Chemistry 11 Part II

167 Pages·2009·7.599 MB·English

by  Various

ANSWERS 7
C
ONTENTS
FOREWORD iii
Unit 8 Redox Reactions 255
8.1 Classical Idea of Redox Reactions-Oxidation and Reduction Reactions 255
8.2 Redox Reactions in Terms of Electron Transfer Reactions 257
8.3 Oxidation Number 259
8.4 Redox Reactions and Electrode Processes 269
Unit 9 Hydrogen 276
9.1 Position of Hydrogen in the Periodic Table 276
9.2 Dihydrogen, H 277
2
9.3 Preparation of Dihydrogen, H 278
2
9.4 Properties of Dihydrogen 278
9.5 Hydrides 280
9.6 Water 281
9.7 Hydrogen Peroxide (H O ) 285
2 2
9.8 Heavy Water, D O 286
2
9.9 Dihydrogen as a Fuel 286
Unit 10 The s-Block Elements 291
10.1 Group 1 Elements: Alkali Metals 292
10.2 General Characteristics of the Compounds of the Alkali Metals 295
10.3 Anomalous Properties of Lithium 296
10.4 Some Important Compounds of Sodium 296
10.5 Biological Importance of Sodium and Potassium 298
10.6 Group 2 Elements : Alkaline Earth Metals 298
10.7 General Characteristics of Compounds of the Alkaline Earth Metals 301
10.8 Anomalous Behaviour of Beryllium 302
10.9 Some Important Compounds of Calcium 302
10.10 Biological Importance of Magnesium and Calcium 304
Unit 11 The p-Block Elements 307
11.1 Group 13 Elements: The Boron Family 309
11.2 Important Trends and Anomalous Properties of Boron 312
11.3 Some Important Compounds of Boron 312
11.4 Uses of Boron and Aluminium and their Compounds 314
8 (viii) CHEMISTRY
11.5 Group 14 Elements: The Carbon Family 314
11.6 Important Trends and Anomalous Behaviour of Carbon 317
11.7 Allotropes of Carbon 317
11.8 Some Important Compounds of Carbon and Silicon 319
Unit 12 Organic Chemistry – Some Basic Principles and Techniques 326
12.1 General Introduction 326
12.2 Tetravalence of Carbon: Shapes of Organic Compounds 327
12.3 Structural Representations of Organic Compounds 328
12.4 Classification of Organic Compounds 331
12.5 Nomenclature of Organic Compounds 332
12.6 Isomerism 340
12.7 Fundamental Concepts in Organic Reaction Mechanism 341
12.8 Methods of Purification of Organic Compounds 348
12.9 Qualitative Analysis of Organic Compounds 354
12.10 Quantitative Analysis 355
Unit 13 Hydrocarbons 365
13.1 Classification 365
13.2 Alkanes 366
13.3 Alkenes 376
13.4 Alkynes 384
13.5 Aromatic Hydrocarbon 388
13.6 Carcinogenicity and Toxicity 395
Unit 14 Environmental Chemistry 398
14.1 Environmental Pollution 398
14.2 Atmospheric Pollution 399
14.3 Water Pollution 406
14.4 Soil Pollution 408
14.5 Industrial Waste 409
14.6 Strategies to control Environmental Pollution 410
14.7 Green Chemistry 410
Answers 414
Index 418
REDOX REACTIONS 255
UNIT 8
REDOX REACTIONS
Where there is oxidation, there is always reduction –
Chemistry is essentially a study of redox systems.
After studying this unit you will be
Chemistry deals with varieties of matter and change of one
able to
kind of matter into the other. Transformation of matter from
••••• identify redox reactions as a class
one kind into another occurs through the various types of
of reactions in which oxidation
and reduction reactions occur reactions. One important category of such reactions is
simultaneously; Redox Reactions. A number of phenomena, both physical
••••• define the terms oxidation, as well as biological, are concerned with redox reactions.
reduction, oxidant (oxidising These reactions find extensive use in pharmaceutical,
agent) and reductant (reducing biological, industrial, metallurgical and agricultural areas.
agent); The importance of these reactions is apparent from the fact
••••• explain mechanism of redox that burning of different types of fuels for obtaining energy
reactions by electron transfer for domestic, transport and other commercial purposes,
process;
electrochemical processes for extraction of highly reactive
••••• use the concept of oxidation metals and non-metals, manufacturing of chemical
number to identify oxidant and
compounds like caustic soda, operation of dry and wet
reductant in a reaction;
batteries and corrosion of metals fall within the purview of
••••• classify redox reaction into redox processes. Of late, environmental issues like
combination (synthesis),
Hydrogen Economy (use of liquid hydrogen as fuel) and
decomposition, displacement
development of ‘Ozone Hole’ have started figuring under
and disproportionation
reactions; redox phenomenon.
••••• suggest a comparative order
8.1 CLASSICAL IDEA OF REDOX REACTIONS –
among various reductants and
OXIDATION AND REDUCTION REACTIONS
oxidants;
••••• balance chemical equations Originally, the term oxidation was used to describe the
using (i) oxidation number addition of oxygen to an element or a compound. Because
(ii) half reaction method; of the presence of dioxygen in the atmosphere (~20%),
••••• learn the concept of redox many elements combine with it and this is the principal
reactions in terms of electrode reason why they commonly occur on the earth in the
processes. form of their oxides. The following reactions represent
oxidation processes according to the limited definition of
oxidation:
2 Mg (s) + O (g) → 2 MgO (s) (8.1)
2
S (s) + O (g) → SO (g) (8.2)
2 2
256 CHEMISTRY
In reactions (8.1) and (8.2), the elements been broadened these days to include removal
magnesium and sulphur are oxidised on of oxygen/electronegative element from a
account of addition of oxygen to them. substance or addition of hydrogen/
Similarly, methane is oxidised owing to the electropositive element to a substance.
addition of oxygen to it. According to the definition given above, the
CH (g) + 2O (g) → CO (g) + 2H O (l) (8.3) following are the examples of reduction
4 2 2 2
processes:
A careful examination of reaction (8.3) in
which hydrogen has been replaced by oxygen 2 HgO (s) 2 Hg (l) + O (g) (8.8)
2
prompted chemists to reinterpret oxidation in
(removal of oxygen from mercuric oxide )
terms of removal of hydrogen from it and,
2 FeCl (aq) + H (g) →2 FeCl (aq) + 2 HCl(aq)
therefore, the scope of term oxidation was 3 2 2
(8.9)
broadened to include the removal of hydrogen
(removal of electronegative element, chlorine
from a substance. The following illustration is
from ferric chloride)
another reaction where removal of hydrogen
can also be cited as an oxidation reaction. CH = CH (g) + H (g) → H C – CH (g) (8.10)
2 2 2 3 3
2 H S(g) + O (g) → 2 S (s) + 2 H O (l) (8.4) (addition of hydrogen)
2 2 2
2HgCl (aq) + SnCl (aq) → Hg Cl (s)+SnCl (aq)
As knowledge of chemists grew, it was 2 2 2 2 4
(8.11)
natural to extend the term oxidation for
(addition of mercury to mercuric chloride)
reactions similar to (8.1 to 8.4), which do not
involve oxygen but other electronegative In reaction (8.11) simultaneous oxidation
elements. The oxidation of magnesium with of stannous chloride to stannic chloride is also
fluorine, chlorine and sulphur etc. occurs occurring because of the addition of
according to the following reactions : electronegative element chlorine to it. It was
soon realised that oxidation and reduction
Mg (s) + F (g) → MgF (s) (8.5)
2 2 always occur simultaneously (as will be
apparent by re-examining all the equations
Mg (s) + Cl (g) → MgCl (s) (8.6)
2 2 given above), hence, the word “redox” was
Mg (s) + S (s) → MgS (s) (8.7) coined for this class of chemical reactions.
Incorporating the reactions (8.5 to 8.7) Problem 8.1
within the fold of oxidation reactions
In the reactions given below, identify the
encouraged chemists to consider not only the
species undergoing oxidation and
removal of hydrogen as oxidation, but also the
reduction:
removal of electropositive elements as
(i) H S (g) + Cl (g) → 2 HCl (g) + S (s)
oxidation. Thus the reaction : 2 2
(ii) 3Fe O (s) + 8 Al (s) → 9 Fe (s)
2K [Fe(CN) ](aq) + H O (aq) →2K [Fe(CN) ](aq) 3 4
4 6 2 2 3 6
+ 2 KOH (aq) + 4Al2O3 (s)
is interpreted as oxidation due to the removal (iii) 2 Na (s) + H2 (g) → 2 NaH (s)
of electropositive element potassium from
Solution
potassium ferrocyanide before it changes to
(i) H S is oxidised because a more
potassium ferricyanide. To summarise, the 2
electronegative element, chlorine is added
term “oxidation” is defined as the addition
to hydrogen (or a more electropositive
of oxygen/electronegative element to a
element, hydrogen has been removed
substance or removal of hydrogen/
from S). Chlorine is reduced due to
electropositive element from a substance.
addition of hydrogen to it.
In the beginning, reduction was
(ii) Aluminium is oxidised because
considered as removal of oxygen from a
oxygen is added to it. Ferrous ferric oxide
compound. However, the term reduction has
REDOX REACTIONS 257
(Fe O ) is reduced because oxygen has For convenience, each of the above
3 4
been removed from it. processes can be considered as two separate
steps, one involving the loss of electrons and
(iii) With the careful application of the
the other the gain of electrons. As an
concept of electronegativity only we may
illustration, we may further elaborate one of
infer that sodium is oxidised and
these, say, the formation of sodium chloride.
hydrogen is reduced.
Reaction (iii) chosen here prompts us to 2 Na(s) → 2 Na+(g) + 2e–
think in terms of another way to define Cl (g) + 2e– → 2 Cl–(g)
2
redox reactions.
Each of the above steps is called a half
reaction, which explicitly shows involvement
8.2 REDOX REACTIONS IN TERMS OF
of electrons. Sum of the half reactions gives
ELECTRON TRANSFER REACTIONS
the overall reaction :
We have already learnt that the reactions
2Na(s) + Cl (g) → 2NaCl (s) (8.12) 2 Na(s) + Cl2 (g) → 2 Na+ Cl– (s) or 2 NaCl (s)
2
2Na(s) + O (g) → Na O(s) (8.13) Reactions 8.12 to 8.14 suggest that half
2 2 reactions that involve loss of electrons are
2Na(s) + S(s) → Na S(s) (8.14)
2 called oxidation reactions. Similarly, the
are redox reactions because in each of these half reactions that involve gain of electrons
reactions sodium is oxidised due to the are called reduction reactions. It may not
addition of either oxygen or more be out of context to mention here that the new
electronegative element to sodium. way of defining oxidation and reduction has
Simultaneously, chlorine, oxygen and sulphur been achieved only by establishing a
are reduced because to each of these, the correlation between the behaviour of species
electropositive element sodium has been as per the classical idea and their interplay in
added. From our knowledge of chemical electron-transfer change. In reactions (8.12 to
bonding we also know that sodium chloride, 8.14) sodium, which is oxidised, acts as
sodium oxide and sodium sulphide are ionic a reducing agent because it donates electron
compounds and perhaps better written as to each of the elements interacting with it and
Na+Cl– (s), (Na+) O2–(s), and (Na+) S2–(s). thus helps in reducing them. Chlorine, oxygen
2 2
Development of charges on the species and sulphur are reduced and act as oxidising
produced suggests us to rewrite the reactions agents because these accept electrons from
(8.12 to 8.14) in the following manner : sodium. To summarise, we may mention that
Oxidation: Loss of electron(s) by any species.
Reduction: Gain of electron(s) by any species.
Oxidising agent : Acceptor of electron(s).
Reducing agent : Donor of electron(s).
Problem 8.2 Justify that the reaction :
2 Na(s) + H (g) → 2 NaH (s) is a redox
2
change.
Solution
Since in the above reaction the compound
formed is an ionic compound, which may
also be represented as Na+H– (s), this
suggests that one half reaction in this
process is :
2 Na (s) → 2 Na+(g) + 2e–
258 CHEMISTRY
and the other half reaction is: At this stage we may investigate the state
H (g) + 2e– → 2 H–(g) of equilibrium for the reaction represented by
2 equation (8.15). For this purpose, let us place
This splitting of the reaction under
a strip of metallic copper in a zinc sulphate
examination into two half reactions
solution. No visible reaction is noticed and
automatically reveals that here sodium is
attempt to detect the presence of Cu2+ ions by
oxidised and hydrogen is reduced,
passing H S gas through the solution to
therefore, the complete reaction is a redox 2
produce the black colour of cupric sulphide,
change.
CuS, does not succeed. Cupric sulphide has
such a low solubility that this is an extremely
8.2.1 Competitive Electron Transfer
sensitive test; yet the amount of Cu2+ formed
Reactions
cannot be detected. We thus conclude that the
Place a strip of metallic zinc in an aqueous
state of equilibrium for the reaction (8.15)
solution of copper nitrate as shown in Fig. 8.1,
greatly favours the products over the reactants.
for about one hour. You may notice that the
Let us extend electron transfer reaction now
strip becomes coated with reddish metallic
to copper metal and silver nitrate solution in
copper and the blue colour of the solution
disappears. Formation of Zn2+ ions among the water and arrange a set-up as shown in
Fig. 8.2. The solution develops blue colour due
products can easily be judged when the blue
colour of the solution due to Cu2+ has to the formation of Cu2+ ions on account of the
reaction:
disappeared. If hydrogen sulphide gas is
passed through the colourless solution
containing Zn2+ ions, appearance of white zinc
sulphide, ZnS can be seen on making the
solution alkaline with ammonia.
The reaction between metallic zinc and the (8.16)
aqueous solution of copper nitrate is : Here, Cu(s) is oxidised to Cu2+(aq) and
Zn(s) + Cu2+ (aq) → Zn2+ (aq) + Cu(s) (8.15) Ag+(aq) is reduced to Ag(s). Equilibrium greatly
In reaction (8.15), zinc has lost electrons favours the products Cu2+ (aq) and Ag(s).
to form Zn2+ and, therefore, zinc is oxidised. By way of contrast, let us also compare the
Evidently, now if zinc is oxidised, releasing reaction of metallic cobalt placed in nickel
electrons, something must be reduced, sulphate solution. The reaction that occurs
accepting the electrons lost by zinc. Copper here is :
ion is reduced by gaining electrons from the zinc.
Reaction (8.15) may be rewritten as :
(8.17)
Fig. 8.1 Redox reaction between zinc and aqueous solution of copper nitrate occurring in a beaker.
REDOX REACTIONS 259
Fig. 8.2 Redox reaction between copper and aqueous solution of silver nitrate occurring in a beaker.
At equilibrium, chemical tests reveal that both However, as we shall see later, the charge
Ni2+(aq) and Co2+(aq) are present at moderate transfer is only partial and is perhaps better
concentrations. In this case, neither the described as an electron shift rather than a
reactants [Co(s) and Ni2+(aq)] nor the products complete loss of electron by H and gain by O.
[Co2+(aq) and Ni (s)] are greatly favoured. What has been said here with respect to
This competition for release of electrons equation (8.18) may be true for a good number
of other reactions involving covalent
incidently reminds us of the competition for
compounds. Two such examples of this class
release of protons among acids. The similarity
of the reactions are:
suggests that we might develop a table in
which metals and their ions are listed on the H (s) + Cl (g) → 2HCl(g) (8.19)
2 2
basis of their tendency to release electrons just and,
as we do in the case of acids to indicate the
CH (g) + 4Cl (g) → CCl (l) + 4HCl(g) (8.20)
strength of the acids. As a matter of fact we 4 2 4
In order to keep track of electron shifts in
have already made certain comparisons. By
chemical reactions involving formation of
comparison we have come to know that zinc
covalent compounds, a more practical method
releases electrons to copper and copper
of using oxidation number has been
releases electrons to silver and, therefore, the
developed. In this method, it is always
electron releasing tendency of the metals is in
assumed that there is a complete transfer of
the order: Zn>Cu>Ag. We would love to make
electron from a less electronegative atom to a
our list more vast and design a metal activity
more electonegative atom. For example, we
series or electrochemical series. The
rewrite equations (8.18 to 8.20) to show
competition for electrons between various
charge on each of the atoms forming part of
metals helps us to design a class of cells,
the reaction :
named as Galvanic cells in which the chemical
0 0 +1 –2
reactions become the source of electrical
energy. We would study more about these cells 2H2(g) + O2(g) → 2H2O (l) (8.21)
in Class XII. 0 0 +1 –1
H (s) + Cl (g) → 2HCl(g) (8.22)
2 2
8.3 OXIDATION NUMBER
–4+1 0 +4 –1 +1 –1
A less obvious example of electron transfer is
CH (g) + 4Cl (g) → 4CCl (l) +4HCl(g) (8.23)
realised when hydrogen combines with oxygen 4 2 4
to form water by the reaction: It may be emphasised that the assumption
2H (g) + O (g) → 2H O (l) (8.18) of electron transfer is made for book-keeping
2 2 2
purpose only and it will become obvious at a
Though not simple in its approach, yet we
later stage in this unit that it leads to the simple
can visualise the H atom as going from a
description of redox reactions.
neutral (zero) state in H to a positive state in
2
H O, the O atom goes from a zero state in O Oxidation number denotes the
2 2
to a dinegative state in H O. It is assumed that oxidation state of an element in a
2
there is an electron transfer from H to O and compound ascertained according to a set
consequently H is oxidised and O is reduced. of rules formulated on the basis that
2 2
260 CHEMISTRY
electron in a covalent bond belongs of oxygen but this number would now be
entirely to more electronegative element. a positive figure only.
It is not always possible to remember or 4. The oxidation number of hydrogen is +1,
make out easily in a compound/ion, which except when it is bonded to metals in binary
element is more electronegative than the other. compounds (that is compounds containing
two elements). For example, in LiH, NaH,
Therefore, a set of rules has been formulated
and CaH , its oxidation number is –1.
to determine the oxidation number of an 2
element in a compound/ion. If two or more 5. In all its compounds, fluorine has an
than two atoms of an element are present in oxidation number of –1. Other halogens (Cl,
the molecule/ion such as Na S O /Cr O2–, the Br, and I) also have an oxidation number
2 2 3 2 7
of –1, when they occur as halide ions in
oxidation number of the atom of that element
their compounds. Chlorine, bromine and
will then be the average of the oxidation
iodine when combined with oxygen, for
number of all the atoms of that element. We
example in oxoacids and oxoanions, have
may at this stage, state the rules for the
positive oxidation numbers.
calculation of oxidation number. These rules are:
6. The algebraic sum of the oxidation number
1. In elements, in the free or the uncombined
of all the atoms in a compound must be
state, each atom bears an oxidation
zero. In polyatomic ion, the algebraic sum
number of zero. Evidently each atom in H ,
2 of all the oxidation numbers of atoms of
O , Cl , O , P , S , Na, Mg, Al has the
2 2 3 4 8 the ion must equal the charge on the ion.
oxidation number zero.
Thus, the sum of oxidation number of three
2. For ions composed of only one atom, the
oxygen atoms and one carbon atom in the
oxidation number is equal to the charge carbonate ion, (CO )2– must equal –2.
on the ion. Thus Na+ ion has an oxidation 3
By the application of above rules, we can
number of +1, Mg2+ ion, +2, Fe3+ ion, +3,
find out the oxidation number of the desired
Cl– ion, –1, O2– ion, –2; and so on. In their
element in a molecule or in an ion. It is clear
compounds all alkali metals have
that the metallic elements have positive
oxidation number of +1, and all alkaline
oxidation number and nonmetallic elements
earth metals have an oxidation number of
have positive or negative oxidation number.
+2. Aluminium is regarded to have an
The atoms of transition elements usually
oxidation number of +3 in all its
display several positive oxidation states. The
compounds.
highest oxidation number of a representative
3. The oxidation number of oxygen in most element is the group number for the first two
compounds is –2. However, we come across groups and the group number minus 10
two kinds of exceptions here. One arises (following the long form of periodic table) for
in the case of peroxides and superoxides, the other groups. Thus, it implies that the
the compounds of oxygen in which oxygen highest value of oxidation number exhibited
atoms are directly linked to each other. by an atom of an element generally increases
While in peroxides (e.g., H O , Na O ), each across the period in the periodic table. In the
2 2 2 2
oxygen atom is assigned an oxidation third period, the highest value of oxidation
number of –1, in superoxides (e.g., KO , number changes from 1 to 7 as indicated below
2
RbO ) each oxygen atom is assigned an in the compounds of the elements.
2
oxidation number of –(½). The second A term that is often used interchangeably
exception appears rarely, i.e. when oxygen with the oxidation number is the oxidation
is bonded to fluorine. In such compounds state. Thus in CO , the oxidation state of
2
e.g., oxygen difluoride (OF ) and dioxygen carbon is +4, that is also its oxidation number
2
difluoride (O F ), the oxygen is assigned and similarly the oxidation state as well as
2 2
an oxidation number of +2 and +1, oxidation number of oxygen is – 2. This implies
respectively. The number assigned to that the oxidation number denotes the
oxygen will depend upon the bonding state oxidation state of an element in a compound.
REDOX REACTIONS 261
Group 1 2 13 14 15 16 17
Element Na Mg Al Si P S Cl
Compound NaCl MgSO AlF SiCl PO SF HClO
4 3 4 4 10 6 4
Highest oxidation +1 +2 +3 +4 +5 +6 +7
number state of
the group element
The oxidation number state of a metal in a The idea of oxidation number has been
compound is sometimes presented according invariably applied to define oxidation,
to the notation given by German chemist, reduction, oxidising agent (oxidant), reducing
Alfred Stock. It is popularly known as Stock agent (reductant) and the redox reaction. To
notation. According to this, the oxidation summarise, we may say that:
number is expressed by putting a Roman
Oxidation: An increase in the oxidation
numeral representing the oxidation number
number of the element in the given substance.
in parenthesis after the symbol of the metal in
Reduction: A decrease in the oxidation
the molecular formula. Thus aurous chloride
number of the element in the given substance.
and auric chloride are written as Au(I)Cl and
Au(III)Cl . Similarly, stannous chloride and Oxidising agent: A reagent which can
3
stannic chloride are written as Sn(II)Cl and increase the oxidation number of an element
2
Sn(IV)Cl . This change in oxidation number in a given substance. These reagents are called
4
implies change in oxidation state, which in as oxidants also.
turn helps to identify whether the species is
Reducing agent: A reagent which lowers the
present in oxidised form or reduced form.
oxidation number of an element in a given
Thus, Hg (I)Cl is the reduced form of Hg(II) Cl .
2 2 2 substance. These reagents are also called as
reductants.
Problem 8.3
Redox reactions: Reactions which involve
Using Stock notation, represent the
change in oxidation number of the interacting
following compounds :HAuCl , Tl O, FeO,
4 2
species.
Fe O , CuI, CuO, MnO and MnO .
2 3 2
Solution
Problem 8.4
By applying various rules of calculating
Justify that the reaction:
the oxidation number of the desired
2Cu O(s) + Cu S(s) → 6Cu(s) + SO (g)
element in a compound, the oxidation 2 2 2
number of each metallic element in its is a redox reaction. Identify the species
compound is as follows: oxidised/reduced, which acts as an
HAuCl → Au has 3 oxidant and which acts as a reductant.
4
Tl O → Tl has 1 Solution
2
FeO → Fe has 2 Let us assign oxidation number to each
Fe O → Fe has 3 of the species in the reaction under
2 3
CuI → Cu has 1 examination. This results into:
CuO → Cu has 2
+1 –2 +1 –2 0 +4 –2
MnO → Mn has 2 2Cu O(s) + Cu S(s) → 6Cu(s) + SO
2 2 2
MnO → Mn has 4
2 We therefore, conclude that in this
Therefore, these compounds may be
reaction copper is reduced from +1 state
represented as:
to zero oxidation state and sulphur is
HAu(III)Cl4, Tl2(I)O, Fe(II)O, Fe2(III)O3, oxidised from –2 state to +4 state. The
Cu(I)I, Cu(II)O, Mn(II)O, Mn(IV)O2. above reaction is thus a redox reaction.