Question

The ‘‘spin-only’’ magnetic moment [in units of Bohr magneton, (mB) of Ni2+ in aqueous solution would be (atomic number of Ni = 28)

• A. 2.84
• B. 4.90
• C. 0
• D. 1.73

Answer 1

Answer: (A)

2.84

Answer 2

Valence shell electron configuration of 28Ni2+ is 3d8 4s0

or

So, number of unpaired electrons (n) = 2

\ m = $\sqrt { n ( n + 2 ) } = \sqrt { 2 ( 2 + 2 ) } = \sqrt { 8 } \approx 2.84$