# 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

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

##### 1 Answer

#### Answer:

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*,

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*, **energy level** on which a specific electron resides.

The *angular momentum quantum number*, **subshell** in which that electron resides. The values of

#l=0 -># the s-subshell#l=1 -># the p-subshell#l=2 -># the d-subshell#l=3 -># the f-subshell

Now, the **number of orbitals** each subshell can hold depends on the possible values of

Without the value of **exactly** which orbital holds the electron.

For example, the fist pair has *fourth energy level*, in the **4s-subshell**.

In this particular case,

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

For **subshell** will be different.

More specifically, this electron will reside in the **4d-subshell**, since it has

#m_l = -2 -># this is the#4d_(x^2-y^2)# orbital#m_l = -1 -># this is the#4d_z^2# orbital#m_l = 0 -># this is the#4d_(xy)# orbital#m_l = 1 -># this is the#4d_(xz)# orbital#m_l = 2 -># this is the#4d_(yz)# orbital

Therefore, a total of **five orbitals** can share the quantum nubmers

Finally, for **second energy level** and the **2p-subshell**.

The values of

this is the#m_l = -1 -># orbital*#2p_x# #m_l = 0 -># this is the#2p_y# orbital#m_l = 1 -># this is the#2p_z# orbital

Therefore, a total of **three orbitals** can share the quantum numbers