Chapter 9, Problem 9.16P

(a)

Interpretation Introduction

Interpretation:

The electronic configuration of benzene anion has to be predicted and its pi bond energy has to be calculated.

Concept Introduction:

Huckel method is used to determine the energy of molecular orbitals that contains pi electrons. This method is applied for conjugated pi electron system.

The pi orbitals of benzene in the increasing order of energy is as follows,

  $\begin{array}{l}1\pi <2\pi <3\pi <4\pi \\ \alpha +\beta \end{array}$

Where $2\pi and3\pi$ are doubly degenerate orbitals which can occupy a maximum of 4 electrons.

$1\pi and2\pi$ are bonding orbitals  & $3\pi and4\pi$ are anti-bonding orbitals.

The energy of bonding pi orbital is $\alpha +2\beta$ and the energy of doubly degenerate bonding pi orbital is $\alpha +\beta .$ The energy of anti-bonding pi orbital is $\alpha -2\beta$ and the energy of doubly degenerate anti-bonding pi orbital is $\alpha -\beta .$

Benzene contain six pi electrons. And the electronic configuration of benzene can be written as follow,

  $1{\pi }^2<2{\pi }^4$

So the pi bonding energy of benzene can be calculated as follows,

  $\begin{array}{c}E=2\left(\alpha +2\beta \right)+4\left(\alpha +\beta \right)\\ \\ =6\alpha +6\beta \end{array}$

(b)

Interpretation Introduction

Interpretation:

The electronic configuration of benzene cation has to be predicted and its pi bond energy has to be calculated.

Concept Introduction:

Huckel method is used to determine the energy of molecular orbitals that contains pi electrons. This method is applied for conjugated pi electron system.

The pi orbitals of benzene in the increasing order of energy is as follows,

$\begin{array}{l}1\pi <2\pi <3\pi <4\pi \\ \alpha +\beta \end{array}$

Where $2\pi and3\pi$ are doubly degenerate orbitals which can occupy a maximum of 4 electrons.

$1\pi and2\pi$ are bonding orbitals and $3\pi and4\pi$ are anti-bonding orbitals.

The energy of bonding pi orbital is $\alpha +2\beta$ and the energy of doubly degenerate bonding pi orbital is $\alpha +\beta .$ The energy of anti-bonding pi orbital is $\alpha -2\beta$ and the energy of doubly degenerate anti-bonding pi orbital is $\alpha -\beta .$

Benzene contain six pi electrons and the electronic configuration of benzene can be written as follow,

$1{\pi }^2<2{\pi }^4$

So the pi bonding energy of benzene can be calculated as follows,

$\pi bondingenergyofbenzene,E=2\left(\alpha +2\beta \right)+4\left(\alpha +\beta \right)=6\alpha +6\beta$