If the electronic configuration of A^{-} is [ | vedclass

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(C) The electronic configuration of $A^{-}$ is $[Ar] \ 4s^2 \ 3d^{10} \ 4p^1$. This corresponds to a total of $18 + 2 + 10 + 1 = 31$ electrons. Since

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Group $- 12$,Period $- 4$

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(C) The electronic configuration of $A^{-}$ is $[Ar] \ 4s^2 \ 3d^{10} \ 4p^1$. This corresponds to a total of $18 + 2 + 10 + 1 = 31$ electrons. Since $A^{-}$ has $31$ electrons,the neutral atom $A$ has $30$ electrons,meaning its atomic number $Z = 30$. The element with $Z = 30$ is Zinc $(Zn)$,which belongs to Group $12$ and Period $4$. Removing electrons to form $A^{+2}$ (i.e.,$Zn^{+2}$) does not change the identity of the element or its position in the periodic table. Therefore,$A^{+2}$ remains in Group $12$ and Period $4$.

If the electronic configuration of $A^{-}$ is $[Ar] \ 4s^2 \ 3d^{10} \ 4p^1$,then $A^{+2}$ will belong to which group and period?

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