IF THERE ARE THREE POSSIBLE VALUES FOR SPIN QUANTUM NUMBER, THEN POSSIBLE NUMBER OF ELEMENTS IN THE 5TH PERIOD OF PERIODIC TABLE WOULD BE- HOW??

how can there be three possible values of spin quantum number...........as there are only two possible values that is clcokwise or anticlockwise.

IF there are three possible values for spin then each orbital will be able to accommodate 3electrons. The total number of orbital in a shell with principle quantum number N is n^2 . Each orbital can hold 3electrons each(normally 2only). So the total no. of electrons in a shell will be 3n^2 . The fifth shell thus can hold 75electrons totally. Now coming to the fifth period of periodic table, the orbitals being filled there are 5s4d5p. As each orbital can hold 3e each(in this case) , a total of 1*3+3*3+5*3 electrons will be filled. Adding of one electron makes one new element. Hence 27 elements will be there in 5th period.

Originally , 

Max orbitals of n = n² and this max electrons = 2n²

However since there are 3 values of 3 each orbital must carry 3 electrons 

This max electrons for n now = 3n²

No. Of electrons max in 5th shell originally = 2n² = 2*5*5 = 50 electrons

Electrons in 5th period now = 3n² = 3*5*5= 75 

No of elements before  in 5th period = 18 

Let no of elements in 5th period now be x 

18:50::x:75 

X = 27 

This there are 27 elements in 5th period now