4.50 mg/Kg 4.5 mg/kg. Our universe – Rows 1 and 2 N=1 L=0 m L =0 m s =1/2 N=1 L=0 m L =0 m s =-1/2 N=2 L=0 m L =0 m s =1/2 N=2 L=0 m L =0 m s =-1/2 N=2.

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Presentation transcript:

4.50 mg/Kg 4.5 mg/kg

Our universe – Rows 1 and 2 N=1 L=0 m L =0 m s =1/2 N=1 L=0 m L =0 m s =-1/2 N=2 L=0 m L =0 m s =1/2 N=2 L=0 m L =0 m s =-1/2 N=2 L=1 m L =-1 m s =1/2 N=2 L=1 m L =-1 m s =-1/2 N=2 L=1 m L =0 m s =1/2 N=2 L=1 m L =0 m s =-1/2 N=2 L=1 m L =1 m s =1/2 N=2 L=1 m L =1 m s =-1/2

There are only 3 quantum numbers. Electrons reside in principal shells labeled n=1, 2, 3, 4… Within each principal shell there are orbitals, call them “λ”, where the allowed λ values are 0, 1, 2….n+1 (You can name these orbitals by the same nomenclature as this universe: λ=0 is an s-orbital, λ=1 is a p-orbital, λ=2 is a d-orbital, etc. ) And the spin quantum numbers are m s =-1/2, 0, +1/2 Assume the Pauli exclusion principal still applies and that the orbitals fill from lowest energy to highest energy. The energy of the shells increases with n, and the energy of the orbitals increases as λ increases. All of the orbitals of a given shell are lower in energy than those of the next higher shell.

How many elements are in the first row of the Periodic Table in Joe’s Universe? How many elements are in the second row of the Periodic Table in Joe’s Universe? What is the electron configuration of calcium (Ca) in Joe’s Universe?

N=1 λ=0 m s =-1/2, 0, +1/2 λ=1 m s =-1/2, 0, +1/2 λ=2 m s =-1/2, 0, +1/2 N=2 λ=0 m s =-1/2, 0, +1/2 λ=1 m s =-1/2, 0, +1/2 λ=2 m s =-1/2, 0, +1/2 λ=3 m s =-1/2, 0, +1/2

n=…λ=…m x =… 10-1/ ½ ½ ½ ½ ½ ½ ½

How many elements are in the first row of the Periodic Table in Joe’s Universe? 9 sets of numbers = 9 electrons How many elements are in the second row of the Periodic Table in Joe’s Universe? 12 sets of numbers = 12 electrons What is the electron configuration of calcium (Ca) in Joe’s Universe? 1s 3 1p 3 1d 3 2s 3 2p 3 2d 3 2f 2