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(D) The spin-only magnetic moment is given by the formula $\mu = \sqrt{n(n+2)} \ B.M.$,where $n$ is the number of unpaired electrons.Given $\mu = 4.9

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$A, B$ and $E$ only

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(D) The spin-only magnetic moment is given by the formula $\mu = \sqrt{n(n+2)} \ B.M.$,where $n$ is the number of unpaired electrons.Given $\mu = 4.9 \ B.M.$,we have $\sqrt{n(n+2)} = 4.9$,which implies $n = 4$.Let us calculate the number of unpaired electrons for each ion:$(A) {}_{24} Cr^{2+} \Rightarrow [Ar] 3d^4$ ($4$ unpaired $e^{-}$)$(B) {}_{26} Fe^{2+} \Rightarrow [Ar] 3d^6$ ($4$ unpaired $e^{-}$)$(C) {}_{26} Fe^{3+} \Rightarrow [Ar] 3d^5$ ($5$ unpaired $e^{-}$)$(D) {}_{27} Co^{2+} \Rightarrow [Ar] 3d^7$ ($3$ unpaired $e^{-}$)$(E) {}_{25} Mn^{3+} \Rightarrow [Ar] 3d^4$ ($4$ unpaired $e^{-}$)Thus,ions $A, B,$ and $E$ have $4$ unpaired electrons and a magnetic moment of $4.9 \ B.M.$

The metal ions that have the calculated spin-only magnetic moment value of $4.9 \ B.M.$ are:
$A. Cr^{2+}$ $B. Fe^{2+}$ $C. Fe^{3+}$ $D. Co^{2+}$ $E. Mn^{3+}$
Choose the correct answer from the options given below:

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The metal ions that have the calculated spin-only magnetic moment value of $4.9 \ B.M.$ are:$A. Cr^{2+}$ $B. Fe^{2+}$ $C. Fe^{3+}$ $D. Co^{2+}$ $E. Mn^{3+}$Choose the correct answer from the options given below: