All you really need in order come answer this question is a variation of the Periodic Table the Elements that reflects the blocks

Now, the principal quantum number, #n#, offers you the energy level on which the electron is located. This is indistinguishable to the period in which the aspect is situated in the routine Table.

In her case, #n=5# designates an facet located in period #5#.

Next, the angular momentum quantum number, #l#, speak you the subshell in i beg your pardon the electron resides. The subshells are identical to the blocks that the periodic Table.

You have

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

In her case, #l=2# designates one electron located in the #d# block that the routine Table.

Now, the #d# block includes a total fo #10# groups, i.e. #10# columns of the routine Table. Each group is equivalent to #1# electron. This way that the #d# block, which is identical to the #d# subshell, deserve to hold a full of #10# electrons.

Therefore, a maximum of #10# electrons can share the 2 quantum numbers

#n=5, l=2#

These electron are located on the fifth energy level, in the d subshell, i.e. In among the #5# d orbitals displayed below

As a next note, friend can discover the number of orbitals that have the right to exist in a subshell by dividing the number of groups in a block by #2#

#"no. That orbitals in a subshell" = "no.

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Of teams in the block"/2#

This is the case due to the fact that an orbital have the right to hold a maximum that #2# electrons as stated by the Pauli exemption Principle.