Do all transition metals have the n2 and (n-1)d orbitals?

[Silhorn]

Does all transition metals have their n2 and ( n-1)d orbitals as valence electrons?
I have been scouring the internet to the answer of this question but have mixed results.
Iron has an electron configuration of: 1s2,2s2,2p6,3s2,3p6,4s2,3d6
Aparantly it has 8 valence electrons because 4s2 + 3d6 = 8 electrons.
Nickel has an electron configuration of: 1s2,2s2,2p6,3s2,3p6,4s2,3d8
Some people say it has 10 valence electrons because 4s2 + 3d8 = 10 electrons.
And some other people say it is wrong, it only has 2 valence electrons because the d orbital is in the 3rd energy level.

And then I come across another thing saying only certain transition metals you count the electrons in the n2 and ( n-1)d orbitals and the rest is just count the electrons in the outside shell as normal. This seems to be the solution but am I right and which ones?

Could someone clear this up for me?

 

[DrDu]

There is no clear-cut difference between valence and non-valence electrons. With d-group elements, you may consider the formal oxidation number as an indication of the number of valence electrons. As d-electrons become more tightly bound to the nucleus the farther right you move in the period and the farther up you move in the column, in the first period, the maximal oxidation number is reached with + VII with manganum which hence uses all its 4s and 3d electrons. Its right neighbour, iron, has a maximal oxidation number of +VI, and not VIII so that it can't use all its 4s and 3d electrons. On the other hand, in the 3rd long period, the maximal oxidation number is reached with Iridium with +IX while its right neighbour, Platin, only reaches +VI.

 

[Silhorn]

Ok, I understand.
Thanks for the help.