perfect chemistry - ii · r – o – r′ xi. aldehydes xii. ketones xiii. carboxylic acids c c c...

Std. XI Sci.

Perfect Chemistry - II

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Edition: July 2014

Prof. Santosh B. Yadav (M. Sc., SET, NET)

Prof. Anil Thomas (M.Sc., Chemistry)

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Std. XI Sci.

Perfect Chemistry - II

Written according to the New Syllabus (2012-2013) published by the Maharashtra State Board of Secondary and Higher Secondary Education, Pune.

TEID : 747

Salient Features :

• Exhaustive coverage of syllabus in Question Answer Format. • Covers answers to all Textual, Intext and NCERT Questions. • Simple and Lucid language. • Neat, Labelled and authentic diagrams. • Quick review for instant revision and summary of the chapter. • Solved & Practice Numericals duly classified. • Multiple Choice Questions for effective preparation.

PREFACE In the case of good books, the point is not how many of them you can get through, but rather how many can get through to you. “Std. XI Sci. : PERFECT CHEMISTRY - II” is a complete and thorough guide critically analysed and extensively drafted to boost the students confidence. The book is prepared as per the Maharashtra State board syllabus and provides answers to all textual and intext questions. Sub-topic wise classified ‘question and answer format’ of this book helps the student to understand each and every concept thoroughly. Neatly labelled diagrams have been provided wherever required. National Council Of Educational Research And Training (NCERT) questions and problems based on Maharashtra board syllabus have been provided along with solutions for a better grasp of the concept and preparing the students on a competitive level. Additional information about a concept is provided in the form of Note. Definitions, statements and laws are specified with italic representation. Formulae are provided in chapters which involve numericals to help the students to tackle difficult problems. Solved problems are provided to understand the application of different concepts and formulae. Quick Review has been provided which gives an overview of the chapters. Additional theory questions have been provided to help the student gain insight on the various levels of theory-based questions. Practice problems and multiple choice questions help the students to test their range of preparation and the amount of knowledge of each topic. The journey to create a complete book is strewn with triumphs, failures and near misses. If you think we’ve nearly missed something or want to applaud us for our triumphs, we’d love to hear from you. Please write to us on : [email protected] A book affects eternity; one can never tell where its influence stops.

Best of luck to all the aspirants!

Yours faithfully Publisher

No. Class of the compound Functional group Representation of the

class (General formula) Example

i. Alkanes (Saturated)

R – H

CH3 − CH3 Ethane

ii. Alkenes (Unsaturated)

R − CH = CH − R H2C = CH2 Ethylene

iii. Alkynes (Unsaturated)

− C ≡ C − R − C ≡ C − R HC ≡ CH

Acetylene

iv. Alkyl halides − X (Halide)

R − X CH3Br Methyl bromide

v. Alkyl cyanides or nitriles − C ≡ N (Cyano or cyanide or

nitrile)

R − C ≡ N CH3 − CN Methyl cyanide or Acetonitrile

vi. Nitroalkanes − NO2 (Nitro)

R − NO2 CH3CH2NO2 Nitroethane

vii. Alcohols

− OH (Hydroxyl)

R − OH CH3OH Methyl alcohol

viii. Phenols

− OH (Phenolic)

Ar − OH C6H5OH Phenol or carbolic acid

ix. Amines

− NH2 (Amino)

R − NH2 C2H5NH2 Ethylamine

x. Ethers

R – O – R′

xi. Aldehydes

xii. Ketones

xiii. Carboxylic acids

C C

C C

C O C

(Ether group)

C

O

HR

C

O

R R′ C

O

CH3 CH3

Acetone

(Carboxyl group)

C OH

O

C

O

R OH C

O

H OH

Formic acid

CH3 − O − CH3

CH3 − O − C2H5

Dimethyl ether

Ethyl methyl ether

C H

O(Formyl or aldehydic)

C

O

HH

Formaldehyde

C

O

HCH3

Acetaldehyde

C

O(Keto or oxo)

xiv. Esters

xv. Amides

xvi. Acid anhydrides

xvii. Acyl chlorides

xviii. Sulphonic acids

Contents

No. Topic Name Page No.

9 Hydrogen 1

10 s - Block Elements 35

11 p - Block Elements (Groups 13 and 14) 65

12 Basic Principles and Techniques in Organic Chemistry

112

13 Alkanes 193

14 Alkenes 236

15 Alkynes 271

16 Aromatic Compounds 295

17 Environmental Chemistry 320

`Chapters 1 to 8 are a part of Std. XI Chemistry - I'

Note: All the Textual questions are represented by * mark All the Intext questions are represented by # mark

C

O(Ester group)

OR C

O

R OR C

O

CH3 OC2H5

Ethyl acetate

C

O(Amido group)

NH2 C

O

R NH2 C

O

CH3 NH2

Acetamide

C

O(Anhydride group)

O CC

O

C

O

O CC

O

R R C

O

O C

O

CH3 CH3

Acetic anhydride

C

O

Cl

(Acyl chloride group)

C

O

ClR C

O

Cl

Acetyl chloride

CH3

SO3H(Sulphonic acid group) SO3HR C2H5 SO3H

Ethane sulphonic acid

Target Publications Pvt. Ltd. Chapter 09: Hydrogen

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09 Hydrogen 9.0 Prominent scientists

Scientists Contributions Aristotle (384 − 322 BC) (Greek philosopher)

i.

Considered water as an element and to be one of the four elementary components of nature along with fire, air and earth.

ii.

Put forward the idea of animated beings coming into existence from inanimate objects.

iii. Was concerned with the transmutation of basemetal into gold.

Henry Cavendish (1731 − 1810) (British scientist)

i. ii.

Established water as a combustion product of hydrogen. Discovered hydrogen in 1766.

Antoine Lavoisier (1743 − 1794) (French chemist)

i. Discovered in 1783, that water is a compound of hydrogen and oxygen.

9.1 Introduction Q.1. Why is hydrogen also called as “dihydrogen”? Ans: i. Hydrogen is the first element in the periodic table, with a simple atomic structure having only one

proton and one electron. Hence, it is the lightest known element. ii. It exists in atomic form only at high temperature and as a diatomic molecule i.e., H2 in the normal

elemental form. Hence, hydrogen is also called as dihydrogen. Note: i. Hydrogen combines with other elements by losing, gaining or sharing of electrons. ii. Hydrogen is the most abundant element in the universe and the third most abundant element on the earth’s

surface. Q.2. *Explain the position of hydrogen in the periodic table on the basis of its electronic configuration.

(NCERT) OR Why is position of hydrogen considered in the periodic table as controversial or “anomalous”? Ans: i. In the periodic table, elements are arranged on the basis of their electronic configuration. ii. Hydrogen with atomic number 1 and electronic configuration 1s1 is the first element of the periodic

table. iii. However, a proper position could not be assigned to hydrogen in the long form of the periodic table

because of the following reasons: a. Resemblance with alkali metals with respect to: 1. Electronic configuration: i. The electronic configuration (1s1) of hydrogen is similar to the outer electronic

configuration (ns1) of alkali metals. ii. So, similar to alkali metals which are placed in the first group of the periodic table,

hydrogen can also be the member of the first group. eg. H(Z = 1) : 1s1 Li(Z = 3) : [He]2s1 Na(Z = 11) : [Ne]3s1 K(Z = 19) : [Ar]4s1.

Target Publications Pvt. Ltd. Std. XI Sci.: Perfect Chemistry

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2. Formation of unipositive ions: Similar to alkali metals, hydrogen can lose one electron to from unipositive ion. eg. H ⎯→ H+ + e− Na ⎯→ Na+ + e− (1s1) (1s0) [Ne]3s1 [Ne] Hydrogen Proton Sodium Sodium atom atom ion 3. Formation of halides, oxides and sulphides: Similar to alkali metals, hydrogen can form corresponding halides, oxides and sulphides. eg.

Halides Sulphides Oxides Alkali metals NaCl KCl

Sodium Potassium chloride chloride

Na2S K2S Sodium Potassium sulphide sulphide

Na2O K2O Sodium Potassium oxide oxide

Hydrogen HCl Hydrochloric acid

H2S Hydrogen sulphide

H2O Water

b. Resemblance with halogens with respect to: 1. Electronic configuration: i. Halogens which are placed in the seventeenth group of periodic table have the

electronic configuration as ns2np5. ii. They require one electron to attain the corresponding noble gas configuration. iii. Hydrogen can be the member of group seventeen, similar to halogens as it is short

of one electron to achieve the corresponding noble gas configuration of helium (1s2).

2. Formation of uninegative ions: Similar to halogens, hydrogen can gain one electron to form uninegative ion. eg. H + e− ⎯→ H− F + e− ⎯→ F−

(1s1) (1s2) [He]2s22p5 [He]2s22p6 Hydrogen Hydride ion Fluorine Fluoride atom atom ion 3. Formation of diatomic molecules and other covalent compounds: Similar to halogens, hydrogen also forms diatomic molecules, hydrides and large number

of covalent compounds. eg. i. Formation of diatomic molecules (like halogens): H2 F2, Cl2, Br2, I2 ii. Formation of hydrides: Na+H− Na+Cl− Sodium hydride Sodium chloride iii. Formation of covalent compounds: CH4, SiH4, GeH4 Compounds of hydrogen CCl4, SiCl4, GeCl4 Compounds of halogens. iv. In terms of ionization enthalpy, hydrogen (which has a very high ionization enthalpy) resembles more

with halogens. a. Ionization enthalpy of hydrogen and alkali metals: H : ∆H = 1312 kJ mol−1 Li: ∆H = 520 kJ mol−1 Na: ∆H = 496 kJ mol−1 K : ∆H = 419 kJ mol−1


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b. Ionization enthalpy of hydrogen and halogens: H : ∆H = 1312 kJ mol−1 F: ∆H = 1681 kJ mol−1 Cl: ∆H = 1255 kJ mol−1 Br : ∆H = 1140 kJ mol−1 v. Also, hydrogen does not possess metallic character under normal conditions unlike the alkali metals

which exhibit metallic character. vi. Hydrogen is very less reactive as compared to halogens. vii. Thus, hydrogen resembles both the alkali metals and the halogens, though in case of certain

properties, it also differs from them. viii. This unique behaviour of hydrogen makes its position in the periodic table controversial or

anomalous. ix. Therefore, hydrogen is best placed separately in the periodic table. Q.3. H+ does not exist freely. Explain. Ans: i. H+ (proton) is formed by loss of electron from hydrogen atom. ii. The nucleus of proton is of extremely small size (−1.5 × 10−3 pm) compared to normal atomic and

ionic sizes (50 to 200 pm). iii. Consequently, H+ does not exist freely and is always associated with other atoms or molecules. *Q.4. Why does hydrogen occur in a diatomic form rather than in monoatomic form under normal

conditions? (NCERT) Ans: i. Hydrogen atom has only one electron in its valence shell having electronic configuration 1s1. ii. It can acquire stable configuration of helium by sharing this electron with another hydrogen atom. iii. Therefore, it shares its single electron with electron of the other H-atom to achieve stable inert gas

configuration of He. iv. Thus, hydrogen readily forms diatomic molecule and exists as H2 rather than in monoatomic form. 9.2 Occurrence of hydrogen (dihydrogen) *Q.5. What is another name of hydrogen? How does hydrogen occur in nature? Ans: i. Hydrogen in the elemental form exists as a diatomic molecule (H2) and hence it is also known as

dihydrogen. ii. Occurrence of hydrogen: a. Hydrogen is the most abundant element in the universe (about 70% of the total mass of the

universe) and is the main element in the solar system. b. Hydrogen forms the major constituent of planets like Jupiter and Saturn. c. In the free state: 1. Hydrogen is present in some volcanic gases and in the outer atmosphere of sun and other

stars of the universe. 2. Earth does not have enough gravitational pull to retain light hydrogen molecules. Thus,

hydrogen in free state is much less abundant (0.15 % by mass) in the earth’s atmosphere. d. In the combined form: 1. It constitutes 15.4% of earth’s crust and is also found in oceans (in the form of water). It is

the third most abundant element in the earth’s crust. 2. It is also found in the plant and animal tissues, carbohydrates, proteins, hydrocarbons,

hydrides and several other compounds. 9.3 Isotopes of hydrogen *Q.6. Mention the isotopes of hydrogen. (NCERT) Ans: Hydrogen has three isotopes namely protium ( 1

1H ), dueterium ( 21 H or D) and tritium ( 3

1 H or T). Q.7. Name the isotope of hydrogen which i. does not contain neutron? ii. is radioactive? Ans: i. Protium ii. Tritium


4

Q.8. What is isotopic effect? Ans: Hydrogen has three isotopes viz., protium, deuterium and tritium. Since the three isotopes have identical

electronic configuration, their chemical properties are similar but they differ in their physical properties due to different masses. This effect is known as isotopic effect.

*Q.9. Write a brief note on isotopes of hydrogen. Ans: Hydrogen has three isotopes: i. Protium ( 1

1 H ): It has 1 electron, 1 proton and no neutron. Hydrogen occuring in nature is mainly composed of atoms of protium.

ii. Deuterium ( 21 H or D): It has 1 electron, 1 proton and 1 neutron. It is also called as “Heavy

Hydrogen”. Terrestrial hydrogen contains 0.0156% of deuterium mostly in the form of HD. iii. Tritium ( 3

1 H or T): It has 1 electron, 1 proton and 2 neutrons. It is the only radioactive isotope of hydrogen and the concentration of tritium is about one atom per 1018 atoms of protium. Since, it is radioactive, it emits low energy β- particles (half life = 12.33 years), hence called β – emitter.

iv. Since, the isotopes have the same electronic configuration, they have similar chemical properties, but differ in their physical properties (isotopic effect).

v. In the study of reaction mechanism, the replacement of H by D or T slows down the reaction rates. Hence, deuterium and tritium are used as tracers.

Q.10. Arrange H2, D2 and T2 in the decreasing order of their: i. boiling points ii. density. Ans: i. T2 > D2 > H2 ii. T2 > D2 > H2. Note: Atomic and physical properties of isotopes of hydrogen are as follows:

Properties Hydrogen Deuterium Tritium Active abundance (%) 99.985 0.0156 10−15 Relative atomic mass (g mol−1) 1.008 2.014 3.016 Melting point (K) 13.96 18.73 20.62 Boiling point (K) 20.39 23.67 25.0 Density (g L−1) 0.09 0.18 0.27 Enthalpy of fusion (kJ mol−1) 0.117 0.197 − Enthalpy of vapourization (kJ mol−1) 0.904 1.226 − Enthalpy of bond dissociation (kJ mol−1 at 298.3 K) 435.88 443.35 − Internuclear distance (pm) 74.14 74.14 − Ionization enthalpy (kJ mol−1) 1312 − − Electron gain enthalpy (kJ mol−1) −73 − − Covalent radius (pm) 37 − − Ionic radius (H−) (pm) 208 − −

9.4 Preparation of dihydrogen Q.11.*Explain the laboratory methods for preparation of di-hydrogen. OR Explain briefly the laboratory preparation for pure hydrogen. Ans: Laboratory method for preparation of hydrogen: i. Zinc on reaction with aqueous sodium hydroxide (NaOH) forms soluble sodium zincate and hydrogen gas. Zn + 2NaOH(aq) ⎯→ Na2ZnO2 + H2(g)↑ ii. Granulated zinc on reaction with dilute hydrochloric acid liberates hydrogen gas. Zn + 2HCl(dil) ⎯→ ZnCl2 + H2(g)↑


5

iii High purity hydrogen gas can be prepared by the action of pure dil. H2SO4 on magnesium ribbon. Mg + H2SO4(dil.) ⎯→ MgSO4 + H2(g)↑ iv Preparation of pure hydrogen gas can also be carried out by action of water on sodium hydride. NaH + H2O ⎯→ NaOH + H2(g)↑ v. Hydrogen can also be prepared by action of KOH on scrap aluminium or silicon. 2Al + 2KOH + 2H2O ⎯→ 2KAlO2 +3H2(g)↑ This method is known as Uyeno’s method and gives very pure hydrogen. Q.12.*Describe the bulk preparation of dihydrogen by electrolytic method. What is the role of an

electrolyte in the process? (NCERT) OR How is dihydrogen prepared by electrolysis of acidified water? Ans: i. Hydrogen can be obtained by electrolysis of acidified water (dilute sulphuric acid) using platinum

electrodes. ii. At cathode, dihydrogen (H2) is liberated whereas at anode, dioxygen (O2) gas is liberated. iii. Role of an electrolyte: The electrolyte used is acidified water. Since, pure water is a poor conductor

of electricity, it is acidified with dilute sulphuric acid to increase its electrical conductivity. iv. During electrolysis, the following reactions take place:

Acidic medium: H2SO4 2H+ + 24SO −

H2O H+ + OH−

At cathode : 2H+ + 2e− ⎯→ 2H; 2H ⎯→ H2(g)↑ At anode: 4OH− ⎯→ 4OH + 4e−; 4OH ⎯→ 2H2O + O2(g)↑ Since, the discharge potential of sulphate ions is much higher than that of hydrogen ions, sulphate

ions are not discharged at the anode. Q.13. How can pure dihydrogen be obtained from barium hydroxide? Ans: Electrolysis of warm aqueous solution of barium hydroxide using nickel electrodes yields dihydrogen of

high purity (> 99.96%). Q.14. How is hydrogen obtained from water gas? OR Describe Bosch process for the preparation of hydrogen. Ans: i. In Bosch process, water gas (CO + H2) is mixed with twice the volume of steam. ii. In the presence of promoter (like Cr2O3 or ThO2), this mixture is then passed over heated catalyst

(Fe2O3) at 773 K. iii. In this reaction, oxidation of CO to CO2 takes place. CO + H2 + H2O Fe O Cr O2 3 2 3

773K+⎯⎯⎯⎯⎯→ CO2(g)↑+ 2H2(g)

(Water gas) Steam iv. Carbon dioxide so formed is removed by dissolving it in water under pressure (20 − 25 atmospheres),

thus leaving behind hydrogen which is collected. Q.15. Explain how dihydrogen is obtained by the electrolysis of brine solution. Ans: i. Electrolysis of brine solution is used for the manufacture of sodium hydroxide and chlorine and

dihydrogen is obtained as a byproduct in the process. ii. Preferentially H+ ions are reduced at cathode liberating hydrogen gas, while Cl− ions are oxidised at

anode liberating Cl2 gas.


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iii. During electrolysis, the following reactions take place: At anode: (aq)2Cl− ⎯→ Cl2(g) + 2e− At cathode: 2H2O + 2e− ⎯→ H2(g) + (aq)2OH−

The overall reaction is, (aq)2Cl− +2H2O(l) ⎯→ Cl2(g) + H2(g) + (aq)2OH−

Q.16. How is hydrogen gas prepared from coke or hydrocarbons? Ans: Reaction of steam on hydrocarbons or coke at high temperatures in the presence of catalyst yields hydrogen. CnH2n+2 + nH2O 1270K

Ni⎯⎯⎯→ nCO(g) + (2n + 1)H2 eg. CH4(g) + H2O(g)

1270KNi⎯⎯⎯→ CO(g) + 3H2(g)

*Q.17.Explain the terms: i Syngas ii. Water gas shift reaction. (NCERT) Ans: i. Syngas: a. Syngas is the mixture of CO and H2. It is also called ‘water gas’. b. It is used for the synthesis of methanol and many hydrocarbons, hence, the name syngas or

‘synthesis gas’. c. It is produced from saw dust or scrap wood. d. The process of producing syngas from coal or coke is called ‘coal gasification’. e. Methanol can be prepared from syngas. CO(g) + 2H2(g) Cobalt

catalyst⎯⎯⎯→CH3 − OH(l) Syngas Methanol (water gas) f. Syngas is also used to prepare hydrogen. CO + H2 + H2O Fe O Cr O2 3 2 3

773K+⎯⎯⎯⎯⎯→ CO2(g) ↑ + 2H2(g)

(Water gas) Steam ii. Water gas shift reaction: a. Since it is difficult to remove carbon monoxide from the water gas (syngas), obtaining

hydrogen from water gas (syngas) also becomes difficult. b. In order to have an increased production of dihydrogen, water gas is mixed with steam and the

resulting gaseous mixture is heated at 673 K in the presence of iron chromate (FeCrO4) catalyst. CO(g) + H2O(g)

4

673KFeCrO⎯⎯⎯→ CO2(g) + H2(g)

Carbon Steam Carbon Hydrogen monoxide dioxide (from syngas mixture) This reaction is called water gas shift reaction. c. The carbon monoxide of the syngas mixture reacts with steam to form carbon dioxide and

hydrogen. d. Carbon dioxide is then removed by scrubbing with sodium arsenite solution. Q.18. What is Lane’s process? OR How is hydrogen prepared from superheated steam? Ans: When superheated steam is passed over iron filings heated to about 1023 − 1073 K, hydrogen is formed. 3Fe + 4H2O

1023 1073KFe−⎯⎯⎯⎯⎯→ Fe3O4(s) + 4H2(g)

Steam


7

*Q.19.Complete the following chemical reactions: i. CO(g) + H2O(g) catalyst

∆⎯⎯⎯→ ii. Zn(s) + NaOH heat⎯⎯⎯→ (NCERT)

iii. CH4(g) + H2O 1270 K⎯⎯⎯→ Ans: i. CO(g) + H2O(g) catalyst

∆⎯⎯⎯→ CO2(g) + H2(g) Carbon Steam Carbon Dihydrogen monoxide dioxide ii. Zn(s) + 2NaOH(aq) heat⎯⎯→Na2ZnO2(aq) + H2(g) Zinc Sodium Sodium Dihydrogen metal hydroxide zincate iii. CH4(g) + H2O(g) Ni

1270K⎯⎯⎯→ CO(g) + 3H2(g) Methane Steam Carbon Dihydrogen monoxide 9.5 Properties of dihydrogen *Q.20.What are the physical properties of dihydrogen? Ans: i. Dihydrogen is a colourless, odourless and tasteless gas. ii. It is combustible, but does not support combustion. iii. It is lighter than air and is insoluble in water. iv. It’s relative abundance is 99.985% while deuterium is 0.0156% and tritium is 10−15 %. v. Melting point of dihydrogen is 13.96 K and its boiling point is 20.39 K. vi. Relative atomic mass of dihydrogen is 1.008 g mol−1 and its density is 0.09 g/L. Q.21. Describe the chemical behaviour of dihydrogen. Ans: i. At room temperature, hydrogen is not very active (relatively inert) due to it’s high H − H bond

dissociation enthalpy (435.88 kJ mol−1 at 298 K). ii. Thus, production of atomic hydrogen is carried out at a high temperature in an electric arc or under

ultraviolet radiations. iii. Hydrogen combines with almost all the elements owing to its incomplete orbital with 1s1 electronic

configuration. iv. Hydrogen exhibits many reactions due to: a. loss of electron to form H+ ion. b. gain of an electron to form H− ion. c. sharing of electrons to form single covalent bond. *Q.22.What is the effect of high enthalpy of H−H bond in terms of chemical reactivity of dihydrogen?

(NCERT) Ans: i. Chemical reactivity of dihydrogen involves the breaking of H−H bond. ii. The bond dissociation enthalpy of H−H bond is very high (435.88 kJ mol−1 at 298 K). iii. Due to high bond enthalpy, it is not very reactive at room temperature. iv. However, at high temperature or in the presence of catalysts, hydrogen combines with many metals

and non-metals to form corresponding hydrides. Q.23. What happens when hydrogen reacts with halogens? Ans: i. Hydrogen reacts with halogens (X2) to give the corresponding hydrogen halides (HX). H2(g) + X2(g) ⎯→ 2HX(g) Hydrogen Halogen Hydrogen halide (X = F, Cl, Br, I)


8

ii. The reaction with iodine requires a catalyst. iii. The reaction between fluorine and hydrogen to form hydrogen fluoride occurs at very low

temperature (63 K) and even in dark due to greater affinity of fluorine towards hydrogen. H2(g) + F2(g) ⎯→ 2HF Hydrogen Fluorine Hydrogen fluoride iv. Chlorine reacts with H2 gas in the presence of diffused sunlight, 673 K, pressure forming HCl. H2(g) + Cl2(g) diffused sunlight

673 K, pressure⎯⎯⎯⎯⎯→ 2HCl(g)

Hydrogen Chlorine Hydrogen chloride Q.24. What happens when dihydrogen reacts with: i. dioxygen ii. dinitrogen (Haber’s process) iii. sodium metal. Ans: i. Dihydrogen reacts with dioxygen: Dihydrogen reacts with dioxygen to form water and this reaction is highly exothermic. 2H2(g) + O2(g) catalyst or heating⎯⎯⎯⎯⎯⎯→ 2H2O(l) ∆H = −285.9 kJ mol−1 Dihydrogen Dioxygen Water ii. Dihydrogen reacts with dinitrogen (Haber’s process): Dihydrogen reacts with dinitrogen at 673K under 200 atm pressure in the presence of iron as a

catalyst to form ammonia. 3H2(g) + N2(g)

673K,Fecatalyst200 atm⎯⎯⎯⎯⎯→ 2NH3(g) ∆H = −92.6 kJ mol−1

Dihydrogen Dinitrogen Ammonia iii. Dihydrogen reacts with sodium metal: Dihydrogen combines with metallic sodium at high temperature to yield sodium hydrides. 2Na + H2 ⎯→ 2NaH Sodium metal Dihydrogen Sodium hydride Q.25. Explain the reducing nature of hydrogen. Ans: i. Dihydrogen reduces some metal ions in aqueous solutions and oxides of metals (less active than iron)

into corresponding metals. H2(g) + Pd2+

(aq) ⎯→ Pd(s) + 2H+(aq)

H2(g) + CuO(s) ⎯→ Cu(s) + H2O(l)

ii. Hydrogen acts as a reducing agent, hence reduces certain oxides of metals below it in the electrochemical series of metals.

ZnO + H2 ⎯→ Zn + H2O Fe3O4 + 4H2 ⎯→ 3Fe + 4H2O Q.26.*Explain the term hydrogenation. (NCERT) OR How does hydrogen react with various organic compounds to give useful, commercially important products? Ans: The reaction in which hydrogen gas reacts with unsaturated organic compounds in the presence of a catalyst

to form hydrogenated (saturated) compounds is called hydrogenation. Hydrogenation gives some useful products of commercial importance. eg. i. Hydrogenation of polyunsaturated vegetable oil gives solid edible fats like margarine and vanaspati

ghee. Vegetable oil + H2

Finelydivided Ni450K,8 10 atm−⎯⎯⎯⎯⎯→ Solid fat

The reaction is also called as hardening of oil wherein liquid oil (containing unsaturated triglycerides)

gets converted into solid fat (containing saturated triglycerides).

(i.e., vanaspati ghee)


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ii. Hydrogenation of olefins gives corresponding aldehydes which further undergo reduction to give corresponding alcohol.

H2 + CO + CH3CH = CH2 ⎯→ CH3CH2CH2CHO Propene Butyraldehyde (Olefin) (Aldehyde) H2 + CH3CH2CH2CHO ⎯→ CH3CH2CH2CH2OH Butyraldehyde n-Butyl alcohol (Butanal) (Butan-1-ol) 9.6 Uses of dihydrogen Q.27.*What are the uses of dihydrogen? OR Give various applications of dihydrogen. Ans: Dihydrogen is used in i. the manufacture of vanaspati fat (ghee) by hydrogenation of polyunsaturated vegetable oils like

soyabean, cotton seed, etc. ii. the synthesis of ammonia, by Haber’s process. iii. the manufacture of methanol, hydrogen chloride and metal hydrides. (i.e., ionic hydrides, covalent

hydrides and interstitial hydrides) iv. fuel cells, for generating electrical energy. v. metallurgy, to reduce heavy metal oxides to metals. vi. atomic hydrogen and oxyhydrogen torches. vii. rocket fuel (mixture of liquid hydrogen and liquid oxygen). Q.28. Mention the benefits of using dihydrogen in fuel cell for generating electrical energy. Ans: i. Using dihydrogen in fuel cell has many advantages over conventional fossil fuels and electric power. ii. In comparison to gasoline and other fuels, dihydrogen releases greater energy per unit mass of fuel

and does not produce pollution. *Q.29.How does the atomic hydrogen or oxy-hydrogen torch function for cutting and welding purposes?

Explain. (NCERT) Ans: i. For cutting and welding of metals, very high temperature is required. ii. Atomic hydrogen is produced when molecular hydrogen (dihydrogen gas) at atmospheric pressure is

passed through an electric arc struck between two tungsten electrodes at 3773−4273 K. H2 Electricarc

3773 4273K−⎯⎯⎯⎯⎯→ 2H ∆H = 435.90 kJ mol−1

Molecular Atomic hydrogen hydrogen iii. The atomic hydrogen atoms thus produced have a very short life time of 0.3 seconds and are

extremely reactive. iv. The atomic hydrogen atoms get immediately converted into molecular form and during this

conversion, large amount of energy is liberated. v. Thus, atomic hydrogen or oxy-hydrogen torches are used for cutting and welding purposes, wherein

atomic hydrogen atoms are allowed to recombine to generate a temperature of 4000 K on the surface to be welded.

9.7 Hydrides (ionic, covalent, interstitial) *Q.30. What are hydrides? Mention different types of hydrides. Ans: i. Hydrides: a. Dihydrogen, under certain reaction conditions combines with almost all elements (except noble

gases) to form binary compounds, called hydrides. b. Hydrides can be represented as EHx (eg. LiH, MgH2) or EmHn (eg. B2H6), where E is the symbol

of an element.


10

ii. Hydrides are classified into three categories: a. Ionic or saline or salt like hydrides: eg. LiH, MgH2 b. Covalent or molecular hydrides: eg. H2O, NH3 c. Metallic or non-stoichiometric or interstitial hydrides: eg. TiH1.8−2.0, LaH2.87 Q.31.*Explain ionic hydrides. OR Write a note on ionic or saline hydrides. Ans: i. Ionic hydrides or saline hydrides are formed by the combination of hydrogen with metals having low

electronegativity values. These hydrides are stoichiometric compounds with highly electropositive character.

ii. These include elements of s-block i.e., of group 1 (alkali metals), group 2 (alkaline earth metals). However, stability of these hydrides decreases down the group.

iii. Ionic hydrides are generally crystalline, non-volatile and non-conducting in solid state. iv. However, in the fused state, they conduct electricity liberating hydrogen at the anode, which indicates

the presence of hydride (H−) ion. 2H−

(melt) anode⎯⎯⎯→ H2(g) + 2e− v. Saline hydrides react violently with water to produce dihydrogen gas. KH(s) + H2O(aq) ⎯→ KOH(aq) + H2(g) vi. Lighter metal hydrides such as LiH, BeH2 and MgH2 exhibit significant covalent character. BeH2 and

MgH2 are polymeric in structure. vii. Lithium hydride is used in the synthesis of other hydrides, as it does not react with Cl2 or O2 at

moderate temperature. 8LiH + Al2Cl6 ⎯→ 2LiAlH4 + 6LiCl 2LiH + B2H6 ⎯→ 2LiBH4 Q.32.*Explain covalent hydrides. OR Write a note on covalent or molecular hydrides. Ans: i. Covalent or molecular hydrides are formed by a reaction of dihydrogen with most of the p–block

elements. eg. H2O, CH4, NH3 and HF. ii. Covalent hydrides are volatile compounds. iii. Molecular or covalent hydrides are further classified according to the relative number of electrons and

bonds in their Lewis structure into a. Electron - deficient hydride: eg. diborane (B2H6), aluminium hydride (AlH3), etc. b. Electron - precise hydride: eg. methane (CH4), silicon hydride (SiH4), etc. c. Electron - rich hydride: eg. ammonia (NH3), water (H2O), hydrogen fluoride (HF), etc. Q.33. *Explain the term electron deficient compounds of hydrogen with suitable examples. (NCERT) OR Write a note on electron deficient hydrides. Ans: i. Hydrides of group 13 elements i.e., the boron family (eg. BH3, AlH3, etc.) do not have sufficient

number of electrons to form normal covalent bonds and hence, are called electron deficient hydrides. ii. These hydrides have very few electrons for writing their conventional Lewis structure. eg. Diborane (B2H6). iii. These hydrides behave as Lewis acids (electron pair acceptors). iv. To make up for this deficiency, they generally exist in polymeric forms such as B2H6, Al2H6, etc.


11

Q.34. *Explain the term electron precise compounds of hydrogen with suitable examples. (NCERT) OR Write a note on electron precise hydrides. Ans: i. Hydrides of group 14 elements i.e., the carbon family (eg. CH4, SiH4, GeH4, SnH4, PbH4) have exact

number of electrons to form covalent bonds and hence, are called electron-precise hydrides. ii. These hydrides have the required number of electrons to write their conventional Lewis structures. iii. All these compounds possess tetrahedral geometry. Q.35. *Explain the term electron rich compounds of hydrogen with suitable examples. (NCERT) OR Write a note on electron rich hydrides. Ans: i. Hydrides of group 15, 16 and 17 elements i.e., the nitrogen family, oxygen family and fluorine family

respectively, have excess electrons which are present as lone pairs and hence, these hydrides are known as electron-rich hydrides.

ii. These hydrides behave as Lewis bases (electron pair donors). eg. Molecules of NH3, H2O and HF contain one, two and three lone pairs of electrons respectively and

thus behave as Lewis bases. iii. The presence of lone pairs of electrons on the highly electronegative atoms like N (3.0), O (3.5) and

F (4.0) in their corresponding hydrides results in hydrogen bond formation of considerable magnitude between the molecules which in turn leads to association of molecules.

iv. Hence, the boiling points of NH3, H2O and HF is higher than the hydrides of their subsequent group members.

#Q.36.Among NH3, H2O and HF, which would you expect to have highest magnitude of hydrogen bonding

and why? (NCERT) Ans: i. Hydrogen bond is formed due to greater difference in the electronegativity of H-atom and the other

atom involved in the bond formation. ii. Among N (3.0), O (3.5) and F (4.0), fluorine is the most electronegative atom. iii. Hence, among the hydrides NH3, H2O and HF, the electronegativity difference between H and F atom

is the greatest. iv. Hence, HF has the highest magnitude of hydrogen bonding (in which H carries partial positive charge

and F carries partial negative charge) than NH3 and H2O.

H H

H N H

H

H FO

Hydrogen bonding in (a) NH3 (b) H2O (c) HF

H

F

F140o

δ−

F

F

δ+H

δ+

δ− δ−

(c)

δ−

δ+H

δ+H

(a)

H Nδ- δ+

H N H Nδ+ δ- δ-

H HH

H H H

O

δ+H

O

HH

HO

H

H

H

O

H

δ-

δ−

δ+

δ−

δ+

δ+

δ+

δ−

δ+

δ+

δ+

(b)


12

Q.37. Distinguish between ionic hydrides and covalent hydrides. Ans:

No. Ionic hydrides Covalent hydrides i. Ionic hydrides are formed by s-block

elements. i. Covalent hydrides are formed by p-block

elements. ii. Ionic hydrides are crystalline, non-volatile

in solid state. ii. Covalent hydrides are gaseous or volatile

liquid compounds. iii. Formed generally by metals. iii. Formed by non-metals.

Q.38. Write a note on metallic hydrides. Ans: i. Most of the d-block elements and f-block elements form metallic hydrides. ii. Only chromium from 6th group forms hydride like CrH whereas Mn group (7th group), Fe group (8th

group) and cobalt group (9th group) do not form hydrides. iii. Metallic or interstitial hydrides conduct heat and electricity, though not as efficiently as their parent

metals. iv. These hydrides are almost always non-stoichiometric being deficient in hydrogen. eg. TiH1.8−2 LaH2.87 YbH2.55 ZrH1.3 − 1. 75 VH0.56 NiH0.6 − 0.7, etc v. In such hydrides, the law of constant proportion does not hold good as the quantity of hydrogen does

not bear an exact stoichiometric ratio to the metal. vi. Earlier it was thought that, hydrogen atoms produce distortion in metal lattices (without any change in

its type) by occupying interstitial positions. Consequently they were termed as interstitial hydrides. vii. However, recent studies have shown that metallic hydrides have different lattice structure from that of

parent metal (Exception: Hydrides of Ni, Pd, Ce and Ac). viii. Transition metals have greater tendency to adsorb hydrogen. This property is widely used in

preparing large number of compounds by catalytic reduction or hydrogenation reactions. ix. Metals like platinum (Pt) and palladium (Pd) are used as a hydrogen storage media and as a source of

energy as they can accommodate a very large volume of hydrogen. Note: Hydrides formed by elements like B (eg. B2H6), Al [eg. (AlH3)x] and Ga (eg. Ga2H6) are called as polynuclear

hydrides. Characteristics of different hydrides:

Type Formed by Physical character Bonding Examples Ionic s-block elements Salt like solids, good

conductors in fused state. Largely ionic

LiH, NaH, CaH2

Molecular or covalent

p-block elements Gaseous or volatile liquids, non-conductors.

Covalent HF, H2O, CH4, H2S

Metallic or interstitial

d and f-block elements

Metallic appearance, conductors/ semi-conductors.

Partly ionic and metallic bond

ZrH1.3−1.75,TiH1.8−2, LaH2.87


13

9.8 Water (physical and chemical properties of water) Q.39. “Chemistry of world is chemistry of water”⎯Justify. Ans: i. Water is the most common, abundant and easily obtainable of all chemical compounds. It is a crucial

compound for survival of life on earth. ii. Human body has about 65% and some plants have 95% water which indicates that a major part of all

living organisms is made up of water. iii. Water is an universal solvent and can exist in solid, liquid and gaseous form. It is used as a principal

cleansing agent, acids in speeding up chemical action and is the basic material needed by almost all industries.

iv. Life on earth would be totally impossible in the absence of water. Thus, it is rightly said that the ‘Chemistry of world is the chemistry of water.’

Note: The distribution of water over the earth’s surface is not uniform. The estimated source of water supply is

given in table below:

Source % of TotalOceans 97.33 Saline lakes and inland sea 0.008 Polar ice and glaciers 2.04 Underground water 0.61 Lakes 0.009 Soil moisture 0.005 Atmospheric water vapour 0.001 Rivers 0.0001

Q.40.What are the physical properties of water? Ans: Physical properties of water: i. Water is a colourless, tasteless liquid. ii. Pure water is a poor conductor of electricity. iii. It has high dielectric constant (78.39). iv. Its boiling point is 373 K and freezing point 273 K. v. Its density is 1.00 g cm−3. vi. There is extensive hydrogen bonding between water molecules. Note: Physical properties of water are given below:

Properties H2O i. Molecular mass (g mol−1) 18.0151 ii. Enthalpy of formation (kJ mol−1) −285.9 iii. Enthalpy of fusion (kJ mol−1) 6.01 iv. Enthalpy of vapourization (kJ mol−1) 40.66 v. Melting point (K) 273 vi. Boiling point (K) 373 vii. Temperature of maximum density (K) 276.98 viii. Density at 298 K (g cm−3) 1.0000 ix. Dielectric constant (C2 / N.m2) 78.39 x. Electrical conductivity at 293 K (ohm−1 cm−1) 5.7 × 10−8

xi. Viscosity (centipoise) 0.8903


14

Q.41. Water shows unusual properties in the condensed phase. Explain. Ans: i. In the condensed phase (i.e., in liquid and solid phase), it has been observed that properties like

freezing point, boiling point, heat of vapourization and heat of fusion of water are abnormally higher than those of hydrides of other elements of group 16. (H2S, H2Se, etc.)

ii. The presence of extensive hydrogen bonding between water molecules results in the unusual properties of water in the condensed phase.

Q.42. Why is water called as universal solvent? Ans: i. Water has a high dielectric constant. Hence, it has high ability to dissolve most of the inorganic

(ionic) compounds. ii. Also most of the organic covalent compounds like alcohols, carboxylic acids, carbohydrates, urea,

etc., are able to form hydrogen bonds with water molecules and thus, they dissolve in water. iii. Water can serve as a solvent for variety of compounds and it is also available in abundance. Thus, water is regarded as a universal solvent. Note: Water is capable of dissolving most of the inorganic (ionic) compounds, except for some ionic compounds

like AgCl, BaSO4, CaF2, AlF3, etc., which are insoluble in it owing to their exceptionally high attractive forces in the lattice.

Q.43. How does unique properties of water help to maintain the biosphere? Ans: i. The high heat of vapourization and heat capacity of water are responsible for moderation of climate

and body temperature of living organisms. ii. Properties like specific heat, thermal conductivity, surface tension, dipole moment and dielectric

constant of water have higher values compared to other liquids. iii. Plant and animal metabolism requires transportation of ions and molecules. Water (being an universal

solvent) serves as an excellent solvent for this purpose. iv. Due to these unique properties, water plays a key role in the biosphere. Q.44. Describe the structure of water. Ans: i. Water molecule has two hydrogen atoms and one oxygen atom. ii. The oxygen atom undergoes sp3 hybridisation and two O−H bonds (bond length 95.7 pm) are formed

by sp-s overlapping. iii. The repulsion forces between two lone pairs present on oxygen atom gives a bent or angular structure

to water molecule with a H − O − H bond angle of 104°35′. iv. Structure of water is as shown in the following figure:

v. It is a highly polar molecule (as shown in figure b). In the liquid phase, water molecules are

associated together with hydrogen bonds. vi. In solid phase, water is present in the crystalline form i.e., ice. Ice crystallizes in the hexagonal form at atmospheric pressure and condenses to cubic form at very

low temperatures.

(a) The bent structure of water molecule (b) The water molecule as a dipole and (c) the sp3-s orbital overlap in a water molecule

lone pair of electrons

O

H H

(a)

104°35′

95.7 pm O2δ−

δ+

HHδ+

(b)

sp3 hybrid orbitals of oxygen

s-orbitals of hydrogen

HH

104°35′

O

Lone pair of electrons

(c)


15

Q.45. Describe the hydrogen bonding in water. Ans: i. Polar nature of water molecules results in the formation of intermolecular hydrogen bonds which hold

the water molecules together. ii. As a result, an extensive three dimensional network is formed as shown in the figure below. iii. In this arrangement, each oxygen is tetrahedrally surrounded by four hydrogen atoms, two linked by

covalent bonds and two by hydrogen bonds. Q.46. Explain the structure of ice. Ans: i. The crystalline form of water is ice and it has a highly ordered

three dimensional hydrogen bonded structure. ii. Ice crystallizes in the hexagonal form at atmospheric pressure

and condenses to cubic form at very low temperatures. iii. X-ray study of ice shows that each oxygen atom is surrounded

tetrahedrally by four hydrogen atoms at a distance of 276 pm. iv. Thus, due to hydrogen bonding, structure of ice appears as an

open cage with wide holes. v. These holes can hold some other molecules of appropriate size

interstitially. Q.47. “Why does ice float on water?” Explain. Ans: i. Liquid water and ice consists of aggregates of varying number of water molecules held together by

hydrogen bonds. ii. Due to the hydrogen bonding, ice has an open cage-like structure having a number of vacant spaces. iii. When ice melts, some of the hydrogen bonds are broken and the water molecules go in between the vacant

spaces in the structure. As a result, the structure of liquid water is less open than structure of ice. iv. Therefore, the density of ice is less than that of water and consequently ice floats over water. Q.48.*What is auto-protolysis of water? What is its significance? (NCERT) OR Write a note on amphoteric nature of water. Ans: i. Auto-protolysis of water means self ionization of water. It may be represented as, H2O(l) + H2O(l) 3 (aq)H O+ + (aq)OH− acid-1 base-2 acid-2 base-1 (Acid) (Base) (Conjugate (Conjugate acid) base)

The structure of ice

= O= H

Hydrogen bonds

O HH

O H

H

OH

OH

HHH

O

O

H

H

H

H

O Hydrogen bondsH


16

ii. Due to auto-protolysis, water is considered as amphoteric in nature i.e., it has the ability to act as an acid as well as base. Water, in accordance with Bronsted concept, acts as a base towards acids stronger than itself (like hydrogen chloride) and as an acid towards bases stronger than itself (like ammonia).

eg. H2O(l) + NH3(aq) ( ) ( )aq 4 aqOH NH− ++ (Water as an acid) acid base

H2O(l) + HCl (aq) ( ) ( )3 aq aqH O Cl+ −+ (Water as a base) base acid Q.49. Explain the redox reactions involving water. OR Show that water can be a source of dihydrogen as well as oxygen. Ans: i. Highly electropositive metals like Na can easily reduce water to dihydrogen. 2H2O(l) + 2Na(s) ⎯→ 2NaOH(aq) + H2(g) Thus, it is a great source of dihydrogen. ii. During, photosynthesis, oxidation of water to O2 takes place. 6CO2(g) +12H2O(l) sunlight

chlorophyll⎯⎯⎯⎯→ C6H12O6 (aq) + 6H2O(l) + 6O2(g) iii. With fluorine, water gets oxidized to O2. 2F2(g) + 2H2O(l) ⎯→ 4H+

(aq) + 4F−(aq) + O2(g)

iv. Thus, water can be a source of both dihydrogen as well as oxygen i.e., it can undergo both reduction as well as oxidation reaction (redox behaviour).

Q.50. What is hydrolysis? Explain the hydrolysis of covalent ionic compounds. Ans: Hydrolysis: Decomposition of a compound in the presence of water is called as hydrolysis. OR Interaction of H+ and OH− ions of H2O with anion and the cation of a salt respectively to give the original

acid and original base is called hydrolysis. Hydrolysis of covalent and ionic compounds: i. Water which is a universal solvent, has very strong solvating tendency due to high dielectric constant. ii. It dissolves many ionic compounds. However, certain covalent and some ionic compounds are

hydrolysed in water. eg. P4O10(s) + 6H2O(l) ⎯→ 4H3PO4(aq) SiCl4(l) + 2H2O(l) ⎯→ SiO2(s) + 4HCl(aq) CaO(s) + H2O(l) ⎯→ Ca(OH)2(aq) CaC2(s) + 2H2O(l) ⎯→ C2H2(s) + Ca(OH)2(aq) Mg3N2(s) + 6H2O(l) ⎯→ 3Mg(OH)2(aq) + 2NH3(g) Na2CO3(s) + 2H2O(l) ⎯→ 2NaOH(aq) + H2CO3(l) *Q.51. Complete the following chemical reactions: i. CaO + H2O ⎯→ ii. AlCl3 + H2O ⎯→ iii. Ca3N2(s) + H2O(l) ⎯→ Ans: i. CaO(s) + H2O(l) ⎯→ Ca(OH)2(aq) ii. AlCl3(s) + 3H2O(l) ⎯→ Al(OH)3(s) +3HCl(aq) iii. Ca3N2(s) + 6H2O(l) ⎯→ 3Ca(OH)2(aq) + 2NH3(g) Q.52. Write a note on formation of hydrates. Ans: i. Hydration means addition of H2O to ions or molecules to form hydrated ions or hydrated salts. ii. Many salts can be crystallized as hydrated salts from their aqueous solution. iii. Such an association of water is of the following different types. a. Co-ordinated water: eg. [Cr(H2O)6]3+3Cl−

b. Interstitial water: eg. BaCl2.2H2O c. Hydrogen-bonded water: eg. [Cu(H2O)4]2+ 2

4 2SO .H O− in CuSO4.5H2O


17

Note: In case of [Cu(H2O)4]2+ 2

4 2SO .H O− , one water molecule, which is outside the bracket (co-ordination sphere), is hydrogen bonded. The other four molecules of water are co-ordinated. *Q.53.What is the difference between the terms hydrolysis and hydration? (NCERT) Ans: i. Interaction of H+ and OH− ions of H2O with anion and the cation of a salt respectively to give the

original acid and original base is called hydrolysis. eg. Na2CO3 + 2H2O ⎯→ 2NaOH + H2CO3 Salt Base Acid ii. Hydration means addition of H2O to ions or molecules to form hydrated ions or hydrated salts. eg. CuSO4(s) + 5H2O(l) ⎯→ CuSO4.5H2O(s) Salt Hydrated salt (Colourless) (Blue) 9.9 Heavy water Q.54. What is heavy water? What are its uses? Ans: i. Heavy water is also called as ‘Deuterium oxide’ (D2O). ii. It is the water which contains deuterium i.e., heavy hydrogen ( 2

1 H ), an isotope of hydrogen. iii. Uses: a. Heavy water slows down the fast moving neutrons and hence, is used as moderator in the

radioactive nuclear reactors. b. It is used as tracer compound in determining the mechanism of many organic reactions. c. It is also used in the preparation of other deuterium compounds. Note: Urey in 1932 discovered heavy water. Q.55. Give an account of the isotopic varieties of water? Ans: i. Hydrogen and oxygen have 3 isotopic forms each and thus, there are 18 isotopic varieties of water

which are possible. Hydrogen has three isotopes namely, protium ( 11 H ), dueterium ( 2

1 H or D) and tritium ( 3

1 H or T). ii. The most common isotope of hydrogen is protium which has only one proton in its nucleus and no

neutron. Water with protium isotope is also known as protium water or ordinary water and is vital for life processes.

iii. A dueterium atom contains one proton and one neutron in its nucleus and when combined with oxygen, it forms heavy water (D2O).

Mass of dueterium atom is about twice the mass of ordinary hydrogen atom (i.e., it is 1.998 times heavier).

Thus, in context of mass, a given volume of heavy water is about 11% heavier compared to the same volume of the ordinary water.

Heavy water differs markedly from ordinary water in properties like freezing point (3.8 °C compared to 0 °C of ordinary water) and boiling point (101.4 °C as compared to 100 °C of ordinary water).

iv. The tritium atom contains one proton and two neutrons in its nucleus and is about three times heavier than ordinary hydrogen atom.

Tritium forms super-heavy water (T2O) by combining with oxygen. The super-heavy water differs from ordinary water in properties like freezing point (T2O freezes at

9 °C), boiling point (T2O boils at 104 °C) and density (at 20 °C, density of T2O is 1.88 g/cm3). v. Oxygen has three isotopic forms 16O, 17O, 18O.


18

Note: 1. Isotopic varieties of water.

H216O

(18) Lightest water

D216O

(20) T2

16O (22)

HD16O (19)

HT16O (20)

DT16O (21)

H217O

(19) D2

17O (21)

T217O

(23) HD17O (20)

HT17O (21)

DT17O (22)

H218O

(20) D2

18O (22)

T218O

(24) Heaviest water

HD18O (21)

HT18O (22)

DT18O (23)

Out of these 18 varieties of isotopic compounds of water, the lightest is H2

16O with a molecular weight of 18 and heaviest is T2

18O with molecular weight of 24. 2. Water can be obtained from different sources like a pond, a pool, a river, a lake, a canal, a well or

even the sea. The abundance of any given isotopic variety of water depends upon the source from which it has been taken.

eg.

Source Abundance of heavy water (in gm per tonne)

Ordinary tap water

150

Pacific ocean

165

This difference is due to a process of isotopic exchange occuring continuously in nature. #Q.56.Knowing the properties of H2O and D2O, do you think that D2O can be used for drinking purpose? Ans: D2O cannot be used for drinking due to the following reasons: i. D2O is about 11% heavier and denser than H2O. ii. It boils at higher temperature i.e., 101.4 °C and freezes at 3.8 °C iii. It is mildly toxic and can interfere with normal functioning of our body. iv. Heavy water slows down the rates of biochemical reactions occurring in plants, animals and human

beings. Hence, D2O is injurious and not suitable for drinking purpose. Note: The physical properties of D2O are as tabulated below:

Property D2O Molecular mass (g mol−1) 20.0276 Melting point (K) 276.8 Boiling point (K) 374.4 Enthalpy of formation (kJ mol−1) −294.4 Enthalpy of vapourization at 373K (kJ mol−1) 41.61 Temperature of maximum density (K) 284.2 Density at 298 K (g cm−3) 1.1059 Viscosity (centipoise) 1.107 Dielectric constant (C2 / N.m2) 78.06 Electrical conductivity at 293 K (ohm−1 cm−1) -


34

32. The volume strength of 1.5 N H2O2 solution is _______.

(A) 4.8 (B) 8.4 (C) 3.0 (D) 8.0 33. Dihedral angle in H2O2 in crystalline state is

_______. (A) 90.2° (B) 109°28′ (C) 120° (D) 94.8° 34. The structure of H2O2 is _______. (A) planar (B) non-planar (C) spherical (D) linear 35. H2O2 is _______. (A) diamagnetic (B) paramagnetic (C) ferromagnetic (D) non-magnetic 36. H2O2 exhibits a dipole moment of _______. (A) 1.2 D (B) 2.1 D (C) 1.0 D (D) 2.0 D 37. H2O2 does NOT act as a _______. (A) reducing agent (B) oxidizing agent (C) dehydrating agent (D) bleaching agent 9.11 Hydrogen as a fuel 38. _______ is present as an impurity in dihydrogen. (A) Nitrogen (B) Oxygen (C) Helium (D) Carbon monoxide 39. Which of the following represents cathode

reaction in fuel cell?

(A) 2OH− ⎯→ H2O + 12

O2 + 2e−

(B) 2H2O + 2e− ⎯→ H2 + 2OH−

(C) 12

O2 + H2O + 2e− ⎯→ 2OH−

(D) H2 + 2OH− ⎯→ 2H2O + 2e− 40. _______ is the main energy currency in

hydrogen economy. (A) Electricity (B) Water (C) Dihydrogen gas (D) Dioxygen gas

Answers to Practice Problems

1. 10 volumes. 2. 3.57 N 3. 0.759 g. 4. 30 volumes.

Answers to Multiple Choice Questions 1. (B) 2. (C) 3. (D) 4. (B) 5. (D) 6. (B) 7. (D) 8. (C) 9. (A) 10. (B) 11. (B) 12. (A) 13. (B) 14. (B) 15. (D) 16. (B) 17. (D) 18. (D) 19. (D) 20. (D) 21. (C) 22. (C) 23. (C) 24. (D) 25. (B) 26. (B) 27. (D) 28. (B) 29. (C) 30. (C) 31. (B) 32. (B) 33. (A) 34. (B) 35. (A) 36. (B) 37. (C) 38. (A) 39. (D) 40. (C)

perfect chemistry - ii · r – o – r′ xi. aldehydes xii. ketones xiii. carboxylic acids c c c...

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