Question

The ‘spin-only’ magnetic moment [in units of Bohr magneton (mb)] of Ni2+ in aqueous solution would be (Atomic number : Ni = 28)

• A. 2.84
• B. 4.90
• C. 0
• D. 1.73

Answer 1

Answer: (A)

2.84

Answer 2

8Ni $\rightarrow$ [Ar]3d8 4s2

Number of unpaired electrons (n) = 2

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