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(C) The magnetic moment is calculated using the formula $\mu = \sqrt{n(n+2)} \, BM$,where $n$ is the number of unpaired electrons.$(i) \, V^{4+} (Z=23

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$(iii) > (ii) > (iv) > (i)$

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(C) The magnetic moment is calculated using the formula $\mu = \sqrt{n(n+2)} \, BM$,where $n$ is the number of unpaired electrons.$(i) \, V^{4+} (Z=23): [Ar] 3d^1$,$n=1$,$\mu = \sqrt{1(1+2)} = \sqrt{3} \approx 1.73 \, BM$.$(ii) \, Mn^{4+} (Z=25): [Ar] 3d^3$,$n=3$,$\mu = \sqrt{3(3+2)} = \sqrt{15} \approx 3.87 \, BM$.$(iii) \, Fe^{3+} (Z=26): [Ar] 3d^5$,$n=5$,$\mu = \sqrt{5(5+2)} = \sqrt{35} \approx 5.92 \, BM$.$(iv) \, Ni^{2+} (Z=28): [Ar] 3d^8$,$n=2$,$\mu = \sqrt{2(2+2)} = \sqrt{8} \approx 2.83 \, BM$.Comparing the values: $5.92 (Fe^{3+}) > 3.87 (Mn^{4+}) > 2.83 (Ni^{2+}) > 1.73 (V^{4+})$.Thus,the order is $(iii) > (ii) > (iv) > (i)$.

Arrange the following ions in the increasing order of their magnetic moments: $(i) \, V^{4+}$,$(ii) \, Mn^{4+}$,$(iii) \, Fe^{3+}$,$(iv) \, Ni^{2+}$. $[V = 23, Mn = 25, Fe = 26, Ni = 28]$

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