Category: Class 11 Chemistry

NCERT Exemplar Class 11 Chemistry Chapter 1 Some Basic Concepts of Chemistry are part of NCERT Exemplar Class 11 Chemistry. Here we have given NCERT Exemplar Class 11 Chemistry Chapter 1 Some Basic Concepts of Chemistry. https://www.cbselabs.com/ncert-exemplar-problems-class-11-chemistry-chapter-1-basic-concepts-chemistry/

NCERT Exemplar Class 11 Chemistry Chapter 1 Some Basic Concepts of Chemistry

Multiple Choice Questions
Single Correct Answer Type

Q1. Two students performed the same experiment separately and each one of them recorded two readings of mass which are given below. Correct reading of mass is 3.0 g. On the basis of given data, mark the correct option out of the following statements.

Results of both the students are neither accurate nor precise.
(b) Results of student A are both precise and accurate.
(c) Results of student B are neither precise nor accurate.
(d) Results of student B are both precise and accurate.
Sol:

For both the students, average value is close to the correct value. Hence, readings of both are accurate. But readings of student A are also close to each other (differ only by 0.02) and also close to the average value, hence readings of A are also precise. But readings of B are not close to each other (differ by 0.1) and hence are not precise. Thus, results of student A are both precise and accurate.

Q2. A measured temperature on Fahrenheit scale is 200°F. What will this reading be on Celsius scale?
(a) 40°C
(b) 94°C
(c) 93.3°C
(d) 30°C

Q3. What will be the molarity of a solution, which contains 5.85 g of NaCl(s) per 500 mL?

(a) 4 mol L^-1
(b) 20 mol L^-1         
(c) 0.2 mol L^-1       
(d) 2 mol L ^-1

Q4. If 500 mL of a 5 M solution is diluted to 1500 mL, what will be the molarity of the solution obtained?
(a) 1.5 M
(b) 1.6 M
(c) 0.017 M
(d) 1.59 M

This all-in-one online Mole to Mass Calculator performs calculations using the formula linking the mass of a substance.

Q5. The number of atoms present in one mole of an element is equal to Avogadro number. Which of the following elements contains the greatest number of atoms?
(a) 4gHe (b) 46gNa (c) 0.40 gCa (d) 12 g He
Sol: (d) As we know that

Hence, 12 g of He contains the greatest number of atoms.

Q6. If the concentration of glucose (C₆H₁₂0₆) in blood is 0.9 g L^-1, what will be the molarity of glucose in blood?
(a) 5 M  
(b) 50 M 
(c) 0.005 M
(d) 0.5 M

Q7. What will be the molality of the solution containing 18.25 g of HCl gas in 500 g of water?
(a) 0.1 m (b) 1 M (c) 0.5 m (d) 1 m

Q8. One mole of any substance contains 6.022 x 10²³ atoms/molecules. Number of molecules of H₂S0₄ present in 100 mL of 0.02 M H₂S0₄ solution is __________
(a)12.044 x 10²⁰ molecules
(b) 6.022 x 10²³ molecules
(c) 1 x 10²³ molecules        
(d) 12.044 x 10²³ molecules

Q9. What is the mass per cent of carbon in carbon dioxide?
(a) 0.034%
(b) 27.27%
(c) 3.4%
(d) 28.7%

Sol: b) Molecular mass of C0₂ =1×12 + 2×16 = 44 g
1 g molecule of C0₂ contains 1 g atom of carbon
44 g of C0₂ contains C = 12 g of carbon
% of C in CO₂ = 12/44 x 100 =27.27%
Hence, the mass per cent of carbon in C0₂ is 27.27%.

Q10. The empirical formula and molecular mass of a compound are CH₂0 and 180g respectively. What will be the molecular formula of the compound?
(a) C₉H₁₈0₉,            
(b) CH₂0                 
(c) C₆H_i20₆             
(d) C₂H₄0₂

Sol: (c) Empirical formula mass = 12 + 2+ 16 = 30

Q11. If the density of a solution is 3.12 g mL¹, the mass of 1.5 mL solution in significant figures is________ 

(a) 4.7 g                                                  
(b) 4680 x 10^-3 g
(c) 4.680 g                                              
(d) 46.80 g

Sol: (a) Density of solution = 3.12 g mL^-1
Volume of solution =1.5 mL
Mass of solution = Volume x Density
= 1.5 mL x 3.12 g mL^-1
= 4.68 g = 4.7 g (up to 2 significant figures)

Q12. Which of the following statements about a compound is incorrect?
(a) A molecule of a compound has atoms of different elements.
(b) A compound cannot be separated into its constituent elements by physical methods of separation.
(c) A compound retains the physical properties of its constituent elements.
(d) The ratio of atoms of different elements in a compound is fixed.
Sol: (c) A compound does not retain the physical or chemical properties of its constituent elements.

Q13. Which of the following statements is correct about the reaction given below? 4Fe(s) + 30₂(g) →2Fe₂0₃(s)

(a) Total mass of iron and oxygen in reactants = total mass of iron and oxygen in product, therefore it follows law of conservation of mass.
(b) Total mass of reactants = total mass of product; therefore, law of multiple proportions is followed.
(c) Amount of Fe₂0₃ can be increased by taking any one of the reactants (iron or oxygen) in excess.
(d) Amount of Fe₂0₃ produced will decrease if the amount of any one of the reactants (iron or oxygen) is taken in excess.
Sol: (a) 4Fe + 30₂ —> 2Fe₂0₃ follows law of conservation of mass since mass of reactants is equal to mass of products.

Q14. Which of the following reactions is not correct according to the law of conservation of mass?
(a) 2Mg(s) + 0₂(g) →2MgO(s)
(b) C₃H₈(g) + 0₂(g) →C0₂(g) + H₂O(g)
(c) P₄(s) + 50₂(g) →P₄O₁₀(s)
(d) CH₄(g) + 20₂(g) → C0₂(g) + 2H₂0(g)

Sol: (b) C₃H₈ + 0₂ → C0₂ + H₂0

Since the reaction is not balanced, hence, mass of reactants and products are different. It is against the law of conservation of mass.

Q15. Which of the following statements indicates that law of multiple proportions is being followed?
(a) Sample of carbon dioxide taken from any source will always have carbon and oxygen in the ratio 1:2.
(b) Carbon forms two oxides namely C0₂ and CO, where masses of oxygen which combine with fixed mass of carbon are in the simple ratio 2:1.
(c) When magnesium bums in oxygen, the amount of magnesium taken for the reaction is equal to the amount of magnesium in magnesium oxide formed.
(d) At constant temperature and pressure 200 mL of hydrogen will combine with 100 mL oxygen to produce 200 mL of water vapour.
Sol: (b) The element, carbon, combines with oxygen to form two compounds, namely, carbon dioxide and carbon monoxide. In C0₂, 12 parts by mass of carbon combine with 32 parts by mass of oxygen while in CO, 12 parts by mass of carbon combine with 16 parts by mass of oxygen.
Therefore, the masses of oxygen combine with a fixed mass of carbon (12 parts) in C02 and CO are 32 and 16 respectively. These masses of oxygen bear a simple ratio of 32 : 16 or 2 : 1 to each other.
This is an example of law of multiple proportion.

More than One Correct Answer Type

Q16. One mole of oxygen gas at STP is equal to____ .
(a) 6. 022 x 10²³ molecules of oxygen
(b) 6.022 x 10²³ atoms of oxygen
(c) 16 g of oxygen .
(d) 32 g of oxygen

Sol:(a_,d) 1 mole of 0₂ gas at STP = 6.022 x 10²³ molecules of 0₂ (Avogadro number) = 32 g of 0₂. Hence, 1 mole of oxygen gas is equal to molecular weight of oxygen as well as Avogadro number.

Q17. Sulphuric acid reacts with sodium hydroxide as follows:

H₂S0₄ + 2NaOH →Na₂SO₄ + 2H₂0
When 1 L of 0.1 M sulphuric acid solution is allowed to react with 1 L of 0.1 M sodium hydroxide solution, the amount of sodium sulphate formed and its molarity in the solution obtained is
(a) 0.1 mol L^-1
(b) 7.10 g
(c) 0.025 mol L^-1
(d) 3.55 g

Sol: (b, c) Moles of H₂S0₄ taken = 0.1 moles
Moles of NaOH taken = 0.1 moles

As H₂S0₄ and NaOH react in ratio 1:2, so 0.1 moles of H₂S0₄ reacts with  0.2 mole of NaOH which we don’t have.

0.1 mole of NaOH reacts with 0.05 mole of H₂S0₄, so NaOH is Limiting reactant. Product is calculated w.r.t limiting reactant so Number of moles of Na₂S0₄ formed will also be equal to 0.05.
Mass of Na₂S0₄ = 0.05 x 142 = 7.1 g

Q18. Which of the following pairs have the same number of atoms?

Q19. Which of the following solutions have the same concentration?
(a) 20 g of NaOH in 200 mL of solution
(b) 0.5 mol of KC1 in 200 mL of solution
(c) 40 g of NaOH in 100 mL of solution
(d) 20 g of KOH in 200 mL of solution

Q20. 16 g of oxygen has the same number of molecules as in 
(a) 16 g of CO
(b) 28 g of N₂            
(c) 14g of N₂
(d) 1.0 gof H₂,

Q21. Which of the following terms are unitless?
(a) Molality
(b) Molarity
(c) Mole fraction
(d) Mass per cent

Q22. One of the statements of Dalton’s atomic theory is given below:
“Compounds are formed when atoms of different elements combine in a fixed ratio”
Which of the following laws in not related to this statement?
(a) Law of conservation of mass
(b) Law of definite proportions
(c) Law of multiple proportions
(d) Avogadro law
Sol:(a, d) The statement is related to law of definite proportions and law of multiple proportions.

Short Answer Type Questions
Q23. What will be the mass of one atom of C-12 ingrams?
Sol: 1 mole of carbon atoms = 12 g = 6.022 x 10²³ atoms. 6.022 x 10²³ atoms of carbon-12 have mass = 12 g

Q24. How many significant figures should be present in the answer of the following calculations?
2.5×1.25×3.5/ 2.01

division, the number of significant figures will be 2.
Since the least number of significant figures from the given figures is 2 (in 2.5 and 3.5) the result should not have more than two significant figures.

Q25. What is the symbol for SI unit of mole? How is the mole defined?
Sol: Symbol for SI unit of mole is mol. A mole is defined as the amount of substance that contains as-many entities as there are atoms in 12 g of carbon – in C-12 isotope.

Q26. What is the difference between molality and molarity?
Sol: Molality is the number of moles of solute present in 1 kg of solvent, whereas molarity is the number of moles of solute dissolved in 1 litre of a solution. Molality is independent of temperature, whereas molarity depends on temperature.

Q27. Calculate the mass per cent of calcium, phosphorus and oxygen in calcium phosphate Ca₃(P0₄)₂.

Q28. 4 L of dinitrogen reacted with 22.7 L of dioxygen and 45.4 L of nitrous oxide was formed. The reaction is given below:
2N₂(g) + 0₂(g) —> 2N₂0(g)
Which law is being obeyed in this experiment? Write the statement of the law.

Sol: Gases are reacting together to form gaseous products. Ratio of volumes of gases:
N₂: 0₂: N₂0 = 45.4 : 22.7 : 45.4                                              –
=2:1:2
Which is a simple whole number ratio. Hence the experiment proves Gay- Lussac’s law of gaseous volumes. This law states that gases combine or are produced in a chemical reaction in a simple whole number ratio by volume provided that all gases are at the same temperature and pressure.

Q29. If two elements can combine to’form more than one compound, the masses of one element that combine with a fixed mass of the other element, are in whole number ratio.
(a) Is this statement true?
(b) If yes, according to which law?
(c) Give one example related to this law
Sol: (a) Yes, the statement is true.
(b) According to law of multiple proportions.
(c) Hydrogen and oxygen react to produce two compounds, water and hydrogen peroxide. Masses of oxygen which combine with fixed mass of hydrogen are in simple ratio.

Masses of oxygen (16 and 32) which combine with fixed mass of hydrogen (2) are in the ratio of 16 : 32 or 1 : 2.

Q31. Hydrogen gas is prepared in the laboratory by reacting dilute HC1 with granulated zinc. Following reaction takes place:

Zn + 2HC1 → ZnCl₂ + H₂
Calculate the volume of hydrogen gas liberated at STP when 32.65 g of zinc reacts with HC1. 1 mol of a gas occupies 22.7 L volume of STP; atomic mass of Zn = 65.3 u.
Sol: Given that, mass of Zn = 32.65 g

1 mole of gas occupies = 22.7 L volume at STP Atomic mass of Zn = 65.3u
The given equation is

Q32. The density of 3 molal solution of NaOH is 1.110 g mL ^-1. Calculate the molarity of the solution.
Sol: 3 molal solution of NaOH means 3 moles of NaOH is dissolved in 1000 g of water.
But 3 moles of NaOH = 3 x 40 g = 120 g
120 g = 1120 g

Q33. Volume of a solution changes with change in temperature, then, will the molality of the solution be affected by temperature? Give reason for your answer.
Sol: No, molality of a solution does not change with temperature since mass remains unaffected by temperature.

Q34. If 4 g of NaOH dissolves in 36 g of H₂0, calculate the mole fraction of each component in the solution. Also, determine the molarity of solution (specific gravity of solution is 1 g mL^-1).

Q35. The reactant which is entirely consumed in reaction is known as limiting reagent. In the reaction 2A + 4B —> 3C + 4D, when 5 moles of A react with 6 moles of B, then

(i) which is the limiting reagent?
(ii) calculate the amount of C formed.

Matching Column Type Questions
36. Match the following :

Column I	Column II
A. 88 g of C02	a. 0.25 mol
B. 6.022 x 1023 molecules of H20	b. 2 mol
C. 5.6 litres of 02 at STP	c. 1 mol
D. 96 g of 02	d. 6.022 x 1023 molecules
E. 1 mole of any gas	e. 3 mol

 

Q37. Match the following physical quantities with units

Column I (Physical)	Column II (Unit)
(i) Molarity	a. g mL-1
(ii) Mole fraction	b. mol
(iii) Mole	c. Pascal
(iv) Molality	d. Unitless
(v) Pressure	e. mol L-1
(vi) Luminous intensity	f. Candela
(vii) Density               .	g. mol kg-1
(viii) Mass	h. N m-[1]
	i. kg

 

Sol: (i →e), (ii → d), (iii → b), (iv →g), (v —> c), (vi —» f), (vii → a), (viii → i)

(i) Molarity = mol L^-1
(ii) Mole fraction = no units
(iii) Mole = mol
(iv) Molality = mol kg ³
(v) Pressure = Pascal or N m ²
(vi)Luminous intensity = Candela
(vii) Density = g mL^-1
(viii) Mass = kg

Assertion and Reason Type Questions
In the following questions a statement of Assertion (A) followed by a statement of Reason (R) is given. Choose the correct option out of the choices given below each question.
38. Assertion (A): The empirical mass of ethene is half of its molecular mass. Reason (R): The empirical formula represents the simplest whole number ratio of various atoms present in a compound.
(a) Both A and R are true and R is the correct explanation of A.
(b) A is true but R is false
(c) A is false but R is true
(d) Both A and R are false.
Sol: (a) Empirical formula of ethene = CH₂
Empirical formula mass of ethene =14 amu

= 1/2 — x molecular mass of ethene .

Empirical formula shows that ethene has (C : H):: 1 : 2

Q39. Assertion (A): One atomic mass unit is defined as one-twelfth of the mass of one carbon-12 atom.
Reason (R): Carbon-12 isotope is the most abundant isotope of carbon and has been chosen as standard.

(a) Both A and R are the true and R is the correct explanation of A.
(b) Both A and R are true but R is not the correct explanation of A.
(c) A is true but R is false.
(d) Both A and R are false.

Sol: (b) 1 amu = —1/12 Mass of one C¹² atom
C¹² isotope is considered as standard for defining the atomic and molecular mass.

Q40. Assertion (A): Significant figures for 0.200 is 3 whereas for 200 it is 1.
Reason (R): Zero at the end or right of a number are significant provided they are not on the right side of the decimal point.
(a) Both A and R are true and R is correct explanation of A.
(b) Both A and R are true but R is not the correct explanation of A.
(c) A is true but R is false.
(d) Both A and R are false.
Sol: (c) Significant figures for 0.200 = 3
Significant figure for 200 = 1
Zeros at the end of a number without decimal- point, may or may not be significant depending on the accuracy of measurement.

 

Q41. Assertion (A): Combustion of 16 g of methane gives 18 g of water.
Reason (R): In the combustion of methane, water is one of the product.
(a) Both A and R are true but R is not the correct explanation of A.
(b) A is true but R is false
(c) A is false but R is true.
(d) Both A and R are false.

Sol: (c) CH₄(g) + 20₂(g) → C0₂(g) + 2H₂0(g)

16 g of CH₄ on complete combustion will give 36 g of water

Long Answer Type Question
Q42. A vessel contains 1.6 g of dioxygen at STP (273.15 K, 1 atm pressure). The gas is now transferred to another vessel at constant temperature, where pressure becomes half of the original pressure. Calculate
(i) volume of the new vessel.
(ii) number of molecules of dioxygen.

Q43. Calcium carbonate reacts with aqueous HC1 to give CaCl₂ and C0₂ according to the reaction given below:

CaC0₃(s) + 2HCl(aq) → CaCl₂(aq) + C0₂(g) + H₂O(l)

What mass of CaCl₂ will be formed when 250 mL of 0.76 M HC1 reacts with 1000 g of CaC0₃? Name the limiting reagent. Calculate the number of moles of CaCl₂ formed in the reaction.

Since HCl on complete consumption gives lesser amount of product hence HC1 will be limiting reagent and the number of moles of CaCl₂ formed will be 0.095 mol.

Q43. Define the law of multiple proportions. Explain it with two examples. How does this law point to the existence of atoms?
Sol: Law of multiple proportions: When two elements combine to form two or more chemical compounds, then the masses of one of the elements which combine with a fixed mass of the other, bear a simple ratio to one another, e.g., carbon combines with oxygen to form two compounds, namely, carbon dioxide and carbon monoxide.

The masses of oxygen which combine with a fixed mass of carbon in C0₂ and CO are 32 and 16 respectively. These masses of oxygen bear a simple ratio of 32 : 16 or 2 : 1 to each other. For example, sulphur combines with oxygen to form two compounds, namely, sulphur trioxide and sulphur dioxide.

The masses of oxygen which combine with a fixed mass of sulphur in S0₃ and S0₂ are 48 and 32 respectively. These masses of oxygen bear a simple ratio of 48 : 32 or 3 : 2 to each other. This law shows that there are constituents which combine in a definite proportion. These constituents may be atoms. Thus, the law of multiple proportions shows the existence of atoms which combine to form molecules.

Q44. A box contains some identical red coloured balls, labeled as A, each weighing 2 grams. Another box contains identical blue coloured balls, labeled as B, each weighing 5 grams. Consider the combinations AB, AB₂, A₂B and A₂B₃ and show that law of multiple proportions is applicable.

Sol:

Combination of A and B	AB	ab2	A,B	A2B3
Mass of A (in g)	2	2	4	4
Mass of B (in g)	5	10	5	15

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NCERT Exemplar Class 11 Chemistry Chapter 10 The S-Block Elements are part of NCERT Exemplar Class 11 Chemistry. Here we have given NCERT Exemplar Class 11 Chemistry Chapter 10 The S-Block Elements. https://www.cbselabs.com/ncert-exemplar-problems-class-11-chemistry-chapter-10-s-block-elements/

NCERT Exemplar Class 11 Chemistry Chapter 10 The S-Block Elements

Multiple Choice Questions
Single Correct Answer Type

Q1. The alkali metals are low melting. Which of the following alkali metals is expected to melt if the room temperature rises to 30°C?
(a) Na (b) K (c) Rb (d) Cs
Sol: (d) Among alkali metals, melting point decreases as the strength of metallic bonding decreases with increasing size of the atom. Thus, Cs has the lowest melting point (28.5°C) and will melt at 30°C.

Q2. Alkali metals react with water vigorously to form hydroxides and dihydrogen. Which of the following alkali metals reacts with water least vigorously?
(a) Li . (b) Na (c) K (d) Cs
Sol: (a) Both melting point and heat of reaction of alkali metals with water decrease down the group from Li to Cs. Although the heat of reaction of Li is the highest, but due to its high melting point, even this heat is not sufficient to melt the metal, which exposes greater surface to water for reaction. As a result, Li has the least reactivity but the reactivity increases as the melting point of alkali metals decreases down the group from Li to Cs.

Q3. The reducing power of a metal depends oh various factors. Suggest the factor which makes Li, the strongest reducing agent in aqueous solution.
(a) Sublimation enthalpy (b) Ionisation enthalpy
(c) Hydration enthalpy (d) Electron-gain enthalpy
Sol: (c) Lithium has highest hydration enthalpy which accounts for its high negative E° value and its high reducing power.

And we can write the lithium oxide formula on the basis of this, which is a balanced oxidation state.

Q4. Metal carbonates decompose on heating to give metal oxide and carbon dioxide. Which of the metal carbonates is most stable thermally?
(a) MgC0₃
(b)CaC0₃
(c)SrCQ₃                            
(d)BaC0₃

 Sol: (d) Thermal stability of metal carbonates increases as the electropositive character of the metal or the basicity of the metal hydroxide increases from Be(OH)₂ to Ba(OH)₂. Thus, BaC0₃ is the most stable.

Q5. Which of the carbonates given below is unstable in air and is kept in C0₂ atmosphere to avoid decomposition.
(a) BeCO₃
(b) MgC0₃
(c) CaC0₃
(d) BaCO₃
Sol: (a) Due to least electropositive character or least basicity of Be, BeC0₃ is less stable and hence decomposes to give BeO and C0₂.
BeC0₃→BeO + C0₂
Since the decomposition reaction is reversible, therefore, to increase the stability of BeC0₃ or to reverse the above equilibrium, BeC0₃ is kept in an atmosphere of C0₂.

Q6. Metals form basic hydroxides. Which of the following metal hydroxides is the least basic?
(a) Mg(OH)₂ (b) Ca(OH)₂               (c) Sr(OH)₂              (d) Ba(OH)₂
Sol: (a) As the ionization enthalpy increases from Mg →Ba, the M – O bond becomes weaker and weaker down the group and hence basicity increases down the group. Thus, Mg(OH)₂ is least basic.

Q7. Some of the Group 2 metal halides are covalent and soluble in organic solvents. Among the following metal halides, the one which is soluble in ethanol is
(a) BeCl₂
(b) MgCl₂                  
(c) CaCl₂                
(d) SrCl₂

Sol: Due to small size, high electronegativity and high ionization enthalpy of Be, BeCl₂ is covalent and hence most soluble in organic solvents such as ethanol.

Q8. The order of decreasing ionization enthalpy in alkali metals is

(a) Na > Li > K > Rb (b) Rb < Na < K < Li

(c) Li > Na > K > Rb (d) K < Li < Na < Rb

Sol: (c) Ionization enthalpy decreases with increase in Size of the atom in a group. Hence, the order is:

Li > Na > K > Rb.

Q9. The solubility of metal halides depends on their nature, lattice enthalpy and hydration enthalpy of the individual ions. Amongst fluorides of alkali metals, the lowest solubility of LiF in water is due to
(a) ionic nature of lithium fluoride. . .
(b) high lattice enthalpy. ‘
(c) high hydration enthalpy for lithium ion.
(d) low ionization enthalpy of lithium atom.
Sol: (b) Due to small size of Li⁺ and F^– ions, lattice enthalpy is much higher than hydration enthalpy and hence LiF is least soluble among alkali metal fluorides.

Q10. Amphoteric hydroxides react with both alkalies and acids. ‘Which of the following group 2 metal hydroxides is soluble in sodium hydroxide?
(a) Be(OH)₂
(b) Mg(OH)₂
(c) Ca(OH)₂
(d) Ba(OH)₂
Sol: (a) Be(OH)₂ reacts with NaOH to give beryllate ion, becoming soluble in it. Be(OH)₂ + 20H^–→[Be(OH)₄]²

Q11. In the synthesis of sodium carbonate, the recovery of ammonia is done by treating NH₄C1 with Ca(OH)₂. The by-product obtained in this process is
(a) CaCl₂
(b) NaCl  
(c) NaOH
(d) NaHC0₃
Sol: (a) Sodium carbonate is synthesised by Solvay or ammonia soda process. The reactions involved are

Q12.When sodium is dissolved in liquid ammonia, a solution of deep blue colour is obtained. The colour of the solution is due to
(a) ammoniated electron                     
(b) sodium ion
(c) sodium amide                                   
(d) ammoniated sodium ion
Sol: (a)M+(x+y)NH₃→ M⁺(NH₃)_x + e^–(NH₃)_y
The colour of solution (deep blue) is due to the ammoniated electron which absorbs energy in the visible region.

Q13. By adding gypsum to cement
(a) setting time of cement becomes less.
(b) setting time of cement increases.
(c) colour of cement becomes light.
(d) shining surface is obtained.
Sol: (b) Raw materials for cement are limestone, clay and gypsum. Cement is a dirty greyish heavy powder containing calcium aluminates and silicates. Gypsum (CaS0₄ -2H₂0) is added to the components to increase the setting time of cement so that it gets sufficiently hardened. Setting of cement is an exothermic process and involves hydration of calcium aluminates and , silicates.

Q14. Dead burnt plaster is

On heating plaster of Paris at 200°C, it forms anhydrous calcium sulphate, i.e., dead plaster which has no setting property as it absorbs water very slowly.

Q15. Suspension of slaked lime in water is known as
(a) lime water                                          
(b) quick lime
(c) milk of lime                                       
(d) aqueous solution of slaked lime
Sol: (c) Suspension of slaked lime in water is known as milk of lime.

Q16. Which of the following elements does not form hydride by direct heating with dihydrogen?
(a) Be                     
(b) Mg                     
(c) Sr                        
(d) Ba
Sol: (a) Due to high ionization enthalpy and small size, Be does not react with hydrogen by direct heating.

Q17. The formula of soda ash is
(a)    NaHCO₃.10H₂O
(b)Na2C0₃.2H₂0
(c) Na₂C0₃.H₂0
(d) Na₂C0₃   

Q18. A substance which gives brick red flame and breaks down on heating to give oxygen and brown gas is
(a)    Magnesium nitrate                       
(b)     Calcium nitrate
(c)     Barium nitrate                               
(d)     Strontium nitrate
Sol: (b) 2Ca(N0₃)₂→2CaO + 4N0₂ + 0₂
N0₂ is a brown gas. Ca²⁺ imparts brick red colour to the flame.

Q19.Which of the following statements is true about Ca(OH)₂?
(a) It is used in the preparation of bleaching powder.
(b) It is a light blue solid.
(c) It does not possess disinfectant property.
(d) It is used in the manufacture of cement.

Q20. A chemical A is used for the preparation of washing soda to recover ammonia. When C0₂ is bubbled through an aqueous solution of A, the solution tons milky. It is used in white washing due to disinfectant nature. What is the chemical formula of A?
(a) Ca(C0₃)₂          
(b) CaO                   
(c) Ca(OH)₂
(d) CaC0₃

Q21. Dehydration of hydrates of halides of.calcium, barium and strontium, i.e., CaCl₂.6H₂0, BaCl₂.2H₂0, SrCl₂.2H₂0, can be achieved by heating. These become wet oh keeping in air. Which of the following statements is correct about these halides?
(a) Act as dehydrating agents.
(b) Can absorb moisture from air.
(c) Tendency to form hydrate decreases from calcium to barium.
(d) All of the above.
Sol: (d) Chlorides of alkaline earth metals are hydrated salts. Due to their hygroscopic nature, they can be used as dehydrating agents to absorb moisture from air.

Extent of hydration decreases from Mg to Ba, i.e., MgCl₂.6H₂0, CaCl₂.6H₂0, BaCl₂ 2H₂0, SrCl₂.2H₂0.

Q22. Metallic elements are described by their standard electrode potential, frision enthalpy, atomic size, etc. The alkali metals are characterized by which of the following properties?
(a) High boiling point. ‘
(b) High negative standard electrode potential.
(c) High density.
(d) Large atomic size.
Sol: (b, d) Alkali metals have high negative standard electrode potential and large atomic size.

Q23. Several sodium compounds find use in industries. Which of the following compounds are used for textile industry?
(a) Na₂C0₃            
(b) NaHC0₃            
(c) NaOH               
(d) NaCl
Sol: (a, c) Na₂C0₃ and NaOH are used in textile industry.         .

Q24. Which of the following compounds are readily soluble in water?
(a) BeS0₄              
(b) MgS0₄              
(c) BaS0₄                
(d) SrS0₄
Sol: (a, b) Solubility decreases down the group because hydration enthalpy decreases more rapidly than lattice enthalpy. Thus, BeS0₄ and.MgS0₄ are soluble.

Q25. When zeolite, which is hydrated sodium aluminium silicate, is treated with hard water, the sodium ions are exchanged with which of the following ion(s)?
(a) H⁺ions                                               
(b) Mg²⁺ions’
(c) Ca²⁺ ions                                           
(d) SO²₄^_ ions
Sol: (b, c) Sodium zeolite removes Ca²⁺ and Mg²⁺ ions from hard water. Na₂Z + M²⁺ → MZ + 2Na⁺ (M = Ca, Mg) where, Z = Al₂Si₂0₈H₂0

Q26. Identify the correct’ formula of halides of alkaline earth metals from the following.
(a) BaCl₂.2H₂0
(b) BaCl₂ .4H₂0
(c) CaCl₂ . 6H₂0
(d) SrCl₂.4H₂0
Sol:(a, c) Tendency to form halide hydrates gradually decreases down the group. The hydrates are MgCl₂.6H₂0, CaCl₂.6H₂0, SrCl₂.6H₂0 and BaCl₂.2H₂0.

Q27. Choose the correct statements from the folio-wing.
(a) Beryllium is not readily attacked by acids because of the presence of an oxide film on the surface of the metal.
(b) Beryllium sulphate is readily soluble in water as the greater hydration enthalpy of Be²⁺ overcomes the lattice enthalpy factor.
(c) Beryllium exhibits coordination number more than four.
(d) Beryllium oxide is purely acidic in nature.
Sol: (a, b) Be does not exhibit coordination number more than four and BeO is amphoteric in nature.

Q28. Which of the following are the correct reasons for anomalous behaviour of lithium?
(a) Exceptionally small size of its atom.
(b) Its high polarizing power.
(c) It has high degree of hydration.
(d) Exceptionally low ionization enthalpy.
Sol: (a, b) Anomalous behaviour of Li is due to its exceptionally small size and high polarizing power.

Short Answer Type Questions
Q29. How do you account for the strong reducing power of lithium in aqueous solution? .
Sol: Electrode potential is a measure of the tendency of an element to lose electrons in the aqueous solution. It mainly depends upon the following three factors
(i) Sublimation enthalpy
(ii) Ionization enthalpy
(iii) Enthalpy of hydration
The sublimation enthalpies of alkali metals are almost similar. Since Li has the smallest size, its enthalpy of hydration is the highest among alkali metals. Although ionization enthalpy of Li is the highest among alkali metals, it is more than compensated by the high enthalpy of hydration. Thus, Li has the most negative standard electrode potential (-3.04 V) and hence Li is the strongest reducing agent in aqueous solution mainly because of its high enthalpy of hydration.

Q30. When heated in air, the alkali metals form various oxides. Mention the oxides formed by Li, Na and K.

Q32. Lithium resembles magnesium in some of its properties. Mention two such properties and give reasons for this resemblance.
Sol:(i) Both Li and Mg are harder and lighter than other elements in their groups.
(ii) Both form ionic nitrides Li₃N and Mg₃N₂ by heating in an atmosphere of nitrogen.
Li resembles Mg due to similar atomic radii and ionic radii.

Q33. Name an element’ from Group 2 which forms an amphoteric oxide and awater soluble sulphate.                   Sol: Due to small size and somewhat high ionization enthalpy of Be, Be(OH)₂ is amphoteric in nature, i.e., it reacts with both acids and bases. Further, due to small size, the hydration enthalpy of Be²⁺ ions is much higher than the lattice enthalpy of BeS0₄. As a result, BeS0₄ is highly soluble in water.

Q34. Discuss the trend of the following:
(i) Thermal stability of carbonates of Group 2 elements.
(ii) The solubility and the nature of oxides, of Group 2 elements.
Sol: (i) All the alkaline earth metals form carbonates (MC0₃). All these carbonates decompose on heating to give C0₂ and metal oxide. The thermal stability; of these carbonates increases down the group, i.e., from Be to Ba,
BeC0₃ < MgC0₃ < CaC0₃ < SrC0₃ < BaC0₃
BeC03 is unstable to the extent that it is stable only in atmosphere of C02. It however shows reversible decomposition in closed container

BeC0₃ ⇌BeO + C0₂

Hence, more is the stability of oxide formed, less will be stability of carbonates. Stability of oxides decreases down the group. Since beryllium oxide is high stable, it makes BeC0₃ unstable.

(ii) All the alkaline earth metals form oxides of formula MO. The oxides are very stable due to high lattice energy and are used as refractory material. Except BeO (predominantly covalent), all other oxides are ionic and their lattice energy decreases as the size of cation increases.
The oxides are basic and basic nature increases from BeO to BaO (due to increasing ionic nature).

BeO dissolves both in acid and alkalies to give salts and is amphoteric.
The oxides of the alkaline earth metals (except BeO and MgO) dissolve in water to form basic hydroxides and evolve a large amount of heat. BeO and MgO possess high lattice energy and thus are insoluble in water.

Q35. Why are BeS0₄ and MgS0₄ readily soluble in water while CaS0₄, SrS0₄ and BaS0₄ are insoluble?
Sol: The hydration enthalpies of BeS0₄ and MgS0₄ are quite high because of small size of Be²⁺ and Mg²⁺ ions. These hydration enthalpy values are higher than their corresponding lattice enthalpies and therefore, BeS0₄ and MgS0₄ are highly soluble in water. However, hydration enthalpies of CaS0₄, SrS0₄ and BaS0₄ are not very high as compared to their respective lattice enthalpies and hence these are insoluble in water.

Q36. All compounds of alkali metals are easily soluble in water but lithium compounds are more soluble in organic solvents. Explain.
Sol: Because of the small size, high electronegativity and high ionization enthalpy, lithium compounds have considerable covalent character while compounds of other alkali metals are ionic in nature. As a result, compounds of lithium are more soluble in organic solvents while those of other alkali metals are more soluble in water.

Q37. In the Solvay process, can we obtain sodium carbonate directly by treating the solution containing (NH₄)₂C0₃ with sodium chloride? Explain.
Sol: No. In Solvay process, ammonium hydrogencarbonate is prepared from ammonium carbonate, which then reacts with sodium chloride to form sodium hydrogencarbonate. Due to low solubility of NaHC0₃, it gets precipitated and decomposes on heating to give Na₂C0₃.
We cannot obtain sodium carbonate directly by treating the solution containing (NH₄)₂C0₃ with sodium chloride as both the products formed on reaction, i.e., Na₂C0₃ and NH₄C1 are soluble and the equilibrium will not shift in forward direction.
(NH₄)₂C0₃ + 2NaCl⇌Na₂C0₃ + 2NH₄Cl

Q39. Why do beryllium and magnesium not impart colour to the flame in the flame test?

Sol: All alkaline earth metals (except Be and Mg) impart a characteristic colour to the Bunsen flame. The different colours arise due to different energies . required for electronic excitation and de-excitation.
Be and Mg atoms, due to their small size, bind their electrons more strongly (because of .higher effective nuclear charge). Hence, they require high excitation energy and are not excited by the energy of the flame with the result that no flame colour is shown by them.

Q40.What is the structure of BeCl₂ molecule in gaseous and solid state? Beryllium chloride has different structures in solid and vapour state. In solid state, it exists in the form of polymeric chain structure in which each Be- atom is surrounded by four chlorine atoms, having two of the chlorine atoms covalently bonded while the other two by coordinate bonds. The resulting bridge structure contains infinite chains.

In vapour state, above 1200 K, it exists as a monomer having linear structure and zero dipole moment. But below 1200 K, it exists as dimer even in the vapour state.

Matching Column Type Questions
In the following questions, more than one option of column I and II may be correlated.
Q41.Match the elements given in Column I with the properties mentioned in Column II.

Column I	Column II
(i) Li	(a) Insoluble sulphate
(ii) Na	(b) Strongest monoacidic base
(iii) Ca	(c) Most negative E° value among alkali metals
(iv) Ba	(d) Insoluble oxalate                                     ‘
	(e) 6s2 outer electronic configuration

 

Sol:(i → c); (ii → b); (iii → d); (iv → a, e)

(i) Li – Most negative E° among alkali metals
[Due to very high hydration energy the resulting E° is most negative].
(ii) Na – Strongest monoacidic base
[Alkalies are more acidic than alkaline earth metals. LiOH has covalent character], –
(iii) Ca – insoluble oxalate
[Calcium oxalate is insoluble in water.]
(iv) Ba – Insoluble sulphate
[Hydration energy decreases as size of cation increases].
6s² outer electronic configuration
₅₆Ba = Is²,2s²,2p,3s², 3p⁶, 3d¹⁰,4s²,4p⁶,4d¹⁰, 5s², 5p⁶,6s²

Q42. Match the compounds given in Column I with their uses mentioned in Column II.

Column I	Column II
(i) CaC03	(a) Dentistry, ornamental work
(ii) Ca(OH)2	(b) Manufacture of sodium carbonate from caustic soda
(iii) CaO	(c) Manufacture-of high quality paper
(iv) CaS04	(d) Used in white washing

Sol: (i → c); (ii →d); (iii → b); (iv→a)
(i) CaC0₃ – Manufacture of high quality paper
(ii) Ca(OH)₂ – Used in white washing
(iii) CaO – Manufacture of sodium carbonate from caustic soda
(iv) CaS0₄ – Dentistry, ornamental work

Q43. Match the elements given in Column I with the colour they impart to the flame given in Column II.

Column I	Column II
(i) Cs	(a) Apple green
(ii) Na	(b) Violet
(iii) K	(c) Brick red

(iv) Ca	(d) Yellow
(v) Sr	(e) Crimson red
(vi) Ba	(f) Blue

 

Sol: (i → f); (ii →d); (iii → b); (iv → c); (v →e); (vi → a)

Elements with the characteristic flame colour are as follows ‘
(i) Cs – Blue
(ii) Na-Yellow
(iii) K- Violet
(iv) Ca-Brick red
(v) Sr – Crimson red
(vi) Ba – Apple green

Flame colours are produced from the movement of the electrons in the metal ions present in the compounds. These movements of electrons (electronic excitation and de-excitation) requires energy.
Each atom has particular energy gap between ground and excited energy ‘ level. Therefore, each of these movements involves a specific amount of energy emitted as light energy, and each corresponds to a particular colour. As we know energy gap between ground and excited state energy level increases, wavelength of light absorbed decreases and complementary colour is observed.

 

Assertion and Reason Type Questions

In the following questions a statement of Assertion (A) followed by a statement of Reason (R) is given. Choose the correct option out of the choices given below each question.
Q44. Assertion (A): The carbonate of lithium decomposes easily on heating to , form lithium oxide and C0₂.
Reason (R): Lithium being very small in size polarizes large carbonate ion leading to the formation of more stable Li₂0 and C0₂.
(a) Both A and R are true and R is the correct explanation of A.
(b) Both A and R are true but R is not the correct explanation of A.
(c) Both A and R are not correct.
(d)A is hot correct but R is correct.
Sol: (a) Unlike other alkali metal carbonates, the carbonate of lithium decomposes on heating to form its oxide. Its oxide is stablised by polarization.

Q44. Assertion (A): Beryllium carbonate is kept in the atmosphere of carbon dioxide.
Reason (R): Beryllium carbonate is unstable and decomposes to give beryllium oxide and carbon dioxide:
(a) Both A and Rare true and R is the correct explanation of A.
(b) Both A and R are true but R is not the correct explanation of A.
(c) Both A and R are not correct.
(d)A is not correct but R is correct.
Sol: (a) BeC0₃ is kept in the atmosphere of C0₂, otherwise it will decompose to give its oxide and carbon dioxide.

Long Answer Type Questions

Q46. The s-block elements are characterized by their larger atomic sizes, lower ionization enthalpies, invariable +1 oxidation state and solubilities of their oxosalts. In the light of these features, describe the nature of their oxides, halides and oxosalts.
Sol: (i) Nature of oxides – Alkali metals form M₂0, M₂0₂ and M0₂ types of oxides. The stability of the peroxide or superoxide increases as the size of metal cation increases. This is due to the stabilization of large anions by larger cations.
(ii) Nature of halides – Alkali metal halides have general formula MX. All halides are soluble in water. LiF is very less soluble in water due to its high lattice energy. Their melting points and boiling points follow the trend – fluoride > chloride > bromide > iodide. This is because with increase in size of the halide ion, lattice energy increases.
(iii) Oxosalts – Oxosalts of alkali metals are generally soluble in water and thermally stable. As electropositive character increases down the group, stability of carbonates and bicarbonates increases.

Q47. Present a comparative account of the alkali and alkaline earth metals with respect to the following characteristics:
(a) Tendency to form ionic/covalent compounds
(b) Nature of oxides and their solubility in water
(c) Formation of oxosalts
(d) Thermal stability of oxosalts

Alkali metals	Alkaline earth metals
(i) All alkali metals except Li form ionic compounds.	(i) All alkaline earth metals except Be form ionic compounds.
(ii) The solubility of oxides of al­kali metals increases down the group. The basic character of the ox­ides increases down the group	(ii) The solubility of oxides of Mg, – Ca, Sr and Ba increases from Mg to Ba. BeO, however, is covalent and insoluble in water. The basic character of oxides in­creases from MgO to BaO. BeO is, however, amphoteric.
(iii) All alkali metals form oxo salts such as carbonates, sul­phates and nitrates.	(iii) All alkaline earth metals form oxo salts such as carbonates, sulphates and nitrates.
(iv) Solubility of carbonates and sul­phates increases down the group.	(iv) Solubility of carbonates and sul­phates decreases down the group.
(v) Carbonates and sulphates of Li decompose on heating while the stability, of carbonates and sulphates of other metals in­creases down the group.	(v) The carbonates and sulphates of alkaline earth metals all decom­pose on heating but the tempera­ture of their decomposition in­creases down the group, i.e., their thermal stability increases.

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NCERT Exemplar Class 11 Chemistry Chapter  14 Environmental Chemistry

Multiple Choice Questions
Single Correct Answer Type

Q1. Which of the following gases is not a greenhouse gas?
(a) CO
(b) 0₃                        
(c) CH₄                    
(d) H₂0 vapour
Sol:(a) The gases which absorb solar energy near the earth’s surface and then radiate it back to the earth are called greenhouse gases.

Q2. Photochemical smog occurs in warm, dry and sunny climate. One of the following is not amongst the components of photochemical smog, identify it.
(a) N0₂
(b) 0₃
(c) S0₂                                                       
(d) Unsaturated hydrocarbon

Sol: (c) The smog which is formed in the presence of sunlight is called photochemical smog. This occurs in the months of summer when N0₂ and hydrocarbons are present in large amounts in the atmosphere. Concentration of 0₃, PAN, aldehydes and ketones increases up in the atmosphere. S0₂ is not responsible for photochemical smog.

Q3. Which of the following statements is not true about classical smog?
(a) Its main components are produced by the action of sunlight on emissions of automobiles and factories.
(b) Produced in cold and humid climate.
(c) It contains compounds of reducing nature.
(d) It contains smoke, fog and sulphur dioxide.
Sol: (a) Classical smog is initiated by a mixture of S0₂, particulates and high humidity in the atmosphere in cold conditions. A fog of H₂S0₄ droplets formed condenses on the particulates to form the smog. It is of reducing nature. The gases released by automobiles and factories are not responsible for classical smog.

Q4. Biochemical Oxygen Demand, /BOD) is a measure of organic material present in water. BOD value less than 5 ppm indicates a water sample to be
Sol: (a) The total amount of oxygen consumed by micro-organisms (bacteria) in decomposing organic matter present in certain volume of a sample of water is called Biochemical Oxygen Demand (BOD) of the water.
Water is considered to be pure if it has BOD less than 5 ppm, whereas highly polluted water has BOD more than 17 ppm. Thus, the water having BOD less than 5 ppm is rich in dissolved oxygen.

Q5. Which .of the following statements is wrong? . ‘
(a) Ozone is not responsible for greenhouse effect.
(b) Ozone can oxidize sulphur dioxide present in the atmosphere to sulphur trioxide.
(c) Ozone hole is thinning of ozone layer present in stratosphere.
(d) Ozone is produced in upper stratosphere by the action of UV rays on oxygen .
Sol: (a) Ozone is also one of the greenhouse gases.

Q6. Sewage containing organic waste should not be disposed in water bodies because it causes major water pollution. Fishes in such a polluted water die because of
(a) large number of mosquitoes.
(b) increase in the amount of dissolved oxygen. .
(c) decrease in the amount of dissolved oxygen in water.
(d) clogging of gills by mud.
Sol: (c) Organic waste consumes oxygen and therefore, dissolved oxygen in water decreases and fish in such polluted water die.

Q7. Which of the following statements about photochemical smog is wrong?
(a) It has high concentration of oxidizing agents.
(b) It has low concentration of oxidizing agent.
(c) It can be controlled by controlling the release of N0₂, hydrocarbons, ozone etc.
(d) Plantation of some plants like pinus helps in controlling photochemical
Sol: (b) Photochemical smog is oxidizing in nature because it contains large concentration of oxidizing agents N0₂ and 0₃.

Q8. The gaseous envelope around the earth is known as atmosphere. The lowest layer of this is extended up to 10 km from sea level. This layer is .
(a) Stratosphere
(b) Troposphere
(c) Mesosphere
(d) Hydrosphere
Sol: (b) The atmosphere is divided into four major regions:
(i) Troposphere (ii) Stratosphere . (iii) Mesosphere and (iv) Thermosphere Troposphere is the lowest region of the atmosphere.

Q9. Dinitrogen and dioxygen are main constituents of air but these do not react with each other to form oxides of nitrogen because____________ .
(a) the reaction is endothermic and requires very high temperature.
(b) the reaction can be initiated only in presence of a catalyst.
(c) oxides of nitrogen are unstable.
(d) N₂ and 0₂ are unreactive,
Sol: (a) Major compounds of atmosphere are dinitrogen, dioxygen and water vapour.
N₂ = 78.08%, 0₂ = 20.95%
Both dinitrogen and dioxygen do not react with each other as nitrogen is an inactive gas. The triple bond in N₂ is very stable and its dissociation energy is very high. Both react with each other at very high temperature.

Q10. The pollutants which come directly in the air from source are called primary pollutants. Primary’pollutants are sometimes converted into secondary pollutants. Which of the following belongs to secondary air pollutants?
(a) CO
(b) Hydrocarbon
(c) Peroxyacetyl nitrate                       
(d) NO
Sol: (c) Hydrocarbons present in atmosphere combine with oxygen atom produced by the photolysis of N0₂ to form highly reactive intermediate called free  radical. Free radical initiates a series of reactions.Peroxyacetyl nitrate is formed, which Gan be said as secondary pollutant.

Q11. Which of the following statements is correct?
(a) Ozone hole is a hole formed in stratosphere from which ozone oozes out.
(b) Ozone hole is a hole formed in the troposphere from which ozone oozes out.                                                                         .
(c) Ozone hole is thinning of ozone layer of stratosphere at some places.
(d) Ozone hole means vanishing of ozone layer around the earth completely.
Sol: (c) Two types of compounds have been found to be the most responsible for depleting the ozone layer. These are:
(i) NO and (ii) Chlorofluorocarbons.

Q12. Which of the following practices will not come under green chemistry?
(a) If possible, making use of soap made of vegetable oils instead of using synthetic detergents.
(b) Using H₂0₂ for bleaching purpose instead of using chlorine based bleaching agents.
(c) Using bicycle for traveling small distances instead of using petrol/diesel based vehicles.
(d) Using plastic cans for neatly storing substances.
Sol: (d) Plastic is non-biodegradable polymer. Hence, it does not come under green chemistry. Green chemistry includes processes which lead to minimum pollution and less harm to the environment.

More than One Correct Answer Type

Q13. Which of the following conditions shows the polluted environment?
(a) pH of rain water is 5.6.
(b) Amount of carbon dioxide in the atmosphere is 0.03%.
(c) Biochemical oxygen demand is 10 ppm.
(d) Eutrophication
Sol: (c, d) Polluted water may contain nutrients for the growth of algae, which covers the water surface and reduces the oxygen concentration in water. This leads to anaerobic condition, accumulation of obnoxious decay and animal death. This is process of eutrophication.

The amount of oxygen required by bacteria to break down the organic matter present in a certain volume of sample of water is called Biochemical Oxygen Demand. Clean water would have BOD value of 5 ppm whereas highly polluted could have BOD value of 17 ppm or more.

Normally rain water has pH of 6 due to H⁺ ion formed by reaction of rain water with carbon dioxide in the atmosphere. When the pH of the rain water drops below 5.6, it is called acid rain.

Q14. Phosphate containing fertilizers cause water pollution. Addition of such compounds in water bodies causes
(a) enhanced growth of algae
(b) decrease in amount of dissolved oxygen in water
(c) deposition of calcium phosphate
(d) increase in fish population
Sol: (a, b) Fertilizers containing phosphate present in water help in the excessive growth of aquatic plants and algae. Micro-organisms which decompose these plants consume oxygen. As a result, the amount of dissolved oxygen in the water decreases.

Q15. The acids present in acid rain are________ .
(a) Peroxyacetylnitrate                        
(b) H₂C0₃
(c) HN0₃                                                  
(d) H₂S0₄

Sol: (b, c, d) C0₂ is slightly soluble in water forming carbonic acid.
 Co₂ + H₂O → H₂Co₃
The oxides of nitrogen undergo oxidation reaction followed by reaction with water vapours to form nitric acid.

2NO + 0₂→ 2N0₂
2N0₂ + H₂0 → HN0₃ + HNO₂

The oxidation of S0₂ into S0₃ occurs in the presence of dust particles or metal ions. The S0₃ then reacts with water vapours to form H₂S0₄.

 

Q16. The consequences of global warming may be .
(a) increase in average temperature of the earth
(b) melting of Himalayan Glaciers –
(c) increased biochemical oxygen demand
(d) eutrophication
Sol: (a, b) The rate at which solar radiation is reaching the earth is constant but the amount of C0₂ in the air is increasing. Consequently, the heat radiated back to the earth will increase and the temperature of the earth surface will also increase.
This increase in temperature will disturb the thermal balance on the earth and could cause glacier  and ice caps to melt.

Short Answer Type Questions
Q17. Greenhouse effect leads to global warming. Which substances are responsible for greenhouse effect?
Sol: The heating of earth due to trapping of radiation is called greenhouse effect. The gases such as C0₂, CH₄, N₂0, CFC1₃, CF₂C1₂,0₃ etc. trap these radiations and are called greenhouse gases.

Q18. Acid rain is known to contain some acids. Name these acids. From where do they come in rain?
Sol: Acid rain contains acids such as HN0₃, H₂S0₄ and H₂C0₃ (along with small amount of HC1).
HN0₃ is formed by the oxidation of NO present in air to N0₂ and N0₃ and subsequent dissolution in water. H₂S0₄ is formed by the oxidation of S0₂ present in air to S0₃ and subsequent dissolution in water.
H₂C0₃ is formed by the dissolution of C0₂ of the air in water.

Q19. Ozone is a toxic gas and is a strong oxidizing agent, even then its presence in the stratosphere is very important. Explain what would happen if ozone from this region is completely removed.
Sol: Ozone layer acts as a protective umbrella and does not allow the harmful UV radiations to reach the earth’s surface. If ozone is completely removed from the stratosphere, the UV radiations will fall directly on the humans causing skin cancer and on the plants affecting plant proteins.
Q20. Dissolved oxygen in water is v£ry important for aquatic life. What processes are responsible for the reduction of dissolved oxygen in water?
Sol: The discharge of human sewage and organic waste from pulp and paper
industry and presence of leaves, grass, trash etc. in water due to run off result ‘ in phytoplankton growth. The microorganisms which decompose this organic matter need oxygen. Hence, the amount of oxygen in water of lakes etc. decreases.

Q21. On the basis of chemical reactions involved, explain how do chlorofluoro- carbons cause thinning of ozone layer in stratosphere.
Sol: Chlorofluorocarbons are stable compounds. They move to stratosphere by random diffusion. These undergo decomposition in the presence of sunlight to release Cl atoms. These Cl atoms cause catalytic chemical reactions and cause significant depletion of ozone layer as shown below:

Since the free radicals use ozone and convert it to oxygen, they cause thinning of ozone layer in stratosphere.

Q22. What could be the harmful effects of improper management of industrial and domestic solid waste in a city?
Sol: If domestic waste in a city is not properly managed, it may find its way into . sewers or may be eaten up by the cattle. The non-biodegradable waste like polythene bags, metal scrap etc. choke the sewers. The polythene bags, if swallowed by the cattle, can result into their death. Similarly, if industrial waste is not properly managed, it will cause pollution of the air, soil and water.

Q23.During an educational trip, a student of Botany saw a beautiful lake in a , village. She collected many plants from that area. She noticed that villagers were washing clothes around the lake and at some places, waste material from houses was destroying its beauty. After few years, she visited the same r lake again. She was surprised to find that the lake was covered with algae, stinking smell was coming out and its water had become unusable. Can you explain the reason for this condition of the lake?
Sol: Disposing of waste material and washing clothes in lake water makes the water rich in nutrients like phosphate. It enhances algae growth. Such profuse ‘ algal growth covers the water surface which reduces oxygen concentration in water. This leads to anaerobic conditions with accumulation of dead and decaying water animals, thus, leaving the water with stinking smell and making it unusable.

Q24. What are biodegradable and non-biodegradable pollutants? ,
Sol: Biodegradable pollutants are those which can be decomposed by bacteria. For example, dust particles, sewage, cow dung etc. Non-biodegradable pollutants are those which cannot be decomposed by bacteria. For example, plastic materials, mercury, aluminium, DDT, etc.

Q25. What are the sources of dissolved oxygen in water?
Sol: Sources of dissolved oxygen in water are (i) Photosynthesis (ii) Natural aeration (iii) Mechanical aeration.

Q26. What is the importance of measuring BOD of a water body?
Sol: BOD is the measure of level of pollution caused by organic biodegradable material in terms of how much oxygen will be required to break down the organic material biologically. Clean water would have BOD values less than 5 ppm while highly polluted water could have a BOD value of 17 ppm or more.

Q27. Why does water covered with excessive algal growth become polluted?
Sol: Presence of excessive algal growth shows that water contains a lot of
phosphate due to inflow of fertilizers, etc from the surroundings. Hence, such a sample of water is polluted.

Q28. A factory was started near a village. Suddenly villagers started feeling the presence of irritating vapours in the village and cases of headache, chest pain, cough, dryness of throat and breathing problems increased. Villagers blamed the emissions from the chimney of the factory for such problems. Explain what could have happened. Give chemical reactions for the support of your explanation.
Sol:The symptoms observed in the villagers show that oxides of nitrogen and sulphur must be coming out of the chimney. This is due to combustion of fossil fuels like coal, oil, natural gas, gasoline, etc. to produce high temperatures at which oxidation of atmospheric nitrogen takes place forming NO and N0₂:

S0₂ is produced due to combustion of sulphur containing coal and fuel oil or roasting of sulphide ores like iron pyrites (FeS₂), copper pyrites (CuFeS₂), etc.

Cu₂S + 0₂→2Cu + S0₂

Q29. Oxidation of sulphur trioxide in the absence of a catalyst is a slow process but this oxidation occurs easily in the atmosphere. Explain how does this happen. Give chemical reactions for the conversion of S0₂ into S0₃.
Sol: The presence of particulate matter in polluted air catalyses the oxidation of S0₂ to S0₃. The reaction is also promoted by ozone and hydrogen peroxide.

Q30. From where does ozone come in the photochemical smog?
Sol: When fossil fuels are burnt, nitric oxide and hydrocarbons from unburnt fuels are produced. In sunlight, nitric oxide is converted to nitrogen dioxide. N0₂  absorbs energy from sunlight and breaks up into NO and free oxygen atoms which are very reactive and combine with 0₂ to form 0₃, which reacts with NO to form N0₂ and 0₂.

Q31. How is ozone produced in stratosphere?
Sol: The formation of ozone in the stratosphere takes place in two steps. In the first step, ultraviolet radiation coming from the sun have sufficient energy to split dioxygen into two oxygen atoms. In the second step, the oxygen atoms react with more of dioxygen to form ozone.

Q32. Ozone is a gas heavier than air. Why does ozone layer not settle down near the earth?
Sol: In stratosphere, the formation of 0₃ gas goes on continuously, but 0₃ is also decomposed by UV radiation between 240-360 nm.

The O-atom reacts with second 0₃ molecule
0₃ + 0 -+ 20₂
Net reaction 20₃ —> 30₂
Thus, the reaction forms a delicate balance in which the rate of 0₃ decomposition matches the rate of 0₃ formation, i.e., a dynamic equilibrium exists and maintains a constant concentration of 0₃.

Q33. Sometime ago formation of polar stratospheric clouds was reported over Antarctica. Why were these formed? What happens when such clouds break up by warmth of sunlight?
Sol: In summer season, nitrogen dioxide and methane react with chlorine monoxide and chlorine atoms forming chlorine sinks, preventing much ozone depletion, whereas in winter, special type of clouds called polar stratospheric clouds are formed over Antarctica. These polar stratospheric clouds provide surface on which chlorine nitrate gets hydrolysed to form hypochlorous acid. It also reacts with hydrogen chloride to give molecular chlorine.

When sunlight returns to the Antarctica in spring, the sun’s warmth breaks up the clouds and HOC1, Cl₂ are photolysed by sunlight.

The chlorine radicals thus formed initiate the chain reaction for ozone depletion.

Q34. A person was using water supplied by Municipality. Due to shortage of water, he started using underground water. He felt laxative effect. What could be the cause?
Sol: The laxative effect is observed only when the concentration of sulphates in water is greater than 500 ppm. Sulphate is harmless at moderate concentration but concentration above 500 ppm produces laxative effects and hypertension.

Matching Column Type Questions

Match the terms given in Column I with the compounds given in Column

Column 1		Column II
(a)	Acid rain	(1)	CHC12-CHF2
(b)	Photochemical smog	(2)	CO
(c)	Combination with haemoglobin	(3)	co2
(d)	Depletion of ozone layer	(4)	so2
		(5)	Unsaturated hydrocarbons

Sol: (a →3,4); (b → 4, 5); (c → 2); (d → 1)
(a) Acid rain is caused due to oxides of carbon, sulphur and nitrogen.
(b) Photochemical smog is formed by unburnt fuel (unsaturated hydrocarbons). *
(c) Carbon monoxide with haemoglobin is poisonous.
(d) Chlorofluorocarbons (CHC1₂ – CHF₂) cause ozone depletion.

Q36. Match the pollutant(s) in Column I with the effect(s) in Column II.

Column I		Column 11
(a)	Oxides of sulphur	(1)	Global warming                        .
(b)	Nitrogen dioxide ,	(2)	Damage to kidney
(c)	Carbon dioxide	(3)	‘Blue baby’ syndrome
(d)	Nitrate in drinking water	(4)	Respiratory diseases
(e)	Lead	(5)	Red haze in traffic and congested areas

Sol: (a → 4); (b → 5); (c →1); (d →3); (e → 2)

(a) Low concentration of sulphur dioxide causes respiratory disease, e.g., asthma, bronchitis etc.
(b) The irritant red haze in traffic and congested places is due to oxides of nitrogen.
(c) The increased amount of C0₂ in air is mainly responsible for global warming.
(d) Excess’nitrate in drinking water cause methemoglobinemia (blue baby syndrome)
(e) Lead can damage kidney, liver, reproductive system etc.

Q37. Match the activity given in Column I with the type of pollution created by it given in Column II.

Column I	Column II
Releasing gases to the atmosphere after burning waste material containing Sulphur.		0)	Water pollution
Using           carbamates  as pesticides.		(2)	Photochemical smog, damage to plant life, corrosion to building material, induce breathing problems, water pollution
Using synthetic detergents for washing clothes.		(3)	Damaging ozone layer
Releasing gases produced by automobiles and factories in the atmosphere.		(4)	May cause nerve diseases in human                                     .
Using chlorofluorocarbon compounds for cleaning computer parts.		(5)	Classical smog, acid rain, water pollution, induce breathing problems, damage to buildings, corrosion of metals

Sol: (a → 5); (b → 4); (c→ 1); (d → 2); (e→ 3)

Q38. Match the pollutants given in Column I with their effects given in Column II.
(a) Sulphur dioxide causes classical smog, acid rain, water pollution, induces breathing problems, causes damage to buildings, corrosion of metals.
(b) Using carbamates as pesticides can cause nerve diseases in humans
(c) Using synthetic detergents for washing clothes causes water pollution.
(d) Unsaturated hydrocarbons and nitrogen oxides produced by automobiles and factories cause photochemical smog, damage to plant life, corrosion to building material, induce breathing problems, water pollution.
(e) Chlorofluorocarbons are believed to be the main reason for ozone layer depletion.

Column I		Column II
(a)	Phosphate fertilizers in water	(1)	BOD level of water increases
(b)	Methane in air	(2)	Acid rain
(c)	Synthetic detergents in water	(3)	Global warming
(d)	Nitrogen oxides in air	(4)	Eutrophication

Sol: (a → 1,4); (b → 3); (c → 1); (d → 2) ‘
(a) Phosphate fertilizers increase growth of algae, increasing BOD level and causing eutrophication.
(b) Methane oxidises to C0₂ which causes global warming.
(c) Use of synthetic detergents increases BOD level of water.
(d) Nitrogen oxides present in air combine with water forming nitric acid.

Assertion and Reason Type Questions
In the following questions a statement of Assertion (A) followed by a statement of Reason (R) is given. Choose the correct option out of the choices given below each question.

Q39. Assertion (A): Greenhouse effect was observed in houses used to grow plants and these are made of green glass.
Reason (R): Greenhouse name has been given because glass houses are made of green glass.
(a) Both A and R are true and R is the correct explanation of A.
(b) Both A and R are true but R is not the correct explanation of A.
(c) Both A and R are not correct.
(d) A is not correct but R is correct.
Sol: (c) There is no scientific relation between greenhouse effect and the given assertion or reason.

Q40. Assertion (A): The pH of acid rain is less than 5.6.
Reason (R): Carbon dioxide present in the atmosphere dissolves in rain water and forms carbonic acid.
(a) Both A and R are true and R is the correct explanation of A.
(b) Both A and R are true but R is not the correct explanation of A.
(c) Both A and R are not correct.
(d) A is not correct but R is correct.
Sol: (b) In acid rain, pH is less than 5.6. Carbon dioxide dissolves in water to form weak acid.

 H₂ + C0₂→H₂C0₃

Q41. Assertion (A): Photochemical smog is oxidizing in nature.
Reason (R): Photochemical smog contains N0₂ and 0₃, which are formed during the sequence of reactions.
(a) Both A and R are true and R is the correct explanation of A.
(b) Both A and R are true but R is not the correct explanation of A.
(c) Both A and R are not correct.
(d) A is not correct but R is correct.
Sol: (a) Photochemical smog contains N0₂ and 0₃; both are oxidizing agents.

Q42. Assertion (A): Carbon dioxide is one of the important greenhouse gases. Reason (R): It is largely produced by respiratory function of animals and plants.
(a) Both A and R are true and R is the correct explanation of A.
(b) Both A and R are true but R is not the correct explanation of A.
(c) Both A and R are not correct.
(d) A is not correct but R is correct.
Sol: C0₂ is produced by respiration of plants and animals; it is a greenhouse gas.

Q43. Assertion (A): Ozone is destroyed by solar radiation in upper stratosphere.
Reason (R): Thinning of the ozone layer allows excessive UV radiations to reach the surface of earth.
(a) Both A and R are true and R is the correct explanation of A.
(b) Both A and R are true but R is not the correct explanation of A.
(c) Both A and R are not correct.
(d) A is not correct but R is correct.
Sol: (d) Solar radiations never destroy ozone layer.

Q44. Assertion (A): Excessive use of chlorinated synthetic pesticides causes soil and water pollution.
Reason (R): Such pesticides are non-biodegradable.
(a) Both A and R are true and R is the correct explanation of A.
(b) Both A and R are true but R is not the correct explanation of A.
(c) Both A and R are not correct.
(d) A is not correct but R is correct.
Sol: (a) Chlorine containinginsecticides and pesticides are non biodegradable and they pollute soil and water.

Q45. Assertion (A): If BOD level of water in a reservoir is less than 5 ppm it is highly polluted.
Reason (R): High biological oxygen demand means low activity of bacteria in water.
(a) Both A and R are true and R is the correct explanation of A.
(b) Both A and R are true but R is not the correct explanation of A.
(c) Both A and R are not correct.
(d) A is not correct but R is correct.
Sol: (c) High BOD means high activity of Bacteria in water.

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NCERT Exemplar Class 11 Chemistry Chapter  13 Hydrocarbons

Multiple Choice Questions
Single Correct Answer Type

Q1. Arrange the following ia decreasing order of their boiling points.
(A) n-Butane
(B) 2-Methylbutane
(C) n-Pentane
(D) 2,2-Dimethylpropane
(a) A > B > C > D
(b) B > C > D > A
(c) D>C>B>A
(d) C >B>D > A
Sol: (d) As the number of carbon atom increases, boiling point increases. Boiling point decreases with branching.

Q2. Arrange the halogens F₂, Cl₂, Br₂ and I₂ in order of their increasing reactivity with alkanes.
(a) I₂ < Br₂ < Cl₂ < F₂
(b) Br₂ < Cl₂ < F₂ < I₂
(c) F₂ < Cl₂ < Br₂ < I₂                               
(d) Br₂ < I₂ < Cl₂ < F₂
Sol:(a) The reactivity order of halogens with alkanes is I₂ < Br₂ < Cl₂ < F₂

Q3. The increasing order of reduction of alkyl halides with zinc and dilute HCl is
(a) R-C1<R-I<R-Br
(b) R-Cl<R-Br<R-I
(c) R -1 < R – Br < R – Cl
(d) R-Br<R-I<R-Cl
Sol:(b) The reactivity of reduction bf alkyl halides with Zn/HCl increases as the strength of the C – X bond decreases, i.e., R – Cl < R – Bf < R -I.

Q4. The correct IUPAC name of the following alkane is

(a) 3,6-Diethyl-2-methyloctane
(b) 5-Isopropyl-3-ethyloctane
(c) 3-Ethyl-5-isopropyloctane
(d) 3-Isopropyl-6-ethyloctane

Q5. The addition of HBr to 1 -butene gives a mixture of products (A), (B) and (C).

(C) CH₃ – CH₂ – CH₂ – CH₂ – Br
The mixture consists of
(a) (A) and (B) as major and (C) as minor products
(b) (B) as major, (A) and (C) as minor products
(c) (B) as minor, (A) andj(C) as major products
(d) (A) and (B) as minor and (C) as major products.
Sol: (a) The alkene is unsymmetrical, hence will follow Markovnikov rule to give major product.

Since I contains a chiral carbon, it exists in two enantiomers (A and B) which are mirror image of each other.

Q6. Which of the following will nofshow geometrical isomerism?

Q7. Arrange the following hydrogen halides in order of their decreasing reactivity with propene.
(a) HC1 > HBr > HI
(b) HBr>HI>HCl
(c) HI > HBr > HCl
(d) HCl>HI>HBr
Sol: (c) The decreasing order of reactivity of hydrogen halides with propene is HI > HBr > HC1. As the size of halogen increases, the strength of H – X bond decreases and hence, reactivity increases.

Q8. Arrange the following carbanions in order of their decreasing stability.
(A) H₃C-C ≡ C^–                                        
(B) H-C ≡ C^–
(C) H₃C – CH^–2
(a) A>B>C
(b) B>A>C
(c) C>B>A
(d) C>A>B

Sol: (b) The order of decreasing stability of carbanions is:

sp-hybridised carbon atom is more electronegative than sp3-hybridised carbon atom and hence, can accommodate the negative charge more effectively. – CH₃ group has +1 effect, therefore, it intensifies the negative charge and, hence, destabilises the carbanion CH₃ →C = C^–.

Q9. Arrange the following alkyl halides in decreasing order of the rate of β -elimination reaction with alcoholic KOH.

is the order of rate of β -elimination with alcoholic KOH.

More the number of β-substituents (alkyl groups), more stable alkene will be formed on β -elimination and more will be the reactivity. Thus, the decreasing order of the rate of β -elimination reaction with alcoholic KOH is: A > C > B.

Q10. Which of the following reactions of methane is incomplete combustion?

Sol: (c) Dining incomplete combustion of alkanes with insufficient amount of air or dioxygen, carbon black is formed which is used in the manufacture of ink, printer ink, black pigments and as filters. Thus,

More than One Correct Answef Type
Q11. Some oxidation reactions of methane are given below. Which of them is/are controlled oxidation reactions?

Reactions in which methane does not undergo complete combustion to give carbon dioxide and water or incomplete combustion to give carbon and water are controlled oxidation reactions.

Q12. Which of the following alkenes on ozonolysis gives a mixture of ketones only?

Sol: (c, d) Alkenes which have two substituents on each carbon atom of the double bond give a mixture of ketones on ozonolysis. Thus, options (c) and (d) give mixture of ketones.

Q13. Which are the correct IUPAC names of the following compound?

(a) 5-Buty 1-4-i sopropyldecane
(b) 5-Ethyl-4-propyldecane
(c) 5-sec-Butyl-4-iso-propyldecane
(d) 4-( 1 -Methylethyl)-5-( 1 -methylpropyl)decane

Sol: (c,d)

Q14. Which are the correct IUPAC names of the following compound

(a) 5-(2′,2′-Dimethylpropyl)decane
(b) 4-Butyl-2,2-dimethylnonane
(c) 2,2-Dimethyl-4-pentyloctane
(d) 5-neo-Pentyldecane

Q15. For an electrophilic substitution reaction, the presence of a halogen atom in the benzene ring ;
(a) deactivates the ring by inductive effect
(b) deactivates the ring by resonance
(c) increases the charge density at ortho and para-positions relative to meta-position by resonance
(d) directs the incoming electrophile to meta-position by increasing the
charge density relative to ortho and para-positions. ‘
Sol: (a, c) For an electrophilic substitution reaction, the presence of halogen atom in the benzene ring deactivates the ring by inductive effect and increases the charge density at ortho- and para-position relative to meta-position by resonance. –
When chlorine is attached to benzene ring, chlorine being more electronegative pulls the electrons because of its -1-effect. The electron cloud of benzene becomes less dense. Thus, chlorine makes the benzene ring in aryl halide somewhat deactivated. But due to resonance, the electron density on ortho- and para-positions is greater than on meta-position.

Q16. In an electrophilic substitution reaction of nitrobenzene, the presence of nitro group ________ ‘
(a) deactivates the ring by inductive effect
(b) activates the ring by inductive effect
(c) decreases the charge density at ortho- and para-positions of the ring relative to meta-position by resonance
(d) increases the charge density at meta-position relative to the ortho and para-positions of the ring by resonance
Sol: (a, c) Nitro group by virtue of-I-effect withdraws electrons from the ring and increases the charge and destabilizes carbocation.

In ortho, para-attack of electrophile on nitrobenzene, we are getting two structures (A) and (B) in which positive charge is appearing on the carbon atom directly attached to the nitro group.
As nitro group is electron withdrawing by nature, it decreases the stability of such product and hence meta attack is more feasible when electron withdrawing substituents are attached.

Q17. Which of the following are correct?
(a) CH₃ – O – CH⁺₂ is more stable than CH₃ – CH⁺₂
(b) (CH₃)₂CH⁺ is less stable than CH₃ – CH₂ – CH⁺₂
(c) CH₂ = CH – CH⁺₂ is more stable than CH₃ – CH₂ – CH⁺₂
(d) CH₂ = CH⁺ is more stable than CH₃ – CH⁺₂

Q18. Four structures are given in options (a) to (d). Examine them and select the aromatic structures.

Q19. The molecules having dipole moment are________ .
(a) 2,2-Dimethylpropane                      
(b) trans-Pent-2-ene
(c) cw-Hex-3-ene            
(d) 2,2,3,3-Tetramethylbutane

Since, the +1 effect of CH₂CH₃ group is higher than that of CH₃ group, therefore, the dipole moments of C-CH₃ and C-CH₂CH₃ bonds are unequal. Although these two dipoles oppose each other, yet they do not exactly cancel out each other and hence trans-2-pentene has small but finite dipole moment.
In cis-hex-3-ene, although the dipole moments of the two C – CH₂CH₃ bond are equal, but they are inclined to each other at an angle of 60° and hence have a finite dipole moment.

Short Answer Type Questions
Q20. Why do alkenes prefer to undergo electrophilic addition reaction while arenes prefer electrophilic substitution reaction? Explain.
Sol: Due to the presence of a π -electron cloud above and below the plane of alkenes and arenes, these are electron rich molecules and, therefore, provide sites for the attack of electrophiles. Hence, they undergo electrophilic reactions. Alkenes undergo electrophilic addition reactions because they are unsaturated molecules. For example,

Arenes, on the other hand, cannot undergo electrophilic addition reactions. This is because benzene has a large resonance energy of 150.4 kJ mol^-1. During electrophilic addition reactions, two new σ-bonds are formed but the aromatic character of benzene gets destroyed and, therefore, resonance energy of benzene ring is lost. Hence, electrophilic addition reactions of arenes are not energetically favourable. Arenes, in contrast, undergo electrophilic substitution reactions in which σ C – H bond is broken and new σ C – X bond is formed: The aromatic character of benzene ring is not destroyed and benzene retains its resonance energy. Hence, arenes undergo electrophilic substitution reactions.

Q21. Alkynes on reduction with sodium in liquid ammonia form trans alkenes. Will butene formed on the reduction of but-2-yne show geometrical isomerism?

Thus, but-2-ene is capable of showing geometrical isomerism.

Q22. Rotation around carbon-carbon single bond of ethane is not completely free. Justify the statement.
Sol: Ethane contains carbon-carbon sigma (σ) bond. Electron distribution of the sigma molecular orbital is symmetrical around the intemuclear axis of the C – C bond which is not disturbed due to rotation about its axis. This permits free rotation around aC-C single bond. However, rotation around a C – C single bond is not completely free. It is hindered by a small energy barrier due to weak repulsive interaction between the adjacent bonds. Such a type of repulsive interaction is called torsional strain. Of all the conformations of ethane, the staggered form has the least torsional strain and the eclipsed form has the maximum torsional strain. The energy difference between the two extreme forms is of the order of 12.5 kJ mol^-1, which is very small. It has not been possible to separate and isolate different conformational isomers of ethane.

Q23. Draw Newman and Sawhorse projections for the eclipsed and staggered conformations of ethane. Which of these conformations is more stable and why? .

In staggered form of ethane, the electron clouds of carbon-hydrogen bonds are as far apart as possible. Thus, there are minimum repulsive forces, minimum energy and maximum stability of the molecule. On the other hand, when the staggered form changes into the eclipsed form, the electron clouds of the carbon-hydrogen bonds come closer to each other resulting in increase in electron cloud repulsions. To check the increased repulsive forces, molecule will have to possess more energy and thus has lesser stability.

Q24. The intermediate carbocation formed in the reactions of HI, HBr and HC1 with propene is the same and the bond energy of HCl, HBr and HI is 430.5 kJ mol^-1,363.7 kJ mol^–1 and 296.8 kJ mol^-1 What will be the other of reactivity of these halogen acids?
Sol: The bond dissociation enthalpy decreases in the order HC1 > HBr > HI, therefore, the order of reactivity is in the reverse order i.e., HI > HBr > HCl.
Q25. What will be the product obtained as a result of the following reaction and why?

Propyl chloride forms CH₃ – CH₂ – CH⁺₂ with anhydrous A1C1₃ which is less stable. This rearranges to a more stable carbocation as:

Q26. How will you convert benzene into
(i) p-nitrobromobenzene (ii) m-nitrobromobenzene

Q27. Arrange the following set of compounds in the order of their decreasing . relative reactivity with an electrophile. Give reason.

Sol: The methoxy group (-OCH₃) is electron releasing group. It increases the electron density in beniene nucleus due to

resonance effect (+R-effect). Hence, it makes anisole more reactive than benzene towards the electrophile.

In case of alkyl halides, the electron density increases at ortho and para positions due to +R effect. However, the halogen atom also withdraws electrons from the ring because of its -I effect. Since the -I effect is stronger than the +R effect, the halogens are moderately deactivating. Thus, overall electron density on benzene ring decreases, which makes further substitution difficult.
-N0₂ group is electron withdrawing group. It decreases the electron density in benzene nucleus due to its strong -R-effect and strong -I-effect. Hence, it makes nitrobenzene less reactive. Therefore, overall reactivity of these three compounds towards electrophiles decreases in the following order:

Q28. Despite their -I effect, halogens are o- andp-direction in haloarenes. Explain.
Sol: In case of aryl halides, halogens are little deactivating because of their strong -I effect. Therefore, overall electron density on the benzene ring decreases. In other words, halogens are deactivating due to -I effect. However, because of the +R-effect, i.e., participation of lone pairs of electrons on the halogen atom with the π-electrons of the benzene ring, the electron density increases more at o- and p-positions than at m-positions.

As a result, halogens are o-, p-directing. The combined result of +R-effect and -I-effect of halogens is that halogens are deactivating but o, p-directing.

Q29. Why does the presence of a nitro group-make the benzene ring less reactive in comparison to the unsubstituted benzene ring? Explain.
Sol: Nitro group is an electron withdrawing group (-R and -I effects). It deactivates the ring by decreasing nucleophilicity for further substitution.

Q30. Suggest a route for the preparation of nitrobenzene starting from acetylene.
Sol: Acetylene when passed through red hot iron tube at 500°C undergoes cyclic polymerisation to give benzene which upon nitration gives nitrobenzene.

Q31. Predict the major product(s) of the following reactions and explain their formation.

Sol: Addition of HBr to unsymmetrical alkenes follows Markonikov rule. It states that negative part of the addendum (adding molecule) gets attached to that carbon atom which possesses lesser number of hydrogen atoms.
Mechanism: Hydrogen bromide provides an electrophile, H⁺, which attacks the double bond to form carbocation as shown below:

Addition reaction of HBr to unsymmetrical alkenes in the presence of peroxide follows anti-Markovnikov rule.
Mechanism: Peroxide effect proceeds via free radical chain mechanism as given below:

Q32. Nucleophiles and electrophiles are reaction intermediates having electron rich and electron deficient centres respectively. Hence, they tend to attack electron deficient and electron rich centres respectively. Classify the following species as electrophiles and nucleophiles.

Q33. The relative reactivity of 1°, 2° and 3° hydrogens towards chlorination is 1: 3.8 : 5. Calculate the percentages of all monochlorinated products obtained from 2-methylbutane.
Solu:

Q34. Write the structures and names of products obtained in the reactions of sodium with a mixture of l-iodo-2-methylpropane and 2-iodopropane.
Solu:

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NCERT Exemplar Class 11 Chemistry Chapter  12 Organic Chemistry: Some Basic Principles and Techniques

Multiple Choice Questions
Single Correct Answer Type

Q1. Which of the following^ the correct IUPAC name?
(a) 3-Ethyl-4,4-dimethylheptane
(b) 4,4-Dimethyl-3-ethylheptane
(c) 5-Ethyl-4,4-dimethylheptane
(d) 4,4-Bis(methyl)-3-ethylheptane
Sol: (a) While writing IUPAC name, the alkyl groups are written in alphabetical order. Thus lower locant 3 is assigned to ethyl. Prefix, di, tri, and tetra are not included in alphabetical order.

Q2. The IUPAC name for

(a) 1-Hydroxypentane-l, 4-dione
(b) 1,4-Dioxopentanol
(c) l-Carboxybutan-3-one
(d) 4-Oxopentanoic acid

When more than one functional group lie in the main chain, nomenclature is done according to that functional group which has higher priority. Carboxylic acid (-COOH) has more priority than the keto group (>C = O).

(a) 1 -Chloro-2-nitro-4-methylbenzene
(b) l-Chloro-4-methyl-2-nitrobenzene
(c) 2-Chloro-1 -nitro-5-methylbenzene
(d) m-Nitro-p-chlorotoluene

Q4. Electronegativity of carbon atoms depends upon their state of hybridization. In which of the following compounds, the carbon marked with asterisk is most electronegative?

Sol: (c) Electronegativity increases as the state of hybridization changes from sp³ to sp² and sp² to sp. Thus, sp hybridized carbon has the highest electronegativity.

Q5. In which of the following, functional group isomerism is not possible?
(a) Alcohols
(b) Aldehydes
(c) Alkyl halides
(d) Cyanides
Sol: (c) Alkyl halides do not show functional isomerism. Alcohols and ethers, aldehydes and ketones, cyanides and isocyanides are functional isomers.

Q6. The fragrance of flowers is due to the presence of some steam volatile organic compounds called essential oils. These are generally insoluble in water at room temperature but are miscible with water vapour in vapour phase. A suitable method for the extraction of these oils from the flowers is
(a) distillation
(b) crystallisation
(c) distillation under reduced pressure
(d) steam distillation.
Sol: (d) Essential oils are insoluble in water, soluble in steam and have high vapour pressure. Therefore, they can be separated by steam distillation.

Q7. During hearing of a court case, the judge suspected that some changes in the documents had been carried out. He asked the forensic department to check  the ink used at two different places. According to you, which technique can give the best results?
(a) Column chromatography (b) Solvent extraction
(c) Distillation (d) Thin layer chromatography
Sol:(d) Thin layer chromatography (TLC) is used to separate the pigments present in ink.

Q8. The principle involved in paper chromatography is
(a) adsorption (b) partition
(c) solubility (d) volatility
Sol:(b) In paper chromatography, separation of the components of a mixture depends upon their partitioning between water held in the stationary phase (i.e. adsorbent paper) and the liquid present in the mobile phase.

Q9. What is the correct order of decreasing stability of the following cations?

(a) 2-Ethyl-3-methylpentane
(b) 3,4-Dimethylhexane
(c) 2-sec-Butylbutane
(d) 2,3-Dimethylbutane

Q11. In which of the following compounds the carbon marked with asterisk is expected to have greatest positive charge?

Q12. Ionic species are stabilized by the dispersal of charge. Which of the following carboxylate ions is the most stable?

Sol: (d) In all the given carboxylate ions, the negative charge is dispersed which stabilizes these ions. Here, the negative charge is dispersed by two factors, i.e., +R-effect of the carboxylate ion (conjugation) and -I-effect of the halogens.

It is evident from the above structures that +R-effect is common in all the four ions. Therefore, overall dispersal of negative charge depends upon the number of halogen atoms and electronegativity. Since F has the highest electronegativity and two F-atoms are present in option (d), thus, dispersal of negative charge is maximum in option (d).

Q13. Electrophilic addition reactions proceed in two steps. The first step involves the addition of an electrophile. Name the type of intermediate formed in the first step of the following addition reaction.
H₃C-HC = CH₂ + H⁺→ ?

(a) 2°Carbanion                                    
(b) 1° Carbocation
(c) 2° Carbocation
(d) l°Carbanion

Sol: (c) When the electrophile attacks CH₃ – CH = CH₂, delocalisation of electrons can take place in two possible ways

As 2° carbocation is more stable than 1° carbocation, the first addition is more feasible.

Q14. Covalent bonds can undergo fission in two different ways. The correct representation involving the heterolytic fission of CH₃ – Br is

Sol: (b) Arrow denotes the direction of movement of electrons

Since, Br is more electronegative than carbon, hence heterolytic fission occurs in such a way that CH₃ gets the positive charge and Br gets the negative charge. Thus, option (b) is correct.

Q15. The addition of HC1 to an alkene proceeds in two steps. The first step is the attack of H⁺ ion to >C = C< portion which can be shown as

Sol: (b) Since double bond is a source of electrons and the charge flows from source of more electron density, therefore, n electrons of the double bond attack the proton.

More than One Correct Answer Type
Q16. Which of the following compounds contain all the carbon atoms in the same hybridization state?

Q17. In which of the following representations given below spatial arrangement of group/atom is different from that given in structure ‘A’?

Sol: (a, c, d) Different groups are present towards and away from the observer.

(a), (c) and (d) have different (anticlockwise) spatial arrangements of atoms or groups of atoms than that given in structure A (clockwise).
Q18. Electrophiles are electron seeking species. Which of the following groups contains only electrophiles?


Q19. Which of the above pairs are position isomers?
(a) I and II
(b) II and III
(c) II and IV
(d) III and IV

Sol: (b) II and III are position isomers as they differ in the position of -C- group.

Q20. Two or more compounds having same the molecular formula but different functional groups are called functional isomers. Which of die following pairs are not functional group isomers?
(a) II and III (b) II and IV
(c) I and IV (d) I and II
Sol: (a, c) (a) II and III have the same functional group.
(c) I and IV have the same functional group.

Q21. Nucleophile is a species that should have
(a) a pair of electrons to donate
(b) positive charge
(c) negative charge
(d) electron deficient species
Sol: (a, c) Nucleophile (nucleus-loving) is a chemical species that donates an, electron pair to an electrophile (electron-loving). Hence, a nucleophile should
have either a negative charge or an electron pair to donate. Thus, options (a) r and (c) are correct.

Q22. Hyperconjugation involves delocalization of .
(a) electrons of carbon-hydrogen σ bond of an alkyl group directly attached to an atom of unsatUrated system.
(b) electrons of carbon-hydrogen σ bond of alkyl group directly attached to the positively charged carbon atom.
(c) π-electrons of carbon-carbon bond.
(d) lone pair of electrons.

Sol: (a, b) Hyperconjugation, also known as sigma-pi conjugation, is the delocalization of sigma electrons. Presence of -H with respect to double bond, triple bond or carbon containing positive charge (in carbonium ion) or unpaired electron (in free radical) is a condition required for hyperconjugation.

Short Answer Type Questions
Note: Consider structures I to VII and answer Questions 23-26.

Q23. Which of the above compounds form pairs of metamers?
Sol: V and VI, VI and VII and V and VII form pairs of metamers because they have different alkyl chains on either side of ethereal oxygen (-O-).

Q24. Identify the pairs of compounds which are functional group isomers.
Sol: Compounds I to IV, i.e., alcohols, and V to VII, i.e., ethers, are functional group isomers with molecular formula C₄H₁₀O and different functional groups (-OH in I to IV and -O- in V to VII).
Hence, I and V, I and VI, I and VII; II and V, II and VI, II and VII; HI and V, IH and VI; IH and VII; IV and V, IV and VI, IV and VH are functional group isomers.

Q25. Identify the pairs of compounds that represent position isomerism.
Sol:  I and II, III and IV, VI and VII are position isomers due to different positions of -OH group and -O- group.

Q26. Identify the pairs of compounds that represent chain isomerism.
Sol: I and III, I and IV, II and III, II and IV.

Q27. For testing halogens in an organic compound with AgN0₃ solution, sodium extract (Lassaigne’s extract) is acidified with dilute HN0₃. What will happen if a student acidifies the extract with dilute H₂S0₄ in place of dilute HN0₃?
Sol: On adding dilute H₂S0₄ for testing halogens in an organic compound with AgN0₃, white precipitate of Ag₂S0₄ is formed. This will interfere with the test of chlorine and this Ag₂S0₄ may be mistaken for white precipitate of chlorine as AgCl. Hence, dilute HN0₃ should be used instead of dilute H₂S0₄.

Q28. What is the hybridization of each carbon in H₇C = C = CH₇?

Q29. Explain how is the electronegativity of carbon atoms related to their state of hybridization in an organic compound?
Sol: Electronegativity increases with increasing s-character. This is because s-electrons are more strongly attracted by the nucleus than p-electrons.
sp³ – 25% s-character, 75% P-character
sp² – 33% s-character, 67% P-character                                             –
sp – 50% s-character, 50% P-character
Hence, the order of electronegativity is sp³ < sp² < sp

Q30. Show the polarization of carbon-magnesium bond in the following structure.

Q31. Compounds with same molecular formula but differing in their structures are said to be structural isomers. What type of structural isomerism is shown by

Sol: They show position isomerism.

Q32. Which of the following selected chains is correct to name the given compound according to IUPAC system?

Sol: The 4 carbon chain is correct*according to the IUPAC system since it contains both the functional groups. The other three carbon chains are incorrect since none of them contains both the functional groups.

Q33. In DNA and RNA, nitrogen atom is present in the ring system. Can Kjeldahl method be used for the estimation of nitrogen present in these? Give reasons.
Sol: No. DNA and RNA have nitrogen in the heterocyclic rings which cannot be removed as ammonia.

Kjeldahl method cannot be used to estimate nitrogen present in rings, azo and nitro groups as nitrogen present in these cannot be converted into ammonium sulphate.

Q34.If a liquid compound decomposes at its boiling point, which method(s) will you choose for its purification? It is known that the compound is stable at low pressure, steam volatile and insoluble in water.
Sol: Steam distillation can be used for its purification. This method is applied to separate substances which are steam volatile and immiscible with water. Note: Answer Questions 35-38 on the basis of information given below: “Stability of carbocations depends upon the electron releasing inductive effect of groups adjacent to positively charged carbon atom, involvement of neighbouring groups in hyperconjugation and resonance.”

Q36. Which of the following ions is more stable? Use resonance to explain your Answer 

Sol: Structure (A) is more stable due to resonance. Structure (B) is non-planar and hence it does not undergo resonance. Double bond is more stable within the ring in comparison to outside the ring.

Q37. The structure of triphenylmethyl cation is given here. This is very stable and some of its salts can be stored for months. Explain the cause of high stability of this cation
Sol: Triphenylmethyl cation is a tertiary carbocation which can show nine possible canonical structures and hence is very stable.

Q38. Write the structures of various carbocations that can be obtained from 2-methylbutane. Arrange these carbocations. in order of increasing stability.
Sol: 2-Methylbutane has four possible carbocations

Stability of carbocation increases in the order 1° < 2° < 3°. Out of I and IV, IV is more stable than I because in IV, CH₃ group is at a-carbon and in I, it is at β-carbon and +I-effect decreases with distance

Q39. Three students, Manish, Ramesh and Rajni, were determining the extra elements present in an organic compound given by their teacher. They prepared the Lassaigne’s extract (LE) independently by the fusion of the compound with sodium metal. Then they added solid FeS0₄ and dilute sulphuric acid to a part of Lassaigne’s extract. Manish and Rajni obtained Prussian blue colour bit Ramesh got red colour. Ramesh repeated the test with the same Lassaigne’s extract, but again got red colour only. They were surprised and went to their teacher and told him about their observation. Teacher asked them to think over the reason for this. Can you help them by giving the reason for this observation? Also, write the chemical equations to explain the formation of compounds of different colours.

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

ferri ferrocyanide.

6NaCN + FeS0₄→ Na₄[Fe(CN)₆] + Na₂S0₄

3Na₄[Fe(CN)₆] + 2Fe₂(S0₄)3 → Fe₄[Fe(CN)₆]₃ + 6Na2S0₄
In compounds containing nitrogen and sulphur together, the sodium metal should be in slight excess otherwise in Lassaigne’s test, sodium thiocyanate (NaCNS) is formed which gives red colour with Fe³⁺ ions and decomposes as follows:

On the basis of above results, it is clear that Ramesh used less sodium and hence NaSCN was formed in the Lassaigne’s extract which gave red colouration due to Fe(SCN)₃ formation while Manish and Rajni used excess sodium and hence, NaCN was formed in the Lassaigne’s extract which gave Prussian blue colour of Fe₄[Fe₄(CN)₆].

Q40. Name the compounds whose line formulae are given below:

Q41. Write structural formulae for compounds named as
(a) 1-Bromoheptane
(b) 5-Bromoheptanoic acid

Q42. Draw the resonance structures of the following compounds:

Q43. Identify the most stable species in the following set of ions giving reasons:

Q44. Give three points of differences between inductive effect and resonance effect.
Sol: 

Inductive effect	Resonance effect
(1) This effect involves displacement of σ-electrons.	The resonance involves displacement of π-electrons or lone pair of electrons.
(2) It operates in saturated compounds.	It operates only in unsaturated conjugated systems.
(3) This effect moves up to three •          carbon atoms and becomes negligible from fourth carbon atom onwards.	This effect moves all along the length of the conjugated system.
(4) This effect causes slight drift of σ-electrons towards the more electronegative atom and hence only partial charges (δ + and δ -) are developed.	This results in complete transfer of electrons and hence full +ve and -ve charges are developed.

Q45. Which of the following compounds will not exist as resonance hybrid? Give reason for your answer.

(a) CH₃OH
(b) R-CONH₂
(c) CH3CH = CHCH₂NH₂
Sol: (a) CH3OH does not contain -electrons, hence, it cannot exist as resonance hybrid.
(b) Due to the presence of -electrons in C = O bond and lone pair of electrons on N, amide can be represented by the following resonating structures.

(c) CH₃CH = CHCH₂NH₂: Since lone pair of electrons on nitrogen atom is not conjugated with the -electrons, therefore, resonance is not possible.

Q46. Why does S0₃ act as an electrophile?
Sol: Three highly electronegative oxygen atoms are attached to sulphur atom. This makes sulphur atom electron deficient. Due to resonance, sulphur also acquires positive charge. Both these factors make S0₃ an electrophile.

Q47. Resonance structures of propenal are given below. Which of these resonating structures is more stable? Give reason for your answer.


Q48. By mistake, an alcohol (boiling point 97°C) was mixed with a hydrocarbon (boiling point 68°C). Suggest a suitable method to separate the two compounds. Explain the reason for your choice.
Sol: Simple distillation can be used because the two compounds have a difference of more than 20° in their boiling points and therefore, both the liquids can be distilled without any decomposition. ‘

Q49. Which of the two structures (A) and (B) given below is more stabilized by resonance? Explain.

Matching Column Type Questions
In the following questions more than one correlation is possible between options of column I and II. Make as many correlations as you can.
Q50. Match the type of mixture of compounds in Column I with the technique of separation/purification given in column II.

Column I	Column II
(a) Two solids which have different solubilities in a solvent and which do not undergo reaction when dissolved in it.	(1) Steam distillation
(b) Liquid that decomposes at its boiling point	(2) Fractional distillation
(c) Steam volatile liquid	(3) Simple distillation
(d) Two liquids which have boiling points close to each other	(4) Distillation under reduced pressure
(e) Two liquids with large difference in boiling points.	(5) Crystallisation

Sol: (a → 5), (b → 4), (c→1), (d → 2), (e →3)

Q51. Match the terms mentioned in Column I with the terms in Column II.

Column I	Column II
(a) Carbocation	(1) Cyclohexane and 1-hexene
(b) Nucleophile	(2) Conjugation of electrons of C – H σbond with empty p-orbital present at adjacent positively charged carbon.
(c) Hyperconjugation	(3) sp2 hybridised carbon with empty p-orbital
(d) Isomers	(4) Ethyne
(e) sp hybridization	(5) Species that can receive a pair of electrons
(f) Electrophile	(6) Species that can supply a pair of electrons

Sol: (a →  3); (b →  6); (c→  2); (d→  1); (e →  4); (f →  5)

Column I	Column II	Explanation
(a) Carbocation	sp2-hybridised carbon with empty p-orbital	H3C+ is carbocation. Loss of e makes its p-orbitals empty (sp2– hybridised carbon)
(b) Nucleophile	Species that can supply a pair of electrons	Nucleus loving, i.e., having negative charge or excess of electrons
(c) Hyperconjugation	Conjugation of electrons of C – H σbond with empty p-orbital present at adjacent positively charged carbon
(d) Isomers	Cyclohexane and 1-hexene	Same molecular formula but different structures
(e) sp hybridization	Ethyne	HC ≡ CH (sp hybridization)
(f) Electrophile	Species that receive a pair of electrons	Electron loving, i.e., positive charge or lack of electrons

Q52. Match Column I with Column II.

Column I	Column II
(a) Dumas method	(1) AgN03
(b) Kjeldahl method	(2) Silica gel
(c) Carius method	(3) Nitrogen gel
(d) Chromatography	(4) Free radicals
(e) Homolysis	(5) Ammonium sulphate

 

Sol: (a → 3); (b → 5); (c→ 1); (d →2); (e → 4)

Column I	Column II	Explanation
(a) Dumas method	Nitrogen gel	Used for N containing compounds
(b) Kjeldahl method	Ammonium sulphate	Nitrogen converts to ammonium sulphate
(c) Carius method	AgN03	Compound is heated in the presence of AgN03
(d) Chromatography	Silica gel	Adsorbent used is silica gel
(e) Homolysis              ‘	Free radicals	Free radicals are formed by homolytic fission

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