Question

By the “vector” technique of example 10.1 , show that the angles between all lobes of the hybrid${\mathbf{sp}}^{\mathbf{3}}$states are $\mathbf{109}\mathbf{.}\mathbf{5}\mathbf{°}\mathbf{.}$.

Short answer

The Angle between all lobes of hybrid ${\mathrm{sp}}^3$states are$109.5°$

Step-by-step solution

Finding Four States that participate in hybrid.

$$\begin{gathered} {\psi }_1={\psi }_{2s}-{\psi }_{2pz} \\ {\psi }_2={\psi }_{2s}+\frac{1}{3}{\psi }_{2pz}-\frac{\sqrt{8}}{3}{\psi }_{2px} \\ {\psi }_3={\psi }_{2s}+\frac{1}{3}{\psi }_{2pz}-\frac{\sqrt{2}}{3}{\psi }_{2px}-\frac{\sqrt{6}}{3}{\psi }_{2px} \\ {\psi }_4={\psi }_3={\psi }_{2s}+\frac{1}{3}{\psi }_{2pz}-\frac{\sqrt{2}}{3}{\psi }_{2px}-\frac{\sqrt{6}}{3}{\psi }_{2px} \end{gathered}$$

Now We will Use Dot product to Find The Angle.

A.B=AB Cos$\mathrm{\theta }$ (Here A and B are vectors).

To Calculate the Dot Product.

Ignoring the 2s state contribution, to calculate the length of each state.

$$\begin{gathered} \left|{\mathrm{\psi }}_1\right|={\left(-1\right)}^2=1 \\ {\left|{\mathrm{\psi }}_1\right|}^2=\left(\frac{1}{3}\right)+{\left(\frac{\sqrt{8}}{3}\right)}^2=1 \\ {\mathrm{\psi }}_{3\mathrm{or}4}{\mathrm{l}}^2={\left(\frac{1}{3}\right)}^2+{\left(\frac{\sqrt{2}}{3}\right)}^2+{\left(\frac{\sqrt{6}}{3}\right)}^2=1 \\ \mathrm{For}\mathrm{Dot}\mathrm{Products}\mathrm{Calculation}.. \\ {\mathrm{\psi }}_{1.}{\mathrm{\psi }}_2=\left(0,0,1\right).(\frac{-\sqrt{8}}{3},0,\frac{1}{3})=\frac{-1}{3} \\ {\mathrm{\psi }}_2.{\mathrm{\psi }}_{3\mathrm{or}4}=\left(0,0,-1\right).(\frac{\sqrt{2}}{3}\pm (\frac{\sqrt{6}}{3},\frac{1}{3})=\frac{-1}{3} \\ {\mathrm{\psi }}_2.{\mathrm{\psi }}_{3\mathrm{or}4}=(\frac{\sqrt{2}}{3}-(\frac{\sqrt{6}}{3},\frac{1}{3}).(\frac{\sqrt{2}}{3},(\frac{\sqrt{6}}{3},\frac{1}{3}) \\ =\frac{2}{9}-\frac{6}{9}+\frac{1}{9} \\ =\frac{-1}{3} \\ \mathrm{In}\mathrm{all}\mathrm{cases}\mathrm{the}\mathrm{Dot}\mathrm{Product}\mathrm{is}\frac{-1}{3}\mathrm{and}\mathrm{the}\mathrm{length}\mathrm{is}\mathrm{always}1. \\ \mathrm{So}, \\ \mathrm{Cos}\theta =\frac{\mathrm{AB}}{\left|\mathrm{AB}\right|} \\ =\frac{-1}{3} \\ \theta ={\cos }^{-1}\left(\frac{-1}{3}\right) \\ =109.5°.. \\ \mathrm{Hence},\mathrm{The}\mathrm{Angle}\mathrm{between}\mathrm{all}\mathrm{lobes}\mathrm{of}\mathrm{hybrid}{\mathrm{sp}}^3\mathrm{states}\mathrm{are}109.5° \end{gathered}$$

For Dot Products Calculation..