Question

Calculate oxidation number of $[Ag\,{(N{H_3})_2}]\,Cl$
A) $+ 2$
B) $+ 1$
C) $+ 3$
D) $0$

Answer 1

For finding out the oxidation state which is also called primary valency you have to learn the charge each type of ligand possess. In the above complex of silver we have two types of ligands attached one is ammonia and second is chloride so ammonia doesn't possess any charge but chloride possesses a negative charge of $- 1$.

Complete step-by-step answer:
We have a complex of silver which is formed when silver gets surrounded by two ammonia ligands and one chloride ion. So in these types of complexes where we have ligands they can be having charge on them or not. Here we have ammonia which is having zero charge on it represented as $\mathop {N{H_3}}\limits^0$ here as we see ammonia get a zero charge on it so whenever it get connected by any coordination metal generally $3d$ metals it will form coordinate bond. Another we have chloride ions which have charge of $- 1$ represented as $C{l^ - }$ .

Now let’s take the oxidation state of metal which is silver here as (x) now putting values of other ligands.
$[Ag\,{(N{H_3})_2}]\,Cl$
$x\,\, + \,(2 \times \,0)\,\, + \,( - 1) = 0$
Solving further we get, $x + 0\, + \,( - 1) = 0$ and value of $x = + 1$
So, we are getting the oxidation state of silver as $+ 1$ , it means silver in its $+ 1$ oxidation state connects with ammonia and chloride ligands and forms a complex.

Hence the correct answer is option ‘B’.

Note: Learning the charge which each ligand can possess is important because by that you only find out the oxidation state of metal in that complex. There are certain monodentate ligands which connect with metal by its one valence while there are bidentate ligands which will connect with the metals by its two valences.