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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). |
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Answer» Solution :(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 ofthe 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. |
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