Chapter 2, Problem 2F.19E

(a)

Interpretation Introduction

Interpretation:

The below two-hybrid orbital are orthogonal has to be determined.

  $\begin{array}{l}{h}_1=\text{s}+{\text{p}}_x+{\text{p}}_y+{\text{p}}_z\\ h_3=\text{s}-{\text{p}}_x+{\text{p}}_y-{\text{p}}_z\end{array}$

Concept Introduction:

Hybrid orbitals are formed by the mixture of atomic orbitals overlapped on each other of different proportions. The four hybrid orbitals of ${\text{sp}}^3$ that are formed by the linear combination of four atomic orbitals are denoted as $h$ and are given as:

  $\begin{array}{l}{h}_1=\text{s}+{\text{p}}_x+{\text{p}}_y+{\text{p}}_z\\ h_2=\text{s}-{\text{p}}_x-{\text{p}}_y+{\text{p}}_z\\ h_3=\text{s}-{\text{p}}_x+{\text{p}}_y-{\text{p}}_z\\ h_4=\text{s}+{\text{p}}_x-{\text{p}}_y-{\text{p}}_z\end{array}$

The two orbitals are orthogonal when there is no interaction between the atomic electrons and another atom.

Explanation of Solution

The two-hybrid orbitals are normalized and for the orbitals to be orthogonal the integral of the wave function should be equal to zero that is the value of I is equal to zero.

  $I={\displaystyle \int {\psi }_1}\cdot {\psi }_3\text{d}\tau$

For the orbitals, $h_1$ and $h_3$ to be orthogonal the value of integral is solved by an expression,

  $I={\displaystyle \int h_1}\cdot h_3\text{d}\tau$        (1)

Substitute $\text{s}+{\text{p}}_x+{\text{p}}_y+{\text{p}}_z$ for $h_1$ and $\text{s}-{\text{p}}_x+{\text{p}}_y-{\text{p}}_z$ for $h_3$ in equation (1)

  $\begin{array}{c}I={\displaystyle \int \left(\text{s}+{\text{p}}_{\text{x}}+{\text{p}}_{\text{y}}+{\text{p}}_{\text{z}}\right)}\cdot \left(\text{s}-{\text{p}}_x+{\text{ p}}_y-{\text{p}}_z\right)\text{d}\tau \\ ={\displaystyle \int \left(\begin{array}{l}{\text{s}}^2-{\text{sp}}_x+{\text{sp}}_y-{\text{sp}}_z+{\text{sp}}_x-{\text{p}}_x^2+{\text{p}}_x{\text{p}}_y-{\text{p}}_x{\text{p}}_z+{\text{sp}}_y-{\text{p}}_x{\text{p}}_y+\\ {\text{p}}_y^2-{\text{p}}_y{\text{p}}_z+{\text{sp}}_z-{\text{p}}_x{\text{p}}_z+{\text{p}}_y{\text{p}}_z-{\text{p}}_z^2\end{array}\right)}\text{d}\tau \end{array}$        (2)

The integral value of product of different orbitals that are hybridized is zero, that is expressed as follows:

  $\begin{array}{l}{\displaystyle \int {\text{sp}}_x}\text{d}\tau =0\\ {\displaystyle \int {\text{sp}}_y}\text{d}\tau =0\\ {\displaystyle \int {\text{sp}}_z}\text{d}\tau =0\\ {\displaystyle \int {\text{p}}_x{\text{p}}_y}\text{d}\tau =0\end{array}$

Further, integral value of product of different orbitals that are hybridized is zero and it is as follows:

  $\begin{array}{l}{\displaystyle \int {\text{p}}_y{\text{p}}_z}\text{d}\tau =0\\ {\displaystyle \int {\text{p}}_x{\text{p}}_z}\text{d}\tau =0\end{array}$

Equation (2) can be modified as follows:

  $I={\displaystyle \int {\text{s}}^2}\text{d}\tau -{\displaystyle \int {\text{p}}_x^2}\text{d}\tau +{\displaystyle \int {\text{p}}_y^2}\text{d}\tau -{\displaystyle \int {\text{p}}_z^2}\text{d}\tau$        (3)

Substitute 1 for ${\displaystyle \int {\text{s}}^2}\text{d}\tau$, 1 for ${\displaystyle \int {\text{p}}_x^2}\text{d}\tau$, 1 for ${\displaystyle \int {\text{p}}_y^2}\text{d}\tau$ and 1 for ${\displaystyle \int {\text{p}}_z^2}\text{d}\tau$ in equation (3).

  $\begin{array}{c}I=1-1+1-1\\ =0\end{array}$

Since the value of I is zero, therefore the two-hybrid orbitals that are $h_1$ and $h_3$ are orthogonal to each other.

(b)

Interpretation Introduction

Interpretation:

The hybrid orbitals those are orthogonal to below hybrid orbitals have to be determined.

  $\begin{array}{l}{h}_1=\text{s}+{\text{p}}_x+{\text{p}}_y+{\text{p}}_z\\ h_3=\text{s}-{\text{p}}_x+{\text{p}}_y-{\text{p}}_z\end{array}$

Concept Introduction:

Refer to part (a).

Explanation of Solution

For the hybrid orbitals $h_1$ and $h_2$.

The other two hybrid orbitals are normalized and orthogonal so the integral of the wave function should be equal to zero that is value of I is equal to zero.

  $I={\displaystyle \int {\psi }_1}\cdot {\psi }_3\text{d}\tau$

For the orbitals, $h_1$ and $h_2$ to be orthogonal the value of integral is solved by an expression as follows:

  $I={\displaystyle \int h_1}\cdot h_2\text{d}\tau$        (4)

Substitute $\text{s}+{\text{p}}_{\text{x}}+{\text{p}}_{\text{y}}+{\text{p}}_{\text{z}}$ for $h_1$ and $\text{s}-{\text{p}}_x-{\text{ p}}_y+{\text{p}}_z$ for $h_2$ in equation (4).

  $\begin{array}{c}I={\displaystyle \int \left(\text{s}+{\text{p}}_{\text{x}}+{\text{p}}_{\text{y}}+{\text{p}}_{\text{z}}\right)}\cdot \left(\text{s}-{\text{p}}_x-{\text{ p}}_y+{\text{p}}_z\right)\text{d}\tau \\ ={\displaystyle \int \left(\begin{array}{l}{\text{s}}^2-{\text{sp}}_x-{\text{sp}}_y+{\text{sp}}_z+{\text{sp}}_x-{\text{p}}_x^2-{\text{p}}_x{\text{p}}_y+{\text{p}}_x{\text{p}}_z+{\text{sp}}_y-{\text{p}}_x{\text{p}}_y-\\ {\text{p}}_y^2+{\text{p}}_y{\text{p}}_z+{\text{sp}}_z-{\text{p}}_x{\text{p}}_z-{\text{p}}_y{\text{p}}_z+{\text{p}}_z^2\end{array}\right)}\text{ d}\tau \end{array}$        (5)

The integral value of product of different orbitals that are hybridized is zero, that is expressed as follows:

  $\begin{array}{l}{\displaystyle \int {\text{sp}}_x}\text{d}\tau =0\\ {\displaystyle \int {\text{sp}}_y}\text{d}\tau =0\\ {\displaystyle \int {\text{sp}}_z}\text{d}\tau =0\\ {\displaystyle \int {\text{p}}_x{\text{p}}_y}\text{d}\tau =0\end{array}$

Further, integral value of product of different orbitals that are hybridized is zero and it is as follows:

  $\begin{array}{l}{\displaystyle \int {\text{p}}_y{\text{p}}_z}\text{d}\tau =0\\ {\displaystyle \int {\text{p}}_x{\text{p}}_z}\text{d}\tau =0\end{array}$

Equation (5) can be modified as follows:

  $I={\displaystyle \int {\text{s}}^2}\text{d}\tau -{\displaystyle \int {\text{p}}_x^2}\text{d}\tau -{\displaystyle \int {\text{p}}_y^2}\text{d}\tau +{\displaystyle \int {\text{p}}_z^2}\text{d}\tau$        (6)

Substitute 1 for ${\displaystyle \int {\text{s}}^2}\text{d}\tau$, 1 for ${\displaystyle \int {\text{p}}_x^2}\text{d}\tau$, 1 for ${\displaystyle \int {\text{p}}_y^2}\text{d}\tau$ and 1 for ${\displaystyle \int {\text{p}}_z^2}\text{d}\tau$. in equation (6).

  $\begin{array}{c}I=1-1-1+1\\ =0\end{array}$

Since the value of I is zero, therefore the two-hybrid orbitals that are $h_1$ and $h_2$ are orthogonal to each other.

For the hybrid orbitals $h_1$ and $h_4$.

The other two hybrid orbitals are normalized and orthogonal so the integral of the wave function should be equal to zero that is value of I is equal to zero.

  $I={\displaystyle \int {\psi }_1}\cdot {\psi }_3\text{d}\tau$

For the orbitals, $h_1$ and $h_4$ to be orthogonal the value of integral is solved by an expression,

  $I={\displaystyle \int h_1}\cdot h_4\text{d}\tau$        (7)

Substitute $\text{s}+{\text{p}}_x+{\text{p}}_y+{\text{p}}_z$ for $h_1$ and $\text{s}+{\text{p}}_x-{\text{p}}_y-{\text{p}}_z$ for $h_4$ in equation (7).

  $\begin{array}{c}I={\displaystyle \int \left(\text{s}+{\text{p}}_{\text{x}}+{\text{p}}_{\text{y}}+{\text{p}}_{\text{z}}\right)}\cdot \left(\text{s}+{\text{p}}_x-{\text{ p}}_y-{\text{p}}_z\right)\text{d}\tau \\ ={\displaystyle \int \left(\begin{array}{l}{\text{s}}^2+{\text{sp}}_x-{\text{sp}}_y-{\text{sp}}_z+{\text{sp}}_x+{\text{p}}_x^2-{\text{p}}_x{\text{p}}_y-{\text{p}}_x{\text{p}}_z+{\text{sp}}_y+{\text{p}}_x{\text{p}}_y-\\ {\text{p}}_y^2-{\text{p}}_y{\text{p}}_z+{\text{sp}}_z+{\text{p}}_x{\text{p}}_z-{\text{p}}_y{\text{p}}_z-{\text{p}}_z^2\end{array}\right)}\text{ d}\tau \end{array}$        (8)

The integral value of product of different orbitals that are hybridized is zero, that is expressed as follows:

  $\begin{array}{l}{\displaystyle \int {\text{sp}}_x}\text{d}\tau =0\\ {\displaystyle \int {\text{sp}}_y}\text{d}\tau =0\\ {\displaystyle \int {\text{sp}}_z}\text{d}\tau =0\\ {\displaystyle \int {\text{p}}_x{\text{p}}_y}\text{d}\tau =0\end{array}$

Further, integral value of product of different orbitals that are hybridized is zero and it is as follows:

  $\begin{array}{l}{\displaystyle \int {\text{p}}_y{\text{p}}_z}\text{d}\tau =0\\ {\displaystyle \int {\text{p}}_x{\text{p}}_z}\text{d}\tau =0\end{array}$

Equation (8) can be modified as follows:

  $I={\displaystyle \int {\text{s}}^2}\text{d}\tau +{\displaystyle \int {\text{p}}_x^2}\text{d}\tau -{\displaystyle \int {\text{p}}_y^2}\text{d}\tau -{\displaystyle \int {\text{p}}_z^2}\text{d}\tau$        (9)

Substitute 1 for ${\displaystyle \int {\text{s}}^2}\text{d}\tau$, 1 for ${\displaystyle \int {\text{p}}_x^2}\text{d}\tau$, 1 for ${\displaystyle \int {\text{p}}_y^2}\text{d}\tau$ and 1 for ${\displaystyle \int {\text{p}}_z^2}\text{d}\tau$ in equation (9).

  $\begin{array}{c}I=1+1-1-1\\ =0\end{array}$

Since, the value of I is zero, therefore the two-hybrid orbitals that are $h_1$ and $h_4$ are orthogonal to each other.