Question

The ion `"X"^(3-)` has `8` neutrons and its electron configuration ends up with `2p^6`. Find its mass number. [I can't find it, can someone please explain it to me?]

Answer 1

`A = 15`

Explanation:

The trick here is to use the ion's net charge and its number of electrons to figure out the atomic number of the neutral atom, i.e. the number of protons located inside any atom fo element `"X"`.

As you know, elements become ions by losing or gaining electrons. In this case, a neutral atom of element `"X"` gained `3` electrons in order to form an `"X"^(3-)` ion and complete its octet.

`"X " + color(white)(.)3"e"^(-) -> " X"^(3-)`

So you know that a neutral atom of `"X"` has `3` electrons fewer than the `"X"^(3-)` ion.

Now, you know that the ion's electron configuration ends in `2p^6`, which means that it looks like this

`"X"^(3+):" " 1s^2 2s^2 2p^6`

You can thus say that an `"X"^(3-)` ion has a total of

`overbrace("2 e"^(-))^(color(blue)("from 1s"^2)) + overbrace("2 e"^(-))^(color(purple)("from 2s"^2)) + overbrace("6 e"^(-))^(color(darkorange)("from 2p"^6)) = "10 e"^(-)`

Therefore, a neutral atom of element `"X"` will have

`"10 e"^(-) - "3 e"^(-) -> "7 e"^(-)`

As you know, neutral atoms have equal numbers of protons inside the nucleus and electrons surrounding the nucleus. This implies that element `"X"` will contain `7` protons inside its nucleus.

The mass number, `A`, tells you the number of protons, `Z`, and neutrons present inside the nucleus of a given atom.

`color(blue)(ul(color(black)(A = Z + "no. of neutrons")))`

In your case, the mass number of the ion, which is equal to the mass number of the neutral atom, will be

`color(darkgreen)(ul(color(black)(A = 7 + 8 = 15)))`

The unknown element is nitrogen, `"N"`, because it has an atomic number equal to `7`. Consequently, the ion is `"N"^(3-)`.