Suppose you take a trip to a distant universe and 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: principal quantum number =1, (as on earth); angular momentum quantum number ℓ=0,1,−1 (as on earth); magnetic quantum number ℓ=0,1,ℓ (only positive integers up to and including ℓ are allowed); spin quantum number =−1,0,+1 (that is, three allowed values of spin). assuming that the pauli exclusion principle remains valid in the distant universe, what is the maximum number of electrons that can populate a given orbital there? maximum number electrons per orbital: write the electronic configuration of the element with atomic number 8 in the periodic table. superscript numbers where appropriate but omit parentheses. electronic configuration:

           1s³ 2s³ 2p²

Explanation:

1) Pauli's exclusion principle.

Pauli's exclusion principle states that none two electrons of an atom may have the same set of quantum numbers.

Since the real rule (in our normal universe) is that the spin quantum number can only have two values (s = +1/2 or -1/2), that implies that only two electrons can populate a given orbital here.

2) Rules in the distant universe.

        - In the Earth: mℓ = from - ℓ to + ℓ

        - In the distant universe: mℓ = from 0 to ℓ

The implication of this is that there will be only two p orbitals in the distant universe, correponding to ℓ = 0 and ℓ = 1,  instead of three p orbitals as in the Earth.

        - In Earth: s = +1/2 or -1/2 (two possibilities)

        - In the distant universe: ms = -1, 0, - 1

Then in each s or p orbital there will be 3 electrons.

3) Electron configuration in the distant universe

Hence, for the element with atomic number 8, which means that the number of electrons is 8, the configuration is:

Conclusion:

             1s³ 2s³ 2p²

7! = 7×6×5×4×3×2

3! = 3× 2

as 7! is multiple of 3! because 7! contains 3! that is 3× 2 as factor

so 7! /(3! ) = 7× 6× 5× 4

= 42 × 20

= 840

its b) 840✌✌✌

dhruv⭐