Question

The spin-only magnetic moment [in units of Bohr magneton, `(mu_(B)` of `Ni^(2+)` in aqueous solution would be (atomic number of `Ni=28)`
A. `2.84`
B. `4.90`
C. `0`
D. `1.73`

Answer 1

Correct Answer - 1
Valence shell electron configuration of `._28Ni^(2+)` is `3d^(8) 4s^(0)`, or

So, number of unpaired electron(n)=2
therefore `mu=sqrtn(n+2)=sqrt2(2+2)=sqrt8approx2.84`