How many elements would be in the IInd period of the Periodic Table if the spin quanturm numbers could have the value +1/2,0,−1/2?
Solution
For IInd period
n=2 hence
| I | m | s |
|---|---|---|
| 0 | 0 | +12,0,−12 |
| 1 | −1 | +12,0,−12 |
| 0 | +12,0,−12 | |
| +1 | +12,0,−12 |
In the modern periodic table, the second period has a principal quantum number is 2
And there are 2s and 2p orbitals are in 2nd period which is going to be filled.
As given that there are 3 spin quantum numbers(-1/2, 0, +1/2) for each suborbital hence 3 electrons will be there in each suborbital.
The total number of sub-orbitals of s and p (px, py, pz) are 1 and 3 respectively
Maximum electrons in orbital= spin quantum no.(s) X azimuthal quantum no.(l)
As given s= 3 and for s orbital l = 1
Maximum electrons in s orbital = 3 X 1 = 3
For p orbital, I = 3 and as given s= 3
Maximum electrons in p orbital = 3 X 3 = 9
Thus there will be 3 + 9 = 12 elements in the 2nd period of the periodic table.
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