1. IIT—JEE Syllabus
Systematic analysis of various cations and anions excluding interfering radicals.
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.
|Brisk effervescence in cold with evolution of colourless and odourless gas||CO32–
|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||SO32–
|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||S2–
|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–
|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–
|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.
|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–
|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–
|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–
|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–
|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||C2O42–
|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 – CO32-, SO32-, S2-, NO2–, CH3COO–, S2O32-
- Carbonate (CO32- ) :
- i) Dilute HCl : gives effervescence, due to the evolution of carbon dioxide
CO32- + 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.
CO32- + Ba2+ ¾® BaCO3 ¯
CO32- + Ca2+ ¾® CaCO3 ¯
iii) Silver nitrate solution : White ppt. of silver carbonate is obtained.
CO32- + 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 (SO32-) :
- i) Dilute HCl or Dilute H2SO4 : decomposes with the evolution of sulphur dioxide
SO32- + 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
iii) Lime water : On passing the gas through lime water, a milky ppt is formed.
SO2 + Ca(OH)2 ¾® CaSO3 ¯ + H2O
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
S2- + 2H+ ¾® H2S
- ii) Turns lead acetate paper black
(CH3COO)2Pb + H2S ¾® PbS ¯ + 2CH3COOH
iii) Gives yellow ppt.with CdCO3
Na2S + CdCO3 ¾® CdS¯ + Na2CO3
Na2S + Na2[Fe(CN)5NO] ®
- iv) Silver nitrate solution: black ppt. of silver sulphide insoluble in cold but soluble in hot dil. nitric aicd.
S2– + 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– + S4O62-
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+ + SO42-
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 + 2SO42- + 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
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
Na2CrO4 + (CH3COO)2 Pb ® 2CH3COONa + PbCrO4 ¯
- Bromide (Br–)
- i) H2SO4 : Gives reddish brown vapours of bromine accompanied by hydrogen bromide
2KBr + H2SO4 ® K2SO4 + 2HBr
2HBr + H2SO4 ® 2H2O + SO2 + Br2
- 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
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 + SO42- + 2H2O + SO2
- 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+ + 7SO42- + 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 + 2S2O32- ¾¾® 2I– + S4O62-.
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 + 2SO42- + 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 : SO42-
- ii) Oxidation and reduction in solution – CrO42 – ,Cr2O, MnO4–
- Sulphate (SO42-): 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.
SO42- + Ba2+ ¾¾® BaSO4 ¯
- ii) Mercury (II) nitrate solution : Gives yellow ppt. of basic mercury (II) sulphate.
SO42- + 3Hg2+ + 2H2O ® HgSO4.2HgO ¯ + 4H+
- Chromate CrO42 –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.
CrO42- + 2Ag+ ¾® Ag2CrO4 ¯
2 Ag2CrO4 + 2H+ ¾® 4 Ag+ + Cr2O72- + H2O
Ag2CrO4 ¯ + 4NH3 ¾® 2[Ag(NH3)2]+ + CrO42-
Ag2CrO4 ¯ + 2Cl– ¾® 2AgCl + CrO42-
A reddish brown ppt. of silver dichromate Ag2Cr2O7 is formed with a conc. solution of a dichromate.
Cr2O72- + 2Ag+ ¾® Ag2Cr2O7
iii) Lead acetate solution : Yellow ppt. of lead chromate PbCrO4 insoluble in acetic acid, but soluble in dil. nitric acid
CrO42- + Pb2+ ¾¾® PbCrO4 ¯.
2PbCrO4 ¯ + 2H+ 2Pb2+ + Cr2O72- + H2O.
- iv) H2O2 : If an acidic solution of a chromate is treated with H2O2 a deep blue solution of chromium penta oxide is obtained.
CrO42- + 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 ´ 10–10 and Hg 2Cl2 = 3.5 ´ 10–18
Let us consider the precipitation of AgCl. From the equilibrium of AgCl in a saturated solution.
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 KsAgCl. KsAgCl 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 KsAgCl 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
but KsMS 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
In presence of NH4Cl which being a strong electrolyte dissociates into
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|
|3.||NH4OH in presence of NH4Cl||Fe3+||Fe(OH)3; Reddish brown|
|4.||H2S in presence of NH4OH||Zn2+||ZnS; Greenish White|
|5||(NH4)2CO3 in presence of NH4OH||Ba2+||BaCO3; White|
|7||NaOH||NH4+||Ammonia gas is evolved|
4.3 Physical Examination of Salt / Mixture
|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
|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 S2- due to common ion effect. Hence decreased conc. of S2- 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 SO42- 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+ + S2-. H+ + OH– ¾¾¾® H2O
Now the excess of S2- 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
PbCrO4 + 4NaOH ® Na2[PbO2] + Na2CrO4 + 2H2O
- B) PbCl2 + 2KI ® PbI2 ¯ + 2KCl
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 ¯
Hg2Cl2 + SnCl2 ® SnCl4 + 2Hg ¯
- 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
When treated with sodium stannite a black ppt. is obtained.
2BiCl3 + 3Na2SnO2 ¾¾® 2Bi ¯ + 3Na2SnO3 + 6NaCl + 3H2O
- iv) Cd+2 ions in solution, with H2S give yellow precipitate.
Cd+2 + H2S ¾® CdS ¯ + 2H+
- Group II B (As3+, As5+, Sb3+, Sb5+, Sn2+, Sn4+)
- i) As+3 ions in solution give yellow precipitate with ammonium molybdate and HNO3.
H3AsO4 +12(NH4)2MoO4 +21HNO3 ¾® (NH4)3 AsMo12O40¯+ 21NH4NO3 + 12H2O
iii) Sn2+ ions in solution as SnCl2 give white ppt. with HgCl2 ,which turns black on standing.
SnCl2 + 2HgCl2 ¾¾® SnCl4 + Hg2Cl2 ¯
Hg2Cl2 + SnCl2 ¾¾®SnCl4 + 2Hg ¯
iii) Sb+3 ions in solution as SbCl3 , on addition of water give white precipitate.
SbCl3 + H2O ® SbOCl¯ + 2HCI
- 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
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
- ii) Zn+2 ions in solution give white precipitate with NaOH, which dissolve in excess of NaOH.
Zn+2 + 2NaOH ® Zn(OH)2 ¯ + 2Na+
Zn(OH)2 + 2NaOH ® Na2ZnO2 + 2H2O
iii) Mn+2 ions in solution give pink precipitate with NaOH turning black or brown on heating.
Mn+2 + 2NaOH Mn(OH)2 + 2Na+
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+
- B) White precipitate with (NH4)2SO4
Ba+2 + (NH4)2 SO4 ® BaSO4¯ +
- C) White precipitate with (NH4)2 C2O4
Ba+2 + (NH4)2C2O4 ® BaC2O4¯ +
- ii) Sr+2 ions give white precipitate with (NH4)2SO4 and (NH4)2C2O4
Sr+2 + (NH4)2SO4 ® SrSO4¯ +
Sr+2 + (NH4)2C2O4 ® SrC2O4¯ +
iii) Ca+2 ions give white precipitate with only (NH4)2 C2O4.
Ca+2 + (NH4)2C2O4 ® CaC2O4 ¯ +
- 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
- B) Paper soaked in CuSO4 solution, becomes deep blue by NH3 due to complex formation
CuSO4 + 4NH3 ¾¾® [Cu(NH3)4]SO4
- C) With Hg2 (NO3)2 , a black colour is obtained
Hg2(NO3)2 + 2NH3 ¾¾® Hg ¯ + Hg(NH2)NO3 ¯ + NH4NO3
- 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
iii) Sodium salts gives heavy white ppt. with potassium dihydrogen antimonate
KH2SbO4 + NaCl ¾¾® NaH2SbO4 ¯ + KCl
- 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|
|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|
|Iron||Brown-Yellow||Pale-Yellow||Bottle Green||Bottle Green|
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|
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
- 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
|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|
|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
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
+ 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)
‘E’ is Cr2(SO4)3 which gives green colour both in oxidising flame and reducing flame
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). 
NaNO2 NO2 (brown gas)
+ NH4Cl + 2H2O
NaNO3 (A) and Na2SO4 (G) impart yellow colour to flame indicating presence of Na.
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 = SO2 D = SO3
E = Fe2O3 F = 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
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) + 2NO2(g) + O2(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
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 S2– 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)
- 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.
- 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)
- 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
- 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)
- Which of the following doesn’t give flame colouration?
(A) MgCl2 (B) BaCl2
(C) CaCO3 (D) SrCO3
- A substance that burns even in atmosphere of CO2, with dazzling light is
(A) Na (B) K
(C) Ca (D) Mg
- 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
(C) Sr (D) Ba
- Which of the following elements (M) reacts with HNO3 to form MO2?
(A) P4 (B) Mg
(C) Zn (D) Sn
- 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
- A monobasic acid of phosphorus, which reduces HgCl2 to black Hg is
(A) Hypophosphorus acid (B) Phosphoric acid
(C) Metaphosphoric acid (D) Pyrophosphoric acid
- FeSO4 gives brown ring with
(A) NO (B) N2O3
(C) NO2 (D) N2O5
- 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
(C) Metabosphate (D) Aresenate
- Phosphine gives black precipitate with
(A) NaCl (B) AgNO3
(C) AlCl3 (D) CuSO4
- 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
(C) Combination of SO2 and I2 (D) Formation of HNO3
- 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
(C) a bromide (D) a sulphide
- A reagent used to distinguish between a chloride and a peroxide is
(A) H2O (B) dil. H2SO4
(C) KOH solution (D) NaCl
- Ferric ion forms a prussian blue coloured precipitate due to the formation of
(A) K4[Fe(CN)6] (B) Fe4[Fe(CN)6]
(C) KMnO4 (D) Fe(OH)3
- 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
(C) increase concentration of Cl– ions (D) increase concentration of NH4+ ions
- H2S gas on passing through an alkaline solution forms a white precipitate. The solution contains ions is
(A) Pb (B) Zn
(C) Cu (D) Ni
- Which salt has its aqueous solution, coloured?
(A) Zn(NO3)2 (B) LiNO3
(C) Co(NO3)2 (D) Potash Alum
- Which gives violet colour with borax
(A) Fe (B) Pb
(C) Co (D)Mn
- Yellow ammonium sulphide solution is a suitable reagent used for the separation of
(A) HgS and PbS (B) PbS and Bi2S3
(C) Bi2S3 and CuS (D) CdS and As2S3
- 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
(C) Chloride (D) Iodide and Bromide
- The silver sulphate solution is used to separate
(A) Nitrate and bromide (B) Nitrate and chlorate
(C) Bromide and iodide (B) Nitrate and Nitrite
- Fe(OH)3 can be separated from Al(OH)3 by addition of
(A) dil. HCl (B) NaCl solution
(C) NaOH solution (D) NH4Cl and NH4OH
- Which sulphide is insoluble in dil. Acids but soluble in alkalies?
(A) PbS (B) CdS
(C) FeS (D) As2S3
- An orange red precipitate obtained by passing H2S through an acidified solution of an inorganic salt indicates the presence of
(A) Cadmium (B) Tin
(C) Antimony (D) Bismuth
- The reagents NH4Cl and aq. NH3 will precipitate
(A) Ca2+ (B) Al3+
(C) Mg2+ (D) Zn2+
- 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
(C) PbCl2 (D) CdSO4
- The presence of NH4+ radical in solution can be detected
(A) Fehling’s solution (B) Benedict’s solution
(C) Schiff’s reagent (D) Nessler’s reagent
- Chloride of which element will be coloured?
(A) Ag (B) Hg
(C) Zn (D) Co
- Conc. NaOH can separate a mixture of
(A) Al3+ and Cr3+ (B) Cr3+ and Fe3+
(C) Al3+and Zn3+ (D) Zn2+ and Pb2+
- Potassium thiocyanate solution reacts with ferric chloride to give
(A) Pink colour (B) Deep blue colour
(C) Green colour (D) Blood red colour
- Ammonium molybdate is used to detect
(A) Fluoride ion (B) Sulphate ion
(C) Phosphate ion (D) Nitrate 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
- C 2. D
- D 4. A
- A 6. C
- D 8. C
- B 10. C
- D 12. D
- B 14. D
Level – I
- Ammonium salt and NH3 gas
- 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.
- 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
iii) CaOCl2 + H2O ¾¾® Ca(OH)2 + Cl2
C2H5OH + Cl2 ¾¾® CH3CHO + 2HCI
CH3CHO + 3Cl2 ¾¾® CCl3CHO CHCl3
- X – AlCl3
Reactions: i) AlCl3 + 3H2O ® Al(OH)3 + 3HCl
- ii) HCl + NH4OH ® NH4Cl + H2O
iii) AlCl3 + 3NH4OH ® Al(OH)3 + 3NH4Cl
Al(OH)3 + NaOH ® NaAlO2 + 2H2O
- 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
¾¾® Na2SO4 + + H2O + S¯
- (A) NH4Cl (B) NH4H CO3
(C) NaHCO3 (D) CO2
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.
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.
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 2CrO42– 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.
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:
ZnCl2 + 2NaOH 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
(C) PbS (D) PbI2 or PbSO4 or PbCrO4
- (A) FeSO4.7H2O (B) FeSO4
(C) [Fe(H2O)5NO]SO4 (D) Fe2O3
(E) SO2 (F) SO3
- (A) NH4NO3 (B) N2O
(C) H2O (D) N2
(E) O2 (F) NH3
- Cation of X – Sn+2
Reactions – i) Sn+2 + H2S SnS ¯
- ii) Sn+2 + 2NaOH ¾® 2Na+ + Sn(OH)2 ¯
Sn(OH)2 + 2 NaOH ¾® Na2SnO2 + 2H2O
iii) Sn+2 + 2Fe+3 ¾¾® Sn+4 + 2Fe+2
- Mixture – FeCl2 and NH4Cl
Reactions i) 2NH4Cl + MnO2 + 2H2SO4 MnSO4+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
(ii) Pb(NO3)2 + 2KI ¾® PbI2 ¯ + 2KNO3
- 5PbO2 + 2Mn(NO3)2 + 4HNO3 ® 4Pb(NO3)2 + Pb(MnO4)2+ 2H2O
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