Which ion has the least value of theoretical | vedclass

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(A) The theoretical magnetic moment is calculated using the formula $\mu = \sqrt{n(n+2)} 	ext{ BM}$,where $n$ is the number of unpaired electrons.$1$

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$Ti^{3+}$

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(A) The theoretical magnetic moment is calculated using the formula $\mu = \sqrt{n(n+2)} 	ext{ BM}$,where $n$ is the number of unpaired electrons.$1$. For $Ti^{3+}$ $(3d^1)$: $n = 1$,$\mu = \sqrt{1(1+2)} = \sqrt{3} \approx 1.73 	ext{ BM}$.$2$. For $Co^{3+}$ $(3d^6)$: In an octahedral field,$n = 4$ (high spin) or $n = 0$ (low spin). Assuming high spin,$n = 4$,$\mu = \sqrt{4(6)} = \sqrt{24} \approx 4.90 	ext{ BM}$.$3$. For $Cr^{3+}$ $(3d^3)$: $n = 3$,$\mu = \sqrt{3(5)} = \sqrt{15} \approx 3.87 	ext{ BM}$.$4$. For $V^{3+}$ $(3d^2)$: $n = 2$,$\mu = \sqrt{2(4)} = \sqrt{8} \approx 2.83 	ext{ BM}$.Comparing the values,$Ti^{3+}$ has the least number of unpaired electrons $(n=1)$,hence it has the least magnetic moment.

Which ion has the least value of theoretical magnetic moment?

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