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

1 Answer
Nov 6, 2015

Answer:

#"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#