Prob. 9E.1DQCh. 9 - Prob. 9E.2DQCh. 9 - Prob. 9E.3DQCh. 9 - Prob. 9E.4DQCh. 9 - Prob. 9E.5DQCh. 9 - Prob. 9E.1AECh. 9 - Prob. 9E.1BECh. 9 - Prob. 9E.2AECh. 9 - Prob. 9E.2BECh. 9 - Prob. 9E.3AECh. 9 - Prob. 9E.3BECh. 9 - Prob. 9E.4AECh. 9 - Prob. 9E.4BECh. 9 - Prob. 9E.6AECh. 9 - Prob. 9E.6BECh. 9 - Prob. 9E.1PCh. 9 - Prob. 9E.2PCh. 9 - Prob. 9E.3PCh. 9 - Prob. 9E.6PCh. 9 - Prob. 9.1IACh. 9 - Prob. 9.2IACh. 9 - Prob. 9.4IA
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