How many electrons in an atom can have the n = 5, l = 2 designation?

Dec 24, 2016

$\text{10 electrons}$

Explanation:

All you really need in order to answer this question is a version of the Periodic Table of Elements that shows the blocks

Now, the principal quantum number, $n$, gives you the energy level on which the electron is located. This is equivalent to the period in which the element is located in the Periodic Table.

In your case, $n = 5$ designates an element located in period $5$.

Next, the angular momentum quantum number, $l$, tells you the subshell in which the electron resides. The subshells are equivalent to the blocks of the Periodic Table.

You have

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

In your case, $l = 2$ designates an electron located in the $d$ block of the Periodic Table.

Now, the $d$ block contains a total fo $10$ groups, i.e. $10$ columns of the Periodic Table. Each group is equivalent to $1$ electron. This means that the $d$ block, which is equivalent to the $d$ subshell, can hold a total of $10$ electrons.

Therefore, a maximum of $10$ electrons can share the two quantum numbers

$n = 5 , l = 2$

These electrons are located on the fifth energy level, in the d subshell, i.e. in one of the $5$ d orbitals shown below

As a side note, you can find the number of orbitals that can exist in a subshell by dividing the number of groups in a block by $2$

$\frac{\text{no. of orbitals in a subshell" = "no. of groups in the block}}{2}$

This is the case because an orbital can hold a maximum of $2$ electrons as stated by the Pauli Exclusion Principle.