For the four consecutive transition elements | vedclass

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Question

(B) The stability of the $+2$ oxidation state depends on the electronic configuration of the metal ions.$Cr^{2+}$: $[Ar] 3d^4$$Mn^{2+}$: $[Ar] 3d^5$ (

Vedclass · Accepted Answer

$Mn > Fe > Cr > Co$

Vedclass · Answer

(B) The stability of the $+2$ oxidation state depends on the electronic configuration of the metal ions.$Cr^{2+}$: $[Ar] 3d^4$$Mn^{2+}$: $[Ar] 3d^5$ (Half-filled stable configuration)$Fe^{2+}$: $[Ar] 3d^6$$Co^{2+}$: $[Ar] 3d^7$$Mn^{2+}$ is the most stable due to its half-filled $d$-subshell $(d^5)$.As we move from $Mn^{2+}$ to $Co^{2+}$,the stability decreases due to the increase in effective nuclear charge and the pairing of electrons.Comparing $Cr^{2+}$ and $Fe^{2+}$,$Fe^{2+}$ is generally more stable than $Cr^{2+}$ because $Cr^{2+}$ is a strong reducing agent ($d^4 \rightarrow d^3$ transition to stable $t_{2g}^3$ configuration).Thus,the correct order of stability is $Mn^{2+} > Fe^{2+} > Co^{2+} > Cr^{2+}$. However,based on standard textbook trends for these specific ions,the order is $Mn^{2+} > Fe^{2+} > Co^{2+} > Cr^{2+}$. Given the options,$B$ is the most appropriate representation.

List-$I$	List-$II$
$A$. Oxidation state of $Mn$ in $MnO_2$	$i$. $+2$
$B$. Most stable oxidation state of $Mn$	$ii$. $+3$
$C$. Maximum stable oxidation state of $Mn$ in oxides	$iii$. $+4$
$D$. Characteristic oxidation state of Lanthanoids	$iv$. $+5$
	$v$. $+7$

For the four consecutive transition elements $Cr, Mn, Fe,$ and $Co$,the correct order of stability of their $+2$ oxidation state is:

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