Which of the following pairs has the same mag | vedclass

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(B) The magnetic moment is determined by the number of unpaired electrons $(n)$ using the formula $\mu = \sqrt{n(n+2)} \ BM$.$1$. $Cr^{3+}$ $([Ar]3d^3

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$Cr^{2+}$ and $Fe^{2+}$

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(B) The magnetic moment is determined by the number of unpaired electrons $(n)$ using the formula $\mu = \sqrt{n(n+2)} \ BM$.$1$. $Cr^{3+}$ $([Ar]3d^3)$: $n = 3$$2$. $Mn^{2+}$ $([Ar]3d^5)$: $n = 5$$3$. $Cr^{2+}$ $([Ar]3d^4)$: $n = 4$$4$. $Fe^{2+}$ $([Ar]3d^6)$: $n = 4$$5$. $V^{2+}$ $([Ar]3d^3)$: $n = 3$$6$. $Sc^{3+}$ $([Ar]3d^0)$: $n = 0$$7$. $Ti^{2+}$ $([Ar]3d^2)$: $n = 2$Since $Cr^{2+}$ and $Fe^{2+}$ both have $n = 4$ unpaired electrons,they have the same magnetic moment.

Which of the following pairs has the same magnetic moment?

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