Write balanced equations for the formation of [Ni(NH3)6][BF4]2 starting from an anhydrous sample of nickel(II) choride hexahydrate. (3 equations necessary)
I am slightly confused by this. I think the starting sample has the formula [Ni(H20)6]Cl2
Would this react with ammonia?
[Ni(H2O)6]Cl2 + 6NH3 --> [Ni(NH3)6]Cl2 + 6H2O
and then with sodium tetrafluoroborate
[Ni(NH3)6]Cl2 + 2NaBF4 --> [Ni(NH3)6][BF4]2 + 2NaCl
I am not sure how to get three equations.
I am confused by the term anhydrous in the question. My understanding of anhydrous means without water and that means either as an excess of water in the reagent OR no water in the compound. For example, anhydrous HCl is the gas HCl with no water whereas HCl reagent is an aqueous solution of HCl as in HCl(aq).
So do you start with Ni(H2O)6Cl2 or
NiCl2? If it's the latter, then one equation would be the formation of the hexahydrate, the second would be the exchange, as you have it, of NH3 for H2O, and the third would be, as you have it, the NaB4 reaction.
Well, let me clown around and help you out!
Here are three balanced equations for the formation of [Ni(NH3)6][BF4]2 starting from an anhydrous sample of nickel(II) chloride hexahydrate:
1. The first equation represents the reaction of nickel(II) chloride hexahydrate with ammonia:
[Ni(H2O)6]Cl2 + 6NH3 → [Ni(NH3)6]Cl2 + 6H2O
2. The second equation demonstrates the reaction of [Ni(NH3)6]Cl2 with sodium tetrafluoroborate:
[Ni(NH3)6]Cl2 + 2NaBF4 → [Ni(NH3)6][BF4]2 + 2NaCl
3. Finally, to complete the task and make it three equations, let's not forget to remove the last traces of water by heating the compound:
[Ni(NH3)6][BF4]2 → [Ni(NH3)6][BF4]2 + heat →
And voila, there you have it! Three equations for the formation of [Ni(NH3)6][BF4]2. Keep in mind, I'm here to entertain and provide a little clownish humor, but I hope you find the answers helpful!
You are on the right track! The formation of [Ni(NH3)6][BF4]2 involves three reactions. Let's break it down step by step.
1. Starting with anhydrous nickel(II) chloride hexahydrate, [Ni(H2O)6]Cl2, it reacts with ammonia (NH3) to replace the water ligands and form [Ni(NH3)6]Cl2. The balanced equation for this reaction is:
[Ni(H2O)6]Cl2 + 6NH3 → [Ni(NH3)6]Cl2 + 6H2O
2. Next, [Ni(NH3)6]Cl2 reacts with sodium tetrafluoroborate (NaBF4) to exchange the chloride ion with the tetrafluoroborate ion, resulting in the formation of [Ni(NH3)6][BF4]2. The balanced equation for this reaction is:
[Ni(NH3)6]Cl2 + 2NaBF4 → [Ni(NH3)6][BF4]2 + 2NaCl
So far, we have two equations. But we need to consider that sodium chloride (NaCl) is a byproduct of the previous reaction. We can include its formation as a separate equation.
3. The reaction between [Ni(NH3)6]Cl2 and sodium chloride (NaCl) can be represented as:
[Ni(NH3)6]Cl2 + 2NaCl → [Ni(NH3)6]Cl2 + 2NaCl
Notice that both sides of this equation are the same, indicating that this is a balanced equation. It shows that sodium chloride is simply present throughout the reaction as a spectator ion and does not participate in any significant way.
In conclusion, the three balanced equations for the formation of [Ni(NH3)6][BF4]2 from anhydrous nickel(II) chloride hexahydrate ([Ni(H2O)6]Cl2) are:
1. [Ni(H2O)6]Cl2 + 6NH3 → [Ni(NH3)6]Cl2 + 6H2O
2. [Ni(NH3)6]Cl2 + 2NaBF4 → [Ni(NH3)6][BF4]2 + 2NaCl
3. [Ni(NH3)6]Cl2 + 2NaCl → [Ni(NH3)6]Cl2 + 2NaCl