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.

2. 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)

6. 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[ [M^III(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. 10H₂O 

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

6. 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) N_­2H_­4
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¯

7. 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

8. 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

2. Reaction with acidified KMnO₄ :

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

3. Reaction with acidified K₂Cr₂O₇ :

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

4. Oxidising property
5. i) 2H₂S + SO₂ ¾¾® 2H₂O + 3S¯
6. 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 (S₂O₃^2-) 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¯

2. Reaction with acidified KMnO₄ :

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

3. 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

2. 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.

3. 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.

9. 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

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

KBr + H₃PO₄ ¾¾® KH₂PO₄ + HBr­

KI + H₃PO_­4 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

2. 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

NCN^2–       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^– > ClO^–₂ > 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 + F₂O­

2Cl₂ + 2HgO HgCl₂××    HgO + Cl₂O­

10. 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

2. On heating it melts to red liquid
3. 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

Cr₂O₇^2- + 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?

 

12. 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 C₂^2– (acetylide ion) or C^4- (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.

13. 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.

2. 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.

14. 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

15. 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

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

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

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

CaOCl₂ + CO₂     ¾¾®     CaCO₃   + Cl₂ ­

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

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

15. 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.

 

17. 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)

                        Ca_­3N₂ + 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)

18. 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

6. Mg₃N₂ when reacted with water gives off NH₃ but HCl is not obtained from MgCl₂ on reaction with water at room temperature. Why?
7. Why calcium cyanide is used as fertilizer?
8. Pure HI kept in a bottle acquires a brown colour after sometime. Explain
9. Why zinc chloride cannot be dehydrated on heating. Explain
10. 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.

3. i)    A black mineral (A) to treatment with dilute NaCN solution in presence of air gives a clear solution of (B) and (C).
4. 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?

 

 

19. 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

2. 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

3. 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

4. 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

5. 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

6. 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

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

(A) S                                                          (B) SO₃

(C) SO₂                                                     (D) H₂

8. A gas which burns with blue flame is –

(A) CO                                                      (B) O₂

(C) N₂                                                  (D) CO₂

9. When lead storage battery is discharged –

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

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

10. 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

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

(A) conc. H₂SO₄                                       (B) P₂O₅

(C) CaCl₂                                                  (D) MgClO₄

12. Which is least basic among the following

(A) NF₃                                                      (B) NCl₃

(C) NBr₃                                                    (D) NI₃

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

(A) N₂O                                               (B) NO

(C) NO₂                                               (D) N₂O₅

14. The inertness of nitrogen is due to its –

(A) high electronegativity                    (B) small atomic radius

(C) high dissociation energy               (D) stable configuration

15. 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

2. 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

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

(A) Na                                                       (B) Sr

(C) K                                                         (D) Zn

4. 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

5. Butter of  tin is

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

(C) SnCl₄. 5H₂O                                       (D) H₂SnCl₄

6. 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

7. 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

8. The material used in solar cell is

(A) Si                                                         (B) Sn

(C) Ti                                                         (D) Cs

9. White phosphorus may be removed from red phosphorus by

(A) sublimation                                          (B) distillation

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

10. Non combustible hydride is –

(A) NH₃                                               (B) PH₃

(C) AsH₃                                             (D) SbH₃

11. Which has S – S bonds

(A) H₂S₂O₃                                               (B) H₂S

(C) H₂S₂O₆                                               (D) S₃O₉

12. 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

13. NO₂ is not obtained when following is heated

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

(C) LiNO₃                                                  (D) KNO₃

14. Acid rain may cause

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

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

15. Following are neutral oxide except

(A) NO                                                      (B) N₂O

(C) CO                                                      (D) NO₂

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

20. 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

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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

3. 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.
4. Since nitric acid is an oxidising agent it oxidises H₂S to S.

3H₂S + 2HNO₃ ¾¾® 2NO + S + 4H₂O

5. 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.
6. ;

 

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

SO₂ + 2H₂O ¾® H₂SO₄ + 2H

Coloured flower + H ¾® Colourless flower

8. The H – S bond is weaker than H – O bond and the proton is easily released when H₂S reacts with water.
9. 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

10. 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

3. Because of the Fajan’s Rule.
4. 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)

5. a) Solubility decreases with increased atomic weight
6. b) This trend is reversed with the fluorides and hydroxides
7. c) Lattice energy decreases as the size of metal increases
8. d) Hydration energy decreases more rapidly than the lattice energy.
9. e) However with fluorides and hydroxides the lattice energy decreases more rapidly.
10. 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

6. 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.
7. 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

8. 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.
9. On heating it reacts with water of crystallisation and forms zinc oxychloride.

3ZnCl₂.2H₂O ¾® Zn₂OCl₂ + 2HCl + 3H₂O

10. 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

2.  Fe₂O₃ + SO₂ + SO₃

FeSO₄ + 2NaOH ¾¾® Fe(OH)₂ + Na₂SO₄

Fe(OH)₂ Fe(OH)₃

3. (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 Al₂Cl_­6.

Al₂Cl_6(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

5. 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® Fe_­2O₃ + SO₂ + SO₃

SO₃ + H₂O  ¾¾® H₂SO₄

————————————————————

¾¾®  + SO₂ + H₂SO₄ + 13H₂O

6. a) CuSO₄ + 2NaOH ¾¾®  + Na₂SO₄

Cu(OH)₂  CuO + H₂O

CuO  +  ¾¾® Cu₂O

4. 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 + H_­2O

ZnCl₂ + 2NH₄Cl ¾¾® ZnCl₂.2NH₄Cl

ZnCl₂.2NH₄Cl   + 2NH₄Cl

7. In silicon d-orbitals are available i.e., it increases its co-ordination number to 6.
8. 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.

9. P₂H₄ , the liquid hydride is always present in impure phosphine. It catches fire as soon as it comes in contact with air.
10. In (SiH_­3)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.

 

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