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,
n=number of unpaired electrons
`mu=2.84` (given)
`therefore 2.84 = sqrt(n(n+2))`B.M
`(2.84)^(2)=n(n+2)`
`8=n^(2)+2n`
`n^(2)+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 unpaired electrons)
`Ti^(3+) = [Ar]3d^(1)4s^(0)` (one unpaired electrons)
`Cr^(3+) = [Ar]3d^(3)`, (three unpaired electrons)
`Co^(2+)=[Ar],3d^(7), 4s^(0)` (three unpaired electrons)
So, only `Ni^(2+)` has 2 unpiared electrons.