CBSE Class 11

Here we are providing Class 11 chemistry Important Extra Questions and Answers Chapter 14 Environmental Chemistry. Chemistry Class 11 Important Questions are the best resource for students which helps in Class 11 board exams.

Class 11 Chemistry Chapter 14 Important Extra Questions Environmental Chemistry

Environmental Chemistry Important Extra Questions Very Short Answer Type

Question 1.
What is the percentage of CO₂ in pure dry air?
Answer:
About 0.03%.

Question 2.
Which gas was responsible for the Bhopal tragedy?
Answer:
Methyl isocyanate (MIC).

Question 3.
What is smog?
Answer:
A combination of smoke and fog.

Question 4.
Name three natural sources of air pollution.
Answer:

1. Volcanic eruptions
2. Forest fires
3. Pollen grains of flowers.

Question 5.
What is that compound formed when CO combines with blood?
Answer:

Question 6.
Name three gases that are major air pollutants.
Answer:

1. CO₂
2. NO₂
3. SO₂.

Question 7.
How does particulate help in cloud formation?
Answer:
The particulates work as nuclei for cloud formation.

Question 8.
Give one main reason for ozone depletion.
Answer:
The release of chlorofluorocarbons (CFCs) also known as freon.

Question 9.
Name two important sinks of CO.
Answer:
Oceans that dissolve it and plants use it for photosynthesis.

Question 10.
Which acids arc present in the added rain?
Answer:

1. H₂SO₄
2. HNO₃
3. HCl.

Question 11.
What is the nature of London smog?
Answer:
It is reducing in nature.

Question 12.
Name two herbicides used to kill weeds.
Answer:

1. Sodium chlorate (NaClO₃)
2. Sodium arsenite (Na₃AsO₃)

Question 13.
Name two gases that form acid rain.
Answer:

1. SO₂
2. NO₂.

Question 14.
What is the composition of photochemical smog?
Answer:
It is a mixture of irritation causing compounds NO₂, O₃, hydrocarbons, acrolein, formaldehyde, peroxyacetyl nitrate (PAN).

Question 15.
Which zone is called the ozonosphere?
Answer:
Stratospheric zone.

Question 16.
Name any four methods of waste management.
Answer:

1. Recycling
2. Burning and incineration
3. Sewage treatment
4. Dumping.

Question 17.
What is BOD?
Answer:
The amount of oxygen consumed by microorganisms in decomposing waste in a sample of sewage water is called BOD (Biochemical Oxygen Demand).

Question 18.
What is the size range of particulates?
Answer:
5 nm to 500,000 nm.

Question 19.
What type of aromatic compounds are present as particulates in the air?
Answer:
Poly cyclic aromatic hydrocarbons [PAH].

Question 20.
In which season and what time of day, there is London smog?
Answer:
In winter during morning hours.

Question 21.
Which disease is caused due to a hole in the ozone layer and why?
Answer:
Ultraviolet rays from the sun will reach the earth after passing through the ozone hole and cause skin cancer.

Question 22.
What are polar stratospherical clouds (PSC)?
Answer:
The special type of clouds present over Antarctica in winter are called polar stratospheric clouds.

Question 23.
What should be the tolerable limit of fluoride ions in drinking water? What happens if it is more than 10 ppm?
Answer:
One ppm or 1 mg dm^-3. The higher concentration is harmful to bones and teeth.

Question 24.
Which main compounds are causing damage to the ozone layer?
Answer:
NO and freons.

Question 25.
What is the composition of London smog?
Answer:
Fog of H₂SO₄ droplets deposited on the particulates.

Question 26.
Why photochemical smog is so called?
Answer:
Because it is formed as a result of the photochemical reaction between oxides of nitrogen and hydrocarbons.

Question 27.
In which season and what time of day, there is photochemical smog?
Answer:
In summer, in the afternoon.

Question 28.
What is the ‘nature of photochemical smog?
Answer:
It is oxidising in nature.

Question 29.
Name three methods used in green chemistry.
Answer:
Use of sunlight and microwaves, use of sound waves and use of enzymes.

Question 30.
Give three examples in which green chemistry has been applied.
Answer:

1. Dry cleaning of clothes
2. Bleaching of paper
3. Synthesising chemicals.

Question 31.
What are the main advantages of using green chemistry?
Answer:
It is a cost-effective approach that involves a reduction in material, energy consumption and waste generation.

Question 32.
What is the troposphere?
Answer:
The troposphere is the lowest region of the atmosphere (~ 10 km) in which man along with other organisms including plants exist.

Question 33.
What is the stratosphere?
Answer:
The stratosphere extends above the troposphere up to 50 km above; sea level.

Question 34.
Name three scientists who got Noble Prize in chemistry in 2005 for their work in reducing hazardous waste in creating new chemicals.
Answer:

1. Yves Chauvin
2. Robert H. Grubbs
3. Richard R. Schrock.

Question 35.
What is meant by metathesis?
Answer:
Metathesis is an example of how important basic science has been applied for the benefit of man, society and the environment.

Question 36.
In what regions of the atmosphere, the temperature increases with altitude and in which regions it decreases?
Answer:
Temperature increases with altitude in the stratosphere and thermosphere and it decreases with altitude in the troposphere and mesosphere.

Question 37.
What do you mean by ‘Inversion temperature’ in different regions of the atmosphere?
Answer:
When we go from one region of the atmosphere to the next adjoining region, the trend of temperature changes from increase to decrease or vice-versa is called inversion temperature.

Question 38.
What is the most important sink of CO pollutant?
Answer:
Soil micro-organisms.

Question 39.
What is the compound formed when CO combines with blood?
Answer:
Carboxyhaemoglobin (HbCO).

Question 40.
What is anoxia starvation in the body (due to CO poisoning)?
Answer:
Actual oxygen starvation in the body (due to CO poisoning).

Question 41.
How are the flue gases from industries freed from oxides of N and S?
Answer:
By scrubbing them with a cone. H₂SO₄ or with alkaline solutions of Ca(OH)₂ & Mg(OH)₂.

Question 42.
What is chlorosis?
Answer:
Slowing down the formation of chlorophyll in plants due to the presence of SO₂ as a pollutant is called chlorosis.

Question 43.
What is the cause of blue baby syndrome?
Answer:
Excess of nitrates in drinking water leads to methemoglobinemia which is commonly called ‘blue baby syndrome.

Question 44.
Give I.U.P.A.C. name of B.H.C.
Answer:
1, 2, 3, 4, 5, 6 Hexa- chloro cyclohexane.

Question 45.
What is meant by P.C.Bs.?
Answer:
P.C.Bs. are polychlorinated biphenyls. They are contaminants of water.

Question 46.
What is P.A.N.?
Answer:
Peroxy acetyl nitrate.

Question 47.
Name the oxidizing agent used in determining C.O.D.
Answer:
Potassium dichromate (K₂C₂O₇).

Question 48.
What are fungicides?
Answer:
The chemicals which check the growth of fungi are called fungicides.

Question 49.
When does rainwater become Acid rain?
Answer:
When the pH of rainwater becomes as low as 2 to 3.5, it forms acid rain.

Question 50.
What is meant by Taj Trapezium?
Answer:
Taj Trapezium is the area that includes the town of Agra, Firozabad, Mathura & Bharatpur. Under this plan, more than 2000 polluting industries lying inside the Trapezium would switch over to the use of natural gas or L.P.G. instead of coal or oil.

Environmental Chemistry Important Extra Questions Short Answer Type

Question 1.
Fish do not grow as well in warm water as in cold water. Why?
Answer:
The amount of dissolved oxygen in warm water is less than in cold water. For the fish to survive in water, the concentration of dissolved O₂ is 6 ppm which decreases in warm water.

Question 2.
Why does rainwater normally have a pH of about 5.6? When does it become acid rain?
Answer:
Normally rain has a pH of about 5.6 due to the dissolution of CO₂ of the atmosphere into it.
CO₂(g) + H₂0O(l) → H₂CO₃(al)
H2CO₃(aq) ⇌ 2H+ + CO₃^2-
When the pH of rain falls below 5.6, it becomes acid rain.

Question 3.
In which season the depletion of ozone in Antarctica takes place and when is it replenished?
Answer:
During spring seasons (i.e., in the months of September and October) depletion of ozone takes place and after spring (in the month of November) it is replenished.

Question 4.
What is the role of CO₂ in the greenhouse effect?
Answer:
The heat from the sun after being absorbed by the earth is remitted by the earth and absorbed by CO₂ and then radiated back to the earth, thereby warming it.

Question 5.
What are viable and non-viable particulates?
Answer:
Viable particulates are small-sized living organisms such as bacteria, fungi, moulds and algae etc. Non-viable particulates are formed by the disintegration of large-sized materials or condensation of small-sized particles or droplets e.g. mist, smoke, fume and dust.

Question 6.
Why there is ozone depletion mainly over Antarctica?
Answer:
This is because, in other parts of the stratosphere, chlorine-free radicals combine away but in Antarctica, the compounds formed are converted back to chlorine-free radicals which deplete the ozone layer.

Question 7.
Explain giving reasons. “The presence of CO reduces the amount of haemoglobin available in the blood for carrying oxygen to the body cells”.
Answer:
CO combines with haemoglobin of the red blood corpuscles (R.B.Cs) about 300 times more easily than oxygen to form carboxyhaemoglobin reversibly as follows:
Hb + CO ⇌ HbCO

Thus it is not able to combine with oxygen to form oxyhaemoglobin and transport of oxygen to different body cells cannot take place.

Question 8.
Why is acid rain considered a threat to the Taj Mahal?
Answer:
Taj Mahal is made of white marble (CaCO₃). The acid rain contains H₂SO₄ in a very dilute form which attacks marble, thereby pitting it, discolouring it and making it lustreless.
CaCO₃(s) + H₂SO₄ (aq) → CaSO₄(aq) + CO₂(g) + H₂O(l)

Question 9.
What are the harmful effects of carbon monoxide?
Answer:
1. Carbon mono-oxide when inhaled passed through the lungs into the blood where it reacts with the haemoglobin of the die R.B.C. to form a complex. Known as Carboxy haemoglobin the latter is not in a position to transport the inhaled oxygen to various parts of the body. This will cause suffocation & will ultimately lead to death.

2. A high concentration of carbon mono-oxide (100 ppm) or more will harmfully affect the plants causing leaf drop reduction in leaf size & pre-mature ageing etc.

Question 10.
What are the harmful effects of oxides of nitrogen?
Answer:

1. A high concentration of NO₂ in the atmosphere is harmful to plants resulting in leaf spotting, retardation of photosynthetic activity & also suppression the vegetation growth.
2. NO₂ results in respiratory problems in human beings & lead to bronchitis.
3. NO₂ has harmful effects on nylon, rayon & cotton yarns & also cause cracks in the rubber.
4. They also react with ozone present in the atmosphere & thus decrease the density of ozone.

Question 11.
What are the harmful effects of oxides of Sulphur?
Answer:

1. SO₂ is a major source of irritation for the eyes & respiratory tract & leads to serious diseases such as bronchitis & lung cancer even when it is about 2.5 ppm. in the atmosphere.
2. London smog & sulphurous acid smog which mainly consists of SO₂(0-40 ppm or above) is known as smog killer.
3. Plants are relatively more sensitive to the harmful effect of SO₂ as compared to animals. Constant exposure to air containing a high level of SO₂ kills leaf tissues, reduces plants productivity & also bleaches leaf pigments.
4. SO₂ has harmful effects on buildings & statues made up of marble & limestone.
5. Air polluted with oxides of sulphur also accelerates the corrosion of metals like copper, zinc, iron etc.

Question 12.
How do detergents cause water pollution?
Answer:

1. Detergents are non-biodegradable & they cause water pollution.
2. They inhibit the oxidation of organic substances present in wastewaters because they form a sort of Envelope around them.
3. They form stable foam in rivers that extend over several hundred meters of the river water.

Question 13.
Why is add rain considered a threat to the Taj Mahal?
Answer:
The air of the city Agra, where the Taj Mahal is located, contains a very high level of Sulphur & nitrogen oxides. The resulting acid rain reacts with the marble of the Taj Mahal. Causing pitting in this wonderful monument that has attracted people from around the world. As a result, the monument is being slowly eaten away & the marble is getting decoloured & lusterless.

Question 14.
Explain giving reasons “The presence of co-reduces the amount of haemoglobin available in the blood for carrying oxygen to the body cells?
Answer:
Carbon mono-oxide binds itself with the haemoglobin of the R.B.Cs. about 200 times more easily than oxygen to form carboxyhaemoglobin reversibly as follows:
Hb + CO ⇌ HbCO

The presence of CO, therefore, reduces the amount of haemoglobin available in the blood for the transport of oxygen to the body cells & therefore, with fewer O₂ levels, normal metabolism is impaired.

Question 15.
What do you understand by the greenhouse effect? What are the major greenhouse gases?
Answer:
The greenhouse effect is the phenomenon in which the earth’s atmosphere traps the heat from the sun, & prevents it from escaping into outer space. The warming of the earth or global warming due to re-emission of sun’s energy absorbed by the earth followed by its adsorption by CO₂ molecules & H₂O vapours present near the earth’s surface & then it’s radiation back to the earth is called “Greenhouse effect”. The major greenhouse gases are:

CO₂, CH₄, C.F.Cs & water vapours.

Question 16.
What do you understand by Mists, Smoke, Fumes & Dust?
Answer:
Mists: Mists are produced by particles of spray liquids & the condensation of vapours in the air. Examples are portions of herbicides & insecticides, that miss their targets & travel through the air to form mists.

Smoke: They are very small soot particles produced by burning & combustion of organic matter. Oil smoke, Tobacco smoke & Carbon smoke are typical examples.

Dust: It consists of fine particle produced during crushing, grinding & attribution of solid materials. Non-viable dust particulates in the atmosphere consist of ground limestone, sand tailing from floatation, pulverised coal, cement, fly ash & silica dust.

Question 17.
What is Pneumo coniosis? How does it occur?
Answer:
The smaller particulate pollutants are more likely to penetrate into the lungs. These fine particles are carcinogens. Inhalation of small particles irritates the lungs & exposure to such particles for long periods of time causes fibrosis of lung lining. This type of disease is termed “Pneumoconiosis”. This occurs due to exposure to such particles for a long time.

Question 18.
What measures should be taken to check pollution by sewage?
Answer:

1. Sewage must be churned by machines so that the large pieces may break into smaller ones & may get mixed thoroughly. The churned sewage is passed into a tank with a gentle slope. Heavier particles settle & the water flowing down is relatively pure.
2. Water must be sterilized with the help of chlorination. Chlorination is very essential, particularly in rainy session.
3. Treatment of water with alum, lime etc.

Question 19.
Discuss the water pollution caused by industrial wastes.
Answer:
The compounds of lead, mercury, cadmium, nickel, cobalt & zinc etc. Which are the products of chemical reactions; carried in the industrial units pollute water to a large extent & are responsible for many diseases. Mercury leads to Minamata disease, & lead poisoning leads to various types of deformities.

Question 20.
What remedial steps should be taken to save a person suffering from co-poisoning?
Answer:
Remedial treatment for CO poisoning:

1. Carry the patient into the fresh air immediately & do not allow him to walk.
2. Lose his clothes & take off his shoes.
3. Give artificial respiration if the patient is not able to breathe properly.
4. In the hospital the patient should be kept in a high-pressure chamber containing oxygen at 2-2.5 atm pressure under pressure, CO of carboxyhaemoglobin is replaced by O₂ & thus the transport of O₂ to different parts of the body starts.
   HbCO + O₂ ⇌ HbO₂ + CO

Question 21.
How can pollution due to nitrogen & Sulphur oxides be controlled?
Answer:
The following steps are taken to control NO₂ & SO₂ pollution:
(i) The catalytic converters should be used in the automobile exhausts which is the first stage of converting the oxides of nitrogen, to free N₂ or to a small amount of NH₃.
(ii) The fumes of gases coming from power plants or industrial, units & containing NO₂ & SO₂ are freed from these gases by scrubbing the fumes with H₂SO₄. The following reactions take place:

• I- step: NO₂ + SO₂ + H₂O → H₂SO₄ + NO
• II- step: NO + NO₂ → N₂O₃
• III- step: N₂O₃ + 2H₂SO₄ → 2NOHSO₄ + H₂O

the fumes of gases are thus freed from NO₂ & SO₂ & are released into the atmosphere. The reaction product NOH₂O₄ is decomposed to get H₂SO₄ which is then used again for scrubbing. As NO_x & SO_x are > acidic oxides, scrubbing can also be done with an alkaline solution such as Ca(OH)₂ & Mg(OH)₂.

Question 22.
What is groundwater pollution? How does it take place?
Answer:
Water below the surface of the earth is called groundwater. Most of the freshwater (> 90%) is present as groundwater. The remaining is present as groundwater. The remaining is present in lakes, ponds, rivers, streams etc. only 2% of water is present as soil moisture above the water table, which is needed for the growth of the plants.

Groundwater collected below the surface of the earth after passing through the pores of earthy materials which acts as a filter for it & is pure. It is for this reason that well water is spring water is used for drinking purpose in rural areas. However, due to the disposal of domestic wastes, industrial effluents, use of fertilizers & pesticides in agriculture, A number of harmful soluble substance dissolve into the rainwater & result in groundwater pollution especially where the water table is high.

Question 23.
How is the pollution of river water caused in India? What measures have been taken by the Government to check river pollution?
Answer:
The main reasons for pollution in rivers are as follows:

1. Industrial waste discharge: It includes those from paper, textiles, fertilizers, rayon, pesticides, detergents, drug industries & refineries.
2. Domestic sewage discharge: The government has enacted laws banning the discharge of industrial or domestic wastes into these rivers. It has started the following plans to clean up the water of these rivers.
   (a) Ganga Action Plan I & II
   (b) Yamuna Action Plan
   (c) Plan to clean Hoogly water.

Question 24.
What are the effects of oil pollution on Sea Water?
Answer:
Effects of oil pollution on Sea Water:

1. oil spills cause heavy damage to fishes oil coating makes them unable to respire & clogs their gill slits. Aromatic compounds present in them are poison for fishes.
2. Emulsified oil goes deep down into the sea damaging aquatic animals & plants.
3. Oil spills result in a reduction of dissolved oxygen.
4. The most affected by oil pollution are the sea birds. Natural, insulating oil & waxes which shield the birds from water are broken down by the spilt oil. As a result due to loss of insulation, they start shivering & are freezing to death, especially in winter.

Question 25.
Which is the permitted safety limit of fluoride & lead concentration with respect to international standards of drinking water?
Answer:
The fluoride concentration should be about 1 ppm in drinking water. This concentration is within agreed safety limits & has been shown to protect teeth against decay. A high concentration of fluoride is poisonous & are harmful to bones & teeth at levels over 10 ppm.

The lead concentration should be about 50 ppm in drinking water.

Question 26.
Define the term pesticides? What are three categories of pesticides?
Answer:
Pesticides are substances that are used to control the reproductive process of unwanted organisms.

Three main categories of pesticides are:

1. Insecticides: These are used to control insects & curb diseases (Malaria, Yellow fever) & protect crops e.g. D.D.T.
2. Herbicides: These are used to kill weeds eg. NaClO₃ & Na₂AsO₃
3. Fungicides: These are used to check the growth of fungi e.g. Methyl mercury.

Question 27.
Write four major pollutants of water, their source & effects.
Answer:

Question 28.
Discuss the mechanism of treatment of industrial wastes.
Answer:
The treatment of industrial waste depends upon the nature of the pollutant present. In order to ascertain it, the pH of a medium is first determined & the waste is then neutralized with the help of suitable acids or alkalies.

The chemical substances present in the industrial waste product, dissolved in water can be precipitated by suitable chemical reactions & removed later on from water. Quite recently, photo-catalysis & ion exchangers have been developed for the treatment of industrial wastes.

Question 29.
What measures are necessary to control soil pollution?
Answer:
In order to control soil pollution, the following measures are necessary:

1. Use of manures: Manures is semi-decayed organic matter which is added to the soil to maintain its fertility. These are mostly prepared from animal dung & another form of wastes. These are much better than the commonly used fertilizers.
2. Use of bio-fertilizers: These are organisms that are inoculated in order to bring about nutrient enrichment of the soil e.g. nitrogen-fixing bacteria & blue-green algae.
3. Proper sewerage System; A proper sewerage system must be employed & sewerage recycling plants must be installed in all town & cities.
4. Salvage & Recycling: Rag-picking removes a large no. of waste articles such s paper, polythene, cardboard rags, empty bottles & metallic, articles. These are subjected to recycling & this helps in checking soil pollution.

Question 30.
What are aerosols? Define the term green chemistry.
Answer:
Very small particles of solids & liquids dispersed in the air are called aerosols. Aerosol particles have a diameter of less than one micrometre (< 1μm). The particles of aerosols are so small that they remain suspended in the air.

Green Chemistry: The branch of chemistry which emphasizes the process & products that reduce or eliminate the use & generation of toxic/hazardous substances is called Green Chemistry.

Environmental Chemistry Important Extra Questions Long Answer Type

Question 1.
What is the difference between London (classical) smog and photochemical (Los Angeles) smog?
Answer:

Question 2.
(a) Mention some of the sources of soil pollution.
Answer:
Some of the sources of soil pollution are:

1. Industrial wastes.
2. Urban wastes
3. Agriculture pollutants include synthetic fertilizers, pesticides, insecticides like DDT, BHC, herbicides like NaClO₃, Na₃AsO₃, fungicides soil conditioners, farm wastes.
4. Radioactive pollutants.

(b) Mention some ways to control environmental pollution.
Answer:
Environmental pollutants like household waste and industrial waste can be controlled in the following manner.

1. Recycling: Used glass bottles, iron scrap, plastic waste, polythene bags, used newspapers and magazines can all be properly recycled.
2. Burning and incineration.
3. Sewage treatment
4. Digesting.
5. Dumping
6. Green chemistry can be used to reduce pollution.

Here we are providing Class 11 chemistry Important Extra Questions and Answers Chapter 13 Hydrocarbons. Chemistry Class 11 Important Questions are the best resource for students which helps in Class 11 board exams.

Class 11 Chemistry Chapter 13 Important Extra Questions Hydrocarbons

Hydrocarbons Important Extra Questions Very Short Answer Type

Question 1.
Give different isomers of C₄H₁₀ with their I.U.P.A.C. names.
Answer:

Question 2.
Give the I.U.P.A.C. name of the lowest molecular weight alkane that contains a quaternary carbon.
Answer:

It is and its I.U.P.A.C. name is 2, 2-Dimethylpropane.

Question 3.
Which of the following has the highest boiling point?
(i) 2-methylpentane
(ii) 2, 3 – dimethylbutane
(iii) 2, 2-dimethylbutane.
Answer:
(i) 2—methyl pentane has the largest surface area and hence has the highest boiling point.

Question 4.
Give the structure of the alkene (C₄H₈) which adds on HBr in the presence and in the absence of peroxide to give the same product C₄H₉Br.
Answer:
2-Butene with structure CH₃ – CH = CH — CH₃ being symmetrical gives the same product, i.e., 2-bromobutane CH₃ CH (Br) CH₂CH₃.

Question 5.
How will you separate propene from propyne?
Answer:
Bypassing the mixture through ammoniacal AgNO₃ solution when propyne reacts while propene passes over.

Question 6.
Name two reagents that can be used to distinguish \ between ethene and ethyne.
Answer:
Tollen’s reagent | Ammoniacal AgNO₃ | and amm. CuCl solution.

Question 7.
How will you detect the presence of unsaturation in an organic compound?
Answer:
Either by Baeyer’s reagent

or by Br, in CC14.

Question 8.
Arrange the following In order of increasing volatility: gasoline, kerosene, and diesel.
Answer:
Diesel, kerosene, gasoline.

Question 9.
Arrange the following: HCl, HBr, HI, HF in order of decreasing reactivity towards alkenes.
Answer:
HF, HCl, HBr, HI.

Question 10.
Out of ethylene and acetylene which is more acidic and why?
Answer:
Acetylene. Ethylene and acetylene have sp², sp hybridized C atoms respectively. Due to the 50% S character of the C – H bond of acetylene rather than the 33% S-Character of the C – H bond in ethene, acetylene is more acidic.

Question 11.
Write the structure of the alkene which on reductive ozonolysis gives butanone and ethanal.
Answer:

Question 12.
Write the I.U.P.A.C. names of

Answer:
3-methylpent-l— en—4-yne.

(ii) CH₂ = CH – CH (CH₃) – CH = CH – CH – CH₂,
Answer:
3-methylhept-1, 4, 6—triene.

Question 13.
Draw the structures of the following:
(i) Dicyclopropyl methane
Answer:

(ii) 2-methyl-3—isopropyl heptane.
Answer:

Question 14.
What effect the branching of an alkane has on its boiling point?
Answer:
Branching decreases the boiling point.

Question 15. Which of the following polymerizes most readily?
(i) Acetylene
(ii) Ethene
(iii) Buta —1, 3—diene
Answer:
(iii) Buta—1, 3—diene polymerizes most readily, being more reactive.

Question 16.
Arrange the following in increasing order of their release of energy on combustion

Answer:
The more the number of C atoms having maximum hydrogen hydrogens, i.e., CH₃ groups, the greater is the heat of combustion. Thus the increasing order of heat of combustion is (iii) < (iv) < (i) < (ii).

Question 17.
Arrangement of the following set of compounds in order of their decreasing relative reactivity with an electrophile E+.
(i) Chloro benzene, 2, 4 —dinitrochlorobenzene p—nitrochlorobenzene.
(ii) toluene, p – CH₃ – C₆H₄ – CH₃, p – CH₃ – C₆H₄ – NO₂, p – O₂N – C₆H₄ – NO₂
Answer:
Electron-donating groups increase the reactivity towards an electrophile E+, while electron-withdrawing groups decrease the reactivity. Thus
(i) Chlorobenzene > p — nitrochlorobenzene > 2. 4—di nitrochlorobenzene.
(ii) p – CH₃ – C₆H₄ – CH₃ > toluene > p – CH₃ – C₆H₄ – NO₂ > p – O₂N – C₆H₄ – NO₂.

Question 18.
What is the order of reactivity of halogen and alkyl groups in the dehydrohalogenation of alkyl halides to give alkenes?
Answer:

• Halogens: Iodine > Bromine > Chlorine
• Alkyl group: Tert > secondary > primary.

Question 19.
What products are formed when zinc reacts with
(i) vicinal C₂H₄Br₂ and
Answer:
CH₂Br – CH₂Br + Zn → CH₂ = CH₂ + ZnBr₂.

(ii) CH3CHBr – CH2Br.
Answer:
CH₃ – CHBr – CH₂Br + Zn → CH₃ — CH = CH₂ + ZnBr₂.

Question 20.
What does L.P.G. stand for?
Answer:
L.P.G. stands for liquefied petroleum gas.

Question 21.
What do the terms (i) CNG and LPG stand for?
Answer:

• CNG: Compressed natural gas
• LPG: Liquefied Petroleum gas.

Question 22.
What are sources to obtain:
(i) LPG
Answer:
LPG is obtained by the fractional distillation of petroleum.

(ii) CNG?
Answer:
CNG is obtained by the fractional distillation of coal tar.

Question 23.
Write down the structures and names of all isomers with a molecular formula of C₅H₁₂.
Answer:

1. (i) CH₃ — CH₂ — CH₂ — CH₂ — CH₃ is n-Pentane or Pentane.
2. 
   is iso-pentane or 2-Methylbutane.
3. 
   is neo-pentane or 2, 2-Dimethylpropane.

Question 24.
The sodium salt of which acid will give ethane on heating with soda-lime? Give reaction.
Answer:
Propionic acid.

Question 25.
What products are obtained by the acidic dehydration of
(i) Ethanol
Answer:
Ethene

(ii) Propan – 2—ol.
Answer:
Propene.

Question 26.
Out of cis-2-butene and trans-2-butene which is polar and which one is non-polar?
Answer:
Cis – 2 – butene is polar (μ = 0.33 D) and trans – 2 – butene is non-polar (μ = 0). .

Question 27.
Arrange the following in the decreasing order of acidic character.
(i) C₂H₄, C₂H₆, C₂H₂
Answer:
H – C ≡ C H > H₂C = CH₂ > H₃C – CH₃

(ii) CH₃ – C = CH, C₂H₂, CH₃ – C = C – CH₃
Answer:
HC = CH > CH₃ — C ≡ CH > > CH₃ – C ≡ C – CH₃.

Question 28.
Name two industrial sources of hydrocarbons.
Answer:

1. Petroleum
2. Coal.

Question 29.
Arrange the following in the increasing order of C – C bond length C₂H₆, C₂H₄, C₂H2₂.
Answer:
C₂H₂ < C₂H₄ < C₂H₆.

Question 30.
What type of hydrocarbons is present in high octane gasoline?
Answer:
Branched-chain aliphatic and/or aromatic hydrocarbons.

Question 31.
What are the chief constituents of light oil fraction?
Answer:
Benzene, toluene, and xylenes.

Question 32.
Which of the following shows geometrical isomerism?
(i) CHCl = CHCl
(ii) CH₂ = CCl₂
(iii) CCl₂ = l. Give the structures of cis-and transforms.
Answer:
(i) HC (Cl) = CH (Cl);

Question 33.
Name the product formed when methyl bromide is treated with sodium and ether.
Answer:

Question 34.
Which of the following shows geometrical isomerism?
But-1—ene or but—2—ene.
Answer:
But-2—ene CH3 – CH = CH – CH3.

Question 35.
Write the structure and I.U.P.A.C. name of Acetonitrile.
Answer:
Acetonitrile is CH₃3 CN. Its I.U.P.A.C. name is Ethane nitrile.

Question 36.
Why does carbon has a larger tendency of catenation than silicon although they have the same number of valance electrons?
Answer:
It is due to the smaller length of the C – C bond which is stronger (335 kJ mol^-1) than the Si-Si bond (225.7 kJ mol^-1).

Question 37.
Give name atm structure to the first organic compound synthesized in the laboratory.
Answer:
Urea

Question 38.
Benzene is highly unsaturated, yet it does not give usual addition reactions readily. Why?
Answer:
Benzene is highly unsaturated, yet, resonance imparts extra-stability to benzene and it does not give additional reactions.

Question 39.
What is Lindlar’s catalyst? What is it used for?
Answer:
Pd/BaSO₄ poisoned with quinoline. It is used for the partial reduction of alkynes to cis-alkenes.

Question 40.
How will you detect the presence of unsaturation in an organic compound?
Answer:
Either by Baeyer’s reagent or by Br2 in CCl₄. The color is discharged.

Question 41.
How can ethylene be converted to ethane?
Answer:

Question 42.
Arrange the following in increasing order of C — C bond length C₂H₆, C₂H₄, C₂H₂.
Answer:C₂H₂ < C₂H₄4 < C₂H₆.

Question 43.
What type of hybridization is involved in
(i) planar and
Answer:
sp²

(ii) linear molecules?
Answer:
sp.

Question 44.
Name the chain isomer of C₅5H₁₂ which has a tertiary hydrogen atom.
Answer:
2-Methyl butane (CH₃), CHCH₂CH₃.

Question 45.
What type of isomerism is shown by butane and isobutane.
Answer:
Chain or nuclear isomerism,

Question 46.
What do you mean by cracking?
Answer:
The thermal decomposition of higher hydrocarbons into lower hydrocarbons in the presence or absence of a catalyst is called cracking.

Question 47.
Complete the reaction HC ≡ CH

Answer:

Question 48.
Out of Octane and is heptane, which has a lower octane number?
Answer:
n-Octane.

Question 49.
What are the main components of LPG?
Answer:
Butane and isobutane.

Question 50.
Write a chemical reaction to illustrate the Saytzeffs rule.
Answer:

Hydrocarbons Important Extra Questions Short Answer Type

Question 1.
The following organic compounds are known by their common names
(i) Neopentane
Answer:
Neopentane is

& its h.U.P.A.C. name is 2, 2—dimethyl propane.

(ii) Acetone
Answer:
Acetone is

and its I.U.P.A.C. name is Propanone.

(iii) Vinyl chloride
Answer:
Vinyl chloride is CPU = CH — Cl and its I.U.P.A.C. name are chloroethene.

(iv) Tert butyl alcohol. Write their structural formulae and I.U.P.A.C. names.
Answer:
Tert; butyl alcohol

and its I.U.P.A.C. name is 2-methyl propan-2-ol.

Question 2.
What are the various products expected when propane reacts with fuming nitric acid?
Answer:

Question 3.
How will you convert methane into
(i) Methanol
Answer:
Conversion of methane into methanol:

(ii) Methanal.
Answer:

Question 4.
What is aromatization? How will you convert ^hexane into benzene?
Answer:
Aromatization. It is the process that involves cyclization, isomerization, and dehydrogenation with the application of heat and catalyst to convert alkanes containing six or more carbon atoms into aromatic hydrocarbons.

Question 5.
Give the different conformations of ethane with their
(i) Sawhorse representation and
(ii) Newmann Projection formulae.
Answer:
Sawhorse representation Newmann projection models

Question 6.
What are the relative stabilities of different conforma¬tions of ethane? Is it possible to isolate these at room temperature?
Answer:
The staggering form of ethane is more stable than the eclipsed form because the force of repulsion between hydrogen atoms on adjacent C atoms is minimum. The energy difference between the staggered form and eclipsed form of ethane is just 12.55 kJ mol^-1. Therefore, it is not possible to separate these two forms of ethane at room temperature.

Question 7.
What is Saytzeff Rule? What are the expected products when 2-Bromobutane is dehydrohalogenation with ale. KOH?
Answer:
Saytzeff Rule. Whatever two alkenes are theoretically possible during a dehydrohalogenation reaction, it is always the more highly substituted alkene that predominates.

Question 8.
What is the order of reactivity of H₂C = CH₂, (CH₃)₂, H₂C = CH₂, CH₃ – CH = CH₂, CH₃ – CH = CH – CH₃, (CH₃)₂ C = C (CH₃)₂, (CH₃)₂ C = CH CH₃ towards electrophilic addition reactions?
Answer:
The order of reactivity of the above alkenes towards electrophilic addition reactions decreases in the order.
(CH₃)₂ C = C (CH₃)₂ > (CH₃3)₂ C = CH CH₃ > (CH₃)₂ C = CH₂ > CH₃ CH – CH – CH₃ > CH₃ – CH = CH, > CH₂ = CH₂.

Question 9.
Define Markownikov rule. Explain it with an example.
Answer:
Markownikov rule states. The negative part of the addendum adding to an unsymmetric alkene goes to that C atom of the double bond which is attached to a lesser number of C atoms.

Question 10.
What is the Peroxide effect/Kharasch Effect? Illustrate with an example.
Answer:
In the presence of peroxides such as benzoyl peroxide, the addition of HBr (but not of HCl or HI) to an unsymmetrical alkene takes place contrary to the Markownikov rule. This is known as the peroxide/Kharsch effect.

Question 11.
An alkene with the molecular formula C₇H₁₄ gives propanone and butanal on ozonolysis. Write down its structural formula and its I.U.P.A.C. name.
Answer:
The structures of the compounds on ozonolysis of C₇H₁₄ and

Remove the oxygen atoms and connect them by a double bond, the structure of the alkene is

Question 12.
How will you prepare propyne and I-Butyne from acetylene;?
Answer:
(i) Preparation of propyne from acetylene

(ii) Preparation of 1-butyne from acetylene

Question 13.
Describe a method to distinguish between ethane, ethene, ethyne.
Answer:
(i) Ethene (C₂H₄) and ethyne (C₂H₂) decolorize bromine in carbon tetrachloride whereas ethane (C₂H₆) does not.

(ii) Ethyne (and not ethane, ethene) reacts with ammoniacal AgNO₃ (Tollen’s reagent) to form white precipitates.

Question 14.
Give the mechanism of an electrophilic addition of chlorine into propene.
Answer:

Cl₄ (chloronium ion) formed by the heterolytic fission of Cl₂ in step (i) being an electrophile attacks propene in (ii) step, propene undergoes electrometric effect combined with + I effect to form carbocation which, is a slow step. In (iii) step Cl ion being a. nucleophile attacks carbocation and forms the product. It is a fast step.

Question 15.
Identify the correct order of reactivity in electrophilic substitution reactions of the following compounds: [I.I.T. 2002]

Answer:
— NO₂ group in structure IV is an electron attracting group, it deactivates the benzene ring largely towards electrophilic substitution reactions. Cl group in III is also a deactivating group, but its deactivation is lower too — NO₂, where—CH₃, a group in II is an electron-releasing group and so activates the benzene ring towards electrophilic substitution. Therefore, the order of reactivity is

Question 16.
What is meant by (i) delocalization
Answer:
Delocalisation: Delocalisation means that pairs of 7t electrons extend over 3 or more atoms. They belong to the whole molecule. For example, 6n electrons present in benzene are delocalized and are spread on the whole of the ring and this imparts extra stability to the molecule.

(ii) resonance energy.
Answer:
Resonance energy: The difference between the energy of the most stable contributing/canonical structural and the energy of the resonance hybrid is known as resonance energy. In the case of benzene, the resonance hybrid has 147 kJ mol^-1 than either A or B below. Thus resonance energy of benzene is 147 kJ mol^-1.

Question 17.
Describe Friedel’s craft reaction with suitable examples.
Answer:
When an alkyl or acid halide is treated with benzene or its derivative in the presence of anhydrous AlCl₃ as a catalyst, we got alkyl or acyl benzene.

Question 18.
Classify the following hydrocarbons into alkanes, alkenes, alkynes, and arenes.
(i) (CH₃)₄C
Answer:
Alkane

(ii) C₂H₂
Answer:
Alkyne

(iii) C₃H₆
Answer:
Alkene

Answer:
Arene.

Question 19.
Write all possible structures of C₅H₈ and give their I.U.P.A.C. names.
Answer:
Structural isomers of C₅H₈ (Pentyne) are
(i) CH₃ – CH₂ – CH₂ – C ≡ CH (Pent-l-yne)
(ii) CH₃ – CH₂ – C ≡ C – CH₃ (Pent-2-yne)
(iii)
(3-Methylbut-l-yne)

Question 20.
How does ethylene undergo polymerization? What is the use to which the polymer obtained is put?
Answer:

Low-density polythene [LDPE] and high-density polyethylene [HDPE] as used for the manufacture of plastic bags, squeeze bottles, refrigerator dishes, toys, pipes, radio, and T.V. cabinets, etc.

Question 21.
What is the action of water on
(i) Ethyne
(ii) Propyne? Name the end products obtained.
Answer:
(i) Action of water on ethyne: When ethyne is warmed with dilute H₂SO₄ at 333K and H₈SO₄ as catalyst ethanol is obtained.

(ii) Action of water on Propyne CH₃ — C ≡ CH + H₂O

Question 22.
Propene reacts with HBr to give Isopropyl bromide but does not give n-propyl bromide. Why?
Answer:
The addition of unsymmetrical addendum (HBr) to unsymmetrical olefines (CH₃CH = CH₂) takes place according to Markownikov rule, the negative part of reagent (i.e. Br-) adds on the carbon atom having a minimum number of hydrogen atoms. Hence Isopropyl bromide will be formed.

Question 23.
An oxidizing agent is needed in the iodination of methane but not in the chlorination or bromination. Give reason.
Answer:
In the iodination of methane, H — I is also formed as the product with iodomethane, since it is a strong reducing agent, it reduces iodomethane back to methane & makes the reaction reversible. In order to destroy HI, an oxidizing agent like HIO₃ (or HNO₃) is needed. But HCl& HBr formed in the chlorination & bromination reactions of methane & not in a position to react with the monosubstituted products (CH₃Cl & CH₃Br) since they are comparatively weak reducing agents.

Therefore, no oxidizing agent is needed for these reactions.
CH₄ + I₂ ⇌ CH₃I + HI
5HI + HIO₃ → 3H₂O + 3I₂

Question 24.
The dipole moment of trans 1, 2-chloroethene is less than the cis isomer. Explain.
Answer:
The structure of the trans isomer is more symmetrical to the cis isomer. In the trans isomer, the dipole moments of the polar C—Cl bonds are likely to cancel with each other & the resultant dipole moment of the molecule is nearly zero. But in the cis isomer, these do not cancel. Therefore, the cis isomer has a specific moment but is zero in the case of the trans isomer.

Question 25.
Write l.U.P.A.C. names of the products obtained by addition reactions of HBr to hex-l-ene.
(a) In the absence of Peroxide
Answer:

(b) In the presence of Peroxide.
Answer:

Question 26.
How will you distinguish between the following:
(a) Butyne-l & Butyne-2 ,
Answer:
Butyne-1 having an acetylene hydrogen atom will give white ppt. With ammoniacal silver nitrate & red ppt. with ammonical cuprous chloride. On the other hand, butyne-2-having no acetylenic hydrogen atom does not respond to either of the two reagents.

(b) Butene-1 & Butene-2
Answer:
Butene-1 & butene-2 can be distinguished either by ozonolysis or by oxidation with acidic KMnO₂ solution with which they give different carbonyl compounds.

Question 27.
Alkynes are less reactive than alkenes towards addition reaction even though they contain 2-7t bond. Give reason.
Answer:
This is due to

1. greater electronegativity of sp-hybridized carbon of alkynes than sp² hybridized carbon atoms of alkenes which holds the π-electrons of alkynes more tightly and
2. greater delocalization of π-electrons in alkynes (because of the cylindrical nature of their n electron cloud) than in alkenes. As a result, n electrons of alkynes are less easily available for addition reactions than those of alkenes.

Consequently, alkynes are less reactive than alkenes towards addition reactions.

Question 28.
Why do addition reactions occur more readily with alkenes & alkynes than with aromatic hydrocarbons?
Answer:
The energy gained by forming two sigma bonds (of four sigma bonds) more than compensates for the loss of one or two n bonds when addition occurs to an alkene or alkyne. However, in aromatic hydrocarbons, the aromatic ring is specially stabilized by the delocalization of n electrons about the ring.

It, therefore, requires substantial activation energy to cause the loss of its aromatic character. The most usual reaction in arenes is thus substitution rather than addition, since substitution does not result in loss of aromatic character.

Question 29.
A Hydrocarbon A, adds one mole of hydrogen in presence of platinum catalyst from n-Hexane. When A is oxidized vigorously with KMnO₄, a single carboxylic acid, containing three carbon atoms is isolated. Give the structure of A & explain.
Answer:

1. Since hydrocarbon A adds one molecule of H₂ in presence of platinum to form n-hexane. A must be a hexene.
2. Since A on vigorous oxidation with KMnO₄ gives a single carboxylic acid containing three carbon atoms, therefore, A must be asymmetrical hexene i.e. hex-3-ene.
   
   Thus, the given hydrocarbon A is hex-3-ene.

Question 30.
How would you carry out the following conversion?
Answer:
Propene to Ethyne

Hydrocarbons Important Extra Questions Long Answer Type

Question 1.
How would you convert the following compounds to benzene?
(i) Acetylene
Answer:
Acetylene into benzene. Ethyne (Acetylene) in passing through a red hot iron tube at 873 K undergoes cyclic polymerization as shown below.

(ii) Benzoic acid
Answer:
Benzoic acid into benzene

(iii) Hexane
Answer:
Hexane into benzene

(iv) Benzene diazonium chloride
Answer:
Benzene diazonium chloride into benzene

Question 2.
How will you convert benzene into
(i) Nitrobenzene
Answer:
Benzene into Nitrobenzene

(ii) Benzene sulphonic acid
Answer:
Benzene into benzene sulphonic acid

(iii) Toluene
Answer:
Benzene into Toluene

(iv) Acetophenone?
Answer:
Benzene into Acetophenone

Question 3.
What is the mechanism of nitration of benzene?
Answer:
Nitration of benzene. It is carried out by treating benzene with a mixture of cones. HNO₃⁺ Cone. H₂SO₄. The various steps involved are:
Step I: Generation of an electrophile, i.e., NOt (nitronium ion)

Step II: Formation of complex or carbocation intermediate

This step is slow and hence is the rate-determining step of the reaction.

Step III: Loss of a proton from the carbocation intermediate

This step is fast and does not affect the rate of the reaction.

Question 4.
(a) What is the general formula of Alkynes?
Answer:
The general formula of alkynes ¡s C_nH_2n-2.

(b) Give the I.U.P.A.C. names and structure of all alkynes having the molecular formula C₂H₈.
Answer:
C5H8 has the following isomers.
(i) CH₃ — CH₂ — CH₂ — C ≡ CH Pent-1-yne
(ii) CH₃ CH, C ≡ C — CH₃ Pent-2-yen
(iii)
3-Methyl but-1-yen

(c) Give any two methods for preparing acetylene.
Answer:
Acetylene can be prepared by the following two methods

(ii) By dehydrohalogenation of dihaloalkanes

(d) Discuss any three chemical properties of acetylene.
Answer:
(i) Addition of Hydrogen:

(ii) On oxidation with alkaline KMnO₄, it gives oxalic acid.

(iii) When acetylene is passed through a red hot iron tube, it trickeries to give benzene

Question 5.
(a) Write structures of different chain isomers of alkanes corresponding to the molecular formula C₆H₁₄. Also, write their I.U.P.A.C. names.
(b) Write structures of different isomeric alkyl groups corresponding to the molecular formula C₅H₁₁. Write IUPAC names of alcohols obtained by attachment of —OH groups at different carbons of the chain.
Answer:

(b) Structures of – C₅H₁₁ group Corresponding alcohols

Question 6.
How will you prepare Alkanes by
(i) Wurtz reaction
Answer:
Methods of preparation of Alkanes
Wurtz reaction. Alkyl halides on treatment with sodium in dry ether give higher alkanes, preferably containing an even number of carbon atoms.

(ii) Decarboxylation of sodium salts of fatty acids
Answer:
Sodium salts of fatty acids on heating with-soda lime (a mixture of NaOH and CaO) give alkanes containing one carbon atom less than the carboxylic acid. The process of elimination of carbon dioxide from a carboxylic acid is known as Decarboxylation

(iii) Kolbe’s electrolytic method.
Answer:
Kolbe’s electrolytic method. An aqueous solution of sodium or potassium salt of a carboxylic acid on electrolysis gives alkanes containing an even number of carbon atoms at the anode.

The probable mechanism for the reaction is

(b) How do alkanes udergo:
(i) substitution reactions with halogens
Answer:
Substitution reactions with Halogens. The order of reactivity of halogens is F₂ > Cl₂ > Br₂ > I₂

Bromination is similar. With fluorine, the reaction is too violent to be controlled. Iodination is very slow and a reversible reaction.
It can proceed in the presence of oxidizing agents like HNO₃, HIO₃.
CH₄ + I₂ ⇌ CH₃I + HI
HIO₃ + 5HI → I₃ + 3H₂O

Halogenation of alkanes proceeds via a free-radical mechanism which consists of three steps.

1. Chain initiation step
2. Chain propagation step
3. Chain termination step

1. Chain initiating step. Cl₂ undergoes hemolysis in the presence of heat and light.

2. Chain propagation step

3. Chain termination step

(ii) oxidation by combustion
Answer:
Oxidation of alkanes by combustion: Alkanes on heating in the presence of air or dioxygen are completely oxidized to carbon dioxide and water with the evolution of a large amounts of heat.

In the presence of an insufficient amount of air or dioxygen, carbon black is formed.

(iii) Isomerisation
Answer:
Isomerization reactions of alkanes, n-Alkanes on heating in the presence of anhydrous aluminum chloride and hydrogen chloride gas isomerize to branched-chain alkanes.

(iv) Aromatisation.
Answer:
Aromatisation. n-Alkanes having six or more C atoms on heating to 773 K at 10-20 atmospheric pressure in. the presence of oxides of vanadium, molybdenum get dehydrogenated and cyclized. This reaction is called aromatization or reforming.

Here we are providing Class 11 chemistry Important Extra Questions and Answers Chapter 12 Organic Chemistry: Some Basic Principles and Techniques. Chemistry Class 11 Important Questions are the best resource for students which helps in Class 11 board exams.

Class 11 Chemistry Chapter 12 Important Extra Questions Organic Chemistry: Some Basic Principles and Techniques

Organic Chemistry: Some Basic Principles and Techniques Important Extra Questions Very Short Answer Type

Question 1.
What type, of hybridisation, is involved in
(i) planar and
Answer:
sp²

(ii) linear molecules?
Answer:
sp.

Question 2.
Arrange the following in increasing order of C – C bond strength:
C₂H₆, C₂H₄C₂H₂.
Answer:
C₂H₆ < C₂H₄ < C₂H₂.

Question 3.
Arrange the following in decreasing order of C — C bond length:
Answer:
C₂H₆ > C₂H₄ > C₂H₂.

Question 4.
What is the type of hybridisation of C atoms in benzene?
Answer:
It is an sp² type of hybridisation.

Question 5.
What are isomers?
Answer:
Compounds having the same molecular formula but different physical and chemical properties are called isomers.

Question 6.
Select electrophiles out of the following:
H⁺ Na⁺, Cl^–, C₂HSOH, AlCl₃, SO₃, CN^–, CH₃CH₂⁺,: CCl₂, R-X.
Answer:
H⁺, Na⁺, A1Cl3, SO3, CH3CH₂⁺,: CCl₂, R-X.

Question 7.
Select nucleophiles from the following.
BF₃ NH₃ OH^–, R-X, C₂H₅OH, H₃O⁺, NO₂, CN^–.
Answer:
NH₃, OH, C₂H₅OH, CN

Question 8.
Give the I.U.P.A.C. names of the following compounds

Answer:
2-Bromo-4 – methyl pent-3- one

Answer:
4-Methyl-2-nitro pent – 3 – one

Answer:
2 – Ethoxy – 4 – methoxypent – 3 – one

Answer:
2-Bromo-4-nitro pent-3-one

(v) (CH₃)₄C
Answer:
2, 2-Dimethylpropane

(vi) (CH₃)₂CHCOOH.
Answer:
2-Methyl propanoic acid.

Question 9.
What is a functional group?
Answer:
The atom or group of atoms present in a molecule that determines its chemical properties is called the functional group.

Question 10.
Arrange the following in increasing order of-I effect.
(i) -NO₂, -COOH, -F, -CN, – I.
Answer:
-I < -F < -COOH < -CN <NO₂.

Question 11.
Arrange the following in decreasing order of + I effect: CH₃-, D, (CH₃)₃C-, (CH₃)₂CH-, CH₃-CH2₂–
Answer:
(CH₃)₃C → (CH₃)₂CH → CH₃-CH₂ → CH₃ → D

Question 12.
Name the alkyl groups derived from isobutane.
Answer:
(CH₃)₂CH – CH₂– isobutyl and (CH₃)₃C – tertiary butyl.

Question 13.
What type of isomerism is shown by butane and isobutane.
Answer:
Chain or nuclear isomerism.

Question 14.
Write the tautomer of acetaldehyde and its I.U.P.A.C. name.
Answer:
CH₂ = CH-OH and its I.U.P.A.C. name is Eth-1-en-1-ol,

Question 15.
Give one example of functional isomerism.
Answer:
CH₃– CH₂– OH and CH₃ – O – CH₃.

Question 16.
Give one example of position isomerism.
Answer:
CH₃ – CH₂ – CH₂OH and CH₃CH(OH) CH₃ .

Question 17.
Draw the structure of the tautomer of phenol and write its I.U.P.A.C name.
Answer:

Question 18.
A compound is formed by the substitution of two chlorine atoms for two hydrogen atoms in propane. What is the number of structural isomers possible?
Answer:
Four
1. CH₃CH₂CHCl₂
2. CH₃CHClCH₂Cl
3. CH₃CCl₂CH₃
4. ClCH₂ – CH₂ – CH₂Cl

Question 19.
Write the metamer of diethyl ether. What is its I.U.P.A.C. name?
Answer:

1. CH₃O CH₃ CH₂ CH₃ Its I.U.P.A.C. name is 1 -Methoxypropane.
2. 

Question 20.
Give the I.U.P.A.C. name of CH₂ = CH-CH (CH₃)₂
Answer:

Question 21.
How many cr and it bonds are present in each of the following molecules?
(a) H-C = CCH=CH-CH₃
Answer:
No. of σ_C-C = 4;
No. of σ_C-H = 6.

Total no. of σ bonds =10.
Total no. of π_C=C bonds = 2 + 1=3.

(b) CH₂=C=CH-CH₃.
Answer:
(b) No. of σ_C-C = 3 bonds = 3,
No. of σ_C-H = 6;

Total No. of , σ-bonds = 3 + 6 = 9
No. of π_C=C bonds = 2.

Question 22.
What is the shape of the following molecules
(a) H₃ C = O
Answer:
HCHO formaldehyde C: sp² hybridised; shape: Trigonal planar

(b) CH₃F
Answer:
CH₃F; C = sp³ hybridised shape: Tetrahedral

(c) HON.
Answer:
H-C ≡ N; C is sp hybridised, HCN is linear.

Question 23.
Write the T.U.P.A.C. name of

Answer:
Compound is

Question 24.
Give the condensed and bond-line structural formula for
(a) Penta-1, 4-dien
Answer:
CH2 = CH – CH2 – CH = CH2

(b) Hexa-1, 3, 5 triene.
Answer:
CH₂ = CH – CH = CH – CH = CH
Its bond line structure is

Question 25.
Write the I.U.P.A.C. names of HOOC-C ≡ C-COOH and

Answer:
But-2-yne – 1,4-dioic acid and 3-Methyl pentanenitrile.

Question 26.
How will you purify essential oils?
Answer:
Essential oils are volatile and are insoluble in water, Therefore, they are purified by steam distillation.

Question 27.
A liquid (1.0 g) has three components. Which technique will you employ to separate them?
Answer:
Column chromatography.

Question 28.
A reaction carried out using aniline as a reactant as well as a solvent. How will you remove unreacted aniline?
Answer:
By steam distillation.

Question 29.
How will you separate a mixture of O-nitrophenol and /Mutrophenol?
Answer:
Steam distillation O-Nitrophenol being volatile distils over along with water while p-nitrophenol being non-volatile remains in the flask.

Question 30.
How will you purify a liquid having non-volatile impurities?
Answer:
Simple distillation will give pure liquid while the non-volatile impurities in the flask as residue.

Question 31.
Suggest a method to purify
(i) Kerosene containing water
Answer:
By solvent extraction using a separating funnel.

(ii) a liquid that decomposes at its boiling point.
Answer:
Distillation under reduced pressure.

Question 32.
Suggest methods for the separation of the following mixtures:
(i) A mixture of liquid A,(B. Pt. =365 K) and liquid B (b.p.356 K)
Answer:
Fractional distillation B.Pts of two liquids differ only by just 9°.

(ii) A mixture of liquid C (b.p. 353 K) and liquid D (b.p. 413 K).
Answer:
Simple distillation, since B.Pts of two liquids, are wide apart.

Question 33.
Name two compounds that do not contain halogen but give a positive Beilstein test.
Answer:
Urea and thiourea give a positive Beilstein test due to the formation of volatile cupric cyanide.

Question 34.
Lassaigne’s test is not shown by diazonium salts. Why?
Answer:
Diazonium salts lose N2 on heating much before they have a chance to react with fused sodium metal.

Question 35.
What type of compounds are purified by sublimation?
Answer:
Substances whose vapour pressures become equal to atmospheric pressure much below their melting point.

Question 36.
How will you separate iodine from sodium chloride?
Answer:
By sublimation.

Question 37.
Name two methods that can be safely used to purify aniline.
Answer:
Vacuum distillation and steam distillation.

Question 38.
Define the term elution as applied to column chromatography.
Answer:
The process of extraction of different adsorbed compounds from the column by means of a suitable solvent is called elution.

Question 39.
Suggest a suitable technique of separating naphthalene from kerosene oil present in a mixture.
Answer:
Simple distillation.

Question 40.
What conclusion would you draw if during Lassaigne’s test a blood-red colouration is obtained?
Answer:
It shows the presence of N & S together in the compound.

Question 41.
What type of organic compounds cannot be Kjeldahlised?
Answer:
A compound containing an N atom in the ring or the presence of – NO₂, (nitro) and — N =N -(azo) in them.

Question 42.
Can we estimate oxygen in the organic compound?
Answer:
No. it is estimated indirectly by subtracting the percentage of all elements present in an organic compound from 100.

Question 43.
Why do we use copper spiral in Duma’s method?
Answer:
To reduce back any oxides of nitrogen formed during combustion to nitrogen.

Question 44.
Give two examples of adsorbents used in chromatography.
Answer:

1. Alumina gel (Al₂O₃) and
2. Silica gel SiO₂.

Question 45.
Is Lassaigne’s extract neutral, acidic or alkaline?
Answer:
It is alkaline.

Question 46.
The empirical formula of a compound is CH₂. It’s one mole has a mass of 42 g. What is its molecular formula?
Answer:
Molecular formula = n × empirical formula
= \(\frac{42}{14}\) × CH₂ = C₃H₆

Question 47.
In which C-C bond of CH₃CH₂CH₂Br, the inductive effect is expected to be the least?
Answer:
The magnitude of the inductive effect diminishes as the number of intervening bonds increases. The effect is least in the C₃-H bond.

Question 48.
Can you use K in place of Na for fusing an organic compound in Lassaigne’s test?
Answer:
No, because K is more reactive than Na.

Question 49.
Which solution is used to absorb CO₂ produced during combustion?
Answer:
KOH solution is used to absorb CO₂ gas.

Question 50.
What is the cause of geometrical isomerism in alkenes?
Answer:
Alkenes have a π-bond & the restricted rotation around the π-bond gives rise to geometrical isomerism.

Organic Chemistry: Some Basic Principles and Techniques Important Extra Questions Short Answer Type

Question 1.
Expand each of the following bond-line formulae to show all the atoms including carbon and hydrogen.

Answer:

Answer:

Answer:

Answer:

Question 2.
For each of the following compounds, write a more condensed and also their bond line formulae.

Answer:
Condensed formulae are
(CH3)2CH CH2 OH

Answer:
CH₃(CH₂)₅ CHBr CH₂ CHO

Answer:
HO(CH₂)₃ CH(CH₃) CH(CH₃)₂

Answer:
HOCH(CN)₂

Question 3.
What is the type of hybridisation of each carbon in the following compounds?
(a) CH₃Cl
Answer:

(b) (CH₃)₂CO
Answer:

(c) CH₃CN
Answer:

(d) HCONH₂
Answer:

(e) CH₃CH = CHCN.
Answer:

Question 4.
What is the shape of the following molecules:
(a) H₂C = O
Answer:
In H₂C = O; C is sp² hybridised, hence its shape is H trigonal planar

(b) CH₃F
Answer:
In CH₃ -F; C is sp³ hybridized
∴ it is tetrahedral

(c) H-C ≡ N?
Answer:
In H-C ≡ N; C is sp-hybridized, hence HCN is linear
H—C ≡ N.

Question 5.
Give the I.U.P. A.C. names of the following compounds:

Answer:
2-Ethylprop-2-en-l-ol

(ii) CH₃ – CH = CH COOH
Answer:
But-2-en-l-oic acid

(iii) (CH₃)₂C = CHCOCH₃
Answer:
4-Methylpent-3-en-2-one

Answer:
3-Chloropropanal

Answer:
3-Methylbutane-l-al

(vi) CH2 = CH – CN.
Answer:
Prop-2-en-1-nitrile.

Question 6.
Write the I.U.P.A.C. names of

Answer:
3-Ethyl-4-methylhept-5-en-2-one

Answer:
2-Ethyl-3-methylpent-2-en-1 -one.

Question 7.
Write 1.U.P.A.C. names of

Answer:
2-[2-methylclo but-l-enyl] ethanal

Answer:
2-(3-Oxobutyl) cyclohexane-1 one

Answer:
Methyl (2-oxo cyclopentane-1-carboxylate

Answer:
Cyclohex-2-en-l-ol

Answer:
2-Ethenyl-3-methyl-cyclohexa-l, 3-diene

Answer:
4-Formyl-2-oxo-cyclohexane-l carboxylic acid.

Question 8.
Draw the structures of
(i) Methyl t-butyl ether
Answer:

(ii) 2-Chloro-1, 1, 1-trifluoro ethane
Answer:
F3C – CH2Cl

(iii) 2-Methyl buta, 1, 3-diene
Answer:

(iv) Pent-2-en-l-ol
Answer:
CH₃ – CH₂ – CH = CH – CH₂OH

(v) Cyclo hex-2-en-l-ol
Answer:

(vi) l-Bromo-3-chloro cyclohex-1-ene
Answer:

Question 9.
Write I.U.P.A.C. names of

Answer:
(2-Isopropyl) benzene

Answer:
1-Phenylpropane-l-one

Answer:
1 -Phenylethan-1 -ol

Answer:
1, 3-Benzene dicarboxylic acid

Answer:
3-Phenylpropanal

Answer:
3-Boromo-4-hydroxybenzoic acid

Answer:
4-Hydroxy-3-methoxy benzaldehyde.

Question 10.
Write the structure of
(i) O-Ethyl anisole
Answer:

(ii) p— nitroaniline
Answer:

(iii) 4-Ethyl-I-fluoro-2-nitrobenzene.
Answer:

Question 11.
Which is more polar bond in the following pairs of molecules
(a) H₃C-H, H₃C-Br
Answer:
C —Br since Br is more electronegative than H.

(b) H₃C-NH₂, H₃C-OH
Answer:
C—O since O is more electronegative than N.

(c) H₃C-OH, H₃C-SH.
Answer:
C — O since O is more electronegative than S.

Question 12.
In which C-C bond of CH₃CH₂CH₂Br, the inductive effect is expected to be the least.
Answer:
The magnitude of the inductive effect decreases with distance from the active centre and hence this effect is least in C₂—C₃ bond.

Question 13.
Write the resonance structures of
(a) CH₃COO- and
Answer:

(b) CH₆H₅NH₂. Show the movement of electrons by curved arrows.
Answer:

Question 14.
Which of the following pairs of structures do not constitute resonance structures?

Answer:
They differ in the position of atoms and hence are not resonance structures.

Answer:
They are a pair of resonance structures as they differ in the position of electrons.

Answer:
They differ in the position of atoms and so are not resonance structures. They are tautomers.

(d) CH₃CH = CHCH₃ and CH₃CH₂CH=CH₂
Answer:
They are not resonance structures as they differ in the position of atoms.

Question 15.
Write the resonance structures of CH₂ = CH-CHO and arrange them in decreasing order of stability.
Answer:

The structure I is most stable as each C & O have octets completed and there is no charge on either of them.
II & III involve charge separation hence both are less stable than I. However II is more stable than III because in II more electronegative oxygen carries a negative charge.

Thus decreasing order of stability is I > II > III.

Question 16.
Using curved arrow notation show the formation, of reactive intermediates when the following covalent bonds undergo heterolytic fission.
(a) CH₃-SH₃
Answer:

(b) CH₃-CN
Answer:

(c) CH₃-CU
Answer:

Question 17.
Giving proper justification categorise the following molecules/ ions as nucleophiles or electrophiles:

Answer:

are all nucleophiles as each one of them has one or more lone pairs of electrons to donate.

are all electrophiles. All these species have a sextet of electrons around positive centres.

Question 18.
A mixture contains two components A and B. The solubilities of A and B in the water near its boiling point are 10 grams per 100 ml and 2g per 100 ml respectively. How will you separate A and B from this mixture?
Answer:
Fractional crystallisation. When the saturated hot solution of this mixture is allowed to cool, the less soluble component B crystallises out first leaving the more soluble component B in the mother liquor.

Question 19.
A mixture containing benzoic acid and nitrobenzene is given to you. Using an appropriate chemical reagent, how will you proceed to separate them?
Answer:
The mixture is shaken with a dilute solution of NaHCO₃ and extracted with-ether or chloroform when nitrobenzene goes into the organic layer; Distillation of this will yield nitrobenzene. The aqueous layer is acidified with dil. HCl and the solution are cooled.

Filteration gives benzoic acid.
C₆H₅COOH + NaHCO₃ → C₆H₅COONa + CO₂ + H₂O.
C₆H₅COONa + HCl (dil.) → C₆H₅COOH + NaCl.

Question 20.
The Rf value of A and B in a mixture determined by TLC in a solvent mixture are 0.65 and 0.42 respectively. If the mixture separated by column chromatography using the same solvent mixture as a mobile phase, which of the two components A or B will elute first? Explain.
Answer:
Since the Rf value of A is 0.65, therefore, it is less strongly adsorbed as compared to component B with an Rf value of 0.42. Therefore an extraction in column chromatography, A will elute first.

Question 21.
Without using column chromatography, how will you separate a mixture of camphor and benzoic acid?
Answer:
Sublimation cannot be used as both camphor and benzoic acid sublime on heating. Therefore, a chemical method using NaHCO₃ solution is used when benzoic acid dissolves leaving camphor behind. The filtrate is then cooled with dilute HCl to get benzoic acid.

Question 22.
0.12g of an organic compound containing phosphorus gave 0.22g of Mg2P207 by the usual analysis. Calculate the percentage of phosphorus in the compound.
Answer:
Mass the substance taken (w) = 0.12g
Wt. of Mg₂P₂O₇ formed (x) = 0.22 g
222 g of Mg₂P₂O₇contain phosphorus = 62 g.

% of phosphorus = \(\frac{62}{222} \times \frac{0.22}{0.12}\) × 100 = 51.20

Question 23.
Compare inductive & mesomeric effects.
Answer:

Question 24.
What is the difference between distillation, distillation under reduced pressure & steam distillation?
Answer:

Question 25.
How will you purify sugar which has impurities of sodium chloride?
Answer:
Sugar may be purified by the crystallization method. This can be purified by shaking the impure solid with hot ethanol at 345K. The sugar will dissolve whereas common salt remains insoluble. The hot solution is filtered, concentrated & allowed to cool when crystals of sugar will separate out. In this case, hot water has been used as a solvent. The purification of sugar would not have been possible since both sugar’& common salt are soluble in water.

Question 26.
Differentiate between Ionic & free radical reactions.
Answer:

Question 27.
For each of the following compounds, write a more condensed formula & also their bond-line formula.

(b) HOCH2CH2CH2CHCH3CHCH3CH3

Answer:

Question 28.
Expand each of the following bond line formulae to show all the atoms including carbon & hydrogen.

Answer:

Answer:

Answer:

Question 29.
Explain why is (CH₃) C+ more stable than CH₃ C⁺ H₂ & C⁺ H₃ is the least stable cation?
Answer:
Hyperconjugation interaction in (CH₃)₃ C⁺ is greater than in C+⁺ H₃ C⁺ H₂ has 9 C -H bonds. In C H₃, the C -H bonds are in the nodal plane of the vacant 2p-orbital & hence cannot overlap with it.
Thus C⁺ H₃ is least stable.

Question 30.
The choice of the solvent is of great importance in crystallizing organic substances. What are the characteristics of a suitable solvent?
Answer:
A suitable solvent must have the following characteristics;

1. The impurities & pure compound must have a large difference in their solubilities.
2. The pure compound must have low solubility at room temperature but high solubility at its boiling point.
3. The impurity should either be insoluble at room temperature or must have high solubility so that crystallization may give a high yield.
4. The solvent should have an average boiling point.
5. The solvent should neither react with the compound nor with impurities.
6. The solvent should not be highly inflammable.

Organic Chemistry: Some Basic Principles and Techniques Important Extra Questions Long Answer Type

Question 1.
Explain the principle of steam distillation.
Answer:
Steam distillation: The process of steam distillation is employed in the purification of substance from non-volatile impurities provided the substance itself is volatile in steam and insoluble in water.

This method is based on the facts that

1. A liquid boils at a temperature when its vapour pressure becomes equal to the atmospheric pressure.
2. The vapour pressure of a mixture of two immiscible liquids is equal to the sum of the vapour pressures of the individual liquids.

In the actual process, steam is continuously passed through the impure organic liquid. Steam heats the liquid and it gets practically condensed to water. After some time mixture of the liquid and water begins to boil, because the vapour pressure of the mixture becomes equal to the atmospheric pressure.

Obviously, this happens at a temperature that is lower than the boiling point of the substance or that of water. Thus an organic compound boils below its boiling points and chances of decomposition avoided. For example, a mixture of aniline (b.p 453 K) with decomposition and water (b.p. 373 K) under normal atmospheric pressure boils at 371K. At this temperature the

Steam Distillation

water boils at 371 K. At this temperature, the vapour pressure of water is 717 mm and that of aniline is 43 mm and therefore the total pressure is equal, to 760 mm. Thus in steam distillation, the liquid gets distilled at a temperature lower than its boiling point and chances of decomposition avoided. The proportion of water and liquid in the mixture that distils over is given by the relation.

\(\frac{w_{1}}{w_{2}}=\frac{P_{1} \times 18}{P_{2} \times M}\)
where w₁ and w₂ stand for the masses of water and liquid that distils over. P₁ and P₂ are vapour pressure of water and of liquid at the distillation temperature and M is the molecular mass of the liquid.

Question 2.
Dehydrobromination of compounds (A) and (B) yield the same alkene (c) Alkene (c) Can regenerate (A) and (B) by the addition of HBr in the presence and absence of peroxide respectively. Hydrolysis of A and B give isomeric products (D) and (E) respectively. 1, 1-Diphenyl ethane is obtained on the reaction of (C) of benzene in the presence of H⁺ ions. Give structures of A to E with reactions.
Answer:
Alkene (C) on reaction with benzene in the presence of H+ ions gives 1, 1-Diphenyl ethane. Therefore C must be styrene as depicted below

Now dehydrobromination of A and B give the same alkene C, i.e.,
styrene.
∴ A and B must be isomeric alkyl bromide.

A and B can be obtained by the addition of HBr in the presence and absence of peroxide to styrene.

Hydrolysis of A and B give isomeric alcohols (D) & (E) as

Question 3.
What are reaction intermediates? How are they generated by bond fission?
Answer:
The species which are generated as a result of bond fission are called reaction intermediates. The important reaction intermediates are:
1. Free Radicals: A free radical may be defined as an atom or group of atoms having an impaired electron. These are obtained as a result of homolytic fission of covalent bonds.

These free radicals are neutral particles, extremely transient, (short-lived) and highly reactive. They get consumed as soon as they are formed. They pair up their electron with another electron from wherever it is available. They occur only as a reaction intermediate. Their presence is felt in reactions, but cannot be isolated in a free state. For example dissociation of Cl2 gas in the presence of Ultraviolet light produces free radicals.

The alkyl free radicals are obtained when free radical: Cl reacts with alkanes.

Free radical may be primary, secondary, tertiary depending upon whether, one, two or three carbon atom attached to the carbon atoms carrying the odd electron.

The stability is CH₃ < 1° < 2° < 3°.

2. Carbocation or carbonium ion: It is defined as a group of atoms that contain positively charged carbon having only six electrons. It is obtained by heterolytic fission of a covalent bond involving a carbon atom.

They are also classified as primary, secondary and tertiary depending upon whether one, two or three carbon atoms are attached to the carbon bearing the positive charge as:

Thus the order of stability if CH₃⁺ < 1° < 2° < 3°.

3. Carbanion: A carbanion may be defined as a species containing a carbon atom carrying a negative charge. These are generated by the atom in which the atom linked to carbon goes without the bonding electrons. As a result of this carbon acquires a negative charge. For example, the removal of hydrogen of methyl part of acetaldehyde molecule as H+ ion leaving both the electron on carbon.

They are also very reactive species. They are also classified as primary, secondary and tertiary depending upon whether one, two or three carbon atoms are attached to the carbon atom bearing negative
charge.

The order of stability is the reverse of free radicals and carbocations
CH₃ ^– > 1° > 2° > 3°.

(iv) Carbenes: The carbenes are reactive neutral species in which carbon atom has six electrons in the valency shell out of which two are shared. The simplest carbene is methylene (CH₂). It is formed when diazomethane is decomposed by the action of light.

It is very reactive. It reacts with alkenes by adding to the double bond forming cyclopropane.

Organic Chemistry: Some Basic Principles and Techniques Important Extra Questions Numerical Problems

Question 1.
0.395 g of an organic compound by various method for the estimation of sulphur gave 0.582g of BaS04. Calculate the percentage of Sulphur.
Answer:
Mass of BaSO₄ = 0.582 g
BaSO₄ = S
233 = 32

233g of BaSO₄ contain sulphur = 32g
0. 582 g of BaSO₄ contains sulphur

Question 2.
0.15g of an organic compound gave 0.12g of AgBr by carius method. Find the percentage of bromine in the compound.
Answer:
Mass of AgBr formed = 0.12g
188 g of AgBr contains bromine = 80g.

Therefore, 0.12g of AgBr will contain bromine
= \(\frac{80 \times 0.12}{188}\) = 0.051 g

Percentage of bromine = \(\frac{0.051}{0.15}\) × 100 = 34%

Question 3.
0.40 g of an organic compound gave 0.3g of AgBr by carius method. Find the percentage of bromine in the compound.
Answer:
Mass of compound = 0.40 g

Now, 188 g of AgBr contains bromine = 80g.

Question 4.
0.15 g of an organic compound gave 0.12g of silver bromide by the carius method. Find out the percentage of bromine in the compound.
Answer:
Here the mass of substance taken = 0.15g
Mass of AgBr formed = 0.12g
Now 1 mole of AgBr = 1 g. atom of Br
or
188 g = 80g of Br

188g of AgBr contains bromine = 80g.
Hence 0.12g of AgBr contain bromine
= \(\frac{80 \times 0.12}{188}\) = 0.051 g

Thus, 0.15g of compound contains 0.051g of Br
Percentage of Br = \(\frac{0.051}{0.15}\) × 100 = 34%

Question 5.
0.2595g of an organic compound when treated with carius method, gave 0.35g of BaSO₄. Calculate the percentage of Sulphur in the compound.
Answer:
Here the mass of substance taken = 0.35g
Mass of BaSO₄ ppt. formed = 0.35g

Now 1 mole of BaSO₄ = 1 g. atom of Sulphur
233 g of BaSO₄ = 32g of S
i. e. 233g of BaS04 contains 32g of Sulphur

Therefore 0.35g of BaSO₄ will contain Sulphur

Question 6.
0.12g of an organic compound containing Phosphorus gave 0.22g of Mg₂P₂O₇ by the usual analysis. Calculate the percentage of Phosphorus in the compound.
Answer:
Here the mass of organic compound taken = 0.12g
Mass of Mg₂P₂O₇ formed = 0.22g .
Now 1 mole of Mg₂P₂O₇ = 222 g of Mg₂P₂O₇
= 62g of Phosphorus

i. e. 222g of Mg₂P₂O₇ contains Phosphorus = 62g
Therefore 0.22g of Mg₂P₂O₇ contains Phosphorus
= \(\frac{62}{222}\) × 0.22

Hence Percentage of Phosphorus = \(\frac{62}{222} \times \frac{0.22}{0.12}\) × 100 = 51.2

Question 7.
In a Dumas nitrogen estimation, 0.303g of an organic compound gave 50 cm3 of nitrogen collected at 300k & 715 m.m. of pressure. Calculate the percentage of nitrogen in the compound, (vapour pressure of water at 300K = 15 m.m.).
Answer:
Vapour pres1 re of the gas = 715 – 15 = 700 mm.
V₁ = 50 cm3, V₂ =?, P₁ = 700 mm, P₂ = 760 mm,
T₁ = 300 K, T₂ = 273 K
Applying \(\frac{\mathrm{P}_{1} \mathrm{~V}_{1}}{\mathrm{~T}_{1}}=\frac{\mathrm{P}_{2} \mathrm{~V}_{2}}{\mathrm{~T}_{2}}\)
or
V₂ = \(\frac{P_{1} V_{1} T_{2}}{P_{2} T_{2}}\)
= \(\frac{700 \times 50 \times 273}{760 \times 300}\)
= 41.9 cm³

22400 cm³ of nitrogen at S.T.P. weighs = 28g.
41.9 cm³ of nitrogen at S.T.P. weighs
= \(\frac{28}{22400}\) × 41.9 = 0.0524 g

Percentage of nitrogen = \(\frac{0.0524}{0.3}\) × 100 = 17.46

Question 8.
In an estimation of Sulphur by various method, 0.2175g of a compound gave 0.5825g barium sulphate. Calculate the percentage of sulphur in the compound.
Answer:
Mass of the compound = 0.2175 g
Mass of barium sulphate = 0.5825 g
Molecular mass of BaS04 = 137 + 32 + 64 = 233 g

233 g of BaSO₄ contains sulphur = 32g
0.5825g of BaSO₄ contains Sulphur
= \(\frac{32}{233}\) × 0.5825g

Percentage of Sulphur = \(\frac{32}{233} \times \frac{0.5825}{0.2175}\) × 100
= 36.78

Question 9.
0.515 g of an organic compound containing Phosphorus give 0.214g of magnesium Pyrophosphate in various method for the estimation of Phosphorus. Calculate the percentage of Phosphorus in the given organic compound.
Answer:
Mass of organic compound = 0.515 g
Mass of magnesium Pyrophosphate = 0.214 g
Molar mass of Mg₂P₂O₇ = 222 g

222 g of Mg₂P₂O₇ obtained from Phosphorus = 62g
0.214g of Mg₂P₂O₇ are obtained from Phosphorus
= \(\frac{62}{222}\) × 100 = 0.0597 g

Percentage of Phosphorus = \(\frac{0.0597}{0.515}\) × 100 = 11.6

Question 10.
In Duma’s method for estimation of Nitrogen 0.3g of an organic compound gave 50 mL of nitrogen collected at 300K temperature & 715 mm pressure. Calculate the percentage composition of nitrogen in the compound (Aqueous tension at 300K = 15 mm)
Answer:
Volume of nitrogen collected at 300K & 715 mm Pressure = 50 mL
Actual pressure = 715 – 15 = 700 mm

Volume of nitrogen at S.T.P = \(\frac{273 \times 700 \times 50}{300 \times 760}\) = 41.9 mL
22400 mL of nitrogen weighs = \(\frac{28 \times 41.9}{22400 g}\)

Percentage of nitrogen = \(\frac{28 \times 41.9 \times 100}{22400 \times 0.3}\) = 17.46

Question 11.
Ammonia produced when 0.75g of a substance was KJeldahlized, neutralized 30 cm³ of 0.25N H₂SO₄. Calculate the percentage of nitrogen in the compound.
Answer:
Mass of organic substance = 0.75g
Volume of H₂SO₄ used up = 30 cm³
Normality of sulphuric acid = 0.25 N 30 cm³

H₂SO₄ of normality 0.25 N = 30 mL of NH₃ solution of normality 0.25N
But 1000 cm³ of ammonia solution of normality 1 contains 14 g of nitrogen

Therefore, 30 cm³ of 0.25N ammonia solution will contain nitrogen
= \(\frac{14}{1000}\) × 30 × 0.25
Percentage of nitrogen = \(\frac{14}{1000}\) × \(\frac{30 \times 0.25}{0.75}\) × 100 = 14

Here we are providing Class 11 chemistry Important Extra Questions and Answers Chapter 11 The p-Block Elements. Chemistry Class 11 Important Questions are the best resource for students which helps in Class 11 board exams.

Class 11 Chemistry Chapter 11 Important Extra Questions The p-Block Elements

The p-Block Elements Important Extra Questions Very Short Answer Type

Question 1.
do boron halides form additional compounds with amines?
Answer:
Boron halides are Lewis acids and hence accept a pair of electrons from amines to form additional compounds.

Question 2.
How does boron interact with NaOH?
Answer:
2B + 6NaOH → 2Na₃BO₃ + 3H₂.

Question 3.
What is the oxidation state of C in
(a) CO
Answer:
+ 2

(b) HCN
Answer:
+ 2

(c) H₂CO₃
Answer:
+ 4

(d) CaC₂.
Answer:
-1.

Question 4.
What is the state of hybridization of C in
(a) CO₃^2-
Answer:
sp²

(b) CCl₄
Answer:
sp³

(c) diamond
Answer:
sp³

(d) graphite?
Answer:
sp²

Question 5.
Give two examples of electron-deficient compounds.
Answer:
BF₃ and B₂H₆.

Question 6.
Arrange the following halides of boron in the increasing order of acidic character.
BF₃, BCl₃, BBr₃, BI₃.
Answer:
BF₃ < BCl₃ < BBr₃ < BI₃.

Question 7.
What is dry ice? Why is it so-called?
Answer:
Solid CO₂ is known as dry ice. It does not wet a piece of paper/cloth and sublimes without melting. Therefore, it is called dry ice.

Question 8.
Write balanced equations to show hydrolysis reactions of CO₃^2- and HCO₃^–.
Answer:
CO₃^2- + H₂O ⇌ OH^– + HCO₃^–
HCO₃^– + H₂O ⇌ OH^– + H₂CO₃.

Question 9.
Why boron does not form B³⁺ ion?
Answer:
Boron has a very high sum of the first three ionisation enthalpies. Hence it cannot lose three electrons to form a B³⁺ ion.

Question 10.
Which oxide of carbon is an anhydride of carbonic acid?
Answer:
CO₂, because H₂CO₃ acid decomposes to give H₂O and CO₂.

Question 11.
What happens when a borax solution is acidified? Write a balanced equation for the reaction.
Answer:
Boric acid is formed.
Na₂B₄O₇ + 2HCl + 5H₂O → 2NaCl + 4H₃BO₃ (boric acid)

Question 12.
By means of a balanced equation show how B(OH)₃ behaves as an acid in water.
Answer:
B(OH)₃ + 2HOH → [B(OH)^– + H₃O⁺.

Question 13.
What happens when boric acid is heated?
Answer:

Question 14.
What are boranes?
Answer:
Stable covalent hydrides of boron like B₂H₆, B₄H₁₀, B₅H₉ on analogy with alkanes are called boranes.

Question 15.
Write a balanced equation for the preparation of boron by reduction of BBr₃.
Answer:
2BBr₃(g) + 3H₂(g) → 2B(s) + 6HBr(g)

Question 16.
What is carborundum? What is its common use?
Answer:
Silicon carbide (SiC). It is used as an abrasive.

Question 17.
What is Freon gas? To what use is it put?
Answer:
Freon gas is dichlorodifluoromethane CCl₂F₂. It is used as a coolant in refrigerators and in air-conditioners.

Question 18.
Give the name of the compound used as a fire extinguisher under the name pyrone.
Answer:
Carbon tetrachloride (CCl₄).

Question 19.
What happens when aluminium metal is dipped in cone, nitric acid?
Answer:
One molecule thick layer of oxide is formed on the surface of A1 and further reaction does not proceed. Al is said to become passive.

Question 20.
What happens when CO is passed over heated nickel?
Answer:
Ni + 4CO → Ni(CO)₄.
Tetra carbonyl nickel (O) is formed.

Question 21.
How does boron react with dinitrogen?
Answer:
2B + N₂ → 2BN (boron nitride)

Question 22.
What is the chemical formula of Borazole or borazine? Why is it called inorganic benzene?
Answer:
Borazole is B₃N₃H₆. Because of its similarity to benzene, it is called inorganic benzene.

Question 23.
Write down the structure of Borazole.
Answer:
B₃N₃H₆.

Question 24.
Why thallium prefers to show an oxidation state of +1 rather than +3?
Answer:
Due to the inert-pair effect.

Question 25.
How do you explain that anhydrous AlCl₃ is covalent, but hydrated AlCl₃ is electrovalent?
Answer:
In the presence of H₂O, Al₂Cl₆ dissociates into hydrated Al³⁺ and Cl^– ions due to the high heat of hydration of these ions.

Question 26.
Why BBr₃ is a stronger Lewis acid than BF₃?
Answer:
This is because the back donation of electrons into empty 2p orbital of boron atom from filled p orbital of Br atom is much less than that by F atom due to larger size of Br atom than F atom.

Question 27.
Why trihalides of group 13 elements fume in the moist air?
Answer:
They are hydrolysed by water forming hydrogen halides.
MX₃ + 3H₂O → M(OH)₃ + 3HX.

Question 28.
Aluminium forms [AlF₆]^3-, but boron does not form [BF₆]^3- why?
Answer:
Boron does not have vacant d-orbitals. Therefore, it cannot expand its coordination number beyond four.

Question 29.
Why boron halides do not exist as dimers whereas AlCl₃ exists as Al₂Cl₆?
Answer:
Boron atom being small iff size is unable to accommodate large-sized halogens around it.

Question 30.
Gold has much higher first ionisation energy than boron, yet gold is metal while boron is non-metal. Explain.
Answer:
This is based on their crystal structure. Gold has a coordination number of 12 while boron has 6 or less than 6.

Question 31.
Why CO₂ is a gas whereas SiO₂ is solid. Explain why?
Answer:
CO₂ forms monomeric linear molecules, while SiO₂ exists as a giant-sized 3-dimensional network structure.

Question 32.
C and Si are almost tetravalent, but Ge, Sn, Pb show divalency. Why?
Answer:
The inert pair effect is shown by Ge, Sn, Pb.

Question 33.
Account for the fact that PbX2 is more stable than PbX₄. [X = Cl, Br]
Answer:
Due to the inert pair effect, Pb shows an oxidation state of +2.

Question 34.
PbCl₄ is less stable than SnCl₄, but PbCl₂ is more stable than SnCl₂. Why?
Answer:
Stability of + 4 oxidation state decreases down the group while that of + 2 oxidation state increases due to inert pair effect.

Question 35.
Why carbon shows maximum catenation in group 14 elements?
Answer:
Due to the strong C-C bond, its bond dissociation energy is the highest among group 14 members,

Question 36.
Pyrosilicates contain which anion?
Answer:
Si₂O₇^6-.

Question 37.
Mention one industrial application of silicones.
Answer:
Silicones are used for making water-proof papers by coating them with a thin layer of silicones.

Question 38.
What is the basic building unit of silicates?
Answer:
SiO₄^4- tetrahedra.

Question 39.
Which element of group 13 forms amphoteric hydroxide?
Answer:
Aluminium.

Question 40.
Which element of group 13 forms the most stable oxidation state of + 1?
Answer:
Thallium.

Question 41.
Explain why silicon shows a higher covalency than carbon?
Answer:
Due to the presence of d-orbitals, silicon shows a higher covalency of 6.

Question 42.
What are silicones?
Answer:
Silicones are synthetic organosilicon compounds containing repeated R₂SiO units held by Si—O—Si linkages.

Question 43.
What are the 2 main uses of zeolites?
Answer:

1. Softening of hard water.
2. Catalysts in petrochemical industries.

Question 44.
Why is boric acid considered a weak acid?
Answer:
Because it is not able to release H⁺ ions on its own. It receives OH^– ions from the water molecule to complete its octet & in turn releases H⁺ ions.

Question 45.
[SiF₆]^-2 is known whereas [SiCl₆]^2- not. Why?
Answer:
The reasons are:

1. Six large chlorine atoms cannot be accommodated around the Silicon atom due to the limitation of its size.
2. Interaction between lone pair of chlorine atom & silicon atom is not very strong.

Question 46.
Diamond is co-valent, yet it has a high M.P. Why?
Answer:
Diamond has a three-dimensional networked structure involving a strong C—C bond, which are very difficult to break & in turn it has a high melting point.

Question 47.
What is the common name of the recently developed allotrope of carbon i.e. C60 molecule?
Answer:
Fullerene.

Question 48.
The M.P. & B.P. of Carbon & Silicon is very high. Why?
Answer:
This is due to the tendency of these elements to form giant molecules.

Question 49.
Why is boron metalloid?
Answer:
Boron resembles both metals & non-metals therefore it is metalloid.

Question 50.
Why does boron resemble Si?
Answer:
Both have a similar charge to radius ratio, i.e. similar polarizing power.

The p-Block Elements Important Extra Questions Short Answer Type

Question 1.
Although boric acid B(OH)₃ contains three hydroxyl groups, yet it behaves as a monobasic acid. Explain.
Answer:

Hydrated species
B(OH)₃ is not a protonic acid.

It behaves as a Lewis acid because it abstracts a pair of electrons from hydroxyl ion.

Question 2.
SiCl₄ forms [SiCl₆]^2- while CCl₄ does not form [CCl₆]^2- Explain.
Answer:
Carbon does not have d-orbitals and hence C.Cl₄ does not combine with Cl^– ions to give [CCl₆]^2-, On the other hand, silicon has vacant 3d-orbitals and thus can expand its covalency from 4 to 6. Therefore SiCl₄ combines with CL ions to form [SiCl₆]^2-

SiCl₄ + 2Cl^– → [SiCl₆]^2-

Question 3.
Why does not silicon form an analogue of graphite?
Or
Why does elemental silicon not form a graphite-like structure as carbon does? Explain.
Answer:
In graphite, C is sp² hybridised and each C is linked to three other C atoms forming hexagonal rings. Thus graphite has a two-dimensional sheet-like structure.

Silicon, on the other hand, does not form an analogue of C because of the following two reasons:

1. Silicon has a much lesser tendency for catenation than C as Si-Si bonds are much weaker than C-C bonds.
2. Silicon because of its larger size than C undergoes sp³ hybridisation.

Question 4.
Why carbon forms covalent compounds whereas lead forms ionic compounds?
Answer:
Carbon cannot lose electrons to form C₄ because the sum of four ionisation enthalpies is very high. It cannot gain four electrons to form C₄ because energetically it is not favourable. Hence C forms only covalent compounds. Down the group 14, ionisation enthalpies decrease, Pb being the last element has so low I.E. that it can lose electrons to form ionic compounds.

Question 5.
How is borax prepared from
(i) Colemanite ore
Answer:
Borax is also called sodium tetraborate decahydrate (Na₂B₄O₇.10H₂O). It can be prepared as follows:

From colemanite: Powdered mineral is boiled with sodium carbonate solution and filtered. The filtrate is concentrated and then cooled when crystals of borax.
Ca₂B₆O₁₁ + 2Na₂CO₃ → Na₂B₄O₇ + 2NaBO₂ + 2CaCO₃.

The mother-liquor which contains sodium meta-borate is treated with a current of C02, to convert it into borax which separates out.
4NaBO₂ + CO₂ → Na₂B₄O₇ + Na₂CO₃

(ii) Tincal.
Answer:
From Tincal: Tincal obtained from dried up lakes is boiled with water. The solution is filtered to get rid of insoluble impurities of clay, sand etc. The filtrate is concentrated to get the crystals of borax.

(iii) Boric acid?
Answer:
From boric acid: Boric acid is neutralised with sodium carbonate and the resulting solution is concentrated and cooled to get the crystals of borax Na₂B₄O₇.10H₂O.
4H₃BO₃ + Na₂CO₃ → Na₂B₄O₇ + 6H₂O + CO₂ ↑

Question 6.
Mention three important uses of borax
Answer:
It is used:

• As a flux soldering and welding in industry.
• In the manufacture of borosilicate glass (or pyrex glass).
• In making enamels and glazes.
• In stiffening of candle wicks.
• In softening of water.
• In a qualitative analysis for borax bead test in the laboratory.

Question 7.
What happens when a borax solution is acidified? Write a balanced equation for the reaction.
Answer:
When acidified aqueous solution of borax (Na₂B₄O₇) is heated, boric acid is formed.

Question 8.
Mention important uses of boric acid.
Answer:
Boric acid is used:

• In the manufacture of enamels and glazes for pottery.
• In making heat-resisting and shock resisting glass called boro glass (or pyrex glass).
• As a mild antiseptic for washing eyes

Question 9.
Mention some important properties of carbon monoxide.
Answer:

1. It is a colourless, odourless gas, slightly soluble in water.
2. It is highly poisonous. It combines with haemoglobin in the red blood cells to form carboxy-haemoglobin which cannot absorb oxygen and thus the supply of oxygen to the body is reduced.
3. It burns with a pale blue flame forming CO₂.
   2CO + O₂ → 2CO₂.
4. It is a reducing agent. It reduces some metal oxides into metals.
   Fe₂O₃ + 3CO → 2Fe + 3CO₂.
5. It combines With transition metals like iron, cobalt, nickel to form their carbonyl compounds.
   

Question 10.
What are halides of carbon? Give few examples.
Answer:
Carbon combines with halogens to form both simple and mixed tetrahalides. In the case of simple halides, all the four expected tetrahalides (e.g. CF₄, CCl₄, CBr₄ and Cl₄) are known to exist. The stability of the simple tetrahalides decreases with the increasing atomic mass of the halogen.
(CF₄ > CCl₄ > CBr₄ > Cl₄)
Amongst the mixed halides the better-known compounds are (CFCl₃, CF₂Cl₂ and C Cl₃Br).

Question 11.
What is allotropy? Give examples of allotropes.
Answer:
Two or more forms of the same element in the same physical state which differ in their physical properties but have the same chemical properties are called allotropic forms or (allotropes) and the phenomenon is called allotropy.

Carbon, phosphorus and sulphur are some elements that exhibit allotropy.

1. Diamond and graphite are allotropic forms of carbon.
2. Red phosphorus and white .phosphorus are allotropes of phosphorus.
3. Rhombic sulphur, monoclinic sulphur and plastic sulphur are allotropic forms of sulphur.

Question 12.
Give the uses of different allotropic forms of carbon.
Answer:

Question 13.
Give reasons:
(i) Graphite is used as a lubricant.
Answer:
Graphite has sp3 hybridized carbon with a layer structure. Due to wide separation & weak interlayer bonds, the two adjacent layers can easily slide over each other. This makes graphite act as a lubricant.

(ii) Cone. HN03 can be transported in an aluminium container.
Answer:
Aluminium on coming in contact with the cone; HNO₃ becomes passive due to the coating of aluminium oxide & thus the cone. HNO₃ can be transported in an aluminium container.

(iii) Aluminium utensils should not be kept in water overnight.
Answer:
Aluminium utensils should not be kept in water overnight because aluminium is readily corroded by water.

Question 14.
Write balanced equations for the reaction of elemental boron with elemental chlorine, oxygen & nitrogen at a high temperature.
Answer:

Question 15.
Why are boron halides & diborane referred to as “Electron deficient compounds”?
Answer:
Boron in its halides has only six electrons in its valence shell, therefore it is short by two electrons to complete its octet. As a result, a molecule of boron halide can accept a pair of electrons from any electron-rich compound that is why boron halides are called electron-deficient compounds. In diborane, the total no. of valence electrons is not sufficient to completely fill the. available orbitals. This gives an electron-deficient character to diborane.

Question 16.
Give reasons:
(i) A mix. of dil. NaOH & aluminium pieces are used to open drains.
Answer:
Aluminium dissolves in dil. NaOH with the evolution of H₂. This H₂ helps to open the drain.

(ii) Diamond is used as an abrasive.
Answer:
Diamond is the hardest substance known & thus used as abrasive & for cutting glass.

(iii) Aluminium wires are used to make transmission cables.
Answer:
Aluminium is cheaply available on a weight to weight basis, the electrical conductivity of aluminium is twice that of copper. Hence Aluminium wires are used to make transmission cables.

Question 17.
Write balanced equations:
(i) B₂H₆ + NH₃ →?
Answer:

(ii) NaH+ B₂H₆ →?
Answer:

(iii) BF₃ + LiH →?
Answer:

Question 18.
Describe briefly three different methods of obtaining elemental boron.
Answer:

1. By reduction of oxides: The reduction of oxides is carried out by electropositive metals like Mg.
   B₂O₃(s) + 3Mg(s) → 2B(s) + 3MgO(s)
2. By reduction of halides: The reduction of the volatile boron halides is carried but by dihydrogen at high temperatures
   
3. By the thermal decomposition of boron hydride
   

Question 19.
All the elements of group 13 except thallium show a +3 oxidation state while it shows a +1 oxidation state. Give reasons.
Answer:
The valence shell of group 13-elements have two electrons in s-subshell & 1-electron in p-subshell Therefore, they are expected to show a +3 oxidation state. In thallium, the electrons of the s-subshell do not take part in bond formation due to the inert pair effect & only one electron of the p-orbital participate in bond formation, thus it shows +1 oxidation state only.

Question 20.
Why carbon does not form ionic compounds?
Answer:
The electronic configuration of carbon atom is 1s², 2s², 2px’, 2py’ & has four valence electrons. In order to form ionic compounds, it has to either lose four electrons or gain four electrons. Since very high energies are involved in doing so. Thus carbon does not form ionic compounds. It completes its octet as a result of electron sharing & forms covalent compounds.

Question 21.
Gallium has higher ionization enthalpy than Aluminium. Explain.
Answer:
Gallium has higher ionization energy than Aluminium because of the higher effective nuclear charge. This is due to additional 3d electrons which do not screen the nuclear charge effectively so that the outer electrons are more strongly held.

Question 22.
Boron forms no compounds in a unipositive state but thallium in a unipositive state is quite unstable. Why?
Answer:
Boron has electronic configuration 2s²2p¹ & therefore forms compounds in a trivalent state. However, thallium prefers to form compounds in the +1 oxidation state rather than in the +3 oxidation state as suggested by its group number. This is due to the inert pair effect. According to this effect, the 6s² electrons in the case of heavy metals preferably do not take part in bonding.

Question 23.
What are silicones? How are they manufactured?
Answer:
Silicones are rubber-like polymers having Si – O-Si linkage & general formula R₂SiO. They are manufactured by hydrolysis of Chloro silicones,

Where R is Me/Ph groups

Question 24.
Mention any two dissimilarities of boron with other elements of group-13.
Answer:

1. All the compounds of boron are covalent in nature because of the non-existence of the B³⁺ ion. It is because of its high charge density.
2. The maximum covalence shown by Boron is 4 while other members of this group show a covalence of six or more.

Question 25.
Why boron trihalides form tetrahedral complexes?
Answer:
A Boron trihalide is an electron-deficient compound having six electrons in its outermost shell. Therefore it has a tendency to form a BX₃L type of complex after accepting an electron pair in the un-hybridized vacant p-orbital from a ligand (L) molecule. Because of this reason the shape of BX₃ changes from planar Triangular to tetrahedral in the BX₃L complex.

Question 26.
Discuss the pattern of variation in the oxidation states of Al & Tl.
Answer:
Aluminium shows a +3 oxidation state only while thallium, the last element of group 3 shows +3 oxidation state & +1 oxidation states. Tl⁺¹ is more stable than Tl³⁺ as is evident by redox potential data
Tl³⁺(aq) + 2e^– → Tl⁺ (aq) E° = +1.25 V

It happens because of the decrease in bond energy with size down the group (i.e. from Al → Tl). Thus the energy required to break the pair of ns² electrons is not compensated by the energy released during the formation of two additional bonds. Thus +1 oxidation state is more stable in thallium.

Question 27.
C—C bond length is shorter in graphite than the C-C bond length of diamond. Explain.
Answer:
Graphite has sp² hybridization & the C-C bond involves sp²—sp² hybridized carbon. Diamond has sp³ hybridization & the C—C bond involves sp³-sp³ hybridization. Furthermore, the more the S character in a hybridized atom, the smaller is the size of the hybridized orbital, resulting in more overlapping which leads to bond length.

Question 28.
[SiF₆]^2- is known whereas [SiCl₆]^2- not. Give reasons.
Answer:

1. Six large chlorine atoms cannot be accommodated around silicon atom due to the limitation of their size.
2. Interaction between lone pair of chlorine atoms.& silicon atom is not very strong.

Question 29.
SiCl. forms [SiCl]^2- while CClL does not form [CCl]^2-. Explain
Answer:
Carbon does not have d-orbitals, hence CCl₄ does not combine with Cl ions to form [CCl₆]^2-. On the other hand, silicon has vacant 3-d orbitals & thus SiCl₄ combines with Cl^– ions to form
[Sicy2-
SiCl₄ + 2Cl^– → [SiCl₆]^2-

In other words, carbon shows a fixed covalency of 4 but silicon exhibits varying covalency from 4 to 6.

Question 30.
Borazine is more reactive than benzene. Why?
Answer:
Both Borazine & Benzene are isoelectronic. In benzene C = C bonds are non-polar while N=B bonds in borazine are polar in nature due to the presence of a co-ordinate bond between N & B atoms. As a result, addiction is quite frequent in borazine while it is less in benzene because of delocalisation of π-electron charge.

The p-Block Elements Important Extra Questions Long Answer Type

Question 1.
(i) What are the different oxidation states exhibited by the group 14 elements? Discuss the stability of their oxidation states.
Answer:
The group 14 elements have four electrons in the outermost shell. The common oxidation states exhibited by these elements are +4 and +2. Since the sum of the first four ionisation enthalpies is very high, compounds in the +4 oxidation state are generally covalent in nature. In heavier members such as Ge, Sn and Pb, the tendency to show +2 oxidation state increases. It is due to the inability of ns2 electrons of the valence shell to participate in bonding.

The relative stabilities of these two oxidation states vary down the group. C and Si mostly show a +4 oxidation state. Ge forms stable compounds in the +4 state and only a few compounds in the +2 state. Sn forms compounds in both oxidation states (Sn in +2 state is a reducing agent).

Lead compounds in the +2 state are stable and in the +4 state are strong oxidising agents. In the tetravalent state, the number of electrons around the central atom in a molecule (e.g., carbon in CCl₄) is eight. Being electron precise molecules, they are normally not expected to act as an electron acceptor or electron donor.

Although carbon cannot exceed its covalence of more than 4, other elements of the group can do so. It is because of the presence of d-orbital in them. Due to this, their halides undergo hydrolysis and have a tendency to form complexes by accepting electron pairs from donor species. For example, the species like SiF₅^–, SiF₆^–, [GeCl₆]^2-, [Sn(OH)₆]^2- exist where the hybridisation of the central atom is sp³d².

(ii) What type of oxides are formed by group 14 elements? Which of them are acidic, neutral or basic?
Answer:
All members when heated in oxygen form oxides. There are -mainly two types of oxides, i. e., monoxide and dioxide of formula MO and MO₂ respectively. SiO only exists at high temperature. Oxides in the higher oxidation state of elements are generally more acidic than those in the lower oxidation state. The dioxides-CO₂, SiO₂ and GeO₂ are acidic, whereas SnO₂ and PbO₂ are amphoteric in nature. Among monoxides, CO is neutral, GeO is distinctly acidic whereas SnO and PbO are amphoteric.

Question 2.
(a) [SiF₆]^2- is known whereas [SiCl₆]^2- is not known. Give reasons
Answer:
(i) [SiF₆]^2- is known whereas [SiCl₆]^2- does not exist.
The main reasons are (i) six large chlorine atoms cannot be accommodated around silicon atom due to the limitation of their size.
(ii) Interactions between lone pairs of a chlorine atom and silicon atom are not very strong.

(b) Select the member (s) of group 14 that
(i) forms the most acidic oxide
Answer:
The most acidic dioxide is formed by carbon (CO₂).

(ii) is commonly found in the +2 oxidation state
Answer:
Lead is mostly found in the +2 oxidation state in its compounds.

(iii) used as a semi-conductor.
Answer:
Silicon and germanium are used as semiconductors.

(c) Explain why a diamond that is covalent has a high melting point?
Answer:
Though diamond has covalent bonding in it, yet it has a high melting point, because a diamond has a 3-dimensional network involving strong C—C bond, which are very difficult to break and in turn, it has a high melting point.

(d) Discuss the reaction of silica with
(i) NaOH
Answer:
SiO₂ reacts with HF as follows:
SiO₂ + 2NaOH → Na₂SiO₃ + H₂O

(ii) HE
Answer:
SiO₂ reacts with HF as follows:
SiO₂ + 4HF → SiF₄ + 2H₂O.

Question 3.
(a) Carbon exhibits catenation, whereas silicon does not. Explain.
Answer:
Carbon shows catenation because of its smaller size, high bond energy of C – C bond, the possibility of sp, sp², sp³ hybridisation and formation of multiple bonds C-C (1σ), C = C (1σ + 1π),- C = C (1σ + 2π). On the other hand, silicon shows only limited catenation because of its large atomic radius, low bond energy of Si-Si bond and absence of multiple bonds between Si atoms.

(b) How does boron differ from aluminium.
Answer:
Difference between boron and aluminium:

1. Boron is a non-metal but aluminium is a metal.
2. Boron is a semi-conductor while aluminium is a good conductor of electricity.
3. Boron forms a number of hydrides called boranes, but Al forms a polymeric hydride.
4. Halides of boron (except BF₃) are readily hydrolysed by water whereas halides of A1 are only partially hydrolysed by water.
5. B₂O₃ is acidic, but Al₂O₃ is amphoteric.
6. Boron hydroxide B(OH)₃ is acidic, but Al(OH)₃ is amphoteric.

(c) Write the similarities between boron and silicon.
Answer:
Similarities between boron and silicon:

1. Both are non-metals.
2. Both are semi-conductors
3. Boron and silicon form a number of covalent hydrides which have similar properties. For example, they spontaneously catch fire on exposure to air and are readily hydrolysed by water.
4. The halides of boron and silicon are readily hydrolysed by water.
5. Boron trioxide (B₂O₃) and silicon dioxide (SiO₂) are acidic in nature. These dissolve in alkali solution forming borates and silicates.

Here we are providing Class 11 chemistry Important Extra Questions and Answers Chapter 10 The s-Block Elements. Chemistry Class 11 Important Questions are the best resource for students which helps in Class 11 board exams.

Class 11 Chemistry Chapter 10 Important Extra Questions The s-Block Elements

The s-Block Elements Important Extra Questions Very Short Answer Type

Question 1
Which element is found in chlorophyll?
Answer:
Magnesium.

Question 2.
Name the elements (alkali metals) which form superoxide when heated in excess of air.
Answer:
Potassium, rubidium and caesium.

Question 3.
Why is the oxidation state of Na and K always + 1?
Answer:
It is due to their high second ionisation enthalpy and stability of their ions [Na⁺, K⁺].

Question 4.
Name the metal which floats on the water without any apparent reaction with water.
Answer:
Lithium floats on the water without any apparent reaction to it.

Question 5.
Why do group 1 elements have the lowest ionisation enthalpy?
Answer:
Because of the largest size in their respective periods, solitary electron present in the valence shell can be removed by supplying a small amount of energy.

Question 6.
Why does the following reaction
C – Cl + MF → C – F + MCI
proceed better with KF than with NaF?
Answer:
Because larger K+ cation stabilises larger anion.

Question 7.
Amongst Li, Na, K, Rb, Cs, Fr which one has the highest and which one has the lowest ionisation enthalpy?
Answer:
Li has the highest and Fr has the lowest ionisation enthalpy.

Question 8.
What is the general electronic configuration of alkali metals in their outermost shells?
Answer:
ns1 where n = 2 to 7.

Question 9.
What is meant by dead burnt plaster?
Answer:
It is anhydrous calcium sulphate (CaS04).

Question 10.
Name three forms of calcium carbonate.
Answer:
Limestone, chalk, marble.

Question 11.
Which member of the alkaline earth metals family has
(i) least reactivity
Answer:
Be

(ii) lowest density
Answer:
Ca

(iii) highest boiling point
Answer:
Be

(iv) maximum reduction potential?
Answer:
Be.

Question 12.
Why does lithium show anomalous behaviour?
Answer:
Due to its small size and high charge/size ratio.

Question 13.
Out of LiOH, NaOH, KOH which is the strongest base?
Answer:
KOH.

Question 14.
Write the balanced equations for the reaction between
(a) Na₂O₂ and water
Answer:
2Na₂O₂ + 2H₂O → 4NaOH + O₂

(b) KO₂ and water
Answer:
2KO₂ + 2H₂O → 2KOH + H₂O + O₂

(c) Na₂O and CO₂.
Answer:
Na₂O₄^– + CO₂ → Na₂CO₃

Question 15.
Arrange the following in order of increasing covalent character MCI, MBr. MF, MI (where M = alkali metal]
Answer:
MF < MCI < MBr < Ml.
With the increasing size of anion, covalent character increases

Question 16.
Name an element that is invariably bivalent and whose oxide is soluble in excess of NaOH and its dipositive ion has a noble gas core.
Answer:
The element is beryllium (Be) which forms a divalent ion and has a noble gas core [He] 2s²
Be²⁺ = 1 s²
BeO + 2NaOH → Na₂BeO₂ + H₂O

Question 17.
Arrange the following in the increasing order of solubility in water:
MgCl₂, CaCl₂, SrCl₂, BaCl₂.
Answer:
BaCl₂ < SrCl₂ < CaCl₂ < MgCl₂.

Question 18.
State the reason for the high solubility of beryllium chloride in organic solvents.
Answer:
Beryllium chloride is a covalent compound.

Question 19.
Which alkali carbonate decomposes on heating to liberate CO₂?
Answer:
Lithium carbonate [Li₂CO₃]

Question 20.
Why is the solution of an alkali metal in ammonia blue?
Answer:
Due to the presence of ammoniated electrons.

Question 21.
Beryllium oxide has a high melting point. Why?
Answer:
Due to its polymeric nature.

Question 22.
Mention the chief reasons for the resemblance between beryllium and aluminium.
Answer:
Both Be²⁺ and Al³⁺ ions have high polarising power.

Question 23.
State any one reason for alkaline earth metals, in general, having a greater tendency to form complexes than alkali metals.
Answer:
Because of their small size and high charge, alkaline earth metals have a tendency to form complexes.

Question 24.
Amongst alkali metals why is lithium regarded as the most powerful reducing agent in aqueous solution?
Answer:
Lithium is the best reducing agent because it has the lowest reduction potential:
Li⁺ + e^– → Li(s) E_Red = – 3.07V.

Question 25.
Name the metal amongst alkaline earth metals whose salt does not impart any colour to a non-luminous flame.
Answer:
Beryllium does not impart colour to a non-luminous flame.

Question 26.
What is the difference between baking soda and baking powder?
Answer:
Baking soda is sodium bicarbonate (NaHCO₃) while baking powder is a mixture of sodium bicarbonate fNaHC03) and Potassium hydrogen tartrate.

Question 27.
Why does table salt get wet in the rainy season?
Answer:
Table salt contains impurities of CaCl₂ and MgCl which being deliquescent compounds absorb moisture from the air during the rainy season.

Question 28.
Sodium readily forms Na⁺ ion, but never Na²⁺ ion. Explain.
Answer:
After the removal of 1 electron from Na, it has been left with a noble gas core [Ne] which is closer to the nucleus and requires more energy.

Question 29.
Which compound of sodium is used
(i) as a component of baking powder
Answer:
NaHCO₃

(ii) for softening hard water.
Answer:
Na₂CO₃.

Question 30.
Anhydrous calcium sulphate cannot be used as Plaster of Paris. Give reason.
Answer:
Because it does not have the ability to set like plaster of Paris.

Question 31.
Which of the following halides is insoluble in water?
CaF₂, CaCl₂, CaBr₂ and Cal₂.
Answer:
CaF₂.

Question 32.
Predict giving reason the outcome of the reaction.
Lil + KF → …………
Answer:
LiI + KF → LiF + KI
Larger cation stabilises larger anion.

Question 33.
Name three metal ions that play important role in performing several biological functions in the animal body.
Answer:
Na⁺, K⁺, Ca²⁺, Mg²⁺, etc.

Question 34.
Calcium metal is used to remove traces of air from vacuum tubes. Why?
Answer:
Calcium has a great affinity for oxygen and nitrogen.

Question 35.
Why is sodium kept in kerosene oil?
Answer:
To prevent its contact with oxygen and moist air, because sodium reacts with them.

Question 36.
What is brine?
Answer:
An aqueous solution of NaCl in water.

Question 37.
Why caesium can be used in photoelectric cells while lithium cannot be?
Answer:
Cs has the lowest & Li highest ionisation enthalpy. Hence Cs can lose electron very easily while lithium cannot.

Question 38.
In an aqueous solution, the Li⁺ ion has the lowest mobility. Why?
Answer:
Li⁺ ions are highly hydrated in an aqueous solution.

Question 39.
Lithium has the highest ionisation enthalpy in group-1, yet it is the strongest reducing agent. Why?
Answer:
This is because Li has the highest oxidation potential.
Li → Li⁺ + e^–
E°_ox = + 3.07

Question 40.
Name one reagent or one operation to distinguish between
(i) BeSO₄ and BaSO₄
Answer:
BeSO₄ is soluble in water while BaSO₄ is not.

(ii) Be(OH)₂ and Ba(OH)₂
Answer:
Be(OH)₂ dissolves in alkali, while Ba(OH)₂ does not.

Question 41.
Which alkaline earth metal hydroxide is amphoteric?
Answer:
Be(OH)₂.

Question 42.
Which alkali metal is radioactive?
Answer:
Francium (Fr).

Question 43.
Which alkaline earth metal is radioactive?
Answer:
Radium (Ra).

Question 44.
Name the alkali metal which shows a diagonal relationship with Mg.
Answer:
Lithium (Li).

Question 45.
Give the chemical formula of carnallite.
Answer:
KCl.MgCl₂.6H₂O.

Question 46.
Arrange CaSO₄, SrSO₄ & BaSO₄ in order of decreasing solubility.
Answer:
The order of decreasing solubility of these sulphates is CaSO₄, SrSO₄ & BaSO₄.

Question 47.
Why is K more reactive than Na?
Answer:
Since the valence electron of K is relatively at a greater distance than Na & it requires lesser energy to remove it.

Question 48.
Why does Beryllium show similarities with Al?
Answer:
Because of their similarity in charge/radius ratio (Be²⁺ = 0.064 & Al³⁺ = 0.660)

Question 49.
What is magnesia?
Answer:
Magnesium oxide (MgO).

Question 50.
How is Potassium extracted?
Answer:
By electrolysis of a fused solution of KOH.

The s-Block Elements Important Extra Questions Short Answer Type

Question 1.
Why the solubility of alkaline metal hydroxides increases down the group?
Answer:
If the anion and the cation are of comparable size, the cationic radius ‘vill influence the lattice energy. Since lattice energy decreases much more than the hydration energy with increasing ionic size, solubility will increases as we go down the group. This is the case of alkaline earth metal hydroxides.

Question 2.
Why the solubility of alkaline earth metal carbonates and sulphates decreases down the group?
Answer:
If the anion is large compared to the cation, the lattice; energy will remain almost constant within a particular group. Since the hydration energies decrease down the group, solubility will decrease as found for alkaline earth metal carbonates and sulphates.

Question 3.
Why cannot potassium carbonate be prepared by the SOLVAY process?
Answer:
Potassium carbonate cannot be prepared by the SOLVAY process because potassium bicarbonate (KHCO₃) is highly soluble in water, unlike NaHCO₃ which was separated as crystals. Due to its high solubility KHCO₃ cannot be precipitated by the addition of ammonium bicarbonate to a saturated solution of KCl.

Question 4.
What are the main uses of calcium and magnesium?
Answer:
Main uses of calcium:

1. Calcium is used in the extraction of metals from oxides which are difficult to reduce with carbon.
2. Calcium, due to its affinity for O₂ and N₂ at elevated temperatures, has often been used to remove air from vacuum tubes.

Main uses of Magnesium:

1. Magnesium forms alloys with Al, Zn, Mn and Sn. Mg-Al alloys being light in mass are used in aircraft construction.
2. Magnesium (powder and ribbon) is used in flashbulbs, powders incendiary bombs and signals.
3. A suspension of Mg(OH)₂ in water is used as an antacid in medicine.
4. Magnesium carbonate is an ingredient of toothpaste.

Question 5.
What is meant by the diagonal relationship in the periodic table? What is it due to?
Answer:
It has been observed that some elements of the second period show similarities with the elements of the third period situated diagonally to each other, though belonging to different groups. This is called a diagonal relationship.

The cause of the diagonal relationship is due to the similarities in properties such as electronegativity, ionisation energy, size etc. between the diagonal elements. For example on moving from left to right across a period, the electronegativity increases, which on moving down a group, electronegativity decreases. Therefore on moving diagonally, two opposing tendencies almost cancel out and the electronegativity values remain almost the same as we move diagonally.

Question 6.
Why is the density of potassium less than that of sodium?
Answer:
Generally, in a group density increases with the atomic number, but potassium is an exception. It is due to the reason that the atomic volume of K is nearly twice Na, but its mass (39) is not exactly double of Na (23). Thus the density of potassium is less than that of sodium.

Question 7.
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.
Answer:
Sodium and potassium ions (Na⁺ and K⁺) are larger than the corresponding Mg²⁺ and Ca²⁺ ions. Due to this lattice energy of Mg(OH)₂, Ca(OH)₂, MgCO₃ and CaCO₃. (Lattice energy is defined as the energy required to convert one mole of the ionic lattice into gaseous ions.

Thus lattices with smaller ions have higher lattice energies). The hydration energies of Mg²⁺ and Ca²⁺ are higher than Na⁺ and K⁺ because of their smaller sizes. But the difference in lattice energies is much more. Therefore, the hydroxides and carbonates of Mg²⁺ and Ca²⁺ are insoluble in water because of their higher lattice energies.

Question 8.
Why is it that the s-block elements never occur in free state/nature? What are their usual modes of occurrence and how are they generally prepared?
Answer:
The elements belonging to the s-block in the periodic table (i.e. alkali and alkaline earth metals) are highly reactive because of their low ionisation energy. They are highly electropositive forming positive ions. So they are never found in a free state.

They are widely distributed in nature in the combined state. They occur in the earth’s crust in the form of oxides, chlorides, silicates and carbonates.

Generally, a group I metals are prepared by the electrolysis of fused solution.
For example:

1.
At cathode: Na⁺ + e^– → Na
At anode: Cl^– → Cl + e^–
Cl + Cl → Cl₂

2. KOH ⇌ K⁺ + OH^–
At cathode: K+ + e^– → K
At anode: 4OH → 4OH + 4e
4OH → 2H₂O + O₂
or
4OH^– → 2H₂O + O₂ + 4e^–
These metals are highly reactive and therefore cannot be extracted by the usual methods, because they are strong reducing agents.

Question 9.
Explain what happens when
(i) Sodium hydrogen carbonate is heated.
Answer:

(ii) Sodium with mercury reacts with water.
Answer:
2Na-Hg + 2H₂O → 2NaOH + H₂ ↑ + 2Hg

(iii) Fused sodium metal reacts with ammonia.
Answer:

Question 10.
What is the effect of heat on the following compounds?
(a) Calcium carbonate
Answer:

(b) Magnesium chloride hexahydrate
Answer:

(c) Gypsum
Answer:

(d) Magnesium sulphate heptahydrate.
Answer:

Question 11.
Why is it that the s-block elements never occur in a free state? What are their usual modes of occurrence?
Answer:
The elements belonging to s-block in the periodic table. These metals (Alkali & alkaline earth metals) are highly reactive because of their low ionization energy. They are highly electropositive forming positive ions. So they are never found in a free state. They are widely distributed in nature in a combined state. They occur in the earth’s crust in the form of oxides, chlorides, silicates & carbonates.

Question 12.
Explain what happens when:
(a) Sodium hydrogen carbonate is heated.
Answer:
Sodium hydrogen carbonate on heating decomposes to sodium carbonate.

(b) Sodium with Mercury reacts with water.
Answer:
When sodium with mercury reacts with water. It produces sodium hydroxide.

Question 13.
What is the effect of heat on the following compounds:
(a) Calcium carbonate
Answer:

(b) Magnesium chloride hexahydrate
Answer:

(c) Gypsum
Answer:

Question 14.
State as to why
(a) An aqueous solution of sodium carbonate gives an alkaline test.
Answer:
Sodium carbonate gets hydrolise by water to form an alkaline solution.
CO₃^2- + H₂O → HCO₃^– + OH
Due to this, it gives an alkaline test.

(b) Sodium is prepared by electrolytic method & not by chemical method.
Answer:
Sodium is a very strong reducing agent. Therefore, it cannot be isolated by a general reduction of its oxides or other compounds. The metal formed by electrolysis will immediately react with water forming hydroxides. So, sodium is prepared by the electrolytic method only.

(c) Lithium on being heated in the air mainly forms mono-oxide & not the peroxides.
Answer:
Lithium is the least reactive but the strongest reducing agent of all the alkali metals. It combines with air, it forms mono-oxide only because it does not react with excess air.

Question 15.
Like Lithium in group-I, beryllium shows anomalous behaviour in group II. Write three such properties of beryllium which makes it anomalous in the group.
Answer:
1. Beryllium has an exceptionally small atomic size, due to high ionization energy & small atomic size it forms compounds that are largely covalent & its salts are hydrolysed easily.

2. Beryllium does not exhibit coordination N. more than four as in its valence shell (n = 2). There are only four orbitals. The remaining member of the group has co-ordination no. six by making use of group have co-ordination no. six by making use of some d-orbitals.

3. The oxides & hydroxides of beryllium quite unlike the other elements, in this group are amphoteric in nature.

Question 16.
Explain which one Na or K has a larger atomic radius?
Answer:
Potassium has a larger atomic size than sodium because, in K, the outermost electron is in the fourth energy state (4s¹) while in Na, the outer-most electron is in the third energy state (3s¹). That is r (K) > r (Na) since the fourth energy state (n = 4) in K is farther away from the nucleus than the third energy state (n = 3) in Na.

Question 17.
The alkali metals (group-1 elements) form only unipositive cations (M⁺) & not bivalent cations (M⁺²), Give reason.

Answer:
Alkali metals have an ns¹ valence shell configuration. They can give up this electron to form univalent cations & attain stable electronic configuration like their nearest inner gas. Now the removal of an electron from the closed-shell configuration requires a very large amount of energy. Which is not available in chemical reactions. Therefore, the alkali metals form a univalent cation (M⁺) & No divalent cations (M²⁺) are formed.

Question 18.
Complete the following equations:
(a) Ca + H₂O →?
Answer:

(b) Ca(OH)₂ + Cl₂ →?
Answer:
2Ca(OH)₂ + l → Ca(OCl)₂ + CaCl₂ + H₂O

(c) BeO + NaOH →?
Answer:
BeO + 2NaOH → Na₂BeO₂ + H₂O

Question 19.
How calcium carbonate can be changed into
(a) Calcium sulphate
(b) Calcium oxide?
Write a balanced chemical equation in each case.
Answer:
(a) When calcium carbonate is heated with dil. sulphuric acid, calcium sulphate is obtained.
CaCO₃(s) + H₂SO₄(aq) → CaSO₄(s) + CO₂(g) + H₂O(l)

(b) Calcium carbonate is heated at a high temperature between 1070 – 1270 K. It decomposes to give calcium oxide & carbon dioxide.

Question 20.
Explain the following phenomenon by means of balanced equations:
(a) When exhaling is made through a tube passing into a solution of lime water, the solution becomes turbid.
Answer:
Ca(OH)₂ (l) + CO₂ (g) → CaCO₃ (s) + H₂O (l)

(b) The turbidity of solution in (a) eventually disappears when continued exhaling is made through it.
Answer:
CaCO₃(s) + H₂O(l) + CO₂(g) → Ca(HCO₃)₂(aq)

(c) When the solution obtained in (B) is heated turbidity re-appears.
Answer:
Ca(HCO₃)₂(aq) → CaCO₃(s) + H₂O(l) + CO₂(g)

Question 21.
Describe the reactivity of alkaline earth metals with water.
Answer:
The reactivity of alkaline earth metals with water increases with increasing atomic numbers.

1. Beryllium does not react readily even with boiling water.
2. Magnesium reacts very slowly with cold water. While it reacts with hot water at an appreciable rate & liberates hydrogen gas.
   
3. Calcium, strontium & Barium react vigorously even with cold water & form hydroxides & liberate hydrogen gas
   M(s) + H₂O(l) → M(OH)₂ + H₂(g)
   [M = Ca, Sr, Ba]

Question 22.
Compare the reactivity of alkali metals with water.
Answer:
All the alkali metals displace hydrogen from water i.e.
2M + 2H₂O → 2MOH + H₂

Lithium reacts slowly with water, sodium reacts violently & Cs reacts so vigorously that the reaction if not controlled can lead to an explosion.
2Li(s) + 2H₂O(aq) → 2LiOH (aq) + H₂(g) Slow
2Na(s) + 2H₂O(aq) → 2NaOH (aq) + H₂(g) Fast
2Cs(s) + 2H₂O(aq) → 2CsOH (aq) + H₂(g) Violent

Question 23.
Alkaline earth metals are harder than alkali metals. Give reason.
Answer:
Alkaline earth metals are harder than alkali metals because:

1. The atomic radius of alkaline metals is small, atomic mass is high & density is larger than those of alkali metals. Alkaline earth metals have closed packed crystal structure.
2. In alkaline earth metals, the metallic bonding in its crystal is very strong as compared to the crystal of an alkali metal. Therefore, the atoms in the crystal of alkaline earth metals are strongly bonded.

Question 24.
Differentiate amongst Quick lime, Lime water & Slaked lime.
Answer:

Question 25.
The mobilities of alkali metal ions in aqueous solution follow the order
Li⁺ < Na⁺ < K⁺ < Rb⁺ < Cs⁺ Give reasons.
Answer:
Smaller ions are expected to have more mobility. But smaller ions such as Li⁺, Na⁺ due to their higher charge density tend to undergo hydration. Hydration increases the mass & effective size of the smaller ion & as a result, the mobility of ions show the trend given.

Question 26.
How is sodium peroxide manufactured? What happens when it reacts with chromium (III) hydroxide? Give it’s two uses.
Answer:
Sodium peroxide is manufactured by heating sodium metal on aluminium trays in the air (free-from CO₂).
2Na + O₂ (air) → Na₂O₂

Sodium Peroxide oxidises chromium (III) hydroxide to sodium chromate.

Uses:

• It is used as a bleaching agent because of its oxidising, power.
• It is used in the manufacture of dyes & many other chemicals such as benzoyl peroxide, sodium perborate etc.

Question 27.
List three properties of Lithium in which it differs from the rest of the alkali metals.
Answer:
(a) Lithium is much harder, its melting & boiling points are higher than the other alkali metals.
(b) Lithium is the least reactive but the strongest reducing agent among all the alkali metals.
(c) LiCl is deliquescent & crystallise as a hydrate, LiCl.2H₂O whereas other alkali metal chlorides do not form hydrates.

Question 28.
Why is LiF almost insoluble in water whereas LiCl is soluble not only in water but also in acetone?
Answer:
The low solubility of LiF is due to its high lattice enthalpy. LiCl has much higher solubility in water. This is due to the small size of Li⁺ ion & much higher hydration energy.

Question 29.
What happens when
(1) Calcium nitrate is heated,
Answer:

(2) Chlorine reacts with slaked lime
Answer:

(3) Quick lime is heated with silica.
Answer:

Question 30.
Arrange the following in order of property mentioned:
(1) Mg(OH)₂, Sr(OH)₂, Ba(OH)₂, Ca(OH)₂
(Increasing ionic solubility in water)
Answer:
Mg(OH)₂, Ca(OH)₂, Sr(OH)₂, Ba(OH)₂

(2) BeO, MgO, BaO, CaO
(Increasing basic character)
Answer:
BeO, MgO, CaO, BaO

(3) BaCl₂, MgCl₂, BeCl,, CaCl₂
(Increasing ionic character)
Answer:
BeCl₂, MgCl₂, CaCl₂, BaCl₂

The s-Block Elements Important Extra Questions Long Answer Type

Question 1.
What raw materials are used for making cement? Describe the manufacture of Portland cement. What is its approximate composition?
Answer:
Raw materials: The raw materials required for the manufacture of cement are limestone, stone and clay, limestone in calcium carbonate, CaCO₃ and it provides calcium oxide, CaO. Clay- is a hydrated aluminium silicate, Al₂O₃ 2Si02.2H20 and it provides alumina as well as silica. A small amount of gypsum, CaS04.2H20 is also required. It is added in calculated quantity in order to adjust the rate of setting of cement.

Manufacture: Cement is made by strongly heating a mixture of limestone and clay in a rotatory kiln. Limestone and clay are finely powdered and a little water is added to get a thick paste called slurry. The slurry is led into a rotatory kiln from the top through the hopper.

The hot gases produce a temperature of about 1770-1870 K in the kiln. At this high temperature, the limestone and clay present in the slurry combine to form cement in the form of small pieces called clinker. This clinker is mixed with 2 – 3 % by weight of gypsum (CaSO₄ .2H₂O) to regulate the setting time and is then ground to an exceedingly fine powder.

Manufacture of cement

When mixed with water the cement reacts to form a gelatinous mass which sets to a hard mass when three-dimensional cross-links are formed between
…………..Si — O — Si……… and …….. Si — O — Al…… chains.

Composition of cement:
CaO = 50 – 60%
SiO₂ = 20 – 25%
Al₂O₃ = 5 – 10%
MgO = 2 – 3%
Fe₂O₃ = 1 – 2%
SO₃= 1 – 2%

For a good quality cement, the ratio of silica (SiO₂) and alumina (Al₂O₃) should be between 2.5 to 4.0. Similarly, the ratio of lime (CaO) to the total oxide mixtures consisting of SiO₂, Al₂O₃ and Fe₂O₃ should be roughly 2: 1. If lime is in excess, the cement cracks during setting. On the other hand, if lime is less than required, the cement is weak in strength. Therefore, the proper composition of cement must be maintained to get cement of good quality.