What is the electron configuration for the aluminum ion? Phosphide ion? Do NOT use shorthand notation!

My answer but when I input it I get it wrong why?

Al: 1s^2,2s^2,2p^6,3s^2,3p^1
P: 1s^2,2s^2,2p^6,3s^2,3p^3

You gave the electron configuration for Al, not Al^3+. You should remove the last three electrons for the Al^3+ ion. Same thing for P^3-. You should add 3e to what you have.

To determine the correct electron configuration for ions, you need to take into account the number of electrons gained or lost by the atom to form the ion.

Aluminum (Al) has an atomic number of 13, which means it has 13 electrons in its neutral state. Aluminum ion (Al3+) loses three electrons to become positively charged. To determine the electron configuration for the aluminum ion:

1. Write out the electron configuration for a neutral aluminum atom:
1s^2, 2s^2, 2p^6, 3s^2, 3p^1

2. Remove three electrons from the highest energy level (3p):
1s^2, 2s^2, 2p^6

Therefore, the electron configuration for the aluminum ion (Al3+) is: 1s^2, 2s^2, 2p^6.

Now for phosphide ion:

Phosphorus (P) has an atomic number of 15, with 15 electrons in its neutral state. The phosphide ion (P3-) gains three electrons to become negatively charged. To determine the electron configuration for the phosphide ion:

1. Write out the electron configuration for a neutral phosphorus atom:
1s^2, 2s^2, 2p^6, 3s^2, 3p^3

2. Add three electrons to the highest energy level (3p):
1s^2, 2s^2, 2p^6, 3s^2, 3p^6

Therefore, the electron configuration for the phosphide ion (P3-) is: 1s^2, 2s^2, 2p^6, 3s^2, 3p^6.