<span>a. the orbital is defined by n,L, mL so (n, L, mL, -1), (n, L,mL, 0) and (n,L,mL, +1) and 3 electrons for any given orbital
b. in (n,L,mL,ms) format the first 12 elements would look like this
(1, 0, 0, +1)
(1, 0, 0, 0)
(1, 0, 0, -1)
(2, 0, 0, +1)
(2, 0, 0, 0)
(2, 0, 0, -1)
(2, 1, 0, +1)
(2, 1, 1, +1)<-----ANSWER
(2, 1, 0, 0)
(2, 1, 1, 0)
(2, 1, 0, -1)
(2, 1, 1, -1)
the idea is we don't pair up electrons until all the mL's have 1 so we wouldn't write
(2, 1, 0, +1)
(2, 1, 0, 0)
(2, 1, 0, -1)
then.
(2, 1, 1, +1)
(2, 1, 1, 0)
(2, 1, 1, -1)
because they would fill
(2, 1, 0, +1)1st
(2, 1, 0, 0)3rd
(2, 1, 0, -1)5th
then.
(2, 1, 1, +1)2nd
(2, 1, 1, 0)4th
(2, 1, 1, -1)6th
to pair (or rather triple up) electrons last
c. ideal gases are when each n level is full...
(1, 0, 0, +1)
(1, 0, 0, 0)
(1, 0, 0, -1)<----- ideal gas 3 electrons so 3 protons and atomic # = 3
(2, 0, 0, +1)
(2, 0, 0, 0)
(2, 0, 0, -1)
(2, 1, 0, +1)
(2, 1, 1, +1)
(2, 1, 0, 0)
(2, 1, 1, 0)
(2, 1, 0, -1)
(2, 1, 1, -1)<----- 2nd ideal gas12 e's so 12 p's and atomic # = 12
continuing on
(3, 0, 0, +1)
(3, 0, 0, 0)
(3, 0, 0, -1)
(3, 1, 0, +1)
(3, 1, 1, +1)
(3, 1, 0, 0)
(3, 1, 1, 0)
(3, 1, 0, -1)
(3, 1, 1, -1)..
(3, 2, 0, +1)
(3, 2, 1, +1)
(3, 2, 2, +1)
(3, 2, 0, 0)
(3, 2, 1, 0)
(3, 2, 2, 0)
(3, 2, 0, -1)
(3, 2, 1, -1)
(3, 2, 2, -1)<--- 3rd nobel gas atomic # = 30
hope it helps
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