How is n=3 holding 18 e-?
- Thread starter prsndwg
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a simple formula to find the maximum number of electrons with a given principal quantum number, n, is 2n^2.
in your case, n=3, so 2n^2 is 18 e-.
if you write out the values for the other quantum numbers for n=3, you get:
l= 0, 1, 2 (these correspond to the s, p, d subshells, respectively)
ml= 0
-1, 0, 1
-2, -1, 0, 1, 2
these ml values represent the number of orbitals you have, and there are two electrons that can occupy each orbital. so you have 9 total orbitals times 2 electrons each, and you have 18 total.
in your case, n=3, so 2n^2 is 18 e-.
if you write out the values for the other quantum numbers for n=3, you get:
l= 0, 1, 2 (these correspond to the s, p, d subshells, respectively)
ml= 0
-1, 0, 1
-2, -1, 0, 1, 2
these ml values represent the number of orbitals you have, and there are two electrons that can occupy each orbital. so you have 9 total orbitals times 2 electrons each, and you have 18 total.
Another way to visualize it.
http://hyperphysics.phy-astr.gsu.edu/HBASE/pertab/imgper/pertabord.gif
It is good idea to have this on formula sheet on the test.
http://hyperphysics.phy-astr.gsu.edu/HBASE/pertab/imgper/pertabord.gif
It is good idea to have this on formula sheet on the test.
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- Pre-Dental
You're reasoning would have been right, but you forgot that the d subshell has an L value of 2, and L values exist from 0 to n-1 (n being 3 in this case). Another way to do it is draw that "step" diagram with all the orbitals and arrows to illustrate Aufbau's Principle.
