Author Archives: Rohit

NCERT Exercises

Question 1.
Define environmental chemistry.
Solution.
The environment around us is made up of chemical species which undergo chemical reactions constantly e.g., photosynthesis in plants, formation of ozone in stratosphere, etc. Environmental chemistry deals with the study of such reactions which take place in the origin, transport and life-cycle of chemical species in the environment.

Question 2.
Explain tropospheric pollution in 100 words.
Solution.
(i) Troposphere is the lowest layer of atmosphere where life exists. The layer gets polluted due to the presence of particulate matter. Such particulate matter may either be:
(a) Solid matter : e.g., dust, mist, fumes, smoke, smog, etc. Smoke particulates consist of solid or mixture of solid and liquid particulates formed during combustion of organic matter. Dust is composed of fine solid particles (over 1 um in diameter), produced during crushing, grinding and attribution of solid materials. Mists are produced by particles of spray liquids and by condensation of vapours in air. Fumes are generally obtained by the condensation of vapours during sublimation, distillation, boiling and several other chemical reactions. The word smog is derived from smoke and fog. There are two types of smog : Classical smog occurs in cool humid climate. It is a mixture of smoke, fog and sulphur dioxide. Photochemical smog occurs in warm, dry and sunny climate. The main components of the photochemical smog result from the action of sunlight on unsaturated hydrocarbons and nitrogen oxides produced by automobiles and factories.
(b) Gaseous matter : e.g., oxides of sulphur and carbon and hydrocarbons, etc. Oxides of sulphur are produced when sulphur containing fossil fuel is burnt. Even a low concentration of sulphur dioxide causes respiratory diseases e.g. asthma, bronchitis, emphysema in human beings. While most of these pollutants are produced by human activities such as mining, burning of fossil fuels, smoke from industries, etc., they may also be produced by natural activities like volcanic eruptions which throw up large quantities of particulate matter or landslides which create an envelope of dust. Hydrocarbons are composed of hydrogen and carbon only and are formed by incomplete combustion of fuel used in automobiles. Hydrocarbons are carcinogens, i.e., they cause cancer. They harm plants by causing ageing, breakdown of tissues and shedding of leaves, flowers and twigs. Carbon monoxide (CO) is one of the most serious air pollutants. It is produced as a result of incomplete combustion of carbon. Carbon dioxide (CO₂) is released into the atmosphere by respiration and burning of fossil fuels for energy. The increased amount of CO₂ in the air is mainly responsible for global warming, thereby, leading to tropospheric pollution.

Question 3.
Carbon monoxide gas is more dangerous than carbon dioxide gas. Why?
Solution.
Carbon monoxide is a colourless, odourless gas which has the tendency to bind to haemoglobin, the oxygen carrying molecule in blood and forms a complex called carboxyhaemoglobin. This complex is 300 times more stable than oxyhaemoglobin complex. When the concentration of carboxyhaemoglobin reaches about 3-4 per cent, the oxygen carrying capacity of blood is greatly reduced. This oxygen deficiency, results into headache, weak eyesight, nervousness and cardiovascular disorder and therefore, if once formed can seriously hamper the body and may cause death. That is why CO is said to be more dangerous than CO₂.

Question 4.
List gases which are responsible for greenhouse effect.
Solution.
Gases responsible for greenhouse effect are :
CO₂, CH₄, N₂O, H₂O_(g), CFCs and O₃. Carbon dioxide molecules trap heat as they are transparent to sunlight but not to the heat radiations. Carbon dioxide is, thus, the major contributor to global warming. Methane is produced naturally when vegetation is burnt, digested or rotted in the absence of oxygen. Large amounts of methane are released in paddy fields, coal mines from rotting garbage dumps and fossil fuels.

Chlorofluorocarbons (CFCs) are man-made industrial chemicals used in air conditioning, etc. CFCs are also damaging the ozone layer. These gases together are responsible for greenhouse effect.

Question 5.
Statues and monuments in India are affected by acid rain. How?
Solution.
Monuments and statues are made of either stone or metal M both of which have a tendency to react with acid and therefore, get affected by acid rain which contains H₂SO₄.

As a result, the monuments are being slowly disfigured.

Question 6.
What is smog? How is classical smog different from photochemical smog?
Solution.
The presence of smoke in fog is called smog. Classical smog occurs in cool and humid climate when smoke and SO₂ are suspended in air. Due to the presence of SO₂ it is found to be reducing in nature. On the other hand photochemical smog occurs in warm, dry and sunny climate. Photochemical smog is produced by the action of light on unsaturated hydrocarbons and NO_x, in air. It is oxdising in nature.

Question 7.
Write down the reactions involved during the formation of photochemical smog.
Solution.
Photochemical smog is produced by action of sunlight on hydrocarbons and nitrogen oxides. NO is converted to NO₂ by the effect of sunlight. This NO₂ further absorbs sunlight and disintegrates into NO and O₂. Thus, a chain reaction begins.

These oxygen atoms react with O₂ to form ozone as :
O_2(g) +O_(g) ⇌ O_3(g)
O_3(g) + NO_(g) ⇌ NO_2(g) + O_2(g)
These two gases formed, react with hydrocarbons to produce other chemicals.

Question 8.
What are the harmful effects of photochemical smog and how can they be controlled?
Solution.
The components of photochemical smog are : O₃, NO, Acrolein, formaldehyde and PAN. Such smog causes damage to human health and property.
(i) Effect on organisms :

•  O₃ and PAN are eye-irritants.
•  O₃ + NO irritate the nose and throat, cause headache, chest pain, cough and breathing problems.
•  Damage to plant life

(ii) Damage to property : These chemicals cause

•  cracking of rubber
•  corrosion of metals, stones, rubber and painted surfaces. If the primary precursors of photochemical smog such as NO₂ and hydrocarbons and the secondary precursors such as ozone and PAN can be controlled, then photochemical smog will be automatically reduced. Catalytic converters are used in automobiles, which prevent the release of nitrogen oxide and hydrocarbons in the atmosphere. Certain plants e.g., Pinus, Juniparus, Quercus, Pyrus and Vitis can metabolise nitrogen oxide and therefore, their plantation could help in reducing photochemical smog.

Question 9.
What are the reactions involved for ozone layer depletion in the stratosphere?
Solution.

1.  The main reason for ozone layer depletion besides other gases are chlorofluorocarbons (CFCs).
2.  Once CFCs are released into the atmosphere, they reach the stratosphere and release free-radicals by the action of sunlight.
   
3.  These radicals further react with ozone and bring about its depletion.
   
4.  Such regeneration of C radicals, causes continuous eating away of the O₃ layer creating ozone holes.

Question 10.
What do you mean by ozone hole? What are its consequences?
Solution.
(i) The depletion of ozone layer is termed as ozone hole. It was first reported by a group of scientists working in Antarctica in 1980s.
(ii) Ozone layer is responsible for prevention of infiltration of UV rays which has the potential to cause serious damage to plants, animals and human life. Due to depletion of this O₃ layer, these harmful rays will find an easy route into the earth’s atmosphere and create problems such as mutation of cells leading to cancer of the skin or increased transpiration in plants and reduced water level in soil. Increase in UV radiations damages paints and fibres causing them to fade faster.

Question 11.
What are the major causes of water pollution? Explain.
Solution.
The major causes of water pollution may be enlisted as:
(i) Pathogens : Pathogens include bacteria and other organisms that enter water from domestic sewage and animal excreta. Human excreta contains bacteria such as Escherichia coli and Streptococcus faecalis which cause gastrointestinal diseases.
(ii) Organic wastes : The other major water pollutant is organic matter such as leaves, grass, trash, etc. They pollute water as a consequence of run off. Excessive phytoplankton growth within water is also a cause of water pollution. These wastes are biodegradable.
(iii) Chemical pollutants : Water soluble inorganic chemicals that include heavy metals such as cadmium, mercury, nickel, etc. constitute an important class of pollutants. These metals then can damage kidneys, central nervous system, liver, etc. Acids (like sulphuric acid) from mine drainage and salts from many different sources including raw salt used to melt snow and ice in the colder climates (sodium and calcium chloride) are water soluble chemical pollutants.

Question 12.
Have you ever observed any water pollution in your area? What measures would you suggest to control it?
Solution.
Yes. The water pollution can be controlled by :

1.  Treatment of sewage :
   1.  by removing impurities,
   2.  by passing chlorine,
   3.  by treatment with alum.
2.  Treatment of industrial waste :
   1.  by precipitating impurities,
   2.  by photocatalysis,
   3.  by using ion exchangers.

Question 13.
What do you mean by Biochemical Oxygen Demand (BOD)?
Solution.
Organic matter in water is biodegradable, i.e., it can be decomposed by the action of bacteria. Now, these bacteria need oxygen to decompose the organic waste. Thus, the amount of oxygen required by bacteria to breakdown the organic matter in a certain volume of a sample of water is called Biological Oxygen Demand (BOD) for that sample.

Question 14.
Do you observe any soil pollution in your neighbourhood? What efforts will you make for controlling the soil pollution?
Solution.
Soil pollution can be controlled by :

1.  using manures
2.  using bio-fertilizers
3.  using proper sewerage system
4.  salvage and recycling waste products.

Question 15.
What are pesticides and herbicides? Explain giving examples.
Solution.
(a) Pesticides : They are chemicals that are used to kill pests. They act by blocking the neurotransmission in the pest. e.g., DDT, BHC, etc.
(b) Herbicides : They are chemicals such as NaClO₃, Na₃AsO₃ which are used to control the population of weeds in the fields.

Question 16.
What do you mean by green chemistry? How will it help decrease environmental pollution?
Solution.
(a) Green chemistry is a production process that would bring about minimum pollution or deterioration to the environment.
Green chemistry aims at :
(a) Use of environment friendly medium for the reaction; and
(b) Use of methods that completely convert the reactants to products such that there are no harmful side-products formed. This is arrived at by working out optimum conditions for synthesis such that there are no harmful side-products formed. Thus, green chemistry produces products which have no adverse impact on the environment.

Question 17.
What would have happened if the greenhouse gases were totally missing in the earth’s atmosphere? Discuss.
Solution.
Greenhouse gases are responsible for trapping heat and thereby warming up earth’s atmosphere to a temperature which is conducive to the formation of molecules of life such as water, amino acids, etc. Had there been no greenhouse gases, the earth would have remained as cold as Mars and life would not have originated.

Question 18.
A large number of fishes are suddenly found floating dead on a lake. There is no evidence of toxic dumping but you find an abundance of phytoplankton. Suggest a reason for the fish kill.
Solution.
Like other organisms, fishes need oxygen for survival. They derive this from the oxygen dissolved in water. But, when large amounts of phytoplankton grow in a pond/lake, they tend to take up all the O₂ dissolved in water. As a result, the amount of O₂ available for fish goes down dramatically and it dies. This is why the fishes in the lake die when large number of phytoplanktons grow therein. Such a lake is called a dead lake.

Question 19.
How can domestic waste be used as manure?
Solution.
Domestic waste consists mainly of two kinds of wastes – Biodegradable and non-biodegradable.
If the biodegradable part is separated and collected from homes it can be dumped in land fills and converted to compost by the action of bacteria present in soil.

Question 20.
For your agricultural field or garden you have developed a compost producing pit. Discuss the process in the light of bad odour, flies and recycling of wastes for a good produce.
Solution.
While a compost pit can be a blessing for the agricultural field since it produces manure, it can also be source of trouble in the form of flies, foul odour and home for growth of bacteria, however, the negative effects can be done away with by making the pit at a place away from the residential area preferably somewhere in the field itself. Equally important is covering the pit with a suitable, lid to prevent foul odour. In such a way one can derive the benefits of a compost pit while negating the disadvantages.

NCERT Exercises

Question 1.
How do you account for the formation of ethane during chlorination of methane?
Solution.
This is how ethane is generated during the chlorination of methane.

Question 2.
Write IUPAC names of the following compounds:

Solution.

Question 3.
For the following compounds, write structural formulae and IUPAC names of all possible isomers having the number of double or triple bond as indicated:
(a) C₄H₈ (one double bond)
(b) C₅H₈ (one triple bond)
Solution.

Question 4.
Write IUPAC names of the products obtained by the ozonolysis of the following compounds:
(i) pent-2-ene
(ii) 3,4-dimethylhept-3-ene
(iii) 2-ethylbut-1-ene
(iv) 1-phenylbut-1-ene
Solution.

Question 5.
An alkene A on ozonolysis gives a mixture of ethanal and pentan-3-one. Write the structure and IUPAC name of A.
Solution.

Question 6.
An alkene A contains three C – C, eight C – H σ bonds and one C – C π bond. A on ozonolysis gives two moles of an aldehyde of molar mass 44 u. Write the IUPAC name of A.
Solution.

Question 7.
Propanal and pentan-3-one are the ozonolysis products of an alkene. What is the structural formula of the alkene?
Solution.

Question 8.
Write chemical equations for combustion reaction of the following hydrocarbons:
(i) Butane
(ii) Pentene
(iii) Hexyne
(iv) Toluene
Solution.

Question 9.
Draw the cis and trans structures of hex-2-ene. Which isomer will have higher boiling point and why?
Solution.

Of the given isomers, the cis isomer has a higher boiling point. This difference arises due to higher dipole moment of the cis isomer which introduces a somewhat ionic character in the compound. In the trans isomer, the dipoles cancel each other resulting in a small dipole moment as the case may be.

Question 10.
Why is benzene extraordinary stable though it contains three double bonds?
Solution.

1.  The extraordinary stability of benzene molecule may be attributed to resonance in the molecule. In benzene, each of the 6 C atoms is sp² hybridised with one p-orbital on each carbon atom left unhybridised. While 2 of the sp² orbitals form bonds with 2 C-atoms, the third one is involved in bonding with hvdrogen atom. Thus, 3 of the valencies of C are satisfied.
2.  This leaves the unhybridised p-orbital containing 1 electron each for bonding. Each of these p-orbitals can overlap with the adjacent C atom and thus, results in bonding.
3.  Since the probability for each p-orbital to overlap with either of the two immediate neighbours is equal. Therefore, it alternately does so.
4.  Thus, the π electrons (in unhybridised p-orbitals) are no more localised between just 2 carbon atoms but these 671 electrons are shared or attracted by all the 6 carbon atoms.
5.  This increased attraction is the reason for the ‘extraordinary’ stability of the benzene molecule with 3 double bonds.
6.  These 3 π-bonds are not localised but are spread over the entire molecule.

Question 11.
What are the necessary conditions for any system to be aromatic?
Solution.
The necessary and sufficient condition for any system to be aromatic is given by Huckel’s rule. As per Huckel’s rule, any system is said to be aromatic if it satisfies the following 3 conditions:

1.  Contains (4n + 2) 71 electrons, where n is any positive integer or 0,
2.  Shows complete delocalisation of π electrons and
3.  The molecule must be planar.

Question 12.
Explain why the following systems are not aromatic?

Solution.
One of the conditions stated by the Huckel’s rule for any system to be aromatic is that of planarity i.e., all atoms of the molecule must be present on the same plane. This rule is violated in structure (i) and (ii). The carbon atom indicated below are sp³ hybridised which disallow planarity (sp³ hybridised carbon is tetrahedral in geometry).

In (iii) the number of n electrons is 8. (2 per double bond). The Huckel’s rule allows 2, 6,10,14, … etc. u electrons for any aromatic system. Since (iii) does not have (4n + 2) 7i electrons, therefore, it is not aromatic.

Question 13.
How will you convert benzene into
(i) p-nitrobromobenzene
(ii) m-nitrochlorobenzene
(iii) p-nitrotoluene
(iv) acetophenone
Solution.

Question 14.

Solution.

Question 15.
What effect does branching of an alkane chain has on its boiling point?
Solution.
As the branching in an alkane increases, the shape of the molecule approaches a sphere and size of the branched chain alkane becomes less than that of its straight chain counterpart. The reduced surface area results in decreased van der Waals’ interaction and finally leads to lower boiling point as compared to straight chain alkanes of comparable molar mass.

Question 16.
Addition of HBr to propene yields 2-bromopropane, while in the presence of benzoyl peroxide, the same reaction yields 1-bromopropane. Explain and give mechanism.
Solution.

Question 17.
Write down the products of ozonolysis of 1,2-dimethylbenzene (o-xylene). How does the result support Kekule structure for benzene?
Solution.

Question 18.
Arrange benzene, n-hexane and ethyne in decreasing order of acidic behaviour. Also give reason for this behaviour.
Solution.

Question 19.
Why does benzene undergo electrophilic substitution reactions easily and nucleophilic substitutions with difficulty?
Solution.
Electrophiles are species that are electron deficient and hence, seek electron rich molecules. Benzene is one such molecule which is rich in electrons. It is so because benzene has 6n electrons delocalised over the entire molecule which acts as a good host for electrophiles.
Another point working in favour of electrophilic substitution reactions is the retention of aromaticity. A benzene molecule is highly stable owing to its aromatic character.
Therefore, it would not want to lose its aromaticity. Upon undergoing electrophilic substitution reaction, this aromaticity is not lost, it is retained and hence, benzene undergoes electrophilic substitution.

Contrast this with a nucleophilic substitution reaction where the nucleophile attacks. A nucleophile (Nu^–) is a species that seeks a positive centre or an electron deficient species. Obviously, benzene is not electron deficient and therefore, will not be a welcome site for a Nu^–. This is the major reason why benzene does not undergo a nucleophilic substitution reaction.
Another reason working against these reactions is the difficulty with which the transition state is formed. The transition state benzyne involved here is formed with great difficulty and hence, these reactions are difficult to bring about.

Question 20.
How would you convert the following compounds into benzene?
(i) Ethyne
(ii) Ethene
(iii) Hexane
Solution.

Question 21.
Write structures of all the alkenes which on hydrogenation give 2-methylbutane.
Solution.

Question 22.
Arrange the following set of compounds in order of their decreasing relative reactivity with an electrophile, E⁺.
(a) Chlorobenzene, 2,4-dinitrochlorobenzene, p-nitrochlorobenzene
(b) Toluene, p-H₃C – C₆H₄ – NO₂, p-O₂N – C₆H₄ – NO₂
Solution.

Question 23.
Out of benzene, m-dinitrobenzene and toluene which will undergo nitration most easily and why?
Solution.

Question 24.
Suggest the name of a Lewis acid other than anhydrous aluminium chloride which can be used during ethylation of benzene.
Solution.
Other Lewis acids besides anhyd. AlCl₃ that may be used during ethylation of benzene is anhy. FeCl₃.

Question 25.
Why is Wurtz reaction not preferred for the preparation of alkanes containing odd number of carbon atoms? Illustrate your answer by taking one example.
Solution.
Wurtz reaction is used for preparation of alkanes. During this reaction, an alkyl halide with half the number of carbon atoms than the desired alkane is made to react with sodium metal in acetone. This leads to the formation of the desired alkane. e.g., If the desired alkane is ethane, methyl iodide is taken.

While this method is highly successful for producing alkanes with even number of carbon atoms but it gives a mixture of alkanes when odd numbered alkanes are to be formed. This happens because two different alkyl halides not only react with each other but also with themselves.

NCERT Exercises

Question 1.
What are hybridisation states of each carbon atom in the following compounds?

Solution.

Question 2.
Indicate the σ and π bonds in the following molecules:

Solution.

Question 3.
Write bond line formulas for: Isopropyl alcohol, 2,3-Dimethylbutanal, Heptan-4-one.
Solution.

Question 4.
Give the IUPAC names following compounds:

Solution.

Question 5.
Which of the following represents the correct IUPAC name for the compounds concerned?

Solution.

Question 6.
Draw formulas for the first five members of each homologous series beginning with the following compounds.
(a) H-COOH
(b) CH3COCH₃
(c) H-CH=CH₂
Solution.

Question 7.
Give condensed and bond line structure formulas and identify the functional groups present, if any, for:
(a) 2,2,4-Trimethylpentane
(b) 2-Hydroxy-1,2,3-propanetricarboxylic acid
(c) Hexanedial
Solution.

Question 8.
Identify the functional groups in the following compounds.

Solution.

Question 9.
Which of the two : O₂NCH₂CH₂O^– or CH₃CH₂O^– is expected to be more stable and why?
Solution.

Question 10.
Explain why alkyl groups act as electron donors when attached to a π system.
Solution.

Question 11.
Draw the resonance structures for the following compounds. Show the electron shift using curved-arrow notation.
(a) C₆H₅OH
(b) C₆H₅NO₂
(c) CH₃CH=CHCHO
(d) C₆H₅-CHO
(e) C₆H₅-CH₂
(f) CH₃CH=CHH₂
Solution.

Question 12.
What are electrophiles and nucleophiles? Explain with examples.
Solution.

Question 13.
Identify the reagents shown in brackets in the following equations as nucleophiles or electrophiles:
(a) CH₃COOH + (HO^–) ➝ CH₃COO^– + H₂O
(b) CH₃COCH₃ + (CN^–) ➝ (CH₃)₂C(CN)(OH)
(c) C₆H₆ + (CH₃ O) ➝ C₆H5COCH₃
Solution.

Question 14.
Classify the following reactions in one of the reaction type studied in this unit.
(a) CH₃CH₂Br + HS^– ➝ CH₃CH₂SH + Br
(b) (CH₃)₂C = CH₂ + HCl ➝ (CH₃)₂CIC-CH₃
(c) CH₃CH₂Br + HO^– ➝ CH₂ = CH₂ + H₂O + Br^–
(d) (CH₃)₃C-CH₂OH + HBr ➝ (CH₃)₂CBrCH₂CH₃ + H₂O
Solution.

Question 15.
What is the relationship between the members of following pairs of structures? Are they structural or geometrical isomers or resonance contributors?

Solution.

Question 16.
For the following bond cleavages, use curved- arrows to show the electron flow and classify each as homolysis or heterolysis. Identify reactive intermediate produced as free radical, carbocation and carbanion.

Solution.

Question 17.
Explain the terms Inductive and Electromeric effects. Which electron displacement effect explains the following correct orders of acidity of the carboxylic acids?

Solution.

Question 18.
Give a brief description of the principles of the following techniques taking an example in each case.
(a) Crystallisation
(b) Distillation
(c) Chromatography
Solution.
(a) Crystallisation : It is based on the difference in solubility of the compound and the impurities in a suitable solvent. While at room temperature, the compound is sparingly soluble and crystallizes out of solution but the impurities do not. As a result, they remain in solution and the compound is obtained as a crystal.
The impure compound is dissolved in a solvent and heated. At elevated temperature the compound dissolves as do the impurities. This solution is then gradually cooled. Being less soluble at room temperature it precipitates out in the form of crystals and pure compound is obtained.
(b) Distillation : This method is used to separate either :
(i) Volatile liquids from non-volatile impurities; and
(ii) Two liquids with different boiling points. The liquid mixture such as that of chloroform and aniline is taken in a round bottom flask fitted with a condenser.
Upon heating, the vapours of lower boiling liquid are formed first and collected through the condenser. The vapours of the higher boiling liquid are formed later. Thus, the two are separated.
(c) Chromatography: (i) It is applicable for the separation of virtually all inorganic and organic materials, except very insoluble polymers.
(ii) In this technique, the mixture of compounds which needs to be separated is applied onto a stationary phase, which may be a solid or a liquid. Another phase which may be a pure solvent, a mixture of solvents or a gas is allowed to more slowly over the stationary phase.
(iii) The components of the mixture which have different solubility in the moving phase, start moving. Since, they have different solubility, they move to different lengths on the stationary phase and become stable there,
(iv) Thus, the different components of the mixture are separated.

Question 19.
Describe the method, which can be used to separate two compounds with different solubilities in a solvent S.
Solution.
Crystallization is the process that may be employed to seperate two compounds with different solubility in a given solvent at room temperature.
Upon heating such a solution to a sufficiently high temperature the solubility of the compound which is insoluble at room-temperature, increases and it dissolves. However, when this solution is cooled down to room temperature the lesser soluble or insoluble component precipitates out and is obtained as crystals while its soluble counterpart remains in solution. Thus, the separation is complete.

Question 20.
What is the difference between distillation, distillation under reduced pressure and steam distillation?
Solution.
Differences between distillation, distillation under reduced pressure and steam distillation may be summarized as :

Question 21.
Discuss the chemistry of Lassaigne’s test.
Solution.
The elements nitrogen, sulphur and halogens are tested in an organic compound by Lassaigne’s test. The organic compound (N, S or halogens) is fused with sodium metal as to convert these elements into ionisable inorganic substances, i.e., nitrogen into sodium cyanide, sulphur into sodium sulphide and halogens into sodium halides.

Once the ions are formed, the inorganic tests can be applied to them and the compound can be analysed.
(i) Test for Nitrogen : The sodium fusion extract is boiled with iron(II) sulphate and then acidified with
acid. The formation of Prussian blue colour confirms the presence of nitrogen.

(ii) Test for Sulphur : The sodium fusion extract is acidified with acetic acid and lead acetate is added to it. Ablack precipitate of lead sulphide indicates the presence of sulphur.

(iii) Test for Halogens : The sodium fusion extract is acidified with nitric acid and then treated with silver nitrate. A white precipitate, soluble in ammonium hydroxide shows the presence of chlorine, a yellowish precipitate, sparingly soluble in ammonium hydroxide shows the. presence of bromine and a yellow precipitate, insoluble in ammonium hydroxide shows the presence of iodine.

(iv) Test for Phosphorus : The compound is heated with an oxidising agent (sodium peroxide). The phosphorus present in the compound is oxidised to phosphate. The solution is boiled with nitric add and then treated with ammonium molybdate. A yellow colouration or precipitate indicates the presence of phosphorus.

Question 22.
Differentiate between the principle of estimation of nitrogen in an organic compound by
(i) Dumas method and
(ii) Kjeldahl’s method.
Solution.

Question 23.
Discuss the principle of estimation of halogens, sulphur and phosphorus present in an organic compound.
Solution.

Question 24.
Explain the principle of paper chromatography.
Solution.
The underlying principle of paper chromatography is that of partition chromatography which is based on continuous differential partitioning of components of mixture between stationary and mobile phases. In paper chromatography, the paper used has water trapped in it which acts as the stationary phase while a suitable solvent or a mixture of solvents is used as a mobile phase. A strip of chromatography paper spotted at the base with the solution of the mixture is suspended in a suitable solvent or a mixture of solvents. As the mobile phase moves over the paper, it carries the mixture with it. Since the different components have different solubility, they travel to different extents on the paper and become stationary at different lengths on the paper and are thus, separated.

Question 25.
Why is nitric acid added to sodium extract before adding silver nitrate for testing halogens?
Solution.

Question 26.
Explain the reason for the fusion of an organic compound with metallic sodium for testing nitrogen, sulphur and halogens.
Solution.
Lassaigne’s test is used for the detection of extra elements such as N, S and X by applying the inorganic tests of analysis to these. Since, in organic compounds, the elements are present in covalent form and .inorganic tests can be applied only to ions, therefore these extra elements are first converted into their inorganic (ionic) forms by fusing with sodium metal.

Question 27.
Name a suitable technique of separation of the components from a mixture of calcium sulphate and camphor.
Solution.
Camphor is sublimable compound while CaSO₄ being ionic is not. Therefore, the two can be separated by the method of sublimation. If a mixture of the two is heated in a China dish covered with a porous paper and an inverted funnel over it, we will find the crystals of camphor forming on the inside walls of the inverted funnel. Thus, pure CaSO4 will be left in the China dish.

Question 28.
Explain, why an organic liquid vaporises at a temperature below its boiling point in its steam distillation?
Solution.
We know that any liquid boils when its vapour pressure is equal to the atmospheric pressure. There are certain liquids such as aniline which need very high temperature in order to start boiling. It is quite likely that at such elevated temperatures the molecules may just disintegrate. Therefore, to prevent this, steam distillation is employed. Here, the mixture of organic liquids containing the high boiling liquid say, aniline is mixed with water and heated. On doing so, at a temperature close to but less than 100°C (b.p. of water) the vapour pressure of water equals the atmospheric pressure and it boils. Since, in the mixture, aniline is present in conjugation with water it vapourises and moves out of the mixture.
The mixture of water and aniline is separated using a separating funnel. Steam distillation is used extensively in perfumery to separate essential oils.

Question 29.
Will CCl₄ give white precipitate of AgCl on heating it with silver nitrate? Give reason for your answer.
Solution.
AgNO₃ solution is ionic in nature. It contains Ag⁺ ions which when react with Ch ions produce a white ppt. of AgCl. In CCl₄ the Cl atoms are covalent in nature. They are not present as ions. Therefore, when AgNO₃ is added it, it does not produce a white ppt. of AgCl.

Question 30.
Why is a solution of potassium hydroxide used to absorb carbon dioxide evolved during the estimation of carbon present in an organic compound?
Solution.
KOH reacts with CO₂ to produce K₂CO3 which is a solid. The K₂CO₃ formed may be weighed and estimated to know the carbon content of the organic compound.

Question 31.
Why is it necessary to use acetic acid and not sulphuric acid for acidification of sodium extract for testing sulphur by lead acetate test?
Solution.
Sulphuric acid cannot be used for acidification of sodium extract because it would oxidize the sulphur to sulphur dioxide which would not give the black ppt. of PbS, which is otherwise obtained upon reaction with lead acetate.

Question 32.
An organiccompound contains 69% carbon and 4.8% hydrogen, the remainder being oxygen. Calculate the masses of carbon dioxide and water produced when 0.20 g of this substance is subjected to complete combustion.
Solution.

Question 33.
A sample of 0.50 g of an organic compound was treated according to Kjeldahl’s method. The ammonia evolved was absorbed in 50 mL of 0.5 M H₂SO₄. The residual acid required 60 mL of 0.5 M solution of NaOH for neutralisation. Find the percentage composition of nitrogen in the compound.
Solution.

Question 34.
0 .3780 g of an organic chloro compound gave 0.5740 g of silver chloride in Carius estimation. Calculate the percentage of chlorine present in the compound.
Solution.

Question 35.
In the estimation of sulphur by Carius method, 0.468 g of an organic sulphur compound afforded 0.668 g of barium sulphate. Find out the percentage of sulphur in the given compound.
Solution.

Question 36.
In the organic compound

Solution.

Question 37.
In the Lassaigne’s test for nitrogen in an organic compound, the Prussian blue colour is obtained due to the formation of

Solution.

Question 38.
Which of the following carbocation is most stable?

Solution.

Question 39.
The best and latest technique for isolation, purification and separation of organic compounds is
(a) Crystallisation
(b) Distillation
(c) Sublimation
(d) Chromatography
Solution.
(d) Chromatography

Question 40.
The reaction:
CH₃CH₂I + K0H_(aq) ➝ CH₃CH₂OH + KI is
classified as
(a) electrophilic substitution
(b) nucleophilic substitution
(c) elimination
(d) addition
Solution.

In the given reaction, the I^– from the alkyl iodide is replaced by the OH^– ion. Thus, it is substitution reaction.
The substitution is brought about by the OH^– ion which is a nucleophile.
∵ The reaction is a nucleophilic substitution reaction.

NCERT Exercises

Question 1.
Discuss the pattern of variation in oxidation states of
(i) B to TI and
(ii) C to Pb.
Solution.
The aspect common in the variation of oxidation state within group 13 (B to Tl) and group 14 (C to Pb) is that of inert pair effect. As we move down the group (13 or 14), we find that the lower oxidation state becomes more stable. In case of group 13 elements, it is +1 oxidation state that becomes stable while in group 14 it is +2 oxidation state that becomes stable. This behaviour can be understood if we consider the fact that as we move from the 2^nd period to the 3^rd a d-subshell is added. Similarly, upon moving further down the group there is an f-subshell which is added, f-subshells have a poor shielding effect. As a result, the s-electrons in the heavier elements of the 13^th and 14^th group are held more tightly and thus, are reluctant to get oxidised. This causes the lower oxidation state to become more stable.

Question 2.
How can you explain higher stability of BCl₃ as compared toTICl₃?
Solution.
The higher stability of BCl3 as compared to TICl₃ can be explained with the help of inert pair effect which becomes more stable as we move down the group. TI is more stable in its +1 oxidation state and hence TICl₃ is not stable as TI shows an oxidation state of +3 in TICl₃.

Question 3.
Why does boron trifluoride behave as a Lewis acid?
Solution.
B has an electronic configuration of 1s²2s²2p¹ and undergoes sp² hybridisation and then bonds with 3 fluorine atoms to yield BF₃.

But, inspite of this bonding it remains electron deficient i.e., it does not have 8 electrons around it in the outermost shell; it has only 6 electrons in the outermost shell.
This electron deficiency coupled with the fact that it has an empty 2p orbital which can accept more electrons making it a Lewis acid. (Lewis acids are compounds that can accept electrons)

Question 4.
Consider the compounds, BCl₃ and CCl₄. How will they behave with water ? Justify.
Solution.
BCl₃ is an electron deficient compound. Also, it has an empty unhybridised p-orbital which can accept electrons. In presence of water, BCl₃ hydrolyses and forms B(OH)₃.

But, when CCl₄ is mixed with water, no reaction takes place because carbon neither has any unhybridised, empty p-orbital where it can accommodate electrons from water nor it has empty d-orbital.
CCl₄ + H₂O ➝ No reaction

Question 5.
Is boric acid a protic acid? Explain.
Solution.
Boric acid – H₃BO₃ or B(OH)₃, is not a protic acid (which are acids that donate protons). But an aqueous solution of boric acid is found to be weakly acidic in nature. This acidic character arises due to the Lewis acid character of boric acid which abstracts a hydroxyl ion from water and leaves free H⁺ ions which make the solution acidic.

Question 6.
Explain what happens when boric acid is heated.
Solution.
On heating orthoboric acid above 370 K, it forms metaboric acid, HBO₂ which on further heating yields boric oxide, B₂O₃.

Question 7.
Describe the shapes of BF₃ and B. Assign the hybridisation of boron in these species.
Solution.
BF₃ has a planar triangular structure which arises from the sp² hybrid orbitals.

These three sp² hybrid orbitals are directed towards the corners of triangle and BF³ has a trigonal structure.

B may be assumed to be made of a central B atom, 3H atoms and one hydride ion H^–.

In order to accommodate the 3H atoms and one H^– ion, B undergoes sp³ hybridisation yielding four orbitals, 3 of which contain one e^– each and one is empty. The fourth, empty orbital accomodates the H^– ion. Thus, the structure of B is tetrahedral.

Question 8.
Write reactions to justify amphoteric nature of aluminium.
Solution.
Amphoteric substances are those that can react with both acids and bases. Aluminium reacts with HCl to liberate H₂ gas as :

Question 9.
What are electron deficient compounds? Are BCl₃ and SiCl₄ electron deficient species? Explain
Solution.

Question 10.
Write the resonance structures of  and .
Solution.
The resonance structures of  and  are :

Question 11.
What is the state of hybridisation of carbon in
(a) 
(b) diamond
(c) graphite?
Solution.

Question 12.
Explain the difference in properties of diamond and graphite on the basis of their structures.
Solution.

Question 13.
Rationalise the given statements and give chemical reactions:
(i) Lead (II) chloride reacts with Cl2 to give PbCI4.
(ii) Lead (IV) chloride is highly unstable towards heat.
(iii) Lead is known not to form an iodide, Pbl4.
Solution.

Question 14.
Suggest reasons why the B—F bond lengths in BF₃ (130 pm) and  (143 pm) differ.
Solution.
The bond length in anv compound is dependent on the hybridisation of the central atom. Boron in BF₃ is sp² hybridised which means that the s-character is 33% and therefore, the bond length is shorter. Also due to similar size of both atoms and vacant p-orbital of B, a pπ-pπ back bonding from F to B occurs causes partial double bond character. This further decreases the bond length of B — F. In , the hybridisation of B is sp³ which means that the s-character is 25% and therefore, a longer bond length.

Question 15.
If B—Cl bond has a dipole moment, explain why BCl₃ molecule has zero dipole moment.
Solution.
The dipole moment of any molecule is the vector sum total of each of the dipole moments. In BCl₃, molecule, although the B-Cl bonds individually are polar, the resultant dipole moment becomes zero.

We can see that the dipole moments of B-¹Cl and B-²Cl produce a resultant which is equal in magnitude but opposite in direction to B-³Cl and hence cancels it out. That is why the net dipole moment of BCl₃, is zero.

Question 16.
Aluminium trifluoride is insoluble in anhydrous HF but dissolves on addition of NaF. Aluminium trifluoride precipitates out of the resulting solution when gaseous BF₃ is bubbled through. Give reasons.
Solution.
AlF₃ is insoluble in anhydrous HF but when little NaF is added the compound becomes soluble. On adding BF₃ to the above solution, AlF₃ is reprecipitated along with sodium tetrafluoroborate.

Question 17.
Suggest a reason as to why CO is poisonous.
Solution.
Carbon monoxide is a colourless, odourless gas which has a tendency to bind to haemoglobin (the oxygen carrying molecule in blood), forming a complex called carboxyhaemoglobin. This complex is 300 times more stable than oxyhaemoglobin complex and therefore, if once, formed can seriously hamper the oxvgen supply to different organs and ultimately can cause death. This is why CO is said to be so dangerous.

Question 18.
Flow is excessive content of CO₂ responsible for global warming?
Solution.
CO₂ molecule is transparent to UV radiation but is opaque to IR radiation. Due to this, it allows the UV radiations of sunlight to pass into earth’s atmosphere but does not allow heat, which is in the form of 1R radiation, to move out of earth’s atmosphere. This causes heating of the atmosphere and produces a greenhouse effect.

Question 19.
Explain structures of diborane and boric acid. Diborane:
Solution.

The four terminal hydrogen atoms and the two boron atoms lie in one plane. Above and below this plane, there are two bridging hvdrogen atoms. The four terminal B – H bonds are regular two centre-two electron bonds (2c – 2e) while the two bridge (B – H – B) bonds are different and can be described in terms of three centre-two electron bonds (3c-2e).

Each B atom uses sp³ hybrids for bonding. Out of the four sp³ hybrids on each B atom, one is without an electron shown in broken lines. The terminal B-H bonds are normal 2-centre-2-electron bonds but the two bridge bonds are 3-centre-2-electron bonds. The 3- centre-2-electron bridge bonds are also referred to as banana bonds.
Boric acid : In boric acid, planar B03 units are joined by hydrogen bonds to give a layered structure.

Question 20.
What happens when:
(a) Borax is heated strongly.
(b) Boric acid is added to water.
(c) Aluminium is treated with dilute NaOFI.
(d) BF₃ is reacted with ammonia.
Solution.

Question 21.
Explain the following reactions :
(a) Silicon is heated with methyl chloride at high temperature in the presence of copper;
(b) Silicon dioxide is treated with hydrogen fluoride;
(c) CO is heated with ZnO;
(d) Hydrated alumina is treated with aqueous NaOH solution.
Solution.

Question 22.
Give reasons:
(i) Conc. HNO₃ can be transported in aluminium container.
(ii) A mixture of dilute NaOH and aluminium pieces is used to open drain.
(iii) Graphite is used as lubricant.
(iv) Diamond is used as an abrasive.
(v) Aluminium alloys are used to make aircraft body.
(vi) Aluminium utensils should not be kept in water overnight.
(vii) Aluminium wire is used to make transmission cables.
Solution.

Question 23.
Explain why is there a phenomenal decrease in ionization enthalpy from carbon to silicon?
Solution.
Large decrease in ionisation potential from C to Si is due to increase in size of the atom and shielding effect.

Question 24.
How would you explain the lower atomic radius of Ga as compared to Al?
Solution.
This can be understood from the variation in the inner core of the electronic configuration. The presence of additional 10 d-electrons offer only poor screening effect for the outer electron from the increased nuclear charge in gallium. Consequently, the atomic radius of gallium (135 pm) is less than that of aluminium (143 pm).

Question 25.
What are allotropes? Sketch the structure of two allotropes of carbon namely diamond and graphite. What is the impact of structure on physical properties of two allotropes?
Solution.
The property due to which an element exists in two or more forms which differ in their physical and some of the chemical properties is known as allotropy and the various forms are called allotropes or allotropic modifications. Carbon exists in two allotropic forms crystalline and amorphous. The crystalline forms are diamond and graphite.

Diamond due to extended covalent bonding is the hardest natural substance on the earth.
with Al₂(SO₄)₃ which is soluble in Water.

Question 26.

Solution.

Question 27.
In some of the reactions thallium resembles aluminium, whereas in others it resembles group I metals. Support this statement by giving some evidence.
Solution.

Question 28.

Solution.

Question 29.
What do you understand by
(a) inert pair effect
(b) allotropy and
(c) catenation?
Solution.
(a) Inert pair effect : The reluctance of ns2 pair in p-block elements having higher atomic number to take part in bond formation is called inert pair effect.
(b) Allotropy : The existence of an element in more than one form having different physical properties but same or slightly different chemical properties is called allotropy.
(c) Catenation : The property by virtue of which a large number of atoms of the same element get linked together through covalent bonds resulting in the formation of long chains, branched chains and rings of different sizes is called catenation.

Question 30.
A certain salt X, gives the following results :
(i) Its aqueous solution is alkaline to litmus.
(ii) It swells up to a glassy material Yon strong heating.
(iii) When conc. H₂SO₄ is added to a hot solution of X, white crystal of acid Z separates out.
Write equations for all the above reactions and identify X, Y and Z.
Solution.

Question 31.
Write balanced equations for:

Solution.

Question 32.
Give one method for industrial preparation and one for laboratory preparation of CO and CO₂ each.
Solution.

Question 33.
An aqueous solution of borax is
(a) neutral
(b) amphoteric
(c) basic
(d) acidic.
Solution.

Question 34.
Boric acid is polymeric due to
(a) its acidic nature
(b) the presence of hydrogen bonds
(c) its monobasic nature
(d) its geometry.
Solution.
(b) the presence of hydrogen bonds

Question 35.
The type of hybridisation of boron in diborane is
(a) sp
(b) sp²
(c) sp³
(d) dsp²
Solution.
(c) sp³

Question 36.
Thermodynamically the most stable form of carbon is
(a) diamond
(b) graphite
(c) fullerenes
(d) coal
Solution.
(b) : Graphite is thermodynamically more stable than diamond and its free energy of formation is 1.9 kJ mol^-1 less than diamond.

Question 37.
Elements of group 14
(a) exhibit oxidation state of +4 only
(b) exhibit oxidation state of +2 and +4
(c) form M^2- and M⁴⁺ ions
(d) form M²⁺ and M⁴⁺ ions.
Solution.
(b, d): The group 14 elements have four electrons in outermost shell. The common oxidation states exhibited by these elements are +4 and +2.
Elements like Sn and Pb are metallic thus, form M²⁺ and M⁴⁺ ions.

Question 38.
If the starting material for the manufacture of silicones is RSiCl₃, write the structure of the product formed.
Solution.

NCERT Exercises

Question 1.
What are the common physical and chemical features of alkali metals ?
Solution.
General characteristics of alkali metals :

1.  Electronic configuration : General electronic configuration of the valence shell is ‘ns¹‘ where ‘n’ gives the number of the outermost shells.
2.  Atomic and ionic sizes : They have the largest size in their respective periods. The atomic size regularly increases upon descending the group.
3.  Ionisation enthalpy : It decreases down the group due to decreasing effective nuclear charge. 2^nd ionisation enthalpy is very high as now an electron has to be removed from stable noble gas configuration.
4.  Melting and boiling points: Due to weak metallic bond strength and large atomic sizes, all alkali metals are soft and have low melting and boiling points.
5.  Metallic character : Alkali metals readily lose their valence electron and form M⁺ ions showing +1 oxidation states. Metallic character increases as we descend.
6.  Flame colouration : They impart characteristic colours to flame.
7.  Photoelectric effect : Due to very low value of ionisation enthalpies, alkali metals exhibit photoelectric effect.
8.  Nature of compounds : Due to highly electropositive character, compounds of alkali metals are ionic in nature.
9.  Reactivity towards air : Their lustrous surface gets tarnished when exposed to air. They burn violently in air or oxygen forming oxides. Li forms monoxide (Li₂O), Na forms peroxide (Na₂O₂) while K, Rb, Cs form superoxides (KO₂, RbO₂, CsO₂).
10.  Reactivity towards water : Alkali metals form hydroxides when they come in contact with water and liberate hydrogen gas.
11.  Reactivity towards hydrogen : They form ionic hydride (M⁺H^–) Ionic character of hydrides increases on descending the group. They act as strong reducing agents.
12.  Reactivity towards halogen : They react with halogens to form ionic halides (M⁺X^–). The ionic character of metal halides increases down the group.
13.  Reducing character : An element which acts as a reducing agent, must have low ionisation energy. Alkali metals act as strong reducing agents as their ionisation energy values are low. Since ionisation energy decreases on moving down from Li to Cs, the reducing property increases in the same order. Thus, Li is weakest reducing agent while Cs is the strongest reducing agent amongst alkali metals in free state.
14.  Solution in liquid ammonia : They dissolve in liq. NH₃ forming blue coloured solutions, which are good reducing agents and good conductors of electricity.

The blue colour of the solution is due to ammoniated electron.

Question 2.
Discuss the general characteristics and gradation in properties of alkaline earth metals.
Solution.

1. Electronic configuration : The valence electronic configuration of atoms of the group II A elements is ns², where ‘n’ is the period number.
2.  Atomic and ionic sizes : The size of the atom increases gradually from Be to Ra. Their ions are also large and size of the ion increases from Be²⁺ to Ra²⁺.
3.  Ionisation enthalpy : The 1^st and 2^nd ionisation energies of these metals decrease from Be to Ba as size increases.
4.  Melting and boiling points : Due to the presence of two electrons in the valence shell and stronger bonding in solid state, they have higher melting and boiling points than corresponding alkali metals.
5.  Metallic character : Due to low ionisation energy values, these metals are highly electropositive and readily form M²⁺ ions.
6.  Flame colouration : Except Be and Mg, other members impart characteristic colours when their salts are introduced in the flame.
7.  Hydration energy : The M²⁺ ions of alkaline earth metals are extensively hydrated to form [M(H₂O)x]²⁺ ions and during hydration a huge amount of energy called hydration energy is released.
   M²⁺ + xH₂O ➝ [M(H₂O)x]²⁺ + Energy
8.  Reactivity towards air or oxygen : They react with air or oxygen slowly on heating. Be, Mg and Ca form normal oxides (MO) while Sr and Ba form superoxides (SrO₂, BaO₂). BeO is amphoteric, MgO is weakly basic and others are distinctly basic.
9.  Reactivity towards water : They have lesser reactivity towards water. Be does not react even with boiling water. Mg forms Mg(OH)₂ liberating H₂ gas with boiling water.
10.  Reactivity towards halogens : All the metals of the group combine with various halogens at appropriate temperature forming halides of the formula MX₂.
11.  Tendency to form complexes : Due to their smaller ionic sizes and greater charge densities Be, Mg metals have highest tendency to form complexes.
    BeF₂ + 2F^– ➝ [BeF₄]^2-
12.  Reducing character : Except Be all other metals of this group are reducing agents.

Question 3.
Why are alkali metals not found in nature ?
Solution.
All the alkali metals have one valence electron, ns¹, outside the noble gas core. The loosely held s-electron in the outermost valence shell of these elements makes them the most electropositive metals. They readily lose an electron to give monovalent M⁺ ions. Thus, due to the reason cited above, alkali metals are never found free in nature but are always found in combined state.

Question 4.
Find out the oxidation state of sodium in Na₂O₂.
Solution.

Question 5.

Explain why is sodium less reactive than potassium.
Solution.
This is mainly due to high ionisation enthalpy of sodium as compared to potassium. Therefore, potassium is more electropositive and a stronger reducing agent than sodium. It also reacts with water more violently than sodium.

Question 6.
Compare the alkali metals and alkaline earth metals with respect to

1.  ionisation enthalpy
2.  basicity of oxides and
3.  solubility of hydroxides.

Solution.

1.  Ionisation Enthalpy: The ionisation energies of alkaline earth elements are higher than those of alkali metals due to higher nuclear charge and smaller radii.
2.  Basicity of oxides : Oxides of alkali metals are stronger bases as compared to those of alkaline earth metals present in the same period, e.g., when Na₂O is dissolved in water, NaOH formed is a stronger base than when MgO is dissolved in water to form Mg(OH)₂. This is due to higher ionisation energies of alkaline earth metals.
3.  Solubility of hydroxides : Alkali metal hydroxides are more soluble in water as compared to the hydroxides of alkaline earth metals present in the same period. This is due to higher lattice energy of the hydroxides of alkaline earth elements as compared to those of alkali metals.

Question 7.
In what ways lithium shows similarities to magnesium in its chemical behaviour?
Solution.

Question 8.
Explain why can alkali and alkaline earth metals not be obtained by chemical reduction methods?
Solution.
Alkali and alkaline earth metals cannot be extracted by the reduction of their oxides and other compounds as they are strong reducing agents themselves and no such reducing agents are there which can reduce them to get pure metal.

Question 9.
Why are potassium and caesium, rather than lithium used in photoelectric cells?
Solution.
The ns¹ electron in K and Cs is so loosely held that even the low energy photons can eject this electron from the metal surface. This property is termed as photoelectric effect. K and Cs are used in photoelectric cells because they have very low ionisation energies. But lithium having very high ionisation energy cannot be used for photoelectric effect.

Question 10.
When an alkali metal dissolves in liquid ammonia the solution can acquire different colours. Explain the reasons for this type of colour change.
Solution.
The alkali metals dissolve in liq. NH3 without evolution of hydrogen. The colour of the dilute solution is blue. The metal atom loses an electron and it combines with ammonia molecule.

The blue colour of the solution is due to the ammoniated electron which absorbs energy in the visible region of light and thus, imparts blue colour to the solution. The solutions are paramagnetic and on standing, slowly they liberate hydrogen resulting in the formation of an amide.

In concentrated solution, the blue colour changes to bronze and becomes diamagnetic.

Question 11.
Beryllium and magnesium do not give colour to flame whereas other alkaline earth metals do so. Why ?
Solution.
In the case of Ca, Sr, Ba and Ra, the electrons can be excited by the supply of energy to higher energy levels. When the excited electrons return to the original level, the energy is released in the form of light. In Be and Mg, the electrons are tightly held and excitation by mere flame is rather difficult, thus, they do not show flame colouration. Ca imparts brick red colour, Sr imparts crimson colour, Ba imparts apple green colour and Ra imparts crimson colour to the flame.

Question 12.
Discuss the various reactions that occur in the Solvay process.
Solution.
Solvay process : It is an industrial method for obtaining sodium carbonate from sodium chloride (NaCl) and limestone (CaCO₃). The raw materials required in this process are common salt, ammonia and limestone. The equations for the complete process may be written as :

Sodium hydrogen carbonate crystal separates out and is heated to give sodium carbonate.

Question 13.
Potassium carbonate cannot be prepared by Solvay process. Why ?
Solution.
Solvay process cannot be extended to the manufacture of K₂CO₃ because KHCO₃ is too soluble to be precipitated by the addition of ammonium hydrogen carbonate to a saturated solution of potassium chloride.

Question 14.
Why is Li₂CO₃ decomposed at a lower temperature whereas Na₂CO₃ at higher temperature?
Solution.

Question 15.

Compare the solubility and thermal stability of the following compounds of the alkali metals with those of the alkaline earth metals.
(a) Nitrates
(b) Carbonates
(c) Sulphates.
Solution.

Question 16.
Starting with sodium chloride how would you proceed to prepare
(i) sodium metal
(ii) sodium hydroxide
(iii) sodium peroxide
(iv) sodium carbonate ?
Solution.

Question 17.
What happens when
(i) magnesium is burnt in air
(ii) quick lime is heated with silica
(iii) chlorine reacts with slaked lime
(iv) calcium nitrate is heated ?
Solution.

Question 18.
Describe two important uses of each of the following:
(i) caustic soda
(ii) sodium carbonate
(iii) quick lime.
Solution.
(i) Caustic soda : It is the commercial name of NaOH. It is used :
(a) in refining of petroleum.
(b) in the manufacture of soap, paper, rayon, drugs and dyes.
(ii) Sodium Carbonate : It is used :
(a) in laundries and in softening of water as washing soda.
(b) in the manufacture of glass, sodium silicate, paper, borax, caustic soda, etc.
(iii) Quick lime : It is used :
(a) in the purification of sugar, manufacture of dye stuffs, bleaching powder, CaC2, mortar, cement, glass, etc.
(b) as a cheap alkali, i.e., as acid neutraliser.

Question 19.
Draw the structure of
(i) BeCl₂ (vapour)
(ii) BeCl₂ (solid).

Solution.

Question 20.
The hydroxides and carbonates of sodium and potassium are easily soluble in water while the corresponding salts of magnesium and calcium are sparingly soluble in water. Explain.
Solution.
The solubility of an ionic compound depends on two factors :
(i) lattice energy and
(ii) hydration energy.
These two factors oppose each other.
If lattice energy is high, the ions will be tightly packed in the crystal and therefore, solubility will be low. If the hydration energy is high, the ions will have greater tendency to be hydrated and therefore, solubility will be high. Alkali metals i.e., sodium or potassium hydroxides will be highly soluble in water due to larger size of Na and K as compared to that of Mg and Ca, the lattice energies of hydroxides and carbonates of sodium and potassium are much lower than those of hydroxides and carbonates of magnesium and calcium. As a result, the hydroxides and carbonates of Na and K are easily soluble in water while the corresponding salts of Mg and Ca are sparingly soluble in water.

Question 21.
Describe the importance of the following :
(i) limestone
(ii) cement
(iii) plaster of Paris.
Solution.
(i) Limestone : It is used,
(a) in the manufacture of quick lime, slaked lime, cement, washing soda and glass,
(b) as a flux in the smelting of iron and lead ores.
(ii) Cement : It is an important building material. It is used in concrete and reinforced concrete, in plastering and in the construction of bridges, dams and buildings.
(iii) Plaster of Paris : The largest use of plaster of Paris is in the building industry as well as plasters. It is used for immobilising the affected part of organ where there is a bone fracture or sprain. It is also employed in dentistry, in ornamental work and for making casts of statues.

Question 22.
Why are lithium salts commonly hydrated and those of the other alkali ions are usually anhydrous?
Solution.
Because of smallest size among alkali metals, Li⁺ can polarise water molecules more easily than the other alkali metal ions and hence get attached to lithium salts as water of crystallisation. For example, lithium chloride crystallises as LiCl.2H₂O but sodium chloride as NaCl.

Question 23.
Why is LiF almost insoluble in water whereas LiCI is soluble not only in water but also in acetone ?
Solution.
Insolubility of LiF in water can be elucidated as follows:
The lithium ion has the highest energy of hydration as it is small in size in comparison to the other alkali metal ions and so it should be highly soluble. But the small Li⁺ and F ions interact very strongly resulting in high lattice energy of LiF which is responsible for its insolubility whereas in LiCl due to the difference in atomic size between Li⁺ and Cl^–, they do not interact very strongly, the lattice energy is comparatively small and the magnitude of hydration enthalpy is quite large. Therefore, LiCI dissolves in water. As LiCl has more covalent character than LiF (Fajan’s rule) thus, it is soluble in organic solvents like acetone.

Question 24.
Explain the significance of sodium, potassium, magnesium and calcium in biological fluids.
Solution.
Biological importance of sodium : Sodium ions are found primarily outside the cells in blood plasma and in the interstitial fluids which surround the cells. These ions participate in the transmission of nerve signals, in regulating the flow of water across cell membranes and in the transport of sugars and amino acids into cells.
Biological importance of potassium : Potassium ions are the most abundant cations within cell fluids, where they activate many enzymes, participate in the oxidation of glucose to produce ATP and with sodium, are responsible for the transmission of nerve signals.
Biological importance of magnesium : All enzymes that utilise ATP in phosphate transfer require magnesium as the cofactor. The main pigment for the absorption of light in plants is chlorophyll which contains magnesium.
Biological importance of calcium: About 99% of body calcium is present in bones and teeth. It also plays important roles in neuromuscular function, maintenance of a regular heartbeat, various aspects of metabolism and blood clotting. The calcium concentration in plasma is maintained by two hormones : calcitonin and parathyroid hormone.

Question 25.
What happens when
(i) sodium metal is dropped in water ?
(ii) sodium metal is heated in free supply of air?
(iii) sodium peroxide dissolves in water ?
Solution.

Question 26.
Comment on each of the following observations:

Solution.
(a) This is attributed to the hydration of the cation in water. As a result, size of the cation increases and its mobility decreases. Due to the smallest size, Li* ion is hydrated to the maximum and has least mobility while Cs+ ion due to least hydration has maximum mobility.
(b) Lithium is a very strong reducing agent. As a result, it directly combines with nitrogen to form its nitride (Li₃N).
(c) E° of any M²⁺/M electrode depends upon three factors :
(i) enthalpy of vaporisation,
(ii) ionisation enthalpy
(iii) enthalpy of hydration. Since the combined effect of these factors is approximately the same for Ca, Sr and Ba, therefore, their electrode potentials are nealy constant.

Question 27.
State as to why
(a) a solution of Na2C03 is alkaline?
(b) alkali metals are prepared by electrolysis of their fused chlorides?
(c) sodium is found to be more useful than potassium?
Solution.
(a) Sodium carbonate being a salt of strong base (NaOH) and weak acid (H₂CO₃) forms an alkaline solution upon hydrolysis.

(b) Alkali metals are prepared by the electrolysis of their fused chlorides as if aqueous solution of their salts are used for extraction by electrolytic method then hydrogen is discharged at cathode instead of an alkali metal as the discharge potentials of alkali metals are high.
(c) Sodium is relatively more abundant than potassium. At the same time, it is also less reactive and its reactions with other substances can be better controlled.

Question 28.
Write balanced equations for reactions between
(a) Na₂O₂ and water
(b) KO₂ and water
(c) Na₂O and CO₂.
Solution.

Question 29.
How would you explain the following observations?
(i) BeO is almost insoluble but BeSO₄ is soluble in water,
(ii) BaO is soluble but BaSO₄ is insoluble in water,
(iii) LiI is more soluble than KI in ethanol.
Solution.

Question 30.
Which of the alkali metal is having least melting point ?
(a) Na
(b) K
(c) Rb
(d) Cs
Solution.
(d) : Cs ➝ Least melting point.
As the size of metal increases, the metallic bonding decreases and melting point decreases.

Question 31.
Which one of the following alkali metals gives hydrated salts ?
(a) Li
(b) Na
(c) K
(d) Cs
Solution.
(a) : Li ➝ Highest number of hydrated salts Due to small size of Li, it has high charge density that attracts water molecules strongly.

Question 32.
Which one of the alkaline earth metal carbonates is thermally the most stable ?
(a) MgCO₃
(b) CaCO₃
(c) SrCO₃
(d) BaCO₃
Solution.
(d) : BaC0₃ ➝ Most thermally stable carbonate.
 being a bigger anion stabilises a bigger cation more efficiently. Thus, Ba²⁺ being bigger in size is the most stable.

NCERT Exercises

Question 1.
Justify the position of hydrogen in the periodic table on the basis of its electronic configuration.
Solution.
Hydrogen has electronic configuration 1s¹. Its electronic configuration is similar to the outer electronic configuration (ns¹) of alkali metals, which belong to the first group of the periodic table.

Question 2.
Write the names of isotopes of hydrogen. What is the mass ratio of these isotopes?
Solution.
Hydrogen has three isotopes :

The mass ratio of hydrogen isotopes is 1 : 2 : 3.

Question 3.
Why does hydrogen occur in a diatomic form rather than in a monoatomic form under normal conditions?
Solution.
In monoatomic form hydrogen atom has only one electron in K-shell (1s¹) while in diatomic form, the K-shell is complete (1s²). This means that in diatomic form, hydrogen (H₂) has acquired the configuration of nearest noble gas, helium. It is therefore, quite stable.

Question 4.
How can the production of dihydrogen, obtained from ‘coal gasification’, be increased?
Solution.
The process of producing ‘syngas’ from coal is called “coal gasification”.

The production of dihydrogen can be increased by reacting carbon monoxide of syngas mixtures with steam in the presence of iron chromate as catalyst.

This is called water-gas shift reaction.

Question 5.
Describe the bulk preparation of dihydrogen by electrolytic method. What is the role of an electrolyte in this process?
Solution.
Dihydrogen is prepared by the electrolysis of water using platinum electrodes in the presence of a small amount of acid or alkali.

During electrolysis, the reactions that take place are :

Question 6.
Complete the following reactions.

Solution.

Question 7.
Discuss the consequences of high enthalpy of H — H bond in terms of chemical reactivity of dihydrogen.
Solution.
Dihydrogen is quite stable and dissociates into hydrogen atoms only when heated at about 2000 K.

Above 2000 K, the dissociation of hydrogen into atoms is only 0.081% which increases to 95.5% at 5000 K. Its bond dissociation energy is very high.

Due to its high bond dissociation energy it is not very reactive.

Question 8.
What do you understand by

1.  electron- deficient,
2.  electron-precise and
3.  electron- rich compounds of hydrogen?

Provide justification with suitable examples.
Solution.

1.  Electron-deficient hydrides : The hydrides of group 13 elements have an incomplete octet and hence are electron deficient molecules. They act as Lewis acids e.g., B₂H₆
2.  Electron-precise hydrides : These compounds have the required number of electrons to write their conventional Lewis structures. All elements of group 14 form such compounds (e.g., CH₄) which have tetrahedral structure.
3.  Electron rich compounds of hydrogen : These compounds have excess electrons which are present as lone pairs, elements of group 15-17 form such compounds, e.g., NH₃ has one lone pair, H₂O has two lone pairs and HF has three lone pairs.

Question 9.
What characteristics do you expect from an electron-deficient hydride with respect to its structure and chemical reactions?
Solution.
It is expected to be a Lewis acid. For example, diborane B₂H₆ (dimer of BH₃) forms complex with LiH which gives H^– ion (Lewis base).

Question 10.
Do you expect the carbon hydrides of the type (C_nH_{2n + 2}) to act as Lewis acid or base? Justify your answer.
Solution.
The type (C_nH_{2n + 2}) is not a Lewis acid or a Lewis base because carbon atoms have a complete octet, therefore, hydrides behave as normal covalent hydrides.

Question 11.
What do you understand by the term ‘non-stoichiometric hydrides’? Do you expect this type of the hydrides to be formed by alkali metals? Justify your answer.
Solution.
Hydrides of d-block (Group No. 3, 4, 5 and 6) and f-block are metallic or interstitial hydrides because in these hydrides, hydrogen occupies some interstitial sites in the metal lattice producing distortion. These hydrides are also called non-stoichiometric hydrides as, the law of constant composition does not hold good for them, e.g., LaH_2.87, YbH_2.55, TiH_1.5-1.8, ZrH_1.3-1.75, vH_0.56, NiH_0.6-0.7, PdH_0.6-0.8 etc. Alkali metals do not form these types of hydrides because alkali metals donate lone pair of electrons to hydrogen which forms H^– ion. As H^– ion is formed by complete transfer of electrons so, alkali metal and H atom are in fixed stoichiometric ratio.

Question 12.
How do you expect the metallic hydrides to be useful for hydrogen storage? Explain.
Solution.
Some metals like palladium (Pd), platinum (Pt) have a tendency to adsorb very large volume of hydrogen and therefore, can be used as its storage media. This property has high potential for hydrogen storage and as a source of energy.

Question 13.
How does the atomic hydrogen or oxy-hydrogen torch function for cutting and welding purposes? Explain.
Solution.
Atomic hydrogen torch : When molecular hydrogen is passed through tungsten electric arc ~ 3000°C, at lowr pressure, it dissociates to form atoms of hydrogen known as atomic hydrogen.

When the atoms of hydrogen produced as above are made to fall at some metal surface, they combine to form molecular hydrogen again along with the liberation of large amount of heat energy. The temperature rises to 4000-5000°C. This is the principle of atomic hydrogen torch which is used for welding purposes.
Oxy-hydrogen torch : When hydrogen is burnt in oxygen, the reaction is highly exothermic in nature. A temperature ranging between 2800°C to 4000°C is generated. This temperature can be employed for welding purpose in the form of oxy-hydrogen torch.

Question 14.
Among NH₃, H₂O and HF, which would you expect to have highest magnitude of hydrogen bonding and why?
Solution.
The strength of H-bonding depends upon the magnitude of the polarity of the bond. Since H-F bond has maximum polarity (F is the most electronegative element) hydrogen bonding is maximum in HF molecules.

Question 15.
Saline hydrides are known to react with water violently producing fire. Can CO₂, a well-known fire extinguisher, be used in this case? Explain.
Solution.
When saline hydrides react with water, the reaction is highly exothermic, the evolved hydrogen catches fire, e.g.,

Question 16.
Arrange the following :
(i) CaH₂, BeH₂ and TiH₂ in order of increasing electrical conductance.
(ii) LiH, NaH and CsH in order of increasing ionic character.
(iii) H – H, D – D and F – F in order of increasing bond dissociation enthalpy.
(iv) NaH, MgH₂ and H₂O in order of increasing reducing property.
Solution.
(i) BeH₂ is significantly covalent, CaH₂ is ionic and TiH₂ is metallic hydride. Hence, increasing electrical conductance : BeH₂<CaH₂<TiH₂
(ii) Electronegativity decreases as Li > Na > Cs. Thus, increasing ionic character :
LiH < NaH < CsH.
(iii) Due to lone pairs of F, bond pairs experience repulsion, hence, F-F has low bond dissociation energy. In D-D, due to higher nuclear attraction bond dissociation energy is geater than H- H. Increasing bond dissociation enthalpy : F-F < H-H < D-D.
(iv) NaH is ionic hydride. MgH₂ and H₂O are covalent hydrides but OH bond in H₂O is more stronger. Hence, increasing reducing power :
H₂O < MgH₂ < NaH.

Question 17.
Compare the structures of H₂O and H₂O₂.
Solution.
In H₂O molecule, the oxygen is sp³-hybridised. Two half-filled sp³-orbitals form O – H σ-bonds, while the other two contain lone pairs of electrons. The expected ∠HOH is 109.5°, but the experimental value is 104.5°. This is because lp-lp repulsion > bp-bp repulsion.

On the other hand, H₂O₂ is a non-planar molecule. The dihedral angle between two planes is 111.5° and ∠OOH is 94.8°. (Gas phase)

Question 18.
What do you understand by the term ‘auto-protolysis’ of water? What is its significance?
Solution.
The reaction in which two water molecules react to give ions with proton transfer is called auto-protolysis of water or k self ionization of water. A proton from one H₂O molecule is transferred to another water molecule leaving behind OH^– ion and forming a H₃O⁺ ion.
H₂O_(l) + H₂O(l) ⇌ H₃O₊(aq) + OH^–(aq)
Significance : Auto-protolysis proves that water is amphoteric in nature.

Question 19.
Consider the reaction of water with F₂ and suggest, in terms of oxidation and reduction, which species are oxidised/reduced?
Solution.
2F_2(g) + 2H₂O_(l) ➝ 4H⁺(aq) + 4F ^–_(aq) + O_2(g)
In this reaction, H₂O is oxidised to oxygen by F₂. Fluorine has been reduced to F^– ion.

Question 20.
Complete the following chemical reactions.

Classify the above into
(a) hydrolysis,
(b) redox and
(c) hydration reactions.
Solution.

Question 21.
Describe the structure of the common form of ice.
Solution.
Water molecule in solid form is in a perfect tetrahedral shape when each oxygen atom is tetrahedrally (sp³ hybridization) surrounded by four hydrogen atoms, two of which are covalently joint with oxygen and rest of the two form hydrogen bonds with oxygen. This gives ice an open cage-like structure in which each oxygen atom is in contact with four hydrogen atoms and each hydrogen atom is attached to two oxygen atoms (covalently with one and through hydrogen bonding by the other).

Question 22.
What causes temporary and permanent hardness of water?
Solution.
Temporary hardness of water is due to the presence of magnesium and calcium hydrogen carbonates. It can be very easily removed by simply boiling the hard water for sometime. Permanent hardness of water is due to the presence of soluble salts of magnesium and calcium in the form of chlorides and sulphates in water. It cannot be easily removed.

Question 23.
Discuss the principle and method of softening of hard water by synthetic ion-exchange resins.
Solution.
Hard water is made soft by a very common ion exchange resin method. In this the hard water is allowed to pass over a zeolite bed [zeolites are sodium aluminium silicates (Na₂Al₂Si₂O₈.xH₂O)] during which the Na⁺ ions from zeolite are replaced by Ca²⁺ and Mg²⁺ ions.
Na₂Z + Ca²⁺➝ CaZ + 2Na⁺
Na₂Z + Mg²⁺ ➝ MgZ + 2Na⁺
When whole of the Na⁺ ions of the zeolite have been exchanged, the zeolite is regenerated by treating it with strong (or saturated) solution of NaCl.
MZ + 2NaCl ➝ Na₂Z + MCl₂ (M = Mg, Ca)

Question 24.
Write chemical reactions to show the amphoteric nature of water.
Solution.
Water is amphoteric in nature because it can behaves both as an acid and base.

Question 25.
Write chemical reactions to justify that
hydrogen peroxide can function as an oxidising as well as reducing agent.
Solution.
H₂O₂ acts as oxidising and reducing agent in both acidic as well as in basic medium.

Question 26.
What is meant by ‘demineralised’ water and how can it be obtained?
Solution.

Question 27.
Is demineralised or distilled water useful for drinking purposes? If not, how can it be made useful?
Solution.
No, demineralised water is not always useful for drinking purposes. It is usually tasteless. Moreover, some ions such as Na⁺, K⁺ and Mg²⁺, etc. are essential to the body.
In order to make demineralised water more useful, proper amount of additional salts of sodium and potassium etc. must be dissolved in it.

Question 28.
Describe the usefulness of water in biosphere and biological systems.
Solution.
A major part of all living organisms is made up of water. It is a crucial compound for the survival of all forms of life. It is a solvent of great importance.
Liquid water is found in water bodies, such as ocean, sea, lake, river, stream, canal, pond, or puddle. The majority of water on the Earth is ocean water. Water is important in both chemical and physical weathering processes at the Earth’s surface. Water exists as vapours in the atmosphere.
All known forms of life depend on water. Water is vital both as a solvent in which many of the body’s solutes dissolve and as an essential part of many metabolic processes within the body.

Question 29.
What properties of water make it useful as a solvent? What types of compound can it
(i) dissolve, and
(ii) hydrolyse?
Solution.
Water is useful as a solvent rather excellent solvent due to the following properties :
(a) It is a liquid over a wide range of temperature (0° to 100°C).
(b) It has high enthalpv of vapourisation and heat capacitv.
(c) It is polar in nature and has a high dielectric constant (78.39).
(i) It can dissolve polar substances and also some organic compounds due to hydrogen bonding.
(ii) Water can hydrolyse oxides, halides, phosphides, nitrides, etc due to interionic attraction between them.

Question 30.
Knowing the properties of H₂O and D₂O, do you think that D2O can be used for drinking purposes?
Solution.
No, heavy water (D₂O) cannot be used for drinking purposes because it is injurious to health due to presence of D⁺ ions. Heavy water of high concentration retards the growth of plants and animals. Heavy water has germicide and bactericide properties.

Question 31.
What is the difference between the term ‘hydrolysis’ and ‘hydration’?
Solution.
When H⁺ and OH^– ions of H₂O interacts with anion and cation of the salt respectively of give the original acid and base then the reaction is called hydrolysis. It results in the change of pH of solution.
When H₂O is added to ion or molecule to give hydrated ion or compound then the reaction is called hydration.

Question 32.
How can saline hydrides remove traces of water from organic compounds?
Solution.
Saline hydrides such as NaH release H^– ions which act as strong Bronsted base whereas H₂O is a weak Bronsted acid. They combine with water to liberate hydrogen gas.
NaH + H₂O ➝NaOH + H₂
As a result, these hydrides can be used to remove traces of water from organic compounds.

Question 33.
What do you expect the nature of hydrides is, if formed by elements of atomic numbers 15, 19, 23 and 44 with dihydrogen? Compare their behaviour towards water.
Solution.
The elements with atomic numbers 15, 19, 23 and 44 are phosphorus (P), potassium (K), vanadium (V) and ruthenium (Ru) respectively. The nature of their respective hydrides is :
PH₃ (Molecular hydride), KH (Ionic), interstitial or metallic hydrides with vanadium (V). Ruthenium being a transition metal of group 8 does not form hydride. PH₃ acts as a weak base with water. KH reacts violenty with water to give KOH and H₂.

Question 34.
Do you expect different products in solution when aluminium(III) chloride and potassium chloride are treated separately with
(i) normal water
(ii) acidified water, and
(iii) alkaline water?
Write equations wherever necessary.
Solution.
Both the compounds are salts and they react differently with water.
Aluminium (III) chloride or A1C13 will react with water as follows :
AlCl₃ + 3H₂O ➝ Al(OH)₃ + 3HCl.
The reaction is known as hydrolysis.
(i) In normal water, both Al(OH)₃ and HCl will be present.

Question 35.
How does H₂O₂ behave as a bleaching agent?
Solution.
H₂O₂ acts as bleaching agent due to the release of nascent oxygen.
H₂O₂ ➝ H₂O + [O]
Thus, the bleaching action of hydrogen peroxide is permanent and is due to oxidation. It oxidises the colouring matter to a colourless product.

Question 36.
What do you understand by the terms:

1.  hydrogen economy
2.  hydrogenation
3.  syngas
4.  water-gas shift reaction
5.  fuel-cell ?

Solution.

1.  Hydrogen economy : The energy is transported and stored in the form of liquid or gaseous hydrogen is the principle due to which hydrogen is considered to be a possible source of clean energy. This proposal to use hydrogen as fuel is called hydrogen economy.
2.  Hydrogenation : The process of addition of hydrogen to unsaturated hydrocarbons is known as hydrogenation.
   
3.  Syngas : The mixture of CO and H₂ is used for the synthesis of methanol and a number of hydrocarbons, it is also called synthesis gas or syngas. Nowadays, syngas is produced from sewage, saw dust, scrap wood, newspapers, etc.
4.  Water-gas shift reaction: The production of dihydrogen can be increased by reacting CO of syngas mixtures with steam in the presence of iron chromate as catalyst.
   
   This is called water-gas shift reaction.
5.  Fuel cell: It is a commercial cell in which the chemical energy produced during the combustion of fuel is converted directly into electricity, e.g., H₂ – O₂ fuel cell. A fuel cell is used as a source of electrical energy in the space vehicles.

NCERT Exercises

Question 1.
Assign oxidation number to the underlined elements in each of the following species :
(a) NaH₂PO₄
(b) NaHSO₄
(c) H₄P₂O₇
(d) K₂MnO₄
(e) CaO₂
(f) NaBH₄
(g) H₂S₂O₇
(h) KAI(SO₄)₂.12H₂O
Solution.
Let the oxidation no. of underlined element in all the given compounds = x

Question 2.
What are the oxidation number of the underlined elements in each of the following and how do you rationalise your results?
(a) KI₃
(b) H₂S₄O₆
(c) Fe₃O₄
(d) CH₃CH₂OH
(e) CH₃COOH
Solution.
(a) In KI₃, since the oxidation number of K is +1, therefore, the average oxidation number of iodine = -1/3. In the structure, K⁺[I – I <— I]^–, a coordinate bond is formed between I2 molecule and I ion. The oxidation number of two iodine atoms forming the I2 molecule is zero while that of iodine ion forming the coordinate bond is -1. Thus, the O.N. of three iodine atoms in KI₃ are 0, 0 and -1 respectively.

Question 3.
Justify that the following reactions are redox reactions:
(a) CuO(s) + H_2(g) ➝ Cu₍₅₎ + H₂O(g)
(b) Fe₂O_3(s) + 3CO_(g) ➝ 2Fe_(s) + 3CO_2(g)
(c) 4BCI_3(g) + 3LiAIH_4(s) ➝ 2B₂H_6(g) + 3LiCI_(s) + 3AICI_3(s)
(d) 2K_(s) + F_2(g) ➝ 2K⁺F^–_(s)
(e) 4NH₃(g) + 5O_2(g) ➝ 4NO_(g) + 6H₂O_(g)
Solution.

Question 4.
Fluorine reacts with ice and results in the change:
H₂O_(s) + F_2(g) ➝ HF_(g) + HOF_(g)
Justify that this reaction is a redox reaction.
Solution.

Question 5.
Calculate the oxidation number of sulphur, chromium and nitrogen in H₂SO₅, Cr₂O₇^2- and NO₃^–. Suggest structure of these compounds. Count for the fallacy.
Solution.

Question 6.
Write formulas for the following compounds:
(a) Mercury(II) chloride
(b) Nickel(II) sulphate
(c) Tin(IV) oxide
(d) Thallium(I) sulphate
(e) Iron(III) sulphate
(f) Chromium(III) oxide
Solution.
(a) HgCl₂
(b) NiSO₄
(c) SnO₂
(d) Tl₂SO₄
(e) Fe₂(SO₄)₃
(f) Cr₂O₃

Question 7.
Suggest a list of the substances where carbon can exhibit oxidation states from -4 to +4 and nitrogen from -3 to +5.
Solution.

Question 8.
While sulphur dioxide and hydrogen peroxide can act as oxidising as well as reducing agents in their reactions, ozone and nitric acid act only as oxidants. Why?
Solution.
The oxidation state of sulphur in sulphur dioxide is +4. It can be oxidised to +6 oxidation state or reduced to +2. Therefore, sulphur dioxide acts as a reducing agent as well as oxidising agent. Similarly, the oxidation state of oxygen in hydrogen peroxide is -1. It can be oxidised to O₂ (zero oxidation state) or reduced to H₂O or OH^– (-2 oxidation state) and therefore, acts as reducing as well as oxidising agents.
However, both ozone and nitric acid can only decrease their oxidation number and therefore, act only as oxidising agents.

Question 9.
Consider the reactions :
(a) 6CO₂(g) + 6H₂O_(l) ➝ C₆H₁₂O_6(ag) + 6O_2(g)
(b) O_3(g) + H₂O_2(l) ➝ H₂O_(l) + 2O_2(g)
Also suggest a technique to investigate the path of the above (a) and (b) redox reactions.
Solution.

Question 10.
The compound AgF₂ is unstable compound. However, if formed, the compound acts as a very strong oxidising agent. Why ?
Solution.
In AgF₂ oxidation state of Ag is +2 which is very unstable. Therefore, it quickly, accepts an electron to form the more stable +1 oxidation state.
Ag²⁺ + e^– ➝ Ag⁺
Therefore, AgF₂, if formed, will act as a strong oxidising agent.

Question 11.
Whenever a reaction between an oxidising agent and a reducing agent is carried out, a compound of lower oxidation state is formed if the reducing agent is in excess and a compound of higher oxidation state is formed if the oxidising agent is in excess. Justify this statement giving three illustrations.
Solution.
(i) C is a reducing agent while O₂ is an oxidising agent. If excess of carbon is burnt in a limited supply of O₂, CO is formed in which oxidation state of C is +2 but when O₂ is in excess CO formed gets oxidised to CO₂ in which oxidation state of C is + 4.

Question 12.
How do you count for the following observations ?
(a) Though alkaline potassium permanganate and acidic potassium permanganate both are used as oxidants, yet in the manufacture of benzoic acid from toluene we use alcoholic potassium permanganate as an oxidant. Why? Write a balanced redox equation for the reaction.
(b) When concentrated sulphuric acid is added toaninorganicmixture containing chloride, we get colourless pungent smelling gas HCI, but if the mixture contains bromide then we get red vapour of bromine. Why?
Solution.
(a) In neutral medium, KMnO₄ acts as an oxidant as follows :
MnO₄^– + 2H₂O + 3e^– ➝ MnO₂ + 40H^–
In laboratory, alkaline KMnO₄ is used to oxidise toluene to benzoic acid.

Question 13.
Identify the substance oxidised, reduced, oxidising agent and reducing agent for each of the following reactions:
Solution.

Question 14.
Consider the reactions:

Why does the same reductant, thiosulphate react differently with iodine and bromine?
Solution.

Question 15.
Justify giving reactions that among halogens, fluorine is the best oxidant and among hydrohalic compounds, hydroiodic acid is the best reductant.
Solution.
The halogens (X₂) have strong electron accepting tendency and have positive standard oxidation potential values. They are therefore, powerful oxidising agents. The decreasing order of oxidising powers of halogens is :

Question 16.
Why does the following reaction occur ?

Solution.

Question 17.
Consider the reactions :

Solution.

Question 18.
Balance the following redox reactions by ion – electron method:
Solution.

Question 19.
Balance the following equations in basic medium by ion-electron method and oxidation number methods and identify the oxidising agent and the reducing agent.

Solution.

Question 20.
What sorts of informations can you draw from the following reaction?
(CN)_2(g) + 2OH^–_(aq) ➝ CN^–_(aq) + CNO^–_(aq) + H₂O_(l)
Solution.

Question 21.
The Mn³⁺ ion is unstable in solution and undergoes disproportionation to give Mn²⁺, MnO₂, and H⁺ ion. Write a balanced ionic equation for the reaction.
Solution.

Question 22.
Consider the elements: Cs, Ne, I and F
(a) Identify the element that exhibits only negative oxidation state.
(b) Identify the element that exhibits only positive oxidation state.
(c) Identify the element that exhibits both positive and negative oxidation states.
(d) Identify the element which exhibits neither the negative nor does the positive oxidation state.
Solution.
(a) F: Fluorine is the most electronegative element and shows only -1 oxidation state.
(b) Cs : Because of the presence of single electron in the valence shell, (alkali metals) Cs exhibits an oxidation state of +1 only.
(c) I: Because of the presence of seven electrons in the valence shell, I shows an oxidation state of-1 in compounds with more electropositive elements (such as H, Na, K, Ca, etc.) and oxidation states of +3, +5, +7 in compounds of I with more electronegative elements (such as, O, F, etc.)
(d) Ne: It is an inert gas (with high ionization enthalpy and highly positive electron gain enthalpy) and hence, it exhibits neither negative nor positive oxidation states.

Question 23.
Chlorine is used to purify drinking water. Excess of chlorine is harmful. The excess of chlorine is removed by treating with sulphur dioxide. Present a balanced equation for this redox change taking place in water.
Solution.

Question 24.
Refer to the periodic table given in your book and now answer the following questions:
(a) Select the possible non metals that can show disproportionation reaction.
(b) Select three metals that can show disproportionation reaction.
Solution.

Question 25.
In Ostwald’s process for the manufacture of nitric acid, the first step involves the oxidation of ammonia gas by oxygen gas to give nitric oxide gas and steam. What is the maximum weight of nitric oxide that can be obtained starting only with 10.00 g of ammonia and 20.00 g of oxygen?
Solution.
The reaction involved in the manufacturing process is :

Question 26.
Using the standard electrode potentials, predict if the reaction between the following is feasible:

Solution.

Question 27.
Predict the products of electrolysis in each of the following:
(i) An aqueous solution of AgNO₃ with silver electrodes.
(ii) An aqueous solution of AgNO₃ with platinum electrodes.
(iii) A dilute solution of H₂SO₄ with platinum electrodes.
(iv) An aqueous solution of CuCl₂ with platinum electrodes.
Solution.

(1) An aqueous solution of AgNO₃ using platinum electrodes :     (P.I.S.A. Based)
Both AgNO₃ and water will ionise in aqueous solution

At cathode : Ag⁺ ions with less discharge potential are reduced in preference to H⁺ ions which will remain in solution. As a result, silver will be deposited at cathode.
Ag⁺ (aq) + e^– → Ag (deposited)
At anode : An equivalent amount of silver will be oxidised to Ag⁺ ions by releasing electrons.

As result of electrolysis, Ag from silver anode dissolves as Ag⁺(aq) ions while an equivalent amount of Ag⁺(aq) ions from the aqueous AgNO₃ solution get deposited on the cathode.

(2) An aqueous solution of AgNO₃ using platinum electrodes :

In the case, the platinum electrodes are the non-attackable electrodes. On passing current the following changes will occur at the electrodes.
At cathode : Ag⁺ ions will be reduced to Ag which will get deposited at the cathode.
At anode : Both NO^–₃ and OH^– ions will migrate. But OH^– ions with less discharge potential will be oxidised in preference to NO₃ ions which will remain in solution.
OH- (aq) → OH + e^–; 4OH → 2H₂O(l) + O₂ (g)
Thus, as a result of electrolysis, silver is deposited on the cathode while O₂ is evolved at the anode. The solution will be acidic due to the presence of HNO₃.

(3) A dilute solution of H₂SO₄ using platinum electrodes :
On passing current, both acid and water will ionise as follows :

At cathode :
H⁺ (aq) ions will migrate to the cathode and will be reduced to H₂.
H+ (aq) + e-→ H ; H + H→ H₂ (g)
Thus, H₂ (g) will evolve at cathode.
At anode : OH ions will be released in preference to SO^2-₄ ions because their discharge potential is less. They will be oxidised as follows :
OH^– (aq) → OH + e^– ; 4OH → 2H₂O(l) + O₂ (g)
Thus, O₂ (g) will be evolved at anode. The solution will be acidic and will contain H₂SO₄.

(4) An aqueous solution of CuCl₂ using platinum electrodes :

The electrolysis proceeds in the same manner as discussed in the case of AgNO₃ solution. Both CuCl₂ and H₂O will ionise as follows :

At cathode :  Cu²⁺ ions will be reduced in preference to H⁺ ions and copper will be deposited at cathode.
Cu²⁺ (aq) + 2e^– → Cu (deposited)
At anode : Cl^– ions will be discharged in preference to OH^– ions which will remain in solution.
Cl^–→Cl^–+ e^– ; Cl + Cl → Cl₂ (g) (evolved)
Thus, Cl₂ will evolve at anode.

Question 28.
Arrange the following metals in the order in which they displace each other from the solution of their salts. Al, Cu, Fe, Mg and Zn.
Solution.

Since a metal with lower electrode potential is a stronger reducing agent, therefore, Mg can displace all the above metals from their aqueous solutions, Al can displace all metals except Mg from the aqueous solutions of their salts. Zn can displace all metals except Mg and Al from the aqueous solutions of their salts while Fe can displace only Cu from the aqueous solution of its salts. Thus, the order in which they can displace each other from the solution of their salts is Mg, Al, Zn, Fe, Cu.

Question 29.
Given the standard electrode potentials,

Solution.

Question 30.
Depict the galvanic cell in which the reaction,
Zn_(s) + 2Ag⁺_(aq) ➝ zn²⁺_(aq) + 2Ag_(s) takes place. Further show:
(i) which of the electrode is negatively charged,
(ii) the carriers of current in the cell and
(iii) individual reaction at each electrode.
Solution.
The given redox reaction is

NCERT Exercises

Question 1.
A liquid is in equilibrium with its vapour in a sealed container at a fixed temperature. The
volume of the container is suddenly increased.
(a) What is the initial effect of the change on vapour pressure?
(b) How do rates of evaporation and condensation change initially?
(c) What happens when equilibrium is restored finally and what will be the final vapour pressure?
Solution.
(a) Vapour pressure decreases since equilibrium is disturbed and the rate of condensation falls considerably.
(b) Rate of evaporation increases while that of condensation falls initially.
(c) On restoration of the equilibrium again, the vapour pressure becomes the same.

Question 2.
What is K_c for the following equilibrium when the equilibrium concentration of each substance is : [SO₂] = 0.60 M, [O₂] = 0.82 M and [SO₂] = 1.90 M
2SO_2(g) + O_2(g) ⇌ 2SO_3(g)
Solution.
2SO_2(g) + O_2(g) ⇌ 2SO_3(g)
Applying law of chemical equilibrium,

Question 3.
At a certain temperature and total pressure of 105 Pa, iodine vapour contains 40% by volume of I atoms. I_2(g) ⇌ 2l_(g)Calculate Kp for the equilibrium.
Solution.

Question 4.
Write the expression for the equilibrium constant, K_c for each of the following reactions:

Solution.

Question 5.
Find out the value of K_cof each of the following equilibria from the value of Kp:

Solution.

Question 6.
For the following equilibrium, K_c= 6.3 x 10¹⁴ at 1000 K. NO_(g) + O_3(g) ⇌ NO_{2(g) + O2(g)} the forward and reverse reactions in the equilibrium are elementary bimolecular reactions. What is K_c, for the reverse reaction?
Solution.
NO_(g) + O_3(g) ⇌ NO_2(g) + O_2(g)

For the reverse reaction,

Question 7.
Explain why pure liquids and solids can be ignored while writing the equilibrium constant expression.
Solution.
For the concentration of pure solid or pure liquid,


Since density of pure solid or liquid is constant at constant temperature and molar mass is also constant therefore, their molar concentrations are constant and are included in the equilibrium constant.

Question 8.
Reaction between N₂ and O₂ takes place as follows:
2N_2(g) + O_2(g) ⇌ 2N₂O_(g)
If a mixture of 0.482 mol of N₂ and 0.933 mol of O₂ is placed in a 10 L reaction vessel and allowed to form N₂O at a temperature for which K_c = 2.0 x 10^-37, determine the composition of equilibrium mixture.
Solution.

Question 9.
Nitric oxide reacts with Br2 and gives nitrosyl bromide as reaction given below:
2NO_(g) + Br_2(g) ⇌ 2NOBr_(g) When 0.087 mol of NO and 0.0437 mol of Br₂ are mixed in a closed container at constant temperature, 0.0518 mol of NOBr is obtained at equilibrium. Calculate equilibrium amount of NO and Br2.
Solution.

Question 10.
At 450 K, K_p = 2.0 x 10¹⁰.bar for the given reaction at equilibrium.
2SO_2(g)+ O_2(g) ⇌ 2SO_3(g)
What is K_c at this temperature?
Solution.

Question 11.
A sample of HI_(g) is placed in a flask at a pressure of 0.2 atm. At equilibrium partial pressure of HI_(g) is 0.04 atm. What is K_p for the given equilibrium ?
2HI_(g)  H_2(g) + I_2(g)
Solution.

Question 12..
AA mixture of 1.57 mol of N₂, 1.92 mol of H₂ and 8.13 mol of NH₃ is introduced into a 20 L reaction vessel at 500 K. At this temperature, the equilibrium constant K_c for the reaction given as follows :
N_2(g) + 3H_2(g)  2NH_3(g) is 1.7 × 10^-2.
Is this reaction at equilibrium ? If not, what is the direction of net reaction ?
Solution.

Question 13.
The equilibrium constant expression for a gas reaction is,

Write the balanced chemical equation corresponding to this expression.
Solution.
Balanced chemical equation for the reaction is
4 NO_(g) + 6 H₂O_(g)  4NH_3(g) + 5 O_2(g)

Question 14.
If 1 mole of H₂O and 1 mole of CO are taken in a 10 litre vessel and heated to 725 K, at equilibrium point 40 percent of water (by mass) reacts with carbon monoxide according to equation,
H₂O_(g) + CO_(g)  H_2(g) + CO_2(g)
Calculate the equilibrium constant for the reaction.
Solution.

Question 15.
At 700 K, the equilibrium constant for the reaction H_2(g) + I_2(g)  2HI_(g) is 54.8. If 0.5 mol L^-1 of HI_(g) is present at equilibrium at 700 K, what are the concentration of H₂(g) and I_2(g) ? Assume that we initially started with HI_(g) and allowed it to reach equilibrium at 700 K.
Solution.

Question 16.
What is the equilibrium concentration of each of the substances in the equilibrium when the initial concentration of ICl was 0.78 M ?
2ICl_(g)  I_2(g) + Cl_2(g) ; K_c = 0.14
Solution.

Question 17.
K_p = 0.04 atm at 898 K for the equilibrium shown below. What is the equilibrium concentration of C₂H₆ when it is placed in a flask at 4 atm pressure and allowed to come to equilibrium.
C₂H_6(g)  C₂H_4(g) + H_2(g)
Solution.

Question 18.
The ester, ethyl acetate is formed by the reaction of ethanol and acetic acid and the equilibrium is represented as :
CH₃COOH_l + C₂H₅OH_l  CH₃COOC₂H_5l + H₂O_l
(i) Write the concentration ratio (concentration quotient) Q for this reaction. Note that water is not in excess and is not a solvent in this reaction.
(ii) At 293 K, if one starts with 1.000 mol of acetic acid and 0.180 mol of ethanol, there is 0.171 mol of ethyl acetate in the final equilibrium mixture. Calculate the equilibrium constant.
(iii) Starting with 0.50 mol of ethanol and 1.000 mol of acetic acid and maintaining it at 293 K, 0.214 mol of ethyl acetate
is found after some time. Has equilibrium been reached ?
Solution.

Question 19.
A sample of pure PCl₅ was introduced into an evacuated vessel at 473 K. After equilibrium was reached, the concentration of PCl₅ was found to be 0.5 x 10^-1 mol L^-1. If K_c is 8.3 × 10^-3, what are the concentrations of PCl₃ and Cl₂ at the equilibrium ?

Solution.

Question 20.
One of the reactions that takes place in producing steel from iron ore is the reduction of iron (II) oxide by carbon monoxide to give iron metal and CO₂
FeO_(s) + CO_(g)  Fe_(s) + CO_2(g); K_p = 0.265 atm at 1050 K
What are the equilibrium partial pressures of CO and CO₂ at 1050 K if the initial pressures are :Pco = 1.4 atm and Pco₂ = 0.80 atm ?
Solution.

Question 21.
Equilibrium constant Kc for the reaction, N_2(g) + 3H_2(g)  2NH_3(g) at 500 K is 0.061. At particular time, the analysis shows that the composition of the reaction mixture is : 3.0 mol L^-1 of N₂ , 2.0 mol L^-1 of H₂,0.50 mol L^-1 of NH₃. Is the reaction at equilibrium ? If not, in which direction does the reaction tend to proceed to reach the equilibrium ?
Solution.

Question 22.
Bromine monochloride (BrCl) decomposes into bromine and chlorine and reaches the equilibrium :
2BrCl_(g)  Br_2(g) + Cl_2(g)
For which K_c is 32 at 500 K. If initially pure BrCl is present at a concentration of 3.3× 10^-3 mol L^-1, what is its molar concentration in the mixture at equilibrium ?
Solution.

Question 23.
At 1127 K and 1 atmosphere pressure, a gaseous mixture of CO and CO₂ in equilibrium with solid carbon has 90.55% by mass.
C_(s) + CO_2(g)  2CO_(g)
Calculate K_c for the reaction at the above temperature.
Solution.

Question 24.

Solution.

Question 25.
Does the number of moles of reaction products increase, decrease or remain same when each of the following equilibria is subjected to a decrease in pressure by increasing the volume ?
(a) PCl_5(g)  PCl_3(g) + Cl₂
(b) CaO_(s) + CO_2(g)  CaCO_3(s)
(c) 3Fe_(s) + 4H₂O_(g)  Fe₃O_4(s) + 4H_2(g)
Solution.
Applying Le Chatelier’s principle, on decreasing the pressure, equilibrium shifts to the direction in which pressure increases, i.e., number of moles of gaseous substances is more. Thus, moles of reaction products will increase in reaction (a), decrease in reaction (b) and remain same (n_p, = n_r gaseous) in reaction (c).

Question 26.
Which of the following reactions will get affected by increase in pressure ? Also mention whether the change will cause the reaction to go in the forward or backward direction.

Solution.

Question 27.
The equilibrium constant for the following reaction is 1.6 x 10⁵ at 1024 K.
H_2(g) + Br_2(g)  2HBr_(g)
Find the equilibrium pressure of all gases if 10.0 bar of HBr is introduced into a sealed container at 1024K.
Solution.

Question 28.
Dihydrogen gas is obtained from the natural gas by partial oxidation with steam as per following endothermic reaction :
CH_4(g) + H₂O_(g)  CO_(g) + 3H_2(g)
(a) Write the expression for K_p for the above reaction
(b) How will the value of K_p and composition of equilibrium mixture be affected by :
(i) increasing the pressure
(ii) increasing the temperature
(iii) using a catalyst ?
Solution.

(b)

1. By Le Chatelier’s principle, on increasing pressure, equilibrium will shift in the backward direction where number of moles decreases.
2. As the given reaction is endothermic, by Le Chatelier ‘sprinciple, equilibrium will shift in the forward direction with increasing temperature.
3. Equilibrium composition will not be disturbed by the presence of catalyst but equilibrium will be attained quickly.

Question 29.
Describe the effect of :
(a) addition of H₂
(b) addition of CH₃OH
(c) removal of CO
(d) removal of CH₃OH on the equilibrium of the reaction:
2H_2(g) + CO_(g)  CH₃OH ?
Solution.
2H_2(g) + CO_(g)  CH₃OH Effect of
(a) addition of H₂ : The equilibrium will shift in the forward direction.
(b) addition of CH₃OH : The equilibrium will shift in the backward direction.
(c) removal of CO : The equilibrium will shift in the backward direction.
(d) removal of CH₃OH : The equilibrium will shift in the forward direction.

Question 30.
At 473 K, the equilibrium constant Kc for the decomposition of phosphorus pentachloride (PCl₅) is 8.3 × 10^-3. If decomposition proceeds as :
PCl_5(g)  PCl_3(g) + Cl_2(g) ; ∆H = + 124.0 kJ mol^-1
(a) Write an expression for K_c for the reaction.
(b) What is the value of K_c for the reverse reaction at the same temperature ?
What would be the effect on K_c if
(i) more PCl₅ is added
(ii) pressure is increased
(iii) the temperature is increased?
Solution.

Question 31.
Dihydrogen gas used in Haber’s process is produced by reacting methane from natural gas with high temperature steam. The first stage of two stage reaction involves the formation of CO and H₂. In second stage, CO formed in first stage is reacted with more steam in water gas shift reaction.
CO_(g) + H₂O_(g)  CO_2(g) + H_2(g)
If a reaction vessel at 400°C is charged with an equimolar mixture of CO and steam so that pco = PH₂O = 4.0 bar, what will be the partial pressure of H₂ at equilibrium ? K_p = 0.1 at 400°C.
Solution.

Question 32.
Predict which of the following will have appreciable concentration of reactants and products :
(a) Cl_2(g)  2Cl_(g) ; K_c = 5 ×10^-39
(b) Cl_2(g) + 2NO_(g)  2NOCl_(g) ; K_c = 3.7 × 10⁸
(c) Cl_2(g) + 2NO₂ (g)  2NO₂Cl_(g) ; K_c = 1.8.
Solution.
(a) Since the value of K_c is very small
∴ There will be appreciable concentration of the reactants.
(b) Since K_c is very large.
∴ There will be appreciable concentration of the products.
(c) Since K_c = 1.8, the concentration of reactants and products will be in comparable amounts.

Question 33.
The value of K_c for the reaction 3O_2(g)  2O_3(g) is 2.0 × 1 x 10^-50 at 25°C. If equilibrium concentration of O₂ in air at 25°C is 1.6 × 10^-2, what is the concentration of O₂ ?
Solution.

Question 34.
The reaction CO_(g) + 3H_2(g)  CH_4(g) + H₂O_(g) is at equilibrium at 1300 K in a 1L flask. It also contain 0.30 mol of CO, 0.10 mol of H₂ and 0.02 mol of H₂O and an unknown amount of CH₄ in the flask. Determine the concentration of CH₄ in the mixture. The equilibrium constant, K_c for the reaction at the given temperature is 3.90.
Solution.

Question 35.
What is meant by conjugate acid-base pair ? Find the conjugate acid/base for the following species :
HNO₂,CN^–,HClO₄,OH^–,CO₃^2-,S^2-.
Solution.

Question 36.
Which of the following are Lewis Acids ?
H₂O, BF₃, H⁺ and NH₄⁺
Solution.
BF₃, H⁺ and NH₄⁺ are Lewis acids because they can accept a lone pair of electrons.

Question 37.
What will be the conjugate bases for the Bronsted acids ? HF, H₂SO₄ and  ?
Solution.

Question 38.
Write the conjugate acids for the following Bronsted acids.
, NH₃ and HCOO^–
Solution.

Question 39.
The species H₂O, HCO₃^–, HSO₄^– and NH₃ can act both as Bronsted acid and base. For each case, give the corresponding conjugate acid and base.
Solution.

Question 40.
Classify the following species into Lewis acids and Lewis bases and show how these can act as Lewis acid/Lewis base ?
(a) OH^–
(b) F^–
(c) H⁺
(d) BCl₃
Solution.
(a) OH^– : OH^– is a Lewis base because it can donate lone pair of electrons.
(b) F^– : F^– is a Lewis base because it can donate lone pair of electrons.
(c) H⁺ : H⁺is a Lewis acid because it can accept lone pair of electrons.
(d) BCl₃ : is a Lewis acid because it is electron deficient and can accept a lone pair of electrons.

Question 41.
The concentration of hydrogen ions in a sample of soft drink is 3.8 × 10^-3 M. What is its pH value ?
Solution.

Question 42.
The pH of soft drink is 3.76. Calculate the concentration of hydrogen ions in it.
Solution.

Question 43.
The ionisation constants of HF, HCOOH and HCN at 298 K are 6.8 x 10^-4, 1-8 × 10^-4 and 4.8 × 10^-9 respectively. Calculate the ionisation constant of the corresponding congugate bases.
Solution.

Question 44.
The ionisation constant of phenol is 1.0 x 10^-10. What is the concentration of phenate ion in 0.05 M solution of phenol and pH of solution ? What will be the degree of ionisation if the solution is also 0.01 M in sodium phenate ?
Solution.

Question 45.
The first dissociation constant H₂S is 9.1 × 10^-8. Calculate the concentration of HS^– ions in its 0.1 M solution and how much will this concentration be affected if the solution is 0.1 M in HCl also. If the second dissociation constant of H₂S is 1.2 × 10^-13, calculate the concentration of S^2- ions under both the conditions.
Solution.

Question 46.
The ionization constant of acetic acid is 1.74 × 10^-5. Calculate the degree of dissociation of acetic acid in 0.05 M solution. Calculate the concentration of acetate ions in solution and the pH of the solution.
Solution.

Question 47.
It has been found that the pH of 0.01 M solution of organic acid is 4.15. Calculate the concentration of the anion, ionization constant of the acid and the pK_a.
Solution.

Question 48.
Assuming complete dissociation, calculate the pH of the following solutions :
(a) 0.003 M HCl
(b) 0.005 M NaOH
(c) 0.002 M HBr
(d) 0.002 M KOH.
Solution.

Question 49.
Calculate the pH of the following solutions :
(a) 2 g of TIOH dissolved in water to give 2 litre of solution.
(b) 0.3 g of Ca(OH)₂ dissolved in water to give 500 mL of solution.
(c) 0.3 g of NaOH dissolved in water to give 200 mL of solution.
(d) 1 mL of 13.6 M HCl diluted with water to give 1 litre of solution.
Solution.

Question 50.
The degree of ionization of 0.1 M bromoacetic acid solution is 0.132. Calculate the pH of the solution and pK_a of bromoacetic acid.
Solution.

Question 51.
The pH of 0-005 M codeine (C₁₈H₂₁NO₃) solution is 9.95. Calculate its ionizationconstant and pK_b.
Solution.

Question 52.
What is the pH of 0.001 M aniline solution ? The ionization constant of aniline is 4.27 × 10^-10. Calculate the degree of ionization of aniline in the solution. Also calculate the ionization constant of the conjugate acid of aniline.
Solution.

Question 53.
Calculate the degree of ionization of 0.05 M acetic acid if its pK_a value is 4.74. How is degree of dissociation affected when the solution contains (a) 0.01 M HCl (b) 0.1 M HCl ?
Solution.

Question 54.
The ionization constant of dimethylamine is 5.4 × 10^-4. Calculate the degree of ionization of its 0.02 M solution. What percentage of dimethylamine is ionized if the solution is also 0.01 M NaOH ?
Solution.

Question 55.
Calculate the hydrogen ion concentration in the following biological fluids whose pH values are as follows :
(a) Human muscles fluid 6.83
(b) Human stomach fluid 1.2.
(c) Human blood 7.38
(d) Human saliva 6.4
Solution.

Question 56.
The pH of milk, black coffee, tomato juice, lemon juice and egg white are 6.8, 5.0, 4.2, 2.2 and 7.8 respectively. Calculate corresponding hydrogen ion concentration in each case.
Solution.

Question 57.
0.561 g KOH is dissolved in water to give 200 mL of solution at 298K. Calculate the concentration of potassium, hydrogen and hydroxyl ions. What is the pH of the solution ?
Solution.

Question 58.
The solubility of Sr(OH)₂ at 298 K is 19.23 g/L of solution. Calculate the concentrations of strontium and hydroxyl ions and the pH of the solution.
Solution.

Question 59.
The ionisation constant of propionic acid is 1.32 × 10^-5. Calculate the degree of ionisation of acid in its 0.05 M solution and also its pH. What will be its degree of ionisation if the solution is 0.01 M in HCl also ?
Solution.

Question 60.
The pH of 0.1M solution of cyanic acid (HCNO) is 2.34. Calculate ionisation constant of the acid and also its degree of dissociation in the solution.
Solution.

Question 61.
The ionisation constant of nitrous acid is 4.5 × 10^-4. Calculate the pH value of 0.04 M NaNO₂ solution and also its degree of hydrolysis.
Solution.

Question 62.
A 0.02 M solution of pyridinium hydrochloride has pH = 3.44. Calculate the ionisation constant of pyridine.
Solution.

Question 63.
Predict if the solutions of the following salts are neutral, acidic or basic :
NaCl, KBr, NaCN, NH₄NO₃, NaNO₂, and KF
Solution.

Question 64.
The ionisation constant of chloroacetic acid is 1.35 × 10^-3. What will be the pH of 0.1 M acid solution and of its 0.1 M sodium salt solution ?
Solution.

Question 65.
The ionic product of water at 310 K is 2.7 × 10^-14. What is the pH value of neutral water at this temperature ?
Solution.

Question 66.
Calculate the pH of the resultant mixtures:
(a) 10 mL. of 0.2 M Ca(OH)₂ + 25 mL of 0.1 M HCl
(b) 10 mL of 0.01 M H₂SO₄ + 10 mL of 0.01 M Ca(OH)₂
(c) 10 mL of 0.1 M H₂SO₄ + 10 mL of 0.1 M KOH.
Solution.

Question 67.
Determine the solubilites of silver chromate, barium chromate and ferric hydroxide at 289 K from their solubility product constants. Determine also the molarities of the individual ions.

Solution.

Question 68.
The solubility product constants of Ag₂CrO₄ and AgBr are 1.1 × 10^-12 and 5.0 × 10^-13 respectively. Calculate the ratio of the molarities of their saturated solutions.
Solution.

Question 69.
Equal volumes of 0.002 M solutions of sodium iodate and copper chlorate are mixed together. Will it lead to the precipitation of copper iodate ? (For copper iodate K_sp = 7.4 × 10^-8).
Solution.

Question 70.
The ionization constant of benzoic acid is 6.46 × 10^-5 and K_sp for silver benzoate is 2.5 × 10^-13. How many times is silver benzoate more soluble in a buffer of pH 3.19 compared to its solubility in pure water ?
Solution.

Question 71.
What is the maximum concentration of equimolar solutions of ferrous sulphate and sodium sulphide so that when mixed in equal volumes, there is no precipitation of iron sulphide ? (For iron sulphide, K_sp = 6.3 × 10^-18).
Solution.

Question 72.
What is the minimum volume of water required to dissolve 1 g of calcium sulphate at 298 K. For calcium sulphate K_sp = 9.1 × 10^-6?
Solution.

Question 73.
The concentration of sulphide ion in 0.1 M HCl solution saturated with hydrogen sulphide is 1.0 × 10^-19 M. If 10 mL of this solution is added to 5 mL of 0.04 M solution of FeSO₄, MnCl₂, ZnCl₂ and CaCl₂, in which solutions precipitation will take place ?
Given K_sp for
FeS = 6.3 × 10^-18,
MnS = 2.5 × 10^-13,
ZnS = 1.6 × 10^-24 and
CdS = 8.0 × 10^-27.
Solution.