# Label the orbital with the following quantum numbers?

## 1)n=5 l=0 2)n=4 l=2 3)n=2 l=1 4)n=3 l=0 5)n=6 l=2 6)n=7 l=1

Nov 21, 2015

Here's what I got.

#### Explanation:

If I understand your question correctly, you need to take those pairs of quantum numbers and specify the orbital that matches those values.

The problem with that lies with the fact that specific orbitals are determined by the value of the magnetic quantum number, ${m}_{l}$, for which you have no values given.

So I assume that you have to name all the orbitals that can share each of those pairs of quantum numbers.

Now, the principal quantum number, $n$, tells you the energy level on which a specific electron resides.

The angular momentum quantum number, $l$, tells you the subshell in which that electron resides. The values of $l$ correspond to the following subshells

• $l = 0 \to$ the s-subshell
• $l = 1 \to$ the p-subshell
• $l = 2 \to$ the d-subshell
• $l = 3 \to$ the f-subshell

Now, the number of orbitals each subshell can hold depends on the possible values of ${m}_{l}$.

Without the value of ${m}_{l}$, you cannot say exactly which orbital holds the electron.

For example, the fist pair has $n = 4$ and $l = 0$. These two quantum numbers correspond to an electron located on the fourth energy level, in the 4s-subshell.

In this particular case, ${m}_{l}$ can only take one possible value, ${m}_{l} = 0$. This means that you electron will be located in the $4 s$ orbitals.

I'll show you the next two pairs, so that you can solve the last three pairs as practice.

For $n = 4$ and $l = 2$, the energy level is the same as before. This time, however, the subshell will be different.

More specifically, this electron will reside in the 4d-subshell, since it has $l = 2$. Now, the possible values of ${m}_{l}$ are

• ${m}_{l} = - 2 \to$ this is the $4 {d}_{{x}^{2} - {y}^{2}}$ orbital
• ${m}_{l} = - 1 \to$ this is the $4 {d}_{z}^{2}$ orbital
• ${m}_{l} = 0 \to$ this is the $4 {d}_{x y}$ orbital
• ${m}_{l} = 1 \to$ this is the $4 {d}_{x z}$ orbital
• ${m}_{l} = 2 \to$ this is the $4 {d}_{y z}$ orbital

Therefore, a total of five orbitals can share the quantum nubmers $n = 4$ and $l = 2$.

Finally, for $n = 2$ and $l = 1$, you have the second energy level and the 2p-subshell.

The values of ${m}_{l}$ for this subshell are

• ${m}_{l} = - 1 \to$ this is the $2 {p}_{x}$ orbital*
• ${m}_{l} = 0 \to$ this is the $2 {p}_{y}$ orbital
• ${m}_{l} = 1 \to$ this is the $2 {p}_{z}$ orbital

Therefore, a total of three orbitals can share the quantum numbers $n = 2$ and $l = 1$.