Is there a formula for the number of elements in any given period of the periodic table?

Question

Cameron Andrews · Accepted Answer

Yes, but the formulas are complicated.

I can't claim credit for these formulas. I found the following formula here: #{("If"color(white)(l)ncolor(white)(l)"is even, "f(n) = ((n+2)^2)/2), ("If"color(white)(l)ncolor(white)(l)"is odd," color(white)(ml)f(n) = ((n+1)^2)/2"):}# where #ncolor(white)(ml) =# the number of the Period and #f(n) =# the number of electrons in that Period A Socratic mathematician combined the two equations into a single formula: #f(n) = ((2n + 3 + ("-1")^n)^2)/8# Check them out! They both give the series #2, 8, 8, 18, 18, 32, 32#.

Henry Antrim · Answer

Here is a "pedestrian" way to do it, starting from knowledge of quantum numbers.

One way to do it is to simply note the number of electronic states in all the subshells in that period.

Hence, for each energy level #n#, there are

#[2(0) + 1] + [2(1) + 1] + . . . + [2l_(max) + 1]#

orbitals, i.e.

#"Max orbitals per energy level"#

#= [2(0) + 1] + [2(1) + 1] + . . . + [2l_(max) + 1]#

#= sum_(l=0)^(l_(max)) [2l + 1]#

The absolute maximum #l# is #n - 1# (which makes sense since we sum from #0# to #n-1#, giving #n# subshells), but we don't always get there.

Unfortunately, the value of #n# is down by #1# for the #d# orbitals, down by #2# for the #f# orbitals, etc., so it gets a little complicated.

(So the value of #l_(max)# varies. For example, we don't get to #d# orbitals until the fourth period, even though #n = 3#.)

As a result, we should add the following qualifications:

#vdots#

Knowing that, the maximum number of electrons in that period, or the number of elements in that period, would be:

#barul|stackrel(" ")(" ""Number of Elements in Period" = 2sum_(l=0)^(l_(max)) [2l + 1]" ")|#

For example, in period #4#, #l_(max) = 2#, so

#color(blue)("Number of Elements in Period 4")#

#= 2sum_(l=0)^(2) [2l + 1]#

#= 2{[2(0) + 1] + [2(1) + 1] + [2(2) + 1]}#

#= 2(1 + 3 + 5)#

#= color(blue)(18)#

Or, in period #6#, #l_(max) = 3#, so:

#color(blue)("Number of Elements in Period 6")#

#= 2sum_(l=0)^(3) [2l + 1]#

#= 2{[2(0) + 1] + [2(1) + 1] + [2(2) + 1] + [2(3) + 1]}#

#= 2(1 + 3 + 5 + 7)#

#= color(blue)(32)#