Question

What is the electron configuration for a phosphorous anion with a charge of `-2`?

Answer 1

`"P"^(2-): ["Ne"] 3s^2 3p^5`

Explanation:

Your starting point here will be the electron configuration of a neutral phosphorus atom.

Phosphorus, `"P"`, is located in period 3, group 15 of the periodic table, and has an atomic number equal to `15`. This means that a neutral phosphorus atom will have `15` electrons surrounding its nucleus.

Therefore, the electron configuration of a neutral phosphorus atom will show `15` electrons

`"P: " 1.2^2 2s^2 2p^6 3s^2 3p^3`

Now, in order for the phosphorus atom to become a phosphorus anion, it must gain two electrons. Since the 3p-subshell only holds `3` electrons in the neutral atom, these incoming electrons will be placed there.

Therefore, the electron configuration of the `"P"^(2-)` anion will be

`"P"^(2-): 1s^2 2s^2 2p^6 3s^2 3p^5`

Using the noble gas shorthand notation, you will get

`"P"^(2-): ["Ne"] 3s^2 3p^5`