Question

The "spin-only" magnetic moment [in units of Bohr
magneton, (μB)] of Ni²⁺ in aqueous solution would be
(At. No. Ni = 28)
(a) 6 (b) 1.73 (c) 2.84 (d) 4.90

Answer 1

The "spin-only" magnetic moment [in units of Bohr

magneton, (μB)] of Ni²⁺ in aqueous solution would be

(At. No. Ni = 28)

(a) 6 (b) 1.73 (c) 2.84 (d) 4.90

HOpe iT HelP YOu DEaR

Answer 2

The "spin-only" magnetic moment [in units of Bohr magneton, (μB)] of Ni²⁺ in aqueous solution would be :

• Atomic mass of Ni = 28

• Electronic configuration of Ni in ground state

= [Ar] 3d^8 4s^2

• Electronic configuration of Ni²⁺ = [Ar] 3d^8

• Hence, no. of unpaired electrons = 2

• Spin only magnetic moment is calculated by formula,

$μB = \sqrt{n \times (n + 2)}$

Where, n = no. of unpaired electrons

• For Ni²⁺,

μB = √2×(2+2) = √2×4 = 2√2 = 2.84 BM

• Correct option : (c)