1
$\begingroup$

It is taught that the orbital shapes derive from wave functions with different numbers of nodes. For example, the "s" orbital comes from a wave that has one node. But what are the waves modeling? A light wave for example is showing a relationship between an electric field and magnetic field. What relationship does an atomic orbital wave show? In other words, if you were to label the axes, what would the labels be?

Thanks!

$\endgroup$
  • $\begingroup$ I believe the waves are the psi function representing the probability of finding an electron at a certain location, but it's been such a long time since I've had to do p-chem I can't remember any more details than that, and may have gotten something wrong. The psi function uses distance from the nucleus, and two angles as inputs, so it's a 3D polar function. $\endgroup$ – user137 Aug 27 '14 at 21:34
  • $\begingroup$ Thanks for your reply! So in the p-oribtal, I would expect to find most electrons in a literal dumbbell shape around the nucleus? $\endgroup$ – Blakeasd Aug 27 '14 at 21:45
  • $\begingroup$ yes. In fact, that may be 100% of the probability, but I can't really remember right now. $\endgroup$ – user137 Aug 27 '14 at 22:01
7
$\begingroup$

In a one-dimensional system (I choose this to keep things as simple as possible) the state of the system is described by the wave function $\psi = \psi(x, t)$, which depends on the coordinate $x$ of the particle at time $t$. Wave functions in general are complex functions of real variables. The symbol $\psi^{∗}(x, t)$ denotes the complex conjugate of $\psi(x, t)$. The quantity $P(x, t) = \psi^{∗}(x, t) \psi(x, t) \mathrm{d}x = | \psi^{∗}(x, t) |^{2} \mathrm{d}x$ gives the probability that at time $t$ the $x$ coordinate of the particle lies in the small interval $[x, x + \mathrm{d}x]$. The probability of the particle being in the interval $(a, b)$ on the $x$ axis is given by $\int_{a}^{b} | \psi^{∗}(x, t) |^{2} \mathrm{d}x$. The probabilistic interpretation of the wave function was proposed by Max Born. By analogy with the formula: mass = density $\times$ volume, the quantity $\psi^{∗}(x, t) \psi(x, t)$ is called the probability density that a particle at time $t$ has position $x$.

Edit: Atomic orbitals

For atoms you have a system that is spherically symmetric. Thus, it is most conveniently expressed in spherical coordinates instead of cartesian ones. The wave function solving the time-independent Schroedinger equation is then expressed in terms of the radial distance $r$, the polar angle $\theta$ and the azimuthal angle $\varphi$. This wave function $\Psi(r, \theta, \varphi)$ can be split into a radial part $R_{n,\ell}(r)$ and an angular part $Y_{\ell,m} (\theta, \varphi )$, so that $\Psi(r, \theta, \varphi) = R_{n,\ell}(r) Y_{\ell,m} (\theta, \varphi )$, where $n$, $\ell$ and $m$ are the Principal, Azimuthal and Magnetic quantum number, respectively, and the functions $R_{n,\ell}(r)$ are the solutions of the radial Schroedinger equation

\begin{equation} \bigg( \frac{ - \hbar^{2} }{ 2 m_{\mathrm{e}} } \frac{ \mathrm{d}^{2} }{ \mathrm{d} r^{2} } + \frac{ \hbar^{2} }{ 2 m_{\mathrm{e}} } \frac{ \ell (\ell + 1) }{ r^{2} } - \frac{ Z e^{2} }{ 2 m_{\mathrm{e}} r } - E \bigg) r R_{n,\ell}(r) = 0 \end{equation}

with the nuclear charge $Z$ and the mass of an electron $m_{\mathrm{e}}$. Those radial wave functions are the part of the atomic wave functions $\Psi$ that determines their "extent" and their radial nodes, i.e. the nodes that grow with the Principal quantum number $n$: for example a $1\mathrm{s}$ orbital has no radial nodes (it is a simple sphere without any sign-change in the wavefunction) while the $2\mathrm{s}$ orbital has 1 radial node (it is a smaller sphere wrapped in a bigger spherical shell whereby the inner sphere has a different sign than the outer spherical shell such that you get one nodal plane between inner and outer part). The angular wave functions $Y_{\ell,m} (\theta, \varphi )$ are the so called Spherical Harmonics. They determine the angular momentum of the atomic wave function $\Psi$ and its characteristic shape, i.e. $\mathrm{s}$ orbitals are spheres while $\mathrm{p}$ orbitals are dumbbell-shaped (side note: the Spherical Harmonics are complex functions which are rather hard to visualize; the shapes that every chemist is used to comes from the usage of the so called real Spherical Harmonics, which are just linear combinations of the complex-valued Spherical Harmonics).

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.