The magnetic moment of 2.84 , B.M. is exhibit | vedclass

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(C) The magnetic moment $\mu$ is given by the formula $\mu = \sqrt{n(n+2)} \, B.M.$,where $n$ is the number of unpaired electrons.Given $\mu = 2.84 \,

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$Ni^{2+}$

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(C) The magnetic moment $\mu$ is given by the formula $\mu = \sqrt{n(n+2)} \, B.M.$,where $n$ is the number of unpaired electrons.Given $\mu = 2.84 \, B.M.$,we have $\sqrt{n(n+2)} = 2.84$,which implies $n \approx 2$.Electronic configurations:$Cr^{2+} (Z=24): [Ar] 3d^4$ $(n=4)$$Co^{2+} (Z=27): [Ar] 3d^7$ $(n=3)$$Ni^{2+} (Z=28): [Ar] 3d^8$ $(n=2)$$Ti^{3+} (Z=22): [Ar] 3d^1$ $(n=1)$Since $Ni^{2+}$ has $2$ unpaired electrons,it corresponds to the magnetic moment of $2.84 \, B.M.$

The magnetic moment of $2.84 \, B.M.$ is exhibited by which of the following ions? (Atomic numbers: $Ni = 28, Ti = 22, Cr = 24, Co = 27$)

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