Magnetic moment 2.83 BM is given by which of | vedclass

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(B) The magnetic moment is given by the formula $\mu = \sqrt{n(n+2)} \ BM$,where $n$ is the number of unpaired electrons.Given $\mu = 2.83 \ BM$,we ha

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$Ni^{2+}$

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(B) The magnetic moment is given by the formula $\mu = \sqrt{n(n+2)} \ BM$,where $n$ is the number of unpaired electrons.Given $\mu = 2.83 \ BM$,we have:$2.83 = \sqrt{n(n+2)}$$(2.83)^2 = n(n+2)$$8.00 \approx n^2 + 2n$$n^2 + 2n - 8 = 0$$(n+4)(n-2) = 0$Since $n$ cannot be negative,$n = 2$.Now,let us check the number of unpaired electrons for each ion:$1. Ti^{3+} (Z=22): [Ar] 3d^1 \rightarrow n = 1$$2. Ni^{2+} (Z=28): [Ar] 3d^8 \rightarrow n = 2$$3. Cr^{3+} (Z=24): [Ar] 3d^3 \rightarrow n = 3$$4. Mn^{2+} (Z=25): [Ar] 3d^5 \rightarrow n = 5$Since $Ni^{2+}$ has $2$ unpaired electrons,it corresponds to a magnetic moment of $2.83 \ BM$.

Magnetic moment $2.83 \ BM$ is given by which of the following ions?
(At. nos. $Ti = 22, Cr = 24, Mn = 25, Ni = 28$)

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Magnetic moment $2.83 \ BM$ is given by which of the following ions?(At. nos. $Ti = 22, Cr = 24, Mn = 25, Ni = 28$)