Magnetic moments `2.84 B.M` is given by :
(At. nos. ni = 28, Ti = 22, Cr = 24, Co = 27).

Options

A. `Ni^(2+)`
B. `Ti^(3+)`
C. `Cr^(3+)`
D. `Co^(2+)`

Answer

A

Explanation

Magnetic moment `mu = sqrt( n(n+2)) BM` where
`therefore 2.84= sqrt(n(n+2)) BM`
`(2.84)^(2) = n(n+2)`
`8=n^(2)+2n`
`n^(2)+2n -8=0`
`n^(2)+4n-2n-8=0`
`n^(2) +4n-2n-8=0`
`n(n+4)-2(n+4)=0`
`n=2`
` Ni^(2+) =[Ar] 3d^(8)4s^(0)`(two uppaired electrons )
` Ti^(3+)=[Ar] 3d^(1) 4s^(0) ` (one unpaired electron )
` Cr^(3+) = [Ar] 3d^(3)` (three unpaired electrons )
` Co^(2+) =[Ar] 3d^(7)4s^(0)` ( three unpaired electrons )
SO , only `Ni ^(2+)` has 2 unpaired elecrons