Qualitative Analysis Theory_E

CHEMISTRY

Qualitative Analysis Introduction : Qualitative analysis involves the detection of cation(s) and anion(s) of a salt or a mixture of salts. The systematic procedure for qualitative analysis of an inorganic salt involves the following steps : (a) Preliminary tests  Physical appearance (colour and smell).  Dry heating test.  Charcoal cavity test.  Charcoal cavity and cobalt nitrate test.  Flame test.  Borax bead test.  Dilute sulphuric acid test.  Potassium permanganate test.  Concentrated sulphuric acid test.  Tests for sulphate, phosphate and borate. (b) Wet tests for acid radicals. (c) Wet tests (group analysis) for basic radicals.

1.

Physical Examination Of the Mixture : The physical examination of the unknown mixture involves the study of colour, smell and density. Table : 1 Physical Examination Ex pe rim e nt

Obse rva tions

Infe re nce

(a ) Colour

Blue or Bluish green

Cu2+ or Ni 2+

Greenish

Ni 2+

Light green

Fe2+

Dark brown

Fe3+

Pink, violet Light pink, flesh colour or dull earthy colour

Co2+ Mn2+

W hite

Shows the absence of Cu2+,Ni 2+,Fe2+,Fe3+ Mn2+, Co2+

Ammonical smell

NH 4+

Vinegar like smell

CH 3COO –

Smell like that of rotten eggs

S 2–

(b) Sm e ll Take a pinch of the salt between your fingers and rub with a drop of water

Salt of Pb 2+ or Ba 2+–

(i) Heavy (.C) De nsity (ii) Light fluffy powder

Carbonate salts (i) If coloured, may be Cu(NO 3 )2,

(d) De lique sce nce

Salt absorbs moisture and becomes paste like

FeCl3 (ii) If colourless, may be Zn(NO 3 )2 , chlorides of Zn2+, Mg 2+ etc.

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CHEMISTRY 2.

Dry Heating Test : This test is performed by heating a small amount of mixture in a dry test tube. Quite valuable information can be generated by carefully performing and noting the observations here. On heating some salts undergo decomposition thus evolving the gases or may undergo characteristic changes in the colour of residue. These observations are tabulated below along with the inferences that you can draw. Table : 2

Observation 1. Gas evolved (a) Colourless and odourless gas CO2 gas – turns lime water milky (b) Colourless gas with odour (i) H2S gas–Smells like rotten eggs, turns lead acetate paper black. (ii) SO2 gas–Characteristic suffocating smell, turns acidified potassium dichromate solution or paper green. (iii) HCl gas – Pungent smell, white fumes with ammonia, white precipitate with silver nitrate solution. (iv) Acetic acid vapours–Characteristic vinegar like smell. (v) NH3 gas– Characteristic smell, turns Nessler's solution brown. (c) Coloured gases – Pungent smell (i) NO2 gas – Reddish brown, turns ferrous sulphate solution black. (ii) Cl2 gas – Greenish yellow, turns starch iodide paper blue. (iii) Br2 vapours – Reddish brown, turns starch paper orange red. (iv) I2 vapours – Dark violet, turns starch paper blue. 2. Sublimate formed (a) White sublimate (b) Black sublimate accompanied by violet vapours. 3. Fusion The mixture fuses.

Inference

CO32– Hydrated S2– SO32–

Cl– CH3COO– NH4+

NO2– or NO3– Cl– Br– –

NH4+ –

Alkali metal salts or salt containing water of crystallisation.

4. Swelling The mixture swells up into voluminous mass.

PO43– , BO33– indicated

5. Residue (i) Yellow when hot, white when cold. (ii) Brown when hot and yellow when cold (iii) Original salt blue becomes white on heating (iv) Coloured salt becomes brown or black on heating.

Zn2+ Pb2+ Hydrated CuSO4 indicated Co2+ , Fe2+, Fe3+ , Cr3+ , Cu2+ , Ni2+ , Mn2+ indicated.

Note :  Use a perfectly dry test–tube for performing this test. While drying a test–tube, keeps it in slanting position with its mouth slightly downwards so that the drops of water which condense on the upper cooler parts, do not fall back on the hot bottom, as this may break the tube.

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CHEMISTRY 

For testing a gas, a filter paper strip dipped in the appropriate reagent is brought near the mouth of the test tube or alternatively the reagent is taken in a gas–detector and the gas is passed through it.

Figure : Detection of gas evolved.

 3.

Do not heat the tube strongly at one point as it may break.

Charcoal Cavity Test : This test is based on the fact that metallic carbonates when heated in a charcoal cavity decompose to give corresponding oxides. The oxides appear as coloured incrustation or residue in the cavity . In certain cases, the oxides formed partially undergo reduction to the metallic state producing metallic beads or scales. Example : (a)

ZnSO4 + Na2CO3  ZnCO3 + Na2SO4 ZnCO3  ZnO (Yellow when hot, white when cold) + CO2 

(b)

CuSO4 + Na2CO3  CuCO3 + Na2SO4 CuCO3  CuO + CO2

CuO + C  Cu (Reddish scales) + CO  Table : 3

Observation Incrusta tion or Re sidue Yellow when hot, white when cold

No characteristic residue W hite residue which glows on heating Black

None

Brown when hot, yellow when cold

4.

Inference Me ta llic be a d None Grey bead which marks the paper Red beads or scales None

Zn 2+ Pb 2+ Cu 2+ Ba ,Ca 2+, Mg 2+ Nothing definite–generally coloured salt 2+

Cobalt Nitrate Test : In case the residue is white in colour after charcoal cavity test, add a drop of cobalt nitrate in the charcoal cavity. A drop of water is then added and the mass is heated in an oxidising flame using blow pipe. It is cooled and one or two drops of cobalt nitrate solution is added and then again heated in the oxidising flame. Different metal salts give different coloured mass as given in the table. To illustrate : ZnSO4 + Na2 CO3  ZnCO3 + Na2 SO4 ; ZnCO3  ZnO + CO2 2Co (NO3)2  2CoO + 4 NO2 + O2 ;

ZnO + CoO  ZnO. CoO (or CoZnO2) (Rinmann's green)

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CHEMISTRY Table : 4

5.

S.No.

Metal

1.

Zinc

Colour of the mass Green

2.

Aluminium

3.

Magnesium

4.

Tin

Blue Pink Bluish - green

Flame test : The chlorides of the metals are more volatile as compared to other salts and these are prepared in situ by mixing the compounds with a little concentrated hydrochloric acid. On heating in a non-luminous Bunsen flame they are volatilized and impart a characteristic colour to the flame as these absorb energy from the flame and transmit the same as light as characteristic colour . Table : 5 Colour of Flame

Inference

Crimson Red / Carmine Red

Lithium

Golden yellow

Sodium

Violet/Lilac

Potassium

Brick red

Calcium

Crimson

Strontium

Apple Green/Yellowish Green

Barium

Green with a Blue centre/Greenish Blue

Copper

Figure : Flame test

6.

Borax Bead test : On Heating borax forms a colourless glassy bead of NaBO2 and B2O3 .   Na2B4O7.10H2O  Na2B4O7  2NaBO2 + B2O3 On heating with a coloured salt , the glassy bead forms a coloured metaborate in oxidising flame. For example, in oxidising flame copper salts give blue bead. CuSO4  CuO + SO3 ; CuO + B2O3  Cu(BO2)2 (blue bead) However, in reducing flame the colours may be different due to different reactions. 2Cu(BO2)2 + C  2CuBO2 + B2O3 + CO 2Cu(BO2)2 + 2C  2Cu (brown red/red and opaque bead) + 2B2O3 + 2CO.

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CHEMISTRY Table : 6 Colour in oxidising flame

Colour in reducing flame

Metal When Hot

When Hot

When Cold

Copper

Green

Blue

Colourless

Brown red

Iron

Brown yellow

Pale yellow/Yellow

Bottle green

Bottle green

Green

Green

Green

Blue

Blue

Blue

Chromium



When Cold

Yellow

Cobalt

Blue

Manganese

Violet/Amethyst

Red/Amethyst

Grey/Colourless

Grey/Colourless

Nickel

Violet

Brown/Reddish brown

Grey

Grey

Non luminous flame is called oxidising flame.



Luminous flame is called reducing flame.

Figure : Borax bead test



All acid radicals which are in JEE syllabus are colourless and diamagnetic. Hence the colour of the salts is only due to the basic radicals. Table : 7 SOLUBILITY CHART

S.No.

Anion

Solubility / Exception

1.

2– 3

CO

Except carbonates of alkali metals and of ammonium, all other normal carbonates are insoluble.

2.

SO32–

Only the sulphites of the alkali metals and of ammonium are water soluble. The sulphite of other metals are either sparingly soluble or insoluble.

3.

S2–

The acid, normal and polysulphide of alkali metals are soluble in water. The normal sulphides of most other metals are insoluble; those of the alkaline earths are sparingly soluble, but are graduallychanged by contact with water into soluble hydrogen sulphides.

4.

NO2– , NO3–

Almost all nitrites and nitrates are soluble in water. AgNO2 sparingly soluble. Nitrates of mercury and bismuth give basic salts on treatment with water. These are soluble in dilute nitric acid.

5.

CH3COO–

Acetates are water soluble except Ag(I) and Hg(II) acetates which are sparingly soluble.

6.

Cl–

Most chlorides are soluble in water. PbCl2 (sparingly soluble in cold but readily soluble in boiling water), Hg2Cl2, AgCl, CuCl, BiOCl, SbOCl and Hg2OCl2 are insoluble in water.

7.

Br–

Silver, mercury(I) and copper(I), bromides are insoluble. Lead bromide is sparinglysoluble in cold but more soluble in boiling water. All other bromides are soluble in water.

8.

I–

Silver, mercury(I), mercury(II), copper(I), lead and bismuth(III) iodides are the least soluble salts. All other iodides are water soluble.

9.

SO42–

The sulphates of barium, strontium and lead are insoluble in water, those of calcium and mercury(II) are slightly soluble. Some basic sulphates of mercury, bismuth and chromium are also insoluble, but these dissolves in dilute hydrochloric or nitric acid.

10

PO43–

The phosphate of the alkali metals, with the exception of lithium and ammonium, are soluble in water ; the primary phosphate of the alkaline earth metals are soluble. All the phosphates of the other metals and also the secondary and tertiary phosphate of the alkaline earth metals are sparingly soluble or insoluble in water.

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CHEMISTRY Analysis of ANIONS (Acidic Radicals) : (A)

Analysis of anions (acidic radicals) can be broadly divided in to two groups. GROUP 'A' RADICALS : It involves those anions which are characterised by volatile products by reaction with HCl/ H2SO4 It is further subdivided in to two groups as given below. (a) Dilute Sulphuric acid/Dilute Hydrochloric acid : The anions of this group liberate gases or acid vapours with dilute sulphuric acid/hydrochloric acid. Table : 8 Inference

Observation Effervescence with the evolution of a colourless and odourless gas which turns lime water milky. Evolution of colourless gas having smell of rotten egg which turns lead acetate paper black. Colourless gas having suffocating odour (like burning sulphur) which turns acidified K2Cr2O7 paper green. Evolution of reddish brown pungent smelling gas which turns (i) FeSO4 solution brownish-black and (ii) wet starch –iodide paper blue. Colourless gas having smell of vinegar. No peculiar gas is evolved.

Gas

Radical

CO2

CO32–

H2S

S2–

SO2

SO32–

NO2

NO2–

HAC(g)

CH3COO–

–

All above are absent

(b) Concentrated Sulphuric acid group : The anions of this group liberate acid vapours or gases with conc. H2SO4. Table : 9 Inference

Observation

(B)

Gas

Radical

Colourless gas with pungent smell which gives dense white fumes with a glass rod dipped in NH4OH.

HCl

Cl–

Reddish brown gas with pungent smell, intensity of reddish brown fumes increases on addition of a pinch of solid MnO2. Also it turns starch paper orange red.

Br2

Br–

Evolution of violet vapours which turns starch paper blue.

I2

I–

Evolution of reddish brown fumes which intensifies on addition of copper turnings or bits of filter paper. Starch iodide paper develops a blue–black spot due to the formation of a I2–starch complex. (NO2 liberated acts as oxidising agent).

NO2

NO3–

GROUP 'B' RADICALS : Anions of this group do not give acid vapours or gases with dilute as well as concentrated H2SO4 but are characterised by their specific reactions in solutions. This group is further sub divided into two groups based on the type of the reactions. (a) Oxidation and reduction in solutions : CrO42–, Cr2O72– etc. (b) Precipitation reactions : These are given by SO42–, PO43– etc. Table : 10 Observation W.E. or S.E. + BaCl2(aq)  White precipitate insoluble in dil. HCl and HNO3. W.E or S.E + conc HNO3 (1–2 mL) + ammonium molybdate and boil  Canary yellow precipitate

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Inference SO42– PO43–

6

CHEMISTRY

 

W.E. = Water extract. (Salt is dissolved in distilled water) S.E. = Sodium carbonate extract

Preparation of sodium carbonate extract : Take 1-2 g of salt/salts mixture and three times the amount of pure solid sodium carbonate in a borosil conical flask. Add 20 mL of distilled water and boil the contents for 10 minutes. Cool the solution and then filter. The Filtrate is termed as "Sodium carbonate extract". Sodium carbonate reacts with the inorganic salt to form water soluble sodium salt of the acid radical. BaCl2 + Na2CO3  BaCO3  (white) + 2NaCl (aq) Cd3 (PO4)2 + 3Na2CO3  3CdCO3  + 2Na3 PO4 (aq) Sodium carbonate extract is used when (a) salt is only partially soluble in water or insoluble (b) cations interfere with the tests for acid radicals or the coloured salt solutions may be too intense in colour that the test results are not too clear.



As sodium carbonate extract contains excess of sodium carbonate, it should be neutralised with a suitable acid before proceeding for analysis of an anion.

Figure : Preparation of sodium carbonate extract

Individual tests : (A) GROUP 'A' RADICALS : (a) DILUTE SULPHURIC ACID/DILUTE HYDROCHLORIC ACID GROUP : 1. CARBONATE ION (CO32–) :  Dilute H2SO4 test : A colourless odourless gas is evolved with brisk effervescence. CaCO3 + H2SO4  CaSO4 + H2O + CO2  

Lime water/Baryta water (Ba(OH)2) test : The liberated gas can be identified by its property of rendering lime water (or baryta water) turbid. CO2 + Ca(OH)2  CaCO3  milky+ H2O On prolonged passage of CO2 the milkiness disappears.

 CaCO3 + CO2 + H2O  Ca(HCO3)2 (soluble)  CaCO3 + H2O + CO2

 



Magnesium sulphate test (for soluble carbonates) : CO32– (aq) + MgSO4 (aq)  MgCO3  (white) + SO42– (aq) Silver nitrate solution : White precipitate is formed CO32– + Ag+  Ag2CO3 White precipitate is soluble in HNO3 and ammonia. The precipitate becomes yellow or brown upon addition of excess reagent owing to the formation of silver oxide ; the same happens if the mixture is boiled. Ag2CO3  Ag2O  + CO2  Phenolphthalein is turned pink by soluble carbonates and colourless by soluble hydrogen carbonates.

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CHEMISTRY



  

Mercury(II) chloride does not form precipitate with hydrogen carbonate ions, while in a solution of normal carbonates a reddish–brown precipitate of basic mercury(II) carbonate (3HgO. HgCO3 = Hg4O3CO3) is formed. CO32– + 4 Hg2+ + 3 H2O  Hg4O3CO3  + 6H+ Lime water milky test is also shown by SO2 but CO2 does not turn the filter paper soaked in acidified K2Cr2O7 green. Soluble bicarbonates give white precipitate with MgSO4 (aq) / MgCl2(aq) only on heating.  Mg2+ + 2HCO3–  Mg(HCO3)2  MgCO3  + H2O + CO2

Action of heat : Bicarbonates : 2NaHCO3  Na2CO3 + H2O + CO2 Carbonates : Except carbonates of Na, K, Rb, Cs ; the Li2CO3 and all alkaline earth metals decompose as given below : Li2CO3  Li2O + CO2 ; MgCO3  MgO + CO2 ; Ag2CO3  2Ag + CO2



 CaCO3  CaO + CO2

2. SULPHITE ION (SO32–) :  Dilute H2SO4 test : Decomposition of salt is more rapidly on warming, with the evolution of sulphur dioxide. CaSO3 + H2SO4  CaSO4 + H2O + SO2  SO2 has suffocating odour of burning sulphur.  Acidified potassium dichromate test : The filter paper dipped in acidified K2Cr2O7 turns green. Cr2O72– + 2H+ + 3SO2  2Cr3+ (green) + 3SO42– + H2O. 



  





Barium chloride/Strontium chloride solution : White precipitate of barium (or strontium) sulphite is obtained. SO32– + Ba2+/Sr2+  BaSO3/SrSO3  (white). White precipitate dissolves in dilute HCl, when sulphur dioxide is evolved. BaSO3  + 2H+  Ba2+ + SO2  + H2O. White precipitate (BaSO3) on standing is slowly oxidised to sulphate which is insoluble in dilute mineral acids. This change is rapidly effected by warming with bromine water, a little concentrated nitric acid or with hydrogen peroxide. 2 BaSO3  + O2  2 BaSO4  BaSO3 + Br2 + H2O  2 BaSO4 + 2 Br– + 2H+ Hence, reddish brown colour of bromine water is decolourised. 3BaSO3  + 2 HNO3  3 BaSO4  + 2NO  + H2O BaSO3  + H2O2  BaSO4  + H2O These reactions are not given by carbonates (distinction from carbonates). Zinc and sulphuric acid test : Hydrogen sulphide gas is evolved. SO32– + 3Zn2+ + 8H+  H2S  + 3Zn2+ + 3H2O Lime water test : A white precipitate is formed. The precipitate dissolves on prolonged passage of the gas, due to the formation of hydrogen sulphite ions. Ca(OH)2 + SO2  CaSO3 (milky) + H2O CaSO3  + SO2 + H2O  Ca(HSO3)2 (soluble) A turbidity is also produced by carbonates ; sulphur dioxide must therefore be first removed when testing for the latter. This may be affected by adding potassium dichromate solution to the test–tube before acidifying. The dichromate oxidizes and destroys the sulphur dioxide without affecting the carbon dioxide. Lead acetate or lead nitrate solution : White precipitate of PbSO3 is obtained. SO32– + Pb2+  PbSO3  White precipitate gets soluble in dil. HNO3 on boiling. The precipitate is oxidized by atmospheric oxygen and PbSO4 is formed. 2PbSO3  + O2  2PbSO4 

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CHEMISTRY 3. SULPHIDE ION (S2–) :  Dilute H2SO4 test : Pungent smelling gas like that of rotten egg is obtained. S2– + 2H+  H2S   Lead acetate test : Filter paper moistened with lead acetate solution turns black. (CH3COO)2Pb + H2S  PbS  (black) + 2CH3COOH.  Sodium nitroprusside test : Purple coloration is obtained. S2– + [Fe(CN)5 (NO)]2–  [Fe(CN)5NOS]4– (violet). It is a ligand exchange reaction not a redox. No reaction occurs with solution of H2S or free gas. If however, filter paper moistened with a solution of the reagent is made alkaline with NaOH or NH3 solution, a purple colouration is produced with free H2S also.



H2S does not provide sufficient concentration of S2– ions so that it does not give sodium nitroprusside



test. Solubility is low 0.1 M and K1 is just 10–7. Cadmium carbonate suspension/ Cadmium acetate solution : Yellow precipitate is formed. Na2S + CdCO3  CdS + Na2CO3



Filter paper moistened with cadmium acetate when brought in contact with evolving gas it turns yellow.





S2– + 2H+  H2S ; H2S + Cd2+  CdS  + 2H+. Silver nitrate solution : Black precipitate is formed which is insoluble in cold, but soluble in hot, dilute nitric acid. Ag+ + S2–  Ag2S  Methylene blue test : NN–Dimethyl–p–phenylenediamine is converted by iron(III) chloride and hydrogen sulphide in strongly acid solution into the water–soluble dyestuff, methylene blue. This is a sensitive test for soluble sulphides and hydrogen sulphide.

4. NITRITE ION (NO2¯ ) :  Dilute H2SO4 test : Solid nitrite in cold produces a transient pale blue liquid (due to the presence of free nitrous acid, HNO2 or its anhydride, N2O3) first and then evolution of pungent smelling reddish brown vapours of NO2 takes place. NO2– + H+  HNO2 ; (2HNO2  H2O + N2O3); 3HNO2  HNO3 + 2NO + H2O ; 2NO + O2  2NO2  

Starch iodide test : The addition of a nitrite solution to a solution of potassium iodide, followed by acidification with acetic acid or with dilute sulphuric acid, results in the liberation of iodine, which may be identified by the blue colour produced with starch paste. A similar result is obtained by dipping potassium iodide–starch paper moistened with a little dilute acid into the solution. 2NO2– + 3I– + 4CH3COOH  I3– + 2NO + 4CH3COO– + 2H2O Starch + I3–  Blue (starch iodine adsorption complex)



Ferrous sulphate test (Brown ring test) : When the nitrite solution is added carefully to a concentrated solution of iron(II) sulphate acidified with dilute acetic acid or dilute sulphuric acid, a brown ring appears due to the formation of [Fe(H2O)5NO]SO4 at the junction of the two liquids. If the addition has not been made slowly and caustiously, a brown colouration results. NO2– + CH3COOH  HNO2 + CH3COO– 3HNO2  H2O + HNO3 + 2NO  Fe2+ + SO42– + NO   [Fe, NO]SO4



Thiourea test : When a dilute acetic acid solution of a nitrite is treated with a little solid thiourea, nitrogen is evolved and thiocyanic acid is produced. The latter may be identified by the red colour produced with dilute HCl and FeCl3 solution. NaNO2 + CH3COOH  HNO2 + CH3COONa HNO2 + H2NCSNH2(s) (thiourea)  N2 + HSCN + 2H2O dil HCl

FeCl3 + 3HSCN   Fe(SCN)3 (blood red colouration) + 3HCl

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CHEMISTRY 



Acidified potassium permanganate solution : Pink colour of KMnO4 is decolourised by a solution of a nitrite, but no gas is evolved. 5 NO2– + 2 MnO4– + 6 H+  5 NO3– + 2 Mn2+ + 3 H2O Silver nitrate solution : White crystalline precipitate of silver nitrite from concentrated solutions. NO2– + Ag+  AgNO2 

5. ACETATE ION (CH3COO¯)  With dilute H2SO4 a vinegar like smell is obtained. (CH3COO)2Ca + H2SO4  2CH3COOH + CaSO4  Neutral ferric chloride test : A deep red/ blood red colouration (no precipitate) indicates the presence of acetate. 6CH3COO– + 3Fe3+ + 2H2O  [Fe3(OH)2(CH3COO)6]+ + 2H+

 

When solution is diluted with water and boiled, brownish red precipitate of basic iron (III) acetate is obtained. Boil [Fe3(OH)2(CH3COO)6]+ + 4H2O   3Fe(OH)2CH3COO ¯ + 3CH3COOH + H+ Silver nitrate solution test : A white crystalline precipitate is produced in concentrated solution in the cold. CH3COO– + Ag+

CH3COOAg 

Precipitate is more soluble in boiling water and readily soluble in dilute ammonia solution. Example-1

Solution

Example-2

Solution

Example-3

Solution

An aqueous solution of salt containing an anion Xn– gives the following reactions : (i) It gives the purple or violet colouration with sodium nitroprusside solution. (ii) It liberates a colourless unpleasant smelling gas with dilute H2SO4 which turns lead acetate paper black. Identify the anion (Xn–) and write the chemical reactions involved. Xn– is S2– because (i) [Fe(CN)5NO]2– + S2–  [Fe(CN)5NOS]4– (purple or violet colouration) (ii) S2– + H2SO4  H2S  (colourless unpleasant smelling) + SO42– H2S + Pb(CH3COO)2  PbS  (black) + 2CH3COOH Sulphite on treatment with dil. H2SO4 liberates a gas which : (A) turns lead acetate paper black (B) burns with blue flame (C) smells like vinegar (D) turns acidified K2Cr2O7 solution green 2– 2– SO3 + H2SO4  SO2 + SO4 + H2O SO2 turns acidified K2Cr2O7 solution green. K2Cr2O7+ H2SO4 + 3SO2  Cr2(SO4)3 (Green) + K2SO4 + H2O Therefore, (D) option is correct. A colourless pungent smelling gas (X) is obtained when a salt is reacted with dil. H2SO4 . The gas (X) responds to the following properties. (A) It turns lime water milky (B) It turns acidified potassium dichromate solution green (C) It gives white turbidity when H2S gas is passed through its aqueous solution. (D) Its aqueous solution in NaOH gives a white precipitate with barium chloride which dissolves in dil. HCl liberating (X). Identify (X) and write the chemical equations involved. As gas X turns lime water milky it may be CO2 or SO2 . But CO2 is colourless and odourless, so 'X' may be SO2 . This is further, confirmed by the following reactions : SO32– + H2SO4  SO42– + SO2 + H2O ; Ca(OH)2 + SO2  CaSO3 (milky) + H2O K2 Cr2O7 + H2SO4 + 3SO2  K2SO4 + Cr2 (SO4)3 (green) + H2O SO2 + 2H2S  3S  (white) + 2H2O ; SO2 + 2NaOH  Na2SO3 + H2O   Na2 SO3 + BaCl2 BaSO3  (white) + 2NaCl ; BaSO3 + 2HCl  BaCl2 (soluble) + SO2 + H2O.

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CHEMISTRY (b) CONC . H2SO4 GROUP : 1. CHLORIDE ION (Cl¯) :  Concentrated H2SO4 test : Colourless pungent smelling gas is evolved which gives fumes of NH4Cl when a glass rod dipped in dil. HCl is brought in contact with evolving gas. Cl– + H2SO4  HCl + HSO4–  NH4OH + HCl  NH4Cl  (white fumes) + H2O.

 

 

 

2NaCl + MnO2 + 2H2SO4 (conc.)  Na2SO4 + MnSO4 + 2H2O + Cl2 Silver nitrate test : Cl– + Ag+  AgCl  (white) With sodium arsenite it is converted into yellow precipitate (distinction from AgBr and AgI) but insoluble in dilute nitric acid. 3AgCl  + AsO33–  Ag3AsO3  + 3Cl–. White precipitate is soluble in aqueous ammonia and precipitate reappears with HNO3. AgCl + 2NH4OH  [ Ag(NH3)2]Cl (Soluble) + 2H2O [Ag(NH3)2]Cl + 2H+  AgCl  + 2NH4+. Chromyl chloride test : 4Cl– + Cr2O72– + 6H+ (conc.)  2CrO2Cl2 (deep red vapours) + 3H2O When deep red vapours are passed into sodium hydroxide solution, a yellow solution of sodium chromate is formed, which when treated with lead acetate gives yellow precipitate of lead chromate. CrO2Cl2 + 4OH–  CrO42– + 2Cl– + 2H2O CrO42– + Pb+2  PbCrO4 (yellow) Heavy metal chlorides such as Hg2Cl2, HgCl2, SnCl2, AgCl, PbCl2 and SbCl3 do not respond to this test as they are partially dissociated. This test is given generally by ionic chlorides. Test should be carried out in a dry test tube otherwise chromic acid will be formed. CrO2Cl2 + 2H2O  H2CrO4 + 2HCl

2. BROMIDE ION (Br¯) :  Concentrated H2SO4 test : First a reddish-brown solution is formed, then reddish-brown bromine vapour accompanies the hydrogen bromide (fuming in moist air) is evolved. 2NaBr + H2SO4  Na2SO4 + 2HBr 2HBr + H2SO4  Br2 + 2H2O + SO2

 

 







2KBr + MnO2 + 2H2SO4  Br2+ K2SO4 + MnSO4 + 2H2O Silver nitrate test : Pale yellow precipitate is formed NaBr + AgNO3  AgBr  + NaNO3 Yellow precipitate is partially soluble in dilute aqueous ammonia but readily dissolves in concentrated ammonia solution. AgBr + 2NH4OH  [Ag(NH3)2] Br + H2O Lead acetate test : Bromides on treatment with lead acetate solution, gives a white crystalline precipitate of lead bromide, which is soluble in boiling water giving colourless solution. 2Br– + Pb+2  PbBr2  Chlorine water test (organic layer test) : When to a sodium carbonate extract of metal bromide containing CCl4, CHCl3 or CS2, chlorine water is added and the content is shaken and then allow to settle down reddish brown colour is obtained in organic layer. 2Br– + Cl2  2Cl– + Br2  . Br2 + CHCl3 / CCl4  Br2 dissolve to give reddish brown colour in organic layer.. With excess of chlorine water, the bromine is converted into yellow bromine monochloride and a pale yellow solution results. Br2 + Cl2   2BrCl Starch paper test : When starch paper is brought in contact with evolving bromine gas orange red spots are produced. Br2 + starch  starch bromine adsorption complex (orange red) Potassium dichromate and concentrated H2SO4 : When a mixture of solid bromide, K2Cr2O7 and concentrated H2SO4 is heated and evolved vapours are passed through water, a orange red solution is obtained. 6KBr + K2Cr2O7 + 7H2SO4  3Br2 + Cr2(SO4)3 + 4K2SO4 + 7H2O

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CHEMISTRY 3. IODIDE ION ¯) :  Concentrated H2SO4 test : Pungent smelling violet vapours are evolved. 2Na + H2SO4  Na2SO4 + 2HI 2HI + H2SO4  I2  (dark violet) + 2H2O + SO2

 



 

Evolution of dark violet fumes intensifies on adding a pinch of MnO2. 3I– + MnO2 + 2H2SO4  I3–  + Mn2+ + 2SO42– + 2H2S Starch paper test : Iodides are readily oxidised in acid solution to free iodine; the free iodine may than be identified by deep blue colouration produced with starch solution. 3I– + 2NO2– + 4H+  I3– + 2NO  + 2H2O. Silver nitrate test : Bright yellow precipitate is formed. I– + Ag+  AgI  Bright yellow precipitate is insoluble in dilute aqueous ammonia but is partially soluble in concentrated ammonia solution. Chlorine water test (organic layer test) : When chlorine water is added to a solution of iodide, free iodine is liberated which colours the solution brown and on shaking with CS2, CHCl3 or CCl4, it dissolves in organic layer forming a violet solution, which settles below the aqueous layer. 2NaI + Cl2  2NaCl + I2 I2 + CHCl3  I2 dissolves to give violet colour in organic layer.. If excess of chlorine water is added, I2 is oxidised to iodic acid (colourless). I3– + 8Cl2 + 9H2O  3 I O3 + 16Cl– + 18 H+







Lead acetate solution : A yellow precipitate is formed which is soluble in hot water forming a colourless solution and yielding golden yellow plates ('spangles') on cooling. 2I– + Pb+2  PbI2  Potassium dichromate and concentrated sulphuric acid : Violet vapours are liberated, and no chromate is present in distillate. 6I– + Cr2O72 + 2H2SO4  3I2 + Cr3+ +7SO42– + 7H2O Action of heat : Most of halides are stable but few decompose as 2FeCl3  2FeCl2 + Cl2 ; MgCl2 . 6H2O  MgO + 2HCl + 5H2O Hg2Cl2  HgCl2 + Hg ; NH4Cl  NH3 + HCl 2CuI2  Cu2I2 + I2 (without heating)

4. NITRATE ION (NO3¯) :  Concentrated H2SO4 test : Pungent smelling reddish brown vapours are evolved. 4NO3– + 2H2SO4  4NO2  + O2 + 2SO42– + 2H2O

 

Addition of bright copper turnings or paper pellets intensifies the evolution of reddish brown gas. 2NO3– + 4H2SO4 + 3Cu  3Cu2+ + 2NO  + 4SO42– + 4H2O ; 2NO  + O2  2NO2  4 C (paper pellet) + 4HNO3  2H2O + 4NO2 + 4CO2. Brown ring test : When a freshly prepared saturated solution of iron (II) sulphate is added to nitrate solution and then concentrated H2SO4 is added slowly from the side of the test tube, a brown ring is obtained at the junction of two layers. NaNO3 + H2SO4  NaHSO4 + HNO3 6FeSO4 + 2HNO3 + 3H2SO4  3Fe2(SO4)3 + 2NO + 4H2O

or

2NO3– + 4H2SO4 + 6Fe2+  6Fe3+ + 2NO  + 4SO42– + 4H2O. Fe2+ + NO  + 5H2O  [Fe(H2O)5 NO+]2+ (brown ring).

Figure : Brown ring test

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CHEMISTRY

 

On shaking and warming the mixture, NO escapes and a yellow solution of iron(iii) ions is obtained. Bromides and iodides interfere in brown ring test as liberated halogens obscure the brown ring. Nitrites also interfere the brown ring test and can be removed by adding a little sulphamic acid, or urea. H2NHSO3 + NO2–  N2  + SO42– + H+ + H2O HCl  HNO2 NO2– + H+ 

CO(NH2)2 + 2HNO2  2N2  + CO2  + 3H2O 



Diphenyl amine test : Blue ring is formed at the junction of two liquids (reagent and nitrate salt solutions). NaNO3 + H2SO4  NaHSO4 + HNO3 2HNO3  H2O + 2NO2 + [O] 2C6H5NHC6H5 + [O]  (C6H5)2 N – N (C6H5)2 (blue ring) + H2O. This test is also given by various oxidising agents like CrO42– , Cr2O72–, ClO3– , BrO3– , IO3–, NO2– etc. TO distinguish Br2 with NO2 (both are reddish brown gases) (a) Br2 + starch–iodide paper  Blue black colour spots do not develop immediately as Br2 is a weaker oxidising agent whereas NO2 being strong oxidising agent develops the blue black colour immediately. (b) Bromine develops orange–red colour spots on starch paper.

(B) GROUP 'B' RADICALS : Group of anions which do not give any gas with dilute as well as concentrated H2SO4 in cold but give precipitate with certain reagents : These acid radicals are identified in inorganic salts by their individual tests as given below 1. SULPHATE ION (SO42–) :  Barium chloride test : W.E. or S.E. + Barium chloride (aq)  White precipitate

 



Na2SO4 + BaCl2  BaSO4 white+ 2NaCl. White precipitate is insoluble in warm dil. HNO3 as well as HCl but moderately soluble in boiling concentrated hydrochloric acid. Lead acetate test : W.E. or S.E. + Lead acetate  white precipitate Na2SO4 + (CH3COO)2Pb  PbSO4 White + 2CH3COONa White precipitate soluble in excess of hot ammonium acetate. PbSO4 + 2CH3COONH4  (CH3COO)2Pb (soluble) + (NH4)2SO4





Match stick test : (a) W.E. or S.E. + Barium chloride  white precipitate Na2SO4 + BaCl2  2NaCl + BaSO4  (white) (b) White precipitate + Na2 CO3(s) mix and apply the paste on the end of the carbonized match stick or a wooden splinter. Put it in the reducing flame. BaSO4 (s) + Na2CO3(s)  Na2SO4 + BaCO3  (white) Na2SO4 + 4C  Na2 S + 4CO (c) Now dip the match stick in sodium nitroprusside solution, purple colour near the fused mass is developed. Na2S + Na2 [Fe(CN)5 NO]  Na4 [Fe(CN)5 NOS] (purple) Mercury II chloride test : Yellow precipitate is formed SO42– + 3Hg2+ + 2H2O  HgSO4. 2H2O (basic mercury II sulphate) + 4H+



Silver nitrate test : White precipitate is obtained. SO42– + 2Ag+  Ag2SO4 

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CHEMISTRY 2. PHOSPHATE ION (PO43– ) :  Ammonium molybdate test : Na2 HPO4 (aq) + 12(NH4)2MoO4 + 23HNO3  (NH4)3PMo12O40  (canary yellow) + 2NaNO3 + 21NH4NO3 + 12H2O

 



Some times ammonium phosphomolybdate is also represented by the formula (NH4)3 PO4 . 12MoO3 Magnesium nitrate or magnesia mixture test : W.E. or S.E + Magnesium nitrate reagent (3-4 mL) and allows to stand for 4-5 minutes, white crystalline precipitate is formed. Na2HPO4 (aq) + Mg(NO3)2 (aq) + NH4OH(aq)  Mg(NH4) PO4  (white) + 2NaNO3 + H2O Magnesia mixture is a solution containing MgCl2, NH4Cl and a little aqueous NH3. PO43– also gives BaCl2 test due to the formation of white precipitate of Ba3 (PO4)2 . So phosphate test should be carried out first and then conclude if PO43– is present or absent before proceeding with the test for SO42–.



Silver nitrate solution : Yellow precipitate is formed which is soluble in dilute ammonia and in dilute nitric acid. PO43– + 3Ag+  Ag3PO4  Ag3PO4 + 6NH3  3[Ag(NH3)2]+ + PO43– ; Ag3PO4 + 2H+  H2PO4– + 3Ag+



Iron (III) chloride solution : Yellowish-white precipitate of FePO4 is obtained HPO42– + Fe3+  FePO4 

3. BORATE ION (BO33–) : Salt (0.2 g) + conc. H2SO4 (1 mL) + Ethyl alcohol (4-5 mL) mix in a test tube and then heat. Ignite the evolved vapours with the help of Bunsen flame, green edged flame is obtained. 2Na3BO3 + 3H2SO4  3Na2 SO4 + 2H3BO3 3C2H5OH + H3BO3  (C2 H5)3 BO3 + 3H2O Example-4

Solution

Example-5 Solution Example-6 Solution

A compound (A) of S, Cl and O has vapour density of 67.5 (approx.). It reacts with water to form two acids and reacts with KOH to form two salts (B) and (C) while (B) gives white precipitate with AgNO3 solution and (C) gives white precipitate with BaCl2 solution. Identify (A), (B) & (C). As mixture give white precipitate with BaCl2 and AgNO3, it should contain SO42– and Cl– ions. As SO2Cl2 when dissolved in water gives, a mixture of H2SO4 & HCl which then react with KOH to form KCl and K2SO4 . Therefore, (A) is SO2Cl2 and (B) & (C) are K2SO4 and KCl respectively. Vapour density of SO2Cl2 = molecular weight / 2. Vapour density of SO2Cl2 = 135 / 2 = 67.2. Bromine vapours turn moist starch iodide paper : (A) brown (B) red (C) blue (D) colourless 2I– + Br2  I2 + 2Br– ; I2 + starch  blue starch iodine adsorption complex. Therefore, (C) option is correct. Na2S2O3 + 2  NaI + .......... [X], [X] is : (A) Na2S4O6 (B) Na2SO4

(C) Na2S

(D) Na3SO4

2Na2S2O3 + 2  2NaI + Na2S4O6 . Therefore, (A) option is correct.

Example-7

Column I and column II contains four entries each. Entries of column I are to be matched with some entries of column II. Each entry of column I may have the matching with one or more than one entries of column II. Column I Column II (A) Colourless gas evolved on addition of dil. H2SO4 (p) Cl– (B) White precipitate on addition of AgNO3 (q) S2– (C) Precipitate with solution containing Pb+2 ions. (r) NO2– (D) Its acidified salt solution decolourises pink KMnO4 solution. (s) SO32–

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CHEMISTRY Solution

(A - p, q, s) ; (B - p, r, s) ; (C - p, q, s) ; (D - p, q, r, s)

(A)

Cl– + H2SO4  HCl (colourless) + HSO4– ; S2– + 2H+  H2S  (colourless) NO2– + 2H+  NO2  (reddish brown) + H2O ; SO32– + 2H+  SO2  (colourless) + H2O

(B)

Ag+ + Cl–  AgCl  (white) ; Ag+ + NO2–  AgNO2  (white) ;

Ag+ + S2–  Ag2S  (black) 2Ag+ + SO32–  Ag2SO3  (white)

(C)

Pb2+ + 2Cl–  PbCl2  (white) ; Pb2+ + NO2–  PbNO2 (soluble) ;

Pb2+ + S2–  PbS  (black) Pb2+ + SO32–  PbSO3  (white)

(D)

2MnO4– + 16HCl  5Cl2 + 2Mn2+ + 6Cl– + 8H2O 2MnO4– + 5H2S + 6H+  Mn2+ + 5S  + 8H2O 2MnO4– + 5NO2– + 6H+  Mn2+ + 5NO3– + 3H2O 2MnO4– + 5SO2 + 2H2O  2Mn2+ + 5SO42– + 4H+

Analysis of CATIONS (Basic Radicals) : Table : 11 Group

Zero

Group reagent

NaOH or Ca(OH)2 , heat if required

Basic radical

NH4+

Composition and colour of precipitate Ammonia gas is evolved.

1.

Dil HCl

Ag+ Hg22+ Pb2+

AgCl ; White Hg2Cl2 ; White PbCl2 ; White

2.(A)

H2S in presence of dil HCl (Insoluble in YAS)

Hg2+ Pb2+ Bi3+ Cu2+ Cd2+

HgS ; PbS ; Bi2S3 ; CuS ; CdS ;

Black Black Black Black Yellow

2.(B)

H2S in presence of dil HCl (Soluble in YAS)

As3+ Sb3+ Sn2+ Sn4+

As2S3 ; Sb2S3 ; SnS ; SnS2 ;

Yellow Orange Brown Yellow

3.

NH4OH in presence of NH4Cl

Fe3+ Cr3+ Al3+

Fe(OH)3 ; Reddish brown Cr(OH)3 ; Green Al(OH)3 ; Gelatinous white

4.

H2S in presence of NH4OH and NH4Cl

Zn2+ Mn2+ Co2+ Ni2+

ZnS ; MnS ; CoS ; NiS ;

White Buff (or Pink) Black Black

5.

(NH4)2CO3 in presence of NH4OH

Ba2+ Sr2+ Ca2+

BaCO3 ; White SrCO3 ; White CaCO3 ; White

6.

Na2HPO4 in presence of NH4OH

Mg2+

Mg(NH4)PO4 ; White



[YAS = Yellow ammonium sulphide. (NH4)2Sx].

There are some important points which should be kept in mind while doing the analysis of cations. 1.

Group 1st radicals (Ag+, Pb2+, Hg22+) are precipitated as chloride because the solubility product of these chlorides (AgCl, PbCl2 , HgCl2) is less than the solubility products of chlorides of all other metal ions, which remain in solution. Lead chloride is slightly soluble in water and therefore, lead is never completely precipitated by adding dilute hydrochloric acid to a sample ; the rest of the lead ions are precipitated with H2S in acidic medium together with the cations of the second group.

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CHEMISTRY 2.

Group 2nd radicals are precipitated as sulphides because of their low solubility products whereas sulphides of other metals remain in solution because of their high solubility products. HCl acts as a source of H+ which decreases the concentration of S2– due to common ion effect. Hence, the concentration of S2– ion is too low that it exceeds only the solubility products of the metal sulphides of IInd group. We can not use H2SO4 inplace of HCl because some cations of higher groups i.e. vth group will also precipitate as their sulphates like BaSO4, SrSO4, CaSO4 etc. HNO3 can't be used in place of HCl. HNO3 is a powerful oxidising agent. HNO3 will oxidize H2S forming sulphur (yellow precipitate) or colloidal solution causing confusion with CdS, As2S3 even though Cd2+ , As3+ will be absent. The colloidal solution is white-yellow and that cannot be filtered causing unnecessary trouble.

3.

Group 3rd radicals are precipitated as hydroxides and the addition of NH4Cl suppresses the ionisation of NH4OH so that only the group 3 cations are precipitated as hydroxides because of their low solubility products. (i) Excess of NH4Cl should not be added, as manganese will precipitate as MnO2.H2O (ii) (NH4)2SO4 cannot be used in place of NH4Cl because the SO42– will also give the precipitate of BaSO4, SrSO4 etc. (iii) While proceeding for 3rd group from 2nd group, the filtrate of 2nd group is boiled off to remove the dissolved H2S and then one drop of concentrated HNO3 is added and again boil so that if Fe2+ is present is oxidised to Fe3+. The Ksp of Fe2+ is higher than Fe3+, therefore, it is partially precipitated and will thus interfere in the analysis of 4th group radicals. In our scheme Fe2+ is not there even if it is present, we shall report only Fe3+ (Fe2+ needs other special tests). (iv) If the medium remains acidic the hydroxides do not precipitate and we would think that Fe3+, Al3+, Cr3+ are absent even though they may be present. (v) In place of NH4OH, NaOH solution can't be used for the precipitation as their hydroxides because in excess of it we get soluble complexes of Al3+ and Cr3+.

4.

In 4th group, ammonium hydroxide increases the ionisation of H2S by removing H+ from H2S as unionised water. H2S 2H+ + S2– ; H+ + OH–  H2O 2– Now the excess of S ions is available and hence the ionic products of group 4th group cations exceeds their solubility products and will be precipitated. In case H2S is passed through a neutral solution, incomplete precipitation will take place due to the formation of HCl, which decreases the ionisation of H2S. For example MnCl2 + H2S  MnS + 2HCl

5.

In 5th group the reagent ammonium carbonate should be added in alkaline or neutral medium. In the absence of ammonia or ammonium ions, magnesium will also be precipitated.

PREPARATION OF ORIGINAL SOLUTION (O.S) : Original solution is used for the analysis of basic radicals except NH4+. It is prepared by dissolving given salt or mixture in a suitable solvent as follows :  H2 O  dil HCl  conc. HCl  Salt or Mixt. + H2O  soluble (then H2O is suitable solvent) If given salt or mixture is insoluble in H2O then it is dissolved in dil HCl.  Salt or Mixt. + dil HCl  soluble (then dil HCl is taken as solvent) If given salt or mixture is insoluble in dilute HCl then it is dissolved in conc. HCl.  Salt or Mixt. + conc. HCl  soluble In this way after selecting suitable solvent, given salt or mixture is dissolved in small quantity in the solvent and filtered. Obtained filtrate is called as original solution (O.S.) and that is used for the detection of basic radicals except NH4+.

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CHEMISTRY ZERO GROUP : 1. AMMONIUM ION (NH4+) : Sodium hydroxide solution : Ammonia gas is evolved on warming the solution containing ammonium salt and sodium hydroxide. NH4Cl + NaOH  NH3 + H2O + NaCl

 — —

The gas can be identified by the following characteristics / reactions. Its characteristics smell. The evolution of the white fumes of ammonium chloride when a glass rod dipped in dilute HCl is held in the vapour. NH3 + HCl  NH4Cl  (white fumes)

—

Its ability to turn filter paper moistened with Hg2(NO3)2 solution black. 2HgNO3 + 2NH3  Hg (NH2 ) NO 3  Hg + NH4NO3  black

— —

Its ability to turns filter paper moistened with CuSO4 solution deep blue. CuSO4 + 4NH3  [Cu(NH3)4]SO4 Filter paper moistened with a solution of manganese (II) chloride and hydrogen peroxide made alkaline with ammonia gives a brown colour due to the oxidation of manganese. 2NH3 + Mn2+ + H2O2 + H2O  MnO(OH)2 + 2NH4+



Nessler's reagent (Alkaline solution of potassium tetraiodidomercurate(II) : Brown precipitate or brown or yellow colouration is obtained according to the amount of ammonia or ammonium ions present. The precipitate is a basic mercury (II) amido–iodide. NH4+ + 2[HgI4]2– + 4OH –  HgO Hg (NH2)I + 7I– + 3H2O



Sodium hexanitrito–N–cobaltate (III) solution : NH4+ ions gives a yellow precipitate with the reagent. 3NH4+ + [Co(NO2)6]3–  (NH4)[Co(NO2)6]



Hexachloridoplatinate (IV) solution (i.e., hexachloroplatinic acid) : NH4+ ions gives a yellow precipitate with the reagent . 2NH4+ + [PtCl6]–  (NH4)2 [PtCl6]yellow



Saturated sodium hydrogen tartrate solution (NaHC4H4O6) : NH4+ ions gives a white precipitate with the reagent . NH4+ + HC4H4O6–  NH4 HC4H4O6 



4-Nitrobenzene - diazonium chloride reagent : NH4+ gives red colouration with the reagent in presence of sodium hydroxide. O 2N

N=N—Cl + NH4+ + OH–  O2N

N=NONH4 + Cl– + H2O

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CHEMISTRY Ist GROUP (Pb2+, Hg22+, Ag+) :

1. LEAD ION (Pb2+) : 







Dilute HCl solution : White precipitate is formed in cold solution. Pb2+ + HCl  PbCI2  (white) + 2H+ White precipitate is soluble in hot water. White precipitate is also soluble in concentrated HCl or concentrated KCl. PbCl2 + 2Cl–  [PbCl4]2– (colourless) Sodium hydroxide solution : White precipitate is formed which is soluble in excess of the reagent. Pb2+ + 2OH–  Pb(OH)2  ; Pb(OH)2 + 2OH–  [Pb(OH)4]2– 2– [Pb(OH)4] + H2O2  PbO2  (black / brownish black) + 2H2O + 2OH– [Pb(OH)4]2– + S2O82–  PbO2  + 2H2O + 2SO42– Potassium iodide solution : A yellow precipitate is formed which is soluble in excess more concentrated (6M) solution of the reagent. Yellow precipitate of PbI2 is moderately soluble in boiling water to give a colourless solution. PbCl2 + 2KI  PbI2  + 2KCl ; Pbl2 + KI K2[PbI4]  Yellow precipitate reappears on dilution with water. Yellow precipitate of PbI2 does not dissolve in excess of dilute solution of KI. Potassium chromate solution (in neutral, acetic acid or ammonia solution) : A yellow precipitate is formed. PbCl2 + K2CrO4  PbCrO4  + 2KCl Yellow precipitate is soluble in sodium hydroxide and HNO3 (nitric acid). 2PbCrO4 + 2H+ 2Pb2+ + Cr2O72– + H2O – PbCrO4 + 4OH [Pb(OH)4]2– + CrO42– Both reversible reactions on buffering the solution with ammonia or acetic acid respectively, PbCrO4 reprecipitates.

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CHEMISTRY 

Ammonia solution : With ammonia solution, Pb2+ gives a white precipitate of lead hydroxide. Pb2+ + 2NH4OH  Pb(OH)2  + 2NH4+



Dilute H2SO4 : White precipitate is formed which is soluble in more concentrated ammonium acetate (6M) solution or ammonium tartrate in the presence of ammonia. PbCl2 + H2SO4  PbSO4  + 2HCl PbSO4  + 4CH3COO–  [Pb(CH3COO)4]2– + SO42– PbSO4  + 2C4H4O62–  [Pb(C4H4O6)2]2– + SO42– Hot concentrated H2SO4 dissolves the precipitate due to the formation of PbHSO4. PbSO4  + H2SO4  Pb2+ + 2HSO4–

2. MERCURY(I) ION (Hg22+) :  Dilute HCl solution : White precipitate is formed in cold solution. Hg22+ + 2HCl  Hg2Cl2  (white) + 2H+  Ammonia solution : A mixture of mercury metal (black precipitate) and basic mercury (II) amido chloride (white precipitate) is formed. 2Hg2Cl2 + 4NH4OH  HgO.Hg(NH2 ) Cl  Hg  + 3NH4Cl + 3H2O    black



Dissolution of white precipitate (Hg2Cl2) in aquaregia : 3Hg2Cl2 + 2HNO3 + 6HCl  6HgCl2 + 2NO+ 4H2O (a) Stannous chloride test : White precipitate is formed which finally turns to black. 2HgCl2 + SnCl2  Hg2Cl2  + SnCl4 ; Hg2Cl2 + SnCl2  2Hg  (black) + 2SnCl4 (b) Potassium iodide test : Scarlet/red precipitate is formed which is soluble in excess of the reagent. HgCl2 + KI  HgI2 + 2KCl ; Hg2 + KI (excess)  K2[Hg4] (soluble) (c) Copper chips test : Shining grey deposition of mercury on copper chips is formed. HgCl2 + Cu  Hg (grey) + CuCl2



Potassium iodide solution : A green precipitate is formed. Hg22+ + 2I–  Hg2I2  Green precipitate in excess of reagent undergoes disproportionation reaction and a soluble [HgI4]2_ ions and black mercury are formed. Hg2I2  + 2I–  [HgI4]2– + Hg (finely divided) Boiling the mercury (I) iodide precipitate with water, disproportionation takes place and a mixture of red mercury (II) iodide precipitate and black mercury is formed. Hg2I2  HgI2  + Hg



Potassium chromate solution : A red crystalline precipitate is formed which turns black when solution of sodium hydroxide is added. Hg22+ + CrO42–  Hg2CrO4 ; Hg2CrO4 + 2OH–  Hg2O + CrO42– + H2O



Potassium cyanide solution : A black precipitate of mercury is obtained Hg22+ + 2CN–  Hg + Hg(CN)2 (soluble).

3. SILVER ION (Ag+) : 

Dilute hydrochloric acid/soluble chlorides : White precipitate is formed. Ag+ + HCl  AgCl  + H+ The precipitate obtained after filtration is soluble in concentrated HCl. AgCl + Cl– [AgCl2]– On dilution with water, the equilibrium shifts back to the left and the precipitate reappears. Dilute ammonia solution dissolves the precipitate forming a soluble complex. AgCl + 2NH3 [Ag (NH3)2 ]+ + Cl– Dilute nitric acid or hydrochloric acid neutralizes the excess ammonia and the precipitate reappears because the equilibrium is shifted backwards. [Ag(NH3)2]Cl + 2HNO3  AgCl (white) + 2NH4NO3.



Potassium iodide solution : A bright yellow precipitate is formed which is insoluble in dilute ammonia but partially soluble in concentrated ammonia. Ag+ + I–  AgI The yellow precipitate is soluble in KCN and in Na2S2O3. AgI + 2CN–  [Ag(CN)2]– + I– ; AgI+ 2S2O32–  [Ag(S2O3)2]3– + I–

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CHEMISTRY 

Potassium chromate solution : Red precipitate is formed which is soluble in dilute HNO3 and in ammonia solution. 2Ag+ + CrO42–  Ag2CrO4  2Ag2CrO4  + 2H+







4Ag+ + Cr2O72– + H2O

2Ag2CrO4  + 4NH3  2[Ag(NH3)2]+ + CrO42– Disodium hydrogen phosphate solution : In neutral solution a yellow precipitate is formed with the reagent. 3Ag+ + HPO42–  Ag3PO4 + H+ The yellow precipitate is soluble in nitric acid and ammonia solution. Hydrazine sulphate (saturated) : With diammineargentate (I) reagent forms finely divided silver which adheres to the cleaned glass walls of the test tube forming an attractive mirror. 4[Ag(NH3)2]+ + H2N–NH2.H2SO4  4Ag + N2 + 6NH4+ + 2NH3 + SO42– Ammonia solution : Brown precipitate is formed. 2Ag+ + 2NH3 + H2O  Ag2O + 2NH4+ Precipitate dissolves in ammonia. Ag2O  + 4NH3 + H2O  2[Ag(NH3)2]+ + 2OH–

Example-8

Solution

A compound on heating with an excess of caustic soda solution liberates a gas (B) which gives white fumes on exposure of HCl. The resultant alkaline solution thus obtained after heating again liberates the same gas (B) when heated with zinc powder. Compound (A) on heating alone gives a neutral oxide of nitrogen not nitrogen gas. Identify (A) and (B) and give the relevant chemical reactions. As NH3 gives white fumes with HCl, therefore, (B) should be NH3 and (A) should be the salt of ammonium. Further we know that nitrite of ammonium gives NH3 with Zn and alkali and when heated alone gives neutral oxide (N2O) not N2. Hence the salt should be ammonium nitrate not ammonium nitrite. NH4 NO3 (A) + NaOH  NaNO3 + H2O + NH3  (B) ; NH3 + HCl  NH4Cl (white fumes) Zn NaOH NaNO + 8[H]  NH NO  N O (neutral) + 2H O   NaOH + 2H O + NH ; 3

Example-9

Solution

Example-10 Solution

2

3

4

3

2

2

A certain metal (A) is boiled with dilute HNO3 to give a salt (B) and an neutral oxide of nitrogen (C). An aqueous solution of (B) gives a white precipitate (D) with brine which is soluble in ammonium hydroxide. An aqueous solution of (B) also gives red / brick red precipitate, (E) with potassium chromate solution. Identify (A) to (E) and write the chemical reactions involved. As solution of (B) gives white precipitate with NaCl (aq) and precipitate is soluble in ammonium hydroxide, it may be of silver salt. Further it gives brick red precipitate with K2CrO4 , therefore, metal (A) may be silver. 3Ag (A) + 4HNO3  3AgNO3 (B) + NO(C) + 2H2O; AgNO3 + NaCl  AgCl  (white) (D) + NaNO3 AgCl + 2NH4OH  [Ag(NH3)]2Cl (soluble) + 2H2O 2AgNO3 + K2CrO4  Ag2CrO4  (red / brick red) (E) + 2KNO3 Which of the following salt will give white precipitate with the solution containing Pb2+ ions ? (A) Na2CO3 (B) NaCl (C) Na2SO3 (D) All of these Pb2+ + CO32–  PbCO3  (white) Pb2+ + 2Cl–  PbCl2  (white) Pb2+ + SO32–  PbSO3  (white) Therefore, (D) option is correct.

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CHEMISTRY IInd Group (Hg2+, Pb2+, Bi3+, Cu2+, Cd2+, As3+, Sb3+, Sn2+) On the basis of the solubility of the precipitates of the sulphides of II group cations in yellow ammonium sulphide, they have been classified into two subgroups as given below : A : HgS, PbS, CuS, Bi2S3, all black but CdS is yellow. All insoluble in yellow ammonium sulphide. B : SnS2, As2S3 are yellow, Sb2S3 is orange & SnS is dark brown All soluble in yellow ammonium sulphide. IIA Group (Hg2+, Pb2+, Bi3+, Cu2+, Cd2+)

1.

MERCURY (II) ION (Hg2+) :  Precipitation with H2S in acidic medium : Black precipitate is formed. Precipitate insoluble in water,

hot dilute HNO3 ,alkali hydroxides, or colourless ammonium sulphide. H

 HgS  + 2H+ Hg2+ + H2S 

Na S (2M) dissolves the precipitate forming soluble complex. 2

HgS + S2–  [HgS2]2–

Aqua regia dissolves the precipitate.

3 HgS + 6 HCl + 2HNO3  3 HgCl2 + 3S  + 2NO  + 4 H2O

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CHEMISTRY Under these circumstance HgCl2 is undissociated. When solution is heated white precipitate of sulphur dissolves forming H2SO4. 2 HNO3 + S   SO42– + 2 H+ + 2NO   Stannous chloride solution : When added in moderate amounts silky white precipitate is formed.

2HgCl2 + SnCl2  SnCl4 + Hg2Cl2  If more reagent is added, Hg (I) chloride is reduced to black precipitate of mercury. Hg2Cl2 + SnCl2  SnCl4 + 2Hg   Potassium iodide solution : On slow addition red precipitate is formed. Hg2+ + 2I–  HgI2  Precipitate dissolves in excess of KI forming colourless soluble complex. HgI2 + 2I–  [HgI4]2–



KCN does not have any effect.  Copper chips , sheet or coin : A black precipitate of mercury is formed. Hg2+ + Cu  Hg  + Cu2+  Sodium hydroxide solution : When added in small quantity brownish–red precipitate of varying composition is formed and in stoichiometric amounts precipitate turns to yellow when Hg (II) oxide is formed. Hg2+ + 2OH–  HgO  + H2O Precipitate is insoluble in excess reagent but dissolves readily in acids and this can be used to differentiate Hg (I) from Hg (II).  Ammonia solution : White precipitate of mixed composition (Mercury (II) oxide + Mercury (II) amido nitrate) is formed with metal nitrate. 2 Hg2+ + NO3– + 4NH3 + H2O  HgO.Hg (NH2)NO3  + 3 NH4+  Cobalt (II) thiocyanate test : When reagent is added to an aqueous solution of Hg2+ ions and the walls of the test tube is stirred with a glass rod, deep–blue crystalline precipitate is formed. Hg2+ + Co2+ + 4 SCN–  Co+2 [Hg(SCN)4]–2 

 2.

In place of Cobalt (II) thiocyanate, Co(CH3COO)2 and NH4SCN can be added to the aqueous solution of Hg2+ ions.

COPPER ION (Cu2+) :  Precipitation with H2S in acidic medium : Black precipitate is formed. H

 CuS  + 2 H+ Cu2+ + H2S  Precipitate is insoluble in boiling dilute (M) H2SO4 (distinction from cadmium), in NaOH, Na2S and (NH4)2S. Precipitate dissolves in hot concentrated HNO3 3 CuS + 8HNO3  3Cu(NO3)2 (blue) + 2NO + 4H2O + 3S When boiled for longer S is oxidised to H2SO4 and a clear solution of Cu(NO3)2 is obtained.



KCN dissolves the precipitate forming a clear solution. 2 CuS  + 8 CN–  2 [Cu(CN)4]3– + S22– (disulphide ion)

 Ammonia solution : When added sparingly a blue precipitate of basic salt (basic copper sulphate) is

formed with CuSO4. 2Cu2+ + SO42– + 2 NH3 + 2 H2O  Cu(OH)2 .CuSO4  + 2NH4+ It is soluble in excess of reagent forming a deep blue colouration. Cu(OH)2 .CuSO4 + 8NH3  2 [Cu(NH3)4]2+ + SO42– + 2 OH–  Sodium hydroxide in cold solution : A blue precipitate is formed. Cu2+ + 2 OH –  Cu (OH)2  Heat Cu(OH)2   CuO(red) + H2O

Potassium iodide : It gives a white precipitate of Cu(I) iodide but the solution is intensely brown because

of the formation of tri–iodide ions (or iodine). 2 Cu2+ + 5 I–  Cu2I2  + I3– The solution becomes colourless and a white precipitate is visible when excess of sodium thiosulphate solution is added. I3– + 2 S2O32–  3 I– + S4O62–

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CHEMISTRY



These reactions are used in quantitative analysis for the iodometric determination of copper.

 Potassium ferrocyanide (Potassium hexacyanidoferrate (II) ) solution : Cu2+ ions gives brown/chocolate

brown precipitate. 2Cu2+ + K4Fe(CN)6  Cu2[Fe(CN)6]  + 4K+



2 [Fe(CN)6]3– + 3 Cu2+  Cu3 [Fe(CN)6]2  (green)

(vi) Potassium cyanide : When added sparingly forms first a yellow precipitate. Cu2+ + 2CN–  Cu(CN)2 (yellow) Precipitate quickly decomposes into CuCN and cyanogen. 2 Cu(CN)2   2CuCN (white) + (CN)2  (highly poisonous) Excess reagent dissolves the precipitate forming a colourless soluble complex. CuCN  + 3 CN–  [Cu(CN)4]3– Complex is so stable that H2S cannot precipitate Cu (I) sulphide (distinction from cadmium).  Potassium thiocyanate solution : The Cu+2 ions solution initially gives a black precipitate which then slowly decomposes to give white precipitate of Cu(I) thiocyanate. Cu2+ + 2 SCN–  Cu(SCN)2  2 Cu(SCN)2   2 CuSCN  + (SCN)2 



Cu (II) thiocyanate can be immediately converted into Cu(I) thiocyanate by adding a suitable reducing

agent like saturated solution of sulphur dioxide. 2 Cu(SCN)2  + SO2 + 2 H2O  2 CuSCN  + 2 SCN– + SO42– + 4H+

3.

BISMUTH ION (Bi3+) :  Precipitation with H2S in acidic medium : Black precipitate is formed which is insoluble in cold dilute

HNO3 and yellow ammonium sulphide. H

 Bi2 S3  (black) + 6H+ 2Bi3+ + 3H2S  Bi2 S3 + 8HNO3  2Bi (NO3)3 + 2NO + 3S + 4H2O



Bi2S3 + 6 HCl (boiling concentrated)  2 Bi3+ + 6 Cl– + 3 H2S

 Sodium hydroxide : White precipitate is formed with the reagent.

Bi3+ + 3 OH–  Bi(OH)3 Very slightly soluble in excess reagent in cold solution but soluble in acids. Bi(OH)3 + 3H+  Bi3+ + H2O



Precipitate on boiling loses water and turns yellowish white which is oxidised to BiO3– by H2O2.

Bi(OH)3  BiO.OH + H2O BiO.OH + H2O2  BiO3– + H+ + H2O  Ammonia solution : White basic salt of variable composition is formed. Approximate chemical reaction is : Bi3+ + NO3– + 2 NH3 + 2 H2O  Bi(OH)2 NO3  + 2 NH4+  Alkaline sodium stannite (Sodium tetrahydroxidostannate (II)) : A black precipitate of metallic bismuth

is obtained. Bi3+ + 3 OH–  Bi(OH)3  2Bi(OH)3  + [Sn(OH)4]2–  2Bi  + 3 [Sn(OH)6]2– (OA) (RA)



The reagent must be freshly prepared and test must be carried out in cold solution.

 Dilution with water : Solution of bismuth salts gives white precipitate when water is added in larger

quantity. Bi3+ + NO3– + H2O  BiO(NO3)  + 2H+ Bi3+ + Cl– + H2O  BiOCl  (bismuth oxychloride or bismuthyl chloride) + 2H+ Soluble in mineral acids (dilute) but insoluble in tartaric acid (distinction from antimony) and in alkali hydroxide (distinction from tin).

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CHEMISTRY  Potassium iodide : When the reagent is added dropwise to a solution containing Bi3+ ions , a black

precipitate is formed. Bi3+ + 3–  Bi3  The precipitate dissolves in excess KI forming orange coloured soluble complex. Bi3 + –

[Bi4]–

On dilution the reaction is reversed and black BiI3 is reprecipitated.



On heating with water black precipitate of BiI3 turns orange. BiI3  + H2O  BiOI  + 2H+ + 2I–

4.

CADMIUM ION (Cd2+) :  Precipitation with H2S in acidic medium : Yellow precipitate is formed which dissolves in hot dilute

HNO3 .

H

 CdS  + 2H+ Cd2+ + H2S  CdS + 8HNO3  3Cd (NO3)2 + 4H2O + 2NO + 3S



Precipitate does not dissolve in KCN.

 Ammonia solution (Dropwise addition) : Ammonium hydroxide first gives white precipitate of Cd(OH)2

which gets dissolve in excess of reagent forming a soluble complex. Cd2+ + 2 NH3 + 2 H2O

Cd(OH)2  + 2 NH4+

Cd(OH)2 + 4 NH3  [Cd(NH3)4]2+ + 2 OH–

.

 Potassium cyanide : Initially a white precipitate of Cd(CN)2 is formed which in excess of reagent dissolves

forming a soluble complex. Cd2+ + 2 CN–  Cd(CN)2  Cd(CN)2  + 2 CN–  [Cd(CN)4]2– The colourless soluble complex is not too stable, therefore, reacts with H2S gas forming a yellow precipitate of CdS. [Cd(CN)4]2– + H2S  CdS  + 2 H+ + 4 CN–



KI forms no precipitate (distinction from Copper)

 Sodium hydroxide : White precipitate is obtained which is not soluble in excess of sodium hydroxide.

Cd2+ (aq) +2NaOH (aq)  Cd(OH)2  + 2Na+

5.

LEAD ION (Pb2+) :  Precipitation with H2S in acidic medium : Black precipitate is formed which is soluble in hot dilute

HNO3 .

Pb2+ + H2S  PbS  (black) + 2H+ 3PbS + 8HNO3  3Pb(NO3)2 + 2NO + 4H2O + 3S  Dilute H2SO4 : White precipitate is formed which is soluble in ammonium acetate. Pb(NO3)2 + H2SO4  PbSO4  (white) + 2HNO3 PbSO4 + 2CH3 COONH4  (CH3COO)2 Pb + (NH4)2 SO4  Potassium iodide : Yellow precipitate is formed which is soluble in excess more concentrated (6M) solution of the reagent. Yellow precipitate of PbI2 is moderately soluble in boiling water to give a colourless solution. (CH3COO)2 Pb + 2KI  2CH3 COOK + PbI2 yellow  Potassium dichromate : Yellow precipitate is formed.

(CH3COO)2 Pb + K2CrO4  2CH3 COOK + PbCrO4 (yellow)

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CHEMISTRY IIB Group (As3+, Sb3+, Sn2+, Sn4+)

1.

ARSENIC ION (As3+) :  Precipitation with H2S in acidic medium : Yellow precipitate is formed which is soluble in warm concentrated nitric acid, sodiumhydroxide solution and yellow ammonium sulphide. H

 As2 S3  + 6H+ 2As3+ + 3H2S  3 As2S3  + 28 HNO3 + 4 H2O  6 AsO43– + 9 SO42– + 36 H+ + 28 NO  As2S3  + 6OH–  AsO33– + AsS33– + 3 H2O As2S3  +4 S22–  2 AsS43– + S32–  Silver nitrate : Yellow precipitate of silver arsenite in neutral solution is formed with AsO33– ions. AsO33– + 3 Ag+  Ag3AsO3  AsO43– + 3 Ag  Ag3AsO4  (brownish red)

The precipitate is soluble in both nitric acid and ammonia.  Magnesia mixture : White crystalline precipitate of magnesium ammonium arsenate Mg(NH4)

AsO4.6 H2O is formed from neutral or ammonical solution. Upon treating the white precipitate with silver nitrate solution containing a few drops of acetic acid, red silver arsenate is formed (distinction from phosphate). H3AsO4 + MgSO4 + 3NH4OH  Mg(NH4)AsO4 + (NH4)2SO4 + 3H2O MgNH4AsO4  + 3 Ag+  Ag3AsO4  + Mg2+ + NH4+  Marsh's test : This test is based on the fact that all soluble compounds of arsenic are reduced by 'nascent'

hydrogen in acid solution to arsine (AsH3), a colourless, extremely poisonous gas with a garlic-like odour. If the gas, mixed with hydrogen, is passed through a heated glass tube, it is decomposed into hydrogen and metallic arsenic, which is deposited as a brownish-black 'mirror' just beyond the heated part of the tube. Zn + H2SO4  ZnSO4 + 2 [H] ; AsCl3 + 3 [H]  AsH3  + 3 [Cl] Heat AsO43– + 4 Zn + 11 H+  AsH3  + 4 Zn2+ + 4H2O ; AsH3    As  + 3/2 H2  3–  Ammonium molybdate test : Solution of AsO4 ion containing considerable excess of HNO3 on boiling with reagent gives a yellow crystalline precipitate. H3AsO4 + 12(NH4)2 MoO4 + 21HNO3  (NH4)3 AsO4 . 12MoO3 + 21NH4NO3 + 12H2O or AsO43– + 12MoO42– + 3NH4+ + 24H+  (NH4)3 AsMo12O40  + 12H2O.

2.

ANTIMONY ION (Sb3+) :  Precipitation with H2S in acidic medium : A orange red precipitate is formed from mildly acidic solutions. H  Sb2S3  (orange) + 6H+ Sb3+ + 3H2S  Precipitate is soluble in warm concentrated HCl, in ammonium polysulphide and in alkali hydroxides. Sb2S3  + 6 HCl  2 Sb3+ + 6 Cl– + 3 H2S  Sb2S3  + 4 S22–  2 SbS43– + S32– Sb2S3  + 4 OH–  SbO2– + 3 SbS2– + 2 H2O

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CHEMISTRY  Dilution with water : When water is poured in a solution of soluble SbCl3, a white precipitate of antimonyl

chloride (SbOCl) is formed, soluble in HCl. With a large excess of water, the hydrated oxide Sb2O3.xH2O is produced. SbCl3 + H2O  SbOCl + 2 HCl



SbO+Cl– is soluble in tartaric acid but BiO+Cl– is insoluble in tartaric acid. 

 Sodium hydroxide or Ammonia solution : White precipitate of hydrate of antimony (III) oxide Sb2O3.xH2O

is soluble in concentrated (5M) solution of caustic alkalies forming antimonites. 2 Sb3+ + 6 OH–  Sb2O3  + 3 H2O ; Sb2O3 + 2 OH–  2 SbO2– + H2O  Potassium iodide solution : Yellow colouration is obtained owing to the formation of a complex salt.

Sb3+ + 6 I–  [SbI6]3–  Reduction with zinc or tin : Sb3+ ions give black precipitate of metal.

2 Sb3+ + 3 Zn  2 Sb  + 3 Zn2+ ; 2 Sb3+ + 3 Sn  2 Sb  + 3 Sn2+ 3.

TIN (II) ION (Sn2+) AND TIN (IV) ION (Sn4+) :  Precipitation with H2S in acidic medium : Brown precipitate is obtained with Sn2+ which is soluble in concentrated HCl and yellow ammonium sulphide forming thiostannate but not in colourless ammonium sulphide. Sn2+ + H2S  SnS (brown) ; SnS  + S22–  SnS32– (thiostannate) Sn4+ + 2H2S  SnS2  (yellow) + 4 H+ Precipitate is soluble in concentrated HCl (difference from As3+ and Hg2+) and in alkali hydroxide, and also in ammonium sulphide and yellow ammonium sulphide.  Sodium hydroxide solution : White precipitate of Sn(OH)2 is formed which is soluble in excess of reagent. Sn2+ + 2OH– Sn(OH)2  ; Sn(OH)2  + 2OH– [Sn(OH)4]2–



With ammonia solution, white tin (II) hydroxide is precipitated, which cannot be dissolved in excess ammonia.  Reduction of stannic chloride to stannous chloride by iron filling or granulated zinc : SnCl4 + Zn  SnCl2 + ZnCl2  Mercuric chloride solution : SnCl2 + 2HgCl2  Hg2Cl2  (silky white) + SnCl4

SnCl2 + Hg2Cl2  2Hg black or grey+ SnCl4



HNO3 is not used because it acts as an oxidising agent and in solution it oxidises H2S to S according to the following reaction. H2S + [O]  H2O + S  (Yellow)

Example-11

Solution

When NaOH solution is mixed with aqueous solution of a salt 'A', and warmed, a black precipitate is formed. Black precipitate is filtered and dissolved in concentrated HNO3 by boiling. The resulting solution gives a chocolate brown coloured precipitate with potassium ferrocyanide solution. The filtrate obtained after filtering off the black precipitate, upon warming with Zn and NaOH evolves an alkaline pungent smelling gas. The resulting solution also responds to the brown ring test. The filtrate does not evolve N2 gas when it is boiled with urea in the presence of H2SO4. Identify the cation and anion present in the salt 'A'. Cu(NO3)2 + NaOH  Cu(OH)2  (blue) + 2NaNO3  Cu(OH)2   CuO  (black) + H2O 4Zn + NaNO3 + 7NaOH  4Na2ZnO2 + 2H2O + NH3  (pungent smelling alkaline gas).  CuO + 2HNO3  Cu(NO3)2 + H2O 2Cu(NO3)2 + K4[Fe(CN)6]  Cu2[Fe(CN)6]  (chocolate brown) + 4KNO3

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CHEMISTRY Example-12

Solution

Select the correct statements from the following. (A) Bi3+ ion gives black precipitate with KI which dissolves in excess of KI forming orange coloured solution. (B) Chloride of Cd2+ is not soluble in excess of NaOH but soluble in excess of aqueous ammonia. (C) Lead sulphate is soluble in ammonium acetate solution. (D) The Co2+ gives green coloured solution with KNO2 and acetic acid. (E) The red colour solution of ferric acetate turns to reddish brown precipitate when it is diluted with water and then boiled. (A) Bi3+ + 3KI  BiI3  (black) + 3K+ ; KI + BiI3  K [BiI4] (orange) (B) CdCl2 + 2NaOH  Cd(OH)2 (white) + 2NaCl ; Cd(OH)2  + NaOH  no reaction CdCl2 + 2NH4OH  Cd(OH)2  (white) + 2NH4Cl Cd(OH)2 + 4NH4OH  [Cd(NH3)4] (OH)2 (soluble) + 4H2O (C) PbSO4 + 2CH3 COO NH4  (CH3 COO)2 Pb (soluble) + (NH4)2 SO4 (D) CoCl2 + 7KNO2 + 2CH3 COOH  K3 [Co(NO2)6]  (yellow) + 2KCl + 2CH3COOK + NO + H2O Boil (E) Fe (CH3COO)3 + 2H2O   Fe (CH3 COO) (OH)2  (reddish brown) + 2 CH3 COOH Therefore, A, B, C and E are correct statements.

Example-13

Solution

Assertion : In aqueous solution, HgCl2 and SnCl2 can exist as Hg2+ and Sn2+ ions. Reason : SnCl2 acts as strong oxidising agent. (A) Both Assertion and Reason are true and Reason is the correct explanation of Assertion. (B) Both Assertion and Reason are true but Reason is not correct explanation of Assertion (C) Assertion is true but Reason is false (D) Assertion is false but Reason is true (E) Assertion and Reason both are false. SnIICl2 (excess) + HgIICl2  Hg0  (black) + SnIVCl4 SnCl2 and HgCl2 can not exist together in aqueous solution because SnCl2 acts as strong reducing agent because Sn show inert pair effect. Therefore, (E) option is correct.

IIIrd Group (Al3+, Cr3+, Fe3+)

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CHEMISTRY

 

Concentrated HNO3 is added to oxidise Fe2+ to Fe3+ if present. Solid NH4Cl should be added in excess other wise cations like Zn, Mn, Mg may be precipitated here. To much excess of NH4Cl should be avoided other wise Cr is not precipitated and Al may form a colloidal solution.

1. ALUMINIUM ION (Al3+) :  Precipitation with NH4OH in presence of NH4Cl : White gelatinous precipitate is formed which is slightly soluble in excess reagent. The solubility is decreased in the presence of ammonium salt. A small portion of the precipitate passes into the solution as colloidal Al(OH)3 (Aluminium hydroxide sol), the sol is coagulated on boiling the solution or upon the addition of soluble salt yielding a precipitate of Al(OH)3, known as Al(OH)3 gel. For complete precipitation, add ammonia solution in excess and the mixture is boiled until, the solution has a slight odour of NH3 . NH4 Cl  Al(OH)3  + 3NH4+ Al3+ + 3NH4OH    Sodium hydroxide : A solution containing Al +3 ions give a white precipitate with the reagent.

Al3+ + 3OH–  Al(OH)3  White precipitate dissolves in excess sodium hydroxide according to following reaction. Al(OH)3 + OH–

[Al(OH)4]–

The reaction is reversible and any reagent, which will reduce the hydroxyl ion concentration sufficiently should cause the reaction to proceed from right to left with the consequent precipitation of aluminium hydroxide. This may be effected with a solution of ammonium chloride (the hydroxyl ion concentration is reduced owing to the formation of the weak base ammonia , which can be readily removed as ammonia gas by heating) or by the addition of acid ; in the latter case, a large excess of acid causes the precipitated hydroxide to redissolve. [Al(OH)4]– + NH4+  Al(OH)3  + NH3 + H2O [Al(OH)4]– + H+ Al(OH)3 +

3H+

Al(OH)3  + H2O (excess)

Al3+ + 3 H2O

 Lake test :

AlCl3 + 3NH4OH  Al(OH)3  (white) + 3 NH4Cl The precipitation of aluminium hydroxide is done in presence of blue litmus. Originally the solution will appear red when the group precipitate is dissolved in HCl as blue litmus turns red in acid medium. On adding large excess of NH4OH, Al(OH)3 will be reprecipitated and will adsorb the litmus as it is flocculent by nature. After a while a blue mass will be seen floating in a colourless solution as all the colour from the solution will adsorbed.  Sodium acetate solution : No precipitate is obtained in cold , neutral solution but on boiling with excess

reagent, a voluminous precipitate of basic aluminium acetate is formed. Al3+ + 3 CH3COO– + 2 H2O  Al(OH)2CH3COO  (white basic salt) + CH3COOH  Disodium hydrogen phosphate solution : A white gelatinous precipitate is formed.

Al3+ + HPO42–

AlPO4  + H+

The reaction is reversible; strong acids dissolves the precipitate. However, the precipitate is insoluble in acetic acid (difference from phosphates of alkaline earths, which are soluble). The precipitate can also be dissolved in sodium hydroxide.  Dry test : Aluminium compounds when heated with sodium carbonate upon charcoal gives a white infusible

solid, which glows when hot. If the residue is moistened with a little cobalt nitrate solution and again heated, a blue infusible mass is obtained. 2 Al2O3 + 2 Co2+ + 4 NO3–  2 CoAl2O4 (thenard blue) + 4NO2  + O2  Use of excess cobalt nitrate solution should be avoided since this will produce black cobalt oxide (Co3O4) upon ignition, which will mask the blue colour.

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CHEMISTRY 2. IRON ION (Fe3+) :  Precipitation with NH4OH in presence of NH4Cl : Gelatinous reddish brown precipitate is formed which is insoluble in excess reagent but soluble in acids. NH 4 Cl Fe3+ + 3NH4OH    Fe(OH)3  + 3NH4+



 

Precipitation of iron(II) hydroxide occurs with ammonia solution. If larger amounts of ammonium ions are present, the dissociation of NH4OH is suppressed and the concentration of OH– ions is lowered to such an extent that solubility product of iron (II) hydroxide, Fe(OH)2 is not attained and precipitation does not occur. Ksp of iron(III) hydroxide is so small (3.8 × 10–38) that complete precipitation take place even in the presence of ammonium salts. Heat Fe(OH)3   Fe2O3 + 3 H2O Oxide is soluble with difficulty in dilute acids but dissolves on vigorous boiling with concentrated HCl. Fe2O3 + 6 H+  2 Fe3+ + 3 H2O

 Sodium hydroxide solution : Reddish brown precipitate is formed insoluble in excess reagent (distinction

from aluminium and chromium). Fe3+ + 3OH–  Fe(OH)3   H2S gas in acidic solution : Fe3+ reduces to Fe2+ . 2 Fe3+ + H2S  2 Fe2+ + 2H+ + S  (milky-white) If a neutral solution of iron (III) chloride is added to a freshly prepared saturated solution of H2S , bluish colouration appears first, followed by precipitation of sulphur. The blue colour is due to a colloidal solution of sulphur of extremely small particle size.



This reaction can be used to test the freshness of H2S solution.

 Ammonium sulphide solution : Black precipitate consisting of Fe(II) sulphide and sulphur is formed.

2 Fe3+ + 3 S2–  2 FeS + S  In HCl, the black precipitate of Fe(II) sulphide dissolves and white precipitate of sulphur becomes visible. FeS  + 2H+  H2S  + Fe2+

 

The damp iron (II) sulphide precipitate, when exposed to air, is slowly oxidised to brown iron(III) hydroxide. 4 FeS  + 6 H2O + 3 O2  4 Fe(OH)3  + 4S 

From alkaline solutions of ammonium sulphide, black iron(III) sulphide is obtained. 2 Fe3+ + 3S2–  Fe2S3  On acidification with hydrochloric acid, iron (III) ions are reduced to iron (II) ions and sulphur is formed. Fe2S3  + 4H+  2 Fe2+ + 2 H2S  + S   Potassium ferrocyanide (Potassium hexacyanidoferrate(II) ) : Intense blue precipitate (Prussian blue) of iron(III) hexacyanidoferrate(II) is formed. 4Fe3+ + 3 [Fe(CN)6]4–  Fe4[Fe(CN)6]3  This is insoluble in dilute acids but decomposes in concentrated HCl. A large excess of the reagent dissolves it partly or entirely, when an intense blue solution is obtained. Sodium hydroxide turns the precipitate red. Fe4 [Fe(CN)6]3  + 12 OH–  Fe(OH)3  + 3[Fe(CN)6]4–



Oxalic acid also dissolves Prussian blue forming a blue-solution.

Important : If iron(III) chloride is added to an excess of potassium hexacyanidoferrate (II), a product with the composition of K Fe[Fe(CN)6] is formed. This tends to form colloidal solutions ("soluble Prussian blue") and can not be filtered.  Potassium ferricyanide (Potassium hexacyanidoferrate(III)) : A brown colouration is formed.

 

Fe3+ + [Fe(CN)6]3–  Fe[Fe(CN)6]

Upon adding hydrogen peroxide or some tin(II) chloride solution, the hexacyanidoferrate(III) part of the compound is reduced and Prussian blue is precipitated. Fe2+ gives dark blue precipitate with potassium ferricyanide. First hexacyanidoferrate(III) ions oxidise iron(II) to iron(III), when hexacyanidoferrate(II) is formed. Fe2+ + [Fe(CN)6]3–  Fe3+ + [Fe(CN)6]4– and these ions combine to form a precipitate called Turnbull's blue. 4 Fe3+ + 3 [Fe(CN)6]4–  Fe4[Fe(CN)6]3

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CHEMISTRY Note : Composition of this precipitate is identical to that of Prussian blue. Earlier the composition suggested was Fe3[Fe(CN)6]2 , hence different name. 3Fe2+ + 2K3 [Fe(CN)6]  Fe3 [Fe (CN)6]2 (ferrous ferric cyanide) + 6K+ Turnbull's blue



Fe(II) in ammonical solution gives red solution with DMG– colouration fades on standing due to the

oxidation of the iron(II) complex. Fe (III) does not give such complex.



In complete absence of air, Fe(II) ions produces white precipitate with potassium hexacyanidoferrate(II).

Fe2+ + 2K+ + [Fe(CN)6]4–  K2Fe [Fe(CN)6]  Under ordinary atmospheric conditions a pale–blue precipitate is formed.  Potassium thiocyanate (Potassium sulphocyanide) : In slightly acidic medium, a deep red colouration is produced due to the formation of a non-dissociated iron(III) thiocyanate complex. Fe3+ + 3 SCN–  Fe(SCN)3 This neutral molecule can be extracted by ether or amyl alcohol.

 

With pure Fe(II) ions no colouration is obtained. Fluorides and Hg(II) ions bleach the red colour. Fe(SCN)3 + 6F–  [FeF6]3– + 3 SCN– 2 Fe(SCN)3 + 3Hg2+  2 Fe3+ + 3 Hg (SCN)2

 Sodium acetate solution : A reddish–brown colouration is obtained owing to the formation of

[Fe3(OH)2(CH3COO)6]+. The reaction becomes complete only if the strong acid which is formed, is removed by the addition of an excess of the reagent, which acts as a buffer. 3 Fe3+ + 6 CH3COOH + 2 H2O

[Fe3(OH)2(CH3COO)6]+ + 2H+

This reddish–brown coloured solution on dilution with water and boiling gives reddish brown precipitate of basic ferric acetate. [Fe3(OH)2(CH3COO)6]+ + 4 H2O  3 Fe(OH)2CH3COO  + 3 CH3COOH + H+ The excess of acetate ion acts as a buffer and the reaction goes to completion.  Disodium hydrogen phosphate solution : A yellowish–white precipitate is formed.  FePO4  + H+ Fe3+ + HPO42–  The reaction is reversible, because a strong acid is formed which dissolves the precipitate. It is advisable to add small amounts of sodium acetate, which acts as a buffer. 3. CHROMIUM ION (Cr3+) :  Precipitation with NH4OH in presence of NH4Cl : A grey-green to green gelatinous precipitate is formed. Cr3+ + 3 NH3 + 3 H2O  Cr(OH)3 + 3 NH4+ Precipitate is slightly soluble in excess of reagent in cold forming pink or violet coloured solution. On boiling the solution, chromium hydroxide is reprecipitated.  [Cr(NH3)6]3+ + 3OH– Cr(OH)3  + 6 NH3   Sodium hydroxide solution : A green precipitate is formed.  Cr(OH)3  Cr3+ + 3 OH–  The reaction is reversible. On addition of acids the precipitate dissolves. In excess of reagent the precipitate dissolves readily forming chromites, the solution is green. The reaction is reversible ; on (slight) acidification and also on boiling green precipitate of chromium(III) hydroxide is obtained. Cr(OH)3 + OH–

 

[Cr(OH)4]–

On adding H2O2 to the alkaline solution of [Cr(OH)4]– , a yellow solution is obtained. 2 [Cr(OH)4]– + 3 H2O2 + 2 OH–  2 CrO42– + 8 H2O The green precipitate of Cr(OH)3 can be dissolved by using any of the following methods. (A) Fusion with fusion mixture (Na2CO3 + KNO3) 2 Cr(OH)3 + 2Na2CO3 + 3 KNO3  2 Na2CrO4 + 3 KNO2 + 2 CO2 + 3 H2O Fused mass on extraction with water gives yellow solution of Na2CrO4 .

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CHEMISTRY (B) 2 Cr(OH)3 + 3 Na2O2  2Na2 CrO4 + 2NaOH + H2O (C) Precipitate is heated with NaOH and bromine water. 2NaOH + Br2  NaOBr + NaBr + H2O ; NaOBr  NaBr + [O] 2Cr(OH)3 + 4NaOH + 3[O]  2Na2 CrO4 + 5H2O. The yellow solution of Na2CrO4 gave the following reactions with lead acetate, barium chloride and silver nitrate solutions. (a) Lead acetate solution : Yellow precipitate is formed. CH3OOH CrO42– + Pb2+   PbCrO4  (b) Barium chloride solution : Yellow precipitate is formed owing to the formation of barium chromate which is insoluble in acetic acid. CrO42– + Ba2+  BaCrO4 

(c) Silver nitrate solution : Red/Brick red precipitate is formed owing to the formation of silver chromate which is soluble in ammonia solution and its acidified solution turns to orange because of the formation of dichromate. Na2CrO4 + 2AgNO3  Ag2CrO4  + 2NaNO3 .  Acidified H2O2 test : On acidifying the yellow solution with dilute sulphuric acid and adding few drops of

ether or amyl alcohol to the mixture and finally adding some hydrogen peroxide, blue colouration is formed which can be extracted into the organic layer by gently shaking. Amyl alcohol Na2CrO4 + H2SO4  Na2SO4 + H2CrO4 ; H2CrO4 + 2H2O2    CrO5 (blue colouration) + H2O Blue colouration fades slowly due to the decomposition of perchromic acid (or chromium peroxide ) with the liberation of oxygen.

4CrO5 + 6H2SO4  2Cr2 (SO4)3 + 7O2 + 6H2O  1,5–Diphenylcarbazide test : In dilute mineral acid solution the reagent produces a soluble violet colour

with chromate. During the reaction, chromate is reduced to chromium (III) and diphenylcarbazone is formed; these reaction products in turn produce a complex with the characteristic colour.

 Phenanthroline test : Green precipitate dissolved in dilute HCl (minimum quantity) gives red colouration

with the reagent.

Example-14

Solution

A black coloured compound (A) on reaction with dil. H2SO4 gives a gas (B) and a yellow colour solution. The gas (B) on passing in a solution of an acid (C) gives a white/yellow turbidity (D). Gas (B) when passed in acidified solution of (E) gives a precipitate (F) soluble in dil HNO3. After boiling this solution when excess of NH4OH is added, a blue coloured compound (G) is formed. To this solution on addition of acetic acid and aqueous potassium ferrocyanide a chocolate coloured precipitate (H) is obtained. On addition of an aqueous solution of BaCl2 to an aqueous solution of (E), a white precipitate insoluble in HNO3 is obtained. Yellow colour solution on reaction with ammonium hydroxide in presence of air gives reddish brown precipitate. Identify (A) to (H). FeS (A) + H2SO4  FeSO4 + H2S (B) HNO3 (C)  NO2 + H2O + O ; H2S + O  H2O + S  (D) CuSO4 (E) + H2S  CuS  (F) + H2SO4. 3CuS + 8HNO3  3Cu(NO3)2 + 2NO  + 4H2O + 3S  Cu2+ + 4NH3  [Cu(NH3)4]2+ (G).

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CHEMISTRY 2Cu2+ + K4Fe(CN)6  Cu2[Fe(CN)6]  (H) + 4K+ Ba2+ + SO42–  BaSO4  (white) Fe2+ + 2H+ + O  Fe3+ + H2O Fe3+ + 3NH3 + 3H2O  Fe(OH)3  (reddish brown) + 3NH4+ Hence, (A) = FeS ; (B) = H2S ; (C) = HNO3 ; (D) = S ; (E) = CuSO4 ; (F) = CuS ; (G) [Cu(NH3)4](NO3)2; (H) Cu2 [Fe(CN)6] Example-15 Solution

Thenard's blue is : (A) CoAl2O4

(B) Fe4(Fe(CN)6)3

(C) Fe2[Fe(CN)6]3

(D) [Ni(NH3)6](OH)2



CoO + Al2O3  CoO . Al2O3 (Thenard's blue or cobalt meta-aluminate) It is cobalt nitrate test. Therefore, (A) option is correct.

IV th GROUP (Zn 2+ , Mn 2+ , Ni 2+ , Co 2+ ) :

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CHEMISTRY 1. MANGANESE ION (Mn2+) :  Precipitation with H2S in presence of NH4OH + NH4Cl : A buff coloured (light pink) precipitate is formed. Mn2+ + S2–  MnS  It is readily soluble in mineral acids (distinction with Ni and Co) and even in acetic acid (distinction with Ni, Co and Zn). MnS  + 2H+  Mn2+ + H2S ; MnS  + 2 CH3COOH  Mn2+ + H2S  + 2 CH3COO–  Ammonia solution : Partial precipitation of white manganese(II) hydroxide takes place. Mn2+ + 2 NH3 + 2H2O Mn(OH)2  + 2 NH4+ The precipitate is soluble in ammonium salts when the reaction proceeds towards left.  Sodium hydroxide solution : Initially white precipitate of Mn(OH)2 is formed which is insoluble in excess reagent and rapidly oxidised on exposure to air, becoming brown. Mn2+ + 2 OH–  Mn(OH)2  Mn(OH)2  + O2  2 MnO(OH)2 (hydrated manganese dioxide).

 

With sodium hydroxide and bromine water initially a white precipitate is formed which immediately turns to black owing to the formation of MnO2 . Mn(OH)2  + Br2 + 2NaOH  MnO2(black) + 2NaBr + 2H2O

Mn(OH)2  + H2O2  MnO (OH)2 (brown) + H2O  Lead dioxide(PbO2) and concentrated nitric acid : On boiling a dilute solution of manganese(II) ions with lead dioxide and a little concentrated nitric acid and allowing the suspended solid containing unattacked lead dioxide to settle, the supernatant liquid acquired a violet-red (or purple) colour due to permanganic acid. 5 PbO2 + 2 Mn2+ + 4H+  2 MnO4– + 5Pb2+ + 2 H2O  Sodium bismuthate (NaBiO3) solution : When sodium bismuthate (NaBiO3) is added to a cold solution of manganese(II) ions in dilute nitric acid or in dilute sulphuric acid and the mixture stirred and then excess reagent filtered off, a violet–red (or purple) solution of permanganate is produced. 2 Mn2+ + 5 NaBiO3 + 14 H+  2MnO4– + 5 Bi3+ + 5 Na+ + 7 H2O





2Mn(NO3)2 + 2Pb3O4 + 26HNO3  2HMnO4 (violet–red/purple) + 15Pb(NO3)2 + 12H2O

 Disodium hydrogen phosphate solution : A pink precipitate of Mn(NH4) PO4 . 7 H2O is formed in

presence of ammonia or ammonium ions. Mn2+ + NH3 + HPO42–  Mn(NH4)PO4  If ammonium salts are absent, pink precipitate of manganese(II) phosphate is formed. 3 Mn2+ + HPO42–  Mn3(PO4)2  + 2 H+ Both precipitates are soluble in acids. 2. ZINC ION (Zn2+) :  Precipitation with H2S in presence of NH4OH + NH4Cl : A white precipitate is formed. In neutral solutions, precipitation is partial as H+ ions concentration produced depressed the ionisation of H2S . Zn2+ + H2S ZnS  The precipitate is soluble in dilute HCl. ZnS + 2H+  Zn2+ + H2S   Sodium hydroxide solution : A white gelatinous precipitate is formed. Zn2+ + 2 OH–  Zn(OH)2  The precipitate is soluble in acids as well as in excess of the reagent. Zn(OH)2 + 2 H+  Zn2+ + 2 H2O Zn(OH)2 + 2 OH–  [Zn(OH)4]2–

 

Thus, zinc hydroxide is amphoteric in nature.

[Zn(OH)4]2– + S2–  ZnS  (white) + 4 OH–  Ammonia solution : A white gelatinous precipitate is formed which is readily soluble in excess of the reagent and in solutions of ammonium salts forming the tetraamminezinc(II). Non–precipitation of Zn(OH)2 by ammonia solution in the presence of NH4Cl is due to the lowering of OH– ion concentration to such a value that the Ksp of Zn(OH)2 is not attained. Zn2+ + 2 NH3 + 2 H2O Zn(OH)2  + 4 NH3

Zn(OH)2  + 2 NH4+ [Zn(NH3)4]2+ + 2 OH–

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CHEMISTRY  Disodium hydrogen phosphate solution : White precipitate of zinc phosphate is formed.

3 Zn2+ + HPO42– Zn3(PO4)2  + 2H+ In presence of ammonium ions zinc ammonium phosphate is formed. Zn2+ + NH4+ + HPO42– Zn(NH4) PO4  + H+ Both precipitates are soluble in dilute acids, when the reactions are reversed. Both precipitates are soluble in ammonia : Zn3(PO4)2 + 12 NH3  3 [Zn(NH3)4]2+ + 2 PO43– Zn(NH4)PO4 + 3 NH3  [Zn(NH3)4]2+ + HPO42–  Potassium ferrocyanide (Potassium hexacyanidoferrate(II)) solution : A white precipitate of variable composition is formed. If excess reagent is added, the composition of precipitate is K2Zn3[Fe(CN)6]2 . 3 Zn2+ + 2 K+ + 2 [Fe(CN)6]4–  K2Zn3[Fe(CN)6]2  The precipitate is insoluble in dilute acids, but dissolves in sodium hydroxide readily. K2Zn3 [Fe(CN)6]2 + 12 OH–  2 [Fe(CN)6]4– + 3 [Zn(OH)4]2– This reaction can be used to distinguish zinc from aluminium.  Ammonium tetrathiocyanatomercurate(II) – copper sulphate test : Test solution faintly acidic with a few drops of MH2SO4 or 2MCH3COOH and 0.1 mL of 0.25 M CuSO4 solution followed by 2 mL of the reagent produces a violet precipitate. Zinc ions alone forms a white precipitate with the reagent in the absence of copper ions. Zn2+ + [Hg(SCN)4]2–  Zn [Hg(SCN)4]  . In the presence of the copper ions, the copper complex coprecipitated with that of zinc and the violet (or blackish–purple) precipitate consists of mixed crystals of Zn[Hg(SCN)4] + Cu[Hg(SCN)4].  Diphenylthiocarbazone (dithizone) test : Dithizone forms complexes with a number of metal ions, which can be extracted with chloroform. The zinc complex, formed in neutral, alkaline or acetic acid solutions, is red in colour. 3. COBALT ION (Co2+) :  Precipitation with H2S in presence of NH4OH + NH4Cl : A black precipitate is formed. The black precipitate of CoS is insoluble in dilute HCl or acetic acid but hot concentrated HNO3 or aquaregia dissolves it and white sulphur remains. On longer heating the mixture becomes clear as because sulphur is oxidised to sulphate. 2[Co(NH3)6]3+ + 3S2–  2CoS  + 12NH3 + S or CoCl2 + H2S 4  CoS  + 2HCl 3 CoS + 2 HNO3 + 6 H+  3 Co2+ + 3 S  + 2NO  + 4 H2O 2CoS + 6HCl + 2HNO3  3CoCl2 + 2NO + 3S + 4H2O NH OH



Solution on evaporation to dryness gives blue residue (CoCl2) which turns pink on adding water.  Potassium nitrite solution : A yellow precipitate is formed from neutral solution of cobalt(II) ions. CoCl2 + 7KNO2 + 2CH3COOH  K3[Co(NO2)6]  + H2O + 2KCl + 2CH3COOK + NO  or Co2+ + 7NO2– + 2H+ + 3K+  K3[Co(NO2)6]  + NO  + H2O.  Ammonium thiocyanate solution : A neutral or acid solution of cobalt(II) gives a blue colouration in amyl alcohol or ether layer when a few crystals of ammonium thiocyanate are added. Co2+ + 4 SCN–  [Co(SCN)4]2–



In amyl alcohol or ether, the free acid H2[Co(SCN)4] is formed and dissolved by the organic solvent (distinction from nickel)  Sodium bicarbonate and bromine water test : To the test solution sodium bicarbonate is added in excess followed by the bromine water. The mixture is slightly heated, the solution turns apple green. CoCl2 + 2NaHCO3  Co(HCO3)2 + 2NaCl Co(HCO3)2 + 4 NaHCO3  Na4 [Co(CO3)3] + 3H2O + 3CO2 Br2 + H2O  2HBr + [O] 2Na4 [Co(CO3)3] + H2O + [O]  2Na3 [Co(CO3)3 ] (green) + 2NaOH.



Green solution of sodium cobalticarbonate is obtained.

 Sodium hydroxide solution : Cobalt(II) nitrate in cold gives a blue basic salt.

Co2+ + OH– + NO3–  Co(OH)NO3  Upon warming with excess alkali, the basic salt is converted into a pink precipitate of Co(II) hydroxide. Co(OH)NO3  + OH–  Co(OH)2  + NO3–

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CHEMISTRY  Ammonia solution : In absence of ammonium salt, its small amount precipitates a blue basic salt.

Co2+ + NH3 + H2O + NO3–  Co(OH)NO3  + NH4+ The excess of the reagent dissolves the precipitate, when hexaamminecobalt(II) ions are formed. Co(OH)NO3  + 6 NH3  [Co(NH3)6]2+ + NO3– + OH–  Potassium cyanide solution :

Co2+ + 2 CN–  Co(CN)2  (reddish brown / buff coloured) In excess of reagent, a brown solution of a soluble complex is formed. Co(CN)2  + 4 CN–  [Co(CN)6]4– On acidification precipitate reappears. [Co(CN)6]4– + 4 H+ (cold and dilute)  Co(CN)2  + 4HCN 



If brown solution is boiled for a longer time in air or some H2O2 is added and solution is heated it turns yellow. 4 [Co(CN)6]4– + O2 + 2 H2O  4 [Co(CN)6]3– (yellow solution) + 4 OH– 2 [Co(CN)6]4– + H2O2  2 [Co(CN)6]3– (yellow solution) + 2 OH–

4. NICKEL ION (Ni2+) :  Precipitation with H2S in presence of NH4OH + NH4Cl :

A black precipitate is formed which is insoluble in cold dilute HCl and CH3COOH but dissolves in hot concentrated HNO3 and in aquaregia. NiCl2 + H2S 4  NiS  (Black) + 2HCl 3 NiS  + 2 HNO3 + 6 H+  3 Ni2+ + 2 NO  + 3 S  + 4 H2O NiS + HNO3 + 3HCl  Ni2+ + S  + NOCl  + 2Cl– + 2H2O. NH OH



Solution on evaporation to dryness gives yellow residue (NiCl2) which turns green on adding water.

 Dimethylglyoxime reagent : A red precipitate is obtained from the solution just made alkaline or acid

solutions buffered with sodium acetate. NiCl2+ 2NH4OH + 2CH3 – C = NOH  (C4H7N2O2)2Ni + 2NH4Cl + 2H2O | (red) CH3 – C = NOH



Fe(II) ions give red colouration, bismuth gives yellow precipitate and cobalt gives brown colouration with DMG in ammonical solutions.

 Bromine water test (alkaline medium) : A black precipitate of Ni2O3 is formed.

NiCl2 + 2NaOH (excess)  Ni(OH)2  (green) + 2 NaCl Green precipitate is formed which gradually turns black on adding bromine water. Br2 + H2O  2HBr + [O] ;

 2Ni(OH)2 + H2O + [O]  2 Ni(OH)3   Ni2O3 + 3H2O

(black)



Ni2+ gives black Co2+ gives

precipitate (Ni2O3) with sodium bicarbonate and bromine water on heating where as

green coloured solution, this is the point of difference.

NiCl2 + 2NaHCO3  NiCO3 + 2NaCl + H2O ; 2NiCO3 + [O]  Ni2O3  (black) + 2CO2  Cyanide test : A green precipitate is formed which dissolves in excess of reagent forming a yellow

coloured solution. NiCl2 + 2KCN  Ni(CN)2  + 2KCl ; Ni(CN)2 + 2KCN  K2[Ni(CN)4] ; Complex on heating with sodium hypobromite (NaOH + Br2 water) solution gives a black precipitate (distinction from cobalt). NaOH + Br2  NaBrO + HBr boil  4NaCNO + 9NaBr + Ni2O3  (black) + 4KCNO +2H2O 2K2[Ni(CN)4] + 4NaOH + 9NaBrO 

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CHEMISTRY Vth Group (Ba2+, Sr2+, Ca2+) : IV Group filtrate  Boil off H2S then add (NH4)2CO3 (aq), NH4OH & NH4Cl (s)

White precipitate (BaCO3, SrCO3 or CaCO3).  Dissolve in CH3COOH and divide into three parts and test in the sequence given below.

I part + K2CrO4.

Filtrate, move for VI group.

II Part + (NH4)2SO4.

 Yellow precipitate (BaCrO4 insoluble in CH3COOH).

 White precipitate (SrSO4).

III part + (NH4)2C2O4.  White precipitate (CaC2 O4 ).

1.BARIUM ION ( Ba2+) :  Precipitation with (NH4)2 CO3 in presence of NH4OH + NH4Cl : A white precipitate is formed which is

soluble in acetic acid and dilute mineral acids BaCl2 + (NH4)2CO3  BaCO3 + 2NH4Cl BaCO3 + 2CH3COOH  Ba (CH3COO)2 + H2O + CO2  Potassium chromate test : A yellow precipitate is formed, practically insoluble in water



Ba(CH3COO)2 + K2CrO4  BaCrO4  + 2CH3COOK Precipitate is insoluble in dilute acetic acid (distinction from strontium and calcium) but readily soluble in mineral acids.



Addition of acid to K2CrO4 changes the yellow colour of the solution to reddish–orange due to the formation of dichromate. 2 H+ + 2 CrO42–



Cr2O72– + H2O

The solubility products of SrCrO4 and CaCrO4 are much larger than for BaCrO4 and hence they require a larger CrO42– ion concentration to precipitate them. The addition of acetic acid to the K2CrO4 solution lowers the CrO42– ion concentration sufficiently to prevent the precipitation of SrCrO4 and CaCrO4 but it is maintained high enough to precipitate BaCrO4 .

2. STRONTIUM ION (Sr2+) : 

Precipitation with (NH4)2 CO3 in presence of NH4OH + NH4Cl : A white precipitate is formed which is soluble in acetic acid. SrCl2 + (NH4)2CO3  SrCO3 + 2NH4Cl SrCO3 + 2CH3COOH  Sr(CH3COO)2 + H2O + CO2



Ammonium sulphate solution : A white precipitate is formed which is slightly soluble in boiling

  

hydrochloric acid. Sr(CH3COO)2 + (NH4)2SO4  SrSO4  + 2CH3COONH4 The solubility of the precipitate in water is low but not negligible. The precipitate is insoluble in ammonium sulphate solution even on boiling (Distinction from calcium– forms a soluble complex (NH4)2[Ca(SO4)2]) and slightly soluble in boiling HCl. It is almost completely converted into the corresponding carbonates by boiling with a concentrated solution of sodium carbonate. SrSO4 + CO32–

SrCO3  + SO42–

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CHEMISTRY 3. CALCIUM ION (Ca2+) :  Precipitation with (NH4)2 CO3 in presence of NH4OH + NH4Cl : A white precipitate is formed. The precipitate is soluble in water which contains excess carbonic acid (e.g freshly prepared soda water) because of the formation of soluble hydrogen carbonate. On boiling precipitate reappears again, as CO2 is removed. The precipitate is also soluble in acetic acid. CaCl2 + (NH4)2CO3  CaCO3  + 2NH4Cl CaCO3 + 2CH3COOH  Ca(CH3COO)2 + H2O + CO2  Ammonium oxalate solution (concentrated) : A white precipitate is formed. The precipitation is facilitated by making the solution alkaline. The precipitate is practically insoluble in water (Ksp = 2.6 × 10–9), insoluble in acetic acid but readily soluble in mineral acids. Ca(CH3COO)2 + (NH4)2C2O4  CaC2O4 + 2CH3COONH4  Potassium hexacyanidoferrate (II) solution : White precipitate of a mixed salt is produced.

Ca2+ + 2 K+ + [Fe(CN)6]4–  K2Ca[Fe(CN)6]  In presence of ammonium chloride the test is more sensitive and in this case K+ is replaced by NH4+ ions in the precipitate.

 

This test can be used to distinguish , calcium from strontium, barium and magnesium ions. The confirmatory test for the basic radicals of (V) group have to be done in the order of Ba2+, Sr2+ then Ca2+ (i.e BSC) because Ba2+ give positive test with all the reagents used in the confirmatory test of these group radicals, K2CrO4, (NH4)2 SO4 & (NH4)2 C2O4. So performs the test for Sr2+ with (NH4)2SO4 only when Ba2+ is absent. Similarly Sr2+ gives the test with both (NH4)2SO4 and (NH4)2C2O4. So proceeds with Ca2+ only when Sr2+ is absent otherwise it will respond to ammonium oxalate test.

Vth GROUP : 1. MAGNESIUM ION (Mg2+) :  Disodium hydrogen phosphate solution : To the filtrate of V group or Mg2+ ions solution add 1 ml

(NH4)2C2O4 solution and heat if white precipitate is formed then filter it. Now to filtrate add a solution of disodium hydrogen phosphate. A white crystalline precipitate is formed in the presence of NH4Cl (prevent precipitation of Mg(OH)2) and ammonia solution. Mg2+ + NH3 + HPO42–  Mg(NH4)PO4 

 

Precipitate is sparingly soluble in water, soluble in acetic acid and in mineral acids. The precipitate separates slowly from dilute solutions because of its tendency to form supersaturated

solution. This may be overcome by cooling and by rubbing the test tube beneath the surface of the liquid with a glass rod. A white flocculent precipitate of magnesium hydrogen phosphate is obtained in neutral solutions. Mg2+ + HPO42–  MgHPO4 



Ammonium phosphate can also be used.

 Ammonia solution : A white gelatinous precipitate is formed.

Mg2+ + 2 NH4OH  Mg(OH)2 + 2 NH4+ The precipitate obtained is sparingly soluble in water but readily soluble in ammonium salts. Mg+2 + 2 OH–

Mg(OH)2

NH4Cl  NH4+ + Cl– ; NH4+ + OH–  NH4OH (weak base) NH4+ ions 'remove' OH– causing the hydroxide to dissolve more. Not possible with NaCl.  Ammonium carbonate solution : A white precipitate of basic magnesium carbonate is obtained in the

absence of NH4+ salts. 5 Mg2+ + 6 CO32– + 7 H2O  2 MgCO3. Mg(OH)2. 5 H2O  + 2HCO3–

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CHEMISTRY In the presence of NH4+ salts no precipitation occurs, because the equilibrium NH4+ + CO32–

NH3 + HCO3–

is shifted towards the formation of HCO3– ions. Ksp of the precipitate being high (Ksp of pure MgCO3 is 1 × 10–5 ), the concentration of carbonate ions necessary to produce a precipitate is not attained.  4–(4–Nitrophenyl azo resorcinol) or Magneson I : Precipitate is dissolved in dilute HCl (minimum

quantity) and to this sodium hydroxide solution is added followed by addition of 0.5 mL of magneson–I reagent. A blue lake is formed. MgCl2 + 2NaOH  Mg(OH)2 + 2NaCl Magneson reagent is p-nitrobenzene-azo resorcinol a dye stuff which is adsorbed over Mg(OH)2 to produce a blue coloured lake.  Titan yellow (a water soluble yellow dyestuff) : It is adsorbed by Mg(OH)2 producing a deep red colour or precipitate. Dissolved the precipitate in dilute HCl (minimum quantity) and to 1 drop of this add 1 drop of NaOH solution (2 M) followed by 1 drop of titan yellow solution a deep red colour solution or precipitate is obtained.



Ba2+ and Ca2+ do not react but intensify the colour.

Example-16 (a) (b) (c) (d)

Explain the following. Reaction between NiCl2 and excess KCN solution. Reaction between cobalt chloride and sodium bicarbonate and bromine water. When manganese nitrate, sodium bismuthate (NaBiO3) solution and concentrated HNO3 is boiled. When precipitate of Zn(NH4)PO4 reacts with ammonia.

Solution

(a) A green precipitate is formed which dissolves in excess of reagent forming a yellow coloured solution. NiCl2 + 2KCN  Ni(CN)2  + 2KCl ; Ni(CN)2 + 2KCN  K2[Ni(CN)4] (b) When sodium bicarbonate is added in excess to cobalt chloride solution followed by the bromine water and then mixture is slightly heated, the solution turns apple green. CoCl2 + 2NaHCO3  Co(HCO3)2 + 2NaCl Co(HCO3)2 + 4 NaHCO3  Na4 [Co(CO3)3] + 3H2O + 3CO2 Br2 + H2O  2HBr + [O] 2Na4 [Co(CO3)3] + H2O + [O]  2Na3 [Co(CO3)3 ] (green) + 2NaOH (c) (d)

Example-17

2 Mn2+ + 5 NaBiO3 + 14 H+  2MnO4– (violet–red/purple) + 5 Bi3+ + 5 Na+ + 7 H2O Zn(NH4)PO4  + 3NH3  [Zn(NH3)4]2+ + HPO42–.

Column I and column II contains four entries each. Entries of column I are to be matched with some entries of column II. Each entry of column I may have the matching with one or more than one entries of column II. Column I Column II (A) Forms coloured precipitate on addition of CrO42– ions but precipitate (p) Ag+ dissolves in ammonia solution. (B) Forms coloured precipitate with ammonia solution but dissolves forming (q) Cu2+ coloured solution in excess of precipitant. (C) Forms coloured precipitate on addition of I– ions but in excess of (r) Ni2+ precipitant precipitate dissolves forming coloured solution. (D) Forms white precipitate, when salt (chloride or nitrate) is poured into a (s) Bi3+ large volume of water.

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CHEMISTRY Solution

(A - p) ; (B - q, r) ; (C - s) ; (D - s) (A)

2Ag+ + CrO42–  Ag2CrO4  (red) Ag2CrO4 + 4NH3  2[Ag(NH3)2]+ + CrO42–

(B)

2Cu2+ + SO42– + 2NH3 + H2O  Cu(OH2)2. CuSO4  (blue) + 2NH4+ Cu(OH)2 . CuSO4  + 8NH3  2 [Cu(NH3)4]2+]SO42– (intense blue) + 2OH– Ni2+ + 2NH3 + 2H2O  Ni(OH)2  (green) + 2NH4+ Ni(OH)2  + 6NH3  [Ni(NH3)6]2+ (deep blue) + 2OH–

(C)

Bi3+ + 3I–  BiI3  (black) BiI3 + I–  [BiI4]– (orange colouration)

(D)

Bi3+ + NO3– + H2O  BiO(NO3)  (white) + 2H+ basic salt

MISCELLANEOUS SOLVED PROBLEMS (MSPs) 1.

Ans.

BaCl2 solution gives a white precipitate with a solution of a salt, which dissolves in dilute hydrochloric acid with the evolution of colourless, pungent smelling gas. The gas as well as the salt both are used as bleaching agent in the textile industries. The salt contains: (A) sulphite (B) sulphide (C) acetate (D) carbonate (A)

Sol.

Ba2+ + SO32–  BaSO3  (white) BaSO3 + 2HCl  BaCl2 + SO2(colourless pungent smelling gas) + H2O SO32– and SO2 both act as bleaching agent.

2. Ans. Sol.

Which of the following precipitate(s) does / do not dissolve in excess of ammonia solution ? (A) Zn(OH)2 (B) Ni(OH)2 (C) Al(OH)3 (D) (B) and (C) both (C) (A) Zn(OH)2 + 4 NH3  [Zn(NH3)4]2+ (colourless solution) + 2OH– (B) Ni(OH)2  + 6NH3  [Ni(NH3)6]2+ (deep blue solution) + 2OH– (C) Al(OH)3 + NH3  No reaction.

3.

Ans. Sol 4.

Ans. Sol.

5.

Ans. Ans.

Chocolate brown precipitate is formed with : (A) Cu2+ ions and [Fe (CN)6]3– (C) Fe2+ ions and [Fe (CN)6]4– (B) (A) Cu3 [Fe(CN)6]2 (green) ; (C) Fe4[Fe(CN)6]3 (Prussian blue) ;

(B) Cu2+ ions and [Fe(CN)6]4– (D) Fe2+ ions and dimethylglyoxime (B) Cu2 [Fe(CN)6]  ( chocolate brown) (D) red solution of iron(II) dimethylglyoxime.

Pink colour of acidified KMnO4 is decolourised but there is no evolution of any gas. This may happen with the compound containing the following acid radical. (A) SO32– (B) NO 2– (C) S2– (D) (A) 5SO32– + 2MnO4– + 6H+  2Mn2+ + 5SO42– + 3H2O (B) 2MnO4– + 5NO2– + 6H+  2Mn2+ + 5NO3– + 3H2O (C) 2MnO4– + H2S + 6H+  2Mn2+ + 5S  + 8H2O

(D) All of these

Which of the following gives a precipitate with Pb(NO3)2 but not with Ba(NO3)2? (A) Sodium chloride (B) Sodium acetate (C) Sodium nitrate (D) Disodium hydrogen phosphate (A) (A) Pb2+ + 2Cl–  PbCl2(white) ; Ba2+ + 2Cl–  BaCl2(water soluble) (B) (CH3COO)2 Pb and (CH3COO)2Ba both are water soluble salts. (C) Nitrates are mostly soluble in water (D) 3Pb2+ + 2HPO42–  Pb3(PO4)2  (white) + 2H+ ; Ba2+ + HPO42–  BaHPO4  (white)

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CHEMISTRY 6. Ans. Sol. 7.

Ans. Sol.

Colour of cobalt chloride solution is : (A) pink (B) black (C) colourless (D) green (A) Anhydrous Co(II) salts are blue in colour while hydrated Co(II) salts are pink/red. A red colouration or precipitate is not obtained when : (A) Fe3+ reacts with potassium thiocyanate (B) Fe2+ reacts with dimethylglyoxime. (C) Hg2+ reacts with potassium iodide. (D) None (D) (A) Fe3+ + 3SCN–  Fe(SCN)3 (red solution) (B) Red solution of iron(II) dimethylglyoxime. (C) Hg2– + 2I–  HgI2  (red).

8.

Ans. Sol. 9.

Ans. Sol.

When H2S gas is passed through an ammonical salt solution X, a slightly white precipitate is formed. The X can be : (A) a cobalt salt (B) a lead salt (C) a zinc salt (D) a silver salt (C) Zn2+ + H2S  ZnS  (white) + 2H+ Consider the following statement : S1 : Cu2+ ions are reduced to Cu+ by potassium iodide and potassium cyanide both, when taken in excess S2 : H2S will precipitate the sulphide of all the metals from the solutions of chlorides of Cu, Zn and Cd if the solution is aqueous. S3 : The presence of magnesium is confirmed in qualitative analysis by the formation of a white crystal line precipitate of MgNH4 PO4. S4 : Calomel on reaction with potassium iodide gives red precipitate. and arrange in the order of true /false. (A) TTFF (B) TFTF (C) TTTT (D) TTTF (D) S1, S2 and S3 are correct statements. S4 : Hg22+ + 2I–  Hg2I2  (green)

10.

Ans. Sol. 11.

Statement - 1 : Addition of NH4OH to an aqueous solution of BaCl2 in presence of NH4Cl (excess) precipitates Ba(OH)2 . Statement - 2 : Ba(OH)2 is water soluble. (A) Both Statement-1 and Statement-2 are true and Statement-2 is the correct explanation of Statement-1. (B) Both Statement-1 and Statement-2 are true but Statement-2 is not correct explanation of Statement-1. (C) Statement-1 is true but Statement-2 is false. (D) Statement-1 is false but Statement-2 is true (D) Ba2+ ions does not give any precipitate with NH4OH solution in excess of NH4Cl because product formed, Ba(OH)2 is soluble in water

Ans.

Statement - 1 : Sodium meta aluminate on boiling with ammonium chloride produces white gelatinous precipitate. Statement - 2 : Aluminium hydroxide is formed which is not soluble in water (A) Both Statement-1 and Statement-2 are true and Statement-2 is the correct explanation of Statement-1. (B) Both Statement-1 and Statement-2 are true but Statement-2 is not correct explanation of Statement-1. (C) Statement-1 is true but Statement-2 is false. (D) Statement-1 is false but Statement-2 is true (A)

Sol.

[Al(OH)4]–

NH4Cl   Al(OH)3  + OH– boil

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CHEMISTRY 12.

Ans. Sol.

Which of the following statement(s) is (are) incorrect? (A) Fe2+ ions give a dark blue precipitate with potassium hexacyanidoferrate (III) solution. (B) Fe3+ ions give intense blue precipitate with potassium hexacyanidoferrate (II) solution. (C) Fe3+ ions give a brown colouration with potassium hexacyanidoferrate (III) solution. (D) Fe2+ ions give a deep red colouration with ammonium thiocyanate. (D) (A) Fe2+ + [Fe(CN)6]3–  Fe3+ + [Fe(CN)6]4– (B) (C) (D)

4Fe3+ + 3 [Fe(CN)6]4–  Fe4[Fe(CN)6]3(turnbull's blue) 4Fe3+ + 3[Fe(CN)6]4–  Fe4[Fe(CN)6]3 (intense blue) Fe3+ + [Fe(CN)6]3–  Fe[Fe(CN)6] (brown colouration) Fe3+ + 4SCN–  Fe (SCN)3 (deep red colouration) Fe2+ + 4 SCN–  No reaction

13. Ans. Sol.

Which of the following pair (s) of ions would be expected to form precipitate when dilute solutions are mixed? (A) NH4+, [Co(NO2)6]3– (B) NH4+, CO32– (C) Fe3+, OH– (D) Ba2+, SO42– (A,C,D) (A) (B)

NH4+ + [Co(NO2)6]3–  (NH4)3 [Co(NO2)6]  (yellow) Ammonium and alkali metal carbonates are water soluble.

(C)

Fe3+ + OH–  Fe(OH)3  (reddish - brown)

(D)

Ba2+ + SO42–  BaSO4  (white)

Comprehension (Q.14 to Q.16)

Aqueous solution of salt (A) NaOH(aq)/warm

Gas(B) Colourless gas which is alkaline in nature

Solution of salt (C) FeSO4 conc. H2SO4

Zn/NaOH /heat

dil.HCI White fumes

Brown ring (D) at the junction of the two layers

Gas(B)

Salt (A) on heating gives a colourless neutral gas which supports combustion. From the aforesaid, flow diagram, answer the following questions. 14. Ans. Sol.

The compound (A) contains the following acid radical. (A) NO2– (B) NO3– (C) Br– (D) SO32– (B) NO3– and NO2– both give brown ring test and reduction of NO3– and NO2– both give ammonia which with dilute HCl gives dense white fumes. if

 NH4NO3  N2O + 2H2O ; N2O supports the combustion

 but NH4NO2  N2 + 2H2O ; Nitrogen does not supports combustion. Hence, the anion is NO3–.

15.

Ans. Sol.

The basic radical of salt (A) and gas B both gives brown precipitate with Nessler’s reagent. The composition of the brown precipitate is : (A) (NH4)2 [HgI4] (B) Hg(NH2) NO3 (C) HgO. Hg (NH2)I (D) (NH4)3[Co(NO2)6] (C) NH4+ + 2[HgI4]2– + 4OH–  HgO. Hg (NH2)I + 7I– + 3H2O Hence the cation is NH4+.

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CHEMISTRY 16.

Ans.

Which of the following statement is correct ? (A) Salt (A) gives yellow precipitate with chloroplatinic acid as well as with sodium cobaltinitrite. (B) The brown ring is formed due to the formation of nitroso ferrous sulphate [Fe(NO)]2+SO4–. (C) Salt ‘C’ reacts with silver nitrate solution to form white precipitate. (D) (A) and (B) both. (D)

Sol.

(A)

2NH4+ + [PtCI6]4–  (NH4)2 [PtCl6]  (yellow) 3NH4+ + [Co(NO2)]3–  (NH4)3 [Co(NO2)6]  (yellow)

(B)

2NO3– + 4H2SO4 + 6Fe2+  6Fe3+ + 2NO  + 4SO42– + 4H2O SO42– + Fe2+ + NO  [Fe(NO)]2+ SO42–

(C)

Ag NO3 + NaNO3  No reaction. If the anion is NO2– then Ag+ + NO2–  Ag NO2  (white)

Reactions : NH4NO3 + NaOH  NH3 + NaNO3 (A) (B) (C) NH3+ HCI  NH4CI (White) NO3– + 4Zn + 7OH– + 6H2O  NH3 + 4[Zn(OH)4]2– True/False : 17. Magnesium is precipitated from its salt solution as only magnesium ammonium phosphate by adding disodium hydrogen phosphate solution in absence of ammonium chloride and aqueous ammonia. Sol. (False) Precipitation is carried out in presence of ammonium chloride and aqueous ammonia as they prevent precipitation of magnesium hydroxide. Mg2+ + NH3 + HPO42–  Mg (NH4) PO4  (white) 18. Sol.

When a solution of nitrite acidified with dilute hydrochloric acid is treated with solid urea, the nitrite is decomposed, and nitrogen and carbon dioxide are evolved. (True) CO(NH2)2 + HNO2  2N2  + CO2  + 3H2O.

19.

Solution of alkali metal cyanide containing freshly prepared iron (II) sulphate solution and dilute H2SO4 on exposure to air produces prussian blue precipitate Sol. (True) Fe2+ + 2 CN–  Fe(CN)2  ; Fe(CN)2  + 4CN–  [Fe(CN)6]4– 4Fe2+ + O2 + 4H+  4Fe3+ + 2H2O ; Fe3+ [Fe(CN)6]4–  Fe4[Fe(CN)6]3  Subjective : 20. What happens when ? (A) Aqueous solution of CrCl3 is added to ammonia solution. (B) Ammonium carbonates reacts with MgCl2 (i) in absence of ammonium salts and (ii) in presence of ammonium salts : Sol. (A) Cr3+ + 3NH3 + 3H2O  Cr(OH)3  (green) + 3NH4+ Cr(OH)3 precipitate formed becomes slightly soluble in excess of precipitant in cold forming a violet or pink solution containing [Cr(NH3)6] 3+ complex ions. Cr(OH)3  + 6 NH3  [Cr(NH3)6]3+ + 3OH– (B)

(i) 5Mg2+ + 6 CO23 + 7H2O  4MgCO3. Mg(OH)2. 5 H2O  + 2HCO3– White precipitate of basic magnesium carbonate is formed. (ii) In presence of ammonium salts no precipitation occurs, because the equilibrium NH4+ + CO32–  NH3+ HCO3– is shifted towards the formation of HCO3– ions.

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CHEMISTRY 21.

Ans. Sol.

Salts given in column (I) reacts with the excess of reagents given in column (II) and form white /coloured precipitates. Select the correct options for the salts given in column (I) with the reagent(s) given in the column (II) Column - I Column - II (A) Zn (NO3)2 (p) Sodium hydroxide (B) Cu (NO3)2 (q) Ammonia solution (C) Fe(NO3)3 (r) Disodium hydrogen phosphate (D) Ag(NO3)2 (s) Potassium ferrocyanide (A - r, s) ; (B - p, r, s) ; (C - p, q, r, s) ; (D - p, q, r, s) (A) Zn2+ + 2OH–  Zn(OH)2 (white) ; Zn (OH)2 + 2OH–  [Zn(OH)4]2– (colourless soluble complex) Zn2+ + 2NH3 + 2H2O  Zn (OH)2  (white) + 2NH4+ Zn(OH)2  + 4NH3  [Zn(NH3)4]2+ (colour less soluble complex) + 2OH– 3Zn2+ + 2HPO42–  Zn3 (PO4)2  (white) + 2H+ (B)

3 Zn2+ + 2K+ + 2[Fe(CN)6]4–  K2 Zn3 [Fe(CN)6]2  (bluish white) Cu2+ + 2OH–  Cu(OH)2  (blue) Cu2+ + 4NH3  [Cu(NH3)4]2+ (deep blue soluble complex) Cu2+ + 2HPO42–  Cu3(PO4)2 (blue) + 2H+

(C)

Cu2+[Fe(CN)6]4–  Cu2[Fe(CN)6]  (chocolate brown) Fe3+ + 3OH–  Fe (OH)3  (reddish brown) Fe3+ + 3NH3 + 3H2O  Fe(OH)3  (reddish brown) + 3NH4+ Fe3+ + HPO42–  FePO4  (yellowish-white) + H+

(D)

4Fe3+ + 3 [Fe(CN)6]4–  Fe4 [Fe(CN)6]3  (intense blue) 2Ag+ + 2OH–  Ag2O  (brown) + H2O 2Ag+ + 2NH3 + H2O  Ag2O (brown) + 2NH4+ 4Ag+ + HPO42–  Ag3PO4  (yellow) + H+ 4Ag+ + 3 [Fe(CN)6]4–  Ag4[Fe(CN)6]  (white)

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