Question

The coordination number of silver in the complex obtained by adding excess of $N{H}_{4}OH$ to $AgN{O}_3$ solution is

• A. 2
• B. 5
• C. 3
• D. 6

Answer 1

The correct option is A 2

This is an interesting question, firstly, we need to figure out what the compound could be when we react $N{H}_{4}OH$ with $AgN{O}_3$.
This is the equilibrium formed:

1)$N{H}_{4}OH\rightleftharpoons N{H}_3+O{H}^{-}$

2)$A{g}^{+}+N{H}_3\rightleftharpoons \left[Ag\right(N{H}_3{)}_2]$


So you need to know two things here:
1) the first equation exists in an equilbrium state primarily to the left (the reaction mainly has reactants), and
2) the second equation also exists primariy to the right (the reaction mixture is marjorly just the products).
meaning that in this system, there is a lot of ammonia and very little silver ions. Most of the final solution would have ammonia, hydroxyl ions and the complex.

about 1) you may already know this from your extensive study of the chapter ionic equilibrium, if you recall, the $K_b$ value of this equation is pretty low which means that the reaction hardly proceeds! You can apply this idea to figure out what happens in 2) now that you have ammonia. So as you can see, this can be figured out using logic. It is a bit difficult but pretty interesting once you figure it out.

The practical application of this:
This is quite a useful way to detect even very small amounts of halide ions in solution. What ammonia does is that it suppresses the presence of silver ions to a very small concentrations, which means that even minute amounts of a halide can be detected as silver halides precipitate out of the solution.
This is going to be quite useful if you need to perform these experiments in a lab.

The answer is 2, since the compound is $\left[Ag\right(N{H}_3{)}_2$]