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(B) The number of oxidation states exhibited by a transition element depends on the number of valence electrons available for bonding.$1$. For $3d^54s

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$3d^54s^2$

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(B) The number of oxidation states exhibited by a transition element depends on the number of valence electrons available for bonding.$1$. For $3d^54s^1$ (Chromium),the total number of valence electrons is $6$,so it can show oxidation states up to $+6$.$2$. For $3d^54s^2$ (Manganese),the total number of valence electrons is $7$,so it can show oxidation states up to $+7$.$3$. For $3d^24s^2$ (Titanium),the total number of valence electrons is $4$,so it can show oxidation states up to $+4$.$4$. For $3d^34s^2$ (Vanadium),the total number of valence electrons is $5$,so it can show oxidation states up to $+5$.Thus,the element with the configuration $3d^54s^2$ exhibits the largest number of oxidation states.

List-$I$ (Property)	List-$II$ (Metal)
$(A)$ Element that can show $+8$ oxidation state	$(1)$ $Mn$
$(B)$ $3d$ series element that can show up to $+7$ oxidation state	$(2)$ $Cr$
$(C)$ $3d$ series element with maximum melting point	$(3)$ $Os$
	$(4)$ $Fe$

Which one of the elements with the following outer orbital configurations may exhibit the largest number of oxidation states?

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