How can we determine how many delocalised electrons every atom of gold contributes to the 'sea of delocalised electrons'? More generally, how can we determine the number of electrons any metal contributes? Must we have thermodynamic data to determine this?
Briefly: valence orbitals on adjacent metal atoms interfere to form "bands" of closely spaced molecular orbitals that spread throughout the metal.
If you have N metal atoms with occupied s orbitals, they interfere to form an "s band" with N molecular orbitals. The lowest energy orbital is fully bonding; the orbital on the top of the band is fully antibonding. The spread of the band will be finite, and since N is a huge number, the molecular orbital energies will be very closely spaced, practically continuous.
If you have two valence s electrons in your metal, the s band will be full (each orbital can hold up to 2 electrons, you have 2N electrons, and N orbitals in the band). If you have only one valence s electron, the s band will be half full (the first N/2 orbitals will be filled, and the remaining orbitals will be empty).
Above the s band you'll have a similar p band. In gold, the p band is unoccupied, while the s band is half full (gold has 6s$^1$). However, IF the bottom of the p band overlaps with the top of the s band, the s electrons can easily move into the bonding MO's at the bottom of the p band. They can also move into any vacant orbitals in the s band that are just above the highest occupied orbital. In either case, the s electron will be delocalized.
To figure out whether the valence electrons will delocalize, then, you'll have to decide what bands are present, where the highest occupied molecular orbital is, and what the width and the spacing of the bands are (does the unoccupied band overlap with the occupied band, or is there a gap between the two?)
Valance electron of s and p orbital take part in delocalisation over the positively charged gold atoms