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 natural 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.
2.1 Physical Examination of Salt / Mixture
|1. Substance is coloured|
|i) Blue||Copper salt|
|ii) Dark green||Chromium salt|
|iii) Green||Salts of Fe(II), Ni, Cu or Cr|
|iv) Light yellow or brown||Salts of Fe(III)|
|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 deliquescent||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|
2.2 Effect of Heating
|1. Substance melts||Salts of alkali metals and salts having water of crystallisation.|
|2. Substance decripitates (crackling 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)2 → 2CuO + 4NO2 + O2
|ii) Br2 – Reddish brown||Bromides
2CdBr2 + O2 → 2CdO + 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|
2.3 Flame Test
3. Classification of Anions
Methods available for the detection of anions are not as systematic as those described for the detection of cations. Furthermore anions are classified essentially on the basis of the process employed.
3.1 Class A
Includes anions that are identified by volatile products obtained on treatment with acids. It is further divided into two sub groups.
- i) Gases evolved with dil. HCl/ dil H2SO4.
- ii) Gases or acid vapours evolved with conc. H2SO4
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+ ⎯→ CO2↑ + H2O
The gas gives turbidity with lime water and baryta water.
CO2 + Ca2+ + 2OH– ⎯→ CaCO3 ↓ + H2O
CO2 + Ba2+ + 2OH– ⎯→ 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 ⎯→ Ca(HCO3)2
The following tests are performed with the aqueous salt solution.
- ii) Barium chloride or Calcium chloride solution: White ppt of barium or calcium carbonate is obtained, which is soluble in mineral acid.
CO32- + Ba2+ ⎯→ BaCO3 ↓
CO32- + Ca2+ ⎯→ CaCO3 ↓
iii) Silver nitrate solution: White ppt of silver carbonate is obtained.
CO32- + 2Ag+ ⎯→ Ag2CO3↓
The ppt so obtained is soluble in nitric acid and in ammonia. The ppt becomes yellow or brown on addition of excess reagent and same may also happen if the mix is boiled, due to the formation of silver oxide
Ag2CO3↓ ⎯→ Ag2O ↓ + CO2 ↑
- Sulphites (SO32-)
- i) Dilute HCl or Dilute H2SO4 : decomposes with the evolution of sulphur dioxide
SO32- + 2H+ ⎯→ SO2 + H2O
The gas has a suffocating odour of burning sulphur.
The following tests are performed with the aqueous salt solution.
- ii) Acidified potassium dichromate solution: Turns filter paper moistened with acidified potassium dichromate solution, green due to the formation of Cr3+
3SO2 + K2Cr2O7 + H2SO4 ⎯→ K2SO4 + Cr2(SO4)3 + H2O
iii) Lime water : On passing the gas through lime water, a milky ppt is formed.
SO2 + Ca(OH)2 ⎯→ CaSO3 ↓ + H2O
Precipitate dissolves on prolonged passage of the gas, due to the formation of hydrogen sulphite ions.
CaSO3 ↓ + SO2 + H2O ⎯→ 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: A colourless gas smelling of rotten eggs (H2S) is evolved.
S2- + 2H+ ⎯→ H2S↑
- ii) The gas turns lead acetate paper black
(CH3COO)2Pb + H2S ⎯→ PbS ↓ + 2CH3COOH
The following tests are performed with the aqueous salt solution.
iii) Gives yellow ppt. with CdCO3
Na2S + CdCO3 ⎯→ CdS↓ + Na2CO3
- iv) Silver nitrate solution: black ppt. of silver sulphide insoluble in cold but soluble in hot dil nitric acid.
S2- + 2Ag+ ⎯→ Ag2S ↓
- v) Sodium nitroprusside solution : Turns sodium nitroprusside solution purple
Na2S + Na2[Fe(CN)5NO] → Na4[Fe(CN)5NOS]
- Nitrites (NO2–)
- i) Dil 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 the air
NO2– + H+ ⎯→ HNO2
2HNO2 ⎯→ H2O + N2O3
3HNO2 ⎯→ HNO3 + 2NO↑ + H2O
2NO↑ + O2↑ ⎯→ 2NO2↑
The following tests are performed with the aqueous salt solution.
- ii) Silver nitrate solution : White crystalline ppt. is obtained
NO2– + Ag+ → AgNO2↓
iii) Turns acidified KI – starch paper blue
2KI + 2NO2 ⎯→ 2KNO2 + I2 ↑
Starch + I2 ⎯→ 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 is formed, due to the formation of [FeNO]SO4 at the junction of the two liquids.
NO2– + CH3COOH ⎯→ HNO2 + CH3COO–
3HNO2 ⎯→ H2O + HNO3 + 2NO↑
FeSO4 + NO ⎯→ [FeNO]SO4.
- Acetate (CH3COO–)
- i) Dilute Sulphuric Acid : Smell of vinegar is observed.
CH3COO– + H+ ⎯⎯→ CH3COOH ↑
The following test is performed with the aqueous salt solution.
- ii) Iron (III) Chloride Solution: Gives deep – red colouration
CH3COONa + FeCl3 ⎯⎯→ (CH3COO)3Fe + 3NaCl
- Thiosulphates ()
- i) Dil Hydrochloric acid: Gives sulphur & sulphur di oxide
S2O+ 2H+ ⎯→ S ↓ + SO2↑ + H2O
- ii) Iodine Solution: Gets decolourised due to reduction of iodine to iodide ions.
I2 + 2S2O⎯→ 2I– + S4O62-
The following tests are performed with the aqueous salt solution.
iii) Barium chloride solution: White ppt. of barium thiosulphate is formed
S2O+ Ba2+ ⎯→ BaS2O3 ↓
But no ppt. is obtained with CaCl2 solution.
- iv) Silver nitrate solution: Gives white ppt. of silver thiosulphate.
S2O+ 2Ag+ ⎯→ Ag2S2O3 ↓
The ppt. is unstable, turning dark on standing, when silver sulphide is formed.
Ag2S2O3↓ + H2O ⎯→ Ag2S + H2SO4
- v) Lead acetate or Lead nitrate solution: Gives white ppt.
+ Pb2+ ⎯→ PbS2O3 ↓
On boiling it turns black due to the formation of PbS.
PbS2O3 ↓ + H2O ⎯→ PbS ↓ + 2H+ + SO42-
Exercise 1: i) You are provided with two test tubes containing an aqueous solution of carbonate and sulphite respectively. How will you identify them?
- ii) A colourless salt on heating, forms a white sublimate on the cooler sides of the test tube. The gas evolved is colourless and has a pungent smell. It gives dense white fumes when a rod dipped in conc. HCl is brought near the mouth of the test tube. Identify the basic radical present in the salt.
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 near 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 coloured gas is obtained.
MnO2 + 2H2SO4 + 2Cl– → Mn2+ + Cl2 ↑+ 2SO42- + 2H2O
The following tests are performed with the aqueous salt solution.
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 salt 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 ⎯⎯→ Na2CrO4 + 2HCl
Na2CrO4 + (CH3COO)2 Pb → 2CH3COONa + PbCrO4 ↓
- Bromide (Br–)
- i) H2SO4 : Gives reddish brown vapours of bromine accompanying the hydrogen bromide.
2KBr + H2SO4 → K2SO4 + 2HBr
2HBr + H2SO4 → 2H2O + SO2 ↑ + Br2↑
- ii) Manganese dioxide and conc. sulphuric acid : When a mix of solid bromide, MnO2 and conc. H2SO4 is heated reddish brown vapours of bromine are evolved.
2KBr + MnO2 + 2H2SO4 → Br2 ↑ + K2SO4 + MnSO4 + 2H2O
The following tests are performed with the aqueous salt solution.
iii) Silver nitrate solution: 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+ ⎯→ AgBr
- iv) Lead acetate solution: White crystalline ppt. of lead bromide which is soluble in
2Br– + Pb+2 ⎯⎯→ PbBr2 ↓
- v) Chlorine water: When chlorine water is added to a mixture of CCl4 and a solution of bromide liberates free bromine is liberated, which colours the organic layer orange – red.
2KBr + Cl2 (water) ⎯→ 2KCl + Br2
Br2 + CCl4 ⎯→ Orange red colour
- vi) Potassium dichromate & conc. H2SO4 : When a mix of solid bromide, K2Cr2O7, and conc. H2SO4 is heated and on 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 ⎯→ I2 + SO42- + 2H2O + SO2 ↑
The following tests are performed with the aqueous salt solution.
- ii) Silver nitrate solution: Yellow ppt. of silver iodide AgI, very slightly soluble in conc. ammonia solution and insoluble in dil nitric acid.
I– + Ag+ ⎯→ AgI
iii) Lead acetate solution: Yellow ppt. of lead iodide soluble in hot water forming a colourless solution & yielding golden yellow plates (spangles) on cooling.
2I– + Pb2+ ⎯→ PbI2↓
- iv) Potassium dichromate & conc. sulphuric acid: Iodine is liberated
6I– + Cr2O7-2 + 7H2SO4 ⎯⎯→ 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 layer is obtained.
2I– + Cl2 ⎯→ I2 + 2Cl–
I2 + chloroform ⎯⎯→ violet coloured layer.
- 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+ ⎯⎯→ 2CuI + I2
I2 + 2S2O32- ⎯⎯→ 2I– + S4O62-.
vii) Mercury (II) chloride solution: Forms scarlet red ppt. of HgI2
2I– + HgCl2 ⎯→ HgI2 ↓ + 2Cl–.
This ppt. dissolves in excess of KI, forming tetraiodo mercurate (II) complex.
HgI2 + 2I– ⎯⎯→ [HgI4]2-
- Nitrate (NO3– )
Action of heat : The result varies with the metal
- Nitrates of sodium and potassium evolve oxygen (test with glowing splint) & leave solid nitrites (brown fumes with dilute acid)
2NaNO3 ⎯→ 2NaNO2 + O2↑.
- Ammonium nitrate yields dinitrogen oxide & steam
NH4NO3 ⎯→ N2O ↑ + 2H2O.
- Nitrates of the noble metals leave a residue of the metal and a mixture of nitrogen dioxide and oxygen is evolved.
2AgNO3 ⎯→ 2Ag + 2NO2 ↑ + O2↑.
- 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 ⎯→ 2PbO + 4NO2↑ + O2↑.
- i) Conc H2SO4 : Gives reddish – brown vapours of nitrogen dioxide
4NO3– + 2H2SO4 ⎯→ 4NO2↑ + 2SO42- + 2H2O + O2↑
The following test is performed with the aqueous salt solution.
- ii) Brown ring test: When freshly saturated solution of iron (II) sulphate is added to nitrate solution and conc. H2SO4 is poured slowly down the side of the test – tube, a brown ring is obtained.
2NO3– + 4H2SO4 + 6Fe2+ ⎯→ 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.
Exercise 2: i) When silver nitrate solution is added to a salt solution of (A), a black precipitate is obtained. (A) decolorises acidified permanganate solution. Identify the acid radical in (A).
- ii) On heating the salt, decripitation takes place. On heating a mixture of salt, manganese dioxide and conc. H2SO4, a greenish yellow gas is evolved. The gas evolved turned moist blue litmus red and then turned it colourless. On performing the flame test, the salt imparted a crimson red colour to the flame. Identify the cation and anion in the salt.
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+ ⎯⎯→ 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 gives 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+ ⎯⎯→ BaCrO4 ↓
Dichromate ion 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 insoluble in acetic acid.
CrO42- + 2Ag+ ⎯→ Ag2CrO4 ↓
2 Ag2CrO4 + 2H+ ⎯→ 4 Ag+ + Cr2O72- + H2O
Ag2CrO4 ↓ + 4NH3 ⎯→ 2[Ag(NH3)2]+ + CrO42-
Ag2CrO4 ↓ + 2Cl– ⎯→ 2AgCl + CrO42-
A reddish brown ppt. of silver dichromate Ag2Cr2O7 is formed with a conc. solution of a dichromate.
Cr2O72- + 2Ag+ ⎯→ Ag2Cr2O7
iii) Lead acetate solution: Yellow ppt. of lead chromate PbCrO4 insoluble in acetic acid, but soluble in dil nitric acid
CrO42- + Pb2+ ⎯⎯→ 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 ⎯→ 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+ ⎯⎯→ 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+ ⎯⎯→ 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 ⎯⎯→ K2MnO4 + MnO2 + O2↑
Exercise 3: i) A salt solution gives a yellow ppt. with BaCl2 which is soluble in dil. HCl. On adding acid to the salt solution, it becomes orange in colour. Identify the acid radical.
- ii) On treating the salt with dil. HCl, a colourless gas turning brown on exposure to air was evolved. On adding FeSO4 solution, the salt solution gives a brown colour. Identify the acid radical present in the salt.
- Classification of Cations
For the purpose of systematic qualitative analysis, cations are classified into five groups on the basis of their behaviour with 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 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 greater than 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 KspAgCl but it becomes smaller and smaller and approaches 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 Cl– 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 precipitate 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 behaves 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, depresses the dissociation of H2S 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 sulphides of Gr.II but insufficient to exceed the comparatively high solubility products of the sulphides of the metals of Gr. III (B). Or in other words with this decreased S– – ions concentrations.
but KspMS of Gr. III (B).
That is why sulphides 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
NH4Cl being a strong electrolyte dissociates into NH4Cl ⎯→
The dissociation of NH4OH is 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 will be 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 concentration 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 ionic 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)PO4 being insoluble is precipitated.
4.2 Separation of Cations
|Group reagent||Ions||Colour & ppt.|
|Group I||dil HCl||Pb2+, Hg+, Ag+||PbCl2, Hg2Cl2, AgCl – white|
Group II A
Group II B
|H2S in dil HCl||
Hg2+, Cu2+, Bi3+, Cd2+
As3+, As5+, Sb3+, Sb5+, Sn2+, Sn4+
As2S5 , SnS2
Black – HgS, CuS, PbS
Orange – Sb2S3, Sb2S5
Brown – Bi2S3 ,SnS
|Group III A||NH4OH in presence of NH4Cl||Fe3+, Al3+, Cr3+||Fe(OH)3, Al(OH)3,Cr(OH)3 Brown White Green|
|GroupIII B||H2S in presence of NH3 & NH4Cl or (NH4)2S.||Ni2+, Co+2, Mn+2, Zn2+||ZnS – white or grey,
Black – CoS, NiS
MnS – Buff (light pink)
|Group IV||(NH4)2CO3 in presence of NH4Cl & NH4OH.||Ba+2, Sr2+, Ca+2||BaCO3, SrCO3, CaCO3 – white|
|Group V||No common group reagent.||Mg+2, Na+, K+, NH4+||⎯|
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 radical only.
- Group III A radicals are precipitated as hydroxides and the NH4Cl suppresses the ionisation of NH4OH so that only the group III A radicals are precipitated because of their low solubility product.
Note: i) Excess of NH4Cl should be added otherwise manganese will be 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– ⎯⎯⎯→ 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 ⎯⎯→ MnS + 2HCl
Exercise 4: i) Though Mg(OH)2 is insoluble, Mg2+does not get precipitated in the III group. Explain.
- ii) When a chloride salt is heated with conc. H2SO4 HCl gas is evolved whereas when a bromide salt is heated with conc. H2SO4, Br2 gas is evolved.
Identification of Basic Radicals – Confirmatory Tests
The confirmatory tests are performed with the salt solution.
- 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 ⎯ Hg ⎯ Cl + Hg↓ + NH4Cl + 2H2O
Amino mercuric Chloride
- Group II A (Hg2+, Cu2+, Bi3+, Cd2+)
- i) Hg+2ions in solution, on addition of SnCl2, give a 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 ⎯→ BiOCl ↓ + 2HCl
When treated with sodium stannite a black ppt. is obtained.
2BiCl3 + 3Na2SnO2 ⎯⎯→ 2Bi ↓ + 3Na2SnO3 + 6NaCl + 3H2O
- iv) Cd+2 ions in solution, with NaOH give a white precipitate.
Cd+2 + 2NaOH ⎯→ Cd(OH)2 ↓ + 2Na+
With ammonium hydroxide, Cd2+ ions give a white precipitate which dissolves in excess.
Cd2+ + 4NH4OH ⎯→ [Cd(NH3)4](OH)2 + 2H2O
- Group II B (As3+, As5+, Sb3+, Sb5+, Sn3+, Sn4+)
- v) As+3 ions in solution give yellow precipitate with ammonium molybdate and HNO3 on heating.
H3AsO4 +12(NH4)2MoO4 +21HNO3 ⎯→ (NH4)3 AsMo12O40↓+ 21NH4NO3 + 12H2O
- vi) Sn2+ ions in solution as SnCl2 give white ppt. with HgCl2 ,which turns black on standing.
SnCl2 + 2HgCl2 ⎯⎯→ SnCl4 + Hg2Cl2 ↓
Hg2Cl2 + SnCl2 ⎯⎯→SnCl4 + 2Hg ↓
vii) 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
iii) Fe(OH)3 is insoluble in NaOH
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
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 5: i) A salt solution on treatment with neutral FeCl3 gives a brown coloured solution. On heating the solution a brown coloured ppt. is obtained. When the solution is treated with KI solution, a straw yellow coloured ppt. is obtained. Identify the salt.
- ii) On passing H2S though a salt solution a brown ppt. is obtained. On adding water to an acidic solution of the salt, turbidity was observed. Identify the cation in the salt.
- 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 (NH4)2 C2O4 only.
Ca+2 + (NH4)2C2O4 → CaC2O4 ↓ +
- Group V (NH4+, Na+, K+, Mg+2)
- i) All ammonium salts on heating with alkali say NaOH give a colourless gas with a pungent smell (NH3)
NH4Cl + NaOH ⎯⎯→ NaCl + NH3 ↑ + H2O
(A) Gas evolved gives white fumes with HCl
NH3 + HCl ⎯→ NH4Cl ↑
(B) Paper soaked in CuSO4 solution, is turned deep blue by NH3 due to complex formation
CuSO4 + 4NH3 ⎯⎯→ [Cu(NH3)4]SO4
(C) With Hg2 (NO3)2 , a black colour is obtained
Hg2(NO3)2 + 2NH3 ⎯⎯→ Hg ↓ + Hg(NH2)NO3 ↓ + NH4NO3
(D) With Nesslers reagent (alkaline solution of potassium tetraiodomercurate(II) ), a brown ppt. is obtained
- ii) Potassium salts give yellow ppt. with sodium cobalt nitrite
Na3[Co(NO2)6] + 3KCl ⎯⎯→ K3[Co(NO2)6] + 3NaCl
iii) Sodium salts give a heavy white ppt. with potassium dihydrogen antimonate
KH2SbO4 + NaCl ⎯⎯→ NaH2SbO4 ↓ + KCl
- iv) Mg2+ gives white ppt. of magnesium hydroxide with sodium hydroxide
Mg2+ + 2NH3 + 2H2O ⎯⎯→ Mg(OH)2 ↓ + 2NH4+
- Solution to Exercises
Exercise 1: i) On addition of acidified KMnO4 to sulphite salt, decolorisation is observed
2MnO4–+ 5SO32– + 6H+⎯→ 2Mn2+ + 5SO42– + 3H2O
Carbonate salt solution does not react, hence there will be no decolorisation.
(Acidified Cr2O72– may also be used instead of acidified MnO4– solution).
- ii) The formation of a white sublimate and the evolution of a colourless pungent smelling gas indicate the presence of ammonium ion in the salt.
NH4+ ⎯→ NH3 +H+
NH3 + conc. HCl ⎯→ NH4Cl (white sublimate compound)
Exercise 2: i) The acid radical must be sulphide
S2– + 2AgNO3 ⎯→ Ag2S + 2NO3–
5S2– + 2MnO4– + 16H+ ⎯→ 5S + 2Mn2+ + 8H2O
- The greenish yellow gas evolved is chlorine
2Cl– + MnO2 + 2H2SO4 ⎯→ MnSO4 + 2H2O + Cl2 + SO42–
Cl2 + H2O ⎯→ 2HCl + (O)
Vegetable dye + (O) ⎯→ Colourless oxidation product
∴ The acid radical in the salt is chloride
Since crimson red colour is imparted, the basic radical is strontium.
Exercise 3: i) The acid radical must be chromate
CrO42– + Ba2+ ⎯→
+ H+ HCrO4– + H2O
- ii) The acid radical must be nitrite
NO2–+ H+ ⎯→ HNO2
2HNO2 ⎯→ H2O + + O
2NO + O2 ⎯→
On adding FeSO4 solution to the nitrite salt, nitroso ferrous sulphate is formed, which is brown in colour.
3HNO2 ⎯→ HNO3 + H2O + 2NO
FeSO4 + NO ⎯→ FeSO4.NO
Exercise 4: i) The III group reagent is saturated NH4Cl and NH4OH. NH4Cl being a strong electrolyte dissociates completely and suppresses the dissociation of the weak electrolyte NH4OH. Hence, the concentration of OH– is reduced and is insufficient to overcome the higher Ksp of Mg(OH)2.
- ii) Cl–+ H2SO4 ⎯→ HCl + HSO4
In the case of a bromide salt, HBr formed is a reducing agent. Conc. H2SO4 oxidises the HBr to Br2
Br– + H2SO4 ⎯→ HBr + H2SO4
2HBr + H2SO4 ⎯→ Br2↑ + SO2 + 2H2O
Exercise 5: i) The given salt is copper acetate
3CH3COO– + FeCl3 ⎯→ + 3Cl–
(CH3COO)3Fe + 2H2O + 2CH3COOH
Hence, the acid radical present in the salt is acetate radical.
2Cu2+ + 4KI ⎯→ Cu2I2 + 4K+ + I2
The white ppt. of Cu2I2 appears straw yellow due to the presence
The basic radical is, therefore, cupric ion.
- ii) The formation of a brown precipitate with H2S indicates Bi3+
The formation of a turbidity on addition of water to an acidic solution confirms the presence of Bi3+
2Bi3+ + 3H2S ⎯→ Bi2S3 + 6H+
BiCl3 + H2O ⎯→ + 2HCl
- Solved Problems
Problem 1: A colourless water soluble solid X on heating gives equimolar quantities of Y and Z. Y gives dense white fumes with HCl and Z does so with ammonia. Y gives brown precipitate with Nessler’s reagent and Z gives white precipitate with nitrates of Ag+, Pb2+ and Hg+ . What is X?
Solution: The reactions reveal following facts
- i) Y is NH3 and (ii) Z is HCl. Hence X must be NH4Cl
Problem2: 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.
Solution: At first Bi(NO3)3 hydrolyses to give nitric acid which, being an oxidising agent, oxidise s 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 ⎯⎯→ [Bi(OH) (NO3)2] + HNO3
+ 4H+ + 3e– ⎯⎯→ NO2 + 2H2O + ] × 2
2I– ⎯⎯→ I2 + 2e–] × 3
+ 8H+ + 6I– ⎯⎯→ 2NO + 4H2O +
KI + I2 ⎯⎯→
Problem 3: 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.
Solution: 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 ⎯⎯→ NH4Cl + HNO2
2HNO2 ⎯⎯→ H2O + 2NO + O
2NO + O ⎯⎯→
Problem4: A compound (X) of S, Cl and O has a 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.
Solution: Reaction of (Y) indicates that it must 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 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 ⎯⎯→ 2HCl + H2SO4
+ 4KOH ⎯⎯→ +
+ AgNO3 ⎯⎯→ AgCl↓ + KNO3
+ BaCl2 ⎯⎯→ BaSO4 ↓ + 2KCl
Problem 5: When a white powder A is strongly heated, it gives off a colourless, odourless gas B which turns lime water milky due to the formation of 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 with the help of the equations.
Solution: (A) ZnCO3 (B) CO2 (C) CaCO3 (D) Ca(HCO3)2 (E) ZnO
The salt must be a carbonate as CO2 gas is evolved or heating. The residue is yellow when hot and white when cold. This indicates ZnO.
CO2 + Ca(OH)2 ⎯→ + H2O
CaCO3 + H2O + CO2 ⎯→
Problem 6: An aqueous solution of gas (X) shows the following reactions :-
- i) It turns red litmus blue.
- ii) When added in excess to copper sulphate solution, a deep blue colour is obtained.
iii) On addition of FeCl3 solution a brown ppt. soluble in dilute nitric acid is obtained.
Identify (X) and give equations for the reactions at step (ii) & (iii)
Solution: X – NH3
Reactions: i) CuSO4 + 4NH4OH ⎯⎯→ [Cu(NH3)4]SO4 + H2O
- ii) FeCl3 + 3NH4OH ⎯→ Fe(OH)3 ↓ + 3NH4Cl
Fe(OH)3 + 3HNO3 ⎯→ Fe(NO3)3 + 3H2O
Problem7: 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.
- iv) It imparts a golden yellow colour to the flame in the flame test.
Identify the compound (X) and give chemical equations for the reactions at steps(i), (ii) and (iii).
Solution: 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 ⎯⎯→ 2NaCl + H2O + SO2 +
- ii) ⎯⎯→ + 2NaI
iii) ⎯⎯→ + 2NaNO3
Ag2S2O3 + H2O ⎯⎯→ + H2SO4
Problem 8: 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 with 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.
Solution: Let us summarise the given facts of the question.
Red vapours of (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 the following four ions. , Fe2,
Equations for the formation of A and B
4NaCl + K2Cr2O7 + 3H2SO4 K2SO4 + 2Na2SO4
+ + 3H2O
3Fe2+ + 2K3[Fe(CN)6] ⎯⎯→
Problem 9: 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 the following reactions.
- i) Addition of silver nitrate solution to (C) gave first a white precipitate which then turns 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 mmole of iodine.
Deduce the formula of A, B and C and explain the reactions.
Solution: 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 ⎯⎯→ + 2NaCl + 2H2O
Na2CrO4 + 2AgNO3 ⎯⎯→ + 2NaNO3
2Na2CrO4 + H2SO4 ⎯⎯→ Na2Cr2O7 + Na2SO4 + H2O
+ 14H+ + 6I– ⎯⎯→ 2Cr3+ + 7H2O + 3I2
The quantitative data is explained in the following manner. From the above reactions we observe that
≡ Na2CrO4 ≡
155 g of CrO2Cl2 require 2 mole of AgNO3
0.155 g of CrO2Cl2 will require = × 0.155 = 0.002 mole = 2 mmole
This coincides with the given data.
Similarly, ≡ 2CrO42– ≡ ≡
Thus 2 × 155 g of CrO2Cl2 liberate 3 moles of I2
0.155g CrO2Cl2 will liberate = = 0.0015 moles = 1.5 mmole
This also coincides with the given data.
Problem 10: 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 NH4Cl 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 with dil H2SO4 give a gas used in II and IV group analysis. What are (A) to (E)? Give balanced chemical equations of the reactions.
Solution: 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 ⎯⎯→
The compound (C) is zinc oxide (ZnO) because it is yellow when hot and white when cold, hence 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.
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 are given below:
ZnCl2 + 2NaOH ⎯⎯→
- Assignments (Subjective Problems)
- An aqueous solution of a gas (X) gives the following reactions:
- i) It turns acidified K2Cr2O7 solution green.
- ii) On boiling with H2O2 , cooling it and then adding an aqueous solution of BaCl2 , a ppt. insoluble in dilute HCl is obtained.
iii) On passing H2S in the solution, turbidity is obtained.
Identify (X) and give equations for the steps (i), (ii), (iii).
- A certain compound (X) is used in laboratory for analysis. Its aq. solution gave the following reactions.
- i) On addition to copper sulphate solution, a brown ppt. is obtained which turns white on addition of excess of Na2S2O3
- ii) On adding Ag+ ions, a yellow ppt. is obtained which is insoluble in NH4 Identify (X), giving reactions
- A solid laboratory reagent (A) gave the following reactions.
- i) It imparted green colour to the flame.
- ii) Its solution did not give ppt. on passing H2
iii) When heated with K2Cr2O7 and conc. H2SO4 a red gas was evolved. The gas when passed in aq. NaOH solution turns it yellow. Identify (A) giving chemical reactions.
- An unknown inorganic compound (A) gave the following reactions.
- i) (A) on heating gave a residue, oxygen and an oxide of nitrogen.
- ii) Aqueous solution of (A) on addition of tap water gave turbidity which did not dissolve in nitric acid.
iii) The turbidity dissolved in ammonium hydroxide.
Identify (A) and give all balanced reactions.
- 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.
- 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).
- A solution of the salt (X) gave a dirty white ppt. when (NH4)2S was passed through it. After adding conc. HNO3 and Co(NO3)2 solution to the salt solution, a filter paper was dipped into the mixture and ignited. Green coloured ash was obtained.
On heating the salt strongly with conc. HNO3 cooling and adding ammonium molybdate a canary yellow precipitate was formed. Identify the salt and give the reactions involved in the above tests.
- 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 forms a black precipitate 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 compound (X) on heating with an excess of NaOH solution gives a gas (Y) which gives white fumes on exposure to HCl. Heating is continued to expel the gas completely. The resultant alkaline solution again liberates the same gas (Y) when heated with Zn powder. However, when the compound (X) is heated alone it does not give nitrogen. Identify (X) and (Y).
10 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) and give the reactions involved.
- 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 colourless inorganic salt (A) decomposes completely at about 2500C to give only two products, (B) and (C), leaving no residue. The oxide (C) is a liquid at room temperature and neutral to moist litmus paper while the gas (B) is a neutral oxide. White phosphorus burns in excess of (B) to produce a strong white dehydrating agent. Write balanced equations for the reactions involved in the above process.
- 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. Write down the equation for the thermal decomposition of X.
- 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 red coloured solid on heating turns blue in colour. When an aqueous solution is treated with ammonium chloride ammonium hydroxide and H2S a black precipitate is obtained on treating with barium chloride solution a white precipitate insoluble in acids is obtained. Identify the solid.
- A white amorphous powder (A) on strong heating gives a colourless non-combustible gas (B) and solid (C). The gas (B) turns lime water milky and the 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) by giving chemical equations.
- (Y) is a pale pink coloured solid. An aqueous solution of Y gave a white ppt. with NaOH which turned black on exposure to air. On boiling the salt solution with conc. HNO3 and PbO2 a violet coloured solution was obtained. On adding lead acetate solution to the salt solution a white precipitate soluble in NaOH was obtained. Identify the salt (Y) and give the equations involved.
- An aq. solution of an inorganic compound (X) shows the following reactions.
- i) It decolorizes an acidified KMnO4 solution accompanied with evolution of O2.
- ii) It liberates I2 from acidified KI solution.
iii) It gives brown ppt. with alkaline KMnO4 solution with evolution of O2.
- iv) It removes black stains from old oil paintings. Identify (X) and give chemical reactions for the steps (i) to (iv).
- 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 white 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).
- i) A black mineral (A) on treatment with dilute sodium cyanide solution in presence of air gives a clear solution of (B) and (C).
- ii) The solution of (B) on reaction with zinc gives a precipitate of metal (D).
iii) (D) is dissolved in dil. HNO3 and the resulting solution gives a white precipitate (E) with dil. HCl.
- iv) (E) on fusion with sodium carbonate gives (D).
- v) (E) dissolves in aqueous solution of ammonia giving a colourless solution of (F). Identify (A) to (F) and give chemical equations for reactions at steps (i) to (v).
- A gaseous mixture containing (X), (Y), (Z) gases were 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 turned milky by absorbing gas (Y). The residual gas when passed through alkaline pyrogallol solution, it turned black. Identify gas (X), (Y), (Z) and explain the reactions involved.
- 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)
- A is binary compound of a univalent metal 1.422 g of A reacts completely with 0.321 g of sulphur in an evacuated and sealed tube to give 1.743 g of a white crystalline solid, B that forms a hydrated double salt, C with Al2(SO4)3. Identify A,B, and C.
- A colourless crystalline solid (A) turns our skin black. It gives two gases (B) and (C) on heating to 950°C along with a residue (D). The gas (B) is soluble in water and gives a compound (E). the residue (D) is soluble in (E). The solid (A) gives white ppt. with NaOH. The white precipitate dissolves in excess NH4OH the clear solution gives a silver mirror with glucose. The solid (A) also gives white ppt. with KCN which is soluble in excess KCN. Identify the different compounds giving equations.
- Identify A,B, C and D in the following sequence of reactions:
- i) A + NaOH ⎯⎯→ NaCl + NH3 + H2O
- ii) NH3 + CO2 + H2O ⎯⎯→ B
iii) B + NaCl ⎯⎯→ C + NH4Cl
- iv) C Na2CO3 + H2O + D
- A certain metal (A) is boiled in dilute nitric acid to give a salt (B) and an oxide of nitrogen (C). An aqueous solution of (B) gives a precipitate (D) with brine which is soluble in ammonium hydroxide. On adding aqueous solution of (B) to hypo solution, a white precipitate (E) is obtained. (E) on standing turns to a black compound (F). Identify (A) to (F).
- 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 a pale green solution (G) is obtained. The solution (E) on treatment with thiocyanate ion gives blood red compound (H). Identify (A) to (H).
- A compound (X) gives a golden yellow flame and shows the following reactions.
- i) Zn powder when boiled with a concentrated solution of (X) dissolves and H2 is evolved
- ii) When an aqueous solution of (X) is added to an aqueous solution of SnCl2 a white ppt. is produced which becomes soluble when (X) is added in excess.
iii) Compound (X) is used for the preparation of washing soap on reaction with fat and oils.
- iv) (X) is not a primary standard hence its standard solution is prepared by titrating against oxalic acid using phenolphthalein indicator.
- v) Aqueous solution of (X) precipitates hydroxides of Al3+ and Cr3+, which dissolves in excess, the former giving a colourless solution while the latter a greenish yellow solution in presence of Br2
Identify (X) giving the different reaction.
- A white solid (A) on heating produces decripitation and brown vapours of (B) are evolved. On adding conc. H2SO4 and Cu turnings, copious brown fumes of (B) and a blue solution of (C) are produced. An aqueous solution of (A) gives a yellow precipitate (D) with potassium chromate solution and a white precipitate (E) with dil. H2SO4. Identify (A) to (E) and give the reactions involved.
LEVEL – III
- 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 an excess of solution of (A), a violet coloured compound (D) is formed. But 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).
Identify (A) to (E) and give chemical equations for the reactions at steps (i) to (iv).
- A black coloured compound (A) on reaction with dil. H2SO4 gives a gas (B) which on passing in a solution of an acid (C) gives a white turbidity (D). Gas (B) when passed in an acidified solution of a compound (E) gives a ppt. (F) soluble in dil. HNO3. After boiling this solution when an excess of NH4OH is added, a blue coloured compound (G) is formed. To this solution on addition of acetic acid and aq. K4[Fe(CN)6] a chocolate brown ppt. (H) is obtained. On addition of an aq. solution of BaCl2 to an aq. solution of (E) a white ppt. insoluble in HNO3 is obtained. Identify the compounds from (A) to (H).
- 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)
- 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.
- An aqueous solution of salt (X) gives a yellow ppt. when H2S is passed through it. On adding NH4OH to the solution a white ppt. soluble in excess was obtained. On treating an aqueous solution of the salt with CCl4 and Cl2 water a violet layer was obtained. Identify (X) and give the equations involved.
- A mixture of two salts was treated as follows
- i) The mixture was heated with MnO2 and conc. H2SO4 when a yellowed green gas was liberated
- ii) The mixture on heating with NaOH solution gave a gas which turned red litmus blue.
iii) Its solution in water gave blue precipitate with K4Fe(CN)6 and red colouration with ammonium thiocyanate.
- iv) The mixture was boiled with KOH and the liberated gas was bubbled through an alkaline solution of K2HgI4 to give a brown ppt. Identify the mixtrue of two salts. Give equation for the reaction involved.
- 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.
- An aqueous solution of salt (A) gives white crystalline ppt. (B) with NaCl solution. The filtrate gives a black ppt. (C) when H2S is passed in it. Compound (B) is dissolved in hot water and the solution gives a yellow ppt. (D) on treating with NaI and cooling. The compound (A) does not give any gas with dil. HCl but liberates reddish brown gas on heating. Identify the compound (A), (B), (C) and (D).
- A pale green amorphous solid responds to the following tests
- i) Its aq. solution gives brown ppt. with Potassium ferrocyanide solution.
- ii) Its aq. solution gives blue colour with NH4OH solution.
iii) The salt produces effervescences with dil. HCl.
Identify the ions present and suggest the formula of the compound.
- 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)
- Salt (A) is a white deliquescent solid. On heating the solid, brown fumes were evolved. On adding NaOH to the salt solution of (A), a white gelatinous ppt. was obtained. On adding, a drop of blue litmus solution to the ppt., a blue lake was obtained. On adding DPA to the salt solution, a deep blue colouration was obtained. Identify the salt and give the equations involved.
- X is a white deliquescent solid which is used as a desiccating agent. An aqueous solution of the salt turned acidified K2Cr2O7 solution green. On treating with AgNO3 solution, a curdy white ppt. was formed.
A paste of (X) in conc. HCl imparted a brick red colour to the flame. On adding ammonium oxalate to the salt solution, a white ppt. in soluble in acetic acid was got. Identify X and also give the reactions involved.
- A scarlet compound (A) is treated with conc. HNO3 to give a chocolate brown precipitate (B). The precipitate is filtered and the filtrate is neutralized 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.
- White solid (X) on heating with conc. HNO3 followed by ammonium molybdate solution gives a canary yellow precipitate. The salt solution on treatment with sodium hydroxide gives a white precipitate which is insoluble in excess. Added cobalt nitrate solution and conc. HNO3 to the above precipitate. Dipped a filter paper into the above solution and ignited. A pink ash was obtained. Identify the salt and give the equation involved.
- Assignments (Objective Problems)
- In qualitative analysis Ca is precipitated in
(A) V group (B) II group
(C) III group (D) IV group
- Al3+, Fe3+, Cr3+ are grouped together for qualitative analysis because their –
(A) carbonates are insoluble in ammonia
(B) hydroxides are insoluble in ammonia
(C) sulphides are soluble in acids
(D) electronic charge is the same
- A colourless solid A dissolves in water. The aqueous solution gives a white precipitate B when NaOH or NH4OH is added. B dissolves in excess of NaOH but not in excess of NH4OH. BaCl2 solution added to a solution of A gives a white precipitateC which is insoluble in dilute HCl. A may be –
(A) Al2(SO4)3 (B) ZnSO4
(C) SnSO4 (D) All the above
- Which of the following ions will liberate iodine when treated with KI?
(A) Cu2+ (B) Fe2+
(C) Pb2+ (D) Sn2+
- In qualitative analysis of I group radicals a white precipitate is formed which is insoluble in boiling water but when treated with NH4OH it turns black. The precipitate may be
(A) PbCl2 (B) AgCl
(C) HgCl2 (D) Hg2Cl2
- On passing CrO2Cl2 in water & then adding (CH3COO)2Pb, the precipitate formed is
(A) PbCrO4 (B) PbCl2
(C) CrCl3 (D) PbSO4
- Which of the following in insoluble in yellow ammonium sulphide?
(A) SnS (B) Sb2S3
(C) Bi2S3 (D) As2S3
- Ammonia (NH3) when treated with Nessler’s reagent in the presence of an alkali gives a brown precipitate of –
(A) NH4HgI4 (B) NH2HgOHgI
(C) KI (D) (NH4)2HgI4
- 9. A salt was first heated with dilute H2SO4 and then with concentrated H2SO4, no action was observed in either case. The salt must be
(A) nitrite (B) sulphide
(C) sulphite (D) sulphate
- Rinmann’s green represents
(A) cobalt nitrate (B) sodium zincate
(C) zinc acetate (D) cobalt zincate
- A solid soluble in water on being heated gives brown gas and when dil. HCl is added to its aqueous solution, a white ppt. is obtained. The solid is –
(A) Ca(NO3)2 (B) Zn(NO3)2
(C) Pb(NO3)2 (D) Ba(NO3)2.
- K4[Fe(CN)6] is used in the identification of –
(A) Cd+2 ions (B) Fe+3 ions
(C) Cu+2 ions (D) All the above
- The deep blue colour obtained on adding K3Fe(CN)6 to ferrous sulphate is due to the formation of
(A) Fe[Fe(CN)6] (B) Fe3[Fe(CN)6]2
(C) Fe3[Fe(CN)6]4 (D) Fe4[Fe(CN)6]3
- The salt used for performing ‘bead’ test in qualitative inorganic analysis is –
(A) K2SO4.Al2(SO4)3.24H2O (B) FeSO4.(NH4)2SO4.6H2O
(C) Na(NH4)HPO4.4H2O (D) CaSO4.2H2O
- When a nitrate is boiled with aluminum turnings and concentrated NaOH solution, the gas liberated is –
(A) NO2 (B) NH3
(C) NO (D) N2O
- A mixture of inorganic substances on treatment with dilute H2SO4 gives out a gas which turns lead acetate paper black while potassium dichromate paper is turned green. The mixture contains –
(A) sulphite (B) sulphide
(C) nitrate (D) nitrite
- Na2HPO4 detects the ion / group –
(A) (B) NO
(C) K+ (D) Mg2+
- When H2S is passed through an ammoniacal salt solution X, a buff coloured precipitate is obtained. Then X can be a –
(A) Co2+ solution (B) Mn2+ solution
(C) Ni2+ solution (D) Zn2+ solution
- The reagent NH4Cl and aq. NH3 will precipitate –
(A) Bi+3 (B) Al+3
(C) Zn+2 (D) Mg+2
- The stopper of a bottle containing an ammonium salt is taken out. A strong smell of ammonia is given out. The salt is –
(A) ammonium chloride (B) ammonium sulphate
(C) ammonium carbonate (D) ammonium nitrate
- Identify the reagent that can be used to detect phosphate radical
(A) Neutral FeCl3 (B) Magnesia mixture
(C) Acidified KMnO4 (D) conc. H2SO4
- A crystalline solid dissolves in water to give colourless solution. On addition of aqueous silver nitrate a yellow precipitate is formed. The solid is most likely to be
(A) copper (II) oxide (B) sodium chloride
(C) sodium iodide (D) zinc sulphate
- Which is the correct order of Ksp for CuS, ZnS and CdS
(A) CuS > ZnS > CdS (B) CdS > CuS > ZnS
(C) ZnS > CdS > CuS (D) ZnS > CuS > CdS
- Magneson reagent is used to detect
(A) Mn2+ (B) Mg2+
(C) PO43– (D) MnO4–
- Which of the following will not produce a precipitate with dilute AgNO3 solution?
(C) I–(aq). (D) F–(aq).
- AgNO3 gives red precipitate with –
(A) NaI (B) KCl
(C) NaNO3 (D) Na2CrO4
- Which of the following pairs of ions cannot be separated by H2S in ammoniacal medium?
(A) Zn2+, Ni2+ (B) Zn2+, Mg2+
(C) Mn2+, Ca2+ (D) Co2+, Ca2+
- On adding conc. H2SO4 to a salt brown colouration was observed on the walls of the test tube. On heating, violet coloured vapours were evolved. The brown colouration is due to the formation of
(A) I2(s) (B) I+
(C) I3– (D) HI
- What is the product obtained when MnSO4 in solution is boiled with sodium bismuthate and concentrated HNO3?
(A) MnO2 (B) HMnO4
(C) Mn3O4 (D) PbMnO4
- When conc. H2SO4 is added to dry KNO3, brown fumes are evolved, These fumes are –
(A) SO3 (B) SO2
(C) NO2 (D) N2O
- Which one among the following pairs of ions cannot be separated by H2S in dilute hydrochloric acid?
(A) Bi+++, Sn++ (B) Al3+, Hg2+
(C) Zn2+, Cu2+ (D) Ni2+, Cu2+
- Sr2+ can be precipitated by
(A) (NH4)2SO4 (B) NH4Cl
(C) (NH4)2C2O4 (D) Both (A) and (C)
- When excess sodium thiosulphate is added to AgBr, the product formed is
(A) Ag2S2O3 (B) Na[Ag(S2O3)]
(C) Na3[AgBr4] (D) Na3[Ag(S2O3)2]
- Nessler’s reagent is
(A) K2HgI4 in excess of KI (B) K2HgI4 in excess of KOH
(C) K2HgI4 in excess of HCl (D) HgI2 in excess of KOH
- A black precipitate is obtained when H2S is passed through a dilute acidic (HCl) aqueous solution containing –
(A) Cu2+ and Pb2+ (B) Zn2+ and Mg2+
(C) Mg2 and Fe3+ (D) Mg2+ and Ni2+
- A solution of a salt in water, on addition of dilute HCl, gives a white precipitate, which is soluble in hot water. The salt contains –
(A) Fe2+ (B) Hg2+
(C) Ag+ (D) Pb2+
- Mg2+ does not impart colour in the flame test because it possesses
(A) High charge (B) High excitation energy
(C) Small size (D) None of the above
- The colour of the solid produced by adding SnCl2 in excess to a solution of HgCl2 is
(A) white (B) black
(C) yellow (D) brown
- Which of the following statements(s) is/are correct with reference to the ferrous and ferric ions?
(A) Fe3+ gives green colour with potassium thiocyanate
(B) Fe2+ gives blue precipitate with potassium ferricyanide
(C) Fe3+ gives red colour with potassium thiocyanate
(D) Fe2+ gives brown colour with ammonium thiocyanate
- Aqua Regia is –
(A) Conc. HCl + conc. HNO3 in 1 : 3 ratio
(B) Dil.HCl & dil.HNO3 in 3 : 1 ratio
(C) Dil. HCl & dil. HNO3 in 3 : 1 ratio
(D) Conc. HCl & conc. HNO3 in 3 : 1 ratio
- Answers to Objective Assignments
LEVEL – I
- A 2. B
- D 4. A
- D 6. A
- C 8. B
- D 10. D
- C 12. D
- B 14. C
- B 16. B
- D 18. B
- B 20. C
- B 2. C
- C 4. B
- D 6. D
- A 8. C
- B 10. C
- A 12. D
- D 14. B
- A 16. D
- C 18. B
- B, C 20. D