91影视

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

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

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

Here, ${\mathit{蔚}}_{\mathbf{0}}$ 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 ${蠄}_{2p_z}鈭�\frac{z}{r}and{蠄}_{2p_x}鈭�\frac{z}{r}$ 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:

$$\begin{gathered} {蠄}_{11}-{蠄}_{2x}鈭�-z \\ {蠄}_{11}-{蠄}_{2x}鈭�\frac{1}{3}z-\frac{\sqrt{8}}{3}x \end{gathered}$$

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

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

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

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

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

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,

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

$鈭�\frac{\sqrt{2}}{3}\sin 胃\cos 蠒-\frac{\sqrt{6}}{3}\sin 胃\sin 蠒$using eq(10-1)

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

Therefore, the lobe III is formed form lobe IV.