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How do electron configurations change for ions?
When you create an ion, you are either adding electrons to, or removing electrons from, the highest occupied energy subshell in the atom.
NONMETALS
Write the electronic structure for the neutral atom.
Then add electrons to the highest occupied subshell.
E.g., for ##"Cl"^"-"##:
##"Cl": "1s"^2 "2s"^2 "2p"^6 "3s"^2 "3p"^5##, but ##"Cl"^"-"## has one more electron
Add it to the 3##p## subshell.
##"Cl"^"⁻": "1s"^2 "2s"^2 "2p"^6 "3s"^2 "3p"^6##
##"s"## and ##"p"## BLOCK METALS
Write the electronic structure for the neutral atom. Then remove electrons from the highest
For ##"Na"^+##:
##"Na": "1s"^2 "2s"^2 "2p"^6 "3s"##, but ##"Na"^+## has one less electron
Take it from the ##"3s"## subshell.
##"Na"^+: "1s"^2 "2s"^2 "2p"^6##
##"d"## BLOCK METALS
Remove ##"s"## electrons before ##"d"## electrons.
E.g., for ##"Cr"^(3+)##
##"Cr": "1s"^2 "2s"^2"2p"^6 "3s"^2"3p"^6 "4s""3d"^5##
##"Cr"^(3+): "1s"^2 "2s"^2"2p"^6 "3s"^2"3p"^6 "3d"^3##
You remove the ##"4s"## electron first, followed by two of the ##"3d"## electrons.
EXAMPLES
Write the electron configurations for ##"O"^"2-", "Ca"^(2+)##, and ##"Zn"^(2+)##.
Solutions
##"O": "1s"^2 "2s"^2"2p"^4##
##"O"^"2-": "1s"^2 "2s"^2"2p"^6##
##"Ca": "1s"^2 "2s"^2"2p"^6 "3s"^2"3p"^6 "4s"^2##
##"Ca"^(2+): "1s"^2 "2s"^2"2p"^6 "3s"^2"3p"^6##
##"Zn": "1s"^2 "2s"^2"2p"^6 "3s"^2"3p"^6 "4s"^2"3d"^(10)##
##"Zn"^(2+): "1s"^2 "2s"^2"2p"^6 "3s"^2"3p"^6 "3d"^(10)##