Preparation & Properties of Compounds



Solution to Subjective Problems 



  1. Water reacts with Na2O2 to produce NaOH which increases the decomposition of H2O2.
  2. Hydrolysis of AlCl3 takes place forming Al2O3

2AlCl3 + 3H2O ⎯⎯→ Al2O3 + 6HCl

  1. AlCl3 lacks back bonding as in BCl3 because of increase in size of aluminium. Aluminium metal atoms complete their octate by coordinate bond forming by chlorine atom between two Al atom, hence AlCl3 exists as dimer  hence, shows anomalous mol.wt.
  2. Because of the Fajan’s Rule.
  3. It is a waste product from steel industry and has properties similar to that of cement
  4. CaSO4 which has no tendency to absorb H2O is called dead burnt plaster.

7 CaO, slaked lime [Ca(OH)2] is formed.

  1. a) K[BrICl]
  2. b) K[BrF4]
  3. Ions are hydrated in solution since Li is very small it is heavily hydrated. This make the radius of the hydrated ions large and hence it move only slowly (although Li+ is very small) and the radius of hydrated Cs+ ion is smaller than the radius of hydrated Li+.
  4. 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.



  1. In silicates pπ – dπ bonding is absent due to non – availability of d orbitals.. Hence in silicates the SiO4 units are held together in polymeric forms by σ bonds. From P to Cl, d – orbitals are available for hybridization and pπ – dπ bonding due to overlap of p orbital of ligand (O etc.) with d – orbital of P,S,Cl etc. As the 3d orbital decreases in the series, the overlap becomes stronger. In phosphates π bonding is strong but it still stronger in the oxyacids of S where pπ – dπ bonding becomes a dominant feature and only a small amount of polymerization occurs. At Cl, pπ – dπ bonding is so strong that no polymerization of oxyanions occurs. 

2: Increase in the polarity of the halides although show the order
B—F> BCl > BBr but the Lewis acidity depends on the availability of 2P orbital of B pπ – pπ back bonding between X(F,Cl, Br) and B occurs. The overlap becomes maximum for B occurs for and decreases in the order BF3 > BCl3 > BBr2 resulting into B—F double bond character and least basicity and follows the order BF3 < BCl3 < BBr3

(BF3 < BCl3 < BBr3)

3: Li forms only oxide, Na forms oxide and peroxides and remaining all alkali metal form oxides, peroxides and super oxides.

Li + O2 ⎯→

4Na + O2 ⎯→

2Na + O2 ⎯→

K + O2 ⎯→

Peroxide [–O–O–)2 has no unpaired electron therefore it is diamagnetic, on the other hand super oxide [O2] has an unpaired electron therefore it is paramagnetic. Super oxides are stronger oxidizing agent then peroxides. Stability of the peroxides and super oxides increases as the metal ions become large. Same if both ions are similar in size the coordination number will be high and this gives a high lattice energy.

KO2 is used in space capsules, breathing masks because it both produces dioxygen and remove carbon dioxide.

4KO2 + 2CO2 ⎯→ 2K2CO3 + 3O2

4KO2 + 4CO2 + 2H2O ⎯→ 4KHCO3 + 3O2

  1. a) Solubility decreases with increased atomic weight
  2. b) This trend is reversed with the fluorides and hydroxides
  3. c) Lattice energy decreases as the size of metal increases
  4. d) Hydration energy decreases more rapidly than the lattice energy.
  5. e) However with fluorides and hydroxides the lattice energy decreases more rapidly.
  6. f) Condition for solubility – hydration energy lattice energy.

Solubility increases









  1. A = NH3 B = CaCO3


  1. ;
  2. Because of the small size of boron atom and presence of only six electrons in its valence shell in B(OH)3, it coordinates with the oxygen atom of the H2O molecule to form a hydrated species.

In this hydrated species, B3+ ion because of its small size has a high polarizing power and hence pulls the σ-electrons of the coordianted oxygen atom towards it. The coordianted O-atom, in turn pull the σ-electrons of the O–H bonds, thereby facilitating the release of a proton.

B(OH)3 + NaOH ⎯→ Na [B(OH)4] Na+ + 2H2O


  1. ion exists as a free ion. Since N is sp2 hybridized, therefore, ion has trigonal planar shape on other hand, does not exist as a free ion. It usually exists in the polymeric form i.e. in which the various units are linked together through P–O–P bonds to form either linear or cyclic structures. Each unit has a tetrahedral (sp3) shape


  1. The minimum oxidation number (O.N.) of S is –2 while its maximum O.N. is +6 in SO2 the O.N. is +4 therefore, it cannot only increase its O.N. but can also reduce
    its O.N.

Due to small size of oxygen the non-bonding electrons on the two oxygen atom is O–O bond strongly repel each other while such repulsion are much less in S–S bond. As a result O – O bond dissociation energy is much less than S –S bond dissociation energy in other words sulphur has much higher tendency for catenation than oxygen. Further, oxygen because of its small size has a high tendency to form P-P multiple bonds but sulphur because of its bigger size does not form P multiple bonds.



  1. a) 2Co(NO3)2 2CoO + 4NO2 + O2

CoO + Al2O3 ⎯→ (Thenard’s blue)

  1. b) (NH4)SO4.Al2(SO4)3.24H2O (NH4)HSO4 + Al2(SO4)3 + 24H2O
  2. c) H2Sn5O44H2O 5SnO2 + 5H2O
  3. d) SnCl4 + Sn ⎯→ 2SnCl2
  4. a) K2Cr2O7 + 14HCl + 3SnCl2 ⎯→ 3SnCl4 + 2KCl + + 7H2O
  5. b) Pb3O4 + 4HNO3⎯→ 2Pb(NO3)2+ PbO2 + 2H2O
  6. c) Pb(CH3COO)2 + Ca(OCl)Cl + H2O ⎯→ PbO2 + CaCl2 + 2CH3COOH
  7. d) PbO2 + 2NaOH ⎯→ Na2PbO3 + H2O
  8. e) 2FeSO4 ⎯→ Fe2O3 + SO2 + SO3
  9. f) 2FeCl3 + 3Na2CO3 + 3H2O ⎯→ 2Fe(OH)3 + 6NaCl + 3CO2

2Fe(OH)3 ⎯→ Fe2O3 + 3H2O

  1. a) 2FeCl3 + 2KI ⎯→ 2FeCl2 + 2KCl + I2
  2. b) 2Fe(OH)3 + 10KOH + Br2 ⎯→ 2K2[FeO4] + 6KBr + 8H2O
  3. c) 2CuSO4 + 4KCN ⎯→ 2CuCN + (CN)2 + 2K2SO4

CuCN + 3KCN ⎯→ K3[Cu(CN)4

(CN)2 i.e. cyanogen gas is complex poisonous and K3 [Cu(CN)4] dissolve

  1. d) CuSO4 + 2NH4CNS ⎯→ Cu(SCN)2 + (NH4)2SO4

2Cu(SCN)2 + H2SO3 + H2O ⎯→ 2HCN + H2SO4 + 2CuSCN

  1. A = (NH4)2Cr2O7 B = Cr2O3

C = N2 D = Mg3N2

E = NH3

  1. a) Al2(SO4)3 + H2O ⎯→ 2Al(OH)3 + 3H2SO4

Because of this hydrolysis of aq. Al2(SO4)3 the fibres when dipped in aq. Al2(SO4)3 solution, Al(OH)3 forms on the surface of the fibres. Dye then combine with Al(OH)3 to form an insoluble lake whichis fast to washing. The process of formation of metallic hydroxides on the fibres before dyeing is known as mordanting

  1. b) 2(AlCl36H2O) Al2O3 + 6HCl + 9H2O

Because of hydrolysis of AlCl3 it never produces anhydrous AlCl3.

  1. c) AlCl3 + 6H2O  [Al(H2O)6]+3 + 3Cl

[Al(H2O)6]+3  [Al(H2O)5OH]+2 + H+

Because of H+ generation in aq. solution AlCl3 shows acidity.


There is Fe – Fe bond in the molecule

  1. a) 4Al(NO3)3 2Al2O3+ 12NO2 + 3O2
  2. b) Al2O3 + C + Cl2 ⎯→ AlCl3 + CO
  3. c) 5Sn + 20HNO3 ⎯→ H2Sn5O114H2O + 20NO2 + 5H2O
  4. d) Sn + 2NaOH + H2O ⎯→ Na2SO4 + 2H2
  5. e) PbO + 2NaOH ⎯→ Na2PbO2 + H2O
  6. f) 2PbO + NaCl ⎯→ Pb2O3 + NaCl
  7. g) 4Fe + 10 HNO3 (cold, dil) ⎯→ 4Fe(NO3)2 + NH4NO3 + 3H2O

4Fe + 10HNO3 (warm dil) ⎯→ 4Fe (NO3)2 + N2O + 5H2O

Fe + 4HNO3 (sp. gr > 1.12) ⎯→ Fe(NO3)2 + NO + 2H2O

  1. h) FeC2O4+ CO + CO2
  2. i) 2FeSO4 ⎯→ Fe2O3 + SO2 + SO3
  3. j) Cu2O + Fe2(SO4)3 + H2SO4 ⎯→ 2CuSO4 + 2FeSO4 + H2O


  1. We known that normality  =

N1 = ; N2 = ; N3 =

After mixing 

N1V1 + N2V2 + N3V3 = NV

N = 1.339

Volume strength of resulting solution = 1.339 × 5.6  = 7.5

  1. a) CuSO4 + 2NaOH ⎯⎯→ + Na2SO4

Cu(OH)2 CuO + H2O

CuO  + ⎯⎯→ Cu2O

  1. b) FeSO4.(NH4)2SO4.6H2O + 4NaOH  Fe(OH)2 + 2NH3 + 2Na2SO4 + 8H2O

Fe(OH)2 + H2SO4 ⎯⎯→ FeSO4 + 2H2O

  1. c) ZnSO4 + 7H2O

ZnSO4 ZnO + SO3

ZnO + 2HCl ⎯⎯→ ZnCl2 + H2O

ZnCl2 + 2NH4Cl ⎯⎯→ ZnCl2.2NH4Cl

ZnCl2.2NH4Cl + 2NH4Cl

  1. + H2SO4 ⎯⎯→ FeSO4 +

+ ⎯⎯→ H2O  +

H2S + ⎯⎯→ + H2SO4

2HNO3 ⎯⎯→ H2O + 2NO2 + [O]

CuS + 2HNO3 + (O) ⎯⎯→ (Cu(NO3)2 + H2O + S

Cu(NO3)2 + K4Fe(CN)6 ⎯⎯→ Cu2[Fe(CN)6] + 4KNO3

+ BaCl2 ⎯⎯→ 

  1. i) (A) is characteristic dimerised compound which sublimes in 180°C and forms monomer if heated to 400°C and thus (A) is (AlCl3)2 or Al2Cl6.

Al2Cl6(s) Al2Cl4(s) 2AlCl3

  1. ii) It fumes with wet air 

Al2Cl4 + 6H2O 2Al(OH)3 + 6HCl 

iii) 2AlCl3 + 6H2O 2Al(OH)3 + 6HCl(s)

  1. iv) (A) gives white ppt. with NH4OH and NaOH, soluble in excess of NaOH

Al2Cl6 + 6NH4OH  ⎯→ 2Al(OH)3 + 6NH4Cl

Al2Cl6 + 6NaOH  ⎯→ 2Al(OH)3 + 6NaCl 2NaAlO2 + 2H2O