What are the quantum numbers for chlorine, iron, and tin?

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Apr 15, 2016

Answer:

#Cl -> 3, 1, 0, +-1/2#
#Fe -> 3, 2, -2, +-1/2#
#Sn -> 5, 1, 0, +-1/2#

Explanation:

The four quantum numbers describe the outermost or valence electron of an atom. They are:

#n# is the distance of the orbital from the nucleus (#1, 2, 3, 4...#),
#l# is the shape of the orbital (#0 -> n-1#),
#m_l# is the orientation of the orbital in space (#-l -> +l#), and
#m_s# is the spin of the electrons (#-1/2# or #+1/2#).

Begin each element by writing out the electronic configuration in terms of #s,p,d,f# orbitals. From this you can directly get #n# and find #l# based on its equivalent orbital. From #l# you can figure the range of #m_l#, and then count one up in the sequence for each electron in the outermost subshell. #m_s# is either #-1/2# or #+1/2#.

Chlorine, #[Ne]3s^2 3p^5#, #n=3# straightaway because that is the highest energy level here, and since the outermost electron is in the #p# orbital, this is the same as #l = 1#. #m_l# can be #-1, 0, +1#, but since there are #5# electrons, you count five, so #-1, 0, +1, -1, 0#, which means the last electron has #m_l = 0#. #m_s# can be either #-1/2# or #+1/2#, it doesn't really matter.

For iron, #[Ar]4s^2 3d^6#, then #n = 3#, #l = 2# (because it is the #d# orbital), #m_l# is #-2# (because you count up one the series from #-l# to #+l# six times, for the six electrons in the #d# subshell) and #m_s# is #+-1/2#.

Tin has the electron configuration of #[Kr]5s^2 4d^10 5p^2#, so #n = 5#, #l = 1#, #m_l = 0# and #m_s = +-1/2#

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