Question

Suppose you take a trip to a distant universe?...
You find that the periodic table there is derived from an arrangement of quantum numbers different from the one on Earth. The rules in that universe are:
1. principal quantum number n = 1, 2, . . . (as on Earth);
2. angular momentum quantum number ℓ = 0, 1, 2,. . . , n – 1 (as on Earth);
3. magnetic quantum number mℓ = 0, 1, 2, . . . , ℓ (only positive integers up to and including ℓ are allowed);
4. spin quantum number ms = –1, 0, 1 (that is, three allowed values of spin).

(a) Assuming that the Pauli exclusion principle remains valid, what is the maximum number of electrons that can populate a given orbital?

(b) Write the electronic configuration of the element with atomic number 8 in the periodic table. (Formatting: superscript numbers where appropriate but omit parentheses.)

(c) What is the atomic number of the second noble gas?

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