Your Cart

Preparation & Properties of Compounds

Loader Loading...
EAD Logo Taking too long?

Reload Reload document
| Open Open in new tab
In this chapter you will study the preparation of various important compounds, their different properties. You will also see the various types of reactions they undergo.

In order to make the study systematic and convenient the chapter has been divided into two parts. We have kept aluminium, Boron and silicon and their related compounds in the first part.

Second part deals with carbon, and its various oxides. Along with this second part also contains various oxides of nitrogen, phosphorus, their fertilizers and various halogen acids.

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

  1. IIT–JEE Syllabus

Alumina, aluminium chloride and alums; oxides and chlorides of tin and lead; ferrous sulphate, Mohr’s salt, ferric oxide and ferric chloride, copper sulphate, oxide and sulphite of zinc, silver nitrate and silver bromide. Hydrogen peroxide; carbon oxides and carbides; nitrogen and phosphorous: oxides and oxyacids, ammonia, fertilisers; sulphur: oxides, sulphurous and sulphuric  acids, sodium thiosulphate and hydrogen sulphide; halogens: hydrohalic acids, oxyacids of chlorine, bleaching powder.

  1. Introduction

This chapter focuses on the various aspects of compounds of aluminum, iron, tin, silver and other important metals along with the various reactions, the metals and their compounds undergo. It also deals with the preparation, properties of non metals like carbon, nitrogen, phosphorous, their various fertilizers widely used in the industry.

   3.    Compounds of Aluminium

 

      3.1       Aluminium oxide or Alumina (Al2O3)

  1. a) a – Al2O3 (rhombic lattice) is dense, hard and resistant to chemical attack.

a – Al2O3 occurs in the earth’s crust as corundum, (Al2O3). Transparent coloured crystals of corundum, viz. Ruby (red due to the presence of Cr) and sapphire (blue due to Ti and Re) are used as precious stones (gems)

  1. b) l – Al2O3 is less dense, soft and has a high surface area. It is used as an excellent selective adsorbent in dehydration, decolourisation and chromatography.

      3.2       Aluminium chloride: (AlCl3)

  1. a) Anhydrous Aluminium chloride (AlCl3)  on heating it sublimes at 180°C and the vapour density corresponds to the formula Al2Cl6.

The dimeric formula is retained in non polar solvents but it broken into [Al(H2O)6]Cl3 on dissolution in water on account of high heat of hydration

      3.3       Aluminium

Aluminium ions crystallize from aqueous solutions, forming double salts. These are called aluminium alums and have the general formula [M¢(H2O)6][Al(H2O)6](SO4)2, M¢ is a singly charged cation such as Na+, K+ or NH4+

Some M3+ ions other than Al3+ also form alums of formula [M¢(H2O)6[ [MIII(H2O)6](SO4)2. Some of important alums are

Potash alum          :     K2SO4. Al2(SO4)3×24H2O

Ammonium alum  :     (NH4)2SO4. Al2(SO4)3×24H2O

Sodium alum         :     Na2SO4.Al2(SO4)3×24H2O

Chrome alum        :     K2SO4.Cr2(SO4)3×24H2O

Ferric alum            :     (NH4)2SO4.Fe2(SO4)3×24H2O

   4.    Compounds of Boron

 

4.1       Boric Acid

  1. a) Orthoboric acid (H3BO3)

a1) i)    Preparation from borax

Na2B4O7 + H2SO4 + 5H2O ¾¾® Na2SO4 + 4H3BO3

  1. ii)   From colemanite

Ca2B6O11 + 2SO2 + 11H2O ¾¾® 2Ca(HSO3)2 + 6H3BO3

Boric acid forms white needle like crystal. It has a layer structure involving triangular BO3 groups joined by hydrogen bonds.

a2) Action of heat

4H3BO3  4HBO2  H2B4O7 2B2O3

Metaboric acid              Tetrabasic acid                           Boron trioxide

a3)       It behaves as a weak monobasic acid

B(OH)3  H3BO3  H+ +    H2O +

Thus on titration with NaOH, it gives sodium metaborate salt

H3BO3 + NaOH  NaBO2  + 2H2O

a4)       B(OH)3 + MO  M – borates

Metal oxide

Where M stands for a bivalent metal

a5)       B(OH)3   BF3

      4.2       Borax (sodium tetraborate) Na2B4O7. 10H2

Borax occurs naturally and is also called Tinacal or suhaga. Tinacal contains 45% of borax

  1. a) Preparation from Boric Acid

4H3BO3 + Na2CO3 ¾¾® Na2B4O7 + 6H2O + CO2

  1. b) Basic Nature:- aqueous solution of borax  is alkaline in nature due to its hydrolysis

Na2B4O7 + 3H2O ¾¾® NaBO2 + 3H3BO3

NaBO2 + 2H2O  NaOH + H3BO3

Strong alkali

  1. c) Action of heat:

Na2B4O7.10H2O  Na2B4O7  2NaBO2 + B2O3

(Anhydroussodium metaborate) anhydride  basic

Transparent glassy mass

When hot glassy mass is brought in contact with a coloured salt and heated again in the flame, B2O3 displaces the volatile oxides and reacts with basic oxides to form metaborates. Metaborates of basic radicals show characteristic colours. This test is known as borax bead test.

Colour of metaborates of Cu             Fe              Co                   Cr                    Ni

Blue     Green        Blue          Green              Brown

   5.    Compounds of Silicon

     

      5.1       Silicones

Silicones are organo – silicon polymers containing Si – O – Si linkages. These are formed  by the hydrolysis of alkyl or aryl substituted chlorosilanes and their subsequent polymerisation.

R3SiCl  R3SiOH

Silanol

¾¾®

Silicone

R2SiCl2 gives rise to straight chain linear polymers

RSiCl3 gives a complex cross linked polomer

These polymers are used in water proofing textiles in glassware,  as lubricants and anti foaming agents.

5.2       Silicon Carbide (Carborundum), SiC

It is obtained when  a mixture of sand, carbon common salt and saw dust is strongly heated in an electric furnace

SiO2 + 3C ¾¾® SiC + 2CO

It is chemically inert and resists the attack of almost all the reagents. It is used as an abrasive to make grind stones knife sharpness, etc.

Illustration 1:    What is etching of glass? How does to happens?

 

Solution:        Glass is chemically attacked by HF

This is known as etching of glass due to corrosive action of HF

Na2SiO3 + 6HF ¾® 2NaF + SiF4­ + 3H2O

CaSiO3 + 6HF ¾® CaF2  SiF4­ + H2O

SiF4 + 2HF ¾® H2SiF6

  1. Compounds of Nitrogen

 

      6.1       Oxides of Nitrogen

6.1.1    Nitrous Oxide (N2O)

 

It is used for the preparation of azides

N2O + NaNH2  ¾¾® NaN3 + H2O

6.1.2    Nitric oxide (NO)

 

The liquid and solid states are diamagnetic, becuause close dimers are formed

 

                        Reaction with Cl2

2NO  + Cl2 ¾¾® 2NOCl

(nitrosyl chloride)

6.1.3    Nitrogen oxide

 

The gas condenses to a brown liquid which turns paler on cooling and eventually becomes a colourless solid, due to dimerization.

2NO2  N2O4

 

Liquid N2O4 is useful as a non – aqueous solvent. It self – ionizes as

N2O4  NO+ + NO3

acid         base

A typical acid base reaction is

NOCl + NH4NO3 ¾¾® NH4Cl + N2O4

Acid       base           salt         solvent

6.1.4    Nitrogen trioxide or nitrogen sesquioxide (N2O3)

 

N2O3 + H2O ¾¾® 2HNO2

(anhydride of nitrous acid)

N2O3 + 2HClO4 ¾¾® 2NO [ClO4]  + H2O

6.1.5    Di nitrogen pentoxide [N2O5]

It is the anhydride of HNO3. It is a strong oxidizing agent

 

   6.2       Oxy Acids of Nitrogen

1.   H2N2O2 Hyponitrous acid
2.   H2 NO2 Hydronitrous acid
3.   HNO2 Nitrous acid  
4.   HNO3 Nitric acid  
5.   HNO4 Per nitric acid

 

Illustration 2:    If starting reagents are elements, what sequence of reactions leads to N2O?

 

Solution:        On passing air through an electric arc, first nitric oxide, NO is formed

N2 + O2 ¾® 2NO

When nitric oxide is allowed to come in contact with moist iron fillings, nitrous oxide. N2O is formed.

2NO + H2O + Fe ¾® N2O + Fe(OH)2

Illustration 3:    How are the following prepared?

  1. i) N­2H­4
  2. ii) NH2OH

                        iii)  H3PO2

 

Solution:  i)    NH3 + NaOCl NH2Cl + NaOH

NH2Cl + NH3 ¾® NH2.NH2 + HCl

NaOH + HCl ¾® NaCl + H2O

  1. ii) NaNO2 + NaHSO3 ¾® HON(SO3Na)2 + Na2SO3 + H2O

HON(SO3Na)2 + H2O ¾® HONH(SO3Na) + NaHSO4

HONH(SO3Na) + H2O ¾® HONH2 + NaHSO4

iii)   8P + 3Ba(OH)2 + 6H2O ¾® 2PH3­ + 3Ba(H2PO2)2

Excess of Ba(OH)2 removed as BaCO3 by passing CO2 into the solution

Ba(H2PO2)2 + H2SO4 ¾® 2H3PO2 + BaSO4¯

  1. Phosphorus Chemistry

 

      7.1       Oxides of Phosphorus                          

(White)                           (Density = 2.1)

(Poisionous)                 (non poisonous polemer of P4 unit)

Structure of Phosphorus pentoxide (P4O10)

P4O10        +   6H2O   ¾¾®             4H3PO4

(anhydride of

phosphoric  acid)

Phosphorus trioxide : [P4O6]

P4O6 + 6H2O 4H3PO3

P4O6 + 6H2O        PH3 +  3H3PO4

7.2       Oxy – Acids of Phosphorus

 

1.       H3PO2 Hypophosphorus acid
2.       H3PO3 Phosphorus acid
3.     H4P2O6 Hypophosphoric acid
4.       H3PO4 Orthophosphoric acid
5. H4P2O7 Pyrophosphoric acid
6.       HPO3 Metaphosphoric acid

 

  1. Sulphur Chemistry

8.1       Sulphur Oxides and Oxy-acid

Sulphur forms several oxides of which sulphur dioxide (SO2) and sulphur trioxide (SO3) are important.

Sulphur dioxide:

Preparation:  It is formed by burning sulphur in air or roasting metal sulphides in the presence of air

S8 + 😯2 ® 8 SO2

4FeS2 + 11O2 ® 2Fe2O3 + 8SO2

Properties:

  1. As reducing agent
  2. i) Action on halogens :

SO2 + Cl2 +2H2O ¾¾® H2SO4 + 2HCl

  1. ii) Action on FeCl3 :

2FeCl3 + SO2 + 2H2O ¾¾® H2SO4 + 2FeCl2 + 2HCl

  1. Reaction with acidified KMnO4 :

      2KMnO4 + 5SO2 + 2H2O ¾¾® K2SO4 + 2MnSO4 + 2H2SO4

  1. Reaction with acidified K2Cr2O7 :

      K2Cr2O7 + 3SO2 + H2SO4 ¾¾® K2SO4 + Cr2(SO4)3 + H2O

  1. Oxidising property
  2. i) 2H2S + SO2 ¾¾® 2H2O + 3S¯
  3. ii) SO2 + 2Mg ¾¾® 2MgO + S¯

Sulphur trioxide:

Preparation:  i)    By catalytic oxidation of sulphur dioxide

 

  1. ii) By dehydration of H2SO4

H2SO4 SO3 + H2O.

      8.2       Oxo Acids of Sulphur

 

H2SO3 Sulphurous aicd
H2S2O5 di-or pyrosuphurous acid
H2S2O4 Dithionous acid
H2S2O3 Thiosulphuric acid
H2S2O7 di or pyrosulphuric acid
H2S2O6 dithionic acid
H2SnO6: Polythionic acid  (n = 1 to 12)
H2SO5 Peroxymonosulphuric acid
H2S2O8 Peroxidisulphuric acid

      8.3       Sulphuric acid or Oil of Vitriol

Preparation: It is prepared by contact process – Sulphur trioxide from the catalytic chamber is passed through sulphuric acid to obtain oleum, H2S2O7. Dilution of oleum with water gives H2SO4 of the desired concentration

SO3 + H2SO4  ¾®    H2S2O7

H2S2O7 + H2O      ¾®    2H2SO4

Properties: Chemical reactions of sulphuric acid are as a result of the following characteristics :

  1. Low volatility
  2. b) Strong acidic character
  3. c) Strong affinity to water
  4. d) Ability to act as an oxidising agent.
  5. i) Low volatility of sulphuric acid is put to use in the manufacture of more volatile acids from their salts.

2MX + H2SO4   ®    2HX + M2SO4.

(X = F, Cl, NO3)

 

  1. ii) sulphuric acid is a strong dehydrating agent. Many wet gases can be dried by passage through a sulphuric acid bubbler provided the gases do not react with the acid. Sulphuric acid removes water from organic compounds as shown by its charring action on carbohydrates.

C12H22O11 + 11H2SO4  ®  12C + 11H2SO4 + 11H2O.

iii)   Hot conc. sulphuric acid is a moderately strong oxidising agent. In this respect, it is intermediate between phosphoric and nitric acids. Both metals and non-metals are oxidised by conc. sulphuric acid, which is reduced to SO2.

C + 2H2SO4    ®  CO2 + 2SO2 + 2H2O

Cu + 2H2SO4  ®  CuSO4 + SO2 + 2H2O

8.4       Sulphurous acid

Preparation:  It is formed when SO2 is dissolved in water

SO2 + H2O ¾¾® H2SO3

Properties :

  1. i) It acts as reducing agent and its chemical properties are similar to those of solution e.g.

2FeCl3 + (SO2 + H2O) + H2O ¾¾® 2FeCl2 + H2SO4 + 2HCl

  1. ii) It reacts with iron, forming ferrous sulphite and ferrous thiosulphate.

2Fe + 2H2SO3 ¾¾® FeSO3 + FeS2O3 + 3H2O

8.5       Sodium thiosulphate (Na2S2O3 .5H2O)

If one of the oxygen atoms in the sulphate ion is replaced by sulphur, the resulting ion (S2O32-) is as known thiosulphate.

Preparation:

  1. i) Sodium thiosulphate is prepared by boiling aq. solution of metal sulphites with elemental sulphur.

Na2SO3 + S8 Na2S2O3.

Hydrated sodium thiosulphate Na2S2O3 .5H2O is known as HYPO

  1. ii) Spring’s reaction may be used for the preparation of sodium thiosulphate.

Na2S + Na2SO3 + l2 ® Na2S2O3 + 2Nal

Properties:

  1. i) Reaction with dilute acids : It reacts with dilute acids to liberate sulphur dioxide gas alongwith precipitate of sulphur

Na2S2O3 + 2HCl ¾® 2NaCl + H2O + S¯ + SO2

  1. ii) Reaction with BaCl2 : It gives white ppt. of barium thiosulphate.

S2O+ Ba2+ ¾® BaS2O3 ¯

white

iii) Reaction with Silver Nitrate Solution : Gives white ppt. which quickly changes to yellow, brown and finally black due to the formation of silver sulphide.

S2O+ 2Ag+ ¾® Ag2S2O3 ¯

white ppt.

Ag2S2O3 + H2O ¾® Ag2S + H2SO4

      With conc. solution of sodium thiosulphate, silver nitrate gives no ppt.

  1. iv) It reacts with silver salts to form sodium argento thiosulphate complex

AgBr + 2Na2S2O3 ¾® Na3 [Ag(S2O3)2]

Sodium argento

thiosulphate complex

  1. v) Thiosulphate () ion is oxidized by iodine I2 to tetrathionate () ion

2Na2S2O3 + I2 ¾¾® Na2S4O6 + 2NaI.

8.6       Hydrogen Sulphide

Preparation: Prepared by the action of dil. HCl or H2SO4 on iron sulphide

FeS + 2HCl (dil) ¾¾® FeCl2 + H2S­.

Properties: It is a colourless, poisonous gas having the smell of rotten eggs.

  1. As reducing agent
  2. i) Action on halogens :

H2S + Cl2 ¾¾® 2HCl + S¯

  1. ii) Action on FeCl3 :

2FeCl3 + H2S  ¾¾® 2FeCl2 + 2HCl + S¯

  1. Reaction with acidified KMnO4 :

2KMnO4 + 3H2SO4 + 5H2S ¾¾® K2SO4 + 2MnSO4 + 8H2O + S¯

  1. Reaction with acidified K2Cr2O7 :

K2Cr2O7 + 4H2SO4 + 3H2S  ¾¾® K2SO4 + Cr2(SO4)3 + 7H2O + S¯

Illustration 4:    What changes occur when sulphur is gradually heated to its boiling point 444°C  and then to 200°C.

 

Solution:

      8.7       Allotropy in Sulphur

Sulphur exists in three important allotropic forms.

  1. Rhombic sulphur
  2. Monoclinic sulphur
  3. Plastic sulphur
  4. Rhombic sulphur or octahedral sulphur or 2-sulphur

It is a bright yellow solid soluble in CS­2 but insoluble in H2O

It exists as S8 molecules

  1. Monochlinic sulphur or b-sulphur

It is dull yellow in colour and is soluble in CS2 but insoluble in H2O. It is stable only above 369 k. Below this temperature, it slowly changes to rhombic sulphur. At 369 k both sulphur can exist. This temperature is called transition temperature.  It also exists as S8 molecules with puckered (non planar) ring structure but symmetry is different.

  1. Plastic sulphur or Amorphous sulphbur or g-sulphur.

It is amorphous form of sulphur. It has rubber like transparent yellow threads and is insoluble in CS2 and H2O Plastic sulphur is regarded as a super cooled liquid. It has random long inverted chains of sulphur atoms.

Illustration  5:   Why does sulphur begin to melt below its actual melting points.

Solution:        Monoclinic sulphur has a true melting point of 392 K. However, it often melts a few degree lower due to the break down of some of the S8 molecules.

Exercise 1:       Oxygen forms diatomic molecules (O2) while sulphur forms octa atomic molecules (S8) explain.

  1. Halogens

 

      9.1 Inter halogen compounds

Halogens react with each other to form a number of compounds called inter-halogen compounds. These compounds are named as halogen halides.

ICl3 Iodine trichloride

IF7 Iodine heptafluoride

All inter-halogen compounds are covalent compounds. These are generally more reactive than the individual halogens. The stability of the inter halogen compounds increases with the size of the central atom.

Shapes of some inter halogen compounds

Type XX¢n (n = 1) (with linear shape) Type XX¢n (n = 3) (with T-shape) XX¢n (n = 5)
(with square pyramidal shape)
XX¢n (n = 7) with pentagonal bipyramidal shape)
CIF ClF3 ClF5  
BrF BrCl BrF3 BrF5  
ICl, IBr, IF ICl3, IF3 IF5 IF7

      9.2 Properties of Hydrogen Halides

Hydrohalic acids (HF, HCl, HBr and HI)

  1. Physical Nature : Except HF, all others HCl, HBr, HI are gases HF is a liquid due to intermolecular hydrogen bonding
  2. Acidic Strength : All act as acids in their aq. solution and acidic strength varies in the order :       HF < HCl < HBr < HI

HX ¾¾® H+ + X

It can be explained in terms of strength of H – X bonds which is in the order :

H – I < H-Br < H-Cl < H-F.

Preparation and Properties:

  1. Less volatile acids displace more volatile acids from their salts.

2NaCl + H2SO4 ¾¾® Na2SO4 + 2HCl ­

HBr and HI cannot be prepared by conc. H2SO4 as they are more powerful reducing agents and reduces conc. H2SO4 to SO2

2KBr + H2SO4 ¾¾® K2SO4 + 2HBr

H2SO4 + 2HBr ¾¾® Br2­ + SO2 + 2H2O

  1. Less volatile, non -oxidising H3PO4 is used to prepare HBr and HI

KBr + H3PO4 ¾¾® KH2PO4 + HBr­

KI + H3PO­4 ¾¾® KH2PO4 + HI ­

Properties:

1    All the three acids are reducing agents  HCl is not attacked by H2SO4.

2HBr + H2SO4 ¾¾® 2H2O + SO2 + Br2 ­

2HI + H2SO4  ¾¾® 2H2O + SO2 + I2

  1. All the three react with KMnO4 and K2Cr2O7

2KMnO4 + 16HCl ¾¾® 2KCl + 2MnCl2 + 8H2O + 5Cl2 ­

K2Cr2O7 + 14HBr ¾¾® 2KBr + 2CrBr3 + 7H2O + 3Br2­

Other reactions are similar.

  1. Dipole moment HI < HBr < HCl < HF
  2. Bond length HF < HCl < HBr < HI
  3. Bond strength HI < HBr < HCl < HF
  4. Thermal stability HI < HBr < HCl < HF
  5. Acid strength HF < HCl < HBr < HI
  6. Reducing power HF < HCl < HBr < HI

Exercise 2:       Dry chlorine does not bleach clothes. Explain why?

Exercise 3:       Explain why?

  1. i)    I2 dissolves more in KI solution than in water?
  2. ii) In the preparation of HI form KI, phosphoric acid is preferred to sulphuric acid.

                              iii)  Bond dissociation energy of F2 is 38 Kcal/  mol and that for Cl2 is 57 Kcal/mol.

 

      9.3       Pseudohalide ions and pseudohalogens

Ions which consist of two or more atoms of which at least one is nitrogen and have properties similar to those of halide ions are called pseudohalide ions. Some of these pseudohalide ions can be oxidised to form covalent dimers comparable to halogens (X­2). Such covalent dimers of pseudohalide ions are called pseudohalogens.

Examples

Pseudohalide ions                      Pseudohalogen

CN, cyanide ion                           (CN)2 cyanogen

OCN, cyanate ion                       (SCN)2 thiocyanogen

SCN, thiocyante ion

SeCN, selenocyanate ion

NCN2–       Cyanamide ion

N3            azide ion

ONC fulminate ion

The best known psuedohalide ion is CN

Some important stable compound of Xenon

XeO3         Pyramidal

XeO4         Tetrahedral

XeOF4       Square pyramidal

XeO2F2       Distorted octahedral

First rare gas compound discovered was Xe+ (PtF6] by Bartlett.

      9.4       Oxyacids of Chlorine

Different oxyacids of chlorine are:

Formula Name Corresponding Salt
HOCl         Hypochlorous acid Hypochlorites
HClO2         Chlorous acid Chlorites
HClO3         Chloric acid Chlorates
HClO4         Perchloric acid         Perchlorates

Acidic Character: Acidic character of the same halogen increases with the increase in oxidation number of the halogen:

HClO4 > HClO3 > HClO2 > HOCl

Reason : It is because the release of H+ ion in each case would result in the formation of
ClO4 , ClO3, ClO2 and ClO ions. Now more is the number of oxygen atoms in the ion greater is the dispersal of the negative charge and hence more is the stability of resulting ion. Since a more stable ion would be formed relatively with more ease, therefore, the ease of formation  of ions would be

ClO4 > ClO3 > ClO2 > ClO

Preparation :

HOCl :            Ca(OCl)2 + 2HNO3  ¾® Ca(NO3)2 + 2HOCl

2HgO + H2O + 2Cl2   HgO.HgCl2 ¯ + 2HOCl

H2O2

(insoluble basic mercury chloride)

HClO2 :     BaO2 + 2ClO2 ¾¾® Ba(ClO2)2 (liquid) + O2

Ba(ClO2)2 + H2SO4(dil.) ¾® BaSO4 ¯ + 2HClO2

HClO3 :     6Ba(OH)2 + 6Cl2  ¾® 5BaCl2 + Ba(ClO3)2 + 6H2O

Ba(ClO3)2 + H2SO4(dil.) ¾® BaSO4 ¯ + 2HClO3

HClO4 :     (a) KClO4 + H2SO4 ¾¾® KHSO4 + HClO4

(b) 3HClO3 ¾¾® HClO4 + 2ClO2 + H2O

 

Illustration 6:    How are F2O and Cl2O prepared?

 

Solution:        2F2 + ¾® 2NaF + H2O + F2

2Cl2 + 2HgO HgCl2××    HgO + Cl2

  1. Compounds of Silver

Silver Nitrate, AgNO3(Lunar Caustic)

Preparation :

It is prepared by dissolving the metal in dilute nitric acid and crystallizing the solution

3Ag + 4HNO3  ¾¾®  3AgNO3 + 2H2O + NO­

Properties:

D
  1. i) On heating, it gives metallic silver and nitrogen dioxide

2AgNO3 2Ag + 2NO2 + O2

  1. ii) It reacts with iodine in two
  2. a) When iodine is in excess

5AgNO3 + 3I2 + 3H2O ¾® HIO3 + 5AgI + 5HNO3

  1. b) When AgNO3 is in excess

6AgNO3 + 3I2 + 3H2O ¾¾® AgIO3 + 5AgI + 6HNO3

iii)   When treated with alkali, it gives precipitate of silver oxide, which dissolves in excess of NH4OH

2AgNO3 + 2NaOH ¾¾® Ag2O ¯ + 2NaNO3 + H2O

brown ppt.

2AgNO3 + 2NH4OH ¾® Ag2O ¯ + 2NH4NO3 + H2O

Ag2O + 4NH4OH ¾® 2[Ag(NH3)2]OH + 3H2O

  1. iv) It gives turbidity with tap water (Cl) and turbidity is soluble in NH4

 

AgCl + 2NH4OH ¾¾® Ag(NH3)2Cl + 2H2O

(Soluble)

Silver Bromide (AgBr)

Preparation:  It is prepared by adding soluble bromide to a silver salt solution

AgNO3 + NaBr ¾® AgBr + NaNO3

Properties:

  1. It is pale- yellowish white solid, insoluble in water and conc. HNO3 but soluble in excess of NH4OH, KCN and Hypo solution.

AgBr + 2NH4OH ¾¾® [Ag(NH3)2] Br + 2H2O

AgBr + 2KCN ¾¾® K[Ag(CN)2] + KBr

AgBr + 2Na2S2O3 ¾¾® Na3[Ag(S2O3)2] + NaBr

  1. On heating it melts to red liquid
  2. Hydrogn Peroxide
  • Hydrogen peroxide :

      Preparation :

Lab Method : It is prepared by the action of cold, dilute sulphuric acid on sodium or barium peroxide

Na2O2 (s) + H2SO4(aq) ¾¾® H2O2(aq) + Na2SO4

BaO2.8H2O + H2SO4(aq) ¾® H2O2 (aq) + BaSO4(s)

Anhydrous barium oxide is not used because the precipitated BaSO4 forms a protective layer on the unreacted barium peroxide and thus prevents its further participation in the reaction. However it can be overcome by using phosphoric acid.

By Electrolysis: It can also be prepared by the hydrolysis of peroxydisulphuric acid which is obtained by the electrolytic oxidation of sulphuric acid

electrolysis

 

 

2H2SO4(aq)  ¾¾¾® H2S2O8 (aq) + H2(g)

2H2O

 

 

H2S2O8(aq) ¾¾¾® 2H2SO4(aq) + H2O2(aq)

By the auto-oxidation of 2-ethyl anthraquinol. The net reaction is a catalytic union of H2 and O2 to yield hydrogen peroxide.

 

 

2-ethyl anthraquinol                    (oxidised product) + H2O2

Properties:

  1. i) Unstable liquid, decomposes to give water and dioxygen and the reaction is slow in the absence of catalyst. It is catalysed by certain metal ions, metal powders and metal oxides.

2H2O2 (l) ¾¾® 2H2O (l) + O2 (g)

  1. ii) It acts as oxidant as well as reductant in both acid and alkaline medium. On the whole, hydrogen peroxide is a very powerful oxidising agent and poor reducing agent. Some typical oxidation and reduction reaction of hydrogen peroxide are as follows :

As oxidising agent

      In acidic medium:             H2O2 + 2H+ + 2e ¾¾® 2H2O

In basic medium :             H2O2 + OH + 2e ¾¾® 3OH-‑

      As reducing agent

      In acidic medium:             H2O2 ¾¾® 2H+ + O2 + 2e

In basic medium :             H2O2 + 2OH ¾¾® 2H2O + O2 + 2e

2Fe2+ + H2O2 + 2H+ ¾¾® 2Fe3+ + 2H2O

2MnO4 + 5H2O2 + 6H+ ¾¾® 2Mn2+ + 8H2O + 5O2

Mn2+ + H2O2 ¾¾® Mn+4 + 2OH

                                                        2Fe3+ + H2O2 + 2OH ¾® 2Fe2+ + 2H2O + O2

The oxidising property of hydrogen peroxide is put to use in the restoration of old paintings, where the original white lead paint has been converted to black PbS by the H2S in the atmosphere. Hydrogen peroxide oxidises the black PbS into white PbSO4.

PbS(s) + 4H2O2 (aq) ¾® PbSO4(s) + 4H2O

black                                                 white

Tests :

  1. a) It liberates iodine from potassium iodide in presence of ferrous sulphate
  2. b) Acidified solution of dichromate ion forms a deep blue colour with H2O2 due to the formation of CrO5. , The blue colour fades away gradually due to decomposition of CrO5 into Cr3+ ions and oxygen

Cr2O72- + 4H2O2 + 2H+ ¾¾® 2CrO5­ +5H2O

  1. c) With a solution of titanium oxide in conc.H2SO4, it gives orange colour due to the formation of pertitanic acid.

Ti4+ + H2O2 + 2H2O  ¾® H2TiO4 + 4H+

pertitanic acid

Exercise 4:       Statues coated with white lead on long exposure to atmosphere turn black and the original colour can be restored in treatment with H2O2 why?

Exercise 5:       Why hydrated barium peroxide is used in the preparation of H2O2 instead of the anhydrous variety?

 

  1. Carbon (Oxide and Carbides)
  • Oxides

Carbon burns in and forms two oxides, carbon monoxide, (CO) and Carbon dioxide (CO2).

Carbon Monoxide :

Preparation  :

  1. i) By heating carbon in limited supply of oxygen.

C +O2 ¾® CO.

  1. ii) By heating oxides of heavy metals e.g. iron, zinc etc with carbon.

Fe2O3 + 3C ¾® 2Fe + 3CO

ZnO + C ¾® Zn + CO

Two important industrial fuels water gas and producer gas contain carbon along with hydrogen and nitrogen, Water gas is obtained by passing steam over hot coke

C + H2O ¾® CO + H2

(water gas)

When air is passed over hot coke, producer gas is obtained.

2C + O2 + 4N2 ¾® 2CO + 4N2

(Producer gas)

Properties :

  1. i) It is a powerful reducing agent and reduces many metal oxides to their corresponding metals.

Fe2O3 + 3CO ¾® 2Fe + 3CO2

CuO + CO ¾® Cu + CO2

  1. ii) It burns in air to give heat and carbon dioxide

CO + O2 ¾® CO2 + heat.

Tests

  1. a) Burns with blue flame
  2. b) A filter paper soaked in platinum or palladium chloride is turned pink, green or black due to reduction of the chloride by carbon monoxide.

Carbon di-oxide:

Preparation:

  1. i) In the lab., it is prepared by the action of acids on carbonates.

CaCO3 + 2HCl ¾® CaCl2 + H2O + CO2

  1. ii) By combustion of carbon

C + O2 ¾® CO2

Properties :

  1. i) It turns lime water milky and milkiness disappears when CO2 is passed in excess

Ca(OH)2 + CO2 ¾® CaCO3 ¯ + H2O,   CaCO3 + H2O + CO2 ¾® Ca(HCO3)2

  1. ii) Solid carbon dioxide or dry ice is obtained by cooling CO2 under pressure. It passes from the soild state straight to gaseous state without liquefying (hence dry ice).

iii)   A burning candle is put out but burning magnesium continues burning in the gas jar.

  • Carbides

Carbon combines with more electropositive elements than itself when heated to high temperature to form carbides. Carbides are of mainly three types.

  1. i) Salt like Carbides : These are the ionic salts containing either C22– (acetylide ion) or C4- (methanide ion)e.g. CaC2, Al4C3, Be2
  2. ii) Covalent Carbides : These are the carbides of non-metals such as silicon and boron. In such carbides, the atoms of two elements are bonded to each other through covalent bonds.

SiC also known as Carborundum.

iii)   Interstitial Carbides : They are formed by transition elements and consist of metallic lattices with carbon atoms in the interstices. e.g. tungsten carbide WC, vanadium  carbide VC.

  1. Ammonia (NH3)

Preparation:

  1. Lab Method :
  2. i) By heating an ammonium salt with a strong alkali ; like NaOH either in solid form or when dissolved in water.

NH4Cl + NaOH ¾¾® NH3­ + NaCl + H2O

  1. ii) By the hydrolysis of magnesium nitride

Mg3N2 + 6H2O  ¾¾® 3Mg(OH)2 + 2NH3.

  1. It is manufactured by Haber’s process

N2(g) + 3H2(g)  2NH3(g).

      Properties:

  1. i) Basic nature : Its aq. solution is basic in nature and turns red litmus blue.

NH3 + H2O              + OH

  1. ii) Reaction with halogens :

Chlorine                       :     8NH3 + 3Cl2 ¾¾® 6NH4Cl + N2

Excess of chlorine      :     NH3 + 3Cl2  ¾® NCl3 + 3HCl

Bromine                      :     8NH3 + 3Br2 ¾¾® 6NH4Br + N2

Excess of bromine      :     NH3 + 3Br2 ® NBr3 + 3HBr

Iodine                          :     2NH3 + 3I2 ¾® NH3.NI3 + 3HI

Nitrogen

tri iodide ammonate

NH3.NI3 explodes in dry state

8NH3.NI3 ¾¾® 6NH4I + 9I2 + 6N2

iii)   Complex formation : Due to the presence of lone pair of electrons on nitrogen, it acts as lewis-base. Thus it forms co-ordinate linkage with metal ions and these ammonia compounds find use in qualitative analysis

Ag+ + NH3 ¾¾® [Ag(NH3)2]+

Cu2+ + 4NH3 ¾¾® [Cu(NH3)4]2+

Cd2+ + 4NH3 ¾¾® [Cd(NH3)4]2+

  1. iv) Precipitation of heavy metal ions from the aq. solution of their salts : Heavy metal ions like Fe3+, Al3+, Cr3+ are precipitated from their aqueous salt solution.

FeCl3 + 3NH4OH ¾¾® Fe(OH)3 + 3NH4Cl

Brown ppt.

AlCl3 + 3NH4OH ¾¾® Al(OH)3 + 3NH4Cl

White ppt.

CrCl3 + 3NH4OH ¾¾® Cr(OH)3 + 3NH4Cl

Green ppt.

  1. Fertilzers

Substances which increase the fertility of soils are known as fertilizers.  They are classified into three categories :

  1. Nitrogeneous fertilizers : These are fertilizers which mainly supply nitrogen to the plants. e.g ammonium sulphate, ammonium nitrate, calcium ammonium nitrate, calcium cyanamide and urea
  2. Phosphatic fertilizers : They supply phosphorus to the plants. e.g. superphosphate of lime Ca(H2PO4)2
  3. Mixed fertilizers : Fertilizers containing more than one elements, namely nitrogen, phosphorus and potassium. They contain a mixture of ammonium salt, ammonium phosphate, superphosphate and potassium salt. It is known as NPK fertilizers

Phosphatic fertilizers such as superphosphate of lime is obtained from phosphatic rocks by treatment with conc. sulphuric acid. In this way, insoluble phosphate rock is rendered soluble in water for use as a source of this essential plant nutrient.

Ca3(PO4)2 + 2H2SO4 + 5H2O ¾¾® Ca(H2PO4)2 H2O + 2CaSO4. 2H2O.

soluble

Treatment of phosphate rock with phosphoric acid leads to the formation of triple superphosphate which is free from calcium sulphate and hence contains a greater percentage of phosphorus.

Ca5(PO4)3F + 7H3PO4 + 5H2O ¾® 5Ca(H2PO4)2.H2O + HF

  1. Bleaching Powder

The exact chemical composition of bleaching powder is not yet known but it behaves as if it contains calcium hypochlorite Ca(OCl)2 and basic calcium chloride, CaCl2.Ca(OH)2.H2O.

Preparation: It is prepared by passing chlorine over slaked lime

Properties:

  1. Reaction with Dilute Acids : With dilute acids, it gives chlorine which is known as available chlorine.

CaOCl2 + 2HCl ¾¾® CaCl2 + H2O + Cl2­

CaOCl2 + H2SO4  ¾¾® CaSO4 + H2O + Cl2­

  1. When treated with water it decomposes into calcium chloride and calcium hypochlorite

2CaOCl2 + H2O ¾® CaCl2 + Ca(OCl)2 + H2O

  1. Bleaching powder reacts with CO2 (atmospheric) and gives chlorine which accounts for its oxidising and bleaching actions.

CaOCl2 + CO2     ¾¾®     CaCO3   + Cl2 ­

  1. Action of Heat : On heating bleaching powder gives a mixture of chlorate and chloride

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

  1. Answers to Exercise

Exercise 1:     The minimum oxidation number (O.N.) of S is –2 while its maximum O.N. is +6 in SO3 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.

Exercise 2:     The bleaching action of chlorine is due to the liberation of nascent oxygen from  water

H2O + Cl2  ¾® 2HCl + [O]

 

Exercise 3:     i)    I2 is a covalent molecule. Thus its solubility is less in polar solvent (water). Potassium iodide combines with iodine to form a polyhalide which is an ionic compound. Being ionic KI3 is more soluble

KI + I2 ¾¾® KI3

  1. ii)   Besides acidic nature of sulphuric acid it acts as an oxidising agent. H2SO4 oxidises HI (reducing agent)  fromed from KI into Iodine. Thus H3PO4 is preferred as it does not oxidise HI.

2KI + 2H2SO4 ¾¾® 2KHSO4 + 2HI

H2SO4 + 2HI ¾¾® I2 + SO2 + 2H2O

iii)   The low dissociation energy of F2 is due to high inter-electronic repulsion between non – bonding electrons in the 2p orbitals as the size of fluorine atom is small. As a result F—F bond is weaker than Cl—Cl bond.

Exercise 4:     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

Exercise 5:     If anhydrous barium peroxide is used in the preparation, the barium sulphate, thus formed  forms an insoluble protective coating on the surface of solid barium peroxide. This prevents the further reaction of the acid, i.e., causing the reaction to stop. If, however, hydrated barium peroxide (in the form of this paste) is used, the water causes to dislodge the insoluble  BaSO4 from the surface of BaO2. BaSO4 thus settles at the bottom of the reaction vessel and the reaction continues without any difficulty.

 

  1. Solved Problems

     

      17.1     Subjective

Problem 1:       A solution of ferric chloride acidified with HCl is unaffected when hydrogen is bubbled through it, but gets reduced when Zn is added to same acidified solution –  why?

Solution:        Molecular hydrogen is not so reactive, Zn reacts with the acid to produce nascent hydrogen which reduces ferric chloride into ferrous chloride.

Problem 2:       Presence of water is avoided in the preparation of H2O2 from Na2O2  – why?

Solution:        Water reacts with Na2O2 to produce NaOH which increases the decomposition of H2O2.

Problem 3:       Boron and aluminium both are in the same group. Yet AlCl3 shows anomalous mol.wt. which BCl3 doesn’t – why?

Solution:        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.

Problem 4:       Calcium burns in nitrogen to produce a white powder which dissolves in sufficient water to produce a gas (A) and an alkaline solution. The solution on exposure to air produces a thin solid layer of (B) on the surface. Identify the compound (A) and (B).

Solution:        Ca + N2 ¾® Ca3N2 (white powder)

                        Ca­3N2 + 6H2O ¾® 3Ca(OH)2 +

Ca(OH)2 + CO3 ¾® + H2O

Problem 5:       Although boric acid B(OH)3 contains three hydroxyl groups yet it behaves as a mono-basic acid. Explain.

Solution:        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 s-electrons of the co-ordinated oxygen atom towards it. The co-ordinated O-atom, in turn pull the s-electrons of the O–H bonds, thereby facilitating the release of a proton.

B(OH)3 + NaOH ¾® Na [B(OH)4]  Na+ + 2H2O

Problem 6:       How do  and  ion differ structurally?

Solution          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

Problem 7:       When a blue litmus is dipped into a solution of hypo chlorous acid, it first turns red and then later gets decolourised . why? 

Solution:        HClO is an acid, thus turns blue litmus into red. HClO is an oxidising agent also and the nascent oxygen given by HClO bleaches the red litmus

Red litmus + O ¾¾® colour less

Problem 8:       The bleaching action of chlorine is permanent while that of sulphur dioxide is temporary – why?

Solution:        Chlorine bleaching action is due to oxidation while that of SO2 is due to reduction. Hence the substances bleached by SO2 is reoxidised by the oxygen of the air to its original state

Problem 9:       Iodine is liberated in the reaction between KI and Cu+ ions. But chlorine is not liberated when KCl is added to Cu+2 ions. Why?

Solution:        I is strong reducing agent it reduces Cu+2 ions to Cu+ ions.   The Cl ion is a weak reducing agent, thus it doesn’t reduce Cu+2 ions.

2Cu+2 + 4KI ¾¾® Cu2I2 + I2 + 4KI

Problem 10:     Give reason for decreasing order of conductivity of following

                        Cs+ > Rb+   > k+ > Na+ > Li+

Solution:        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+.

      17.2     Objective

Problem 1:       H2O2 is not

                        (A) a reducing agent                       (B) an oxidising agent

                        (C) a dehydrating agent                  (D) a bleaching agent

Solution:        (C)

Problem 2:       A chloride dissolves appreciably in cold water when placed on a Pt wire in Bunsen flame, no distinctive colour is noticed. The cation is 

                        (A) Mg+2                                          (B) Ba+2

                        (C) Pb+2                                          (D) Ca+2

Solution:        Magnesium does impart colour to flame

                        \ (A)

Problem 3:       Which of the following is obtained by oxidation of phosphorous by HNO3?

                        (A) H3PO4                                        (B) H3PO3

                        (C) H4P2O7                                       (D) H3PO2

Solution:        P4 + 20HNO3 ¾® 4H3PO4 + 20NO3 +  4H2O

                        \ (A)

Problem 4:       The blue colour mineral lapis lazuli which is used as a semi precious stone is a mineral of the following class –

                        (A) Sodium alumino silicate           (B) Zinc cobaltate

                        (C) Basic copper carbonate            (D) Prussian blue

Solution:        (A)

Problem 5:       Hydrolysis of PI3 yield –

                        (A) Monobasic acid and a salt       

                        (B) Monobasic acid and a dibasic acid

                        (C) A Monobasic base and a dibasic acid

                        (D) A Monobasic acid and tribasic acid

Solution:        (B)

Problem 6:       Decomposition of H2O2 is slowed down by addition of –

                        (A) Alcohol                                     (B) MnO2

                        (C) Alkali                                        (D) Pt

Solution:        (A)

Problem 7:       Which is/are true statement(s)?

                        (A) all halogens form oxy acids

                        (B) all halogens show –1, +1, +3, +5 and +7 oxidation states

                        (C) HF is a dibasic acid and attack glass

                        (D) oxidising power is in order F2 < Cl2 < Br2 < I2

Solution:        (C)

Problem 8:       The metal (s) soluble in aqua regia is (are)

                        (A) Pt                                             (B) Au

                        (C) Ag                                            (D) All the above

Solution:        (D)

Problem 9:       By burning NH3 in oxygen, it gives ________ and H2O –

                        (A) NO                                            (B) N2

                        (C) NO2                                           (D) N2O

Solution:        (B)

Problem 10:     P4  + 5O2 ¾® X Y, Y is

                        (A) H3PO4                                        (B) P2O5

                        (C) H3PO3                                        (D) H4P2O7

Solution:        P4 + 5O2 ¾® P4O10 4H3PO4 

                        \(A)

  1. Assignment (Subjective Problems)

 

LEVEL – I

  1. The size of d orbital, decrease : Si > P > S > Cl but p bonding increases in the same order. Explain.
  2. Explain the stability of oxides of alkali metals.
  3. BaO2 is a peroxide but PbO2 is not a peroxide why?
  4. Why nitric acid cannot be used to prepare H2S.
  5. Ferric iodide is very unstable but ferric chloride is stable. Why?
  6. Element A burns in nitrogen to give an ionic compound B. Compound B reacts with water to give C and D. A solution of C becomes milky on bubbling carbon dioxide. Identify A, B, C and D.
  7. Why dry SO2 does not bleach dry flowers?
  8. Why H2S is stronger acid than H2O?
  9. In the manufacture of sulphuric acid by the contact process, sulphur trioxide is not directly dissolved in water. Why?
  10. Conc. H2SO4 cannot be used for drying H2. Explain

 

 

 

 

 

 

 

 

 

 

 

 

 

LEVEL – II

  1. Complete the following reactions.
  2. a) KBr + ICl ¾®
  3. b) KF + BrF3 ¾®
  4. Anhydrous AlCl3 cannot be prepared by heating hydrated AlCl3.6H2O – why?
  5. Why AlF3 is ionic while AlCl3 is covalent?
  6. The polarity of B—X bonds is in the order B—F > B—Cl > B—Br but Lewis  acidity order is BF3< BCl3 < BBr3. Explain.
5.         Explain the solubility of salts of alkali and alkali earth metals
  1. Mg3N2 when reacted with water gives off NH3 but HCl is not obtained from MgCl2 on reaction with water at room temperature. Why?
  2. Why calcium cyanide is used as fertilizer?
  3. Pure HI kept in a bottle acquires a brown colour after sometime. Explain
  4. Why zinc chloride cannot be dehydrated on heating. Explain
  5. Explain why a solution of ferric chloride gives brown precipitate.

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

LEVEL – III

  1. Hydrogen peroxide acts both as an oxidising and as a reducing agent in alkaline solution towards certain first row transitional metal ions. Illustrate both these properties of H2O2 using chemical equations.
  2. Compound (A)
  3. i) On strong heating gives two oxides of sulphur
  4. ii) On adding aqueous NaOH solution to its aqueous solution, a dirty green ppt is obtained which starts turning brown on exposure to air.

Identify (A) and give chemical equations involved.

  1. i)    A black mineral (A) to treatment with dilute NaCN solution in presence of air gives a clear solution of (B) and (C).
  2. ii) The solution of (B) on reaction with Zn gives precipitate of a metal (D).

iii)   (D) is dissolved in dil. HNO3 and the resulting solution gives a white ppt. (E) with dil. HCl.

  1. iv) (E) on fusion with Na2CO3 gives (D).
  2. v) (E) dissolves in aqueous solution of NH3 giving a colourless solution of (F)
  3. vi) Identify (A) to (F)
  4. An inorganic compound (A) shows the following reaction.
  5. i) It is white solid and exists as dimer; gives fumes of (B) with wet air.
  6. ii) It sublines in 180°C and form monomer if heated to 400°

iii)   Its aqueous solution turns blue litmus to red.

  1. iv) Addition of NH4OH and NaOH seperately to a solution of (A) gives white ppt. which is however soluble in excess of NaOH.
  2. What happens when:
  3. a) Ferric  chloride is added to potassium ferrocyanide
  4. b) Iron reacts with cold dilute nitric acid
  5. c) Green vitrol is strongly heated
  6. How will you prepare
  7. a) Copper oxide from copper sulphate
  8. b) Ferrous sulphate from Mohr’s salt
  9. c) Anhydrous ZnCl2 from white vitriol
  10. Carbon tetrachloride is not affected but silicon tetrachloride is hydrolysed by water. Why?
  11. Explain the structure of C3O2 is term of bonding in the molecule.
  12. Pure PH3 does not burn in air but impure sample of PH3 burns in air. Why?
  13. (SiH3)3N is a weaker base than (CH3)3N, Why?

 

 

  1. Assignment (Objective Problems)

Level – I

  1. Which of the following oxides of nitrogen combines with Fe (II) ions to form a dark brown complex?

(A) N2O                                                     (B) NO

(C) NO2                                                     (D) N2O5

  1. Of the following acids

I : hypo phosphorous acid                        II: hydroflouric acid

III: oxalic acid                                            IV: glycine

(A) I, II are monobasic, III dibasic acid and IV amphoteric

(B) II monobasic, I, III dibasic acid, IV amphoteric

(C) I monobasic, II, III dibasic, IV amphoteric

(D) I, II, III dibasic, IV amphoteric

  1. In the following statements, select the correct statement :

(A) N(CH3)3 has pyramidal structure

(B) N(SiH3)3 shows planar arrangement

(C) both correct

(D) none is correct

  1. A solution of sodium in liquid ammonia is strongly reducing agent due to the presence of

(A) Na atoms                                            (B) Sodium hydride

(C) sodium amide                                     (D) solvated electron

  1. Amongst sodium halides NaF has the highest m.p. because it has

(A) highest oxidising power                       (B) lower polarity

(C) minimum ionic character                    (D) maximum ionic character

  1. Crude common salt is hygroscopic because of impurities of

(A) CaSO4 and MgSO4                            (B) CaCl2 and MgCl2

(C) CaBr2 and MgBr2                               (D) Ca(HCO3)2 and Mg(HCO3)2

  1. Which of the following is obtained by the reaction of Cu and Conc. H2SO4?

(A) S                                                          (B) SO3

(C) SO2                                                     (D) H2

  1. A gas which burns with blue flame is –

(A) CO                                                      (B) O2

(C) N2                                                  (D) CO2

  1. When lead storage battery is discharged –

(A) SO2 is evolved                              (B) lead sulphate is consumed

(C) lead is formed                               (D) sulphuric acid is consumed

  1. In white phosphorus the incorrect statement is

(A) Six P – P single bonds are present

(B)  Four P – P single bonds are present

(C) Four lone pairs of electrons are present

(D) PPP bond angle is 60

  1. Anhydrous _______ is a very effective dessiccant (water absorber used in dry battery. It is

(A) conc. H2SO4                                       (B) P2O5

(C) CaCl2                                                  (D) MgClO4

  1. Which is least basic among the following

(A) NF3                                                      (B) NCl3

(C) NBr3                                                    (D) NI3

  1. Which of the following is the strongest oxidising agent?

(A) N2O                                               (B) NO

(C) NO2                                               (D) N2O5

  1. The inertness of nitrogen is due to its –

(A) high electronegativity                    (B) small atomic radius

(C) high dissociation energy               (D) stable configuration

  1. Nitrolim is –

(A) CaC2                                             (B) CaCN2 + C

(C) CaC2N2                                         (D) CaC2 + CaCN

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

LEVEL – II

  1. Molecular formula of Glauber’s salt is

(A) MgSO4,7H2O                                      (B) FeSO4,7H2O

(C) CuSO4,5H2O                                      (D) Na2SO4, 10H2O

  1. A solution of Na2SO4 in water is electrolysed using inert electrodes. The products at the cathode and anode are respectively

(A) H2,O2                                                  (B) O2,H2

(C) O2,Na                                                  (D) O2,SO2

  1. Which of the following does not give flame test?

(A) Na                                                       (B) Sr

(C) K                                                         (D) Zn

  1. In the electrolysis of alumina, cryolite is added to alumina to

(A)                                                             Lower the m.p. of alumina

(B)                                                             Increase the electrical conductivity

(C) Minimise the anode effect

(D) Remove impurities from alumina

  1. Butter of  tin is

(A) (NH4)2SnCl6                                        (B) SnCl2 + Sn(OH)2

(C) SnCl4. 5H2O                                       (D) H2SnCl4

  1. The metal X is prepared by the electrolysis of fused chloride. It reacts with hydrogen to form a colourless solid from which hydrogen is released on treatment with water. The metal is

(A) Al                                                         (B) Ca

(C) Cu                                                       (D) Zn

  1. Which of the following statement about anhydrous AlCl3 is corect?

(A) It exists as AlCl3 molecules

(B) It is a strong Lewis base

(C) It is sublimes at 100°C under  vacuum

(D) it is not easily hydrolysed

  1. The material used in solar cell is

(A) Si                                                         (B) Sn

(C) Ti                                                         (D) Cs

  1. White phosphorus may be removed from red phosphorus by

(A) sublimation                                          (B) distillation

(C) dissolving in CS2                                 (D) heating with an alkali solution

  1. Non combustible hydride is –

(A) NH3                                               (B) PH3

(C) AsH3                                             (D) SbH3

  1. Which has S – S bonds

(A) H2S2O3                                               (B) H2S

(C) H2S2O6                                               (D) S3O9

  1. While testing , there is a green – edged flame on heating the salt with conc. H2SO4 and CH3OH, green colour is of

(A) (CH3)3B                                               (B) (CH3O)3B

(C) B2O3                                                   (D) H3BO3

  1. NO2 is not obtained when following is heated

(A) Pb(NO3)2                                            (B) AgNO3

(C) LiNO3                                                  (D) KNO3

  1. Acid rain may cause

(A) rusting easier                                       (B) stone  cancer in Taj Mahal

(C) non – fertility of soil                             (D) all are correct

  1. Following are neutral oxide except

(A) NO                                                      (B) N2O

(C) CO                                                      (D) NO2

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

  1. Answers to Objective Assignments

 

Level  – I

 

  1. B                                                               2.         A
  2. C                                                               4.         D
  3. D                                                               6.         B
  4. C                                                               8.         A
  5. D                                                               10.       B
  6. D                                                               12.       A
  7. D                                                               14.       C
  8. B

 

level –II

 

  1. D                                                               2.         A
  2. D                                                               4.         B
  3. C                                                               6.         B
  4. C                                                               8.         A
  5. B                                                               10.       A
  6. C                                                               12.       B
  7. D                                                               14.       D
  8. D

6

Preparation & Properties of Compounds

 

Subjective Problems

 

Level – I

  1. In silicates pp – dp bonding is absent due to non – availability of d orbitals.. Hence in silicates the SiO4 units are held together in polymeric forms by s bonds. From P to Cl, d – orbitals are available for hybridization and pp – dp 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 p bonding in strong but it still stronger in the oxyacids of S where pp – dp bonding becomes a dominant feature and only a small amount of polymerization occurs. At Cl, pp – dp bonding is so strong that no polymerization of oxyanions occurs.
  2. Li forms only oxide, Na forms oxide and peroxides and remaining all alkali metal form oxides, peroxides and super oxides.

Li + O2 ¾®

4Na + O2 ¾®

2Na + O2 ¾®

K + O2 ¾®

Peroxide [–O–O–)2 has not 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 peroxide and super oxide are larger. 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 ¾® 2K2CO3 + 3O2

4KO2 + 4CO2 + 2H­2O ¾® 4KHCO3 + 3O2

  1. 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.
  2. Since nitric acid is an oxidising agent it oxidises H2S to S.

3H2S + 2HNO3 ¾¾® 2NO + S + 4H2O

  1. I is a strong reducing agent. Iodide thus reduces ferric ion into ferrous ion. Cl is a weak reducing agent hence doesn’t reduces Fe+3 to Fe+2.
  2. ;

 

  1. Nascent hydrogen is responsible for bleaching, according to the following reaction

SO2 + 2H2O ¾® H2SO4 + 2H

Coloured flower + H ¾® Colourless flower

  1. The H – S bond is weaker than H – O bond and the proton is easily released when H2S reacts with water.
  2. SO3 reacts with water to form H2SO4 which is not easily condensed and escapes in the atmosphere. It is therefore, absorbed in conc. H2SO4 to form oleum which may be diluted to form H2SO4.

H2SO4 + SO3 ¾® H2S2O7

H2S2O7 + H2O ¾® 2H2SO4

  1. Hydrogen, sometimes, ignites the large amount of heat released when water is absorbed by acid.

 

Level – II

  1. a) K[BrICl]
  2. b) K[BrF4]
  3. Hydrolysis of AlCl3 takes place forming Al2O3

2AlCl3 + 3H2O ¾¾® Al2O3 + 6HCl

  1. Because of the Fajan’s Rule.
  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 pp – pp 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 > BBr3 resulting into B—F double bond character and least basicity and follows the order BF3 < BCl3 < BBr3

(BF3 < BCl3 < BBr3)

  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 ³ lattice energy.
  MF

 

½

½

¯

Solubility increases

MCl

 

½

½

¯

Decreases

MBr

 

½

½

¯

Decreases

MI

 

½

½

¯

Decreases

Li
Na
K
  1. MgCl2 is a salt of strong acid and a strong base, so its hydrolysis is not possible, but Mg3N2 is a salt of strong base Mg(OH)2 and a weak acid NH3, so it get hydrolysed to give NH3.
  2. In soil, it is converted into urea, which then decomposes into ammonia. This is assimilated by the plants.

CaCN2 + H2O + CO2 ¾®  + CaCO3

                CN.NH2 + H2O ¾®

NH2.CONH2+ H2O ¾® 2NH3 + CO2

  1. HI is a strong reducing agent. It is even oxidised by oxygen of the air. The iodine is liberated which is dissolved imparting a brown colour to solution.
  2. On heating it reacts with water of crystallisation and forms zinc oxychloride.

3ZnCl2.2H2O ¾® Zn2OCl2 + 2HCl + 3H2O

  1. Hydrolysis of ferric chloride occurs giving finally brown precipitate

FeCl3 + 3H2O  Fe(OH)3 + 3HCl

2Fe(OH)3 ¾®  + 3H2O

Level – IIi

 

  1. As oxidising agent

H2O2 ¾¾® H2O + O

 

As reducing agent:

2K3Fe(CN)6 + 2KOH ¾® 2K4Fe(CN)6 + H2O + O

H2O2 + O ¾® H2O + O2

———————––––––––––—————————————

2K3Fe(CN)6 + 2KOH + H2O2 ¾®2K4Fe(CN)6 + 2H2O + O2

  1.  Fe2O3 + SO2 + SO3

FeSO4 + 2NaOH ¾¾® Fe(OH)2 + Na2SO4

Fe(OH)2 Fe(OH)3

  1. (A) Ag2S (a black mineral)

(B) Na[Ag(CN)2]

(C) Na2S

(D) Ag

(E) AgCl

(F) Ag(NH3)2Cl

The reaction are

  1. Ag2S + 4NaCN ¾® 2Na[Ag(CN)2] + Na2S
  2. 2Na[Ag(CN)2] + Zn  ¾®  Na2[Zn(CN)4] + 2Ag
  3. 3Ag + 4 HNO3 ¾® 3AgNO3 + NO + 2H2O
  4. AgNO3 AgCl + HNO3
  5. AgCl + 2NH3 ¾® Ag(NH3)2Cl
  6. 4AgCl + 2Na2CO3 ¾® 4Ag + 4NaCl + 2CO2 + O2
  7. 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) Al2Cl6 2AlCl3

  1. ii) It fumes with wet air

Al2Cl6 + 6H2O 3Al(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

Al2Cl4 + 6NH4OH  ¾® 2Al(OH)3 + 6NH4Cl

Al2Cl6 + 6NaOH  ¾® 2Al(OH)3 + 6NaCl        2NaAlO2 + 2H2O

  1. a) On addition of FeCl3 to K4Fe(CN)6 a Prussian blue coloured is formed.

4FeCl3 + 3K4Fe(CN)6 ¾¾®  + 12KCl

  1. b) Ammonium nitrate is formed

[Fe + 2HNO3 ¾¾® Fe(NO3)2 + 2H] ´ 4

HNO3 + 8H ¾¾® NH3 + 3H2O

NH3 + HNO3 ¾¾® NH4NO3

———————————————————

4Fe+ 10HNO3 ¾¾® 4Fe(NO3)­2 + NH4NO3 + 3H2O

  1. c) When heated strongly, a mixture of gases consisting SO2 and SO3 is evolved and a red residue, Fe2O3 is formed

FeSO4.7H2O ¾¾® FeSO4 + 7H2O ] ´ 2

2FeSO4 ¾¾® Fe­2O3 + SO2 + SO3

SO3 + H2O  ¾¾® H2SO4

————————————————————

¾¾®  + SO2 + H2SO4 + 13H2O

  1. a) CuSO4 + 2NaOH ¾¾®  + Na2SO4

Cu(OH)2  CuO + H2O

CuO  +  ¾¾® Cu2O

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

Fe(OH)2 + H2SO4 ¾¾® FeSO4 + 2H2O

  1. c)   ZnSO4 + 7H2O

ZnSO4 ZnO + SO3

                        ZnO + 2HCl ¾¾® ZnCl2 + H­2O

ZnCl2 + 2NH4Cl ¾¾® ZnCl2.2NH4Cl

ZnCl2.2NH4Cl   + 2NH4Cl

  1. In silicon d-orbitals are available i.e., it increases its co-ordination number to 6.
  2. In carbon suboxide, following linear structure is found

O = C = C = C = O

Each carbon lies in sp hybrid state. The p electrons are delocalised from one end to other. The delocalisation does not alter the geometry of molecule.

  1. P2H4 , the liquid hydride is always present in impure phosphine. It catches fire as soon as it comes in contact with air.
  2. In (SiH­3)3N, the lone pair of electrons is used up in pp-dp bonding. Such a pp – dp bonding is not possible in (CH3)3N due to absence of d-orbitals in carbon. This account for more basic nature of (CH3)3N than (SiH3)3N.

 

6

 

 

Leave a Reply

Your email address will not be published.

X