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(D) The electronic configuration of Manganese $(Z = 25)$ is $[Ar] 3d^5 4s^2$.After the removal of two electrons,the configuration of $Mn^{2+}$ becomes

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Manganese $(Z = 25)$

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(D) The electronic configuration of Manganese $(Z = 25)$ is $[Ar] 3d^5 4s^2$.After the removal of two electrons,the configuration of $Mn^{2+}$ becomes $[Ar] 3d^5$.This $3d^5$ configuration is exceptionally stable due to the half-filled $d$-subshell.Therefore,removing the third electron requires a significantly higher amount of energy,making the third ionization enthalpy of Manganese the highest among the given elements.

List-$I$	List-$II$
$(1)$ $Ti^+$	$(A)$ $(n-1)d^4ns^2$
$(2)$ $_{30}Zn$	$(B)$ $(n-1)d^{10}ns^1$
$(3)$ $_{24}Cr$	$(C)$ $(n-1)d^2ns^0$
$(4)$ $_{29}Cu$	$(D)$ $(n-1)d^5ns^1$
	$(E)$ $(n-1)d^0ns^2$
	$(F)$ $(n-1)d^{10}ns^2$

Four successive members of the first row transition elements are listed below with their atomic numbers. Which one of them is expected to have the highest third ionization enthalpy?

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Match the elements in List-$I$ with their electronic configurations in List-$II$.

List-$I$ List-$II$

$(1)$ $Ti^+$ $(A)$ $(n-1)d^4ns^2$

$(2)$ $_{30}Zn$ $(B)$ $(n-1)d^{10}ns^1$

$(3)$ $_{24}Cr$ $(C)$ $(n-1)d^2ns^0$

$(4)$ $_{29}Cu$ $(D)$ $(n-1)d^5ns^1$

$(E)$ $(n-1)d^0ns^2$

$(F)$ $(n-1)d^{10}ns^2$

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