Transition Metal Ions

Hii,

Guys, I'm just confused because I can't find any straightforward answers according to my exam board, but I was wondering how transition metal complexes/ions react with silver nitrates? Like what colour precipitates do they make? What complexes are normally formed?

I'd appreciate if someone could tell me the answers in relation to Fe (III), Fe(II) and Cu(II) and Al (III) (I'm away Al is not a trasnsitoin metal) transition metal complexes, as those are the ones covered by my exam board.

I really apprecaite the help!!

Reply 1

charco

Study Forum Helper

Original post by MlightOr

Hii,
Guys, I'm just confused because I can't find any straightforward answers according to my exam board, but I was wondering how transition metal complexes/ions react with silver nitrates? Like what colour precipitates do they make? What complexes are normally formed?
I'd appreciate if someone could tell me the answers in relation to Fe (III), Fe(II) and Cu(II) and Al (III) (I'm away Al is not a trasnsitoin metal) transition metal complexes, as those are the ones covered by my exam board.
I really apprecaite the help!!

Nitrates do not make precipitates with any cations. All nitrates are soluble.

Silver makes precipitates with many anions. The only soluble silver salts are the nitrate and ethanoate.
Silver salts are usually white, but silver iodide is yellow, silver bromide is creamy coloured (off-white). Silver chromate is red and sometimes used as the indication of the end of a titration in a chloride determination.

(edited 1 year ago)

Reply 2

Pigster

Original post by MlightOr

Hii,
Guys, I'm just confused because I can't find any straightforward answers according to my exam board, but I was wondering how transition metal complexes/ions react with silver nitrates? Like what colour precipitates do they make? What complexes are normally formed?
I'd appreciate if someone could tell me the answers in relation to Fe (III), Fe(II) and Cu(II) and Al (III) (I'm away Al is not a trasnsitoin metal) transition metal complexes, as those are the ones covered by my exam board.
I really apprecaite the help!!

I am very sleepy, so I might have the wrong end of your stick, but...

If you add Fe2+(aq) to AgNO3, you'll end up with a redox equilibrium making some Fe3+(aq) and Ag - which will appear as a (probably) black ppt. Ag+ is not a strong enough oxidising agent to push any of the other ions you stated into higher oxidation states.

Reply 3

UtterlyUseless69

Original post by MlightOr

Hii,
Guys, I'm just confused because I can't find any straightforward answers according to my exam board, but I was wondering how transition metal complexes/ions react with silver nitrates? Like what colour precipitates do they make? What complexes are normally formed?
I'd appreciate if someone could tell me the answers in relation to Fe (III), Fe(II) and Cu(II) and Al (III) (I'm away Al is not a trasnsitoin metal) transition metal complexes, as those are the ones covered by my exam board.
I really apprecaite the help!!

This sounds like you are doing AQA.

For Ag(I), you only need to know the colours of the halide precipitates (which Charco has given you) and that in excess ammonia (concentration dependent), some silver(I) compounds dissolve to form the linear, colourless [Ag(NH3)2]^+ complex.

For Fe(II), Fe(III), Cu(II) and Al(III), you need to know their reactions with NaOH, NH3 (aq) and Na2CO3.

With NH3 and NaOH, Fe(II) forms a green precipitate (Fe(H2O)4(OH)2) that is insoluble in excess. With Na2CO3, it forms green FeCO3.

With NH3 and NaOH, Fe(III) forms an orange/brown precipitate (Fe(H2O)3(OH)3) that is insoluble in excess. With Na2CO3, it also forms iron(III) hydroxide, but liberates CO2 as Fe(III) forms a fairly acidic hexaaqua complex.

With NaOH, Cu(II) forms a pale blue precipitate of Cu(OH)2(H2O)4 that is insoluble in excess. The same precipitate initially forms with NH3, but it dissolves in excess to give the royal blue [Cu(NH3)4(H2O)2]^2+ complex. Na2CO3 leads to the precipitation of blue green “copper carbonate” (at A level, you learn the formula as CuCO3 - in reality, that isn’t quite true).

With NaOH, Al(III) initially forms a white precipitate (Al(OH)3(H2O)3), but in excess, this dissolves to give the colourless, tetrahedral [Al(OH)4]^- complex. With NH3, it forms also aluminium hydroxide, but it does not dissolve in excess. Na2CO3 similarly gives aluminium hydroxide, but as with Fe(III), due to the acidity of the hexaaqua complex, CO2 is liberated.

You should also be aware of the colours of different vanadium ions, too:

V(II) is lilac
V(III) is green
VO^2+ is blue
VO3^- and VO2^+ are yellow

(edited 1 year ago)

Reply 4

MlightOr