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The spin-only magnetic moment [in units of Bohr magneton, `(mu_(B)` of `Ni^(2+))` in aqueous solution would be (atomic number of `Ni=28)`
A. `2.84`
B. `4.9`
C. 0
D. `1.73`

1 Answer

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Best answer
Correct Answer - A
`mu_(B) = sqrt(n(n+2))Ni^(+2) = 3d^(8)`
Number of unpaired electrons = 2

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