# Inorganic Chemistry/Qualitative Analysis/Tests for anions
## Sulphides
### Sodium Nitroprusside

Use sodium nitroprusside. In the presence of sulphide ions a deep violet coloration is formed. [Fe(CN)$_5$NO]$^{4-}$ complex anion formed in the presence of sulphide ions is the cause for this coloration, colouration is important.

### Silver Nitrate

If a sulfide solution is reacted with silver nitrate, it forms a black precipitate of Ag$_2$S.

### Lead Ethanoate

If the substance is soluble use lead acetate solution, if solid add dilute hydrochloric acid and test the gas with lead ethanoate paper. It is prepared by dipping a filter paper for some time in Lead acetate solution.If the salt in question is a sulphide an acrid smell of rotten eggs will be observed and the lead ethanoate will turn black, due to formation of solid black Lead sulfide powder. (CH$_3$COO)$_2$Pb + H$_2$S -> PbS + 2(CH$_3$COOH)

### H$_2$S

The following two points do not fall in place here, kindly arrange them wherever they are best fit:
* When H2S is bubbled through a solution of Sodium sulphite or Sodium bisulphite, a white turbidity of sulfur is seen. What happens is, the S atom in sulphite/bisulphite (Oxidation state +4) and in H2S (Oxidation state -2) have coproportionated to give molecular sulfur in zero oxidation state. The reaction can be viewed as the reverse of a disproportionation reaction.
* When H2S is bubbled through acidified KMnO4 solution, it decolorises and a white turbidity is observed. This is because, the powerful oxidizing agent permanganate ion has oxidized the sulfide ion (-2) to zero oxidation state. In doing so, the permanganate ions themselves reduce to Mn$^{2+}$ ions, thus the purple/violet/pink color of the solution is diluted and eventually fades away as the manganous ions do not impart any color to the solution.

## The nitrite ion

The nitrite ion gives several reactions based on its redox properties.

### I$_3$-

When it reacts with the I$_3$- anion, the nitrite ion reduces itself to nitric oxide. The student should check the oxidation states of nitrogen in both the cases.
NO$_2$$^-$ + 2I$_3$$^-$ + 2H$^+$ --> NO(g) + 3I$_2$(aq) + H$_2$O(l)
The yellow color of the I$_3$- ions changes to a greenish tinge which later turns violet. This observation is made in the presence of an oxidising agent which oxidizes iodine from -⅓ to 0. The O.A. in our case is of course, the nitrite ion.
Before turning violet, a greenish tinge is observed because the increasing concentration of violet color and the decreasing concentration of yellow color during the reaction impart a greenish color.

### Lead/Silver Nitrates

Most lead and silver compounds show much the same characteristics in analysis. However, lead nitrite is soluble while silver nitrite is an insoluble white solid.
Therefore, to test for the presence of nitrite ions we can use silver and lead nitrate. No precipitate will be observed with lead nitrate, but a white precipitate will form when silver nitrate is added.
2NO$_2$$^-$ + Pb$^{2+}$ -> Pb(NO$_2$)$_2$(aq)
NO$_2$$^-$ + Ag$^+$ -> AgNO$_2$(s)

## The cyanide ions
### Ferric ions

When ferric ions are added to a solution of cyanide ions, no specific change in color is observed. However, if we add the ferric ions to the solution after adding thiosulphate or S$_2$$^{2-}$, we get a blood red coloration of ferric thiocyanate.
Ferric thiocyanate is Fe(SCN)$_3$.
Actually, this is a characteristic test for the presence of ferric ions and for thiocyanate ions. Cyanide ions are nowhere directly involved in the formation of ferric(III) thiocyanate. If we review the test procedure for the detection of cyanide ions, we find that we add, according to our choice, either thiosulphate ions or S$_2$$^{2-}$ ions. These ions disproportionate in the presence of cyanide ions, forming thiocyanate ions. It is these thiocyanate ions that eventually react with the added ferric ions to give the blood red complex formation
CN$^-$ + S$_2$O$_3$$^{2-}$ -> SCN$^-$ + SO$_3$$^{2-}$
CN$^-$ + S$_2$$^{2-}$ -> SCN$^-$ + S$^{2-}$
Fe$^{3+}$ + SCN$^-$ -> Fe(SCN)$^{2+}$

### Metallic copper

It is a popular fact that most metals react with either water or dilute acid solution to liberate hydrogen. This is a redox reaction in which the metal acts as a reducing agent, reducing the proton or hydrogen in water to hydrogen gas, oxidizing itself to a metal cation.
A classic example of this type of redox reaction is the reaction of metallic sodium with water. The reaction takes place and liberates enough heat to melt the sodium which then floats on a cushion of hydrogen gas. e.g., 2Na + 2H$_2$O -> H$_2$ + 2NaOH. In terms of the ions involved in the reaction Na -> Na$^{1+}$ + e$^{1-}$ and then 2H$^{1+}$ + 2e$^{1-}$ -> H$_2$. The same reaction occurs with any of the alkali metals in exactly the same way.
However, there are some metals like copper which will not react with either water or dilute acids in the same way. Quantitatively, Standard Electrode Potential is lower than that of hydrogen ions, meaning that more energy is required to form copper ions than the hydrogen ions can provide. [Note to editors --> Is this any better?]
However, if we add metallic copper to a solution containing cyanide ions, copper gives into the greed of forming that ultrastable cyano complex of itself. That is, equilibrium favours the formation of tetracyanocomplexcuprate(I) ions. This is the point - in presence of cyanide ions, copper oxidizes itself to form the +1 metal complex. But then, if an oxidation has occurred so must have a reduction. What is the reduction reaction? The formation of hydrogen (editors - formation from what? proton or water molecule?) is the involved reduction.
We conclude that an aqueous solution of KCN dissolves metallic copper, generally administered as copper turnings through, the formation of tetracyanocuprate(I) ions.
(Note to editors --> Explain this anamlolous behaviour of copper through the quantitative aspects of electrochemistry and chemical equilibrium, short enough to maintain interest.)

#### Note on the tetracyano complex of copper

When H$_2$S is bubbled through the cyano complex of ferrous/ferric/cuprous/cupric ions, no sulphide is precipitated. While with most other metal cations, metal sulphide does get precipitated.
e.g. tetracyanocadmiumate ions give CdS with H$_2$S and dicyanoargentate (silver complex) give Ag$_2$S.
This can be explained on the basis of equilibrium constants - solubility products and formation constants of the concerned sulphides and cyanocomplexes. (note to editors --> plz explain in a little more detail)

## The hypochlorite ion

Adding a solution containing lead ions, such as lead nitrate, to a hypochlorite solution produces a brown precipitate of lead(IV) oxide.
2ClO$^-$ (aq) + Pb$^{2+}$ (aq) → PbO$_2$ (s) + 2Cl$^-$ (aq)
Adding hydrogen peroxide to sodium hypochlorite produces sodium chloride, water and oxygen.
ClO$^-$ (aq) + H$_2$O$_2$ (l) → O$_2$ (g) + H$_2$O (l) + Cl$^-$ (aq)

## The azide ion

On adding triiodide ions, the azide ion oxidizes to nitrogen and decolorisation is observed. . Triiodide ions are reduced to iodide ions, and nitrogen is produced. The colour of solution changes from yellow to colourless and a brisk effervescence is noted.
2N$_3$$^-$ -> 3N$_2$ + 2e$^-$
I$_3$$^-$ + 3e$^-$ -> 3I$^-$

## The acetate ion
### Acidification

Adding strong acid (such as HCl) to a concentrated acetate solution produces acetic acid, which emits a strong smell of vinegar. However, even if acetate is present, the vinegar smell may not be noticeable if the acetate is too dilute.

### Esterification

When a drop of concentrated sulphuric acid and ethanol are added, the distinctive smell of an ester can be observed. This applies to all carboxylic acids but is particularly useful in this case because the two reactants, acetic acid and ethanol, have distinctive smells which are replaced completely by different smell of the ester; in this case, ethyl acetate, which smells not unlike glue or nail polish remover.
CH$_3$COOH + CH$_3$CH$_2$OH -> CH$_3$COOCH$_2$CH$_3$ + H$_2$O
(The sulphuric acid is not shown as it is merely a catalyst, specifically a dehydrating agent.)

### Ferric Chloride

When added to neutral FeCl$_3$ acetates produce a red coloration.
FeCl$_3$ + 3CH$_3$COOH -> (CH$_3$COO)$_3$Fe + 3HCl
Neutral FeCl$_3$ is prepared by taking about an mL of FeCl$_3$ in a test tube and adding about few drops of dil. Na$_{OH}$ to give a permanent reddish brown precipitate.Ferric chloride is added to the test tube till the precipitate has JUST dissolved. The resultant solution is added to the salt solution containing acetate to give a red colouration. The acetate ion is confirmed by adding water to this mixture and warming till formation of red precipitate.A-LEVEL EXPERIMENT.

## Thiosulphates

Thiosulphates are reducing agents. They can reduce I$_3$$^-$ ions to iodide ions which are colorless. However, it should be noted that this is not a confirmatory test as any reducing agent can do this.
2S$_2$O$_3$$^{2-}$ -> S$_4$O$_6$$^{2-}$ + 2e
I$_3$$^-$ +2e -> 3I$^-$

### Hydronium ions

Adding dilute acid to a solution with thiosulphate ions yields a solution with precipitate, elemental sulfur, and release a burnt rubber smell (gaseous sulphur dioxide)
2H$^+$ + S$_2$O$_3$$^{2-}$ -> S$_{(s)}$ + SO$_{2(g)}$ + H$_2$O$_{(l)}$

### Ferric ions

It also produces a momentary violet coloration with ferric ions. --->Expand<--

### Copper Sulphate

With copper sulphate, a white ppt. is obtained which dissolves in excess of thiosulphate to produce a colorless solution through the formation of a stable complex.
CuSO$_4$ + Na$_2$S$_2$O$_3$ -> CuS$_2$O$_3$ + Na$_2$SO$_4$
This test has to be carried out in such a way that initially the concentration of thiosulphate is low which then gradually increases enough to form the complex. The solution containing thiosulphate should be poured into the copper sulphate to prevent the changes being missed.

### Silver Nitrate

This is a very peculiar reaction, because on the surface it seems as if the precipitate obtained is changing colors!
What actually happens is that when we add silver nitrate, a white ppt of silver thiosulphate is formed. This silver thiosulphate disproportionates to give silver sulphide and sulphuric acid, wherein as we know, silver sulphide is black. That is, on adding silver nitrate a white precipitate is obtained which changes to brown and finally to black.
S$_2$O$_3$$^{2-}$ + 2AgNO$_3$ -> 2NO$_3$$^-$ + Ag$_2$S$_2$O$_3$
Ag$_2$S$_2$O$_3$(s) + H$_2$O(l) -> H$_2$SO$_4$(aq) + Ag$_2$S
Furthermore, in the presence of excess of thiosulphate ions the white ppt of Silver thiosulphate dissolves giving a colorless complex.

## Thiocyanate ions

SCN$^-$ ions have been discussed once in the testing of cyanide ions.
In fact, one of the characteristic tests for thiocyanate ions is the same reaction that we discussed over there.

### Ferricthiocyanate

With ferric ions, thiocyanate ions give a blood red coloration of Fe(SCN)$_3$. This test is NOT given by ferrous ions - and hence is confirmatory for the presence of both ferric ions and for thiocyanate ions.
-->Check lingo<--

### Vogel's test

If Co$^{2+}$ ions are added to the thiocyanate ions a blue solution of tettrathiocyanatocobaltate is obtained. This complex on addition mercuric ions gives a Blue crystalline precipitate.
This ppt is NOT Mercuric tetrathiocyanatocobaltate! It is interesting to note that the actual precipitate is an isomer of the expected ppt - it is Cobalt(II) Tetrathiocyanatomercurate(II), that is the metal cations have switched places.
If no effervescence is obtained with dilute HCl, heat the salt with Conc. H$_2$SO$_4$
The following ions belong to this list:

## Fluorides
### Evolution of HF

When we add concentrated sulphuric acid to fluoride ions, they are protonated to give hydrogen fluoride. This is because hydrogen fluoride is a weak acid, its anion F$^-$ can be protonated back to the gas.
F$^-$ + H$_2$SO$_4$ -> HF + HSO$_4$$^-$
The extent to which hydrogen fluoride is produced is dependent on the initial concentration of the sulphuric acid. By Le Chatelier's equilibrium principle, a larger amount of the reactant H$_2$SO$_4$, coming from an increase in concentration, would drive the reaction towards the production of more gaseous hydrogen fluoride.
The question is, how do we know that HF has evolved? When we hold a glass rod at the mouth of the testtube, a white waxy deposit is obtained on the glass rod. This is because of the reaction of HF with Silica. (Glass rod - is of course nearly all silica!).
HF(g) + SiO$_2$(s) -> SiF$_4$(g) + H$_2$0(l)
The silicon fluoride gas is easily hydrolyzed by the water released, and we get hydrogen silicofluoride and hydrogen silicate - the mixture which is the white waxy deposit in question.
3SiF$_4$(g) + 4H$_2$0(l) -> 2H$_2$(SiF$_6$)(s) + H$_4$SiO$_4$(s)
Question: Is is possible to detect HF or F$^-$ using Ca(NO$_3$)$_2$?
2 F$^-$$_{(aq)}$ + Ca(NO$_3$)$_2$$_{(aq)}$ -> CaF$_{2(s)}$ + 2 NO$_3$$^-$$_{(aq)}$

### Oxidation of fluorides

In this particular test, We added conc. H2SO4. H2SO4 is an oxidizing agent. When it is added to a fluoride salt, does it oxidize to molecular fluorine? Let us think for a second. Yes, we accept that Conc. H2SO4 is a pretty strong oxidizing agent, but the question is, is it strong enough to oxidize fluoride ions? The product of oxidation of fluoride ions is molecular fluorine - something which is rarer than anything. Molecular fluorine reacts with almost anything. It causes mayhem wherever it goes. So our answer is no. H2SO4 just can't oxidize fluoride ions into molecular fluorine.
-->Note to editors: This is the purpose of the text - to acquaint the reader with the various facts of chemistry and making them realize their importance in the real life. While this point doesn't exactly point out the high reactivity of fluorine, half a dozen scattered discussions can do enough to hammer this point!
Possible insertions - the problems involved in the electrolytic extraction of fluorine.<--

## Chlorides
### Conc. H$_2$SO$_4$

When concentrated H$_2$SO$_4$ is added, chloride salts evolve colorless HCl gas.
Now the presence of this HCl gas can be detected in two ways. The first is to hold an ammonium hydroxide test tube near the mouth of the original solution. The HCl gas, if evolved, will react with NH4OH to give dense white fumes of NH4Cl.
Otherwise, we can use MnO2. Manganese in +4 state will be reduced to +2, oxidizing the chlorine in HCl to molecular chlorine gas - Cl2. Pale green (pale because of low concentration) Chlorine gas confirms that HCl was evolved. Chlorine gas has a bleaching action and should not be inhaled.

### The Silver Nitrate test

AgCl is a precipitate. So, if a silver nitrate solution is added to a sample of the being tested solution, a white ppt. is obtained. This precipitated Agcl is soluble in an excess of Ammonium Hydroxide solution because of the formation of [Ag(NH3)2]$^+$ complex ions.
if we note carefully, the ammine complex is the same one that is used in the silver mirror test (Tollen's test) for the detection of aldehydes. The reaction, if we recall, was a redox reaction in which the silver (I) is reduced to metallic (0) silver mirror, and the aldehyde being oxidized to an ate ion.

### The Chromyl Chloride test

It should be noted that this test is not for a solution containing the solvated ions. Chromyl chloride test is performed on the salt in solid state.
Solid salt is mixed with Solid Potassium Dichromate and Conc. H$_2$SO$_4$. If the salt is a chloride, on heating deep red vapors of Chromyl Chloride (CrO$_2$Cl$_2$) are evolved.
Some chemists still bear a doubt, because the deep red vapors bear some resemblance to NO$_2$ and Br$_2$ vapors. They can be distinguished, though.
If these Chromyl Chloride vapors are passed through a dilute NaOH solution, it turns yellow. Most of us might have already guessed that the color is due to the formation of the Chromate ions. Indeed, the guess is true. This is then how we resolve our doubts and say that the salt is a chloride.
It is not unwise to note here that most chromates are yellow in color, with the notable exceptions being Ag$_2$CrO$_4$ and Hg$_2$CrO$_4$. Both are red!
Predominantly covalent chlorides like HgCl$_2$ and alkyl chlorides do not offer this test.
The real confirmation is adding acetic acid followed by lead acetate to the yellow solution which gives a pale yellow precipitate of lead chromate.

## Bromides/Iodides
### Conc. H2SO4

With conc. H2SO4, the florides/chlorides gave HF/HCl respectively. However, the bromides and iodides are oxidized to molecular bromine and iodine respectively!
So, in case of the reaction of iodides/bromides with Conc. H2SO4, bromides give deep brown vapors of Br2, while iodides give violet vapors of iodine.

### Silver nitrate solution

With silver nitrate solution, bromides give a pale yellow ppt. of AgBr which is partially soluble in NH4OH. Iodides give a yellow ppt of AgI which is insoluble in ammonium hydroxide.
-->Explain this point with reference to Solubility Product and formation constant of the concerned compounds. We need not give a numerical explanation! We just want to emphasize that Inorganic Chemistry is not plain mugging up.

### The Chloroform test

Take the solution and add an equal amount of chloroform/CHCl3 to it. Add chlorine water and shake vigorously. Watch the change in color of the non-aqueous layer.
If iodide/bromide ions are present, they would be oxidized to Iodine/Bromine respectively. Iodine would impart a violet color to the non-aqueous layer while if bromine is created it will give out a reddish color to the non-aqueous layer.
The same reaction will also occur if dichloromethane is added in replacement of the chloroform.

## Oxalate ions

The reactions for oxalate ions- 2MnO$_4$- + 16H+ + 5C$_2$O$_4$$^{-2}$ ---------> 2Mn$^{+2}$ +8H$_2$O + 10 CO$_2$ Oxalate ions are C$_2$O$_4$$_{--}$

### Conc. H$_2$SO$_4$

Concentrated sulphuric acid decomposes oxalate ions into carbon monoxide and carbon dioxide. Readers shouldn't take long to realize that the oxalate ions have decomposed through a disproportionation reaction. If the evolution of the two gases is confirmed, we can say that sulphuric acid reacted with oxalate ions.
If the evolved gaseous mixture is passed through limewater, it will turn milky and finally colourless. Carbon Monoxide in the mixture will burn with a blue flame to produce carbon dioxide. The two gases are confirmed in this way, confirming the presence of the oxalate ion.

### Acidified Potassium Permanaganate solution

Acidified KMnO4 is an oxidizing agent. It will oxidize the oxalate ions into carbon dioxide. In the process, it will decolorize because of the reduction of the violet manganate ions into colorless manganous ions.
Now the question is, how do we know that the decolorization was due to oxalate ions? KMnO4 solution is decolorized each time the manganate ions oxidize ANYTHING.
The answer lies again, in the detection of the carbondioxide gas. Once the gas is confirmed, oxalate ions have been virtually confirmed.

### Calcium chloride solution

Aqueous solution of oxalate ions give a white precipitate of calcium oxalate which is insoluble in acetic acid.

## Nitrates
### Decomposition by Heat or Strong Acid

All nitrate salts are decomposed by heat or dilute sulphuric acid to produce reddish brown fumes of nitrogen dioxide.
2Cu(NO$_3$)$_2$(aq) -> 2CuO(aq) + 4NO$_2$(g) + O$_2$(g)

### Copper chips

To the Aqueous solution of the salt when we added conc. sulphuric acid, and then copper chips are added, Dark brown solution forms. It forms due to acid of sulphur.

### The brown ring test

When a freshly prepared acidified solution of ferrous sulphate is added to a solution of nitrate ions, in the presence of concentrated H$_2$SO$_4$, a deep brown ring is formed.
This deep brown ring is due to the formation of the complex sphere [Fe(H$_2$O)$_5$NO]$^{2+}$. In this form, the Fe ion is in both its +1 and +2 oxidation states, and oscillates between the two. This means that even a slight disturbance in the medium will disturb the complex. It is for this reason that the acidified ferrous sulphate is added drop-wise and allowed to roll down the walls of the test tube, this minimises any disturbance that would destroy the ring. It reacts upon contact with the medium, and forms the mentioned complex, but only at the edges, forming the characteristic brown ring.
The oscillation occurs in the following two ways
i) The NO ion loses an electron to the Fe2+ ion forming Fe$^{1+}$ and NO$^{1+}$ NO -> NO$^{1+}$ + e$^{1-}$ Fe$^{2+}$ + e$^{1-}$ -> Fe$^{1+}$
ii) The Fe$^{1+}$ returns the electron to the NO$^{1+}$ forming NO and Fe$^{2+}$ Fe$^{1+}$ -> Fe$^{2+}$ + e$^{1-}$ NO$^{1+}$ + e$^{1-}$ -> NO
It is worth noting that it is very rare to find iron in its (I) oxidation state, (II) and (III) are much more common

### Reduction to Ammonia

The NItrate ion can easily be reduced to ammonia with either Devardas Alloy or Aluminium Foil. The aluminium is a very powerful reducing agent, and this combined with heating causes the nitrate ions to form ammonia gas. This can be tested for by holding a piece of damp red litmus paper over the end of the test tube. The ammonia will form alkaline ammonium ions in the water and turn the paper blue.
4NO$_3$$^-$(aq) + 6H$_2$0(l) -> 4NH$_3$(g) + 9O$_2$(g)
Aluminium powder is not shown as it merely catalyses the reaction.

## Sulphates
### Barium Chloride

The addition of a sulphate to a barium chloride solution produces a white precipitate of barium sulphate. This precipitate is very insoluble, not dissolving in any common laboratory reagents.
Magnesium Sulphate + Barium Chloride -> Barium Sulphate + Magnesium Chloride
MgSO$_4$(aq) + BaCl$_2$(aq) -> BaSO$_4$(s) + MgCl$_2$(aq)

### Lead Nitrate/Lead Acetate

A similar reaction to the above, a white precipitate of lead sulphate is formed. Again this is very difficult to dissolve in common reagent like water, but dissolves in Sodium hydroxide and to a much extent in Ammonium acetate solution.
PbNO$_3$(aq) + MgSO$_4$(aq) -> PbSO$_4$(s) + MgNO$_3$(aq)
(CH$_3$COO)$_2$Pb(aq) + K$_2$SO$_4$(aq) -> PbSO$_4$(s) + 2CH$_3$COOK(aq)

## Manganate Ion
### H$_2$S test

The oxidising properties of potassium manganate allow it to oxidise hydrogen sulphide rapidly from oxidation state (II) to 0, reducing itself to manganese ions. These do not impart any colour to the solution, yet the sulphur precipitate forms a white cloud as the purple colouration of manganate ions disappears.
2S$^{2-}$ -> S$_2$ + 4e
MnO$_4$$^{2-}$ + 4e$^-$ -> Mn$^{2+}$ + 2O$_2$

## Borates
### Ethyl Alcohol

on adding ethyl alcohol we get ethyl borate on expose to flame it gives greenish yellow colour flame.

## Ortho-Phosphates
### Ammonium Molybdate

When added to phosphate ions, warmed with nitric acid, forms a canary yellow precipitate of ammonium phosphorous molybdate.The same yellow canary colour is also given by arsenic salts after treating with Nitric acid. Confirmatory test for presence of phosphates. PO$_{43-}$ + 12(NH$_4$)$_2$MoO$_4$ + 21HNO$_3$ + 3H$^+$ -> (NH$_4$)$_3$PO$_4$·12MoO$_3$↓ + 21NH$_4$NO$_3$ + 12H$_2$O

## Carbonates and Bicarbonates

In the presence of carbonate (CO$_3$$^{2-}$) or bicarbonate (HCO$_3$$^-$) ion, the addition of concentrated strong acid (e.g. HCl) causes the evolution of CO$_2$ gas, resulting in fizzing or bubbling. To determine which is present, MgSO$_4$ is added to the salt solution. A white precipitate at low temperatures indicates carbonate while precipitation upon heating confirms bicarbonate.