If Br^{-} configuration is [Ar], 3d^{10},4s^2 | vedclass

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Question

(B) The electronic configuration of $Br$ (atomic number $35$) is $[Ar]\, 3d^{10}\,4s^2\,4p^5$.$Br^{-}$ is formed by adding one electron to $Br$,so $Br

Vedclass · Accepted Answer

$As$

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(B) The electronic configuration of $Br$ (atomic number $35$) is $[Ar]\, 3d^{10}\,4s^2\,4p^5$.$Br^{-}$ is formed by adding one electron to $Br$,so $Br^{-}$ is $[Ar]\, 3d^{10}\,4s^2\,4p^6$.$Br^{+2}$ is formed by removing two electrons from the neutral $Br$ atom. Electrons are removed from the outermost shell ($4p$ orbital first).$Br^{+2}$ configuration is $[Ar]\, 3d^{10}\,4s^2\,4p^3$.The element with atomic number $33$ is Arsenic $(As)$,which has the configuration $[Ar]\, 3d^{10}\,4s^2\,4p^3$.Therefore,the configuration of $Br^{+2}$ is identical to $As$.

If $Br^{-}$ configuration is $[Ar]\, 3d^{10}\,4s^2\,4p^6$,then $Br^{+2}$ configuration will be identical to which element?

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