Analysis of Acidic Radicals
Analysis of Acidic RadicalsI have often faced queries related to this topics and this forms a very important topic from engineering entrance perspective. And there has not been a good resource to master this topic .So i will be writing posts on this topic .Earlier a full fledged question was asked in IIT JEE and often it was the backbone of all chemistry fundoo guys .
So lets start with this topic . Queries are invited if you have any problems . You all can discuss in the organic chemistry forum . Now IITJEE forms are out and students should cover all the easy topics.
Acidic radicals are categorised into three groups. There is no as such scheme which permits the separation of the common anions into major groups. The classification may be studied in two parts :
1. Those involving the identification by volatile products obtained on treatment with acids, and
2. Those dependent upon reactions in solution.
The part (1) is subdivided into
(i) Gases evolved with dilute HCl or dil H2SO4 and
(ii) Gases or vapours evolved with conc. H2SO4.
The part (2) is subdivided into
(i) Precipitation reactions and
(ii) Oxidation and reduction in solution
Group Category of Acid Radicals
Group I
Radicals which are detected by dilute H2SO4 or dilute HCl, by liberating a gas
(i) Carbonate (ii) Sulphite
(iii) Sulphide (iv) Nitrite
(v) Acetate
Group II
Radicals which are detected by concentrated H2SO4
(i) Chloride (ii) Bromide
(iii) Iodide (iv) Nitrate
(v) Oxalate
Group III
Radicals which do not give any characteristic gas with dilute and concentrated H2SO4
(i) Sulphate (ii) Phosphate
(iii) Borate (iv) Fluoride
Identification of Acid Radicals
Group : I
Group acidic radicals : CO32–, SO32–, S–2, NO2–
Group reagent : dil. HCl or dil H2SO4
Procedure : Take some quantity of substance and add dil. HCl or dil. H2SO4.
Observation :
(i) Brisk efferrescence in cold with evolution of colourless and odourless gas – Carbonate ion (CO32–)
(ii) A colourless gas with suffocating odour having smell of burning sulphur – Sulphite ion (SO32–)
(iii) A colourless gas with smell of rotten eggs – Sulphide ion (S–2)
(iv) A light brown gas – Nitrite ion (NO2–)
3HNO2 —® H2O + 2NO + HNO3
3NO + O2 —® 2NO2 (Brown)
Confirmatory Test :
1. For Carbonate ion (CO32–) : If CO2 gas formed during above procedure, is passed through lime water; the lime water turns milky with the formation of calcium carbonate, confirms CO32– ion.
Salt (Na2CO3) + H2SO4 —® Na2SO4 + H2O + CO2
Further addition of excess of CO2 through lime water, milky solution turns colourless due to the formation of calcium bicarbonate.
2. For Sulphite ion (SO32–) : If SO2 gas is a result of group analysis then it gives suffocating smell of burning sulphur. When acidified potassium dichromate paper is exposed to the gas it attains green colour due to the formation of chromic sulphate, confirms SO32– ion.
Salt (Na2SO3) + H2SO4 —® Na2SO4 + H2O + SO2
3.For Sulphide ion (S–2) : If dil. H2SO4 on reaction with salt gives H2S gas, which on exposure to this gas the lead acetate paper turns black due to the formation of lead sulphide, confirms S–2 ion.
Salt (Na2S) + H2SO4 —® Na2SO4 + H2S
4. For Nitrite ion (NO2–) : When salt containing nitrite ion is treated with dilute H2SO4, it yields a colourless gas (NO) which in contact with oxygen of the air becomes brown due to the formation of NO2 gas. This NO gas in dil. FeSO4 solution gives brown coloured salt.
And when a mixture of iodide and nitrite is acted upon by dil. H2SO4, the iodide is decomposed giving violet vapours of iodine.
2NaNO2 (Salt) + H2SO4 —® Na2SO4 + 2HNO2
2KI + H2SO4 —® K2SO4 + 2HI
5.For Acetate ion (CH3COO–) : If salt is with acetate ion then on reaction with dil. H2SO4 decomposes to give acetic acid vapours which possess smell of vinegar. These acetate salt (soluble in water) when treated with neutral FeCl3 solution gives blood red colouration of ferric acetate.
2CH3COONa (Salt) + H2SO4 —® 2CH3COOH + Na2SO4
Group : II
Group acidic radicals : Cl–, Br–, I–, NO3–, C2O42–
Group reagent : Conc. H2SO4
Procedure : Take salt and add conc. H2SO4 and warm gently.
Observation :
(i)Colourless gas with pungent smell which fumes in air. Addition of pinch of MnO2 in the solution gives pale green gas. – Chloride ion (Cl–)
NaCl (Salt) + H2SO4 —® NaHSO4 + HCl
MnO2 + 4HCl —® MnCl2 + Cl2 + 2H2O
(ii) Reddish brown fumes which intensify on addition of MnO2. Bromide ion (Br–)
NaBr (Salt) + H2SO4 —® NaHSO4 + HBr
2HBr + H2SO4 —® Br2 + 2H2O + SO2
2NaBr (Salt) + MnO2 + 3H2SO4 —® 2NaHSO4 + MnSO4 + 2H2O + Br2
(iii) Violet pungent fumes evolved which may condense as black spots on the cooler part of test tube. These violet fumes intensity on addition of MnO2 – Iodine ion (I–)
2KI (Salt) + 2H2SO4 —® 2KHSO4 + 2HI
2HI + H2SO4 —® I2 + SO2 + 2H2O
2KI (Salt) + MnO2 + 3H2SO4 —® 2KHSO4 + MnSO4 + 2H2O + I2
(iv) Light brown fumes of NO2 which intensify on adding copper turning – Nitrate ion (NO3–)
NaNO3 (Salt) + H2SO4 —® NaHSO4 + HNO3
4HNO3 —® 2H2O + 4NO2 + O2
Cu + 4HNO3 —® Cu(NO3)2 + 2NO2 + 2H2O
(v) Colourless, odourless gas which burns with blue flame at mouth of tube – Oxalate ion (C2O42–)
Na2C2O4 (Salt) + H2SO4 —® Na2SO4 + H2C2O4
H2C2O4 —® CO + CO2 + H2O
Confirmatory Test
1. For Chloride ion (Cl–) : If salt contains chloride ion than it evolves HCl gas on reaction with conc. H2SO4.
(i) The HCl gas so evolved when reacted with NH4OH gives fumes of ammonium chloride.
(ii) HCl gas also gives curdy white precipitate of silver chloride with silver nitrate solution.
(iii) When chloride salt is directly heated with conc. H2SO4 in presence of K2Cr2O7, deep red vapours of chromyl chloride are evolved. This test is known as chromyl chloride test.
NaCl (Salt) + H2SO4 —® NaHSO4 + HCl
K2Cr2O7 + 2H2SO4 —® 2KHSO4 + 2CrO3 + H2O
When vapours of chromyl chloride are passed through NaOH solution, solution becomes yellow due to the formation of sodium chromate, which on reaction with CH3COOH and Pb(CH3COO)2 gives a yellow precipitate of lead chromate.
Note : Test is not given by HgCl2, SnCl2, AgCl, PbCl2.
2. For bromide ion (Br–) :
(i) The bromide salt which is under analysis gives pale yellow precipitate of AgBr on reaction with NH4OH. If NH4OH is taken in excess then the soluble complex of silver is formed.
(ii) Bromide salt when treated with dil. H2SO4, CHCl3 (or CCl4) and chlorine water, chlorine is able to displace bromine due to which Br2 gas is liberated. This Br2 gas dissolves in CHCl3 or CCl4 forming brown layer.
2KBr (Salt) + Cl2 —® 2KCl + Br2
Br2 + Chloroform —® Brown layer
3. For Iodide ion (I–) :
(i) The salt if with iodide ion when treated with conc. H2SO4 evolves iodine vapour which on reaction with starch produce blue colour.
I2 + Starch —® Blue colour
(ii) The iodide salt solution gives yellow precipitate of AgI with AgNO3 solution which is insoluble in NH4OH.
(iii) The iodide salt is treated with chlorine water, chlorine replaces iodine from salt. The iodine thus evolved dissolves in CHCl3 or CCl4 layer giving violet colour.
2KI (Salt) + Cl2 —® 2KCl + I2
I2 + CHCl3 —® Violet layer
4. For Nitrate ion (NO3–) :
(i) Brown ring test : If salt under analysis contains nitrate ion, to confirm it, aqueous solution of the salt is treated with freshly prepared solution of ferrous sulphate and conc. H2SO4, a brown ring complex is formed.
(NaNO3) Salt + H2SO4 —® NaHSO4 + HNO3
6FeSO4 + 2HNO3 + 3H2SO4 —® 3Fe2(SO4)3 + 4H2O + 2NO
(ii) Salt of nitrate when reacted with Zn or Al in presence of conc. NaOH evolves NH3 gas which can be detected by its characteristic odour.
Zn + 2NaOH —® Na2ZnO2 + 2H
or Al + NaOH + H2O —® NaAlO2 + 3H
NaNO3 (Salt) + 8H —® NaOH + 2H2O + NH3
Group :III
Group acidic radicals : SO42–, BO33–, PO43– and F–
Group reagents
(i) BaCl2 – (For SO42–)
(ii) C2H5OH and conc. H2SO4 – (For BO33–)
(iii) Conc. HNO3 and ammonium molybdate – (For PO43–)
(iv) Sand and conc. H2SO4 – (For F–)
Procedure & Observation : Take salt and add above reagents one by one to observe the result.
(i) For Sulphate ion : Take salt and add BaCl2, if salt containing SO42– then precipitate of Barium sulphate (BaSO4) is obtained which is insoluble in conc. HNO3.
Note : If sulphate salt is of silver or lead then on reaction with BaCl2 it will give AgCl or PbCl2. So to avoid this BaNO3 can be used for instead of BaCl2 to analyse SO42– ion .
In next article i will deal with Analysis of Base Radicals.

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so difficult
REALLY BRAINSTORMING……
it’s been long since you have said that you will be writing on for the basic radicals!
will do it soon
superb blog & writing skills. you make this look easy lol. Keep up the great work I’ll be back to read more of your posts later my friend!
thanking you to provide this for us.
thanx it was really helpful