Question #add1d

1 Answer
Stefan V.
Sep 6, 2017

#9#

Explanation:

The number of orbitals present in an energy level denoted by the principal quantum number #n# is given by

#color(blue)(ul(color(black)("no. of orbitals" = n^2)))#

In your case, you have

#n = 3#

and so

#"no. of orbitals" = 3^2 = 9#

To verify this, use the fact that the angular momentum quantum number, #l#, which denotes the energy subshell in which an electron is located inside an atom, can take the possible values

#l = {0, 1, 2,..., n-1}#

In your case, the third energy level has a total of #3# subshells

#l = {0, 1, 2}#

Now, each subshell can hold a specific number of orbitals as given by the possible values of the magnetic quantum number, #m_l#

#m_l = {-l, -(l-1),..., -1, 0 ,1, ..., (l-1), l}#

This means that you have

  • #l = 0 implies m_l = {0} -> # the #s# subshell contains #1# orbital
  • #l = 1 implies m_l = {-1, 0 ,1} -># the #p# subshell contains #3# orbitals
  • #l = 2 implies m_l = {-2,-1,0,1, 2} -># the #d# subshell contains #5# orbitals

You can thus say that the third energy level contains a total of

#overbrace("1 orbital")^(color(blue)("in the 3s subshell")) + overbrace("3 orbitals")^(color(blue)("in the 3p subshell")) + overbrace("5 orbitals")^(color(blue)("in the 3d subshell")) = overbrace("9 orbitals")^(color(blue)("on the third energy level"))#

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