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

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.