- 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) α – Al2O3 (rhombic lattice) is dense, hard and resistant to chemical attack.
α – 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) λ – 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 ⎯⎯→ Na2SO4 + 4H3BO3
- ii) From colemanite
Ca2B6O11 + 2SO2 + 11H2O ⎯⎯→ 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 Tetraboric 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 NH4BF4 BF3
Ammonium boro fluoride
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 ⎯⎯→ Na2B4O7 + 6H2O + CO2
- b) Basic Nature:- aqueous solution of borax is alkaline in nature due to its hydrolysis
Na2B4O7 + 3H2O ⎯⎯→ 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 ⎯⎯→ 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.
- Compounds of Nitrogen
6.1 Oxides of Nitrogen
Nitrous Oxide (N2O)
It is used for the preparation of azides
N2O + NaNH2 ⎯⎯→ NaN3 + H2O
Nitric oxide (NO)
The liquid and solid states are diamagnetic, because close dimers are formed
Reaction with Cl2
2NO + Cl2 ⎯⎯→ 2NOCl
The gas condenses to a brown liquid which turns pale 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 ⎯⎯→ NH4Cl + N2O4
Acid base salt solvent
Nitrogen trioxide or nitrogen sesquioxide (N2O3)
N2O3 + H2O ⎯⎯→ 2HNO2
(anhydride of nitrous acid)
N2O3 + 2HClO4 ⎯⎯→ 2NO [ClO4] + H2O
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|
- Phosphorus Chemistry
7.1 Oxides of Phosphorus
(White) (Density = 2.1)
(Poisonous) (non poisonous polymer of P4 unit)
Structure of Phosphorus pentoxide (P4O10)
P4O10 + 6H2O ⎯⎯→ 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 Oxyacid
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 ⎯⎯→ H2SO4 + 2HCl
- ii) Action on FeCl3 :
2FeCl3 + SO2 + 2H2O ⎯⎯→ H2SO4 + 2FeCl2 + 2HCl
- Reaction with acidified KMnO4
2KMnO4 + 5SO2 + 2H2O ⎯⎯→ K2SO4 + 2MnSO4 + 2H2SO4
- Reaction with acidified K2Cr2O7
K2Cr2O7 + 3SO2 + H2SO4 ⎯⎯→ K2SO4 + Cr2(SO4)3 + H2O
- Oxidising property
- i) 2H2S + SO2 ⎯⎯→ 2H2O + S↓
- ii) SO2 + 2Mg ⎯⎯→ 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)|
O3 is an extremely powerful oxidizing agent, second only to F2 in oxidizing power, and reacts much more readily than dioxygen.
3PbS + 4O3 ⎯⎯→ 3PbSO4
2NO2 + O3 ⎯⎯→ N2O5 + O2
S + H2O + O3 ⎯⎯→ H2SO4
2KOH + SO3 ⎯⎯→ 2KO3 + SO2 + H2O
Potassium ozonide KO3 is an orange coloured solid and contains the paramagnetic ion.
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 conc.
SO3 + H2SO4 ⎯→ H2S2O7
H2S2O7 + H2O ⎯→ 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 ⎯⎯→ H2SO3
- i) It acts as reducing agent and its chemical properties are similar to those of solution e.g.
2FeCl3 + (SO2 + H2O) + H2O ⎯⎯→ 2FeCl2 + H2SO4 + 2HCl
- ii) It reacts with iron, forming ferrous sulphite and ferrous thiosulphate.
2Fe + 2H2SO3 ⎯⎯→ 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 as 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. It consists in treating a mixture of sodium sulphide and sodium sulphite with calculated quantity of iodine.
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 ⎯→ 2NaCl + H2O + S↓ + SO2
- ii) Reaction with BaCl2 : It gives white ppt. of barium thiosulphate.
S2O+ Ba2+ ⎯→ 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+ ⎯→ Ag2S2O3 ↓
Ag2S2O3 + H2O ⎯→ 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 ⎯→ Na3 [Ag(S2O3)2]
Sodium argento thiosulphate complex
- v) Thiosulphate () ion is oxidized by iodine I2 to tetrathionate () ion
2Na2S2O3 + I2 ⎯⎯→ Na2S4O6 + 2NaI.
8.6 Hydrogen Sulphide
Preparation: Prepared by the action of dil. HCl or H2SO4 on iron sulphide
FeS + 2HCl (dil) ⎯⎯→ FeCl2 + H2S↑.
Properties: It is a colourless, poisonous gas having the smell of rotten eggs.
- As reducing agent
- i) Action on halogens :
H2S + Cl2 ⎯⎯→ 2HCl + S↓
- ii) Action on FeCl3 :
2FeCl3 + H2S ⎯⎯→ 2FeCl2 + 2HCl + S↓
- Reaction with acidified KMnO4 :
2KMnO4 + 3H2SO4 + 5H2S ⎯⎯→ K2SO4 + 2MnSO4 + 8H2O + S↓
- Reaction with acidified K2Cr2O7 :
K2Cr2O7 + 4H2SO4 + 3H2S ⎯⎯→ K2SO4 + Cr2(SO4)3 + 7H2O + S↓
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 β-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 γ-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 1: 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.
- Magnesium Chemistry
Illustration 2: Identify the missing chemicals A to K
Solution: A = MgSO4 B = Hydrogen (H2)
C = Mg(OH)2 D = Na2SO4
E = MgO F = H2O
G = MgCl2 H = NaCl
J = CO2 K = NO2
K = O2
10.1 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′1 (n = 1) (with linear shape)||Type XX′3 (n = 3) (with T-shape)||XX′5 (n = 5)
(with square pyramidal shape)
|XX′7 (n = 7) with pentagonal bipyramidal shape)|
|ICl, IBr, IF||ICl3, IF3||IF5||IF7|
10.2 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 ⎯⎯→ 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 ⎯⎯→ 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 ⎯⎯→ K2SO4 + 2HBr
H2SO4 + 2HBr ⎯⎯→ Br2↑ + SO2 + 2H2O
- Less volatile, non -oxidising H3PO4 is used to prepare HBr and HI
KBr + H3PO4 ⎯⎯→ KH2PO4 + HBr↑
KI + H3PO4 ⎯⎯→ KH2PO4 + HI ↑
1 All the three acids are reducing agents HCl is not attacked by H2SO4.
2HBr + H2SO4 ⎯⎯→ 2H2O + SO2 + Br2 ↑
2HI + H2SO4 ⎯⎯→ 2H2O + SO2 + I2
- All the three react with KMnO4 and K2Cr2O7
2KMnO4 + 16HCl ⎯⎯→ 2KCl + 2MnCl2 + 8H2O + 5Cl2 ↑
K2Cr2O7 + 14HBr ⎯⎯→ 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 < BI
- Reducing power HF < HCl < HBr < HI
10.3 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 (SCM)2 thiocyanogen
SCN–, thiocyante ion
SeCN–, selenocyanate ion
NCN2– Cyanamide ion
N3– azide ion
OMC– 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.
10.4 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– > ClO2 > ClO–
HOCl : Ca(OCl)2 + 2HNO3 ⎯→ Ca(NO3)2 + 2HOCl
2HgO + H2O + 2Cl2 HgO.HgCl2 ↓ + 2HOCl
(insoluble basic mercury chloride)
HClO2 : BaO2 + 2ClO2 ⎯⎯→ Ba(ClO2)2 (liquid) + O2
Ba(ClO2)2 + H2SO4(dil.) ⎯→ BaSO4 ↓ + 2HClO2
HClO3 : 6Ba(OH)2 + 6Cl2 ⎯→ 5BaCl2 + Ba(ClO3)2 + 6H2O
Ba(ClO3)2 + H2SO4(dil.) ⎯→ BaSO4 ↓ + 2HClO3
HClO4 : (a) KClO4 + H2SO4 ⎯⎯→ KHSO4 + HClO4
(b) 3HClO3 ⎯⎯→ HClO4 + 2ClO2 + H2O
- 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 ⎯⎯→ 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 ⎯→ HIO3 + 5AgI + 5HNO3
- b) When AgNO3 is in excess
6AgNO3 + 3I2 + 3H2O ⎯⎯→ AgIO3 + 5AgI + 6HNO3
iii) When treated with alkali, it gives precipitate of silver oxide, which dissolves in excess of NH4OH
2AgNO3 + 2NaOH ⎯⎯→ Ag2O ↓ + 2NaNO3 + H2O
2AgNO3 + 2NH4OH ⎯→ Ag2O ↓ + 2NH4NO3 + H2O
Ag2O + 4NH4OH ⎯→ 2[Ag(NH3)2]OH + 3H2O
- iv) It gives turbidity with tap water (Cl–) and turbidity is soluble in NH4
AgCl + 2NH4OH ⎯⎯→ Ag(NH3)2Cl + 2H2O
Silver Bromide (AgBr)
Preparation: It is prepared by adding soluble bromide to a silver salt solution
AgNO3 + NaBr ⎯→ 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 ⎯⎯→ [Ag(NH3)2] Br + 2H2O
AgBr + 2KCN ⎯⎯→ K[Ag(CN)2] + KBr
AgBr + 2Na2S2O3 ⎯⎯→ 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) ⎯⎯→ H2O2(aq) + Na2SO4(s)
BaO2.8H2O + H2SO4(aq) ⎯→ 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) H2S2O8 (aq) + H2(g)
H2S2O8(aq) 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) ⎯⎯→ 2H2O (l) + O2 (g)
- ii) It acts as oxidant as well as reluctant 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– ⎯⎯→ 2H2O
In basic medium : H2O2 + OH– + 2e– ⎯⎯→ 3OH–
As reducing agent
In acidic medium: H2O2 ⎯⎯→ 2H+ + O2 + 2e–
In basic medium : H2O2 + 2OH– ⎯⎯→ 2H2O + O2 + 2e–
2Fe2+ + H2O2 + 2H+ ⎯⎯→ 2Fe3+ + 2H2O
2MnO4– + 5H2O2 + 6H+ ⎯⎯→ 2Mn2+ + 8H2O + 5O2
Mn2+ + H2O2 ⎯⎯→ Mn+4 + 2OH–
2Fe3+ + H2O2 + 2OH– ⎯→ 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) ⎯→ 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+ ⎯⎯→ 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 ⎯→ H2TiO4 + 4H+
- Carbon (Oxide and Carbides)
Carbon burns in and forms two oxides, carbon monoxide, (CO) and Carbon dioxide (CO2).
- i) By heating carbon in limited supply of oxygen.
C + O2 ⎯→ CO.
- ii) By heating oxides of heavy metals e.g. iron, zinc etc with carbon.
Fe2O3 + 3C ⎯→ 2Fe + 3CO
ZnO + C ⎯→ 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 ⎯→ CO + H2
When air is passed over hot coke, producer gas is obtained.
2C + O2 + 4N2 ⎯→ 2CO + 4N2
- i) It is a powerful reducing agent and reduces many metal oxides to their corresponding metals.
Fe2O3 + 3CO ⎯→ 2Fe + 3CO2
CuO + CO ⎯→ Cu + CO2
- ii) It burns in air to give heat and carbon dioxide
CO + O2 ⎯→ 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 ⎯→ CaCl2 + H2O + CO2
- ii) By combustion of carbon
C + O2 ⎯→ CO2
- i) It turns lime water milky and milkiness disappears when CO2 is passed in excess
Ca(OH)2 + CO2 ⎯→ CaCO3 ↓ + H2O, CaCO3 + H2O + CO2 ⎯→ 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 ⎯⎯→ NH3↑ + NaCl + H2O
- ii) By the hydrolysis of magnesium nitride
Mg3N2 + 6H2O ⎯⎯→ 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 ⎯⎯→ 6NH4Cl + N2
Excess of chlorine : NH3 + 3Cl2 ⎯→ NCl3 + 3HCl
Bromine : 8NH3 + 3Br2 ⎯⎯→ 6NH4Br + N2
Excess of bromine : NH3 + 3Br2 → NBr3 + 3HBr
Iodine : 2NH3 + 3I2 ⎯→ NH3.NI3 + 3HI
tri iodide ammonate
NH3.NI3 explodes in dry state
8NH3.NI3 ⎯⎯→ 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 ⎯⎯→ [Ag(NH3)2]+
Cu2+ + 4NH3 ⎯⎯→ [Cu(NH3)4]2+
Cd2+ + 4NH3 ⎯⎯→ [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 ⎯⎯→ Fe(OH)3 + 3NH4Cl
AlCl3 + 3NH4OH ⎯⎯→ Al(OH)3 + 3NH4Cl
CrCl3 + 3NH4OH ⎯⎯→ 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 ⎯⎯→ 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 ⎯→ 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 ⎯⎯→ CaCl2 + H2O + Cl2↑
CaOCl2 + H2SO4 ⎯⎯→ CaSO4 + H2O + Cl2↑
- When treated with water it decomposes into calcium chloride and calcium hypochlorite
2CaOCl2 + H2O ⎯→ CaCl2 + Ca(OCl)2 + H2O
- Bleaching powder reacts with CO2 (atmospheric) and gives chlorine which accounts for its oxidising and bleaching actions.
CaOCl2 + CO2 ⎯⎯→ CaCO3 + Cl2 ↑
- Action of Heat : On heating bleaching powder gives a mixture of chlorate and chloride
- 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: Hydrogen peroxide acts both as an oxidising and as a reducing agent in alkaline solution towards certain first now transitional metal ions. Illustrate both these properties of H2O2 using chemical equations.
Solution: As oxidising agent
H2O2 ⎯⎯→ H2O + O
As reducing agent:
2K3Fe(CN)6 + 2KOH ⎯→ 2K4Fe(CN)6 + H2O + O
H2O + O ⎯→ H2O + O2
2K3Fe(CN)6 + 2KOH + H2O2 ⎯→2K4Fe(CN)6 + 2H2O + O2
Problem 3: 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.
Solution: Fe2O3 + SO2 + SO3
FeSO4 + 2NaOH ⎯⎯→ Fe(OH)2 + Na2SO4
Problem 4: Why hydrated barium peroxide is used in the preparation of H2O2 instead of the anhydrous variety?
Solution: 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 vessle and the reaction continues without any difficulty.
Problem 5: 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 ⎯⎯→ colour less
Problem 6: 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 7: 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 ⎯⎯→ Cu2I2 + I2 + 4KI
Problem 8: Dry chlorine does not bleach clothes. Explain why?
Solution: The bleaching action of chlorine is due to the liberation of nascent oxygen from water.
H2O + Cl2 ⎯→ 2HCl + [O]
Problem 9: The gas liberated on heating a mixture of two salts with NaOH gives a raddish brown precipitate with an alkaline solution of K2HgI4. The aqueous solution of the mixture on treatment with BaCl2 gives a white precipitate which is sparingly soluble in concentrated HCl. On heating the mixture with K2Cr2O7 and conc. H2SO4, red vapours (A) are produced. The aqueous solution of the mixture gives a deep blue colouration (B) with K3[Fe(CN)6] solution. Identify the radicals in the given mixture and write the balanced equations for the formation of (A) and (B).
Solution: i) The gas which is liberated on heating the mixture with NaOH gives red ppt. With K2HgI4 so gas is NH3 and mixture contains NH4+ ion.
- ii) The aqueous solution gives white ppt. with BaCl2, so mixture contains SO4–2
iii) Mixture on heating with K2Cr2O7 and H2SO4 gives red vapours (of CrO2Cl2) so mixture contains Cl– ions.
- iv) Aqueous solution of mixture gives blue colour with K3[Fe(CN)6] and thus, it contains Fe+2
- v) Thus mixture has NH4+, Fe+2, SO4–2, Cl–
- i) NH4+ + NaOH ⎯→ NH3 + Na+ + H2O
iii) SO4–2 + BaCl2 ⎯→
- iv) 4Cl– + K2Cr2O7 + 3H2SO4 ⎯→
- v) 3Fe+2+ 2K3[Fe(CN)6] ⎯→
Problem 10: 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
Identify (A) to (F)
Solution: (A) Ag2S (a black mineral)
The reaction are
- Ag2S + 4NaCN ⎯→ 2Na[Ag(CN)2] + Na2S
- 2Na[Ag(CN)2] + Zn ⎯→ Na2[Zn(CN)4] + 2Ag
- 3Ag + 4 HNO3 ⎯→ 3AgNO3 + NO + 2H2O
- AgNO3 AgCl + HNO3
- AgCl + 2NH3⎯→ Ag(NH3)2Cl
- 4AgCl + 2Na2CO3 ⎯→ 4Ag + 4NaCl + 2CO2 + O2
Problem 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
Problem 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
Problem 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
Problem 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
Problem 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
Problem 6: Molecular formula of Glauber’s salt is
(A) MgSO4,7H2O (B) FeSO4,7H2O
(C) CuSO4,5H2O (D) Na2SO4, 10H2O
Problem 7: 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
Problem 8: Which of the following does not give flame test?
(A) Na (B) Sr
(C) K (D) Zn
Problem 9: 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
Problem 10: Butter of tin is
(A) (NH4)2SnCl6 (B) SnCl2 + Sn(OH)2
(C) SnCl4. 5H2O (D) H2SnCl4
- Assignment (Subjective Problems)
LEVEL – I
- Presence of water is avoided in the preparation of H2O2 from Na2O2 – why?
- Anhydrous AlCl3 cannot be prepared by heating hydrated AlCl3.6H2O – why?
- Boron and aluminium both are in the same group. Yet AlCl3 shows anomalous mol.wt. which BCl3 doesn’t – why?
- Why AlF3 is ionic while AlCl3 is covalent?
- What is fly ash?
- What is dead burnt plaster?
- What is quick lime? What happens when we add water to it?
- Complete the following reactions.
- a) KBr + ICl ⎯→
- b) KF + BrF3 ⎯→
- Give reason for decreasing order of conductivity of following
Cs+ > Rb+ > k+ > Na+ > Li+
- BaO2 is a peroxide but PbO2 is not a peroxide why?
LEVEL – II
- The size of d orbital, decrease : Si > P > S > Cl but π bonding increases in the same order. Explain.
2: 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.
- Explain the stability of oxides of alkali metals.
4. Explain the solubility of salts of alkali and alkali earth metals
- 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).
- 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.
- Although boric acid B(OH)3 contains three hydroxyl groups yet it behaves as a mono-basic acid. Explain.
- How do and ion differ structurally?
- Oxygen forms diatomic molecules (O2) while sulphur forms octa atomic molecules (S8) explain.
- 10. Give reasons.
- a) Ammonia is more soluble in aqueous ammonium chloride than in pure water.
- b) Solid ammonium fluoride and ice are miscible in all proportions.
LEVEL – III
What happens when?
- a) Al2O3 is treated with Co(NO3)2
- b) Ammonium alum is heated
- c) Metastannic acid is heated
- d) SnCl4 is treated with Sn
- a) K2Cr2O7is treated with SnCl2 in acidic media.
- b) Pb3O4 is trerated with conc. HNO3
- c) Pb(CH3COO)2 is treated with bleaching powder.
- d) PbO2 is treated with NaOH
- e) Ferrous sulphate is ignited
- f) FeCl3 is treated with sodium carbonate. The resulting precipitate is further heated.
- a) FeCl3is heated
- b) Fe(OH)3 is treated with Br2 water in alkaline media
- c) CuSO4 is treated with KCN
- d) CuSO4 is treated with NH4CNS
- An orange solid (A) on heating gives a green residue (B), a colourless gas (C) and water vapours. The dry gas (C) on passing over heated Mg gave a white solid (D). (D) on reaction with water give a gas (E) which formed dense white fumes with HCl. Identify (A) to (E) and give the reactions involved.
- Explain the following
- a) Al2 (SO4)3 is used as a mordant in textile industry
- b) AlCl3.6H2O on heating never produces anhydrous AlCl3.
- c) AlCl3 in water shows acidity.
- d) Why red colour Rousin’s salt [Fe(NO)2S]2 is diamagnetic. How do you account for the phenomenon.
- Write the reactions given below
- a) Al(NO3)3 is heated
- b) Al2O3 is heated with carbon in presence of dry Cl2
- c) Sn is treated with hot and conc. HNO3
- d) Sn is treated with alkali
- e) PbO is treated with NaOH
- f) PbO is treated with NaCl
- g) Fe is treated with (Cold, dilute), (hot, concentrated), concentrated HNO3
- h) Ferrous oxalate is heated
- i) Ferrous sulphate is ignited
- j) Cuprous oxide is treated with ferric sulphate.
- The three samples of H2O2 labeled as 10 volume, 15 volume and 20 volume. Half litre of each sample are mixed and then diluted with equal volume of water. What is the volume strength of resultant solution?
- How will you prepare
- a) Copper oxide from copper sulphate
- b) Ferrous sulphate from Mohr’s salt
- c) Anhydrous ZnCl2 from white vitriol
- A black coloured compound (A) on reaction with dilute H2SO4 gives a gas (B) which on passing in a solution of acid (C) gives a white turbidity (D). Gas (B) when passed in an acidified solution of compound (E) gives a ppt. (F) soluble in dil. HNO3. After boiling this solution when an excess of NH4OH is added, a blue coloured compound (G) is formed. To this solution on addition of acetic acid and aqueous potassium ferrocyanide a chocolate precipitate (H) is obtained. On addition of an aqueous solution of barium chloride to an aqueous solution of (E), a white precipitate insoluble in HNO3 is obtained. Identify from A to H.
- 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 sublimes 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 separately to a solution of (A) gives white ppt. which is however soluble in excess of NaOH.
- Assignment (Objective Problems)
LEVEL – I
- The oxide that gives H2O2 on treatment with a dilute acid is
(A) PbO2 (B) MnO2
(C) Na2O2 (D) TiO2
- H2O2 is not
(A) a reducing agent (B) an oxidising agent
(C) a dehydrating agent (D) a bleaching agent
- A mixture of hydrazine and H2O2 is
(A) antiseptic (B) rocket fuel
(C) germicide (D) insecticide
- Hydrolysis of one mole of peroxy dissulphuric acid produces
(A) two moles of sulphuric acid
(B) two moles of peroxy monosulphuric acid
(C) one mole of sulphuric acid and one mole of peroxy mono sulphuric acid
(D) one mole of sulphuric acid one mole of peroxy mono sulphuric acid and one mole of H2O2.
- 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
- 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
- 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
- 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
- The largest bond angle is in
(A) AsH3 (B) NH3
(C) H2O (D) PH3
- White phosphorus may be removed from red phosphorus by
(A) sublimation (B) distillation
(C) dissolving in CS2 (D) heating with an alkali solution
- Which of the following complex will give white precipitate with BaCl2(aq)?
(A) [Co(NH2)4SO4]NO2 (B) [Cr(NH3)5SO4]Cl
(C) [Cr(NH3)5Cl]SO4 (D) Both (B) and (C)
- Which of the following is obtained by oxidation of phosphorous by HNO3?
(A) H3PO4 (B) H3PO3
(C) H4P2O7 (D) H3PO2
- Which of the following is obtained by the reaction of Cu and Conc. H2SO4?
(A) S (B) SO3
(C) SO2 (D) H2
- 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
- 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
- Non combustible hydride is –
(A) NH3 (B) PH3
(C) AsH3 (D) SbH3
- 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
- Decomposition of H2O2 is slowed down by addition of –
(A) Alcohol (B) MnO2
(C) Alkali (D) Pt
LEVEL – II
- 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
- 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
- The metal (s) soluble in aqua regia is (are)
(A) Pt (B) Au
(C) Ag (D) Cu
- 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
- Anhydrous is a very effective dessiccant (water absorber used in dry battery. It is
(A) conc. H2SO4 (B) P2O5
(C) CaCl2 (D) MgClO4
- NH3 cant be obtained by
(A) heating NH4NO3 or NH4NO2
(B) heating of NH4Cl or (NH4)2CO3
(C) heating of glycine
(D) reaction of AlN or Mg3N2 or CaCN2 with H2O
- Which is least basic among the following
(A) NF3 (B) NCl3
(C) NBr3 (D) NI3
- 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
- By burning NH3 in oxygen, it gives ________ and H2O –
(A) NO (B) N2
(C) NO2 (D) N2O
- In Holmes signals the compound used is –
(A) Ca3P2 + CaC2 (B) CaC2
(C) Ca3P2 (D) CaCN2 + Ca3P2
- Which of the following is the strongest oxidising agent?
(A) N2O (B) NO
(C) NO2 (D) N2O5
- Marsh test is applied for detection of –
(A) N (B) P
(C) As (D) S
- Caliche is –
(A) NaNO3 – NaIO3 (B) NaIO
(C) NaNO2 – NaIO3 (D) NaNO3 – I
- Muriatijc acid is
(A) HF (B) HCl
(C) HBr (D) HI
- The inertness of nitrogen is due to its –
(A) high electronegativity (B) small atomic radius
(C) high dissociation energy (D) stable configuration
- Conc. HNO3 oxidises phosphorus –
(A) H3PO4 (B) P2O5
(C) H3PO3 (D) H4P2O7
- Nitrolim is –
(A) CaC2 (B) CaCN2 + C
(C) CaC2N2 (D) CaC2 + CaCN
- Answers to Objective Assignments
LEVEL – I
- C 2. C
- D 4. C
- B 6. B
- A 8. C
- A 10. B
- B 12. B
- A 14. C
- A 16. A
- D 18. A
- B 20. A
- B 2. C, D
- A, B, C, D 4. C
- B 6. D
- D 8. A, C
- A 10. D
- D 12. B
- A 14. D
- C 16. A
- B 18. C
- A 20. B