Find the total spin and spin magnetic moment of following ion.
(i) `Fe^(+3)` (ii) `Cu^(+)`

Answer

(i) `+5//2` or `-5//2`, spin magnetic moment `=sqrt(35)` B.M. (ii) `0, 0`

Explanation

(i) `_(26)Fe^(3+): 1s^(2)2s^(2)2p^(6)3s^(2)3p^(6)3d^(5)`
It contains `5` unpaired electrons `:. n=5`
`:.` Total spin `= +-n/2= +-5/2`
Magnetic moment `=sqrt(n(n+2))=sqrt(5(5+2))=sqrt(35) BM`.
(ii) `_(29)Cu^(+): 1s^(2)2s^(2) 2p^(6)3s^(2)3p^(6)3d^(10)`
It contains `0` unpaired electron
`:.` Total spin =`0`
`:.` spin magnetic Moment `=0`