The "spin-only" magnetic moment [in units of Bohr magneton, $(\mu_{\mathrm{B}})$] of $\mathrm{Ni}^{2+}$ in aqueous solution would be (Atomic number of Ni = 28)

JEE Main · 2006 · 30 Apr 2006 · Offline · d and f Block Elements

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

(μ_{B})

] of

Ni^{2 +}

in aqueous solution would be (Atomic number of Ni = 28)

Answer: (A)

Nickel, atomic number 28, gives

N i^{2 +}

the configuration

[Ar] 3 d^{8}

. With water acting as a weak-field ligand the

d^{8}

ion keeps two unpaired electrons. The spin-only moment is

μ = n (n + 2)

Bohr magnetons, and with

n = 2

this is

2 \times 4 = 8 \approx 2.84 μ_{B}

JEE Main · 2006 · 30 Apr 2006 · Offline · Single correct option