Writing net ionic equation:

A solution of diamminesilver(I) chloride is treated with dilute nitric acid.

So what I did was..

[Ag(NH3)2]Cl(aq) + HNO3(aq) -> __________

The answer is
[Ag(NH3)2]+ + Cl- + 2H+ -> AgCl + 2NH4+

I would guess that the oxidation #s for diamminesilver(I) chloride would be:
Ag +1
NH3 0
Cl -1

Can someone please show me the steps in between?
I don't really know how to break the [Ag(NH3)2]Cl apart and predict/form the products.

Question

07-03-2018
Chemistry

contestada

Writing net ionic equation:

A solution of diamminesilver(I) chloride is treated with dilute nitric acid.

So what I did was..

[Ag(NH3)2]Cl(aq) + HNO3(aq) -> __________

The answer is
[Ag(NH3)2]+ + Cl- + 2H+ -> AgCl + 2NH4+

I would guess that the oxidation #s for diamminesilver(I) chloride would be:
Ag +1
NH3 0
Cl -1

Can someone please show me the steps in between?
I don't really know how to break the [Ag(NH3)2]Cl apart and predict/form the products.

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What the correct answer

ChemistryHelper2024 ChemistryHelper2024 · Answer 1 · 2018-03-20T18:58:50+01:00

The net ionic equation is:
[Ag(NH₃)₂]⁺ + Cl⁻ + 2H⁺ → AgCl + 2 NH₄⁺
As i can see that you balanced it correctly also you calculated the charge of each part correctly but you need to know more details about complexes:
complex is formed between Metal cations as Ag⁺, Cu⁺², Fe⁺³ and ligands which may be neutral as NH₃ or H₂O or negatively charged as CN⁻ or halide
The number of ligands attached to metal is the co-ordination number, in our case Ag has coordination number = 2 so it must attached to two ligands which are NH₃ here (neutral ligands)
- The charge of complex is the sum of metal and ligand charges +1 for Ag and 0 for NH₃ so the overall charge of complex is +1
so the complex must be attracted to negatively charged anion which is Cl⁻ here and both react with H⁺ from acid (HNO₃) to give AgCl precipitate and NH₄⁺ ion.