An element has [Ar] 3d^1 configuration in its | vedclass

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(A) The electronic configuration of the element in its $+2$ oxidation state is $[Ar] 3d^1$.To find the neutral state configuration,we add the two elec

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period-$4$,group-$3$

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(A) The electronic configuration of the element in its $+2$ oxidation state is $[Ar] 3d^1$.To find the neutral state configuration,we add the two electrons back to the $4s$ orbital,which is filled before the $3d$ orbital.Thus,the neutral state configuration is $[Ar] 3d^1 4s^2$.The principal quantum number of the outermost shell is $n=4$,which indicates the element belongs to period-$4$.The total number of valence electrons is $1 (d) + 2 (s) = 3$,which indicates the element belongs to group-$3$ (Scandium).Therefore,the correct position is period-$4$,group-$3$.

An element has $[Ar] 3d^1$ configuration in its $+2$ oxidation state. Its position in the periodic table is

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