Making use of the Hund rules, find the numbers of electrons in the only partially filled subshell of the atom whose basic term is (a) `.^(3)F_(2)` (b) `.^(2)P_(3//2)`, (c ) `.^(6)S_(5//2)`.

Explanation

(a) `.^(3)F_(2)` The maximum value of spin is `S=1` here. This means there are `2` electrons. `L= 3` so `p` electrons are ruled out. This is simplest plssiblility is `d` electrons. This is the correct choice for if we were considering `f` electrons, the maximum value of `L` allowed by Pauli priciple will be `L=5` (maximum value of the magnitude of megnetic quantum number will be `3+2=5`).
Thus the atom has two `d` electrons in the unfilled shell.
(b) `.^(2)P_(3//2)` Here `L=1,s=(1)/(2)` and `J=(3)/(2)`
Since `J=L+S`, Hund's rule implies the shell is more than half full. Thsi means one electron less than a full shell. On the basis of hole picture it is easy to see that we have `p` electrons. Thus the atom has `5p` electrons.
(c )`.^(6)S_(5//2)` Here `S=(5)/(2), L=0` we either have five electrons of five holes. The angular part is antisymmertic. For five `d` electrons, the maximum value of the quantum number cosistent with Pauli exclution principle is `(2+1+0-1-2)=0` so `L=0` for `f` or `g` electrions `L gt 0` whether the shell has five electrohns or five holes. Thus the atom has five `d` electrons.