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Chemistry Subjective Questions and Solutions Module Paper 3

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Illustration 1:          Among maleic acid and fumaric acid which is a stronger  acid and  why.

Solution:              Maleic acid is nothing but cis-butene dioic acid while fumaric acid is the trans isomer. The conjugate base of maleic acid  i.e., the maleate ion is stabilized due to intramolecular hydrogen bonding which is not possible in the fumarate ion. Now more than stable the conjugate base is the more more the acidity. Therefore  maleic acid is stronger  than fumaric acid.

 

Illustration 2:          The ionisation constants of benzoic acid, p-nitrobenzoic acid and p-hydroxy benzoic acids are 6.3 ´ 10–5, 40 ´ 10–5 and 2.4 ´ 10–5 respectively. Explain.

Solution:              Factors dispersing the cahrge increase acidity. NO2 – group being electron attracting, produces a positive charge at the para-C bonded to COO, with the result negative cahrge on –COO group is dispersed. On the other hand, –OH group, being electron releasing, concentrates the negative charge on the para – C bonded to –COOwhich increases basicity of anion and decrease acidity of its conjugate acid.

Illustration 3:          Among N,N dimethyl aniline and N,N,2,6 – tetramethyl aniline which  one is a stronger base and why.

Solution:

In N,N-2,6 – tetramethyl aniline the methyl groups on nitrogen and the ortho position are very close to each other resulting in a steric crowding. Now to avoid steric crowding the C—N bond rotates and becomes perpendicular to the benzene ring. In this process the lone pair on nitrogen becomes perpendicular to the p-orbitals of benzene ring thereby inhibiting resonance. But in N,N dimethyl aniline there is no steric hindrance, so the lone pair is in the same plane as the benzene ring and undergoes resonance. Therefore the lone pair on  the tetramethyl derivative is more available and hence it is more basic.

Illustration 4:          Arrange the following in decreasing basic order with proper reasoning. RNH2,

                             

Solution:

                              The lone pair of electrons on N is accomodated by sp3 hybrid atomic orbital in RNH2, sp2 hybrid atomic orbital in RCONH2 and two sp2 in  succinidime.

Illustration 5:          On the basis of H-bonding explain that the second ionization constant K2 for fumaric acid is greater than for maleic acid.

Solution:              We know that H-bonding involving acidic H has an acid weakening effect and H-bonding in conjugate base has an acid strengthening effect.

Both dicarboxylic acids have two ionisable hydrogen atoms.  Considering second ionization step.

Since the second ionisable H of the Maleate participates in H-bonding more energy is needed to remove this H because the H-bond must be broken. The maleate mono anion is, therefore, the weaker acid.

Illustration 6:          BaCO3 has got a higher decomposition temperature than MgCO3.

Solution:              BaCO3 has got a higher decomposition temperature the MgCO3 which illustrates its high thermal stability. The thermal stability depends on the polarising power of the cation which in turn is inversely proportional to the radius Ba2+ having a larger radius than Mg2+ has got less polarising power and hence thermaly more stable.

Illustration 7:          Dipole moment of BF3 is zero.

Solution:              BF3 is a planar molecule with the bond angles being 120°.

                              In each B-F bond the bond dipole is projected towards Fluorine. Now the direction of the resultant bond dipoles of these two bonds is shown in the diagram as R. If R is equal to the dipole of the 3rd B-F bond then R and B-F dipole will neutralize each other as equal dipoles acting in opposite direction cancel each other. Now let’s see the magnitude of R. As dipole moment is a vector quantity, so the resultant of two dipoles can be obtained from the law of vector addition. Suppose the bond dipole is m1.

R   =                  a & b are the individual vectors

=               q = angle between vectors

=    cos120° =  =      =   m1 = mB–F

\ R is equal in magnitude to B-F bond dipole. Similar is the case with the other two resultants. So the net dipole of BF3 is zero.

Illustration 8:          Compare the dipole moments of CH3Cl, CH2Cl2, CHCl3, CCl4

Solution:              All the molecules have tetrahedral geometry with CCl4 having symmetrical structure resulting in a zero dipole.

The net dipole of the molecules is the resultant of the bond dipoles.  In CHCl3 due to the interaction of the non bonded electrons of chlorine the bond angle increases whereas in CH3Cl this interaction in less and so the bond angle is less compared to CHCl3. Now the resultant dipole is dependent on cosq. Now higher the value of q less is cosq and therefore less is the dipole moment. On that basis CH3Cl has highest dipole and CHCl3 has lowest with CCl4 having zero dipole.

Illustration 9:          SO2 is angular and CO2 is linear why?

Solution:              This is because, by the method discussed above, S in SO2 having sp2 hybridisation with one lone pair of S occupying the sp2 hybridised orbital gives angular shape whereas C in CO2 is sp hybridised and is thus linear.

 

Illustration 10:        AlF3 is ionic while AlCl3 is covalent.

Solution:              Since F is smaller in size, its polarisability is less and therefore it is having more ionic character. Whereas Cl being large in size is having more polarisability and hence more covalent character.

 

Illustration 11:        Though copper, silver and gold have completely filled sets of a d-orbitals yet they are considered as transition metals. Why?

Solution:              These metals in their common oxidation states have incompletely filled d-orbitals e.g. Cu2+ has 3d9 and Au3+ has 5d8 configuraton.

Illustration 12:        Draw structure of

  1. I) Zeise’s salt anion [PtCl3(h2 – C2H4)]
  2. II) [Al(CH3)3]2
Solution:

                             

Illustration 13:        ives the IUPAC names of the following Co-ordination compounds :

  1. i) [Co(NH3)6]Cl3                             ii)   K3[Fe(C2O4)3]
  2. ii) [CoCl(NH3)5]Cl2                                                   iv)  [Co(NH3)5Br]SO4
  3. v) Na2[Fe(CN)5NO]                         vi) [Ir(Ph3P)2(CO)Cl]     

                              vii) (NH4)3[Co(C2O4)3]                       viii)      [Al(OH)(H2O)5]SO4

  1. ix) [V(H2O)6]Cl3                               x)  [Co(NH3)4Cl2][Cr(CN)6]          

Solution:              i)    Hexammine Cobalt (III) chloride

  1. ii) Potassium trioxalate ferrate (III)

iii)   Pentammine chlorocoablt (III) chloride

  1. iv) Pentammine bromocobalt (III) sulphate
  2. v) Sodiumpentacyanonitrosylferrate(III)
  3. vi) Carbonylchlorobis(triphenylphosphine)iridium(I)

vii) Ammoniumtrisoxalatocobaltate(III)

viii) Pentaquohydroxoaluminimum(III)sulphate

  1. ix) Hexaaquavanadium(III) chloride
  2. x) Tetraamminedichlorocobalt(III) hexacyanochromate(V)

Illustration 14:        Arrange the following compounds in order of increasing molar conductivity:

                               (a) K[Co(NH3)2(NO2)4]                      (b) [Cr(NH3)3(NO2)3]

                               (c) [Cr(NH3)5 (NO2]3 [Co(NO2)6]2        (d) Mg[Cr(NH3) (NO2)5]

Solution:              The larger the number of ions and the larger the charge on each, the larger the conductivity. The compounds from lowest conductivity to highest conductivity are:

b < a < d < c

Illustration 15:        Write the formula of

  1. a) Tetrachlorocuprate(II)ion
  2. b) Dichlorotetraaquochromium(III)chloride
  3. c) Bromo Chlorotetra ammine Cobalt(III) sulphate
  4. d) Diammine Silver (I) hexacyano ferrate (II)
  5. e) Dichlorobis (ethylenediammine) chromium (III) tetrachloro palladate (II)
  6. f) Aluminium tetrachloro aurate (III)

Solution:              a)   [CuCl4]2–

  1. b) [Cr(H2O)4Cl2]Cl
  2. c) [Co(NH3)4BrCl2]SO4
  3. d) [Ag(NH3)2]4[Fe(CN)6]
  4. e) [Cr(en)2Cl2]2[PdCl4]
  5. f) Al[AuCl4]3

 

Illustration 16:        i)    A black coloured compound (B) is formed an passing hydrogen sulphide through the solution of a compound (A) in NH4OH.

  1. ii) (B) on treatment with hydrochloric acid and KClO3 gives (A)

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

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

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

Solution:              The compound (A) is cobalt salt (COCl2)

  1. i)
  2. ii) CoS + 2HCl + (O) (from KCIO3)

iii)   CoCl2 + 2KCN

Co(CN)2 + 4KCl

  1. v) CoCl2 + 2NaHCO3 ¾® Na4[Co(CO3)3+2NaCl + 3H2O + 3CO2

2Na2[Co(CO3)3] + H2O + O

Illustration 17:        A mixture of two salts was treated follows:

  1. i) The mixture was heated with maganese dioxide and concentrated H2SO4 when yellowish green gas was obtained.
  2. ii) The mixture on heating with NaOH solution gave a gas which turned red litmus blue.

                              iii)  Its solution in water gave blue precipitate with potassium ferricynate and red coluoration with NH4CNS

  1. iv) The mixture was boiled with potassium hydroxide and the liberated gas was bubbled through an alkaline solution of K2HgI4 to give brown precipitate

                              Identify the two salts. Given ionic reactions involved in the tests (i), (ii) and (iii).

 

Solution:              Tests (ii) and (iv) show the presence of ammonium radical. Test (i) shows the presence of Cl ion and test (iii) shows the presence of Fe2+ and Fe3+ ion.

2Cl + 2MnO2 + H2SO4 + 2H+ ¾® MnSO4 + H2O + Cl­2 (yellowish green gas)

NH4+ + OH ¾®

3Fe2+ + 2[Fe(CN)6]3–

Fe3++ 3CNS

Thus, the mixture consists FeCl2 and NH4Cl. Some of the FeCl2 has undergone oxidation into FeCl3 with atmospheric oxygen.

Illustration 18:        In steel manufacture, manganese metal is used as a scavenger to reduce traces of iron oxide and iron sulphide. Suggest two reasons why manganese is effective for this purpose?

Solution:              (I) Manganese is active enough to react with iron compounds of sulphur and oxygen (ii) Small quantities of the products of these reaction dissolve in the metal without disruption of the lattice. Any excess manganese acts as cathodic protection for the iron.

Illustration 19:        Write balanced chemical equations for the following

  1. a)   Dilute nitric acid is slowly reacted with metallic tin 
  2. b)   Heating of ZnCl2.2H2O
  3. c)   Iron reacts with cold dilute nitric acid
  4. d)   Gold is dissolved in aqua regia 
  5. e)   Zinc oxide is treated with excess of sodium hydroxide solution

Solution:             a)   Tin reduces dilute nitric acid to ammonia

4Sn+NO3+10H+ ¾¾® 4Sn2++NH4++3H2O

  1. b) ZnCl2.2H2O ZnCl(OH)+HCl+H2O
  2. c)   4Fe+10HNO3 ¾® 4Fe(NO3)2+NH4NO3+3H2O
  3. d)   Au+4HCl + 3HNO3 ¾® HAuCl4+3NO2+3H2O
  4. e)   ZnO+2NaOH ¾¾® Na2ZnO2+H2O

Illustration 20:        Why does AgNO3 produce a black stain on the skin?

Solution:              In presence of organic matter (skin) and light, AgNO3 decomposes to produce a black stain of metallic silver

2AgNO3 ¾¾® 2Ag+2NO2+O2

Illustration 21:        Although aluminium is above hydrogen in the electro chemical series, it is stable in air and water. Give reason

Solution:             It is due to the production of oxide layer

Illustration 22:        Magnesium oxide is used for lining of steel making furnace. Why?

Solution:              MgO is basic in nature. It removes acidic impurities present in cast iron used for making steel

MgO+SiO2 ¾¾® MgSiO3(slag)

MgO+P2O5 ¾¾® Mg3(PO4)2

MgO is also a refractory material as it can tolerate very high temperature of the furnace.

Illustration 23:        BaO2 is a peroxide but PbO2 is not a peroxide why?

Solution:              Metallic oxides which on treatment with dilute acids produce hydrogen peroxide are called peroxides. All peroxides contain a peroxide ion (O­2)2– having the structure – O – O – PbO2 does not contain a peroxide ion (O2)2 and it can not be called as peroxides.

Illustration 24:        Statues coated with white lead on long exposure to atmosphere turn black and the original colour can be restered in treatment with H2O2 why?

Solution:              On long exposure to atmosphere, white lead is converted into black PbS due to the action of H2S present in the atmosphere.  As a result statues turn black.

PbO2 + 2H2S ¾® PbS + 2H2O

On treatment of these blackened statues with H2O2, the black PbS gets oxidised to white PbSO4 and the colour is restored

PbS + 4H2O2 ¾® PbSO4 + 4H2O

Illustration 25:        How would you prepare the following?

  1. a)  Micro cosmic salt from disodium hydrogen phosphate
  2. b)   Sodium thiosulphate from sodium carbonate

Solution:              a)   It is prepared by dissolving NH4Cl in disodium hydrogen phosphate in molecular proportion in hot water

NH4Cl+Na2HPO4 ¾¾® NaNH4HPO4 + NaCl

  1. b)   Sodium carbonate is first converted to sodium sulphate by passing sulphur dioxide

Na2CO3+SO2 ¾® Na2SO3 + CO2

The solution of Na2SO3 is then boiled with sulphur for two hours

Na2SO3+S ¾¾-® Na2S2O3

Illustration 26:        Complete and balance the following reactions 

  1. a)   Copper reacts with HNO3 to give NO and NO2 in molar ratio of 2:1
  2. b)   Phosphorus reacts with conc. HNO3 to give phosphoric acid

Solution:              a)   7Cu+20HNO3 ¾® 7 Cu(NO3)2+4NO+2NO2+10H2O

  1. b)   P4+20HNO3 ¾¾ 4H3PO4+20NO2+4H2O

Illustration 27:        Give reasons for the following

  1. a) Why conc. H2SO4, anhyd. CaCl2 or P4O10 cannot be used as dehydrating agents for ammonical
  2. b)   Nitric acid acts as oxidizing agent while nitrous acid can act both as oxidizing and reducing agent

Solution:              a)   Conc. H2SO4 anhyd. CaCl2 and P4O10 directly react with ammonia

H2SO4+2NH3 ¾¾® (NH4)2 SO4

CaCl2+8NH3 ¾¾® CaCl2.8NH3

P4O10 + 12 NH3+6H2O ¾¾® 4(NH4)3PO4

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

Concepts of Acids and Bases

  1. Why CHCl3 is more acidic than CHF3?
  2. Why F–– is most basic among the halogen anion
  3. Arrange the following in the increasing order of acidic strength.

 

  1. Why pyrole is less basic than pyridine
  2. Arrange the following compounds in the increasing order of their basic strength.

CH3NHNa+, C2H5NH2, (iso-C3H7)3N and CH3CONH2

  1. What should be the order of acidic strength in the series H3PO4, H3PO3 and H3PO2?
  2. Arrange the following in correct order of acidity

 

  1. In dilute benzene solutions, equimolar additions of (C4H9)3N and HCl produce a substance with a dipole moment. In the same solvent, equimolar additions of (C4H9)3N and SO3 produce a substance having an almost identical dipole moment. What is the nature of the polar substances formed and what is the unifying feature of HCl and SO3?
  2. Arrange according to increasing Lewis acid character,

B(n–Bu)3,  B(t–Bu)3

  1. Arrange according to increasing Lewis acid character,

SiF4, SiCl4, SiBr4, Sil4

Chemical Bonding

  1. The product of hydrolysis of NCl3 and PCl3 are different. Why?
  2. SF6 is possible but SH6 is not possible. Why?
  3. KHF2 is possible but not KHCl2. Why?
  4. Explain why the solubility in water of halides of Aluminium follows the order

AlF3 > AlCl3 > AlBr3 > AlI3

  1. SnF4 boils at 705°C while SnCl4 boils at 114°.
  2. In ICl5, the four chlorine atoms on square planar base and the central atom I does not remain in same plane. Justify the statement.
  3. NF3 is pyramidal but BF3 is planar.
  4. Dipole moment of KCl is 3.336 ´ 10–29 coulomb metre which indicates that it is highly polar molecule. The interatomic distance between k+ and Cl is 2.6 ´ 10–10m. Calculate the dipole moment of KCl molecule if there were opposite charges of one fundamental unit located at each nucleus. Calculate the percentage ionic character of KCl.
  5. PbCl4 exists but PbI4 is not possible.
  6. PCl5 exists in the solid state in the form of [PCl4]+ [PCl6] yet it is a non conductor of electricity.

Transition Element

  1. Why the pale red colour of [Co(H2O)6]2+ ion changes into intense blue colour by the additon of chloride ion?
  2. Which tripositive lanthanoid ion(s) is (are) most likely to react with chromium (II) chloride?
  3. Arrange the following compounds in order of increasing molar conductivity:

(a) K[Co(NH3)2(NO2)4]                                   (b) [Cr(NH3)3(NO2)3]

(c) [Cr(NH3)5 (NO2]3 [Co(NO2)6]2      (d) Mg[Cr(NH3) (NO2)5]

  1. Ethylenediamine tetracetic acid, EDTA in the from of its calcium dihydrogen salt, is administered as an antidote for lead poisioning. Explain why this reagent might be an effective medicine. Why is the calcium salt administrated rather than the free acid?
  2. Explain the following:
  3. a) The complex [CuCl4]2– exists, but [CuI4]2– does not.
  4. b) gold is not attacked by common acids but dissolves in aqua regia.
  5. c) Copper dissolves in aqueous KCN solution with the evolution of hydrogen.
  6. Square planar complexes with co-ordiantion number of four exhibit geometrical isomerism whereas tetrahedral complexes do not. Why?
  7. FeSO4 solution mixed with (NH4)2SO4 solution (in the molar ratio 1:1) gives the test of Fe2+ ion but CuSO4 solution mixed with liquid NH3. (the molar ration 1 :4) does not give the test of Cu2+. Explain why?
  8. A coordination compound has the formula CoCl3.4NH­3. It does not liberate ammonia but precipitates chlorides ions as silver chloride. Give the IUPAC name of the complex and write is structural formula.
  9. [NiCl4]2– is paramagnetic while (Ni(CO)4] is diamagnetic. Why?
  10. Account for the following
  11. i) Co(II) is stable is aqueous solution but in the presence of strong ligand and air, it can get oxidized to Co(II).
  12. ii) [Ni(CN)4]2– is square planar and diamagnetic whereas (NiCl4]2– is tetrahedral and paramagnetic

Qualitative Analysis

  1. A yellow solid (A) is unaffected by acids and bases. It is not soluble in water. It dissolves slowly in hot conc. HNO3 and brown gas (B) is released. The solid (A) dissolves only in a boiling solution of sodium sulphite giving a clear solution (C). Acidification of solution (C) causes a colourless gas (D) to be liberated, accompanied by the appearance of a milky precipiate (E) in the solution. Identify (A) to (E)
  2. A is binary compound of a univalent metal 1.422 g of A reacts completely with 0.321 g of sulphur in an evacuated and sealed tube to give 1.743 g of a white crystalline solid, B that forms a hydrated double salt, C with Al2(SO4). Identify A,B, and C.
  3. Identify A,B, C and D in the following sequence of reactions:
  4. i) A + NaOH  ¾¾® NaCl + NH3 + H2O
  5. ii) NH3 + CO2 + H2O ¾¾® B

iii)   B + NaCl ¾¾® C + NH4Cl

  1. iv) C Na2CO3 + H2O + D
  2. A certain metal (A) is boiled in dilute nitric acid to give a salt (B) and an oxide of nitrogen (C). an aqueous solution of (B) with brine gives a precipitate (D) which is soluble in ammonium hydroxide. On adding aqueous solution of (B) to hypo solution, a white precipitate (E) is obtained. (E) on standing turns to a black compound (F).
    Identify (A) to (F).
  3. An aqueous solution of salt (A) gives white crystalline ppt. (B) with NaCl solution. The filtrate gives a black ppt. (C) when H2S is passed in it. Compound (B) is dissolved in hot water and the solution gives a yellow ppt. (D) on treating with NaI and cooling. The compound (A) does not give any gas with dil. HCl but liberates reddish brown gas on heating. Identify the compound (A), (B), (C) and (D).
  4. An aqueous solution of salt (A) gives a white crystalline precipitate (B) with NaCl solution. The filterate gives a black precipitate (C) when H2S gas is passed through it. Compound (B) dissolves in hot water and the solution gives yellow precipitate (D) on treatment with KI and cooling, orange ppt. with K2CrO4 solution and white ppt. with dil. H2SO4 solution which is insoluble in C2H5OH. The compound (A) does not evolve any gas with dil. HCl, liberates a reddish brown gas on heating. Identify the compounds (A) to (D) and explain the reactions involved.
  5. Compound (A) is a light green crystalline solid. It gives the following tests:
  6. i) It dissolves in dilute H2SO4 without evolving any gas
  7. ii) A drop of KMnO4 is added to the above solution. The pink colour disappears.

iii)   Compound (A) is heated strongly. Gases (B) and (C) with pungent smell came out a brown residue (D) is left behind.

  1. iv) The gas mixture (B) and (C) is passed into dichromate solution. The solution turns green
  2. v) The green solution from step (iv) gives a white ppt. (E) with a solution of Ba(NO3)2.
  3. vi) Residue (D) from (v) is heated on charcoal in reducing flame. It gives a magnetic substance. Identify compounds (A) to (E) and predict all the equations.
  4. A mixture of two salts was treated as follows
  5. i) The mixture was heated with MnO2 and conc. H2SO4 when a yellowed green gas was liberated
  6. ii) The mixture on heating with NaOH solution gave a gas which turned red litmus blue.

iii)   Its solution in water gave blue precipitate with K­4Fe(CN)6 and red colouration with ammonium thiocyanate.

  1. iv) The mixture was boiled with KOH and the liberated gas was bubbled through an alkaline solution of K2HgI4 to give a brown ppt. Identify the mixtrue of two salts. Give equation for the reaction involved.
  2. A hydrated metallic salt (A), light green in colour, gives a white anhydrous residue (B) after being heated gradually. (B) is soluble in H2O and its aqueous solution reacts with NO to give a dark brown compound (C). (B) on strong heating gives a brown residue and a mixture of two gases (E) and (F). The gaseous mixture, when passed through acidified permanganate, discharges the pink colour and when passed through acidified BaCl2 solution, gives a white precipitate. Identify  (A), (B), (C), (D), (E) and (F). Explain the reactions involved.
  3. A compound (X) gives a golden yellow flame and shows the following reactions.
  4. i) Zn powder when boiled with a concentrated solution of (X) dissolves and H2 is evolved
  5. ii) When an aqueous solution of (X) is added to an aqueous solution of SnCl2 a white ppt. is produced which becomes soluble when (X) is added in excess.11

iii)   Compound (X) is used for the preparation of washing soap on reaction with fat and oils.

  1. iv) (X) is not a primary standard hence its standard solution is prepared by titrating against oxalic acid using phenolphthalein indicator.
  2. v) Aqueous solution of (X) precipitates hydroxides of Al3+ and Cr3+, which dissolves in its excess, the former giving colourless solution while the latter a yellow solution in presence of Br2

Identify (X) giving different reaction .

  1. An inorganic compound (A) when heated, decomposes completely to give only two gases (B) and (C). (B) is a neutral gas, fairly soluble in H2O and itself decomposes on heating to different gases (D) and (E).

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

  1. A light bluish green crystalline solid responds the following tests :
  2. i) Its aq. solution gives brown ppt. or colour with alkaline K2HgI4
  3. ii) Its aq. solution gives blue colour with K3Fe(CN)6

iii)   Its solution in HCl gives white ppt. with BaCl2 solution.

Identify the ions present and suggest the formula of the compound.

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

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

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

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

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

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

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

Ores & Metallurgy

  1. Iron oxide is reduced to pig iron with carbon. The excess carbon is oxidised as a part of the steel making process. Later still carbon is added to make steel. Explain why middle step is necessary and why it is not combined with the first or the third step?
  2. What characteristics are desirable for a furnace lining? State under which circumstances SiO2 would be preferred over CaO or MgO as a furnace lining.
  3. Carbon monoxide is more effective reducing agent for metallic oxides than carbon below a temperature about 980K, but above this temperature reverse is true. How do you account for this?
  4. Why do most of elements occurs as oxides or sulphides? In what form do tin and lead occur and why?
  5. Why are sulphides ores generally roasted to oxides before subjecting them to reduction with carbon?
  6. Describe the properties of an ore which is to be concentrated by (a) leaching with alkali (b) leaching with acid (c) floatation (d) panning
  7. What is the principal ore of aluminum? How is the ore purified, and how is the metal extracted? What is the process called? What is the function of cryolite in the process?
  8. Write the name of principal ores of copper. Write various steps in preparation of metallic copper from chalcopyrite. What is the use of silica? Write formula of slag that is removed.
  9. Complete the following reactions
  10. a) Al(OH)4 +CO2 ¾¾® A+B+C
  11. b) Ag++Na2S2O3 ¾¾® D+E
  12. c) MnO4+H2S  F+G
  13. A sulphide ore (A) on roasting leaves a residue (B). (B) on heating with chlorine gives (C), soluble in water. Addition of excess potassium iodide to a solution of (C) gives a solution (D). A brown precipitate (E) is formed when a solution of ammonium sulphate is added to an alkaline solution of (D). Identify (A) to (E)
  14. Complete and balance the following equations
  15. a) Zn+Fe2(SO4)3 ¾® A+B
  16. b) CuSO4  C+D
  17. c) Al+conc.H2SO4 ¾¾® X+Y+Z
  18. d) Al2O3+C+N2  D+E
  19. e) Pb+O2+H2O ¾¾® W
  20. In moist air copper corrodes to produce a green layer on the surface. Explain
  21. Excess of carbon is added in zinc metallurgy. Explain
  22. CuS is not precipitated by passing H2S through copper sulphate solution containing KCN. Why?
  23. The colour of mercurous chloride changes from white to black when treated with ammonia solution. Explain

Preparation & Properties

61.

Identify (A) to (H) and explain the reactions.

  1. ClO2 is a free radical with one unpaired electron but has no tendnecy to dimerise like NO2.
  2. Identify (A) and (B)

Br2 + OH (hot) ¾® (A) + (B)

  • + (B) + H+¾® Br2

(A) gives yellow ppt. with AgNO3

64.

Identify (A) to (D) in the above.

  1. Identify (A) to (E) in the following

Mg + air (A) + (E)

(A) + H2O ¾® (B) (C)

 

1 mol of (B) neutralises 2 mol of HCl

  1. Identify (A) to (C) in the following

 

  1. Identifying (A) to (E)

 

68.

(C) and (D) both decolourise acidified KMnO4. Identify (A), (B), (C), (D), (E) and explain reactions.

  1. Aq. FeSO4 gives test for Fe+2 and SO4–2 but after excess of KCN is added, solution does not give test for Fe+2 – Explain.
  2. When H2S gas is passed into FeCl3 solution, yellow colour of FeCl3 charges to light green – Explain.
  3. Colourless salt (A) gives white ppt. (B_) with NaOH; ppt. (B) dissolves in excess of NaOH to form (C). On passing H2S into (C) white ppt. (D) appears. (A) also gives white ppt. (E) with AgNO3 solution. Identify (A) to (E).
  4. Complete the reactions
  5. a) Fe2O3 ×H2O + NaOCl ¾®
  6. b) Fe2O3 (rust) + H2C2O4  ¾®
  7. When K2HgI­4 reacts with NH3, brown ppt. is formed. Explain the formation of brown ppt.
  8. Explain why a green solution of potassium manganate (VI) [K2MnO4] turns purple and a brown solid is precipitated when CO2 is bubbled into the solution.
  9. The mercurous ion is written as Hg2+2 while the cuprous ion is written as Cu+ – Explain.

 

 

 

Concepts of Acids and Bases

  1. Cl3C: is less basic than F3C: because fluroine can disperse charge only by an inductive effect. White Cl (having empty 3d orbitals) disperses charge by inductive effect as well as by pp – pp bonding delocalisatioin. Fluorine is a second period element with no 2d orbital.
  2. Among the halogens, fluorine has the smallest size hence it has availability of electron most.
  3. I > V > IV > II > III
  4. Pyrole uses lone pair of electron on nitrogen atom in delocalisation hence less electron are available for protonation, hence less basic than pyridine.
  5. CH3CONH2 < (iso-C3H7)3N, C2H5NH2, CH3ONHNa+
  6. It can be seen that hydrogens in these molecules are not all bonded to oxygens. It is clear from their structures,

                             

that the number of terminal oxygen atoms is 1 in all three acids. The electronegativities of P and H are almost the same. Thus no much difference in acidity is expected.

7.         IV > III > I > II

  1. Both HCl and SO3 are Lewis acids and can react with amine base to form polar substances which could presumably undergo ionic dissociation in a solvent sufficiently more polar than benzene. The reactioins may be represented as follows (R = C­4H9):

Sulphur is a third element  which can expand its octet due to availability of vacant d-orbitals. Thus sulphur expands its number of valance electron by attaching to the lone pair on the nitrogen. The N – S bond will be polar because of the big difference in  electronegativity between N and S.

 ¾¾®

            Here the proton of the HCl attaches to the lone pair on the N. The connection between N – H and Cl is designated by (—-) symbolizing an electron pair on the Cl connected to nitrogen by a hydrogen bond.

  1. The highly branched tertiary butyl group involve appreciable back – strain (B-strain) when the boron atom changes to pyramidal environment on adduct formation. This destabilizes the adduct. Hence the order is

B(t–Bu)3 < B(n–Bu)3

  1. The order in this case is the reverse of that for BX3. p-conjugation from the halogen p-orbital to the Si-d orbital is not as intense as in the case of BX3 and the order of acidity follows the increase in electron withdrawing power of the halogen from I to F. Hence the order is

SiI4 < SiBr4 < SiCl4 < SiF4

Chemical Bonding

  1. NCl3 + 4H2O ¾® 3HOCl + NH4OH

PCl3 + 3H2O ¾® 3HCl + H3PO3

In NCl3, H2O attacks vacant d-orbital of Cl where as in PCl3, H2O attacks the vacant d-orbital of P.

  1. It is because when S combines with strongly electronegative atoms like ‘F’ their d-orbital contracts in size and becomes lower in energy and thereafter they undergo hybridization with s and p-orbital to give sp3d2 hybridized orbitals. But with ‘H’ it can not occur.
  2. Since ‘F’ is having very high electronegativity it can form strong hydrogen bond with HF as F¼ Hd+ –– Fd– but due to lower electronegativity ‘Cl’, HCl cannot do it with Cl can not do it.
  3. This is because, as the size of anion increases, its plolarisatibility increases and hence % of covalent character increases.
  4. SnCl4 is more covalent in nature compared to SnF4 due to the high polarisability of Cl ion and hence boiling is less.
  5. Because of one lone pair, which repel the four Cl atoms slightly up, making them and the central I, non-coplanar.
  6. Boron in BF3 is sp2 hybridised giving a planar geometry whereas nitrogen in NF3 is sp3 hybridised with a lone pair, which account for its pyramidal geometry.
  7. Dipole moment m = e ´ d coulomb metre

For KCl d = 2.6 ´ 10–10 m

For complete separation of unit charge

e = 1.602 ´ 10–19 C

Hence m = 1.602 ´ 10–19 ´ 2.6 ´ 10–10 = 4.1652 ´ 10–29 Cm

mKCl = 3.336 ´ 10–29 Cm

\ % ionic character of KCl =  = 80.09%

  1. Cl being more electronegative, contracts the d-orbital of Pb, and allows hybridisation involving d-orbital and thus brings about higher oxidation state but ‘I’ being less electronegative can’t do it.
  2. The shape of PCl4+ is tetrahedral while that of PCl6 is octahedral. Now conduction of electricity is possible whenever the ions can move. Here due to strong coulombic attraction between the tetrahedra and octahedra the ions cannot move and hence do not conduct electricity.

Transition Element

  1. On addition, the pale red coloured octahedral complex [Co(H2O)6]2+ changes into tetrahedral [CoCl4]2– ion. In Tetrahedral complexes, the d–d absorption bands are considerably more intense than in octahedral complexes.
  2. The lanthanoids which will most likely react with this strong reducing agent are those with stable +2 oxidation states to which they can be reduced –Sm, Eu and yb.
  3. The larger the number of ions and the larger the charge on each, the larger the conductivity. The compounds from lowest conductivity to highest conductivity are:

b < a < d < c

  1. EDTA coordinates lead ion in the body, in which form it is passed out of the body without harmful effects. The calcium salt is used so that any excess EDTA will not remove calcium ions from the body.
  2. a) CuII is reduced to CuI by I but not by Cl.
  3. b) The oxidation is aided by the complexation of the product gold (III) as
  4. c) The stability of  ion is so great that the E value for oxidation is reduced to a point where H2O can oxidise the copper.
  5. Tetrahedral complexes do not show geometrical isomerism because the relative positions of the ligands attached to the central metal atom are same with respect to each other.
  6. FeSO4 does not form any complex with (NH4)2SO4 instead, they form a double salt FeSO4 (NH4)2SO4.6H2O which dissociates completely into ions but CuSO4 combines with NH3 to from the complex [Cu(NH3)4]SO4 in which complex ion, [Cu(NH3)4]2+ does not dissociate to give Cu2+ ions.
  7. Remembering that co-ordination number of CO is 6 the formula of the complex will be [CoCl2(NH3)4]Cl. The name will be tetraminedichlorocobalt (III) chloride.
  8. In [Ni(CO)4], Ni is in zero oxidation state whereas is (NiCl4]2–, Ni is in +2 oxidation state. In the presence of ligand CO, the unpaired electrons of Ni pair up but Cl being a weak ligand is unable to pair up the unpaired electrons.
  9. i) CO(II) has the configuration 3d7 i.e. it has three unpaired electrons. Water being a weak ligand, the unpaired electrons do not pair up. In the presence of strong ligands and air, two unpaired electrons in 3d pair up and the third unpaired electron shifts to higher energy sub shell from where it can be easily lost and hence shows an oxidation state of III.

Qualitative Analysis

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

¾¾® H2SO4 +  + 2H2O

Na2SO3 ¾¾®

¾¾® Na2SO4  +  + H2O + S¯

  1. As the solid B forms a hydrated salt C with Al2(SO4)3; B should be sulphate of a monovalent cation, i.e., M2SO4.

Now since sulphate of a monovalent cation contains one sulphur atom per mol, weight of metal sulphate obtained by 32.1 g (at.wt. of S) should be the molecular weight of the metal sulphate. Thus,

0.321 g of sulphur is present in 1.743 g of B

32.1 g of sulphur is present in =  ´ 32.1  = 174.3 g

Thus mol.wt. of B (M2SO4) = 174.3 g mol–1

Thus 2x + 32.1 + 64  = 174.3

2x = 78.2

x = 39.1

Atomic weight 39.1 corresponds to metal potassium K

Thus B is ­K2SO4, and C is K2SO4.Al2(SO4)3.24H2O

            Nature of compound. As since A is a binary compound of potassium and it reacts with sulphur to form K2SO4 is must be oxide of potassium, probably potassium superoxide (KO2) which is supported by the given data.

+ S ¾¾®

32.1 g of S reacts with 142.2 g of KO2

0.321 g of S reacts with  =  = 1.422 g

Similarly,

32.1 g of S gives 174.3 g of K2SO4

0.321 g of S give  =  ´ 0.321  = 1.743 g

Both these datas are also given in the problem. Thus a is KO2

  1. (A) NH4Cl (B) NH4H CO3

(C) NaHCO3                           (D) CO2

  1. Given fact that the reaction of aqueous solution of the compound (B) with brine (sodium chloride) gives a precipitate (D) soluble in ammonium hydroxide indicates that (D) is AgCl and hence the metal (A) and compound (B) must be Ag and AgNO3 respectively. This is further supported by the fact that (B), AgNO3, when added to hypo (sodium thiosulphate solution) gives a white precipitate (E) of Ag2S2O3, which turns to a black compound (F) of the formula Ag2S. Thus (A) to (F) can be written as below.
Ag AgNO3 NO
Silver (A) Silver nitrate (B) Nitric oxide (C)
AgCl Ag2S2O3 Ag2S
Silver chloride (D) Silver thisulphate (E) Silver sulphide (F)
  1. (A) is Pb(NO3)2, (B) PbCl2

(C) PbS                                   (D) PbI2 or PbSO4 or PbCrO4

  1. (A) FeSO4 (B) SO2

(C) SO3                                   (D) Fe2O3

(E) BaSO4

  1. The given unknown mixture contains , Fe++ and Cl ions or NH4Cl and FeCl2

 

  1. (A) FeSO4.7H2O (B) FeSO4

(C) [Fe(H2O)5NO]SO4            (D) Fe2O3

(E) SO2                                   (F) SO3

  1. Thus initial compound (X) is NaOH (caustic soda)

 

  1. (A) NH4NO3 (B) N2O

(C) H2O                                   (D) N2

(E) O2  (F) NH3

  1. Ions present –  NH4+, Fe2+ , SO42-

            Formula – FeSO4 . (NH4)2SO4 . 6H2O

 

H+
  1. Cation of X     –   Sn+2

Reactions –    (i)   Sn+2 + H2S ¾® SnS ¯

brown ppt.

(ii)  Sn+2 + 2NaOH ¾® 2Na+ +  Sn(OH)2 ¯

white ppt.

Sn(OH)2  + 2 NaOH  ¾® Na2SnO2 + 2H2O

Soluble

(iii)       Sn+2 + 2Fe+3 ¾¾® Sn+4 + 2Fe+2

D
  1. Mixture – FeCl2 and NH4Cl
D

                  Reactions –      (i)       2NH4Cl + MnO2 + 2H2SO4 ® MnSO4+Cl2­ + 2H2O +(NH4)2SO4

(ii)      NH4Cl + NaOH ¾¾¾® NH3 + H2O

Blue Colour

(iii)     3FeCl2 + 2K3[Fe(CN)6] ¾¾¾® Fe3[Fe(CN)6]2  + 6KCl

  1. (A) – Pb3O4

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

(B)

(ii)  Pb(NO3)2 + 2KI ¾® PbI2 ¯ + 2KNO3

(C)

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

(D)

Ores & Metallurgy

  1. The percent carbon in the final product may be controlled more precisely by removing all the carbon initially present and then adding measure quantities.
  2. The furnace lining must be very high melting. It should be acidic or basic, depending on the type of impurities which are to be removed. Two remove acidic impurities, such as SO2 or P4O10 a basic lining is preferable – MgO or CaO. Two remove basic impurities, such an metaloxides an acidic lining SiO2 is used.
  3. a) 2Al(OH)4+CO2 ¾® 2Al(OH)3+CO32-+H2O
  4. b) Ag++2Na2S2O3 ¾® Na3[Ag(S2O3)2]+Na+
  5. c) MnO4+H2S ¾¾¾® Mn2++S
  6. HgS

(A)                                                                                       (B)                    (C)                    (D)                                                                                             (E) (brown ppt)

  1. a) Zn+Fe2(SO4)3 ¾® ZnSO4+2FeSO4
  2. b) CuSO4   CuO+SO3
  3. c) 2Al +conc.6H2SO4 ¾¾® Al2(SO4)3+3SO2+6H2O
  4. d) Al2O3+2C+N2  2AlN + 3CO
  5. e) 2Pb+O2+2H2O ¾¾® 2Pb(OH)2
  6. In presence of moist air a thin layer of green basic copper carbonate is formed on its surface and hence copper corrodes

2Cu+O2+H2O+CO2 ¾¾® CuCO3.Cu(OH)2

                                                                 green

  1. Carbon has to play two roles
  2. a) It reduces zinc oxide to zinc

ZnO+C ¾¾¾® Zn+CO

ZnO+CO ¾¾¾® Zn+CO2

  1. b) It reduces CO2 into CO which is used as a fuel

CO2+C ¾¾® 2CO

  1. CuSO4 forms a complex with KCN

CuSO4+2KCN ¾¾® Cu(CN)2+K2SO4

2Cu(CN)2 ¾¾® Cu2(CN)2+(CN)2

Cu2(CN)2+6KCN ¾¾® 2K3Cu(CN)4  6K++2[Cu(CN)4]3-

K3Cu(CN)4 complex does not furnish Cu2+ ions. Hence, no ppt of CuS is formed, when H2S is passed through solution.

  1. Mg2Cl2 reacts with NH3 to form a mixture of mercury and mercuric amino chloride which is a

black substance

Hg2Cl2+NH3 ¾® Hg+Hg(NH2)Cl+HCl

White                    Black

Preparation & Properties of Compounds

  1. A :     NH3                     D   :     N2O3               G   :     Mg3N2

B    :     NO                        E    :     HNO2              H   :     CaCN2

C   :     NO2                 F    :     N2

  1. A :     Br                   B    :     BrO3
  2. A :     CrCl3                    C   :     Na2CrO4

B    :     CrO2Cl2           D   :     PbCrO4

  1. A :     MgO                D   :     NH3

B    :     Mg(OH)2         E    :     Mg3N2             C   :     MgCO3

 

  1. A :     K2SO4               B    :     BaSO4

C   :     K2SO4 × Al­2(SO4)3 × 24H2O

  1. A :     NaHCO3         D   :     CO2

B    :     ZnCO3             E    :     Na2ZnO2         C   :     ZnO

  1. A :     CO                        D   :     COOH

|

COOH

B    :     HCOONa        E    :     CaC2O4

C   :     COONa

|

COONa

  1. A :     ZnCl2               D   :     ZnS

B    :     Zn(OH)2          E    :     AgCl

C   :     Na2ZnO2

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