Chapter 9, Problem 9E.1P

Interpretation Introduction

Interpretation:

The Huckel secular equations for $\pi$ electrons of ${\text{CO}}_3^{2-}$ ion have to be set up and solved.  The energies have to be expressed in the form of Coulomb integrals ${\alpha }_{\text{O}}$ and ${\alpha }_{\text{C}}$ and resonance integral $\beta$.  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.

Answer to Problem 9E.1P

The Huckel secular equations for $\pi$ electrons of ${\text{CO}}_3^{2-}$ ion are set as shown below.

    $\begin{array}{c}\left(\left(E-{\alpha }_{\text{O}}\right)\left(E-{\alpha }_{\text{C}}\right)-3{\beta }^2\right)=0\\ E^2-E\left({\alpha }_{\text{O}}+{\alpha }_{\text{C}}\right)+{\alpha }_{\text{O}}{\alpha }_{\text{C}}-{\beta }^2=0\end{array}$

The energies are expressed in the form of Coulomb integrals ${\alpha }_{\text{O}}$ and ${\alpha }_{\text{C}}$ and resonance integral $\beta$ as shown below.

    $\begin{array}{c}{E}_{+}=\frac{1}{2}\left(\left({\alpha }_{\text{O}}+{\alpha }_{\text{C}}\right)+\left({\alpha }_{\text{O}}-{\alpha }_{\text{C}}\right)\sqrt{1+\frac{12{\beta }^2}{{\left({\alpha }_{\text{O}}-{\alpha }_{\text{C}}\right)}^2}}\right)\\ E_{-}=\frac{1}{2}\left(\left({\alpha }_{\text{O}}+{\alpha }_{\text{C}}\right)-\left({\alpha }_{\text{O}}-{\alpha }_{\text{C}}\right)\sqrt{1+\frac{12{\beta }^2}{{\left({\alpha }_{\text{O}}-{\alpha }_{\text{C}}\right)}^2}}\right)\\ E_{\pm \left(\text{local}\right)}=\frac{1}{2}\left[\left({\alpha }_{\text{O}}+{\alpha }_{\text{C}}\right)\pm \left({\alpha }_{\text{O}}-{\alpha }_{\text{C}}\right)\sqrt{1+\frac{4{\beta }^2}{{\left({\alpha }_{\text{O}}-{\alpha }_{\text{C}}\right)}^2}}\right]\end{array}$

The delocalization energy of ${\text{CO}}_3^{2-}$ is given below.

    $E_{\text{delocal}}=\frac{4{\beta }^2}{\left({\alpha }_{\text{O}}-{\alpha }_{\text{C}}\right)}$

Explanation of Solution

In ${\text{CO}}_3^{2-}$ ion $4$ atoms are involved in delocalization.

Therefore, the secular determinant for the compound is of the order $4\times 4$.

The secular determinant for ${\text{CO}}_3^{2-}$ ion is written as shown below.

  $\left|\begin{array}{cccc}{\alpha }_{\text{O}}-E& \beta & 0& 0\\ \beta & {\alpha }_{\text{O}}-E& \beta & 0\\ 0& \beta & {\alpha }_{\text{O}}-E& \beta \\ 0& 0& \beta & {\alpha }_{\text{C}}-E\end{array}\right|=0$

The above determinant is solved as shown below.

    $\begin{array}{c}\left|\begin{array}{cccc}{\alpha }_{\text{O}}-E& \beta & 0& 0\\ \beta & {\alpha }_{\text{O}}-E& \beta & 0\\ 0& \beta & {\alpha }_{\text{O}}-E& \beta \\ 0& 0& \beta & {\alpha }_{\text{C}}\end{array}\right|=0\\ \left({\alpha }_{\text{O}}-E\right)\left|\begin{array}{ccc}{\alpha }_{\text{O}}-E& \beta & 0\\ \beta & {\alpha }_{\text{O}}-E& \beta \\ 0& \beta & {\alpha }_{\text{C}}\end{array}\right|-\beta \left|\begin{array}{ccc}\beta & \beta & 0\\ 0& {\alpha }_{\text{O}}-E& \beta \\ 0& \beta & {\alpha }_{\text{C}}\end{array}\right|+0+0=0\\ \left({\alpha }_{\text{O}}-E\right)\left[\begin{array}{l}\left[\left({\alpha }_{\text{O}}-E\right){\left(\left({\alpha }_{\text{O}}-E\right)\times {\alpha }_{\text{C}}-\left(\beta \right)\right)}^2\right]-\\ \beta \left(\beta {\alpha }_{\text{C}}\right)\end{array}\right]-\beta \left[\left[\left({\alpha }_{\text{O}}-E\right){\alpha }_{\text{C}}-{\beta }^2\right]-\beta \left(0\right)\right]=0\\ {\left(E-{\alpha }_{\text{O}}\right)}^2\times \left(\left(E-{\alpha }_{\text{O}}\right)\left(E-{\alpha }_{\text{C}}\right)-3{\beta }^2\right)=0\end{array}$

The above equation is solved as shown below.

    $\begin{array}{c}\left(\left(E-{\alpha }_{\text{O}}\right)\left(E-{\alpha }_{\text{C}}\right)-3{\beta }^2\right)=0\\ E^2-E\left({\alpha }_{\text{O}}+{\alpha }_{\text{C}}\right)+{\alpha }_{\text{O}}{\alpha }_{\text{C}}-3{\beta }^2=0\\ E_{\pm }=\frac{\left({\alpha }_{\text{O}}+{\alpha }_{\text{C}}\right)\pm \sqrt{{\left({\alpha }_{\text{O}}+{\alpha }_{\text{C}}\right)}^2+12{\beta }^2}}{2}\\ E_{\pm }=\frac{1}{2}\left(\left({\alpha }_{\text{O}}+{\alpha }_{\text{C}}\right)\pm \left({\alpha }_{\text{O}}-{\alpha }_{\text{C}}\right)\sqrt{1+\frac{12{\beta }^2}{{\left({\alpha }_{\text{O}}-{\alpha }_{\text{C}}\right)}^2}}\right)\end{array}$

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

    $\begin{array}{l}{E}_{+}=\frac{1}{2}\left(\left({\alpha }_{\text{O}}+{\alpha }_{\text{C}}\right)+\left({\alpha }_{\text{O}}-{\alpha }_{\text{C}}\right)\sqrt{1+\frac{12{\beta }^2}{{\left({\alpha }_{\text{O}}-{\alpha }_{\text{C}}\right)}^2}}\right)\\ E_{-}=\frac{1}{2}\left(\left({\alpha }_{\text{O}}+{\alpha }_{\text{C}}\right)-\left({\alpha }_{\text{O}}-{\alpha }_{\text{C}}\right)\sqrt{1+\frac{12{\beta }^2}{{\left({\alpha }_{\text{O}}-{\alpha }_{\text{C}}\right)}^2}}\right)\end{array}$

The localization energy of $\pi$ bond is calculated using the expression given below.

  $\left|\begin{array}{cc}{\alpha }_{\text{O}}-E& \beta \\ \beta & {\alpha }_{\text{C}}-E\end{array}\right|=0$

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

    $\begin{array}{l}\left|\begin{array}{cc}{\alpha }_{\text{O}}-E& \beta \\ \beta & {\alpha }_{\text{C}}-E\end{array}\right|=0\\ \left({\alpha }_{\text{O}}-E\right)\left({\alpha }_{\text{C}}-E\right)-{\beta }^2=0\\ \left(E-{\alpha }_{\text{O}}\right)\left(E-{\alpha }_{\text{C}}\right)-{\beta }^2=0\\ E^2-E\left({\alpha }_{\text{O}}+{\alpha }_{\text{C}}\right)+{\alpha }_{\text{O}}{\alpha }_{\text{C}}-{\beta }^2=0\end{array}$        (1)

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

    $\begin{array}{c}{E}_{\pm \left(\text{local}\right)}=\frac{\left({\alpha }_{\text{O}}+{\alpha }_{\text{C}}\right)\pm \sqrt{{\left({\alpha }_{\text{O}}+{\alpha }_{\text{C}}\right)}^2-\left(4\times 1\times {\alpha }_{\text{O}}{\alpha }_{\text{C}}-{\beta }^2\right)}}{2}\\ =\frac{\left({\alpha }_{\text{O}}+{\alpha }_{\text{C}}\right)\pm \sqrt{{\alpha }_{\text{O}}{}^2+{\alpha }_{\text{C}}{}^2-2{\alpha }_{\text{O}}{\alpha }_{\text{C}}-\left(4\times 1\times {\alpha }_{\text{O}}{\alpha }_{\text{C}}-{\beta }^2\right)}}{2}\\ =\frac{\left({\alpha }_{\text{O}}+{\alpha }_{\text{C}}\right)\pm \sqrt{{\left({\alpha }_{\text{O}}-{\alpha }_{\text{C}}\right)}^2+4{\beta }^2}}{2}\\ =\frac{1}{2}\left[\left({\alpha }_{\text{O}}+{\alpha }_{\text{C}}\right)\pm \left({\alpha }_{\text{O}}-{\alpha }_{\text{C}}\right)\sqrt{1+\frac{4{\beta }^2}{{\left({\alpha }_{\text{O}}-{\alpha }_{\text{C}}\right)}^2}}\right]\end{array}$        (2)

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

    $\begin{array}{l}{E}_{{+}_{\text{local}}}=\frac{1}{2}\left(\left({\alpha }_{\text{O}}+{\alpha }_{\text{C}}\right)+\left({\alpha }_{\text{O}}-{\alpha }_{\text{C}}\right)\sqrt{1+\frac{4{\beta }^2}{{\left({\alpha }_{\text{O}}-{\alpha }_{\text{C}}\right)}^2}}\right)\\ E_{{-}_{\text{local}}}=\frac{1}{2}\left(\left({\alpha }_{\text{O}}+{\alpha }_{\text{C}}\right)-\left({\alpha }_{\text{O}}-{\alpha }_{\text{C}}\right)\sqrt{1+\frac{4{\beta }^2}{{\left({\alpha }_{\text{O}}-{\alpha }_{\text{C}}\right)}^2}}\right)\end{array}$

The number of $\pi$ 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.

    $E_{\text{delocal}}=2E_{+}-2E_{{+}_{\text{local}}}$

Substitute the value of $E_{+}$ and $E_{{+}_{\text{local}}}$ in above equation as shown below.

    $\begin{array}{l}{E}_{\text{delocal}}=\left[\begin{array}{l}2\left[\frac{1}{2}\left(\left({\alpha }_{\text{O}}+{\alpha }_{\text{C}}\right)+\left({\alpha }_{\text{O}}-{\alpha }_{\text{C}}\right)\sqrt{1+\frac{12{\beta }^2}{{\left({\alpha }_{\text{O}}-{\alpha }_{\text{C}}\right)}^2}}\right)\right]\\ -2\left[\frac{1}{2}\left(\left({\alpha }_{\text{O}}+{\alpha }_{\text{C}}\right)+\left({\alpha }_{\text{O}}-{\alpha }_{\text{C}}\right)\sqrt{1+\frac{4{\beta }^2}{{\left({\alpha }_{\text{O}}-{\alpha }_{\text{C}}\right)}^2}}\right)\right]\end{array}\right]\\ E_{\text{delocal}}=\left({\alpha }_{\text{O}}-{\alpha }_{\text{C}}\right)\left(\sqrt{1+\frac{12{\beta }^2}{{\left({\alpha }_{\text{O}}-{\alpha }_{\text{C}}\right)}^2}}-\sqrt{1+\frac{4{\beta }^2}{{\left({\alpha }_{\text{O}}-{\alpha }_{\text{C}}\right)}^2}}\right)\end{array}$

The value of $12{\beta }^2\gg {\left({\alpha }_{\text{O}}-{\alpha }_{\text{C}}\right)}^2$, and the value of $4{\beta }^2\gg {\left({\alpha }_{\text{O}}-{\alpha }_{\text{C}}\right)}^2$.

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

    $\sqrt{1+\frac{12{\beta }^2}{{\left({\alpha }_{\text{O}}-{\alpha }_{\text{C}}\right)}^2}}=1+\frac{12{\beta }^2}{2{\left({\alpha }_{\text{O}}-{\alpha }_{\text{C}}\right)}^2}$

    $\sqrt{1+\frac{4{\beta }^2}{{\left({\alpha }_{\text{O}}-{\alpha }_{\text{C}}\right)}^2}}=1+\frac{4{\beta }^2}{2{\left({\alpha }_{\text{O}}-{\alpha }_{\text{C}}\right)}^2}$

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

    $\begin{array}{c}{E}_{\text{delocal}}=\left({\alpha }_{\text{O}}-{\alpha }_{\text{C}}\right)\left(1+\frac{12{\beta }^2}{2{\left({\alpha }_{\text{O}}-{\alpha }_{\text{C}}\right)}^2}-\left(1+\frac{4{\beta }^2}{2{\left({\alpha }_{\text{O}}-{\alpha }_{\text{C}}\right)}^2}\right)\right)\\ =\left({\alpha }_{\text{O}}-{\alpha }_{\text{C}}\right)\left(\frac{6{\beta }^2}{{\left({\alpha }_{\text{O}}-{\alpha }_{\text{C}}\right)}^2}-\frac{2{\beta }^2}{{\left({\alpha }_{\text{O}}-{\alpha }_{\text{C}}\right)}^2}\right)\\ =\left({\alpha }_{\text{O}}-{\alpha }_{\text{C}}\right)\left(\frac{4{\beta }^2}{{\left({\alpha }_{\text{O}}-{\alpha }_{\text{C}}\right)}^2}\right)\\ =\frac{4{\beta }^2}{\left({\alpha }_{\text{O}}-{\alpha }_{\text{C}}\right)}\end{array}$

Therefore, the delocalization energy of ${\text{CO}}_3^{2-}$ is calculated below.

    $E_{\text{delocal}}=\frac{4{\beta }^2}{\left({\alpha }_{\text{O}}-{\alpha }_{\text{C}}\right)}$