How to find oxidation no of Ag in [(NH3)2] [Ag(CN)2]
To determine the oxidation number of any coordination compound two things are taken into consideration,
1. The charge that each ligand attached to the metal carries , because the ligand may be either neutral like (NH3) or negative (Cl- ) too, and also the charge may be 1,2,3 etc.
2. The overall charge of the compound itself inside the brackets as the the charge inside bracket = charge on the bracket/charge of the element or group outside the bracket.
Thus, we take both the things and calculate the oxidation no.
Hence,: in case of [(NH3)2 ][Ag(CN)2]
Here, charge on charge on (NH3) = 0, charge on CN = -1, overall charge on bracket = 0
So, oxidation no. of Ag= > 0 x 2 + x + (-1)x2 = 0
hence, oxidation no. of Ag = +2