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QUESTION

# 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 3p 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)