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The "spin-only" magnetic moment [in units of Bohr magneton, (μB)] of Ni2+ in aqueous solution would be (At. No. Ni = 28)

1. 6
2. 1.73
3. 2.84
4. 4.90

Correct answer: 2.84

Solution

The spin-only magnetic moment for transition metal ions can be calculated using the formula μₛ = √(n(n+2)), where n is the number of unpaired electrons. For Ni2+, which has 2 unpaired electrons, the calculation yields a magnetic moment of approximately 2.84 μB.

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