As we have learnt in a previous post, a wave function is used to describe the probability of finding an electron at a particular location in the atom. It uses three quantum numbers, i.e. principal quantum number *n*, angular quantum number *l*, magnetic quantum number *m*, to define an orbital which is a possible region of electrons showing up.

So how does an orbital look like? Let's take a look.

*s* orbital

Let's start with the simplest orbital, the *s* orbital. In each energy level, there is only one *s* orbital because when *l* is zero, *m* can only be zero too. Let's take a hydrogen atom, H, as an example as it has only one electron occupying the *1s* orbital. So when we examine the hydrogen atom under a microscope, we may get an image of the *1s* orbital shown by the electron (see the photo below).

We could simplify it using the illustration below.

So what are these dots in the electron cloud? Well, those are the footprints of the electron in the hydrogen atom. As the electron is showing a wave characteristic, it would appear briefly before disappearing, and then appear at another location again. Each appearance of the electron would leave a dot in the image above. By staggering all the dots in one image, we could get a cloud of electrons, although only one electron is present in a hydrogen atom.

You may not see a clear boundary outside of which there is no electron at all. In fact, electrons may show up at any location despite of an extremely low probability. Hence, conventionally we would artificially draw a boundary to enclose 90% of the dots, and we define that as the boundary of the orbital.

So the electron cloud could be replaced by the illustration below to show the *s* orbital, which is a **spherical shape**.

### Radial probability distribution

Let's take another look at the *1s* orbital from a two-dimensional perspect. One thing to remind you is that an orbital is not a single line or trajectory that electrons would travel along. It is indeed a region in space where electrons may appear. Therefore, we'll plot the probability of finding electrons against the radius according to the wave functions, shown in the illustration below.

The probability of finding an electron at atomic radius zero (\( r = 0 \)) is also zero because that's where the nucleus is. There will be no electron in the nucleus.

As we moving outwards from the nucleus, you could find electrons with a high probability, which soon reaches the peak indicating that electrons in this particular orbital would most likely appear at this particular radius. It is still possible to find electrons farther away from the nucleus, but with a very low probability.

The radial probability distribution clearly shows that an orbital is a region in space as electrons could show up at various locations. This distribution is calculated using the wave function, which tells us the relationship between the probability and the radius.

As the quantum numbers increase their values (e.g. for a *3d* orbital), the radial probability distribution curve may show a different shape, meaning for that particular wave function, the radius may lead to a different probability than in the case of a *1s* orbital. So don't be surprised to see a distribution curve of another shape.

*p* orbital

For *p* orbitals, \( l = 1 \) and *m* could be assigned values of \( -1, 0, +1 \). Therefore, there are in total 3 *p* orbitals. This could be reflected in the spacial arrangements of the orbitals.

As shown above, there are 3 *p* orbitals indicated by different colors. The orange, yellow, and green **dumbbell** shaped *p* orbitals are named as \( p_x \), \( p_y \), and \( p_z \) orbitals respectively, because they are positioned along the respective axis.

Please take note that each *p* orbital has *two lobes* on the same axis. Hence there're only 3 orbitals instead of 6 in the illustration above.

### Nodes

While a *1s* orbital is relatively simple as a sphere, *p* orbitals and *s* orbitals in higher energy levels are a little more complicated as they have one or more nodes. A node is a region of zero electron probability.

There are two kinds of nodes, *angular nodes* and *radial nodes*. The number of angular nodes for a specific orbital is determined by its angular quantum number *l*.

For example, a *p* orbital has an angular quantum number *1*, therefore, it would have *1 angular node* as well. This is observed in the illustration above, where there is no electron between the two lobes of the *p* orbitals. That is an *angular node*.

Orbitals may have radial nodes as well. This may not be shown easily. We'll use the radial probability distribution curve of a *2s* orbital to illustrate this concept.

The green line in the illustration above shows the radial probability distribution of a *2s* orbital. Again, as we mentioned earlier, the distribution curve is calculated based on the wave functions with the quantum numbers corresponding to individual orbitals. Therefore, a *2s* orbital will have a different distribution curve as compared to a *1s* orbital.

You may have noticed in the distribution curve of the *2s* orbital that there're two peaks with a minima in between. That minima tells us there is a zero probability of finding electrons at that particular radial, hence a *radial node*.

Although we still say a *2s* orbital has a spherical shape for simplicity, it is more like a Ferrero chocolate whereby it has multiple layers with some gaps between the layers.

Do take note that a *2s* orbital does not have an angular node because it has an *l* value of zero. On the other hand, a *2p* orbital does not have a radial node but it does have an angular node (\( l = 1 \)). The rule is that:

- Total number of angular and radial nodes is \( n - 1 \) whereby
*n*is the principal quantum number - The number of angular nodes is the same as the angular quantum number
*l* - Hence the number of radial nodes is calculated as \( n - 1 - l \)

*d* orbital

The shape of *d* orbitals is more complex than that of *p* orbitals. They are **double dumbbell** shaped with four lobes as illustrated below.

As *d* orbitals, the angular quantum number *l* being *2*, they would show *2 angular nodes* in each *d* orbital.

For example, in the \( d_{xy} \) orbital, there're in fact *two nodal planes*, i.e. the *x-z* and *y-z* planes. On these two nodal planes, there is absolutely zero probability of electrons. Same for \( d_{xz} \) and \( d_{yz} \) orbitals. These three orbitals have their lobes sitting between the axes.

The \( d_{x^2-y^2} \) orbital is a little different. The four lobes are *on* the *x* and *y* axes. It still has two nodal planes though.

A very unique shape is observed with the \( d_{z^2} \) orbital. There are two lobes sitting on the *z* axis while a ring on the *x-y* plane is surrounding them.

Please take note that all the four lobes are belonging to *one single orbital* instead of making up four orbitals. Similarly, the dumbbell and the ring from the \( d_{z^2} \) orbital are not two different orbitals either.

## Orbital penetration

We won't show you the *f* orbitals as they are more relavant to professional chemists or researchers. Most of us won't go into that level of details in chemistry. But there is one thing that we do want to share with you. Although you probably won't be asked or tested on that, it's good for you to know the basics as it would answer one question that you might have when you read our previous posts.

That question is: *why does a 4s orbital have a lower energy than a 3d orbital*?

The answer is **orbital penetration**. Let's take a look at the radial probability distribution of *2s* and *2p* orbitals.

As you may see, the highest peak of the *2s* orbital is at a larger radius than that of the *2p* orbital. But when we write electron configurations, *2s* is filled up with electrons before *2p*. Why is that?

The reason is the small peak of the *2s* orbital that is seen before the large peak of the *2p* orbital.

Although the highest probability of finding electrons in a *2s* orbital is at a larger radius, there is a small but high enough probability of finding electrons in the *2s* orbital at a radius much closer to the nucleus. This is called orbital penetration, the ability of an electron to penetrate through other orbitals to approach the nucleus.

This is the same as what we observe for a *4s* orbital. Despite of being an orbital in energy level 4 and the fact that the highest probability of finding electrons occurs at a large radius, a *4s* orbital shows a small but high enough peak at a radius much closer to the nucleus as compared to the *3d* orbital. This allows electrons in the *4s* orbital to stay closer to the nucleus for a significant amount of time.

So here's the answer to our question. We fill up the *4s* orbital before the *3d* orbital because *electrons in the 4s orbital could stay closer to the nucleus with a lower energy as compared to those in the 3d orbital*.

Nevertheless, in practical, you could just follow the periodic table to write the electron configuration (still remember? or read again). Next, we'll talk more about the periodic table. Let's go.