Question

What is the electron configuration of d-block ions, for example V+?

Answer 1

EXAMPLE: TYPICAL TRANSITION METAL

If we start from vanadium atom, it is easier to think about.

Since `"V"` is atomic number `23`, it is on the third column (and fourth period) of the transition metal series and has three `3d` electrons.

Using the noble gas configuration, we reference the previous noble gas, which is `"Ar"`, and get

`color(green)([Ar]3d^3 4s^2)`.

Since the `3d` orbital is lower in energy to a significant extent on the first-row transition metals after `"Sc"`, we would remove electrons from the `4s` orbital first upon ionization of `"V"` into `"V"^(+)`.

In fact, the orbital potential energies are `"V"_(4s) = -"7.32 eV"` and `"V"_(3d) = -"10.11 eV"`, and clearly, `"V"_(3d) < "V"_(4s)`.

The ionization can be written as follows:

`"V" -> "V"^(+) + e^(-)`

Thus, the electron configuration for `"V"^(+)` is reasonably predicted to be

`color(blue)([Ar]3d^3 4s^1)`,

but it is not reasonable to predict:

`color(red)([Ar]3d^2 4s^2)`.

The actual observed result is `[Ar] 3d^4`. (It may be because the removal of the `4s` electron reduces repulsions enough that the `4s` orbital sufficiently lowers in energy, so that the electron favors being in the `3d` orbital for this particularly unpaired configuration.)

EXAMPLE: ATYPICAL TRANSITION METAL

In general you should be asked for electron configurations of transition metals that follow the trend of the `ns` being higher in energy than the `(n-1)d`, but be aware that that is not always the case.

For instance, yttrium has the electron configuration

`[Kr]5s^2 4d^1`,

and in fact, the `4d` orbital is higher in energy than the `5s`. `"V"_(5s) = -"6.70 eV"` and `"V"_(4d) = -"6.49 eV"`, so the ionization of `"Y"` is likely to give you the configuration for `"Y"^(+)` to be

`color(blue)([Kr]5s^2)`.

And we indeed see that if we look here for reference.