Question

Suppose you take a trip to a distant universe?...

Answer 1

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

Answer 2

Answer:

(a) 3

(b) $1s^{3} 2s^{3}2p^{2}$

(c) 12

Explanation:

• Pauli exclusion principal says no two electrons can have same set of quantum number since for our distant universe three such electrons exist hence the correct answer for this is 3.
• since we have 3 spin quantum number the configuration will be 1s3 2s3 2p2.
• A noble gas has all orbital filled, first noble gas will have configuration 1s3 2s3 2p6 which means that its atomic number is 3+3+6 = 12