What is the electron configuration for the phosphide anion?

You gots #P^(3-)#, i.e. for #Z=15# we gots 18 electrons to distribute....and thus we have an electronic configuration equivalent to that of #Ar#...i.e. #1s^(2)2s^(2)2p^(6)3s^(2)3p^6#...reduction fills the valence shell of phosphorus, whose atomic configuration was #1s^(2)2s^(2)2p^(6)3s^(2)3p^3#...

For the phosphide anion there are 8 valence electrons.....

If you use the periodic table to determine this, it's not that tricky; should be an argon configuration.

Well, I assume you mean #"P"^(3-)# ion. Phosphorus has an atomic number of #15#, which means that provided the atom is neutral, it must also have #15# electrons.

Being on the 3rd row (#n = 3#) of the periodic table and in the #p#-block, it has:

• #1s#, #2s#, and #2p# core orbitals
• #3s# and #3p# valence orbitals
• Some #3d# orbitals to use if needed

Its electron configuration reveals its valence electrons as a neutral atom:

#"P": " " 1s^2 2s^2 2p^6 color(red)(3s^2 3p^3)#

As the ion, #"P"^(3-)#, since the electron has a #-1# relative charge, #"P"^(3-)# has three more electrons relative to #"P"#.

Thus, its electron configuration (that is, for the ion!) is:

#"P"^(3-): " " 1s^2 2s^2 2p^6 color(red)(3s^2 3p^6)#

Since the previous noble gas is #"Ne"#, with #10# electrons, we can rewrite this in terms of the so-called noble-gas core:

#"P"^(3-): " " ["Ne"] color(red)(3s^2 3p^6)#

And this then shows the #bb8# valence electrons of #"P"^(3-)#, just like those of #"Ar"#.