Question

What is the electron configuration for the phosphide anion?

Answer 1

Of `"phosphide anion...."`, `P^(3-)`....?

Explanation:

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.....

Answer 2

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

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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"`.