Comprehensive Study Notes

The hydrogenlike wave functions are one-electron spatial wave functions and so are hydrogenlike orbitals (Section 6.5). These functions have been derived for a one-electron atom, and we cannot expect to use them to get a truly accurate representation of the wave function of a many-electron atom. The use of the orbital concept to approximate manyelectron atomic wave functions is discussed in Chapter 11. For now we restrict ourselves to one-electron atoms.

There are two fundamentally different ways of depicting orbitals. One way is to draw graphs of the functions; a second way is to draw contour surfaces of constant probability density.

First consider drawing graphs. To graph the variation of $\psi$ as a function of the three independent variables $r, \theta$, and $\phi$, we need four dimensions. The three-dimensional nature of our world prevents us from drawing such a graph. Instead, we draw graphs of the

FIGURE 6.10 Polar graphs of the $\theta$ factors in the $s$ and $p_{z}$ hydrogen-atom wave functions.

factors in $\psi$. Graphing $R(r)$ versus $r$, we get the curves of Fig. 6.8, which contain no information on the angular variation of $\psi$.

Now consider graphs of $S(\theta)$. We have (Table 5.1)

\(
S{0,0}=1 / \sqrt{2}, \quad S{1,0}=\frac{1}{2} \sqrt{6} \cos \theta
\)

We can graph these functions using two-dimensional Cartesian coordinates, plotting $S$ on the vertical axis and $\theta$ on the horizontal axis. $S{0,0}$ gives a horizontal straight line, and $S{1,0}$ gives a cosine curve. More commonly, $S$ is graphed using plane polar coordinates. The variable $\theta$ is the angle with the positive $z$ axis, and $S(\theta)$ is the distance from the origin to the point on the graph. For $S{0,0}$, we get a circle; for $S{1,0}$ we obtain two tangent circles (Fig. 6.10). The negative sign on the lower circle of the graph of $S{1,0}$ indicates that $S{1,0}$ is negative for $\frac{1}{2} \pi<\theta \leq \pi$. Strictly speaking, in graphing $\cos \theta$ we only get the upper circle, which is traced out twice; to get two tangent circles, we must graph $|\cos \theta|$.

Instead of graphing the angular factors separately, we can draw a single graph that plots $|S(\theta) T(\phi)|$ as a function of $\theta$ and $\phi$. We will use spherical coordinates, and the distance from the origin to a point on the graph will be $|S(\theta) T(\phi)|$. For an $s$ state, $S T$ is independent of the angles, and we get a sphere of radius $1 /(4 \pi)^{1 / 2}$ as the graph. For a $p{z}$ state, $S T=\frac{1}{2}(3 / \pi)^{1 / 2} \cos \theta$, and the graph of $|S T|$ consists of two spheres with centers on the $z$ axis and tangent at the origin (Fig. 6.11). No doubt Fig. 6.11 is familiar. Some texts say this gives the shape of a $p{z}$ orbital, which is wrong. Figure 6.11 is simply a graph of the angular factor in a $p{z}$ wave function. Graphs of the $p{x}$ and $p_{y}$ angular factors give tangent spheres lying on the $x$ and $y$ axes, respectively. If we graph $S^{2} T^{2}$ in spherical coordinates, we get surfaces with the familiar figure-eight cross sections; to repeat, these are graphs, not orbital shapes.

FIGURE 6.11 Graph of $\left|Y{1}^{0}(\theta, \phi)\right|$, the angular factor in a $p{z}$ wave function.

Now consider drawing contour surfaces of constant probability density. We shall draw surfaces in space, on each of which the value of $|\psi|^{2}$, the probability density, is constant. Naturally, if $|\psi|^{2}$ is constant on a given surface, $|\psi|$ is also constant on that surface. The contour surfaces for $|\psi|^{2}$ and for $|\psi|$ are identical.

For an $s$ orbital, $\psi$ depends only on $r$, and a contour surface is a surface of constant $r$, that is, a sphere centered at the origin. To pin down the size of an orbital, we take a contour surface within which the probability of finding the electron is, say, $95 \%$; thus we want $\int_{V}|\psi|^{2} d \tau=0.95$, where $V$ is the volume enclosed by the orbital contour surface.

Let us obtain the cross section of the $2 p_{y}$ hydrogenlike orbital in the $y z$ plane. In this plane, $\phi=\pi / 2$ (Fig. 6.5), and $\sin \phi=1$; hence Table 6.2 gives for this orbital in the $y z$ plane

\(
\begin{equation}
\left|2 p_{y}\right|=k^{5 / 2} \pi^{-1 / 2} r e^{-k r}|\sin \theta| \tag{6.123}
\end{equation}
\)

where $k=Z / 2 a$. To find the orbital cross section, we use plane polar coordinates to plot (6.123) for a fixed value of $\psi ; r$ is the distance from the origin, and $\theta$ is the angle with the $z$ axis. The result for a typical contour (Prob. 6.44) is shown in Fig. 6.12. Since $y e^{-k r}=y \exp \left[-k\left(x^{2}+y^{2}+z^{2}\right)^{1 / 2}\right]$, we see that the $2 p{y}$ orbital is a function of $y$ and $\left(x^{2}+z^{2}\right)$. Hence, on a circle centered on the $y$ axis and parallel to the $x z$ plane, $2 p{y}$ is constant. Thus a three-dimensional contour surface may be developed by rotating the cross section in Fig. 6.12 about the $y$ axis, giving a pair of distorted ellipsoids. The shape of a real $2 p$ orbital is two separated, distorted ellipsoids, and not two tangent spheres.

Now consider the shape of the two complex orbitals $2 p_{ \pm 1}$. We have

\(
\begin{gather}
2 p{ \pm 1}=k^{5 / 2} \pi^{-1 / 2} r e^{-k r} \sin \theta e^{ \pm i \phi} \
\left|2 p{ \pm 1}\right|=k^{5 / 2} \pi^{-1 / 2} e^{-k r} r|\sin \theta| \tag{6.124}
\end{gather}
\)

and these two orbitals have the same shape. Since the right sides of (6.124) and (6.123) are identical, we conclude that Fig. 6.12 also gives the cross section of the $2 p_{ \pm 1}$ orbitals in the $y z$ plane. Since [Eq. (5.51)]

\(
e^{-k r} r|\sin \theta|=\exp \left[-k\left(x^{2}+y^{2}+z^{2}\right)^{1 / 2}\right]\left(x^{2}+y^{2}\right)^{1 / 2}
\)

we see that $2 p_{ \pm 1}$ is a function of $z$ and $x^{2}+y^{2}$; so we get the three-dimensional orbital shape by rotating Fig. 6.12 about the $z$ axis. This gives a doughnut-shaped surface.

Some hydrogenlike orbital surfaces are shown in Fig. 6.13. The $2 s$ orbital has a spherical node, which is not visible; the $3 s$ orbital has two such nodes. The $3 p_{z}$ orbital has a spherical node (indicated by a dashed line) and a nodal plane (the $x y$ plane).

FIGURE 6.12 Contour of a
$2 p_{y}$ orbital.

FIGURE 6.13 Shapes of some hydrogen-atom orbitals.

The $3 d{z^{2}}$ orbital has two nodal cones. The $3 d{x^{2}-y^{2}}$ orbital has two nodal planes. Note that the view shown is not the same for the various orbitals. The relative signs of the wave functions are indicated. The other three real $3 d$ orbitals in Table 6.2 have the same shape as the $3 d{x^{2}-y^{2}}$ orbital but have different orientations. The $3 d{x y}$ orbital has its lobes lying between the $x$ and $y$ axes and is obtained by rotating the $3 d{x^{2}-y^{2}}$ orbital by $45^{\circ}$ about the $z$ axis. The $3 d{y z}$ and $3 d_{x z}$ orbitals have their lobes between the $y$ and $z$ axes and between the $x$ and $z$ axes, respectively. (Online three-dimensional views of the real hydrogenlike orbitals are at www.falstad.com/qmatom; these can be rotated using a mouse.)

FIGURE 6.14 Probability

densities for some hydrogenatom states. [For accurate stereo plots, see D. T. Cromer, J. Chem. Educ., 45, 626 (1968).]

Figure 6.14 represents the probability density in the $y z$ plane for various orbitals. The number of dots in a given region is proportional to the value of $|\psi|^{2}$ in that region. Rotation of these diagrams about the vertical $(z)$ axis gives the three-dimensional probability density. The $2 s$ orbital has a constant for its angular factor and hence has no angular nodes; for this orbital, $n-l-1=1$, indicating one radial node. The sphere on which $\psi_{2 s}=0$ is evident in Fig. 6.14.

Schrödinger's original interpretation of $|\psi|^{2}$ was that the electron is "smeared out" into a charge cloud. If we consider an electron passing from one medium to another, we find that $|\psi|^{2}$ is nonzero in both mediums. According to the charge-cloud interpretation, this would mean that part of the electron was reflected and part transmitted. However, experimentally one never detects a fraction of an electron; electrons behave as indivisible entities. This difficulty is removed by the Born interpretation, according to which the values of $|\psi|^{2}$ in the two mediums give the probabilities for reflection and transmission. The orbital shapes we have drawn give the regions of space in which the total probability of finding the electron is $95 \%$.