Which of the following transition metal ions | vedclass

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

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$Ti^{3+}$

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(B) The magnetic moment $(\mu)$ is calculated using the formula $\mu = \sqrt{n(n+2)} \, BM$, where $n$ is the number of unpaired electrons.$1$. For $Sc^{3+}$ $(3d^0)$: $n = 0$, so $\mu = 0 \, BM$.$2$. For $Ti^{3+}$ $(3d^1)$: $n = 1$, so $\mu = \sqrt{1(1+2)} = \sqrt{3} \approx 1.73 \, BM$.$3$. For $Cu^{+}$ $(3d^{10})$: $n = 0$, so $\mu = 0 \, BM$.$4$. For $Zn^{2+}$ $(3d^{10})$: $n = 0$, so $\mu = 0 \, BM$.Thus, $Ti^{3+}$ has a definite magnetic moment of $1.73 \, BM$.

Which of the following transition metal ions has a definite magnetic moment?

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