Question: Referring to equations(10-2), lobe I of the hybrid states combines the spherically symmetric s state with the state that is oriented along thez-axis. and thus sticks out in the direction (see Exercises 28 and 33), If Figure is a true picture, then in a coordinate system rotated counterclockwise about they-axis by the tetrahedral angle, lobe II should become lobe. In the new frame. -values are unaffected. but what had been values in the 2x -plane become values in the -plane. according to$\mathit{x}\mathbf{=}{\mathit{x}}^{\mathbf{\text{'}}}\mathit{c}\mathit{o}\mathit{s}\mathit{\alpha }\mathbf{+}{\mathit{z}}^{\mathbf{\text{'}}}\mathit{s}\mathit{i}\mathit{n}\mathit{\alpha }$ and ${\mathit{z}}^{\mathbf{\text{'}}}\mathit{c}\mathit{o}\mathit{s}\mathit{\alpha }\mathbf{-}{\mathit{x}}^{\mathbf{\text{'}}}\mathit{s}\mathit{i}\mathit{n}\mathit{\alpha }$, where$\mathit{\alpha }\mathbf{=}\mathbf{}\mathbf{109}\mathbf{.}{\mathbf{5}}^{\mathbf{o}}$ is$\mathit{c}\mathit{o}{\mathit{s}}^{\mathbf{-}\mathbf{1}}\left(-\frac{1}{3}\right)$ , or .

(a) Show that lobe II becomes lobe I. Note that since neither the 2s state nor the radial part of the p states is affected by a rotation. only the angular parts given in equations (10-1) need be considered.

(b) Show that if lobe II is instead rotated about thez-axis by simply shifting $\mathit{\phi }\mathbf{}\mathit{b}\mathit{y}\mathbf{}\mathbf{\pm }{\mathbf{120}}^{\mathbf{0}}$ . it becomes lobes III and IV.

The value of x in the $z^{\text{'}}{x}^{\text{'}}$ plane is $x^{\text{'}}\cos \alpha +z^{\text{'}}\sin \alpha$.

The value of z in the $z^{\text{'}}{x}^{\text{'}}$ plane is $z^{\text{'}}\cos \alpha -x^{\text{'}}\sin \alpha$.

The potential energy between the two protons is given by a relation, $\mathit{U}\mathbf{=}\frac{\mathbf{1}}{\mathbf{4}\mathbf{\pi }{\mathbf{\epsilon }}_{\mathbf{0}}}\frac{{\mathbf{e}}^{\mathbf{2}}}{\mathbf{a}}$.

Here, ${\mathit{\epsilon }}_{\mathbf{0}}$ is permittivity of the free-space, e is charge on each proton, and a is the distance between the two protons.

(a)

The expression for the coordinate ${\psi }_{2p_z}\propto \frac{z}{r}and{\psi }_{2p_x}\propto \frac{z}{r}$ as the radius remains the same before and after the y-axis is rotated, the expression when the omit dependence is omitted is given as:

$$\begin{gathered} {\psi }_{11}-{\psi }_{2x}\propto -z \\ {\psi }_{11}-{\psi }_{2x}\propto \frac{1}{3}z-\frac{\sqrt{8}}{3}x \end{gathered}$$

Given that the rotational transformation of z-x plane is represented by

$$\begin{gathered} x=x^{\text{'}}\cos \alpha +z^{\text{'}}\sin \alpha n \\ z=z^{\text{'}}\cos \alpha -x^{\text{'}}\sin \alpha \\ \alpha ={\cos }^{-1}\left(\frac{-1}{3}\right)=109.5^0 \end{gathered}$$

The above expression shows that lobe II in new coordinate has the same expression as lobe I in the old coordinates.

Therefore, the lobe II in new coordinate has the same expression as lobe / in the old coordinates.

(b)

The expression for the coordinate when the 2s states are not considered is given as:

The four identical lobes of Sp³ are given by

$$\begin{gathered} {\Psi }_I={\Psi }_{2s}-{\Psi }_2{P}_z \\ {\Psi }_{II}={\Psi }_{2s}+\frac{1}{3}{\Psi }_2{P}_z-\frac{\sqrt{8}}{3}{\Psi }_2{P}_x \\ {\Psi }_{III}={\Psi }_{2s}+\frac{1}{3}{\Psi }_2{P}_z+\frac{\sqrt{2}}{3}{\Psi }_2{P}_x-\frac{\sqrt{6}}{3}{\Psi }_2{P}_y{\Psi }_{IV} \\ {\Psi }_{III}={\Psi }_{2s}+\frac{1}{3}{\Psi }_2{P}_z+\frac{\sqrt{2}}{3}{\Psi }_2{P}_x+\frac{\sqrt{6}}{3}{\Psi }_2{P}_y \end{gathered}$$

$$\begin{gathered} {\Psi }_{III}\propto \frac{1}{3}\left(\frac{z}{r}\right)-\frac{\sqrt{8}}{3}\left(\frac{x}{r}\right) \\ \propto \frac{z}{3}-\frac{\sqrt{8}x}{3} \end{gathered}$$

Substituting the transformation,

role="math" localid="1660191414868" $$\begin{gathered} z=z^{\text{'}}\cos (109.5^0)-x^{\text{'}}\sin (109.5^0) \\ z=-\frac{z^{\text{'}}}{3}-\frac{\sqrt{8}}{3}{x}^{\text{'}} \\ and \\ x=x^{\text{'}}\cos (109.5^0)-z^{\text{'}}\sin (109.5^0) \\ x=-\frac{x^{\text{'}}}{3}+\frac{\sqrt{8}}{3}{z}^{\text{'}} \end{gathered}$$

Consider,

$$\begin{gathered} {\Psi }_{III}={\Psi }_{2s}+\frac{1}{3}{\Psi }_{P_z}+\frac{\sqrt{2}}{3}{\Psi }_{2_{P_x}}-\frac{\sqrt{6}}{3}{\Psi }_{2P_y} \\ {\Psi }_{III}-{\Psi }_{2s}-\frac{1}{3}{\Psi }_{2_{P_y}}=\frac{\sqrt{2}}{3}{\Psi }_{2_{P_x}}-\frac{\sqrt{6}}{3}{\Psi }_{P_y} \end{gathered}$$

$\propto \frac{\sqrt{2}}{3}\sin \theta \cos \varphi -\frac{\sqrt{6}}{3}\sin \theta \sin \varphi$using eq(10-1)

under transformation ofrole="math" localid="1660191565333" $\phi =\phi -120$ behaves like

Therefore, the lobe III is formed form lobe IV.