The sum of oxidation states of two silver ions in AgNH$$_{3_{2}}$$Ag(CN)$$_{2}$$ complex is _________.

First, we recognise that the given species $$\text{AgNH}_3{}_2\;\text{Ag(CN)}_2$$ actually represents two separate complex ions that stay together in the crystalline solid. They are:

$$$[\text{Ag}(\text{NH}_3)_2]^+ \quad\text{and}\quad [\text{Ag}(\text{CN})_2]^-.$$$

Each of these ions contains one silver atom. Hence we must determine the oxidation state of silver in each complex ion and then add the two values.

We begin with the cation $$[\text{Ag}(\text{NH}_3)_2]^+.$$ The general rule is:

$$$\text{Sum of oxidation states of all atoms in an ion} = \text{Charge on that ion}.$$$

Now, we note that ammonia, $$\text{NH}_3,$$ is a neutral ligand. Therefore, for each ammonia molecule:

$$\text{Oxidation state of NH}_3 = 0.$$

There are two ammonia molecules attached to silver, so the total contribution from ligands is:

$$0 + 0 = 0.$$

Let the oxidation state of silver in this cation be $$x.$$ Applying the oxidation‐state sum rule, we write:

$$x + 0 = +1.$$

Solving, we have:

$$x = +1.$$

So, in $$[\text{Ag}(\text{NH}_3)_2]^+,$$ the oxidation state of the silver ion is $$+1.$$

Next, we consider the anion $$[\text{Ag}(\text{CN})_2]^-.$$ Again, we apply the same rule for ionic charge. Here the cyanide ligand, $$\text{CN}^-,$$ carries a charge of $$-1$$ each. There are two such ligands, so their total charge contribution is:

$$(-1) + (-1) = -2.$$

Let the oxidation state of silver in this anion be $$y.$$ Equating the sum to the overall charge on the anion, we obtain:

$$y + (-2) = -1.$$

Rearranging,

$$y = -1 + 2 = +1.$$

Thus, in $$[\text{Ag}(\text{CN})_2]^-, $$ the oxidation state of the silver ion is also $$+1.$$

Finally, we add the two individual oxidation states to find the required sum:

$$\text{Sum} = (+1) + (+1) = +2.$$

Hence, the correct answer is Option 2.