Preparation & Properties of Compounds
- 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.
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)
- 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)
- 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)
- 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
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
- a) Orthoboric acid (H3BO3)
a1) i) Preparation from borax
Na2B4O7 + H2SO4 + 5H2O ¾¾® Na2SO4 + 4H3BO3
- 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
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
- a) Preparation from Boric Acid
4H3BO3 + Na2CO3 ¾¾® Na2B4O7 + 6H2O + CO2
- b) Basic Nature:- aqueous solution of borax is alkaline in nature due to its hydrolysis
Na2B4O7 + 3H2O ¾¾® NaBO2 + 3H3BO3
NaBO2 + 2H2O NaOH + H3BO3
- 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
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.
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
- 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
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.
Liquid N2O4 is useful as a non – aqueous solvent. It self – ionizes as
N2O4 NO+ + NO3–
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?
- i) N2H4
- ii) NH2OH
Solution: i) NH3 + NaOCl NH2Cl + NaOH
NH2Cl + NH3 ¾® NH2.NH2 + HCl
NaOH + HCl ¾® NaCl + H2O
- 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¯
- 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
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
- Sulphur Chemistry
8.1 Sulphur Oxides and Oxy-acid
Sulphur forms several oxides of which sulphur dioxide (SO2) and sulphur trioxide (SO3) are important.
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
- As reducing agent
- i) Action on halogens :
SO2 + Cl2 +2H2O ¾¾® H2SO4 + 2HCl
- ii) Action on FeCl3 :
2FeCl3 + SO2 + 2H2O ¾¾® H2SO4 + 2FeCl2 + 2HCl
- Reaction with acidified KMnO4 :
2KMnO4 + 5SO2 + 2H2O ¾¾® K2SO4 + 2MnSO4 + 2H2SO4
- Reaction with acidified K2Cr2O7 :
K2Cr2O7 + 3SO2 + H2SO4 ¾¾® K2SO4 + Cr2(SO4)3 + H2O
- Oxidising property
- i) 2H2S + SO2 ¾¾® 2H2O + 3S¯
- ii) SO2 + 2Mg ¾¾® 2MgO + S¯
Preparation: i) By catalytic oxidation of sulphur dioxide
- ii) By dehydration of H2SO4
H2SO4 SO3 + H2O.
8.2 Oxo Acids of Sulphur
|H2S2O5||di-or pyrosuphurous acid|
|H2S2O7||di or pyrosulphuric acid|
|H2SnO6:||Polythionic acid (n = 1 to 12)|
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 :
- Low volatility
- b) Strong acidic character
- c) Strong affinity to water
- d) Ability to act as an oxidising agent.
- 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–)
- 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
- 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
- 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.
- 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
- ii) Spring’s reaction may be used for the preparation of sodium thiosulphate.
Na2S + Na2SO3 + l2 ® Na2S2O3 + 2Nal
- 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
- ii) Reaction with BaCl2 : It gives white ppt. of barium thiosulphate.
S2O+ Ba2+ ¾® BaS2O3 ¯
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 ¯
Ag2S2O3 + H2O ¾® Ag2S + H2SO4
With conc. solution of sodium thiosulphate, silver nitrate gives no ppt.
- iv) It reacts with silver salts to form sodium argento thiosulphate complex
AgBr + 2Na2S2O3 ¾® Na3 [Ag(S2O3)2]
- 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.
- As reducing agent
- i) Action on halogens :
H2S + Cl2 ¾¾® 2HCl + S¯
- ii) Action on FeCl3 :
2FeCl3 + H2S ¾¾® 2FeCl2 + 2HCl + S¯
- Reaction with acidified KMnO4 :
2KMnO4 + 3H2SO4 + 5H2S ¾¾® K2SO4 + 2MnSO4 + 8H2O + S¯
- 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.
8.7 Allotropy in Sulphur
Sulphur exists in three important allotropic forms.
- Rhombic sulphur
- Monoclinic sulphur
- Plastic sulphur
- 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
- 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.
- 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.
9.1 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)|
|ICl, IBr, IF||ICl3, IF3||IF5||IF7|
9.2 Properties of Hydrogen Halides
Hydrohalic acids (HF, HCl, HBr and HI)
- Physical Nature : Except HF, all others HCl, HBr, HI are gases HF is a liquid due to intermolecular hydrogen bonding
- 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:
- 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
- Less volatile, non -oxidising H3PO4 is used to prepare HBr and HI
KBr + H3PO4 ¾¾® KH2PO4 + HBr
KI + H3PO4 ¾¾® KH2PO4 + HI
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
- 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.
- Dipole moment HI < HBr < HCl < HF
- Bond length HF < HCl < HBr < HI
- Bond strength HI < HBr < HCl < HF
- Thermal stability HI < HBr < HCl < HF
- Acid strength HF < HCl < HBr < HI
- Reducing power HF < HCl < HBr < HI
Exercise 2: Dry chlorine does not bleach clothes. Explain why?
Exercise 3: Explain why?
- i) I2 dissolves more in KI solution than in water?
- 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.
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
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:
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–
HOCl : Ca(OCl)2 + 2HNO3 ¾® Ca(NO3)2 + 2HOCl
2HgO + H2O + 2Cl2 HgO.HgCl2 ¯ + 2HOCl
(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
- Compounds of Silver
Silver Nitrate, AgNO3(Lunar Caustic)
It is prepared by dissolving the metal in dilute nitric acid and crystallizing the solution
3Ag + 4HNO3 ¾¾® 3AgNO3 + 2H2O + NO
- i) On heating, it gives metallic silver and nitrogen dioxide
2AgNO3 2Ag + 2NO2 + O2
- ii) It reacts with iodine in two
- a) When iodine is in excess
5AgNO3 + 3I2 + 3H2O ¾® HIO3 + 5AgI + 5HNO3
- 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
2AgNO3 + 2NH4OH ¾® Ag2O ¯ + 2NH4NO3 + H2O
Ag2O + 4NH4OH ¾® 2[Ag(NH3)2]OH + 3H2O
- iv) It gives turbidity with tap water (Cl–) and turbidity is soluble in NH4
AgCl + 2NH4OH ¾¾® Ag(NH3)2Cl + 2H2O
Silver Bromide (AgBr)
Preparation: It is prepared by adding soluble bromide to a silver salt solution
AgNO3 + NaBr ¾® AgBr + NaNO3
- 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
- On heating it melts to red liquid
- Hydrogn Peroxide
- Hydrogen peroxide :
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
2H2SO4(aq) ¾¾¾® H2S2O8 (aq) + H2(g)
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
- 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)
- 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
- a) It liberates iodine from potassium iodide in presence of ferrous sulphate
- 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
- 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+
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?
- Carbon (Oxide and Carbides)
Carbon burns in and forms two oxides, carbon monoxide, (CO) and Carbon dioxide (CO2).
Carbon Monoxide :
- i) By heating carbon in limited supply of oxygen.
C +O2 ¾® CO.
- 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
When air is passed over hot coke, producer gas is obtained.
2C + O2 + 4N2 ¾® 2CO + 4N2
- 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
- ii) It burns in air to give heat and carbon dioxide
CO + O2 ¾® CO2 + heat.
- a) Burns with blue flame
- 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.
- i) In the lab., it is prepared by the action of acids on carbonates.
CaCO3 + 2HCl ¾® CaCl2 + H2O + CO2
- ii) By combustion of carbon
C + O2 ¾® CO2
- 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
- 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.
Carbon combines with more electropositive elements than itself when heated to high temperature to form carbides. Carbides are of mainly three types.
- i) Salt like Carbides : These are the ionic salts containing either C22– (acetylide ion) or C4- (methanide ion)e.g. CaC2, Al4C3, Be2
- 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.
- Ammonia (NH3)
- Lab Method :
- 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
- ii) By the hydrolysis of magnesium nitride
Mg3N2 + 6H2O ¾¾® 3Mg(OH)2 + 2NH3.
- It is manufactured by Haber’s process
N2(g) + 3H2(g) 2NH3(g).
- i) Basic nature : Its aq. solution is basic in nature and turns red litmus blue.
NH3 + H2O + OH–
- 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
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+
- 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
AlCl3 + 3NH4OH ¾¾® Al(OH)3 + 3NH4Cl
CrCl3 + 3NH4OH ¾¾® Cr(OH)3 + 3NH4Cl
Substances which increase the fertility of soils are known as fertilizers. They are classified into three categories :
- 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
- Phosphatic fertilizers : They supply phosphorus to the plants. e.g. superphosphate of lime Ca(H2PO4)2
- 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.
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
- 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
- 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
- When treated with water it decomposes into calcium chloride and calcium hypochlorite
2CaOCl2 + H2O ¾® CaCl2 + Ca(OCl)2 + H2O
- Bleaching powder reacts with CO2 (atmospheric) and gives chlorine which accounts for its oxidising and bleaching actions.
CaOCl2 + CO2 ¾¾® CaCO3 + Cl2
- Action of Heat : On heating bleaching powder gives a mixture of chlorate and chloride
- 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
- 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.
- Solved Problems
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+.
Problem 1: H2O2 is not
(A) a reducing agent (B) an oxidising agent
(C) a dehydrating agent (D) a bleaching agent
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
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
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
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
Problem 6: Decomposition of H2O2 is slowed down by addition of –
(A) Alcohol (B) MnO2
(C) Alkali (D) Pt
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
Problem 8: The metal (s) soluble in aqua regia is (are)
(A) Pt (B) Au
(C) Ag (D) All the above
Problem 9: By burning NH3 in oxygen, it gives ________ and H2O –
(A) NO (B) N2
(C) NO2 (D) N2O
Problem 10: P4 + 5O2 ¾® X Y, Y is
(A) H3PO4 (B) P2O5
(C) H3PO3 (D) H4P2O7
Solution: P4 + 5O2 ¾® P4O10 4H3PO4
- Assignment (Subjective Problems)
LEVEL – I
- The size of d orbital, decrease : Si > P > S > Cl but p bonding increases in the same order. Explain.
- Explain the stability of oxides of alkali metals.
- BaO2 is a peroxide but PbO2 is not a peroxide why?
- Why nitric acid cannot be used to prepare H2S.
- Ferric iodide is very unstable but ferric chloride is stable. Why?
- 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.
- Why dry SO2 does not bleach dry flowers?
- Why H2S is stronger acid than H2O?
- In the manufacture of sulphuric acid by the contact process, sulphur trioxide is not directly dissolved in water. Why?
- Conc. H2SO4 cannot be used for drying H2. Explain
LEVEL – II
- Complete the following reactions.
- a) KBr + ICl ¾®
- b) KF + BrF3 ¾®
- Anhydrous AlCl3 cannot be prepared by heating hydrated AlCl3.6H2O – why?
- Why AlF3 is ionic while AlCl3 is covalent?
- 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
- Mg3N2 when reacted with water gives off NH3 but HCl is not obtained from MgCl2 on reaction with water at room temperature. Why?
- Why calcium cyanide is used as fertilizer?
- Pure HI kept in a bottle acquires a brown colour after sometime. Explain
- Why zinc chloride cannot be dehydrated on heating. Explain
- Explain why a solution of ferric chloride gives brown precipitate.
LEVEL – III
- 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.
- Compound (A)
- i) On strong heating gives two oxides of sulphur
- 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.
- i) A black mineral (A) to treatment with dilute NaCN solution in presence of air gives a clear solution of (B) and (C).
- 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.
- iv) (E) on fusion with Na2CO3 gives (D).
- v) (E) dissolves in aqueous solution of NH3 giving a colourless solution of (F)
- vi) Identify (A) to (F)
- An inorganic compound (A) shows the following reaction.
- i) It is white solid and exists as dimer; gives fumes of (B) with wet air.
- ii) It sublines in 180°C and form monomer if heated to 400°
iii) Its aqueous solution turns blue litmus to red.
- iv) Addition of NH4OH and NaOH seperately to a solution of (A) gives white ppt. which is however soluble in excess of NaOH.
- What happens when:
- a) Ferric chloride is added to potassium ferrocyanide
- b) Iron reacts with cold dilute nitric acid
- c) Green vitrol is strongly heated
- How will you prepare
- a) Copper oxide from copper sulphate
- b) Ferrous sulphate from Mohr’s salt
- c) Anhydrous ZnCl2 from white vitriol
- Carbon tetrachloride is not affected but silicon tetrachloride is hydrolysed by water. Why?
- Explain the structure of C3O2 is term of bonding in the molecule.
- Pure PH3 does not burn in air but impure sample of PH3 burns in air. Why?
- (SiH3)3N is a weaker base than (CH3)3N, Why?
- Assignment (Objective Problems)
Level – I
- 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
- 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
- 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
- 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
- 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
- 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
- Which of the following is obtained by the reaction of Cu and Conc. H2SO4?
(A) S (B) SO3
(C) SO2 (D) H2
- A gas which burns with blue flame is –
(A) CO (B) O2
(C) N2 (D) CO2
- When lead storage battery is discharged –
(A) SO2 is evolved (B) lead sulphate is consumed
(C) lead is formed (D) sulphuric acid is consumed
- 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
- Anhydrous _______ is a very effective dessiccant (water absorber used in dry battery. It is
(A) conc. H2SO4 (B) P2O5
(C) CaCl2 (D) MgClO4
- Which is least basic among the following
(A) NF3 (B) NCl3
(C) NBr3 (D) NI3
- Which of the following is the strongest oxidising agent?
(A) N2O (B) NO
(C) NO2 (D) N2O5
- The inertness of nitrogen is due to its –
(A) high electronegativity (B) small atomic radius
(C) high dissociation energy (D) stable configuration
- Nitrolim is –
(A) CaC2 (B) CaCN2 + C
(C) CaC2N2 (D) CaC2 + CaCN
LEVEL – II
- Molecular formula of Glauber’s salt is
(A) MgSO4,7H2O (B) FeSO4,7H2O
(C) CuSO4,5H2O (D) Na2SO4, 10H2O
- 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
- Which of the following does not give flame test?
(A) Na (B) Sr
(C) K (D) Zn
- 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
- Butter of tin is
(A) (NH4)2SnCl6 (B) SnCl2 + Sn(OH)2
(C) SnCl4. 5H2O (D) H2SnCl4
- 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
- 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
- The material used in solar cell is
(A) Si (B) Sn
(C) Ti (D) Cs
- White phosphorus may be removed from red phosphorus by
(A) sublimation (B) distillation
(C) dissolving in CS2 (D) heating with an alkali solution
- Non combustible hydride is –
(A) NH3 (B) PH3
(C) AsH3 (D) SbH3
- Which has S – S bonds
(A) H2S2O3 (B) H2S
(C) H2S2O6 (D) S3O9
- 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
- NO2 is not obtained when following is heated
(A) Pb(NO3)2 (B) AgNO3
(C) LiNO3 (D) KNO3
- Acid rain may cause
(A) rusting easier (B) stone cancer in Taj Mahal
(C) non – fertility of soil (D) all are correct
- Following are neutral oxide except
(A) NO (B) N2O
(C) CO (D) NO2
- Answers to Objective Assignments
Level – I
- B 2. A
- C 4. D
- D 6. B
- C 8. A
- D 10. B
- D 12. A
- D 14. C
- D 2. A
- D 4. B
- C 6. B
- C 8. A
- B 10. A
- C 12. B
- D 14. D
Preparation & Properties of Compounds
Level – I
- 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.
- 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
- 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.
- Since nitric acid is an oxidising agent it oxidises H2S to S.
3H2S + 2HNO3 ¾¾® 2NO + S + 4H2O
- 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.
- Nascent hydrogen is responsible for bleaching, according to the following reaction
SO2 + 2H2O ¾® H2SO4 + 2H
Coloured flower + H ¾® Colourless flower
- The H – S bond is weaker than H – O bond and the proton is easily released when H2S reacts with water.
- 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
- Hydrogen, sometimes, ignites the large amount of heat released when water is absorbed by acid.
Level – II
- a) K[BrICl]
- b) K[BrF4]
- Hydrolysis of AlCl3 takes place forming Al2O3
2AlCl3 + 3H2O ¾¾® Al2O3 + 6HCl
- Because of the Fajan’s Rule.
- 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)
- a) Solubility decreases with increased atomic weight
- b) This trend is reversed with the fluorides and hydroxides
- c) Lattice energy decreases as the size of metal increases
- d) Hydration energy decreases more rapidly than the lattice energy.
- e) However with fluorides and hydroxides the lattice energy decreases more rapidly.
- f) Condition for solubility – hydration energy ³ lattice energy.
- 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.
- 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
- 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.
- On heating it reacts with water of crystallisation and forms zinc oxychloride.
3ZnCl2.2H2O ¾® Zn2OCl2 + 2HCl + 3H2O
- Hydrolysis of ferric chloride occurs giving finally brown precipitate
FeCl3 + 3H2O Fe(OH)3 + 3HCl
2Fe(OH)3 ¾® + 3H2O
Level – IIi
- 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
- Fe2O3 + SO2 + SO3
FeSO4 + 2NaOH ¾¾® Fe(OH)2 + Na2SO4
- (A) Ag2S (a black mineral)
The reaction are
- Ag2S + 4NaCN ¾® 2Na[Ag(CN)2] + Na2S
- 2Na[Ag(CN)2] + Zn ¾® Na2[Zn(CN)4] + 2Ag
- 3Ag + 4 HNO3 ¾® 3AgNO3 + NO + 2H2O
- AgNO3 AgCl + HNO3
- AgCl + 2NH3 ¾® Ag(NH3)2Cl
- 4AgCl + 2Na2CO3 ¾® 4Ag + 4NaCl + 2CO2 + O2
- 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
- ii) It fumes with wet air
Al2Cl6 + 6H2O 3Al(OH)3¯ + 6HCl
iii) 2AlCl3 + 6H2O 2Al(OH)3¯ + 6HCl(s)
- 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
- a) On addition of FeCl3 to K4Fe(CN)6 a Prussian blue coloured is formed.
4FeCl3 + 3K4Fe(CN)6 ¾¾® + 12KCl
- 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
- 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
- a) CuSO4 + 2NaOH ¾¾® + Na2SO4
Cu(OH)2 CuO + H2O
CuO + ¾¾® Cu2O
- b) FeSO4.(NH4)2SO4.6H2O + 4NaOH ® Fe(OH)2¯ + 2NH3 + 2Na2SO4 + 8H2O
Fe(OH)2 + H2SO4 ¾¾® FeSO4 + 2H2O
- c) ZnSO4 + 7H2O
ZnSO4 ZnO + SO3
ZnO + 2HCl ¾¾® ZnCl2 + H2O
ZnCl2 + 2NH4Cl ¾¾® ZnCl2.2NH4Cl
ZnCl2.2NH4Cl + 2NH4Cl
- In silicon d-orbitals are available i.e., it increases its co-ordination number to 6.
- 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.
- P2H4 , the liquid hydride is always present in impure phosphine. It catches fire as soon as it comes in contact with air.
- 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.
Leave a Reply