How many transition metals actually have used more than 8 valence electrons? Does $"Ag"$ have 11 valence electrons or 1 or what?

In terms of actual , there is literally only one example. You can consider that their oxidation states have never gotten higher than $+8$, except for Iridium:

In terms of electrons listed after the , note that that's not a good indication of how many there would be in a particular transition metal.

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SILVER

No, silver does not have any more than $1$ valence electron.

We only count valence electrons as those electrons that are important in chemical (here, the one $5s$ electron):

$[Kr] 4d^10 5s^1$

We never see silver with an oxidation state higher than $+1$ (which would in principle require transferring one valence electron completely).

This is because the $4d$ orbitals are lower in energy by about $"5.2 eV"$ which is significant (about $"501.7 kJ/mol"$'s worth of energy, about five times the average strength of a chemical covalent bond!).

Hence, it only uses one valence electron in most if not all cases.

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IRIDIUM

Iridium is a strange example that has $bb9$ valence electrons, and can use ALL $9$ of them, such as in the iridium (IX) oxide cation,

$[stackrel(color(blue)(+9))"Ir"stackrel(" "color(blue)(-2))("O"_4)]^(+)$.

The atomic of $"Ir"$ is actually:

$[Xe] 6s^color(blue)(2) 4f^14 5d^color(blue)(7)$

Don't get fooled --- there are not $23$ valence electrons here; its $4f$ orbital energies are quite core-like.

However, its $6s$ and $5d$ energies are very close together (about $"1.39 eV"$ apart, or only $"134.11 kJ/mol"$!), with the $6s$ higher in energy, so the $6s$ and $5d$ orbitals hold the $9$ valence electrons.

Using all $9$ is a little hard, as it normally maxes out at $8$, but $9$ is possible.

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Endnotes

$""^([1])$ (Appendix B.9)

$""^([2])$