Why is AuCl4 an anion?

[Redriq1]

Hello there can anybody provide an explanation of why AuCl4 is an anion when Au [III] is dative covalent bonded with 4 Chlorine ligands, i don't understand why AuCl4 is an anion when there is no ionic bonding happening here. Thanks!!!

 

[Borek]

No idea why you think it should be otherwise, you start with somethin that is +3 charged and combine it with 4 -1 charges, no way it can produce anything but anion.

Lack of ionic bonding doesn't matter - SO₄^2- is also an anion despite not having ionic bonding.

Or is your question why it doesn't end at neutral AuCl₃? Well, it can, it is called auric chloride then. But even then gold is surrounded by four ligands, that's they way coordination chemistry works, it is just that the fourth ligand is either shared with a neighbor, or replaced by water molecule, in both cases it ends with a zero charge.

I feel like there is some misconception here.

 

[Redriq1]

No idea why you think it should be otherwise, you start with somethin that is +3 charged and combine it with 4 -1 charges, no way it can produce anything but anion.

Lack of ionic bonding doesn't matter - SO₄^2- is also an anion despite not having ionic bonding.

Or is your question why it doesn't end at neutral AuCl₃? Well, it can, it is called auric chloride then. But even then gold is surrounded by four ligands, that's they way coordination chemistry works, it is just that the fourth ligand is either shared with a neighbor, or replaced by water molecule, in both cases it ends with a zero charge.

I feel like there is some misconception here.

Ohhh i get it now, yeah i think i misunderstood it earlier.
I do have one more question though why was the Gold [III] was attracting 4 chlorine in the first place why not just 3 to even the charge? (im talking about AuCl4)
Is it because it has something to do about Gold being a transition metal?

 

[Mayhem]

Ohhh i get it now, yeah i think i misunderstood it earlier.
I do have one more question though why was the Gold [III] was attracting 4 chlorine in the first place why not just 3 to even the charge? (im talking about AuCl4)
Is it because it has something to do about Gold being a transition metal?

Both AuCl₃ and [AuCl₄]^- exist.

Chemistry principles get funky when you get to the third row metals (where Au sits) so I can't comment on why exactly, but it would appear that [AuCl₄]^- is prepared under rather extreme chemical conditions (aqua regia), where an excess of Cl^- and H⁺ are present. So it is "neutral" in that sense, as it nominally speciates as H[AuCl₄].

 

[symbolipoint]

As if from chloroauric acid? HAuCl₄

You can account the the "oxidation" states of every element in the formula and find what is that of Au part.
(I might or might not do that myself and post.)
(In fact, I do that now here.)

I could need some help with the formatting, but I try to show:
H AuCl₄

H Au Cl ₄
+1 x -1*4=-4


1+x-4=0

x=4-1

x=3


The amount of charge on the Au is +3.

edit: Let me try to put better spacing in there and hope make it easier to read.. if not I will then go back later and try something else.

 

[Astronuc]

Adding to what others mentioned, Cl is an anion, so one has a cation (Au) surrounded by 3 or 4 Cl atoms. H⋅Au(III)Cl₄, has 4 Cl anions surrounding an Au cation. Any one of the 4 Cl atoms can lose an electron such that the complex becomes negative.

Something I just learned: "Au(III)Cl₄ is used in the Miller process and Wohlwill process for refining gold. It is a key gold compound produced by dissolving gold in aqua regia (HNO₃ + 3 HCl)."
https://en.wikipedia.org/wiki/Chloroauric_acid
https://en.wikipedia.org/wiki/Aqua_regia
https://en.wikipedia.org/wiki/Miller_process
https://en.wikipedia.org/wiki/Wohlwill_process
I could not find any information on fluoroauric acid, which would be very nasty.

I could not find a comparable Ag compound, but there is a Cu compound (2-),
Tetrachlorocuprate(II) (CuCl₄^2-} with common salts of ammonium or potassium.

 

[Mayhem]

As if from chloroauric acid? HAuCl₄

You can account the the "oxidation" states of every element in the formula and find what is that of Au part.
(I might or might not do that myself and post.)
(In fact, I do that now here.)

I could need some help with the formatting, but I try to show:
H AuCl₄

H Au Cl ₄
+1 x -1*4=-4


1+x-4=0

x=4-1

x=3


The amount of charge on the Au is +3.

Granted, this approach is only strictly valid if you have knowledge of the bonding, but it is typically always valid for complexes of monodentate ligands.