Qualitative Analysis

September 6, 2022 admin Off Uncategorized

Taking too long?

Reload document

Open in new tab

Qualitative Analysis

Welcome to this colourful zone. Visualize colours varying from white to red, black to golden and, what not. Be ready to face undesirable odour and corrosion too. Yes, we are talking about this chapter Qualitative Analysis where the radicals are ready to decode themselves after very planned and systematic activity.

In this chapter you will find all the clues and confirmatory test that will help you detect the radicals and hence the salt.

1. IIT—JEE Syllabus

Systematic analysis of various cations and anions excluding interfering radicals.

2. Introduction

The purpose of chemical analysis is to establish the composition of naturally or artificially manufactured substances. For the purpose of systematic qualitative analysis cations are classified into five groups on the basis of their behaviour with some reagents. Similarly the anions are also classified as class A or B depending on their behaviour with certain reagents.

3. Classification Of Anions

Methods available for the detection of anions are not much systematic as described for the detection of cations. Furthermore anions are classified essentially on the basis of process employed.

3.1 Class A

Take a small amount of the substance and add a little volume of dil. HCl (or) dil. H₂SO₄. Observe the reaction in cold. Warm the contents gently and infer the reactions.

Observation	Inference	Confirmatory Tests
Brisk effervescence in cold with evolution of colourless and odourless gas	CO₃^2– (carbonate)	Pass the gas in a test tube containing small quantity of lime water. It turns milky.
A colourless gas with suffocating odour having smell of burning sulphur	SO₃^2– (sulphite)	Moisten a piece of filter paper with acidified K₂Cr₂O₇ and put it on the mouth of the test tube. It turns green.
A colourless gas with smell of rotten eggs	S^2– (sulphide)	Moisten a piece of filter paper with lead acetate solution and place it on the mouth of the test tube. It turns black.
A light brown gas	NO₂^– (nitrite)	a) Pass the evolved gas through FeSO₄ solution. It turns brown. b) Mix the given salt (or) mixture with a little of KI and dil. H₂SO₄. Evolution of violet vapours c) Place a piece of filter paper on the mouth of test tube moistened with solutions of starch, KI and acetic acid. It turns blue.
Colourless vapours with the smell of vinegar	CH₃CO₂^– (Acetate)	a) To the aqueous solution of substance add neutral FeCl₃ solution. A blood red colour is formed. b) Rub the moistered salt (or) mixture with dry oxalic acid. Smell of vinegar is obtained.

Take a small amount of the substance and add some conc. H₂SO₄ warm gently, observe the changes and draw interference as follows.

Observation	Inference	Confirmatory Tests
Colourless gas with pungent smell which fumes in air. Add a pinch of MnO₂ in the solution and a pale green gas is evolved.	Cl^– (chloride)	a) Bring a glass rod dipped in NH₄OH on the mouth of the test tube; white fumes are formed. b) Bring a glass rod dipped in AgNO₃ solution on the mouth of the test tube; white curdy precipitate is formed on the rod. c) To the substance in a dry test tube add three times its weight of powdered K₂Cr₂O₇ and conc. H₂SO₄. Heat the contents. Red vapours are evolved. Pass the vapour in a test tube containing NaOH solution. A yellow precipitate appears.
Reddish brown fumes which intensify on addition of MnO₂. Vapours passed in water make it yellow	Br^–(Bromide)	a) Take aqueous extract of the substance (or) extract with dil. HNO₃ and add. AgNO₃ solution. A light yellow precipitate appears. b) To a small amount of the substance add dil. H₂SO₄. Warm and cool. Add 1 mL of CHCl₃ (or) CCl₄ and then chlorine water with constant shaking. The chloroform layer becomes orange brown.
Violet pungent fumes evolved which may condense as black specks on the cooler parts of the test tube. The violet fumes intensify on addition of MnO₂.	I^– (Iodide)	a) Place a piece of filter paper moistered with starch solution on the mouth of the test tube. The paper turns blue. b) Take aqueous extract of the substance and add AgNO₃ solution. Yellow precipitate is formed which is insoluble in NH₄OH solution. c) To the small amount of the substance add dil. H₂SO₄ and 1 mL of either CHCl₃ (or) CCl₄ and then Cl₂ water with constant shaking. Chloroform layer attains violet colour.
Light brown vapours having pungent smell. It intensifies on adding copper turnings.	NO₃^– (Nitrate)	Take an aqueous extract of the substance in a test tube and add freshly prepared FeSO₄ solution. Add conc. H₂SO₄ by the side of the test tube without disturbing the solution. A brown ring is formed at the junction of two liquids.
Colourless, odourless gas which burns with blue flame at the mouth of the test tube and turns time water milky	C₂O₄^2– (Oxalate)	Take the substance, add dil. H₂SO₄ and heat till there are no more effervescences. Now add MnO₂ (solid) and brisk effervescences is obtained.

Let us study the reactions in detailed now.

Class A (i): Anions which evolve gases on reaction with dil. HCl/dil. H₂SO₄.

It includes – CO₃^2-, SO₃^2-, S^2-, NO₂^–, CH₃COO^–, S₂O₃^2-

Carbonate (CO₃^2-) :
i) Dilute HCl : gives effervescence, due to the evolution of carbon dioxide

CO₃^2- + 2H⁺ ¾® CO₂ + H₂O

The gas gives white turbidity with lime water and baryta water

CO₂ + Ca²⁺ + 2OH^– ¾® CaCO₃ ¯ + H₂O

CO₂ + Ba²⁺ + 2OH^– ¾® BaCO₃ ¯ + H₂O

On prolonged passage of carbon dioxide in lime water, the turbidity slowly disappears due to the formation of soluble hydrogen carbonate.

CaCO₃ ¯ + CO₂ + H₂O ¾® Ca(HCO₃)₂

ii) Barium chloride or Calcium chloride solution : White ppt of barium or Calcium carbonate is obtained, which is soluble in mineral acid.

CO₃^2- + Ba²⁺ ¾® BaCO₃ ¯

CO₃^2- + Ca²⁺ ¾® CaCO₃ ¯

iii) Silver nitrate solution : White ppt. of silver carbonate is obtained.

CO₃^2- + 2Ag⁺ ¾® Ag₂CO₃¯

The ppt. so obtained is soluble in nitric acid and in ammonia and the ppt. becomes yellow or brown on addition of excess reagent and same may also happen if the mixture is boiled, due to the formation of silver oxide

Ag₂CO₃¯ ¾® Ag₂O ¯ + CO₂

Sulphites (SO₃^2-) :
i) Dilute HCl or Dilute H₂SO₄ : decomposes with the evolution of sulphur dioxide

SO₃^2- + 2H⁺ ¾® SO₂ + H₂O

The gas has suffocating odour of burning sulphur.

ii) Acidified potassium dichromate solution : Turns filter paper moistened with acidified potassium dichromate solution, green due to the formation of Cr³⁺

SO₂ + K₂Cr₂O₇ + H₂SO₄ ¾® K₂SO₄ + Cr₂(SO₄)₃ + H₂O

green

iii) Lime water : On passing the gas through lime water, a milky ppt is formed.

SO₂ + Ca(OH)₂ ¾® CaSO₃ ¯ + H₂O

milky

Precipitate dissolves on prolonged passage of the gas, due to the formation of hydrogen sulphite ions.

CaSO₃ ¯ + SO₂ + H₂O ¾® Ca(HSO₃)₂.

iv) Barium chloride or Strontium chloride solution : Gives white ppt of barium or strontium sulphite.

Sulphide (S^-2) :
i) HCl or Dil. H₂SO₄ : Pungent smelling gas (smell of rotten eggs) H₂S is evolved

S^2- + 2H⁺ ¾® H₂S

ii) Turns lead acetate paper black

(CH₃COO)₂Pb + H₂S ¾® PbS ¯ + 2CH₃COOH

black

iii) Gives yellow ppt.with CdCO₃

Na₂S + CdCO₃ ¾® CdS¯ + Na₂CO₃

Yellow

Na₂S + Na₂[Fe(CN)₅NO] ®

iv) Silver nitrate solution: black ppt. of silver sulphide insoluble in cold but soluble in hot dil. nitric aicd.

S^2– + 2Ag⁺ ¾® Ag₂S¯

v) Sodium nitroprusside solution: Turns sodium nitroprusside solution purple

Na₂S + Na₂[Fe(CN)₅NO] ¾®

Nitrites (NO₂^–):
i) HCl and Dil. H₂SO₄ : Adding to solid nitrite in cold yields pale blue liquid (due to the presence of free nitrous acid (HNO₂) or its anhydride N₂O₃) & the evolution of brown fumes of nitrogen dioxide, the latter being largely produced by combination of nitric oxide with the oxygen of air

NO₂^– + H⁺ ¾® HNO₂

2HNO₂ ¾® H₂O + N₂O₃

3HNO₂ ¾® HNO₃ + 2NO + H₂O

2NO + O₂ ¾® 2NO₂

ii) Silver nitrate solution : White crystalline ppt. is obtained

NO₂^– + Ag⁺ ® AgNO₂¯

iii) Turns acidified KI – starch paper blue

2KI + 2NO₂ ¾® 2KNO₂ + I₂

Starch + I₂ ¾® Blue Colour

iv) Brown ring test : When the nitrite solution is added carefully to a conc. solution of Iron(II) sulphate acidified with dil. acetic acid or with dilute sulphuric acid, a brown ring, due to the formation of [FeNO]SO₄ at the junction of the two liquids.

NO₂^– + CH₃COOH ¾® HNO₂ + CH₃COO^–

3HNO₂ ¾® H₂O + HNO₃ + 2NO

Fe²⁺ + NO ¾® [FeNO]²⁺

Acetate (CH₃COO^–) :
i) Dilute Sulphuric Acid : Smell of vinegar

CH₃COO^– + H⁺ ¾¾® CH₃COOH

ii) Iron (III) Chloride Solution : Gives deep – red colouration

CH₃COONa + FeCl₃ ¾¾® (CH₃COO)₃Fe + 3NaCl

Blood red colour

Thiosulphates () :
i) Dil Hydrochloric acid : Gives sulphur & sulphur di oxide

S₂O+ 2H⁺ ¾® S ¯ + SO₂ + H₂O

ii) Iodine Solution : Decolourise due to the formation of tetrathionate ion

I₂ + 2S₂O¾® 2I^– + S₄O₆^2-

iii) Barium chloride solution : White ppt. of barium thiosulphate is formed

S₂O+ Ba²⁺ ¾® BaS₂O₃ ¯

But no ppt. is obtained with CaCl₂ solution.

iv) Silver nitrate solution : Gives white ppt. of silver thiosulphate.

S₂O+ 2Ag⁺ ¾® Ag₂S₂O₃ ¯

The ppt. is unstable, turning dark on standing, when silver sulphide is formed.

Ag₂S₂O₃¯ + H₂O ¾® Ag₂S + H₂SO₄

v) Lead acetate or Lead nitrate solution : Gives white ppt.

+ Pb²⁺ ¾® PbS₂O₃ ¯

On boiling it turns black due to the formation of PbS.

PbS₂O₃ ¯ + H₂O ¾® PbS ¯ + 2H⁺ + SO₄^2-

Class A (ii): Gases or acid vapours evolved with conc. Sulphuric acid

It includes – Cl^–, Br^–, I^–, NO₃^–.

Chloride (Cl^–) :
i) H₂SO₄ : decomposes with the evolution of HCl.

Cl^– + H₂SO₄ ® HCl + HSO

Gas so produced (1) Turns blue litmus paper red

2) Gives white fumes of NH₄Cl when a glass rod moistened with ammonia solution is brought to the mouth of test tube.

ii) Manganese dioxide and conc. sulphuric acid :When a solid chloride is treated with MnO₂ and conc. H₂SO₄, yellowish green colour is obtained.

MnO₂ + 2H₂SO₄ + 2Cl^– ® Mn²⁺ + Cl₂ + 2SO₄^2- + 2H₂O

iii) Silver nitrate solution : White, curdy ppt. of AgCl insoluble in water & in dil. nitric acid, but soluble in dilute ammonia solution.

Cl^– + Ag⁺ ® AgCl ¯

AgCl ¯ + 2NH₃ ® [Ag(NH₃)₂]Cl

Ag(NH₃)₂Cl + 2H⁺ ® AgCl + 2NH₄⁺.

iv) Lead acetate solution : White ppt. of lead chloride is formed

2Cl^– + Pb⁺² ® PbCl₂ ¯

v) Chromyl chloride test : When a mixture containing chloride ion is heated with K₂Cr₂O₇ and conc. H₂SO₄ orange red fumes of chromyl chloride (CrO₂Cl₂) are formed.

K₂Cr₂O₇ + 4NaCl + 6H₂SO₄ ® 2KHSO₄ + 4NaHSO₄ + 2CrO₂Cl₂ + 3H₂O

orange – red

fumes

Chlorides of mercury, owing to their slight ionization, do not respond to this test and only partial conversion to CrO₂Cl₂ occurs with the chlorides of lead, silver, antimony and tin.

When chromyl chloride vapours are passed into sodium hydroxide a yellow solution of sodium chromate is formed which when treated with lead acetate gives yellow ppt. of lead chromate.

CrO₂Cl₂ + 2NaOH ¾¾® Na₂CrO₄ + 2HCl

Yellow solution

Na₂CrO₄ + (CH₃COO)₂ Pb ® 2CH₃COONa + PbCrO₄ ¯

(yellow ppt.)

Bromide (Br^–)
i) H₂SO₄ : Gives reddish brown vapours of bromine accompanied by hydrogen bromide

2KBr + H₂SO₄ ® K₂SO₄ + 2HBr

2HBr + H₂SO₄ ® 2H₂O + SO₂ + Br₂

(reddish brown)

ii) Manganese dioxide and conc. sulphuric acid : When a mixture of solid bromide, MnO₂ and conc. H₂SO₄ is heated, reddish brown vapours of bromine are evolved.

2KBr + MnO₂ + 2H₂SO₄ ® Br₂ + K₂SO₄ + MnSO₄ + 2H₂O

iii) Silver nitrate solution : Curdy pale yellow ppt. of silver bromide is obtained. This ppt. is sparingly soluble in dil. but readily soluble in conc. ammonia solution and insoluble in dil. HNO₃.

Br^– + Ag⁺ ¾® AgBr

iv) Lead acetate solution : White crystalline ppt. of lead bromide which is soluble in
boiling water.

2Br^– + Pb⁺² ¾¾®

v) Chlorine water : When this solution is added to a solution of bromide it liberates free bromine, which colours the solution orange – red, if CHCl₃ is added.

2KBr + Cl₂ (water) ¾® 2KCl + Br₂

Br₂ + Chloroform ¾® Orange red colour

vi) Potassium dichromate & conc. H₂SO₄ : When a mix of solid bromide, K₂Cr₂O₇, and conc. H₂SO₄ is heated and passing the evolved vapours into water, a yellowish brown solution is obtained.

2KBr + K₂Cr₂O₇ + 7H₂SO₄ ® 3Br₂ + Cr₂(SO₄)₃ + 4K₂SO₄ + 7H₂O.

Iodide (I^–) :
i) H₂SO₄ : Gives violet vapours of iodine

2I^– + 2H₂SO₄ ¾® I₂ + SO₄^2- + 2H₂O + SO₂

violet vapours

ii) Silver nitrate solution : Yellow, curdy ppt. of silver iodide AgI, very slightly soluble in conc. ammonia solution and insoluble in dil. nitric acid.

I^– + Ag⁺ ¾® AgI

iii) Lead acetate solution : Yellow, curdy ppt. of lead iodide soluble in much hot water forming a colourless solution & yielding golden yellow plates (spangles) on cooling.

2I^– + Pb²⁺ ¾® PbI₂¯

iv) Potassium dichromate & conc. sulphuric acid : Iodine is liberated

6I^– + Cr₂O₇^-2 + 7H₂SO₄ ¾¾® 3I₂ + 2Cr³⁺ + 7SO₄^2- + 7H₂O.

v) Chlorine water : Iodine is liberated, by the dropwise addition of chlorine water to iodide, and on addition of CHCl₃ violet solution is obtained.

2I^– + Cl₂ ¾® I₂ + 2Cl^–

I₂ + chloroform ¾¾® violet coloured compound.

vi) Copper sulphate solution : Gives brown ppt. consisting of a mixture of copper (I) iodide & iodine and on additon of hypo solution brown ppt changes to white ppt.

4I^– + 2Cu²⁺ ¾¾® 2CuI + I₂

I₂ + 2S₂O₃^2- ¾¾® 2I^– + S₄O₆^2-.

vii) Mercury (II) chloride solution : Forms scarlet ppt. of HgI₂

2I^– + HgCl₂ ¾® HgI₂ ¯ + 2Cl^–.

This ppt. dissolves in excess of KI, forming tetraiodo mercurate (II) complex.

HgI₂ + 2I^– ¾¾® [HgI₄]^2-

Nitrate (NO₃^–) :
i) Conc H₂SO₄ : – Gives reddish – brown vapours of nitrogen dioxide

4NO₃^– + 2H₂SO₄ ¾® 4NO₂ + 2SO₄^2- + 2H₂O + O₂

ii) Brown ring test : When freshly prepared saturated solution of iron (II) sulphate is added to nitrate solution & on addition of conc. H₂SO₄ slowly down the sides of the test – tube, a brown ring is obtained.

2NO₃^– + 4H₂SO₄ + 6Fe²⁺ ¾® 6Fe³⁺ +2NO + 4SO₄^–2 + 4H₂O

Fe²⁺ + NO ® [Fe(NO)]²⁺

On shaking and warming the mix, the brown colour disappears, nitric oxide is evolved and a yellow solution of Iron(III) ions remains.

Action of heat : The result varies with the metal

1) Nitrates of sodium and potassium evolve oxygen (test with glowing splint) & leave solid nitrites (brown fumes with dilute acid)

2NaNO₃ ¾® 2NaNO₂ + O₂.

2) Ammonium nitrate yields dinitrogen oxide & steam

NH₄NO₃ ¾® N₂O + 2H₂O.

3) Nitrates of the noble metals leave a residue of the metal and a mixture of nitrogen dioxide and oxygen is evolved.

2AgNO₃ ¾® 2Ag + 2NO₂ + O₂.

4) Nitrates of other metals, such as those of lead and copper, evolve oxygen and nitrogen dioxide and leave a residue of the oxide.

2Pb(NO₃)₂ ¾® 2PbO + 4NO₂ + O₂.

3.2 Class B

Includes anions that are identified by their reaction in solutions. It is subdivided into two groups:

(i) Precipitation reactions (ii) Oxidation and reduction in solution

Class B i) Precipitation reaction : SO₄^2-

ii) Oxidation and reduction in solution – CrO₄^{2 –} ,Cr₂O, MnO₄^–
Sulphate (SO₄^2-): All sulphates except those of Ba, Pb, Sr are soluble in water. Sulphates of calcium and mercury(II) are slightly soluble.
i) Barium chloride solution : White ppt. of barium sulphate BaSO₄ insoluble in warm dil. hydrochloric acid and in dilute nitric acid, but moderately soluble in boiling, conc. hydrochloric acid.

SO₄^2- + Ba²⁺ ¾¾® BaSO₄ ¯

ii) Mercury (II) nitrate solution : Gives yellow ppt. of basic mercury (II) sulphate.

SO₄^2- + 3Hg²⁺ + 2H₂O ® HgSO₄.2HgO ¯ + 4H⁺

Chromate CrO₄^{2 –}and Dichromate (Cr₂O)

Metallic chromates give yellow solution when dissolved in water. In the presence of H⁺ chromates are converted into dichromates (orange-red solution).

2CrO + 2H⁺ Cr₂O + H₂O

Cr₂O + 2OH^– 2CrO + H₂O

It may also be expressed as :

2CrO+ 2H⁺ 2HCrO₄^– Cr₂O₇^-2 + H₂O

i) Barium chloride solution : Pale – yellow ppt. of barium chromate soluble in dilute mineral acids but insoluble in water and acetic acid.

CrO + Ba²⁺¾¾® BaCrO₄ ¯

Dichromate ions also gives the same ppt. but due to the formation of strong acid precipitation is partial.

Cr₂O+ 2Ba²⁺ + H₂O 2 BaCrO₄ ¯ + 2H⁺

If sodium hydroxide or sodium acetate is added, precipitation becomes quantitative.

ii) Silver nitrate solution

Brownish – red ppt. of silver chromate Ag₂CrO₄ which is soluble in dil. nitric acid & in ammonia solution, but is insoluble in acetic acid.

CrO₄^2- + 2Ag⁺ ¾® Ag₂CrO₄ ¯

2 Ag₂CrO₄ + 2H⁺ ¾® 4 Ag⁺ + Cr₂O₇^2- + H₂O

Ag₂CrO₄ ¯ + 4NH₃ ¾® 2[Ag(NH₃)₂]⁺ + CrO₄^2-

Ag₂CrO₄ ¯ + 2Cl^– ¾® 2AgCl + CrO₄^2-

A reddish brown ppt. of silver dichromate Ag₂Cr₂O₇ is formed with a conc. solution of a dichromate.

Cr₂O₇^2-+ 2Ag⁺ ¾® Ag₂Cr₂O₇

iii) Lead acetate solution : Yellow ppt. of lead chromate PbCrO₄ insoluble in acetic acid, but soluble in dil. nitric acid

CrO₄^2- + Pb²⁺ ¾¾® PbCrO₄ ¯.

2PbCrO₄ ¯ + 2H⁺ 2Pb²⁺ + Cr₂O₇^2- + H₂O.

iv) H₂O₂ : If an acidic solution of a chromate is treated with H₂O₂ a deep blue solution of chromium penta oxide is obtained.

CrO₄^2- + 2H⁺ + 2H₂O₂ ¾® CrO₅ + 3H₂O

CrO₅ is unstable and it decomposes yielding oxygen and a green solution of a Cr⁺³ Salt.

Permanganate MnO
i) Hydrogen peroxide : It decolourises acidified potassium permanganate solution

2MnO₄^– + 5H₂O₂ + 6H⁺ ¾¾® 5O₂ + 2Mn²⁺ + 8H₂O.

ii) Iron (II) sulphate, in the presence of sulphuric acid, reduces permanganate to manganese (II). The solution becomes yellow because of the formation of iron (III) ions

MnO₄^– + 5Fe²⁺ + 8H⁺ ¾¾® 5Fe³⁺ + Mn²⁺ + 4H₂O

iii) Action of heat : On heating a black residue of potassium manganate K₂MnO₄ and manganese dioxide remains behind. Upon extracting with water and filtering, a green solution of potassium manganate is obtained.

2KMnO₄ ¾¾® K₂MnO₄ + MnO₂ + O₂.

Exercise 1: A compound A forms an unstable pale blue coloured solution in water which rapidly decomposes even in the cold. The solution acts as reducing agent and decolourises bromine water and an acidified solution of KMnO₄. It also oxidises SnCl₂ in dil. HCl solution. Identify it.

Classification of Cations

For the purpose of systematic qualitative analysis, cations are classified into five groups on the basis of their behaviour against some reagents and classification is based on whether a cation reacts with these reagents by the formation of precipitates or not (solubility difference)

4.1 Basis of Classification

The division of different cations in different analytical groups depends on the principle of solubility product. These divisions into groups are altogether different from the groupings in the Periodic Table. To understand fully the application of the principle of solubility product in analytical chemistry we should consider the principles of the Law of Mass Action and Common Ion Effect.

Precipitation of Pb⁺⁺, Ag⁺ and Hg₂²⁺ ions as their chlorides in group (I): The metals of the Gr. I are precipitated by adding chloride ions (from dil HCl). Solubility products at laboratory temperature are

PbCl₂ = 2.4 ´ 10^–4, AgCl = 1.6 ´ 10^–10and Hg₂Cl₂ = 3.5 ´ 10^–18

Let us consider the precipitation of AgCl. From the equilibrium of AgCl in a saturated solution.

AgCl

By adding dil. HCl, which dissociates into H⁺ and Cl^– ions, the concentration of Cl^– is increased, thus momentarily. becomes greater than the solubility product of AgCl and hence some Ag⁺ ions and Cl^– ions combine together to form solid AgCl until the product of the concentrations of Ag⁺ and Cl^– ions, i.e., . becomes equal to Ks_AgCl. Ks_AgCl is a constant quantity, the larger is the concentration of Cl^– ions the smaller is the concentration of Ag⁺ ions and hence by adding excess of Cl^– ions it is possible to remove Ag⁺ ions as AgCl. (That is why it is always necessary to add a slight excess of the precipitating reagent). It should be remembered that by increasing the amount of Cl^– ions the value of never becomes equal to zero since the product of the concentrations is always equal to Ks_AgCl but it becomes smaller and smaller and approaches towards zero).

Similarly by adding dil. HCl i.e., by increasing Cl^– ions concentration, the products of the concentrations of Pb⁺⁺ and Cl^– ions and and 2Cl^– ions exceed the respective solubility product of PbCl₂ and Hg₂Cl₂ and hence PbCl₂ and Hg₂Cl₂ are precipitated in group I. But this increased concentration of Cl^– ions cannot exceed the solubility product of the chlorides of the metals of subsequent analytical Groups, and that is why their chlorides remain in solution.

Precipitation of Sulphides of Group (II) and (IIIB): The sulphides of Gr. II are precipitated by passing H₂S in dil. HCl medium whereas those of Gr. III (B) are precipitated by passing H₂S in NH₄OH medium. Let us select for the sake of convenience, a representative member from two groups, CuS from Gr. II and ZnS from Gr. III (B). Others will follow similarly.

The solubility product of CuS = 1 ´ 10^–14 and that of ZnS = 1 ´ 10^–23. According to our previous discussions, sulphides of the metals can occur only when the product of the concentration of metal ions and sulphide ions i.e.,. exceeds the solubility product of the metal sulphide, MS.

Now H₂S in aqueous solution behave like a weak acid, feebly dissociating into H⁺ ions and S^{– –} ions. H₂S 2H⁺+S^{– –}

This dissociation is still further depressed by the addition of HCl, which being a strong electrolyte dissociates as HCl H⁺ + Cl^–.

H⁺ ions, being common to both HCl and H₂S, depress the dissociation i.e., the concentration of sulphide ions is decreased by the addition of common ions H⁺ from the strong electrolyte HCl. But this decreased value of sulphide ion concentrations is sufficient to exceed the low solubility products of the sulphide of Gr.II but insufficient to exceed the comparatively high solubility products of the sulphide of the metals of Gr. III (B). Or in other words with this decreased S^{– –} ions concentrations

of Gr.II

but Ks_MS of Gr. III (B).

That is why sulphide of Gr.(III) (B) are not precipitated in presence of dil. HCl.

In the presence of NH₄OH, which dissociates into NH₄OH + (OH)^– the (OH) ions combine with H⁺ ions from H₂S giving un-ionised H₂O and since some H⁺ ions are removed in this way from the equilibrium the reaction will proceed in the forward direction, i.e., more H₂S will dissociate giving H⁺ and S^{– –} ions and the concentration of S^{– –} ions will be eventually increased. This increased value of the concentration of sulphide ions is now sufficient to exceed the solubility product of the sulphides of Gr. III (B) and hence the precipitation of the sulphides of Gr. III (B) occurs in the presence of NH₄OH.

Precipitation of Hydroxides of Gr. (III A): The hydroxides of Gr. III (A) metals are precipitated by adding NH₄Cl and NH₄OH. The latter being a weak base dissociates to a small extent according to

NH₄OH

In presence of NH₄Cl which being a strong electrolyte dissociates into

NH₄Cl

The dissociation of NH₄OH is still further depressed due to common ion effect (NH₄⁺ ion being common) i.e., some OH^– ions and NH₄⁺ ions recombine together to form undissociated NH₄OH thereby decreasing the (OH)^– concentrations. With this decreased value of the (OH)^– ion concentration the solubility product of the hydroxides of Al. Fe and Cr alone is reached.

i.e., and under this condition hydroxides of Gr. III (A) are precipitated. But with this low value of the comparatively high value of the solubility product of the hydroxides of the metals of the subsequent groups e.g. Zn, Mn, Ni, Co, Mg etc. are not reached and hence these are not precipitated as their hydroxides in presence of NH₄Cl. If NH₄Cl were not added, then with the undiminished value of (as a result of dissociation of NH₄OH) all these metals of Grs. III (A), III (B) and IV were precipitated as their hydroxides because the solubility products of their hydroxides were reached. This is why NH₄Cl is added to decrease the (OH)^– concentration due to common ion effect so as to reach the solubility products of the hydroxides of Gr. III (A) only but not to reach the solubility product of the hydroxides of Gr. III (B) IV and V.

Precipitation of Carbonates of Gr. (IV): NH₄OH and (NH₄)₂CO₃ are added to precipitate the carbonates of Gr. IV (Ba, Sr and Ca). By the addition of NH₄Cl and NH₄OH such conditions are maintained so as to render the ion concentrations to be approximately equal to 1.6 ´ 10^–3 and the lowest limit of concentration of the metal ion necessary for obtaining an appreciable ppt. is 10^–4(M), the solubility product of the metal carbonate will be 1.6 ´ 10^–3 ´ 10^–4 = 1.6 ´ 10^–7. This value is greater than the solubility product of the carbonates of Ba, Ca and Sr but less than the solubility product of MgCO₃. Hence, the carbonates of Ba, Ca and Sr are precipitated in Gr. IV.

Mg⁺⁺ remains in solution. By the addition of Na₂HPO₄. Mg(NH₄)PO₃ being insoluble is precipitated.

4.2 Separation of Cations

Group reagent: Hydrochloric acid, hydrogen sulphide, ammonium sulphide and ammonium carbonate.

Group	Group reagent	Basic Radical	Composition and Colour of Precipitate
1	dil. HCl	Ag⁺	AgCl; White
		Pb²⁺	PbCl₂; White
		Hg₂²⁺	Hg₂Cl₂; White
2	H₂S	Hg²⁺	HgS; Black
		Pb²⁺	PbS: Black
		Bi³⁺	Bi₂S₃; Black
		Cu²⁺	CuS; Black
		Cd²⁺	CdS; Yellow
		As³⁺	As₂S₃; Yellow
		Sb³⁺	Sb₂S₃; Orange
		Sn²⁺	SnS; Brown
		Sn⁴⁺	SnS₂; Yellow
3.	NH₄OH in presence of NH₄Cl	Fe³⁺	Fe(OH)₃; Reddish brown
		Cr³⁺	Cr(OH)₃; Green
		Al³⁺	Al(OH)₃; White
4.	H₂S in presence of NH₄OH	Zn²⁺	ZnS; Greenish White
		Mn²⁺	MnS; Buff
		Co²⁺	CoS; Black
		Ni²⁺	NiS; Black
5	(NH₄)₂CO₃ in presence of NH₄OH	Ba²⁺	BaCO₃; White
		Sr²⁺	SrCO₃; White
		Ca²⁺	CaCO₃; White
6	Na₂HPO₄	Mg²⁺	Mg(NH₄)PO₄; White
7	NaOH	NH₄⁺	Ammonia gas is evolved

4.3 Physical Examination of Salt / Mixture

Observation	Inference
1. Substance is coloured
i) Blue	Copper salt
ii) Dark green	Chromium salt
iii) Green	Salts of Fe, Ni, Cu or Cr
iv) Light yellow or brown	Salts of Fe(ic)
v) Dark brown	PbO₂,Bi₂S₃
vi) Light pink	Salts of Mn
vii) Pink	Salts of Co
viii) Red	HgO, HgI₂,Pb₃O₄
ix) Orange red	Sb₂S₃
2. Substance is wet	CaCl₂,ZnCl₂,MgCl₂, MnCl₂, nitrites, nitrates
3. Substance is heavy	Salts of Pb, Hg and Ba
4. Substance is light	Carbonates of Bi, Mg, Al, Zn, Ca, Sr

4.4 Effect of Heating

Observation	Inference
1. Substance melts	Salts of alkali metals and salts having water crystallisation.
2. Substance decrepitates (craking noise)	NaCl, KI, Pb(NO₃)₂ and Ba(NO₃)₂
3. Substance swells (due to loss of water of crystallisation)	Alums, borates and phosphates
4. The substance sublimes and the colour of sublimate is
i) White	HgCl₂, Hg₂Cl₂,NH₄X, AlCl₃, As₂O₃, Sb₂O₃
ii) Yellow	As₂S₃ and HgI₂ (turns red when rubbed with glass rod).
iii) Blue black and violet vapours	Iodides
5. A residue (generally oxide) is left and its colour is
i) Yellow (hot) and white (cold)	ZnO
ii) Reddish brown (hot); yellow (cold)	PbO
iii) Black (hot); Red (cold)	HgO, Pb₃O₄
iv) Black (hot); Red brown (cold)	Fe₂O₃
6. Gas is evolved
(A) Colourless and odourless
i) O₂ – rekindles a glowing splinter	Alkali nitrates (2KNO₃ ® 2KNO₂ + O₂)
ii) CO₂ – turns lime water milky	Carbonates and oxalates (CaCO₃ ® CaO + CO₂)
iii) N₂	Ammonium nitrite (NH₄NO₂ ® N₂ + 2H₂O)
(B) Colourless gas with odour
i) NH₃ – Turns red litmus blue and mercurous nitrate paper black	Ammonium salts (NH₄)₂SO₄ ® NH₄HSO₄ + NH₃
ii) SO₂ – Smell of burning sulphur, turns acidified K₂Cr₂O₇ paper green	Sulphites and thiosulphates CaSO₃ ® CaO + SO₂
iii) HCl – Pungent smell, white fumes with ammonia	Hydrated chlorides CaCl₂.6H₂O ® Ca(OH)₂ + 4H₂O + 2HCl
iv) H₂S – smell of rotten eggs, turns lead acetate paper black	Sulphides Na₂S + 2H₂O ® 2NaOH + H₂S
(C) Coloured gas
i) NO₂ – Brown, turns starch iodide paper blue	Nitrites and nitrates of heavy metals 2Cu(NO₃) ® 2CuO + 4NO₂ + O₂
ii) Br₂ – Reddish brown	Bromides 2CdBr₂ + O₂ ® CdO + 2Br₂
(A) Turns starch paper yellow
(B) turns starch iodide paper blue
iii) I₂ – Violet, turns starch paper blue	Iodides 2CdI₂ + O₂ ® 2CdO + 2I₂
iv) Cl₂ – Greenish yellow	Chlorides
(A) bleaches moist litmus paper	CuCl₂ + H₂O ® CuO + 2HCl CuO + 2HCl ® Cu + H₂O + Cl₂
(B) bleaches indigo solution
(C) turns starch iodide paper blue

* Points to Remember

Group I radicals (Ag⁺, Pb⁺² Hg₂²⁺) are precipitated as chlorides because the solubility product of these chlorides (AgCl, PbCl₂, Hg₂Cl₂) is less than the solubility products of all other chlorides which remain in solution.
Group II radicals are precipitated as sulphides because sulphides of other metals remain in solution because of their low solubility products, HCl acts as a source of H⁺ and thus decreases the conc. of S^2- due to common ion effect. Hence decreased conc. of S^2- is only sufficient to precipitate the Group II metal only.
Group III A radicals are precipitated as hydroxides and the NH₄Cl suppresses the ionisation of NH₄OH so that only the group III A metal are precipitated because of their low solubility product.

Note : i) Excess of NH₄Cl should be added otherwise manganese will ppt. as MnO₂.H₂O.

ii) (NH₄)₂SO₄ can’t be used in place of NH₄Cl because the SO₄^2- will ppt. barium as BaSO₄.

iii) NH₄NO₃ can’t be used in place of NH₄Cl because NO₃^– ions will oxidise Mn²⁺ to Mn³⁺ and thus Mn(OH)₃ will be precipitated in III A group.

iv) Only Al(OH)₃ is soluble in excess of NaOH followed by boiling to form sodium meta aluminate while Fe(OH)₃ and Cr(OH)₃ are insoluble.
Ammonium hydroxide increases the ionisation of H₂S by removing H⁺ from H₂S as unionised water

H₂S 2H⁺ + S^2-. H⁺ + OH^–¾¾¾® H₂O

Now the excess of S^2- ions are available and hence the ionic product of Group III B exceed their solubility product and ppt. will be obtained.In case H₂S is passed through a neutral solution, incomplete precipitation will take place due to the formation of HCl which decreases the ionization of H₂S.

MnCl₂ + H₂S ¾¾® MnS + 2HCl

Identification of Basic Radicals

Group I (Pb²⁺, Ag⁺, Hg⁺)
A) PbCl₂ gives a yellow ppt. with K₂CrO₄. The ppt. is insoluble in acetic acid but soluble in NaOH

PbCl₂ + K₂CrO₄ ® PbCrO₄ ¯ + 2KCl

Yellow ppt.

PbCrO₄ + 4NaOH ® Na₂[PbO₂] + Na₂CrO₄ + 2H₂O

B) PbCl₂ + 2KI ® PbI₂ ¯ + 2KCl

(Yellow)

PbCl₂ + 2KI (excess) ® K₂[PbI₄]

AgCl is soluble in NH₄OH forming a complex while Hg₂Cl₂ forms a black ppt. with NH₄OH.

AgCl + 2NH₄OH ® Ag(NH₃)₂Cl + 2H₂O

Hg₂Cl₂ + 2NH₄OH ® H₂N ¾ Hg ¾ Cl + Hg¯ + NH₄Cl + 2H₂O

Amino mercuric chloride

Group II A (Hg²⁺, Cu²⁺, Bi³⁺, Cd²⁺)
i) Hg⁺²ions in solution, on addition of SnCl₂, give white precipitate turning black.

2Hg⁺² + SnCl₂ ® Sn⁺⁴ + Hg₂Cl₂ ¯

White

Hg₂Cl₂ + SnCl₂ ® SnCl₄ + 2Hg ¯

Black

ii) Cu⁺² ions in solution give deep blue colour with excess of NH₄OH

Cu⁺² + 4NH₄OH ® [Cu(NH₃)₄]⁺² + 4H₂O

Deep blue in colour

Cu⁺² ions give chocolate precipitate with K₄Fe(CN)₆.

2Cu⁺² + K₄Fe(CN)₆ ® Cu₂[Fe(CN)₆] + 4K⁺

iii) Bi⁺³ ions in solution of HCl on addition of water give white cloudy precipitate.

BiCl₃ + H₂O ¾® BiOCl ¯ + 2HCl

White ppt.

When treated with sodium stannite a black ppt. is obtained.

2BiCl₃ + 3Na₂SnO₂ ¾¾® 2Bi ¯ + 3Na₂SnO₃ + 6NaCl + 3H₂O

black

iv) Cd⁺² ions in solution, with H₂S give yellow precipitate.

Cd⁺² + H₂S ¾® CdS ¯ + 2H⁺

Yellow

Group II B (As³⁺, As⁵⁺, Sb³⁺, Sb⁵⁺, Sn²⁺, Sn⁴⁺)
i) As⁺³ ions in solution give yellow precipitate with ammonium molybdate and HNO₃.

As³⁺ As⁵⁺

H₃AsO₄ +12(NH₄)₂MoO₄ +21HNO₃ ¾® (NH₄)₃ AsMo₁₂O₄₀¯+ 21NH₄NO₃ + 12H₂O

Yellow ppt.

iii) Sn²⁺ ions in solution as SnCl₂ give white ppt. with HgCl₂ ,which turns black on standing.

SnCl₂ + 2HgCl₂ ¾¾® SnCl₄ + Hg₂Cl₂ ¯

White

Hg₂Cl₂ + SnCl₂ ¾¾®SnCl₄ + 2Hg ¯

Black

iii) Sb⁺³ ions in solution as SbCl₃, on addition of water give white precipitate.

SbCl₃ + H₂O ® SbOCl¯ + 2HCI

White

Group III A (Al³⁺, Fe³⁺, Cr³⁺)
i) White precipitate of Al(OH)₃ is soluble in NaOH

Al(OH)₃ + NaOH ® NaAlO₂ + 2H₂O

ii) Precipitate of Cr(OH)₃ is soluble in NaOH + Br₂ water and addition of BaCl₂ to this solution gives yellow precipitate.

Br₂ + H₂O ® 2HBr + (O)

2Cr(OH)₃ + 4NaOH + 3(O) ® 2Na₂CrO₄ + 5H₂O

Na₂CrO₄ + BaCl₂ ® BaCrO₄ ¯ + 2NaCl

Yellow ppt.

Fe(OH)₃ is insoluble in NaOH

iii) Brown precipitate of Fe(OH)₃ is dissolved in HCl and addition of KCNS to this solution gives blood red colour.

Fe(OH)₃ + 3HCl ® FeCl₃ + 3H₂O

FeCl₃ + 3KCNS ® Fe(CNS)₃ + 3KCl

blood red colour

Also on addition of K₄Fe(CN)₆ to this solution, a prussian blue colour is obtained.

FeCl₃ + 3K₄Fe(CN)₆ ® Fe₄[Fe(CN)₆]₃ + 12KCl

prussian blue colour

Group III B (Ni²⁺, Co²⁺, Mn²⁺, Zn⁺²)
i) Ni⁺² and Co⁺² ions in solution, on addition of KHCO₃ and Br₂ water give apple green colour if Co⁺² is present and black precipitate if Ni⁺² is present.

CoCl₂ + 6KHCO₃ ® K₄[Co(CO₃)₃] + 2KCl + 3CO₂ + 3H₂O

2K₄[Co(CO₃)₃] + 2KHCO₃ + [O] ® 2K₃[Co(CO₃)₃] + 2K₂CO₃ + H₂O

Apple green colour

NiCl₂ + 2KHCO₃ ® NiCO₃ + 2KCl + H₂O + CO₂

2NiCO₃ + 4NaOH + [O] ® Ni₂O₃ ¯ + 2Na₂CO₃ + 2H₂O

Black ppt.

ii) Zn⁺² ions in solution give white precipitate with NaOH, which dissolve in excess of NaOH.

Zn⁺² + 2NaOH ® Zn(OH)₂ ¯ + 2Na⁺

White

Zn(OH)₂ + 2NaOH ® Na₂ZnO₂ + 2H₂O

Soluble

iii) Mn⁺² ions in solution give pink precipitate with NaOH turning black or brown on heating.

Mn⁺² + 2NaOH Mn(OH)₂ + 2Na⁺

Pink

Mn(OH)₂ + [O] MnO₂ + H₂O

Brown or black

Exercise 2: An inorganic compound A in its aqueous solution produced a white precipitate with NaOH, which dissolves in excess of NaOH. The aqueous solution of A also produced white precipitate with NH₄OH which also dissolves in excess NH₄OH. Also its aqueous solution produced light yellow precipitate with AgNO₃ solution, soluble in dill. HNO₃. Identify A.

Exercise 3: An aqueous solution of a compound a when treated with BaCl₂ solution gives a white precipitate insoluble in conc. HCl. Another sample of A when treated with NaOH gave a white precipitate first which dissolved in excess of NaOH; yielding a colourless solution. When H₂S was passed into the solution, white precipitate was obtained. Identify A.

Group IV (Ba²⁺, Sr²⁺, Ca²⁺)
i) Ba⁺² ions in solution give
A) Yellow precipitate with K₂CrO₄

Ba⁺² + K₂CrO₄ ® BaCrO₄ ¯ + 2K⁺

Yellow

B) White precipitate with (NH₄)₂SO₄

Ba⁺² + (NH₄)₂ SO₄ ® BaSO₄¯ +

White

C) White precipitate with (NH₄)₂ C₂O₄

Ba⁺² + (NH₄)₂C₂O₄ ® BaC₂O₄¯ +

White

ii) Sr⁺² ions give white precipitate with (NH₄)₂SO₄ and (NH₄)₂C₂O₄

Sr⁺² + (NH₄)₂SO₄ ® SrSO₄¯ +

White ppt.

Sr⁺² + (NH₄)₂C₂O₄ ® SrC₂O₄¯ +

White

iii) Ca⁺² ions give white precipitate with only (NH₄)₂ C₂O₄.

Ca⁺² + (NH₄)₂C₂O₄ ® CaC₂O₄ ¯ +

White

Group V (NH₄⁺, Na⁺, K⁺, Mg⁺²)
i) All ammonium salts on heating with alkali say NaOH give smell of NH₃

NH₄Cl + NaOH ¾¾® NaCl + NH₃ + H₂O

A) Gas evolve gives white fumes with HCl

NH₃ + HCl ¾® NH₄Cl

White fumes

B) Paper soaked in CuSO₄ solution, becomes deep blue by NH₃ due to complex formation

CuSO₄ + 4NH₃ ¾¾® [Cu(NH₃)₄]SO₄

deep blue

C) With Hg₂(NO₃)₂ , a black colour is obtained

Hg₂(NO₃)₂ + 2NH₃ ¾¾® Hg ¯ + Hg(NH₂)NO₃ ¯ + NH₄NO₃

black

D) With Nesslers reagent(alkaline solution of potassium tetraiodomercurate(II) ), a brown ppt. is obtained

ii) Potassium salts gives yellow ppt. with sodium cobalt nitrite

Na₃[Co(NO₂)₆] + 3KCl ¾¾® K₃[Co(NO₂)₆] + 3NaCl

yellow

iii) Sodium salts gives heavy white ppt. with potassium dihydrogen antimonate

KH₂SbO₄ + NaCl ¾¾® NaH₂SbO₄ ¯ + KCl

White ppt.

iv) Mg²⁺ gives white gelatinous ppt. of magnesium hydroxide with sodium hydroxide

Mg²⁺ + 2NH₃ + 2H₂O ¾¾® Mg(OH)₂ ¯ + 2NH₄⁺

ppt. obtained is sparingly soluble in water but readily soluble in ammonium salt.

Exercise 4: A colourless crystalline solid A, deliquescent in nature is obtained from kieserite. It loses 6H₂O at 150°C and becomes anhydrous at 200°C. On strong heating it gives a white residue and a suffocating gas which is purgative. Identify A.

Dry Tests

Dry tests are of great importance as these tests give clear indications of the presence of certain radicals. The following tests are performed in dry state.

i) Flame test
ii) Borax-bead test

iii) Micro-cosmic bead test

iv) Charcoal cavity test

5.1 Flame Test

Some volatile salts impart characteristic colour to the non-luminous flame. The chlorides of the metals are more volatile in comparison to other salts. The metal chloride volatilises and its thermal ionisation takes place.

NaCl Na⁺+ Cl^–

CaCl₂ Ca²⁺ + 2Cl^–

The cations impart a characteristic colour to the flame as these absorb energy from the flame and transmit the same as light as characteristic colour.

Colour of flame	Inference
Golden Yellow	Sodium
Violet	Potassium
Brick Red	Calcium
Crimson Red	Strontium
Apple Green	Barium
Green with a blue centre	Copper

Exercise 5: A white compound A gave a golden yellow flame colour on performing flame test. The aqueous solution of A produced a precipitate with AgNO₃ solution. On heating with conc. H₂SO₄ dense brown fumes come out. Identify A.

5.2 Borax Bead Test

On heating borax the colourless glassy bead formed consist of sodium metaborate and boric anhydride.

On heating with a coloured salt, the glassy bead forms a coloured metaborate in oxidising flame.

CuSO₄ ¾® CuO + SO₃

CuO + B₂O₃ ¾®

The metaborates posses different characteristic colours. The shade of the colour gives a clue regarding the presence of the radical. However in reducing flame the colours may be different due to different reactions. For example copper metaborate may be reduced to colourless cuprous metaborate or to metallic copper, which appears red and opaque.

2Cu(BO₂)₂ + C ¾® 2CuBO₂ + B₂O₃ + CO

2Cu(BO₂)₂ + 2C ¾® 2Cu + 2B₂O₃ + 2CO

Metal	Colour of Flame in
	Oxidizing Flame		Reducing Flame
	Hot	Cold	Hot	Cold
Copper	Green	Blue	Colourless	Brown-Red
Iron	Brown-Yellow	Pale-Yellow	Bottle Green	Bottle Green
Chromium	Green	Green	Green	Green
Cobalt	Blue	Blue	Blue	Blue
Manganese	Violet	Amethyst Red	Grey	Grey
Nickel	Violet	Brown	Grey	Grey

5.3 Microcosmic Salt Bead Test

This test is similar to borax bead test. When microcosmic salt is heated, a colourless transparent bead of sodium metaphosphate is formed.

Na(NH₄)HPO₄ ¾® + NH₃ + H₂O

Sodium metaphosphate combines with metallic oxides to form ortho phosphates which are usually coloured. The shade of the colour gives a due regarding the presence of metal. Like borax bead test, colours are noted both in oxidizing and reducing flames in hot and cold conditions.

Metal	Colour of the bead in
	Oxidizing Flame		Reducing Flame
	Hot	Cold	Hot	Cold
Copper	Green	Blue	Colourless	Red
Iron	Yellow	Yellow	Yellow	Colourless
Chromium	Green	Green	Green	Green
Cobalt	Blue	Blue	Blue	Blue
Manganese	Violet	Violet	Colourless	Colourless
Nickel	Brown	Brown	––	Grey

5.4 Charcoal Cavity Test

This test is carried out on a charcoal block in which a small cavity has been made by a knife. When a metallic salt is heated with Na₂CO₃, the metal carbonate is formed which decomposes into oxide. The carbon of the block reduces the oxide into metal.

CuCl₂ + Na₂CO₃ ¾® CuO₃ + 2NaCl

CuCO₃ ¾® CuO + CO₂

CuO + C ¾® Cu + CO

Observation Inference

Formation of metallic bead
a) Lustruous white, malleable Ag
b) Greyish white, marks paper Pb
c) White, does not mark paper Sn
d) Red Cu
In crustation with metal
a) White incrustation, brittle metal Sb
b) Yellow incrustation, brittle metal Bi
c) Yellow incrustation, malleable metal Pb
In crustation without metal
a) White and yellow when hot ZnO, SnO
b) Yellow and orange when hot BiO
c) Brown CdO
d) White As₂O₃

Some Important Observations During Qualitative Analysis

Residue and Its Colour

*S. No.*	*Colour*	*Residue*
i)	Yellow (hot) and white (cold)	ZnO
ii)	Reddish brown (hot) and yellow (cold)	PbO
iii)	Black (hot) and Red (cold)	HgO, Pb₃O₄
iv)	Black (hot) and Red brown (cold)	Fe₂O₃

Gases

S. No.	Nature	Gases
i)	Colourless and odourless gases	O₂, CO₂, N₂
ii)	Colourless gases with odour	NH₃, SO₂, HCl, H₂S
iii)	Coloured gases	NO₂ (brown), Br₂, (reddish brown), I₂ (violet) Cl₂ (greenish yellow)

Answers to Exercise

Exercise 1: HNO₂

Exercise 2: AlBr₃

Exercise 3: Al₂(SO₄)₃

Exercise 4: MgSO₄.7H₂O

Exercise 5: NaBr

Solved Problems

7.1 Subjective

Problem 1: A black mineral (A) on heating in presence of air gives a gas (B). The mineral (A) on reaction with dilute H₂SO₄ gives a gas (C) and a solution of a compound (D). On passing the gas (C) into an aqueous solution of (B) a white turbidity is obtained. The aqueous solution of (D) on reaction with potassium ferricyanide gives a blue compound (E). Identify (A) to (E) and give chemical equations for the reactions involved.

Solution: A = FeS

B = SO₂

C = H₂S

D = FeSO₄

E = Fe₃[Fe(CN)₆]₂

Problem 2: i) An alloy containing two metals ‘A’ & B is treated with dilute HCl. ‘A’ dissolves with evolution of hydrogen leaving behind B. B is separated from the solution C.

ii) The residue B dissolves in concentrated nitric acid giving a blue solution D.

iii) Mercuric chloride solution with solution C gives a silky white precipitate which turns grey on addition of excess C.

iv) Addition of NH₃ solution to D gives a blue precipitate which dissolves in excess of NH₃ giving a deep blue coloration. Name two metals A & B and alloy. Give equations for the reaction (i) to (iv).

Solution: i) CuSn + 2HCl ¾¾®

ii) Cu + 4HNO₃ ¾¾®

iii) SnCl₂ + 2HgCl₂ ¾¾®SnCl₄ + Hg₂Cl₂ (silky white ppt.)

Hg₂Cl₂ + SnCl₂ ¾¾® + SnCl₄

iv) Cu(NO₃)₂ + 2NH₄OH ¾¾® + 2NH₄NO₃

Cu(NO₃)₂ + 4NH₄OH ¾¾® [Cu(NH₃)₄(NO₃)₂] + 4H₂O

Problem 3: A compound (A) when treated with KI gives a scarlet red precipitate (B) which dissolves in excess of KI. This solution made akaline with NaOH and gave brown precipitate (C) when NH₃ gas is passed through it. When (A) is added with small amount of SnCl₂, it gives a white precipitate (D) but gives grey precipitate with excess amount of SnCl₂. When H₂S gas is passed through an acidic solution of (A) a black precipitate (E) is obtained.

Identify (A) to (E) and write the reactions involved.

Solution: A = HgCl₂ B = HgI₂

D = Hg₂Cl₂, E = HgS

reactions are

+ 2KI ¾¾®

HgI₂ + ¾¾®

¾¾® SnCl₂ +

HgCl₂ + ¾¾® SnCl₄ +

HgCl₂ + H₂S ¾¾® + 2HCl

Problem 4: A finely powdered mineral of calcium ‘A’ was boiled with Na₂CO₃ solution and the precipitate was filtered off. The filtrate was cooled and after cooling another precipitate ‘B’ produced which normally exists in the hydrated form. The aqueous solution of ‘B’ is alkaline. When ‘B’ is strongly heated, it gives a compound ‘C’ along with a glassy bead ‘D’. ‘C’ was also present initially into filtrate. ‘D’ is fused with one of the metal sulphate ‘E’ to give green colour both in oxidising flame and reducing flame. Identify A to E with reaction and give the explanation.

Solution: Clearly ‘B’ is Na₂B₄O₇ which gives the glassy bead of B₂O₃ on strong heating and Na₂B₄O₇ absorbs water to form borax, Na₂B₄O₇, 10H₂O. Borax is also less soluble in cold water but highly soluble in hot water. So after cooling a precipitate of borax appears also the solution of ‘B’ is alkaline.

So, A = Ca₂B₆O₁₁, B = Na₂B₄O₇, C = NaBO₂

D = B₂O₃, E = Cr₂(SO₄)₃ F = Na₂[(OH)₂B(O—O)₂B(OH)₂].6H₂O

So the reaction are

‘B’ normally exists into hydrated from Na₂B₄O₇.10H₂O (borax)

Na₂B₄O₇

‘E’ is Cr₂(SO₄)₃ which gives green colour both in oxidising flame and reducing flame

Cr₂O₃

Cr₂O₃ + ¾¾¾®

Problem 5: A crystalline inorganic compound (A) when comes in contact with skin leaves a black stain. (A) is freely soluble in water and when to this solution some sodium chloride is added, a white ppt. appears which is insoluble in HNO₃ but soluble in NH₄OH. When hydrogen sulfide is passed through a solution of (A), a black ppt. appears. When potassium chromate is added to the solution of (A), a brick red ppt. results. To another portion of solution (A) mixed with ferrous sulfate in a test tube concentrated H₂SO₄ is added, a brown ring results. Identify the compound and give equations for the various reactions involved.

Solution: The compound can be AgNO₃

Solution of A + NaCl ¾¾¾® A white ppt. insoluble in HNO₃ but dissolves in NH₄OH, indicating the presence of Ag⁺ ions in solution.

Solution of A + H₂S ¾¾¾® Ag₂S ¾¾¾® a black ppt. soluble in HNO₃

Solution of A + K₂CrO₄ ¾¾¾® Ag₂CrO₄ ¾¾¾® a brick red ppt.

Presence of Ag⁺ is confirmed

Solution of A + FeSO₄ + conc. H₂SO₄ ¾¾¾® brown ring indicates presence of nitrate

The compound in therefore identified as AgNO₃

2AgNO₃ + H₂SO₄ ¾¾¾® Ag₂SO₄ + 2HNO₃

6FeSO₄ + 3H₂SO₄ + 2HNO₃ ¾¾¾® 3Fe₂(SO₄)₃ + 4H₂O + 2NO

FeSO₄ + NO + 5H₂O ¾¾¾®

Problem 6: A colorless solid (A) on heating gives a white solid (B) and a colorless gas (C). (B) gives off reddish brown fumes on treatment with dilute acids. Also when B is heated with NH₄Cl a colorless gas (D) and a residue (E) are obtained. When (A) is heated with (NH₄)₂SO₄ a colorless gas (F) is obtained along with a white residue (G). Both (E) and (G) impart color to the Bunsen flame. The gas (C) reacts with heated magnesium. The colorless gas (D) reacts with heated calcium and the product on hydrolysis gives ammonia. Identify the compounds (A) to (G). [8]

Solution:

NaNO₂ NO₂ (brown gas)

+ NH₄Cl + 2H₂O

¾®

NaNO₃ (A) and Na₂SO₄ (G) impart yellow colour to flame indicating presence of Na.

2MgO

Ca₃N₂

Ca₃N₂ + 6H₂O ¾¾® 3Ca(OH)₂ + 2NH₃

Problem 7: A certain inorganic compound (A) on heating loses water of crystallization. On further heating, a blackish brown powder (B) and two oxides of sulphur (C) and (D) are obtained. The powder (B) on boiling with HCl acid gives a yellow solution (E). when H₂S is passed in (E), a white turbidity (F) and an apple green solution (G) is obtained. The solution (E) on treatment with thiocyanate ion gives blood red compound (H). Identify (A) to (H).

Solution: A = FeSO₄.7H₂O

B = Fe₂O₃

C = SO₂

D = SO₃

E = FeCl₃

F = S

G = FeCl₂

H = Fe(CNS)₃

Problem 8: i) An inorganic compound (A) is formed on passing a gas (B) through a concentrated liquor containing sodium sulphide and sodium sulphite.

ii) On adding (A) into a dilute solution of silver nitrate, a white precipitate appears which quickly changes into a black coloured compound (C).

iii) On adding two or three drops of ferric chloride into the excess of solution of (A), a violet coloured compound (D) is formed this colour disappears quickly.

On adding a solution of (A) into the solution of cupric chloride, a white precipitate is first formed which dissolves on adding excess of (A) forming a compound (E).

Identify (A) to (E) and give chemical equations for the reactions at steps (i) to (iv).

Solution: Let us summarise the given facts .

i) Na₂SO₃ + Na₂S + Gas(B) ¾¾® A
ii) A + AgNO₃ ¾¾® White ppt. ¾¾®

iii) A + FeCl₃ ¾¾® ¾¾® Disappears

iv) A + CuCl₂ ¾¾® White ppt.

The given set of reactions indicate that the compound is sodium thiosulphate which can be formed from equation (i) when the gas (B) is either SO₂ or I₂ and thus the various reactions can be written as below:

i) ¾¾®

or Na₂SO₃ + Na₂S + ¾¾® + 2NaI

ii) Na₂S₂O₃ + 2AgNO₃ ¾¾® + 2NaNO₃

Ag₂S₂O₃ + H₂O ¾¾® + H₂SO₄

iii) + 2FeCl₃ ¾¾® + 6NaCl

2Fe³⁺ + ¾¾®

iv) + 2CuCl₂ ¾¾® 2CuCl + Na₂S₄O₆ + 2NaCl

¾¾® + 2NaCl ] ´ 3

3Cu₂S₂O₃ + 2Na₂S₂O₃ ¾¾®

Problem 9: A compound (A) is greenish crystalline salt, which gave the following results.

i) Addition of BaCl₂ solution to the solution of (A) results in the formation of a white ppt. (B) which is insoluble in dil. HCl.
ii) On heating (A), water vapours and two oxides of sulphur (C) and (D) are liberated leaving a red brown residue (E).

iii) (E) dissolves in warm conc. HCl to give a yellow solution (F).

iv) With H₂S the solution (F) yields a pale yellow ppt. (G), Which when filtered, leaves a greenish filtrate(H).
v) Solution (F) on treatment with thiocyanate ions gives blood red coloured compound (I). Identify the substances from (A) to (I).

Solution: A = FeSO₄.7H₂O B = BaSO₄

C = SO₂D = SO₃

E = Fe₂O₃F = FeCl₃

Reaction: i) 4FeS + 7O₂ ¾¾® 2Fe₂O₃ + 4SO₂

ii) FeS + H₂SO₄ ¾¾® FeSO₄ + H₂S

iii) 2H₂S + SO₂ ¾¾® 2H₂O + 3S

iv) 3FeSO₄ + 2K₃[Fe(CN)₆] ¾¾® Fe₃[Fe(CN)₆]₂ + 3K₂SO₄

Problem 10: A colourless solid A on heating gives a white solid B and a colourless gas C; B gives off reddish brown fumes on treatment with dilute acids. On heating with NH₄Cl, B gives a colourless gas D and a residue E.

The compound A also gives a colourless gas F on heating with ammonium sulphate and white residue G. Both E and G impart bright yellow colour to bunsen flame. The gas C forms white powder with strongly heated magnesium metal. The white powder forms magnesium hydroxide with water. The gas D, on the other hand, is absorbed by heated calcium which gives of ammonia on hydrolysis.

Identify the substances A to G and give reactions for the changes involved.

Solution: It is advisable to summarise the given facts in the form of a chart

+ Mg ¾¾® White powder Mg(OH)₂

+ Ca(heated) NH₃

The above reactions lead to the following conclusions.

a) Gas D reacts with calcium forming a compound which on hydrolysis gives ammonia indicating that D must be nitrogen.
b) Residues E and G give yellow flame on burning indicating that these are sodium salts. Hence compounds B ( which give E) and A (which give G) must be sodium salt.
c) The colourless solid B gives reddish brown fumes with dilute acids, the reddish brown fumes are probably of NO₂. Hence compound B must be nitrite (recall that ions are not attacked by dil. acids). Consequently, A must be which can give (B) on heating.

Thus compound A is NaNO₃ which explains all the given reactions as below.

i) ¾¾® +
ii) + H₂SO₄(dil.) ¾¾® Na₂SO₄ + 2HNO₂

3HNO₂ ¾¾® HNO₃ + H₂O + 2NO

2NO + O₂ ¾¾®

iii) + NH₄Cl + + 2H₂O

iv) ¾¾® + + 2NHO₃
v) 2Mg + ¾¾®

MgO + H₂O ¾¾® Mg(OH)₂

vi) 3Ca + ¾¾®

Ca₃N₂ + 6H₂O ¾¾® 3Ca(OH)₂ + 2NH₃

Thus substance (A) to (G) can be represented as

(A) NaNO₃ (B) NaNO₂ (C) O₂

(D) N₂(E) NaCl (F) NH₃

(G) Na₂SO₄

7.2 Objective

Problem 1: A free metal is formed by the thermal decomposition of

(A) KNO₃ (B) Pb(NO₃)₂

(C) Mg(NO₃)₂ (D) AgNO₃

Solution: 2AgNO₃ 2Ag_(s) + 2NO_2(g) + O_2(g)

Problem 2: The formula of the compound which gives violet colour in Lassiagne’s test for sulphur with sodium nitroprusside is

(A) Na₄[Fe(CN)₅NOS] (B) Na₃[Fe(CN)₅NOS]

(C) Na₂[Fe(CN)₅S] (D) Na₄[Fe(CN)₄S]

Solution: Na₂S + Na₂[Fe(CN)₅NO] ¾®Na₄[Fe(CN)₅NOS]

Problem 3: When magnesium ammonium phosphate is heated, it is converted into

(A) Magnesium pyrophosphate (B) Magnesium oxide

(C) Magnesium phosphate (D) Magnesium nitride

Solution:

Problem 4: An acidic solution contains Cu²⁺, Pb²⁺ and Zn²⁺. If H₂S_(g) is passed through the solution the precipitate will contain

(A) CuS and ZnS (B) PbS and ZnS

(C) CuS and PbS (D) CuS, PbS and ZnS

Solution: When H₂S is passed in acidic solution the ionisation of H₂S is suppressed, because of the common ions, furnished by the strong acid. The conc. of S^2– ion is not sufficient for the precipitation of Zn²⁺ as ZnS. Only Cu²⁺ and Pb²⁺ are precipitated because their solubility products are less.

Problem 5: When a small amount of HCl is added to an aq. Solution of BiCl₃ a white precipitate is formed. This is due to

(A) Bi(OH)₃ (B) Bi₂O₃

(C) BiOCl (D) None of the above

Solution: BiCl₃ + 3H₂O ¾® BiOCl + 2HCl

Problem 6: White phosphorus when boiled with concentrated solution of KOH produces

(A) Na₃P (B) Na₃PO₄

(C) PH₃ (D) Red P

Solution: P₄ + 3KOH + 3H₂O ¾®PH₃ + 3KH₂PO₂

Problem 7: When a fluoride is heated with conc. H₂SO₄ in a glass tube and if a drop of water is held at the mouth of the glass tube, a white deposit formed is of

(A) H₂SiF₆ (B) SiO₂

(C) H₂SiO₃ (D) SiF₄ + H₂F₂

Solution: CaF₂ + H₂SO₄ ¾® CaSO₄ + H₂F₂

SiO₂ + 2H₂F₂ ¾® SiF₄ + 2H₂O

3SiF₄ + 3H₂O ¾® + 2H₂SiF₆

Problem 8: Which of the following compounds is formed, when a mixture of K₂Cr₂O₇ and NaCl is heated with conc. H₂SO₄?

(A) CrO₂Cl₂ (B) CrCl₃

(C) Cr₂(SO₄)₃ (D) Na₂CrO₄

Solution: K₂Cr₂O₇ + 2H₂SO₃ ¾® KHSO₄ + 2CrO₃ + H₂O

KCl + H₂SO₄ ¾® KHSO₄ + HCl

CrO₃ + 2HCl ¾®

Problem 9: A green mass is formed in charcoal cavity test when a colourless salt(X) is fused with cobalt nitrate, (X) may contain

(A) Al (B) Cu

(C) Ba (D) Zn

Solution: ZnSO₄ + Na₂CO₃ ¾® ZnCO₃ + Na₂SO₄

ZnCO₃ ¾® ZnO + CO₂

2Co(NO₃)₂ ¾® 2CaO + 4NO₂ + O₂

CaO + ZnO ¾®

Problem 10: Ammonium dichromate on heating gives

(A) NO (B) N₂O

(C) NO₂ (D) N₂

Solution: (NH₄)₂Cr₂O₇ N₂ + Cr₂O₃ + 4H₂O

Assignments (Subjective Problems)

Level- I

A mineral popularly known as appetite, is used to prepare a fertilizer, which provides phosphorus element to the soil:
i) The fertilizer is obtained by treating appetite with H₂SO₄
ii) When heated with silica and cake, it yields white phosphorus and calcium silicate. Suggest formula for appetite
The action of dil. Acid on an iron salt produces a gas with a strong odour. The gas burns with a blue flame to produce moisture, a yellow residue and traces of another gas, which has a familiar smell. Identify the gas produced from iron salt and dil. acid.
A white solid A was dissolved in HNO₃. Its solution produced a white precipitate with dil. HCl. The precipitate is insoluble in dil. HNO₃ but is soluble in aqua-regia. Its solution produced grey precipitate with copper turnings and a white precipitate with aq. AgNO₃. Identify the compound A.
Sodium hydroxide is hygroscopic, that is, it absorbs moisture when exposed to the atmosphere. A student placed a pellet of NaOH on a watch glass. A few days later, she noticed that the pellet was covered with a white solid. What is the identity of this solid?
A white solid was added to sodium hydroxide solution and warmed. A colourless gas with a characteristic odour evolved on heating, which turned moist red litmus to blue. Identify the solid and gas.
A white compound A gave a golden yellow flame colour on performing flame test. The aqueous solution of A produced a precipitate with AgNO₃ solution. On heating with conc. H₂SO₄ dense brown fumes come out. Identify A.
When a white powder A is strongly heated, it gives off a colourless, odourless gas B which turns lime water milky C and if the passage of this gas is continued, the milkiness disappears and gives a solution D. The solid residue E is yellow when hot but turns white on cooling. Identify A to E.
Calcium burns in nitrogen to produce a white powder which dissolves in sufficient water to produce a gas (A) and an alkaline solution. The solution on exposure to air produces a thin solid layer of (B) on the surface. Identify the compounds (A) and (B).
A white amorphous powder (A) on strongly heating gives a colourless non-combustible gas (B) and solid (C). The gas (B) turns lime water milky and turbidity disappears with the passage of excess of gas.The solution of (C) in dilute HCl gives a white ppt. with an aqueous solution of K₄[Fe(CN)₆]. The solution of (A) in dilute HCl gives a white ppt. (D) on passing H₂S in presence of excess of NH₄OH. Identify (A) to (D).
A gaseous mixture containing (X), (Y), (Z) gases, when passed into acidified K₂Cr₂O₇ solution, gas (X) was absorbed and the solution was turned green. The remainder gas mixture was then passed through lime water, which turns milky by absorbing gas (Y). The residual gas when passed through alkaline pyrogallol solution, it turned black. Identify gas (X), (Y), (Z).
When 20.02 g of a white solid (X) is heated, 4.4 g of an acid gas (A) and 1.8 g of a neutral gas (B) are evolved leaving behind a solid residue (Y) of weight 13.8 g. (A) turns lime water milky and (B) condenses into a liquid which changes anhydrous copper sulphate blue. The aqueous solution of (Y) is alkaline to litmus and gives 19.7 g of white ppt. (Z) with barium chloride solution. (Z) gives carbon dioxide with an \acid. Identify (A), (B), (X), (Y) and (Z).
When gas A is passed through dry KOH at low temperature, a deep red coloured compound, B and a gas C are obtained. The gas A, on reaction with but -2-ene, followed by treatment with Zn/H₂O yields acetaldehyde. Identify A,B and C.
To a solution containing Ca²⁺, Ag⁺, Cu²⁺ and K⁺, 2M HCl is added when a white ppt. (A) is obtained. After filteration H₂S is passed through the filterate, a black ppt.(B) is formed. On removing (B) by filteration it gave white ppt. (C) with conc. Na₂CO₃ solution. Identify (A), (B) and (C).

14 An orange solid (A) on heating gave a green residue (B), a colourless gas (C) and water vapour. The dry gas (C) on passing over heated Mg gave a white solid (D). (D) on reaction with water gave a gas (E) which formed white fumes with HCl. Identify (A) to (E).

When 16.8 gm of a solid X were heated, 4.4 gm of a gas A that turned lime water milky was driven off together with 1.8 gm of a gas B which condensed to a colourless liquid. The solid that remained Y, dissolved in water to give an alkaline solution, which with excess of barium chloride solution gave white ppt. Z. The ppt. effervesced with acid giving off carbon di-oxide. Identify A, B, X and Y.

Level- II

A solution of a salt (A) gives white precipitate with AgNO₃. When treated with calculated quantity of NaOH, gave a green coloured precipitate (B) which dissolves in excess of NaOH. (B) acts as a weak base and loses water on heating to give a green powder (C). The green powder is used as refractory material. When (C) is fused with alkali in presence of air (or) oxidising agent, a yellow coloured solution (D) is obtained. Identify the compounds (A) to (D).
A solution of white solid (A) gave white precipitate (B) with water. On treatment with HCl the precipitate (B) produced (A). When solution of (A) was treated with sodium stannite and NaOH, it produced a black precipitate (C). With conc. H₂SO₄ the compound A produced a colourless gas (D). The gas is soluble in H₂O and its aqueous solution produced a white precipitate (E) with Hg₂(NO₃)₂ but no precipitate with Hg(NO₃)₂. Identify (A) to (E).
A waxy crystalline solid (A) with a garlic odour is obtained by burning white phosphorus in a steam of air. (A) reacts vigorously with hot water forming gas (B) and an acid (C). Gas (B) has unpleasant odour of rotten fish and is almost neutral towards litmus. When passed through CuSO₄ solution, gas (B) produced a black precipitate (D). Identify the compounds (A) to (B).
A metal chloride (A) gives white precipitate (B) in presence of NH₄OH and (NH₄)₃PO₄. (B) on heating gives (C) and a pungent smelling gas (D) which turns red litmus blue. Identify (A) to (D).
A solid (A), orange in colour, is fairly soluble in water. It evolves a gas (B) and a residue (C) resembling green tea, on heating. When NaOH is added to the solution of (A), a gas (D) is produced, which turns red litmus blue and the solution turns yellow. Identify (A), (B), (C) and (D).
A brown black solid (A) on fusion with KNO₂ in NaOH gives a green compound (B) and a colurless gas (C). Gas (C) on exposure in air gives a brown gas (D), which dissolves in water to produce nitric acid. Green compound (B) in aqueous solution on electrolytic oxidation gives a violet compound (E), KOH and H₂ gas. An aqueous solution of (E) in cold reacts with ethylene to give glycol and its pink colour is discharged due to precipitation of (A). What are (A) to (E)?
Gradual addition of KI solution to Bi(NO₃)₃ solution initially produces a dark brown precipitate which dissolves in excess of KI to give a clear yellow solution. Write chemical equations for the above reactions.
A colourless solid A liberates a brown gas B on acidification, a colourless alkaline gas C on treatment with NaOH, and a colourless non-reactive gas D on heating. If heating of the solid A is continued, it completely disappears. Identify A to D.
A compound (X) of S, Cl and O has vapour density of 67.5. It reacts with water to form two acids, and reacts with KOH to form two salts (Y) and (Z). While (Y) gives white precipitate with AgNO₃ solution, (Z) gives white precipitate with BaCl₂ solution. Identify X.
i) An inorganic iodide (A) on heating with a solution of KOH gives a gas (B) and the solution of a compound (C).
ii) The gas (B) on ignition in air gives a compound (D) and water

iii) Copper sulphate is finally reduced to the metal on passing (B) through its solution

iv) A precipitate of compound (E) is formed on reaction of (C) with copper sulphate solution.

Identify (A) to (E) and give chemical equations for reactions at steps (i) to (iv)

A certain salt (X) gives the following tests:
i) Its aqueous solution is alkaline to litmus
ii) On strongly heating it swells to give a glassy material

iii) When concentrated sulphuric acid is added to a hot concentrated solution of (X) white crystals of a weak acid separate out.

Identify (X) and write down the chemical equations for reactions at steps (i), (ii)
and (iii)

A certain compound (X) shows the following reactions :
i) When KI is added to an aq. suspension of (X) containing acetic acid, iodine is liberated
ii) When CO₂ is passed through an aq. suspension of (X) the turbidity transforms to a ppt.

iii) When a paste of (X) in water is heated with ethyl alcohol a product of anaesthetic use is obtained.

Identify (X) and write down chemical equations for reactions at step (i), (ii) and (iii).

An inorganic Lewis acid (X) shows the following reactions :
i) It fumes in moist air.
ii) The intensity of fumes increases when a rod dipped in NH₄OH is brought near it.

iii) An acidic solution of (X) on addition of NH₄Cl and NH₄OH gives a precipitate which dissolves in NaOH solution.

iv) An acidic solution of (X) does not give a precipitate with H₂ Identify (X) and give chemical equation for steps (i) to (iii).
A yellow solid (A) is unaffected by acids and bases. It is not soluble in water. It dissolves slowly in hot conc. HNO₃ and brown gas (B) is released. The solid (A) dissolves only in a boiling solution of sodium sulphite giving a clear solution (C). Acidification of solution (C) causes a colourless gas (D) to be liberated, accompanied by the appearance of a milky precipiate (E) in the solution. Identify (A) to (E)
Identify A,B, C and D in the following sequence of reactions:
i) A + NaOH ¾¾® NaCl + NH₃ + H₂O
ii) NH₃ + CO₂ + H₂O ¾¾® B

iii) B + NaCl ¾¾® C + NH₄Cl

iv) C Na₂CO₃ + H₂O + D

Level- III

An unknown inorganic compound (X) loses its water of crystallisation on heating and its aqueous solution gives the following reactions.
i) It gives a white turbidity with dilute hydrochloric acid solution.
ii) It decolourises a solution of iodine in potassium iodide.

iii) It gives a white precipitate with silver nitrate solution which turns black on standing.

Identify the compound (X) and give chemical equations for the reactions at steps(i), (ii) and (iii).

The gas liberated, on heating a mixture of two salts with NaOH, gives a reddish brown precipitate with an alkaline solution of K₂HgI₄. The aqueous solution of the mixture on treatment with BaCl₂ gives a white precipitate which is sparingly soluble in conc. HCl. On heating the mixture of K₂Cr₂O₇ and conc. H₂SO₄ red vapours of A are produced. The aqueous solution of the mixture gives a deep blue colouration B with potassium ferricyanide solution. Identify the radicals in the given mixture and write the balanced equations for the formation of A and B.
When an orange coloured crystalline compound (A) was heated with common salt and concentrated sulphuric acid an orange – yellow coloured gas (B) was evolved. The gas (B) when passed through caustic soda solution gave a yellow solution (C) which in turn gave following reactions.
i) Addition of silver nitrate solution to (C) gave first a white precipitate and then red. Quantitatively, 0.155 g of the gas (B) required 2.0 m moles of AgNO₃ to produce the first trace of red colour.
ii) Acidification of the solution (C) with dil. H₂SO₄ gave an orange solution which contained chromium in +6 oxidation state. The solution liberated iodine from aqueous potassium iodide, leaving a green solution containing chromium in +3 oxidation state. Quantitatively 0.155 g of the gas (B) liberated 1.5 m mole of iodine.

Deduce the formula of A, B and C and explain the reactions.

A white amorphous powder (A) when heated, gives a colourless gas (B), which turns lime water milky (which dissolves on passing excess of gas (B) and the residue (C) which is yellow while hot but white when cold. The residue (C) dissolves in dilute HCl and the resulting solution gives a white precipitate on addition of potassium ferrocyanide solution. (A) dissolves in dil. HCl with the evolution of a gas which is identical in all respects to gas (B). The solution of (A) in dil. HCl gives a white ppt. (D) on addition of excess of NH₄OH and on passing H₂S gas. Another portion of this solution gives initially a white ppt. (E) on addition of NaOH solution which dissolves in excess of NaOH, the solution on passing again H₂S gives back the white ppt. of (E), the white ppt. on heating solution on passing again H₂S gas gives back the white ppt. of (E), the white ppt. on heating with dil H₂SO₄ give a gas used in II and IV group analysis. What are (A) to (E)? Give balanced chemical equation of the reaction.
A yellow fuming liquid (A) gives the following observations:
i) Its vapour undergoes partial dissociation into (B) and a gas chlorine.
ii) An passing H₂S in aqueous solution of (B) in acidic medium gives an orange precipitate (C).

iii) The orange precipitate © is soluble in yellow ammonium sulphide but solution on treatment with dil. HCl again liberates (C).

iv) (C) on boiling with conc. HCl again produces (B) and a gas (D) which gives violet colour with sodium nitroprusside and black precipitate with lead acetate solution.
v) (B) forms an unstable complex (E) with ammonium oxalate on passing H₂S in solution of (E), (B) is precipitate.
vi) (B) reacts with Fe powder to give black precipitate of a metalloid (F) which on boiling with conc. HCl gives (B) an yellow liquid.

Identify (A) to (F)

A colourless solid (A) on heating gives a white solid (B) and a colourless gas (C). (B) gives a reddish brown gas on treatment with dilute acids. When (B) is heated with solid NH₄Cl a colourless gas (D) and a residue (E) are obtained. When (A) is heated with (NH₄)₂SO₄ a colourless gas (F) is obtained alongwith a white residue (G). Both (E) and (G) imports yellow colour to flame. The gas (C) reacts with Mg to give white powder. The gas (D) is heated with calcium and the product on hydrolysis gives NH₃. Identify (A) to (G).

i) An inorganic compound (A) is formed on passing a gas (B) to a concentrated liquor containing sodium sulphate.
ii) On adding (A) into a dilute solution of silver nitrate, a white precipitate appears which quickly changes into a black coloured compound (C).

iii) On adding two (or) three drops of ferric chloride into excess of solution (A), a violet coloured compound (D) is formed. This colour disappears quickly.

iv) On adding a solution of (A) into the solution of cupric chloride, a white precipitate is first formed which dissolves on adding excess of (A) forming a compound (E).

(A) is binary compound of a univalent metal 1.422g of (A) reacts completely with 0.321 g of sulphur in an evacuated and sealed tube to give 1.743g of a white crystalline solid (B) that formed a hydrated double salt (C) with Al₂(SO₄)₃. Identify (A), (B) and (C).
i) A black coloured compound (B) is formed on passing hydrogen sulphide through the solution of a compound (A) in NH₄OH.
ii) (B) on treatment with hydrochloric acid and potassium chlorate gives (A)

iii) (A) on treatment with potassium cyanide gives a buff coloured precipitate which dissolves in excess of this reagent forming a compound (C).

iv) The compound (C) is changed into a compound (D) when its aqueous solution is boiled.
v) The solution of (A) was treated with excess of sodium bicarbonate and then with bromine water. On cooling and shaking for some time, a green colour of compound (E) formed. No change is observed on heating.

Identify (A) to (E) and give chemical equations for the reactions at steps (i) to (iv)

An aqueous solution of salt (A) gives a white crystalline precipitate (B) with NaCl solution. The filterate gives a black precipitate (C) when H₂S gas is passed through it. Compound (B) dissolves in hot water and the solution gives yellow precipitate (D) on treatment with KI and cooling, orange ppt. with K₂CrO₄ solution and white ppt. with dil. H₂SO₄ solution which is insoluble in C₂H₅OH. The compound (A) does not evolve any gas with dil. HCl, liberates a reddish brown gas on heating. Identify the compounds (A) to (D) and explain the reactions involved.
A hydrated metallic salt (A), light green in colour, gives a white anhydrous residue (B) after being heated gradually. (B) is soluble in H₂O and its aqueous solution reacts with NO to give a dark brown compound (C). (B) on strong heating gives a brown residue and a mixture of two gases (E) and (F). The gaseous mixture, when passed through acidified permanganate, discharges the pink colour and when passed through acidified BaCl₂ solution, gives a white precipitate. Identify (A), (B), (C), (D), (E) and (F). Explain the reactions involved.
An inorganic compound (A) when heated, decomposes completely to give only two gases (B) and (C). (B) is a neutral gas, fairly soluble in H₂O and itself decomposes on heating to different gases (D) and (E).

Compound (A) when warmed with NaOH gives another gas (F) which turns mercurous nitrate paper black. After sometime the gas (F) ceases to evolve, however its supply is restored by treating the residual solution with Al powder. Identify (A) to (F) giving the equations involved.

A certain inorganic compound (X) shows the following reactions :
i) On passing H₂S through an acidified solution of (X), a brown ppt. is obtained.
ii) The ppt. obtained in first step dissolves in excess of yellow ammonium sulphide.

iii) On adding an aq. solution of NaOH to solution of (X), first a white ppt. is obtained which dissolves in excess of NaOH.

iv) The aq. solution of (X) reduces ferric chloride

Identify the cation of X and give chemical equations for the steps (i), (ii), (iii) and (iv).

A mixture of two salts was treated as follows :
i) The mixture was heated with manganese dioxide and conc. sulphuric acid, when yellowish green gas was liberated
ii) The mixture on heating with sodium hydroxide solution gave a gas which turned red litmus blue.

iii) Its solution in water gave blue ppt. with potassium ferricyanide and red colouration with ammonium thiocyanate.

iv) The mixture was boiled with potassium hydroxide and the liberated gas was bubbled through an alkaline solution of K₂HgI₄ to give brown ppt. Identify the two salts. Give ionic equations for reactions involved in the tests (i), (ii) and (iii).
A scarlet compound (A) is treated with conc. HNO₃ to give a chocolate brown precipitate (B). The precipitate is filtered and the filtrate is neutralised with NaOH. Addition of KI to the resulting solution gives a yellow precipitate (C). The precipitate (B) on warming with conc. HNO₃ in the presence of Mn(NO₃)₂ produces a pink – coloured solution due to the formation of (D). Identify (A), (B), (C) and (D). Write the reaction sequence.
Assignments (Objective Problems)

Level- I

Which of the following doesn’t give flame colouration?

(A) MgCl₂ (B) BaCl₂

A substance that burns even in atmosphere of CO₂, with dazzling light is

(A) Na (B) K

The colours emitted by excited atoms are characteristic of the element. The element famous for the red emission of fireworks and warning flares is

(A) Pb (B) Mg

Which of the following elements (M) reacts with HNO₃ to form MO₂?

(A) P₄ (B) Mg

When a small amount of HCl is added to an aq. Solution of BiCl₃ a white precipitate is formed. This is due to

(A) Bi(OH)₃ (B) Bi₂O₃

A monobasic acid of phosphorus, which reduces HgCl₂ to black Hg is

(A) Hypophosphorus acid (B) Phosphoric acid

FeSO₄ gives brown ring with

(A) NO (B) N₂O₃

A substance which gives a yellow precipitate when boiled with an excess of HNO₃ and ammonium molybdate and a red precipitate with silver nitrate is

(A) Orthophosphate (B) Pyrophosphate

Phosphine gives black precipitate with

(A) NaCl (B) AgNO₃

When SO₂ is passed through an aq. Solution of I₂, it becomes colourless. This is due to

(A) Bleaching action of SO₂ (B) Formation of HI

A reddish brown gas, obtained on heating an inorganic compound with K₂Cr₂O₇ and conc. H₂SO₄ was bubbled through dil. NaOH. The alkaline solution yielded a yellow precipitate on addition of lead acetate. The inorganic compound is likely to be

(A) a chloride (B) a nitrate

A reagent used to distinguish between a chloride and a peroxide is

(A) H₂O (B) dil. H₂SO₄

Ferric ion forms a prussian blue coloured precipitate due to the formation of

(A) K₄[Fe(CN)₆] (B) Fe₄[Fe(CN)₆]

In the precipitation of the iron group in qualitative analysis ammonium chloride is added before adding ammonium hydroxide to

(A) decrease concentration of OH^– ions (B) prevent interference by phosphate ions

H₂S gas on passing through an alkaline solution forms a white precipitate. The solution contains ions is

(A) Pb (B) Zn

Level- II

Which salt has its aqueous solution, coloured?

(A) Zn(NO₃)₂ (B) LiNO₃

Which gives violet colour with borax

(A) Fe (B) Pb

Yellow ammonium sulphide solution is a suitable reagent used for the separation of

(A) HgS and PbS (B) PbS and Bi₂S₃

When Cl₂ water is added to an aq. Solution of potassium halide in the presence of chloroform, a violet colour is formed. On adding more of Cl₂ water, the violet colour disappears and a colourless solution is obtained. The test confirms the presence of ____________ in solution

(A) Iodide (B) Bromide

The silver sulphate solution is used to separate

(A) Nitrate and bromide (B) Nitrate and chlorate

Fe(OH)₃ can be separated from Al(OH)₃ by addition of

(A) dil. HCl (B) NaCl solution

Which sulphide is insoluble in dil. Acids but soluble in alkalies?

(A) PbS (B) CdS

An orange red precipitate obtained by passing H₂S through an acidified solution of an inorganic salt indicates the presence of

(A) Cadmium (B) Tin

The reagents NH₄Cl and aq. NH₃ will precipitate

(A) Ca²⁺ (B) Al³⁺

A white crystalline substance dissolves in hot water. On passing H₂S in this solution, a black precipitate is obtained. The black precipitate dissolves completely in hot HNO₃. On adding a few drops of conc. H₂SO₄ a white precipitate is obtained. This substance is

(A) BaSO₄ (B) SrSO₄

The presence of NH₄⁺ radical in solution can be detected

(A) Fehling’s solution (B) Benedict’s solution

Chloride of which element will be coloured?

(A) Ag (B) Hg

Conc. NaOH can separate a mixture of

(A) Al³⁺ and Cr³⁺ (B) Cr³⁺ and Fe³⁺

Potassium thiocyanate solution reacts with ferric chloride to give

(A) Pink colour (B) Deep blue colour

Ammonium molybdate is used to detect

(A) Fluoride ion (B) Sulphate ion

Answers to Objective Assignments

Level – I

A 2. D
C 4. D
C 6. A
A 8. D
B 10. C
A 12. B
B 14. A
B

level –II

C 2. D
D 4. A
A 6. C
D 8. C
B 10. C
D 12. D
B 14. D
C

Qualitative Analysis

Subjective Problems

Level – I

Ca₃(PO₄)₂
H₂S
Hg₂Cl₂
Na₂CO₃
Ammonium salt and NH₃ gas
NaBr
7. A = ZnCO₃

B = CO₂

C = CaCO₃

D = Ca(HCO₃)₂

E = ZnO

A = NH₃

B = CaCO₃

A = ZnCO₃

B = CO₂

C = ZnO

D = ZnS.

X = SO₂

Y = CO₂

Z = O₂

A = CO₂ B = H₂O

X = KHCO₃ Y = K₂CO₃

Z = BaCO₃

The reaction of gas (A) with but-2-ene followed by treatment with Zn/H₂O gives CH₃CHO. This shows that the gas (A) is ozone (O₃).
A = AgCl B = CuS, C = CaCO₃
(A) (NH₄)₂Cr₂O₇ (B) Cr₂O₃ (C) N₂ (D) Mg₃N₂ (E)NH₃

A – CO₂ B – H₂O Y – Na₂CO₃

Level – II

A = CrCl₃ B = Cr(OH)₃

C = Cr₂O₃ D = Na₂CrO₄

A = BiCl₃ B = BiOCl

C = Bi D = HCl

E = Hg₂Cl₂

A = P₂O₃ B = PH₃

C = H₃PO₄ D = Cu₃P₂

A = MgCl₂ B = Mg(NH₄)PO₄.6H₂O

C = Mg₄P₂O₇ D = NH₃

A = (NH₄)₂Cr₂O₇ B = N₂

C = Cr₂O₃ D = NH₃

A = MnO₂ B = K₂MnO₄

C = NO D = NO₂

E = KMnO₄

At first Bi(NO₃)₃ hydrolyses to give nitric acid which, being an oxidising agent, oxidises potassium iodide liberating free iodine responsible for dark brown precipitate. Iodine dissolves in excess of potassium iodide forming soluble KI₃ imparting yellow colour to solution.

Bi(NO₃)₃ + H₂O ¾¾® [Bi(OH) (NO₃)₂] + HNO₃

+ 4H⁺ + 3e^– ¾¾® NO₂ + 2H₂O ´ 2

2I^– ¾¾® I₂ + 2e^–] ´ 3

—————————————————————————

+ 8H⁺ + 6I^– ¾¾® 2NO + 4H₂O +

KI + I₂ ¾¾®

Treatment of compound A with NaOH to give alkaline gas C(NH₃) indicates that A is an ammonium salt. Further, heating of A to give non-reactive gas D(most probably N₂) indicates that A is nitrite. Thus the compound A is ammonium nitrite and its different reactions are represented as below.
i) + 2H₂O
ii) NH₄NO₂ + NaOH NaNO₂ + H₂O +

iii) NH₄NO₂ + HCl ¾¾® NH₄Cl + HNO₂

2HNO₂ ¾¾® H₂O + 2NO + O

2NO + O ¾¾®

Reaction of (Y) indicates it to be a chloride, Cl^–, while reaction of (Z) indicates it to be a sulphate, . Since both of them are derived by the action of KOH on (X), containing S, Cl and O.

and O containing compound +

(X) should be sulphuryl chloride, SO₂Cl₂ (Mol.wt. = 32 + 32 + 71 = 135) which coincides with its vapour density. All the given properties of the compound may be explained as below.

+ 2H₂O ¾¾® 2HCl + H₂SO₄

+ 4KOH ¾¾® +

+ AgNO₃ ¾¾® AgCl¯ + KNO₃

+ BaCl₂ ¾¾® BaSO₄ ¯ + 2KCl

Gas (B) when passed through copper sulphate solution reduces the latter to copper metal; this indicates that the gas (B) is phosphine (PH₃). Thus the inorganic iodide must be phosphonium iodide (PH₄I). The various reactions can be written as below.
i) PH₄I + KOH ¾¾® + + H₂O
ii) + 4O₂ ¾¾® + 3H₂O

iii) 3CuSO₄ + 2PH₃ ¾¾® Cu₃P₂ + 3H₂SO₄

iv) 2CuSO₄ + 4KI ¾¾® Cu₂I₂¯ + 2K₂SO₄ + I₂
Since the salt (X) swells on heating to give a glassy material, it seems to be borax, Na₂B₄O₇, a well-known compound showing this property. This is also in accordance with the fact that its aqueous solution is alkaline to litmus and its reaction with conc. H₂SO₄ to give crystals of a weak acid, H₃BO₃ (boric acid). Thus (X) is Na₂B₄O₇.10H₂O.

Chemical reactions.

i) + 7H₂O ¾¾®
ii) Na₂B₄O₇

iii) Na₂B₄O₇ + H₂SO₄ + 5H₂O ¾¾® Na₂SO₄ +

X – CaOCl₂

Reactions: i) CaOCl₂ +2CH₃COOH ¾¾® Ca(CH₃COO)₂ + Cl₂ + H₂O

2KI + Cl₂ ¾¾® 2KCl + I₂

ii) CaOCl₂(aq) + CO₂ ¾¾® CaCO₃ + Cl₂

white ppt.

iii) CaOCl₂ + H₂O ¾¾® Ca(OH)₂ + Cl₂

C₂H₅OH + Cl₂ ¾¾® CH₃CHO + 2HCI

CH₃CHO + 3Cl₂ ¾¾® CCl₃CHO CHCl₃

Anaesthetic

X – AlCl₃

Reactions: i) AlCl₃ + 3H₂O ® Al(OH)₃ + 3HCl

fumes

ii) HCl + NH₄OH ® NH₄Cl + H₂O

White fumes

iii) AlCl₃ + 3NH₄OH ® Al(OH)₃ + 3NH₄Cl

White ppt.

Al(OH)₃ + NaOH ® NaAlO₂ + 2H₂O

Soluble

Properties of the given compound, especially its solubility in boiling solution of Na₂SO₃ indicates that (A) is sulphur which explains all the given reactions.

¾¾® H₂SO₄ + + 2H₂O

Na₂SO₃ ¾¾®

¾¾® Na₂SO₄ + + H₂O + S¯

(A) NH₄Cl (B) NH₄H CO₃

Level – III

Since the compound X decolourises a solution of iodine in potassium iodide, it should contain thiosulphate ion, which also coincides with the two other given facts, i.e., (i) and (iii). Hence the compound X is sodium thiosulphate, Na₂S₂O₃.5H₂O which explains the given reactions as below .

Na₂S₂O₃.5H₂O Na₂S₂O₃ + 5H₂O

i) Na₂S₂O₃ + 2HCl ¾¾® 2NaCl + H₂O + SO₂ +
ii) ¾¾® + 2NaI

iii) ¾¾® + 2NaNO₃

Ag₂S₂O₃ + H₂O ¾¾® + H₂SO₄

Let us summarise the given facts of the question.

Red vapours oPf (A) Mixture of two salts Gas Reddish brown ppt.

Deep blue colour, (B) Aq. solution of the mixture White ppt.

The given reactions lead to following conclusions.

i) Heating of mixture with NaOH to give NH₃ gas (indicated by reddish brown ppt. with alkaline solution of K₂HgI₄) indicates the presence of ion in the mixture
ii) Heating of mixture with K₂Cr₂O₇ and conc. H₂SO₄ to give red vapours (of chromyl chloride) indicates the presence of Cl^– ion in the mixture.

iii) Reaction of aqueous solution of the mixture with barium chloride solution to give white ppt. (of BaSO₄) sparingly soluble in conc. HCl indicates the presence of ions in the mixture.

iv) Reaction of aqueous solution of the mixture with potassium ferricyanide solution to give deep blue colour indicates the presence of Fe²⁺ ions in the mixture.

Hence the mixture contains following four ions. , Fe²⁺and Cl^–

Equations for the formation of A and B

4NaCl + K₂Cr₂O₇ + 3H₂SO₄ K₂SO₄ + 2Na₂SO₄ + + 3H₂O

3Fe²⁺ + 2K₃[Fe(CN)₆] ¾¾®

Let us summarise the given reactions.

I₂ +

The above set of reactions indicates that compound A is K₂Cr₂O₇, B is chromyl chloride gas and C is sodium chromate which explains all the given reactions as below.

®+ 3H₂O

CrO₂Cl₂ + 4NaOH ¾¾® + 2NaCl + 2H₂O

Na₂CrO₄ + 2AgNO₃ ¾¾® + 2NaNO₃

2Na₂CrO₄ + H₂SO₄ ¾¾® Na₂Cr₂O₇ + Na₂SO₄ + H₂O

+ 14H⁺ + 6I^– ¾¾® 2Cr³⁺ + 7H₂O + 3I₂

The quantitative datas are explained in the following manner. From the above reactions we observe that

º Na₂CrO₄ º

155 g of CrO₂Cl₂ require 2 mole of AgNO₃

0.155 g of CrO₂Cl₂ will require = ´ 0.155 = 0.002 mole = 2 m mole

This coincides with the given data

Similarly, º 2CrO₄^2– º º

Thus 2 ´ 155 g of CrO₂Cl₂ liberate 3 moles of I₂

0.155g CrO₂Cl₂ will liberate = = 0.0015 moles = 1.5 mole

This also coincides with the given data.

a) +
b) a solution a white ppt.
c) A solution + B

From observations for section (a), one may conclude that the colourless gas is CO₂ because it turns lime water milky, due to formation of insoluble CaCO₃.

¾¾®

CaCO₃ is soluble in excess of CO₂ due to formation of soluble calcium bicarbonate

CaCO₃ + CO₂ + H₂O ¾¾®

The compound (C) is zinc oxide (ZnO) because it is yellow in hot and white in cold, hence due the initial compound (A) is zinc carbonate (ZnCO₃).

From section (b), it is inferred that (C) is a salt of Zn(II) which dissolves in dil. HCl and white ppt. obtained after addition of K₄Fe(CN)₆ is due to zinc ferrocyanid, a test of Zn⁺⁺ cation.

The data of collection (C) proved that (A) is ZnCO₃ because on treatment with dil. HCl it gives gas (B), i.e., CO₂, while Zn (II) goes in solution, i.e., ZnCl₂, on passing H₂S gas in presence of NH₄OH, it gives a white ppt. of ZnS(D). ZnS on heating with dil. H₂SO₄ evolves H₂S, which is used for the precipitation of sulphides of group II in acidic medium and of IV group in alkaline medium. ZnCl₂, reacts with NaOH to give a ppt. of Zn(OH)₂ which dissolves in NaOH, as Zn(OH)₂ is amphoteric in nature. The solution Na₂ZnO₂ again gives ZnS on passing H₂S gas into it. Different chemical equations concerned to the above numerical are given below:

¾¾®

ZnCl₂ + 2NaOH ¾¾®

¾¾®

A = SbCl₅ B = SbCl₃

C = Sb₂S₃ D = H₂S

E = (NH₄)₃[Sb(C₂O₄)₃] F = Sb

A = NaNO₃ B = NaNO₂

C = O₂ D = N₂

E = NaCl F = N₂O

G = Na₂SO₄

i) Na₂S + Na₂SO₃ + ¾® + 2NaI
ii) Na₂S₂O₃ + AgNO₃ ¾® + 2NaNO₃

Ag₂S₂O₃ + H₂O ¾® + 6NaCl

iii) 3Na₂S₂O₃ + 2FeCl₃ ¾® + 6NaCl

Fe₂(S₂O₃)₃ ¾® 2Fe²⁺ + S₄O₆^{– –}

iv) Na₂S₂O₃ + CuCl₂ ¾® CuS₂O₃ + 2NaCl

CuS₂O₃ + Na₂S₂O₃ ¾® + 4Na₂S₄O₆

3Cu₂S₂O₃ + 2Na₂S₂O₃ ¾®

(B) forms double salt with Al₂(SO₄)₃ and thus may be K₂SO₄

(A) + S ¾® (B) K₂SO₄

1.743g K₂SO₄ is obtained by 1.433g of A

174g K₂SO₄ is obtained by = 142g of A

174g K₂SO₄ requires 32g S

1.743g K₂SO₄ requires = 0.321g of S

Thus given data confirms (B) is K₂SO₄

2(A) + S ¾® K₂SO₄

Molecular weight of A ´ 2 = 142

Molecular weight of A = 71

Since, (A) is a potassium salt,

Molecular weight of left component = 71– 34 = 32

Thus salt is KO₂

The formation of black coloured compound (B) by passing H₂S through the alkaline solution of the compound (A) indicate that (A) is a salt of the IV group radicals (Co²⁺, Ni²⁺, Mn²⁺ or Zn²⁺). However, the given reactions especially reaction (iii) indicates what compound (A) is a cobalt salt (CoCl₂) which explains all the given reactions
i) + 2NH₄OH + H₂S ¾¾®
ii) CoS + 2HCl + ¾¾® CoCl₂ + H₂S

2KICO₃ ¾¾® 2KCl + 3O₂

iii) CoCl₂ + 2KCN ¾¾® + 2KCl

Co(CN)₂ + 4KCN ¾¾®

iv) 2K₄[Co(CN)₆] + O + H₂O ¾¾®
v) CoCl₂ + 6NaHCO₃ ¾¾® Na₄[Co(CO₃)₃] + 2NaCl + 3CO₂ + 3H₂O

2Na₄[Co(CO₃)₃] + 2NaHCO₃ + O ¾¾® + 2Na₂CO₃ + H₂O

(A) is Pb(NO₃)₂, (B) PbCl₂

(A) FeSO₄.7H₂O (B) FeSO₄

(E) SO₂ (F) SO₃

(A) NH₄NO₃ (B) N₂O

(E) O₂ (F) NH₃

Cation of X – Sn⁺²

Reactions – i) Sn⁺² + H₂S SnS ¯

brown ppt.

ii) Sn⁺² + 2NaOH ¾® 2Na⁺ + Sn(OH)₂ ¯

white ppt.

Sn(OH)₂ + 2 NaOH ¾® Na₂SnO₂ + 2H₂O

Soluble

iii) Sn⁺² + 2Fe⁺³ ¾¾® Sn⁺⁴ + 2Fe⁺²

Mixture – FeCl₂ and NH₄Cl

Reactions i) 2NH₄Cl + MnO₂ + 2H₂SO₄MnSO₄+Cl₂ + 2H₂O +(NH₄)₂SO₄

ii) NH₄Cl + NaOH NH₃ + H₂O

iii) 3FeCl₂ + 2K₃[Fe(CN)₆] ¾¾¾® + 6KCl

(A) – Pb₃O₄

Reactions : (i) Pb₃O₄ + 4HNO₃ ¾® 2Pb(NO₃)₂ + PbO₂¯ + H₂O

(B)

(ii) Pb(NO₃)₂ + 2KI ¾® PbI₂ ¯ + 2KNO₃

(C)

5PbO₂ + 2Mn(NO₃)₂ + 4HNO₃ ® 4Pb(NO₃)₂ + Pb(MnO₄)₂+ 2H₂O

(D)

About The Author

admin

Editable Study Material for JEE mains, JEE advanced, Foundation, CBSE. Online Tuition, Online Tutors, Home Tutors, Coaching, Online Classes by Study Innovations IIT JEE study material, NEET Study Material, JEE mains Study Material, JEE advanced study material, AIIMS Study Material, IIT JEE Foundation study material, NEET Foundation study material, CBSE Study Material, Test Series, Question Bank, ICSE Study Material, School Exams study material, board exams study material, XII board exams Study Material, X board exams Study Material, Study Material, JEE mains, JEE advanced, Video Lectures, Study Innovations, online tuition, home tuition, online tutors, coaching & tutorials for English, Mathematics, Science, Physics, Chemistry, Biology, soft copy study material, editable study material, study material provider,customised study material,customised study material provider for coaching institutes,study material for coaching classes, how to make study material for coaching institute, jee neet study material,

1. IIT—JEE Syllabus

2. Introduction

3. Classification Of Anions

Level- I

Level- II

Level- III

Level- I

Qualitative Analysis

Subjective Problems

Level – I

Level – II

Level – III

About The Author

admin

Quick Links