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For $n=2$, \begin{equation} \psi_0=R_{20}(r)/4\pi \quad \psi_1=R_{21}(r)\sqrt{3/4\pi} x/r \quad \psi_2=R_{21}(r)\sqrt{3/4\pi} y/r \quad \psi_3=R_{21}(r)\sqrt{3/4\pi} z/r \end{equation}

When I compute $\langle \psi_0 | \vec{r} |\psi_0\rangle=0 $. I am getting 0 also in $\langle \psi_1 | \vec{r} |\psi_1\rangle=0 $ and $\langle \psi_2 | \vec{r} |\psi_2\rangle=0 $, so I am doing something wrong.

I want to find three orthonormal states such that they have expectation value of $\vec{r}$ in the directions $\hat{x}$, $-\frac{1}{2}\hat{x}+\frac{\sqrt{3}}{2}\hat{y}$ and $-\frac{1}{2}\hat{x}-\frac{\sqrt{3}}{2}\hat{y}$.

Thanks!

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  • $\begingroup$ Did you include the angular distributions while evaluating expectation values? Because $\vec{r}$ will depend on $\theta$, $\phi$. $\endgroup$ – negligible_singularity Nov 13 '15 at 5:50
  • $\begingroup$ I did the integrals in cartesian coordinates. So $\vec{r}=(x,y,z)$ and $r=| \vec{r} |$. $\endgroup$ – Peter Nov 13 '15 at 6:23
  • $\begingroup$ What you are writing above are the $2s$, $2p_x$, $2p_y$ and $2p_z$ orbitals of the hydrogen atom. These have all probability densities for the electron which are symmetrical with respect to the origin of your coordinate system (position of the nucleus), therefore expectation values of $\vec r$ have to be 0. If you know some group theory I could provide you with a simple derivation of the formulas for $sp^2$ hybrid orbitals. $\endgroup$ – LLang Nov 13 '15 at 6:27
  • $\begingroup$ LLang, can you show me the formula of these three orthonormal states such a linear combination of $\psi_i$? $\endgroup$ – Peter Nov 13 '15 at 6:46
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Your orbitals are not $sp^2$ hybridized orbital. They are just real spherical harmonics $s, p_x, p_y, p_z$. Their expectation value of $ \vec{r}$ is zero, because they are symmetrical function.

The three $sp^2$ hybridized orbitals are

$ \phi_0 = \frac{1}{\sqrt 3 }s - \frac{1}{\sqrt 6 }p_x + \frac{1}{\sqrt 2 }p_y$

$\phi_1 =\frac{1}{\sqrt 3 }s - \frac{1}{\sqrt 6 }p_x - \frac{1}{\sqrt 2 }p_y$

$\phi_2 =\frac{1}{\sqrt 3 }s + \frac{2}{\sqrt 6 }p_x$

I think you can get correct result with this orbital.

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  • $\begingroup$ If I want to express a four state $\phi_3$ orthogonal to $\phi_i$ $i=0,1,2.$ such that $<\phi_3 | \vec{r} | \phi_3 >= \hat{z}$, what should be the expression of that state? ($sp^3 hibridisation$). $\endgroup$ – Peter Nov 14 '15 at 19:13
  • $\begingroup$ If I want to express a four state $\phi_3$ orthogonal to $\phi_i$ $i=0,1,2.$ such that $<\phi_3 | \vec{r} | \phi_3 >= \hat{z}$, what should be the expression of that state? ($sp^3$ hibridisation). I think it's gonna be like $\alpha s + \beta p_z$. $\endgroup$ – Peter Nov 14 '15 at 19:47

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