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# is this reaction equation correct? watch

1. [Ni(H2O)6]Cl2 + 6NH3 + 6NH4BF4 --> [Ni(NH3)6](BF4)6 + 6H2O + 2NH4Cl

I feel like the ammonium chloride part is wrong bc i cant seem to balance it! I have 4 extra ammoniums on the LHS
2. (Original post by kiiten)
[Ni(H2O)6]Cl2 + 6NH3 + 6NH4BF4 --> [Ni(NH3)6](BF4)6 + 6H2O + 2NH4Cl

I feel like the ammonium chloride part is wrong bc i cant seem to balance it! I have 4 extra ammoniums on the LHS
You're trying to do two things at once: ligand subs and it looks like redox.

The ligand subs first would be simply:
[Ni(H2O)6]Cl2 + 6NH3 --> [Ni(NH3)6]Cl2 + 6H2O

The 2xCl are both 1- ions, to balance the Ni2+ ion.
Your version of the equation involves (BF4)6, which would change the Ni to 6+, but there is no reduction anywhere to counter it - apart from the loss of 4NH4+ ions.
3. (Original post by Pigster)
You're trying to do two things at once: ligand subs and it looks like redox.

The ligand subs first would be simply:
[Ni(H2O)6]Cl2 + 6NH3 --> [Ni(NH3)6]Cl2 + 6H2O

The 2xCl are both 1- ions, to balance the Ni2+ ion.
Your version of the equation involves (BF4)6, which would change the Ni to 6+, but there is no reduction anywhere to counter it - apart from the loss of 4NH4+ ions.
Hm im a bit stuck then.... I understand the ligand subs part but not the redox. Because if i add 4NH4+ on the RHS the charges dont balance.The LHS has 2- and 2+ and the RHS would have 6+ and 4+ = 10+ ???

But I need the overall eq to calculate yield
4. Ahhh i think i got it - made a typo in the number of BF6 ligands. Does this make sense now?

[Ni(H2O)6]Cl2 + 6NH3 + 2NH4BF4 à [Ni(NH3)6](BF4)2 + 6H2O + 2NH4Cl
5. (Original post by kiiten)
Ahhh i think i got it - made a typo in the number of BF6 ligands. Does this make sense now?

[Ni(H2O)6]Cl2 + 6NH3 + 2NH4BF4 à [Ni(NH3)6](BF4)2 + 6H2O + 2NH4Cl
Now it is just ligand subs (with no redox) and a change in anion.
6. (Original post by kiiten)
Ahhh i think i got it - made a typo in the number of BF6 ligands. Does this make sense now?

[Ni(H2O)6]Cl2 + 6NH3 + 2NH4BF4 à [Ni(NH3)6](BF4)2 + 6H2O + 2NH4Cl
This looks highly improbable as ligand substitution. The nickel complex ion in the product would have to have a coordination number of 8.

As it is written, it seems to be simply:

[Ni(H2O)6]2+ + 6NH3 --> [Ni(NH3)6]2+ + 6H2O

The other species there are just balancing ions. It looks like you have used a solution of ammonia /ammonium borofluoride for the sole purpose of effecting the ligand substitution.
7. (Original post by charco)
This looks highly improbable as ligand substitution. The nickel complex ion in the product would have to have a coordination number of 8.

As it is written, it seems to be simply:

[Ni(H2O)6]2+ + 6NH3 --> [Ni(NH3)6]2+ + 6H2O

The other species there are just balancing ions. It looks like you have used a solution of ammonia /ammonium borofluoride for the sole purpose of effecting the ligand substitution.
Hm i actually did this reaction in the lab and the goal was to synthesise [Ni(NH3)6](BF4)2 but the equation wont balance without the other species on the LHS

And as for the second part of the experiment im even more confused . Not sure if HCl is supposed to be on the LHS or not. And for the RHS i have 2 extra OHs??

[Ni(NH3)6](BF4)2 + 2HDMG + 2NH4OH + HCl --> 2NH4BF4 + Ni(HDMG)2 + 6NH3 + 2OH
8. (Original post by kiiten)
Hm i actually did this reaction in the lab and the goal was to synthesise [Ni(NH3)6](BF4)2 but the equation wont balance without the other species on the LHS
When you synthesise a compound you can write a reaction including the balancing ions, but they do nothing for the chemistry of the process.

If you put in the balancing ions it will be ok, but it's not necessary for the understanding of the process.

And as for the second part of the experiment im even more confused . Not sure if HCl is supposed to be on the LHS or not. And for the RHS i have 2 extra OHs??

[Ni(NH3)6](BF4)2 + 2HDMG + 2NH4OH + HCl --> 2NH4BF4 + Ni(HDMG)2 + 6NH3 + 2OH
Once again this is an impossible equation as you have both ammonium hydroxide (a base) and hydrochloric acid on the LHS. They would immediately combine to form a salt and water.

This is a ligand substitution used to test gravimetrically for nickel using DMG.

The nickel complexes with dimethylglyoxime (DMG) and presumably the extra hydrogen ions then combine with the hydroxyl ions that you have "left over" to make water.
9. (Original post by charco)

Once again this is an impossible equation as you have both ammonium hydroxide (a base) and hydrochloric acid on the LHS. They would immediately combine to form a salt and water.

This is a ligand substitution used to test gravimetrically for nickel using DMG.

The nickel complexes with dimethylglyoxime (DMG) and presumably the extra hydrogen ions then combine with the hydroxyl ions that you have "left over" to make water.
Ohh yes ofc. So ive removed HCl from the overall equation. The method i was given says the product is Ni(HDMG)2 unless there is a typo? So there wouldnt be any extra H+. So im not sure if i can leave it like this??

[Ni(NH3)6](BF4)2 + 2HDMG + 2NH4OH --> 2NH4BF4 + Ni(HDMG)2 + 6NH3 + 2OH
10. (Original post by kiiten)
Ohh yes ofc. So ive removed HCl from the overall equation. The method i was given says the product is Ni(HDMG)2 unless there is a typo? So there wouldnt be any extra H+. So im not sure if i can leave it like this??

[Ni(NH3)6](BF4)2 + 2HDMG + 2NH4OH --> 2NH4BF4 + Ni(HDMG)2 + 6NH3 + 2OH
This equation needs adjusting because you have two negative (hydroxide) ions on the RHS and no charges on the LHS.

DMGH is the anionic form and hence carries a negative charge.
11. (Original post by charco)
This equation needs adjusting because you have two negative (hydroxide) ions on the RHS and no charges on the LHS.

DMGH is the anionic form and hence carries a negative charge.
im confused idk how to "adjust" it . If you say HDMG is negative then the charges balance????

I just need the overall equation to calculate yield
12. (Original post by kiiten)
im confused idk how to "adjust" it . If you say HDMG is negative then the charges balance????

I just need the overall equation to calculate yield
No, to calculate yield yo just need to know the relationship between the nickel complexes on both sides.

One nickel on the LHS makes one nickel on the RHS
13. (Original post by charco)
No, to calculate yield yo just need to know the relationship between the nickel complexes on both sides.

One nickel on the LHS makes one nickel on the RHS
Oh thats good .... but after looking through my notes i need a reaction summary of the whole equation to complete my lab report.

Im just confused about the products side. I know there should be H2O and Ni(HDMG)2 but im not sure about the other products :/
14. (Original post by kiiten)
Oh thats good .... but after looking through my notes i need a reaction summary of the whole equation to complete my lab report.

Im just confused about the products side. I know there should be H2O and Ni(HDMG)2 but im not sure about the other products :/
Perhaps if you posted a method people could break it down for you...
15. (Original post by charco)
Perhaps if you posted a method people could break it down for you...

Weigh out [Ni(NH3)6](BF4)2 (0.25 g) using a crude top loading balance and transfer into a beaker (either 400 cm3 or 600 cm3). Dissolve the sample in deionised water (100 cm3), add concentrated hydrochloric acid (3 cm3 [CARE - FUME HOOD]), then make the total volume of the solution up to ca. 200 cm3 with more deionised water. Cover the beaker with a watch glass (convex side down) and heat the solution to about 70°C using a hotplate. At this temperature, REMOVE THE BEAKER FROM SOURCES OF IGNITION then add dimethylglyoxime (1% solution in ethanol, 30 cm3 [CARE - FLAMMABLE]). Stir the solution continuously with a glass rod and add dropwise dilute (3 M) ammonium hydroxide. The bis(dimethylglyoxime) complex of nickel(II) will begin to precipitate. When you think that precipitation is complete, check by using Universal Indicator Paper to measure the pH, which should be in the range 7-8. Do not dip the paper into the solution (contamination) but instead touch the edge of the paper on the wet side of the glass rod that has been dipped in the solution. If necessary, continue to add dilute ammonium hydroxide until the pH is in the range 7-8. Err on the side of high rather than low pH. Cover the solution with a watch glass (convex side down) and heat the beaker on a hot plate for 15 minutes to obtain a more filterable precipitate.

Cool the beaker and contents in an ice bath and then collect the precipitate by filtration using a Büchner funnel and Büchner flask. Wash the precipitate with deionised water (3 x 25 cm3) then dry with suction. The product may form a sticky filter cake adhered to the filter paper, in which case place the filter paper onto a watch glass and dry briefly in the 60°C oven. Record the weight of product and calculate the percentage yield. Submit a labelled sample (Quote: the sample formula; your name; the date) for assessment. When you have completed the experiment, ensure that all the equipment is clean - soaking in dilute nitric acid should remove all traces of the precipitate.
16. (Original post by kiiten)
Weigh out [Ni(NH3)6](BF4)2 (0.25 g) using a crude top loading balance and transfer into a beaker (either 400 cm3 or 600 cm3). Dissolve the sample in deionised water (100 cm3), add concentrated hydrochloric acid (3 cm3 [CARE - FUME HOOD]), then make the total volume of the solution up to ca. 200 cm3 with more deionised water. Cover the beaker with a watch glass (convex side down) and heat the solution to about 70°C using a hotplate. At this temperature, REMOVE THE BEAKER FROM SOURCES OF IGNITION then add dimethylglyoxime (1% solution in ethanol, 30 cm3 [CARE - FLAMMABLE]). Stir the solution continuously with a glass rod and add dropwise dilute (3 M) ammonium hydroxide. The bis(dimethylglyoxime) complex of nickel(II) will begin to precipitate. When you think that precipitation is complete, check by using Universal Indicator Paper to measure the pH, which should be in the range 7-8. Do not dip the paper into the solution (contamination) but instead touch the edge of the paper on the wet side of the glass rod that has been dipped in the solution. If necessary, continue to add dilute ammonium hydroxide until the pH is in the range 7-8. Err on the side of high rather than low pH. Cover the solution with a watch glass (convex side down) and heat the beaker on a hot plate for 15 minutes to obtain a more filterable precipitate.

Cool the beaker and contents in an ice bath and then collect the precipitate by filtration using a Büchner funnel and Büchner flask. Wash the precipitate with deionised water (3 x 25 cm3) then dry with suction. The product may form a sticky filter cake adhered to the filter paper, in which case place the filter paper onto a watch glass and dry briefly in the 60°C oven. Record the weight of product and calculate the percentage yield. Submit a labelled sample (Quote: the sample formula; your name; the date) for assessment. When you have completed the experiment, ensure that all the equipment is clean - soaking in dilute nitric acid should remove all traces of the precipitate.
The HCl in the first part is just to encourage the loss of ammonia ligands. It is not an important part of the procedure, in that it can be shown as a side reaction. The final addition of ammonia is to remove the acidity of the solution (excess HCl), and remove a pair of hydrogen ions from the dimethylglyoxime

You reaction can be shown as a simple ligand substitution.

[Ni(NH3)6]2+ + 2DMGH + 2OH- --> [Ni(DMG)2] + 6NH3 + 2H2O

The ammonia reacts with the HCl as it leaves the complex ion:

NH3 + HCl --> NH4Cl

All of the other ions are irrelevant, except when calculating the moles of starting material and product.

You can't squash all of this into one equation as there are several things going on.
17. That makes sense, thank you...now the only thing im confused about is that the title of that experiment is:

LIGAND REPLACEMENT REACTION TO FORM A NICKEL(II) BIS(DIMEHTYLGLYOXIME) COMPLEX Ni(HDMG)2

It says the product is Ni(HDMG)2 and not DMG so the equation is wrong UNLESS they made a typo in the title? Im not sure

I know you mentioned the other ions being irrelevant but does it make sense if I keep the BF4 and NH4 in my equation (it helps when i calculate the moles used for each reactant etc.)

So the equation become: (this is assuming the final complex has DMG ligands and not HDMG)

[Ni(NH3)6](BF4)2 + 2HDMG + 2NH4OH --> 2NH4BF4 + Ni(DMG)2 + 6NH3 + 2H2O
18. (Original post by kiiten)
That makes sense, thank you...now the only thing im confused about is that the title of that experiment is:

LIGAND REPLACEMENT REACTION TO FORM A NICKEL(II) BIS(DIMEHTYLGLYOXIME) COMPLEX Ni(HDMG)2

It says the product is Ni(HDMG)2 and not DMG so the equation is wrong UNLESS they made a typo in the title? Im not sure

I know you mentioned the other ions being irrelevant but does it make sense if I keep the BF4 and NH4 in my equation (it helps when i calculate the moles used for each reactant etc.)

So the equation become: (this is assuming the final complex has DMG ligands and not HDMG)

[Ni(NH3)6](BF4)2 + 2HDMG + 2NH4OH --> 2NH4BF4 + Ni(DMG)2 + 6NH3 + 2H2O
The compound itself is usually abbreviated as dmgH2

In this complexation the base removes one hydrogen ions from each dimethylglyoxime to give dmgH-

It is this that forms a planar bis(dmgH) complex with nickel(II)

The negative charge on the deprotonated oxygen then hydrogen bonds to the hydrogen of the other ligand, This happens in two places in the complex, giving a pseudo tetradentate complex which is very stable and very insoluble.

19. Thanks for your help, much appreciated

@charco do you recommend any sites where i can read up on DMG? In m report i need to say why the DMG causes the complex to precipitate (any why its insoluble in aqueous solution). Apparently this is due to a "macrocyclic effect" but i havent heard of that term so need to read up on that too.
20. (Original post by kiiten)
Thanks for your help, much appreciated

@charco do you recommend any sites where i can read up on DMG? In m report i need to say why the DMG causes the complex to precipitate (any why its insoluble in aqueous solution). Apparently this is due to a "macrocyclic effect" but i havent heard of that term so need to read up on that too.

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