How to find oxidation no of Ag in [(NH3)2] [Ag(CN)2]

Dear student!

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

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oxidaton no of Ag in this compound is +2..as oxidation no of NH3 would be 0..so we wil only consider (CN) that have oxidation no of -1..

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