Using the spin-only formula, the magnetic moment of Ni^2+ (Z=28) in Bohr magnetons, to two decimal places, is

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Using the spin-only formula, the magnetic moment of $Ni X^{2 +}$ ( $Z = 28$ ) in Bohr magnetons, to two decimal places, is

Numerical answer type, answer in BM, tolerance 0.02.

Ni X^{2 +}

3 d^{8}

with 2 unpaired electrons.

μ = 2 \times 4 = 8 = 2.83 BM