Which one of the following has a magnetic mom | vedclass

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(D) The magnetic moment $(\mu)$ is calculated using the formula: $\mu = \sqrt{n(n+2)} \ BM$,where $n$ is the number of unpaired electrons.For $Ti^{3+}

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$Ti^{3+}$

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(D) The magnetic moment $(\mu)$ is calculated using the formula: $\mu = \sqrt{n(n+2)} \ BM$,where $n$ is the number of unpaired electrons.For $Ti^{3+}$ $(3d^1)$,$n = 1$. Thus,$\mu = \sqrt{1(1+2)} = \sqrt{3} \approx 1.73 \ BM$.For $V^{3+}$ $(3d^2)$,$n = 2$. Thus,$\mu = \sqrt{2(2+2)} = \sqrt{8} \approx 2.82 \ BM$.For $Cr^{3+}$ $(3d^3)$,$n = 3$. Thus,$\mu = \sqrt{3(3+2)} = \sqrt{15} \approx 3.87 \ BM$.For $Fe^{3+}$ $(3d^5)$,$n = 5$. Thus,$\mu = \sqrt{5(5+2)} = \sqrt{35} \approx 5.91 \ BM$.Therefore,$Ti^{3+}$ has a magnetic moment of approximately $1.73-1.75 \ BM$.

Which one of the following has a magnetic moment of $1.75 \ BM$?

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