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(B) The spin-only magnetic moment is calculated using the formula $\mu = \sqrt{n(n+2)} 	ext{ B.M.}$, where $n$ is the number of unpaired electrons.$1

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

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(B) The spin-only magnetic moment is calculated using the formula $\mu = \sqrt{n(n+2)} 	ext{ B.M.}$, where $n$ is the number of unpaired electrons.$1$. $Ti^{3+}$ $(Z=22)$: $[Ar] 3d^1$, $n=1$, $\mu = \sqrt{1(3)} = 1.73 	ext{ B.M.}$$2$. $Zn^{2+}$ $(Z=30)$: $[Ar] 3d^{10}$, $n=0$, $\mu = 0 	ext{ B.M.}$$3$. $Cr^{2+}$ $(Z=24)$: $[Ar] 3d^4$, $n=4$, $\mu = \sqrt{4(6)} = 4.90 	ext{ B.M.}$$4$. $Fe^{2+}$ $(Z=26)$: $[Ar] 3d^6$, $n=4$, $\mu = \sqrt{4(6)} = 4.90 	ext{ B.M.}$$5$. $Cr^{3+}$ $(Z=24)$: $[Ar] 3d^3$, $n=3$, $\mu = \sqrt{3(5)} = 3.87 	ext{ B.M.}$$6$. $Cu^{2+}$ $(Z=29)$: $[Ar] 3d^9$, $n=1$, $\mu = 1.73 	ext{ B.M.}$Comparing the values, $Cr^{2+}$ and $Fe^{2+}$ both have $n=4$ and thus the same magnetic moment.

Which of the following pairs of elements in their respective oxidation states develops the same value of calculated spin-only magnetic moment?

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