The angular momentum quantum number (ℓ) tells the number of angular nodes in an orbital.
Let’s pull that apart and try to make sense out of it.
Let’s start by imagining an s orbital. This is a simple sphere. We can think of this like the sound wave around a violinist that was discussed here. An s orbital looks something like this:
For the sake of this discussion, we need to impose a set of axes on this picture.
Nodes, you recall, are places where the wave has no measurable value. An angular node is a node that can be defined by an angle. These are often planes. For instance, a node could be defined as all points that are 90o away from the z axis. That would be the xy plane.
To see what a node on the xy plane would do to the orbital, imagine placing a band around a balloon and tightening it until it is impossibly small (like the belt that a wasp would wear). This would break the orbital into 2 parts creating a p orbital.
A second angular node would cut the orbital into 4 pieces and would produce a d orbital.
So, when ℓ=0, there are no angular nodes and the orbital is an s orbital. When ℓ=1, there is one angular node and the orbital is a p orbital. When ℓ=2, there are 2 angular nodes and the orbital is a d orbital. Likewise, ℓ=3 produces an f orbital.
There is a limitation on the value of ℓ, that is the value of n. Remember, that n tells us the number of nodes present in an orbital and that at the lowest energy level (n=1) there is no node at all. In fact we said that the number of nodes was equal to (n-1). Since ℓ is the number of angular nodes, it can never be more than the number of nodes, therefore ℓ can be any whole number from 0 up to (n-1). So if n=3, ℓ could be 0, or 1, or 2. That means that on level 3, an orbital can be an s orbital, a p orbital, or a d orbital.
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