1. Nitrogen and its Compound
Chapter 43
Nitrogen and its Compound
43.1 Introduction
43.2 Unreactive Nature of Nitrogen
43.3 Direct Combination of Nitrogen and Oxygen leading to Formation of Nitrogen Oxides
43.4 Ammonia
43.5 Nitric(V) Acid
43.6 Nitrates(V)

2. Nitrogen (first member of Group VA):
43.1 Introduction (SB p.110)
Nitrogen (first member of Group VA):
Electronic configuration: 1s22s22p3
Complete octet by forming diatomic molecules N  N
Non-metal, colourless and odourless gas
Very low melting and boiling points
Slightly soluble in water and does not support combustion
Some information about nitrogen
Covalent radius (nm)
0.074
Melting point (°C)
–210
Boiling point (°C)
–196
Bond enthalpy (kJ mol–1)
+944
First ionization enthalpy (kJ mol–1)
+1 400
Electron affinity (kJ mol–1) +3
Electronegativity
3.0

3. Mainly as free N2 molecules in the atmosphere (78% by volume)
43.1 Introduction (SB p.110)
Nitrogen
Mainly as free N2 molecules in the atmosphere (78% by volume)
Combine with other elements in the form of proteins in all living things
Liquid N2 is used as coolant
Raw material for Haber process (manufacture of ammonia)
Ammonia is the major component of nitrogenous fertilizers

4. Nitrogen in gaseous state
43.2 Unreactive Nature of Nitrogen (SB p.111)
Nitrogen in gaseous state
As diatomic molecules (N2) which are held by weak van der Waals’ forces
2 atoms are joined by extremely strong triple covalent bonds
Bond enthalpy of the triple bond = +944 kJ mol–1
Due to extremely strong covalent bonds and absence of bond polarity
 Nitrogen molecule is very unreactive

5. Bond enthalpy (kJ mol–1) Bond enthalpy (kJ mol–1)
43.2 Unreactive Nature of Nitrogen (SB p.111)
Bond enthalpies of some common covalent bonds
Bond
Bond enthalpy (kJ mol–1)
Bond enthalpy (kJ mol–1)
N  N
O = O
H – H
C – C
+944
+496
+436
+348
S – S
Cl – Cl
P – P
F – F
+264
+242
+172
+158

6. 43.2 Unreactive Nature of Nitrogen (SB p.111)
Reactions involving nitrogen usually have high activation energies and unfavourable equilibrium constants
e.g. At 25°C
N2(g) + O2(g) 2NO(g) Kc = 4.5  10–31
The presence of catalyst and high temperature and pressure may be required for nitrogen to react
N2(g) + 3H2(g) 2NH3(g)
400 – 500°C, 300 – 1000 atm
Fe as catalyst

7. N2 will not react at room temperature due to high bond enthalpy
43.3 Direct Combination of Nitrogen and Oxygen leading to Formation of Nitrogen Oxides (SB p.112)
N2 will not react at room temperature due to high bond enthalpy
At high temperature, N2 shows some reactions with other elements
∵ sufficient energy to break N  N triple bond
At high temperature,
N2(g) + O2(g) 2NO(g) 

8. 43. 3 Direct Combination of Nitrogen and Oxygen leading to
43.3 Direct Combination of Nitrogen and Oxygen leading to Formation of Nitrogen Oxides (SB p.112)
The electric discharge in lightning provides sufficient energy to break the N  N triple bond and then react with O2
N2(g) + O2(g)  2NO(g)
2NO(g) + O2(g)  2NO2(g)
lightning
colourless
Reddish brown
(poisonous)

9. The above reactions are very important in nature
43.3 Direct Combination of Nitrogen and Oxygen leading to Formation of Nitrogen Oxides (SB p.112)
The above reactions are very important in nature
The NO2 formed dissolves in rainwater to produce nitric(V) acid and nitric(III) acid
2NO2(g) + H2O(l)  HNO3(aq) + HNO2(aq)

10. The NO formed will be oxidized to NO2
43.3 Direct Combination of Nitrogen and Oxygen leading to Formation of Nitrogen Oxides (SB p.112)
At high temperatures in car engines, N2 & O2 react to form NO(g) which emitted into air with exhausted gas
The NO formed will be oxidized to NO2
The NO2 absorbs sunlight and breaks down into NO and O atom
NO2(g)  NO(g) + O(g)
These leads to formation of photochemical smog
sunlight

11. NO is formed and followed by NO2
43.3 Direct Combination of Nitrogen and Oxygen leading to Formation of Nitrogen Oxides (SB p.113)
In laboratory, we use the apparatus shown on the right to convert N2 into NO2
When current is switched on, electric discharges occur in the gap between the electrodes
NO is formed and followed by NO2

12. Other than NO and NO2, N2 can form other oxides
43.3 Direct Combination of Nitrogen and Oxygen leading to Formation of Nitrogen Oxides (SB p.113)
Other than NO and NO2, N2 can form other oxides
e.g. 2 NO2 molecules (brown) can combine to form a N2O4 molecule (yellow)
NO2 & N2O4 exist in equilibrium in gas phase
2NO2(g) N2O4(g) H = –58 kJ mol–1
brown
yellow

13. The formation of N2O4 is exothermic
43.3 Direct Combination of Nitrogen and Oxygen leading to Formation of Nitrogen Oxides (SB p.113)
NO2 (left) and N2O4 (right) predominate in hot water and ice water respectively
The formation of N2O4 is exothermic
 N2O4 predominantes at low temperatures
 NO2 predominantes at high temperatures
 the colour of mixture fades on cooling, darkens on warming

14. Check Point 43-1 (a) Draw the structures of the following compounds.
43.3 Direct Combination of Nitrogen and Oxygen leading to Formation of Nitrogen Oxides (SB p.114)
Check Point 43-1
(a) Draw the structures of the following compounds.
(i) Dinitrogen monoxide
(ii) Nitrogen monoxide
(iii) Dinitrogen trioxide
(iv) Nitrogen dioxide
(v) Dinitrogen tetraoxide
(vi) Dinitrogen pentaoxide
(a) (i) Dinitrogen monoxide
(ii) Nitrogen monoxide
Answer

15. (iii) Dinitrogen trioxide
43.3 Direct Combination of Nitrogen and Oxygen leading to Formation of Nitrogen Oxides (SB p.114)
(iii) Dinitrogen trioxide
(iv) Nitrogen dioxide

16. (v) Dinitrogen tetraoxide
43.3 Direct Combination of Nitrogen and Oxygen leading to Formation of Nitrogen Oxides (SB p.114)
(v) Dinitrogen tetraoxide
(vi) Dinitrogen pentaoxide

17. 43. 3 Direct Combination of Nitrogen and Oxygen leading to
43.3 Direct Combination of Nitrogen and Oxygen leading to Formation of Nitrogen Oxides (SB p.114)
Check Point 43-1
(b) Arrange N2, O2 and F2 in an ascending order of reactivity. Explain the order briefly.
(b) The ascending order of reactivity is: N2 < O2 < F2.
The reactivity of diatomic molecules depends on the bond enthalpy of covalent bonds. The bond enthalpy of N  N is greater than that of O = O, which in turn is greater than that of F – F. Therefore, the breakage of N  N bond requires the greatest amount of energy, whereas the breakage of F – F bond requires the least amount of energy.
Answer

18. colourless, pungent gas polar molecules
43.4 Ammonia (SB p.114)
Ammonia
colourless, pungent gas
polar molecules
trigonal pyramidal shape with a lone pair of electrons on nitrogen
extremely soluble in water and easy to condense to liquid due to hydrogen bonds
good solvent for ionic compounds
weakly alkaline
NH3(aq) + H2O(l) NH4+(aq) + OH–(aq)
Kb = 1.8  10–5 mol dm–3

19. one of the most fundamental raw materials for modern industries
43.4 Ammonia (SB p.114)
Ammonia
one of the most fundamental raw materials for modern industries
important source of fertilizers and 85% of ammonia is used to make nitrogenous fertilizers (e.g. (NH4)2SO4, NH4NO3)
making fibres and plastics (rayon, nylon)
making nitric(V) acid (used to make fertilizers, dyes)
making household cleaners
making detergents

20. 43.4 Ammonia (SB p.115)
Percentages of ammonia used in different industries

21. Manufacture of Ammonia by the Haber Process
43.4 Ammonia (SB p.115)
Manufacture of Ammonia by the Haber Process
NH3 is manufactured industrially by the Haber Process, named after the German chemist Fritz Haber
The process involves direct combination of N2 and H2 under special conditions
N2(g) + H2(g) 2NH3(g) H = –92 kJ mol–1
Fritz Haber (1868 – 1934)

22. 43.4 Ammonia (SB p.116)
Flow diagram for the Haber process

23. Process of Haber Process:
43.4 Ammonia (SB p.116)
Process of Haber Process:
N2 is obtained from fractional distillation of liquid air
H2 is obtained from methane, naphtha or mixture by steam reforming
CH4(g) + H2O(g)  CO(g) + 3H2(g)
CH4(g) + air  CO(g) + 2H2(g) + N2(g) Ni
900°C

24. CO(g) + H2O(l)  CO2(g) + H2(g)
43.4 Ammonia (SB p.116)
Mixture of CO & H2O is mixed with steam and passed over a heated catalyst
CO(g) + H2O(l)  CO2(g) + H2(g)
The CO2 formed is dissolved in water under pressure
The gases (N2 & H2) are purified before proceeding to the next stage
∵ Compounds of oxygen and sulphur will poison the catalyst

25. Purified N2 & H2 are mixed in ratio of 3 : 1 by volume
43.4 Ammonia (SB p.116)
Purified N2 & H2 are mixed in ratio of 3 : 1 by volume
 Compressed to 200 – 1000 atm and heated in the heat exchanger
 Hot gaseous mixture is passed over iron in the catalytic chamber
 Gases contain 10 – 15% of NH3 and unreacted N2 and H2 when leaving the chamber
 The gases are cooled after passing through the heat exchanger
 NH3 is liquefied under pressure and unreacted gases are recycled

26. Physico-chemical principles:
43.4 Ammonia (SB p.117)
Physico-chemical principles:
Synthesis of ammonia is an exothermic and reversible reaction
N2(g) + H2(g) 2NH3(g) H = –92 kJ mol–1
According to Le Chatelier’s principle,
(1) high pressure will increase the yield
(2) low temperature will increase the yield

27. Apart from increasing yield, the reaction rate should be fast
43.4 Ammonia (SB p.117)
Apart from increasing yield, the reaction rate should be fast
 Low temperatures would lower the rate of reaction
∴ optimum temperature is around 500°C which is high enough for reaction to proceed quickly but low enough to give satisfactory yield
Catalyst is used to increase the reaction rate
 poisoned by CO, CO2, H2S
 Gases entering the catalytic chamber should have high purity!!

28. Chemical Properties of Ammonia
43.4 Ammonia (SB p.118)
Chemical Properties of Ammonia
As a base
NH3 partly ionizes in water to give NH4+ and OH– ions
∴ NH3(aq) is alkaline
NH3(g) + H2O(l) NH4+(aq) + OH–(aq)
Kb = 1.8  10–5 mol dm–3
∴ NH3(aq) is a weak base

29. NH3 neutralizes acids to give ammonium salts
43.4 Ammonia (SB p.118)
Reaction with Acids
NH3 neutralizes acids to give ammonium salts
e.g. 2NH3(aq) + H2SO4(aq)  (NH4)2SO4(aq)
NH3(aq) + HNO3(aq)  NH4NO3(aq)
ammonium sulphate(VI)
ammonium nitrate(V)
Filter paper soaked with NH3
Filter paper soaked with HCl
Formation of NH4Cl by reacting NH3 with HCl
NH3(aq) + HCl(aq)  NH4Cl(s)

30. Reaction with Metal Salts
43.4 Ammonia (SB p.118)
Reaction with Metal Salts
NH3 precipitates the hydroxide of many metals from solutions of their salts
CaSO4(aq) + 2NH3(aq) +2H2O(l)  Ca(OH)2(s) + (NH4)2SO4(aq)
ZnSO4(aq) + 2NH3(aq) +2H2O(l)  Zn(OH)2(s) + (NH4)2SO4(aq)
Pb(NO3)2(aq) + 2NH3(aq) +2H2O(l)  Pb(OH)2(s) + 2NH4NO3(aq)
CuSO4(aq) + 2NH3(aq) +2H2O(l)  Cu(OH)2(s) + (NH4)2SO4(aq)
FeSO4(aq) + 2NH3(aq) +2H2O(l)  Fe(OH)2(s) + (NH4)2SO4(aq)
Fe2(SO4)3(aq) + 6NH3(aq) +6H2O(l)  2Fe(OH)3(s) + 3(NH4)2SO4(aq)
white
blue
dirty green
reddish brown

31. 43.4 Ammonia (SB p.119)
Pb(OH)2(s)
Cu(OH)2(s)
Fe(OH)2(s)
Fe(OH)3(s)

32. Zn(OH)2(s) + 4NH3(aq)  [Zn(NH3)4]2+(aq) + 2OH–(aq)
43.4 Ammonia (SB p.119)
Some metal hydroxides (e.g. Zn(OH)2 & Cu(OH)2) redissolve in excess NH3 solution and form complex compounds
Zn(OH)2(s) + 4NH3(aq)  [Zn(NH3)4]2+(aq) + 2OH–(aq)
Cu(OH)2(s) + 4NH3(aq)  [Cu(NH3)4]2+(aq) + 2OH–(aq)
colourless
deep blue
A solution containing Cu2+(aq)
Cu(OH)2(s)
[Cu(NH3)4]2+(aq)

33. Silver(I) ions also form a complex with ammonia
43.4 Ammonia (SB p.120)
Silver(I) ions also form a complex with ammonia
AgCl is insoluble in water and acids, but dissolves in excess NH3 forming soluble complex ion [Ag(NH3)2]+(aq)
AgCl(s) Ag+(aq) + Cl–(aq)
Ag+(aq) + 2NH3(aq) [Ag(NH3)2]+(aq)
AgCl(s) dissolves
Addition of excess NH3(aq)
water
AgCl(s)

34. NH3 does not burn in air or support combustion
43.4 Ammonia (SB p.120)
As a Reducing Agent
Reaction with Oxygen
NH3 does not burn in air or support combustion
It burns in O2 with a yellow flame, forming N2 and water vapour
Laboratory set-up for oxidation of ammonia
4NH3(g) + 3O2(g)  2N2(g) + 6H2O(g) 

35. This is called catalytic oxidation of ammonia
43.4 Ammonia (SB p.120)
In the presence of catalyst (red hot spiral coil of platinum at 800 – 900°C), NH3 is oxidized to NO by O2
Laboratory set-up for catalytic oxidation of ammonia
4NH3(g) + 5O2(g)
 4NO(g) + 6H2O(g)
This is called catalytic oxidation of ammonia
Key reaction in the preparation of HNO3  Pt

36. Reaction with Copper(II) Oxide
43.4 Ammonia (SB p.121)
Reaction with Copper(II) Oxide
When dry NH3 is passed over heated black CuO, NH3 is oxidized to N2 and H2O
The CuO turns from black to reddish brown as it is reduced to Cu
2NH3(g) + 3CuO(s)  3Cu(s) + N2(g) + 3H2O(g) 
Laboratory set-up for oxidation of ammonia by copper(II) oxide

37. 43.4 Ammonia (SB p.121)
Check Point 43-2
(a) Write chemical equations to show how hydrogen is produced from
(i) the reaction of natural gas (mainly methane) with water;
(ii) the reaction of coal (mainly carbon) with water.
(a) (i) CH4(g) + H2O(g)  CO(g) + 3H2(g)
(ii) C(s) + H2O(l)  CO(g) + H2(g) 
Answer

38. 43.4 Ammonia (SB p.121)
Check Point 43-2 (cont’d)
(b) Consider the following reversible reaction:
N2(g) + 3H2(g) 2NH3(g) H = –92 kJ mol–1
Discuss how each of the following factors affects the above equilibrium:
(i) increase in temperature
(ii) decrease in pressure
(iii) addition of a suitable catalyst
Answer

39. 43.4 Ammonia (SB p.121)
(b) (i) The forward reaction is exothermic. According to Le Chatelier’s principle, exothermic reactions are favoured at low temperatures. Therefore, an increase in temperature will favour the backward reaction, and thus decrease the yield of ammonia.
(ii) According to Le Chatelier’s principle, a high pressure will increase the yield of ammonia as the forward reaction is accompanied by a decrease of volume from four to two volumes of the gas. Therefore, a decrease in pressure will decrease the yield of ammonia.
(iii) Addition of a suitable catalyst will increase the rate of both forward and backward reactions to the same extent. As it does not change the position of the equilibrium, the yield of ammonia remains constant.

40. 43.4 Ammonia (SB p.121)
Check Point 43-2 (cont’d)
(c) Ammonia reacts with oxygen in two different ways. Give equations for both of these reactions and explain how one of them is used industrially to produce nitric(V) acid.
Answer

41. 4NH3(g) + 3O2(g)  2N2(g) + 6H2O(g)
43.4 Ammonia (SB p.121)
(c) In the absence of catalyst, ammonia burns to give molecular nitrogen and water vapour.
4NH3(g) + 3O2(g)  2N2(g) + 6H2O(g)
Industrially, in the presence of red hot platinum-rhodium at about 850°C, ammonia is catalytically oxidized to nitrogen monoxide.
4NH3(g) + 5O2(s) 4NO(g) + 6H2O(g)
The nitrogen monoxide formed then reacts with oxygen from the air to give nitrogen dioxide.
2NO(g) + O2(g)  2NO2(g)
The nitrogen dioxide reacts with excess air and water to produce aqueous nitric(V) acid.
4NO2(g) + O2(g) + 2H2O(l)  4HNO3(aq)
H2O(l) + 3NO2(g)  2HNO3(aq) + NO(g)
The NO(g) is recycled and subsequently combines with more oxygen and water to give more nitric(V) acid. Finally, the product is distilled to give concentrated nitric(V) acid (containing 68% HNO3).
Pt – Rh
850°C

42. 4HNO3(l)  4NO2(aq) + 2H2O(l) +O2(g)
43.5 Nitric(V) Acid (SB p.121)
Nitric(V) acid
a very strong acid
turns yellow on storage as the formation of dissolved NO2 from decomposition of some acid
4HNO3(l)  4NO2(aq) + 2H2O(l) +O2(g)
keep in brown bottles as light will speed up decomposition
used to make explosives, nylon, fertilizers and dyestuff synthesis

43. Manufacture of Nitric(V) Acid from the Catalytic Oxidation of Ammonia
43.5 Nitric(V) Acid (SB p.122)
Manufacture of Nitric(V) Acid from the Catalytic Oxidation of Ammonia
Most of the ammonia formed is converted to nitric(V) acid by Ostward process
Ostward process is divided into 3 stages:
1. Mixture of ammonia and excess air is passed over Pt-Rh catalyst at around °C under low pressure
4NH3(g) + 5O2(g) 4NO(g) + 6H2O(g)
Pt-Rh
850°C

44. 2. The NO formed then reacts with O2 to form NO2
43.5 Nitric(V) Acid (SB p.122)
2. The NO formed then reacts with O2 to form NO2
2NO(g) + O2(g)  NO2(g)
3. The NO2 reacts with excess air and water to give aqueous HNO3
4NO2(g) + O2(g) + 2H2O(l)  4HNO3(aq)

45. Nitric(V) Acid as an Oxidizing Agent
43.5 Nitric(V) Acid (SB p.122)
Nitric(V) Acid as an Oxidizing Agent
HNO3 is a strong oxidizing agent, especially when concentrated
NO3– acts as an electron acceptor when H+ ions are present
HNO3 can be reduced to different nitrogen compounds with different oxidation states, depending on
1. the conc. of HNO3
2. nature of substance being oxidized

46. If concentrated HNO3 is reduced, NO2 will be formed
43.5 Nitric(V) Acid (SB p.122)
If dilute or moderately concentrated HNO3 is reduced, NO will be formed
4HNO3(aq) + 3e–  3NO3 –(aq) + 2H2O(l) + NO(g) or NO3–(aq) + 4H+(aq) + 3e–  NO(g) + 2H2O(l)
If concentrated HNO3 is reduced, NO2 will be formed
2HNO3(aq) + e–  NO3 –(aq) + NO2(g) + H2O(l) or NO3–(aq) + 2H+(aq) + e–  NO2(g) + H2O(l)
The electrons are supplied by the reducing agent in the reaction

47. Cu reacts with warm dilute HNO3 to give NO 3Cu(s) + 8HNO3(aq) 
43.5 Nitric(V) Acid (SB p.123)
Reaction with Copper
Cu reacts with warm dilute HNO3 to give NO 3Cu(s) + 8HNO3(aq) 
3Cu(NO3)2(aq) + 4H2O(l) + 2NO(g)
The NO formed reacts with atmospheric O2 to give NO2 2NO(g) + O2(g)  2NO2(g)

48. Cu(s) + 4HNO3(aq)  Cu(NO3)2(aq) + 2H2O(l) + 2NO2(g)
43.5 Nitric(V) Acid (SB p.123)
Conc. HNO3 (~14 M) reacts with Cu to give NO2 and a blue solution of Cu(NO3)2
Cu(s) + 4HNO3(aq)  Cu(NO3)2(aq) + 2H2O(l) + 2NO2(g)

49. Reaction with Iron(II) Ion
43.5 Nitric(V) Acid (SB p.123)
Reaction with Iron(II) Ion
Conc. HNO3 oxdizes green Fe2+ ions to brown Fe3+ ions while itself reduced to NO
3Fe2+(aq) + NO3–(aq) + 4H+(aq)  3Fe3+(aq) + NO(g) + 2H2O(l)
The NO formed reacts with atmospheric O2 to form NO2
2NO(g) + O2(g)  NO2(g)

50. 43.5 Nitric(V) Acid (SB p.123)
Reaction with Sulphur
Hot concentrated HNO3 oxidizes sulphur to give sulphuric(VI) acid and brown fumes of NO2
S(s) + 6HNO3(aq) 
H2SO4(aq) + 6NO2(g) + 2H2O(l)

51. 43.5 Nitric(V) Acid (SB p.123)
Check Point 43-3
Account for the following observation by giving a balanced equation.
(a) Nitrogen monoxide turns brown when exposed to air.
Answer
(a) Nitrogen monoxide reacts with atmospheric oxygen to give brown nitrogen dioxide gas.
2NO(g) + O2(g)  2NO2(g)

52. 43.5 Nitric(V) Acid (SB p.123)
Check Point 43-3 (cont’d)
Account for the following observation by giving a balanced equation.
(b) Nitric(V) acid turns yellowish brown on standing.
Answer
(b) Nitric(V) acid turns yellowish brown on standing because of the dissolved nitrogen dioxide formed from the decomposition of some of the acid.
4HNO3(aq)  4NO2(aq) + 2H2O(l) + O2(g)

53. 43.5 Nitric(V) Acid (SB p.123)
Check Point 43-3 (cont’d)
Account for the following observation by giving a balanced equation.
(c) Silver dissolves in dilute nitric(V) acid, yielding a colourless gas.
Answer
(c) Dilute nitric(V) acid is reduced by silver to form colourless nitrogen monoxide gas.
3Ag(s) + 4HNO3(aq)
 3Ag+(aq) + 3NO3–(aq) + 2H2O(l) + NO(g)

54. 43.5 Nitric(V) Acid (SB p.123)
Check Point 43-3 (cont’d)
Account for the following observation by giving a balanced equation.
(d) The nitrate of a metal ion decomposed on heat to give the metal.
Answer
(d) Both mercury nitrate(V) and silver nitrate(V) decompose on heating to give the corresponding metal.
Hg(NO3)2(s)  Hg(s) + 2NO2(g) + O2(g)
2Ag(NO3)2(s)  2Ag(s) + 2NO2(g) + O2(g) 

55. e.g. Mg(s) + 2HNO3(aq)  Mg(NO3)2(aq) + H2(g)
43.6 Nitrates(V) (SB p.124)
Metal nitrates(V) can be prepared by reacting very dilute nitric(V) acid with metals, metal oxides, hydroxides or carbonates
e.g. Mg(s) + 2HNO3(aq)  Mg(NO3)2(aq) + H2(g)
CuO(s) + 2HNO3(aq)  Cu(NO3)2(aq) + H2O(l)
NaOH(aq) + HNO3(aq)  NaNO3(aq) + H2O(l)
Na2CO3(aq) + 2HNO3(aq)  2NaNO3(aq) + H2O(l) + CO2(g)
very dilute

56. Mg(NO3)2(aq) + 2H2O(l) + 2NO2(g)
43.6 Nitrates(V) (SB p.124)
Metal nitrates(V) can be prepared by reacting metals with concentrated nitric(V) acid
e.g. Mg(s) + 4HNO3(aq) 
Mg(NO3)2(aq) + 2H2O(l) + 2NO2(g)
concentrated

57. Action of Heat on Nitrates(V)
43.6 Nitrates(V) (SB p.124)
Action of Heat on Nitrates(V)
When metal nitrates(V) in solid form are strongly heated, they decompose differently according to their thermal stability
Metal oxide, nitrogen dioxide and oxygen
2Ca(NO3)2(s)  2CaO(s) + 4NO2(g) + O2(g)
2Mg(NO3)2(s)  2MgO(s) + 4NO2(g) + O2(g)
4Al(NO3)3(s)  2Al2O3(s) + 12NO2(g) + 3O2(g)
2Zn(NO3)2(s)  2ZnO(s) + 4NO2(g) + O2(g)
2Fe(NO3)2(s)  2FeO(s) + 4NO2(g) + O2(g)
2Pb(NO3)2(s)  2PbO(s) + 4NO2(g) + O2(g)
2Cu(NO3)2(s)  2CuO(s) + 4NO2(g) + O2(g)
Calcium
Magnesium
Aluminium
Zinc
Iron
Lead
Copper
Metal nitrate(III), oxygen
2KNO3(s)  2KNO2(s) + O2(g)
2NaNO3(s)  2NaNO2(s) + O2(g)
Potassium
Sodium
Product
Reaction
Nitrate of 

58. Dinitrogen oxide and water NH4NO3(s)  N2O(g) + 2H2O(l) Ammonium ion
43.6 Nitrates(V) (SB p.124)
Dinitrogen oxide and water
NH4NO3(s)  N2O(g) + 2H2O(l)
Ammonium ion
Metal, nitrogen dioxide and oxygen
Hg(NO3)2(s)  Hg(s) + 2NO2(g) + O2(g)
2AgNO3(s)  2Ag(s) + 2NO2(g) + O2(g)
Mercury(II)
Silver
Product
Reaction
Nitrate of
Cont’d 

59. Brown Ring Test for Nitrate(V) Ions
43.6 Nitrates(V) (SB p.125)
Brown Ring Test for Nitrate(V) Ions
The brown ring test is used to detect nitrate(V) ions in aqueous solutions
Procedure:
1. Mix a freshly prepared FeSO4 solution with a solution suspected of containing nitrate(V) ions in a test tube
2. Conc. H2SO4 is added carefully along the side to the bottom of the test tube with the test tube tilted
Laboratory set-up for brown ring test

60. 43.6 Nitrates(V) (SB p.125)
Formation of a brown ring confirms the presence of nitrate(V) ions in the solution

61. Reactions involved in the brown ring test:
43.6 Nitrates(V) (SB p.125)
Reactions involved in the brown ring test:
Nitrate(V) ions react with conc. H2SO4 to give HNO3
NO3–(aq) + H2SO4(l)  HNO3(aq) + HSO4–(aq)
The nitric(V) acid oxidizes some FeSO4 to Fe2(SO4)3 and is itself reduced to NO
HNO3(aq) + 3Fe2+(aq) + 3H+(aq)
 NO(g) + 3Fe3+(aq) + 2H2O(l)
Finally, NO reacts with unreacted FeSO4 to form a brown complex
FeSO4(aq) + NO(g)  FeSO4 • NO(aq)
Brown complex

62. 43.6 Nitrates(V) (SB p.125)
Check Point 43-4
Give the name of the ion responsible for the following observation.
(a) An ion produces a blue precipitate with ammonia solution. The blue precipitate redissolves in excess ammonia solution to give a clear deep blue solution.
Answer
(a) Copper(II) ion

63. 43.6 Nitrates(V) (SB p.125)
Check Point 43-4 (cont’d)
Give the name of the ion responsible for the following observation.
(b) An ion produces a dirty green precipitate with ammonia solution.
Answer
(b) Iron(II) ion

64. 43.6 Nitrates(V) (SB p.125)
Check Point 43-4 (cont’d)
Give the name of the ion responsible for the following observation.
(c) An ion gives a positive result in the brown ring test.
Answer
(c) Nitrate(V) ion

65. The END