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 –COO^–which 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 = m_B–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

2. b) ZnCl₂.2H₂O ZnCl(OH)+HCl+H₂O
3. c)   4Fe+10HNO₃ ¾® 4Fe(NO₃)₂+NH₄NO₃+3H₂O
4. d)   Au+4HCl + 3HNO₃ ¾® HAuCl₄+3NO₂+3H₂O
5. 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.

 

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

CH₃NH^–Na⁺, C₂H₅NH₂, (iso-C₃H₇)₃N and CH₃CONH₂

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

 

8. 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?
9. Arrange according to increasing Lewis acid character,

B(n–Bu)₃,  B(t–Bu)₃

10. Arrange according to increasing Lewis acid character,

SiF₄, SiCl₄, SiBr₄, Sil₄

Chemical Bonding

11. The product of hydrolysis of NCl₃ and PCl₃ are different. Why?
12. SF₆ is possible but SH₆ is not possible. Why?
13. KHF₂ is possible but not KHCl₂. Why?
14. Explain why the solubility in water of halides of Aluminium follows the order

AlF₃ > AlCl₃ > AlBr₃ > AlI₃

15. SnF₄ boils at 705°C while SnCl₄ boils at 114°.
16. In ICl₅, the four chlorine atoms on square planar base and the central atom I does not remain in same plane. Justify the statement.
17. NF₃ is pyramidal but BF₃ is planar.
18. 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.
19. PbCl₄ exists but PbI₄ is not possible.
20. PCl₅ exists in the solid state in the form of [PCl₄]⁺ [PCl₆]^– yet it is a non conductor of electricity.

Transition Element

21. Why the pale red colour of [Co(H₂O)₆]²⁺ ion changes into intense blue colour by the additon of chloride ion?
22. Which tripositive lanthanoid ion(s) is (are) most likely to react with chromium (II) chloride?
23. 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]

24. 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?
25. Explain the following:
26. a) The complex [CuCl₄]^2– exists, but [CuI₄]^2– does not.
27. b) gold is not attacked by common acids but dissolves in aqua regia.
28. c) Copper dissolves in aqueous KCN solution with the evolution of hydrogen.
29. Square planar complexes with co-ordiantion number of four exhibit geometrical isomerism whereas tetrahedral complexes do not. Why?
30. 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?
31. 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.
32. [NiCl₄]^2– is paramagnetic while (Ni(CO)₄] is diamagnetic. Why?
33. Account for the following
34. i) Co(II) is stable is aqueous solution but in the presence of strong ligand and air, it can get oxidized to Co(II).
35. ii) [Ni(CN)₄]^2– is square planar and diamagnetic whereas (NiCl₄]^2– is tetrahedral and paramagnetic

Qualitative Analysis

31. 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)
32. A is binary compound of a univalent metal 1.422 g of A reacts completely with 0.321 g of sulphur in an evacuated and sealed tube to give 1.743 g of a white crystalline solid, B that forms a hydrated double salt, C with Al₂(SO₄)_3­. Identify A,B, and C.
33. Identify A,B, C and D in the following sequence of reactions:
34. i) A + NaOH  ¾¾® NaCl + NH₃ + H₂O
35. 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 .

41. 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.

42. A light bluish green crystalline solid responds the following tests :
43. i) Its aq. solution gives brown ppt. or colour with alkaline K₂HgI₄
44. 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.

43. A certain inorganic compound (X) shows the following reactions :
44. i) On passing H₂S through an acidified solution of (X), a brown ppt. is obtained.
45. 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).

44. A mixture of two salts was treated as follows :
45. i) The mixture was heated with manganese dioxide and conc. sulphuric acid, when  yellowish green gas was liberated
46. 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

46. 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?
47. What characteristics are desirable for a furnace lining? State under which circumstances SiO₂ would be preferred over CaO or MgO as a furnace lining.
48. 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?
49. Why do most of elements occurs as oxides or sulphides? In what form do tin and lead occur and why?
50. Why are sulphides ores generally roasted to oxides before subjecting them to reduction with carbon?
51. Describe the properties of an ore which is to be concentrated by (a) leaching with alkali (b) leaching with acid (c) floatation (d) panning
52. 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?
53. 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.
54. Complete the following reactions
55. a) Al(OH)₄^– +CO₂ ¾¾® A+B+C
56. b) Ag⁺+Na₂S₂O₃ ¾¾® D+E
57. c) MnO₄^–+H₂S  F+G
58. 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)
59. Complete and balance the following equations
60. a) Zn+Fe₂(SO₄)₃ ¾® A+B
61. b) CuSO₄  C+D
62. c) Al+conc.H₂SO₄ ¾¾® X+Y+Z
63. d) Al₂O₃+C+N₂  D+E
64. e) Pb+O₂+H₂O ¾¾® W
65. In moist air copper corrodes to produce a green layer on the surface. Explain
66. Excess of carbon is added in zinc metallurgy. Explain
67. CuS is not precipitated by passing H₂S through copper sulphate solution containing KCN. Why?
68. 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.

62. ClO₂ is a free radical with one unpaired electron but has no tendnecy to dimerise like NO₂.
63. 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.

65. 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

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

 

67. Identifying (A) to (E)

 

68.

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

69. 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.
70. When H₂S gas is passed into FeCl₃ solution, yellow colour of FeCl₃ charges to light green – Explain.
71. 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).
72. Complete the reactions
73. a) Fe₂O₃ ×H₂O + NaOCl ¾®
74. b) Fe₂O₃ (rust) + H₂C₂O₄  ¾®
75. When K₂HgI_­4 reacts with NH₃, brown ppt. is formed. Explain the formation of brown ppt.
76. 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.
77. 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, CH₃ONH^–Na⁺
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

8. 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.

9. 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

10. 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

11. 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.

12. 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.
13. Since ‘F’ is having very high electronegativity it can form strong hydrogen bond with HF as F^{– ¼} H^d+ –– F^d– but due to lower electronegativity ‘Cl’, HCl cannot do it with Cl^– can not do it.
14. This is because, as the size of anion increases, its plolarisatibility increases and hence % of covalent character increases.
15. SnCl₄ is more covalent in nature compared to SnF₄ due to the high polarisability of Cl^– ion and hence boiling is less.
16. Because of one lone pair, which repel the four Cl atoms slightly up, making them and the central I, non-coplanar.
17. 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.
18. 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

m_KCl = 3.336 ´ 10^–29 Cm

\ % ionic character of KCl =  = 80.09%

19. 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.
20. 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

21. 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.
22. 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.
23. 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

24. 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.
25. a) Cu^II is reduced to Cu^I by I^– but not by Cl^–.
26. b) The oxidation is aided by the complexation of the product gold (III) as
27. 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.
28. 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.
29. 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.
30. 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.
31. 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.
32. 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

31. 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¯

32. 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

33. (A) NH₄Cl (B) NH₄H CO₃

(C) NaHCO₃                           (D) CO₂

34. 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)

36. (A) is Pb(NO₃)₂, (B) PbCl₂

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

37. (A) FeSO₄ (B) SO₂

(C) SO₃                                   (D) Fe₂O₃

(E) BaSO₄

38. The given unknown mixture contains , Fe⁺⁺ and Cl^– ions or NH₄Cl and FeCl₂

 

39. (A) FeSO₄.7H₂O (B) FeSO₄

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

(E) SO₂                                   (F) SO₃

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

 

41. (A) NH₄NO₃ (B) N₂O

(C) H₂O                                   (D) N₂

(E) O₂  (F) NH₃

42. Ions present –  NH₄⁺, Fe²⁺ , SO₄^2-

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

 

H+

43. 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

44. 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

45. (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

46. The percent carbon in the final product may be controlled more precisely by removing all the carbon initially present and then adding measure quantities.
47. 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.
48. a) 2Al(OH)₄^–+CO₂ ¾® 2Al(OH)₃+CO₃^2-+H₂O
49. b) Ag⁺+2Na₂S₂O₃ ¾® Na₃[Ag(S₂O₃)₂]+Na⁺
50. c) MnO₄^–+H₂S ¾¾¾® Mn²⁺+S
51. HgS

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

56. a) Zn+Fe₂(SO₄)₃ ¾® ZnSO₄+2FeSO₄
57. b) CuSO₄   CuO+SO₃
58. c) 2Al +conc.6H₂SO₄ ¾¾® Al₂(SO₄)₃+3SO₂+6H₂O
59. d) Al₂O₃+2C+N₂  2AlN + 3CO
60. e) 2Pb+O₂+2H₂O ¾¾® 2Pb(OH)₂
61. 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

58. Carbon has to play two roles
59. 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

59. 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.

60. 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

61. A :     NH₃                     D   :     N₂O₃               G   :     Mg₃N₂

B    :     NO                        E    :     HNO₂              H   :     CaCN₂

C   :     NO₂                 F    :     N₂

63. A :     Br^–                   B    :     BrO₃^–
64. A :     CrCl₃                    C   :     Na₂CrO₄

B    :     CrO₂Cl₂           D   :     PbCrO₄

65. A :     MgO                D   :     NH₃

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

 

66. A :     K₂SO₄               B    :     BaSO₄

C   :     K₂SO₄ × Al­₂(SO₄)₃ × 24H₂O

67. A :     NaHCO₃         D   :     CO₂

B    :     ZnCO₃             E    :     Na₂ZnO₂         C   :     ZnO

68. A :     CO                        D   :     COOH

|

COOH

B    :     HCOONa        E    :     CaC₂O₄

C   :     COONa

|

COONa

71. A :     ZnCl₂               D   :     ZnS

B    :     Zn(OH)₂          E    :     AgCl

C   :     Na₂ZnO₂