Question

How would you write the electron configuration for the following ions?

`K^(+1)`
`Br^-1`
`Al^(+3)`
`As^-3`?

Answer 1

`1s^2 2s^2 2p^6 3s^2 3p^6`
`1s^2 2s^2 2p^6 3s^2 3p^6 4s^2 3d^10 4p^6`
`1s^2 2s^2 2p^6`
`1s^2 2s^2 2p^6 3s^2 3p^6 4s^2 3d^10 4p^6`

Explanation:

The electron configuration for these elements with a +0 charge is:

K: `1s^2 2s^2 2p^6 3s^2 3p^6 4s^1`
Br: `1s^2 2s^2 2p^6 3s^2 3p^6 4s^2 3d^10 4p^5`
Al: `1s^2 2s^2 2p^6 3s^2 3p^1`
As: `1s^2 2s^2 2p^6 3s^2 3p^6 4s^2 3d^10 4p^3`

For `K^+`, remove one electron; `4s^1`, since it is the only electron in the 4th energy level. the configuration will be:
`1s^2 2s^2 2p^6 3s^2 3p^6`

For `Br^-`, add one electron. This will be a `4p` electron to satisfy the octet rule.
`1s^2 2s^2 2p^6 3s^2 3p^6 4s^2 3d^10 4p^6`

For `Al^(3+)`, remove 3 electrons. They will be the `3p` and `3s` electrons, since they are the valence electrons.
`1s^2 2s^2 2p^6`

For `As^(3-)`, add 3 electrons. These will be `4p` electrons to satisfy the octet rule.
`1s^2 2s^2 2p^6 3s^2 3p^6 4s^2 3d^10 4p^6`

Answer 2

Refer to the explanation.

Explanation:

`"K"^(+):` `["Ne"]"3s"^2"3p"^6"`

`"Br"^(-):` `["Ar"]"3d"^10"4s"^2"4p"^6`

`"Al"^(3+):` `"1s"^2"2s"^2"2p"^6"` or `["He"]"2s"^2"2p"^6"`

`"As"^(3-):` `["Ar"]"3d"^10"4s"^2"4p"^6`

The positively charged cations have the electron configuration of the noble gas in the previous period. The negatively charged anions have the electron configuration of the noble gas at the end of the period in which they occur. Atoms gain or lose electrons to form ions with filled valence shells, having an octet, except for hydrogen.