Category: Class 11 Chemistry

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

NCERT Exemplar Class 11 Chemistry Chapter 11 The p-Block Elements

Multiple Choice Questions
Single Correct Answer Type
P-Block NCERT Exemplar Class 11

Q1. The element which exists in liquid state for a wide range of temperature and can be used for measuring high temperature is
(a) B
(b) A1
(c) Ga
(d) In
Sol: (c) The melting point of gallium is 30°C and boiling point is 2240°C. Thus, the element exists in liquid state for a wide range of temperature.

Q2. Which of the following is a Lewis acid?
(a) AlCl₃
(b) MgCl₂                 
(c) CaCl₂                 
(d) BaCl₂
Sol: (a) A1C1₃ is electron-deficient and hence acts as a Lewis acid.

NCERT Exemplar P Block Class 11

Q3. The geometry of a complex species can be understood from the knowledge of type of hybridisation of orbitals of central atom. The hybridisation of orbitals of central atom in [B(OH)₄]- and the geometry of the complex are respectively
(a) sp³, tetrahedral
(b) sp³, square planar                          
(c) sp³d², octahedral                              
(d) dsp², square planar

Sol: Boron has die electronic configuration:
ls²2s²2p_x¹2p⁰_y2p°_z
In the excited state, 2s-orbital electrons are impaired and one electron is shifted to a p-orbital. Now, hybridisation occurs between one s-and three p-orbitals to give sp³ hybridisation and tetrahedral geometry.

P Block Elements Class 11 NCERT Exemplar

Q4. Which of the following oxides is acidic in nature?
(a) B₂0₃
(b) A1₂0₃                  
(c) Ga₂0₃                   
(d) In₂0₃
Sol: (a) B₂0₃ is acidic in nature. It reacts with basic oxides to form metal borates. Acidic nature decreases on moving down the group.

Q5. The exhibition of highest co-ordination dumber depends on the availability of vacant orbitals in the central atom. Which of the following elements is not likely to act as central atom in MF₆^3- ?
(a) B
(b) AI
(c) Ga
(d) In
Sol: (a) The element M in the complex ion MF₆^3-  has a coordination number of six. Since B has only s- and p-orbitals and no d – orbitals, therefore, at the maximum it can show a coordination number of 4. Thus, B cannot form complex of the type MF₆^3-, i.e., option (a) is correct.

Q6. Boric acid is an acid because its molecule
(a) contains replaceable H⁺   ion
(b) gives up a proton.
(c)accepts OH^–from water releasing proton.
(d) combines with proton from water molecule.
Sol: (c) Because of the small size of boron atom and presence of only six electrons in its valence shell, B(OH)₃ accepts a pair of
electrons from OH^– ion of H₂0, releasing a proton.

Q7.  Catenation, i.e., linking •of similar atoms depends on size and electronic  configuration of atoms. The tendency of catenation in Group 14 elements follows the order
(a) C > Si > Ge > Sn
(b) C » Si > Ge = Sn
(c) Si > C > Sn > Ge
(d) Ge > Sn > Si > C
Sol: (b) The decrease in catenation property is linked with M – M bond energy which decreases from carbon to tin.

Q8. Silicon has a strong tendency to form polymers like silicones. The chain length of silicone polymer can be controlled by adding f (a) MeSiCl₃ (b) Me₂SiCl₂ (c) Me₃SiCl (d) Me₄Si
Sol: (c) The chain length of the polymer can be controlled by adding (CH₃)₃SiCl  which blocks the ends.

Q9. Ionisation enthalpy (∆ _tH₁ kJ mol^-1) for the elements of Group 13 follows the order.
(a) B > A1 > Ga > In > T1
(b) B < A1 < Ga< In <T1
(c) B < A1 > Ga < In < T1                    
(d) B > A1 < Ga > In < T1

Sol: (d) On moving down the group from B to Tl, a regular decreasing trend in the ionisation energy values is not observed.

In Ga, there are ten d-electrons in the penultimate shell which screen the nuclear charge less effectively and thus, outer electron is held firmly. As a result, the ionisation energy of both A1 and Ga is nearly the same. The increase in ionisation energy from In to Tl is due to poor screening effect of 14f electrons present in the inner shell.

Q10. In the structure of diborane
(a) all hydrogen atoms lie in one plane and boron atoms lie in a plane perpendicular to this plane.
(b) 2 boron atoms and 4 terminal hydrogen atoms lie in the same plane and 2 bridging hydrogen atoms lie in the perpendicular plane.
(c) 4 bridging hydrogen atoms and boron atoms lie in one plane and two terminal hydrogen atoms lie in a plane perpendicular to this plane.
(d) all the atoms are in the same plane.
Sol: (b) Four terminal hydrogen atoms and two boron atoms lie in the same plane and two hydrogen atoms forming bridges lie in a plane perpendicular to the rest of the molecule.

Q11. A compound X, of boron reacts with NH₃ on heating to give another compound Y which is called inorganic benzene. The compound X can be prepared by treating BF₃ with lithium aluminium hydride. The compounds X and Y are represented by the formulas.
(a) B₂H₆,B₃N₃H₆
(b) B₂0₃, B₃N₃H₆
(c) BF₃, B₃N₃H₆
(d) B₃N₃H₆ , B₂H₆

Q12. Quartz is extensively used as a piezoelectric material, it contains
(a) Pb
(b) Si
(c) Ti
(d) Sn
Sol: (b) Quartz is a crystalline form of silica.

Q13. The most commonly used reducing agent is

(a) AlCl₃
(b) PbCl₂
(c) SnCl₄
(d) SnCl₂
Sol: d) +4 oxidation state of Sn is more stable than +2 oxidation state. Therefore, Sn²⁺ can be easily oxidised to Sn⁴⁺ and hence SnCl₂ acts a reducing agent.
SnCl₂ + 2Cl → SnCl₄ + 2e^–

Q14. Dry ice is(a) Solid NH₃ (b) Solid S0₂ (c) solid C0₂  (d) solid N₂
Sol: (c) Solid C0₂ is known as dry ice.

Q15. Cement, the important building material is a mixture of oxides of several elements. Besides calcium, iron and sulphur, oxides of elements of which of the group(s) are present in the mixture?
(a) group 2                                               
(b) groups 2,13 and 14
(c) groups 2 and 13                                
(d) groups 2 and 14
Sol: (b) Cement contains elements of group 2 (Ca), group 13 (Al) and group 14 (Si).

More than One Correct Answer Type

Q16. The reason for small radius of Ga compared to Al is_________ .
(a) poor screening effect of d and f orbitals
(b) increase in nuclear charge
(c) presence of higher orbitals
(d) higher atomic number
Sol: (a, b) The additional 10 d-electrons offer poor screening effect for the outer electrons from the increased fluclear charge in Gallium. Hence, atomic radius of Gallium is less than that of aluminium.

Q17. The linear shape of C0₂ is due to  ______ .
(a) sp³ hybridisation of carbon
(b) sp hybridisation of carbon
(c) pπ-pπ bonding between carbon and oxygen
(d) sp² hybridisation of carbon
Sol: (b, c) The linear shape of C0₂ is due to pπ-pπ bonding between carbon and oxygen and sp hybridisation of carbon.

Q18. Me₃SiCl is used during polymerisation of organo silicones because
(a) the chain length of organo silicone polymers can be controlled by adding Me₃
(b) Me₃SiCl blocks the end terminal of silicone polymer.
(c) Me₃SiCl improves the quality and yield of the polymer. –
(d) Me₃SiCl acts as a catalyst during polymerization.
Sol: (a, b) The chain length of the polymer can be controlled by adding Me₃SiCl which blocks the ends of the silicon polymer.
Q19. Which of the following statements are correct?
(a) Fullerenes have dangling bonds.
(b) Fullerenes are cage-like molecules.
(c) Graphite is thermodynamically most stable allotrope of carbon.
(d) Graphite is slippery and hard and therefore used as a dry lubricant in
Sol: (b, c) Fullerenes are cage-like (soccer or rugby ball) molecules and graphite is thermodynamically most stable allotrope of carbon. Thus, options (b) and (c) are correct

Q20. Which of the following statements are correct? Answer on the basis of figure.

(a) The two bridged hydrogen atoms and the two boron atoms lie in one plane.
(b) Out of six B – H, bonds two bonds can be described in terms of 3 centre 2-electron bonds.
(c) Out of six B – H bonds four B – H bonds can be described in terms of 3 centre 2 electron bonds.
(d) The four terminal B – H bonds are two centre-two electron regular bonds.
Sol: (a, b, d) Each of the two boron atoms is in sp³-hybrid state. Of the four hybrid orbitals, three have one electron each while the fourth is empty. Two of the four orbitals of each, of the boron atom overlap with two terminal hydrogen atoms forming two normal B – H σ-bonds. One of the remaining hybrid orbitals (either empty or singly occupied) of one of the boron atoms, 15-orbital of H (bridge atom) and one of hybrid orbitals of the other boron atom overlap to form a delocalized orbital covering the three nuclei with a pair of electrons. This is three centre two electron bond. Similar overlapping occurs with the second hydrogen atom (bridging) forming three centre two electrons bond.

Q21. Identify the correct resonance structures of carbon dioxide from the ones given below:

Short Answer Type Questions

Q22. Draw the structure of BC1₃.NH₃ and AlCl₃ (dimer).
Sol: In BCl₃, the central B atom has six electrons in the valence shell. It is, therefore, an electron deficient molecule and needs two more electrons to ‘ complete its octet. In other words, BCl₃ acts as a Lewis acid. NH₃, on the other hand, has a lone pair of electrons which it can donate easily. Therefore, NH₃ acts as a Lewis base. The Lewis acid (BC1₃) and the Lewis base (NH₃) combine together to form an adduct as shown below:

In A1C1₃, A1 has six electrons in the valence shell. Therefore, it is an electron deficient molecule and needs two more electrons to complete its octet.
Chlorine, on the other hand, has three lone pairs of electrons. Therefore, to complete its octet, the central A1 atom of one molecule accepts a lone pair of electrons from Cl atom of the other molecule forming a dimeric structure as shown below.

Q23. Explain the nature of boric acid as a Lewis acid in water.
Sol:Boric acid is a weak monobasic acid and acts as a Lewis acid by accepting electrons from a hydroxyl ion.
B(OH)₃ + 2H₂0 →[B(OH)₄]^– + H₃0⁺

Q24. Draw the structure of boric acid showing hydrogen bonding. Which species is present in water? What is the hybridisation of boron in this species?

[B(OH)₄]^– units are present in water. Boron has sp³ hybridisation in [B(OH)₄]^– unit.

Q25. Explain why the following compounds behave as Lewis acids?
(i) BC1₃
(ii) AICI3
Sol:(i) BCl₃: Boron has 6 electrons in its outermost orbital and has a vacant p orbital. Thus, it is an electron deficient compound. Hence, it acts as Lewis acid and accepts a lone pair of electrons.

(ii) AlCl3 is also an electron deficient compound and acts as Lewis acid. It generally forms a dimer to achieve stability.

Q26. Give reasons for the following:
(a) CCl₄ is immiscible in water, whereas SiCl₄ is easily hydrolysed.
(b) Carbon has a strong tendency for catenation compared to silicon.
Sol: (i) CC1₄ is a covalent compound while H₂0 is a polar compound. Therefore, it is insoluble in water. Alternatively, CCl₄ is insoluble in water because carbon does not have (/-orbitals to accommodate the electrons donated by oxygen atom of water molecules. As a result, there is no interaction between CC1₄ and water molecules and hence CC1₄ is insoluble in water. On the other hand, SiCl₄ has d-orbitals to accommodate the lone pair of electrons donated by oxygen atom of water molecules. As a result, there is a strong interaction between SiCl₄ and water molecules. Consequently, SiCl₄ undergoes hydrolysis by water to form silicic acid.

(b)  The bond dissociation energy decreases rapidly as the atomic size increases. Since the atomic size of carbon is much smaller (77 pm) as compared to that of silicon (118 pm), therefore, carbon-carbon bond dissociation energy is much higher (348 kJ mol^-1) than that of silicon-silicon bond (297 kJ mol^-1). Hence, because C – C bonds are much stronger as compared to Si-Si bonds, carbon has a much higher tendency for catenation than silicon.

Q27. Explain the following:
(i) C0₂ is a gas whereas Si0₂ is a solid.
(b) Silicon forms SiF₆^2- ion whereas corresponding fluoro compound of carbon is not known.
Sol: (a) Because of its small size and good π-overlap with other small atoms, carbon forms strong double bonds with two oxygen atoms to give discrete C0₂ molecules.                                            –
Silicon atom, on account of large size, does not have good π -overlap with other atoms. It uses its four valence electrons to form four single bonds directed towards the four apices of a tetrahedron (sp³-hybridisation). Each oxygen is linked with two silicon atoms, i.e., a giant three dimensional structure comes into existence which is very stable. Thus, C0₂ is a gas and Si0₂ is a solid.

(b) Silicon has 3d-orbitals in the valence shell and thus expands its octet giving sp³d² hybridisation while d-orbitals are not present in the valence shell of carbon. It can undergo ip³-hybridisation only. Thus, carbon is unable to form CF₆^2- anion.

Q28. The+1 oxidation state in group 13 and +2 oxidation state in group 14 becomes more and more stable with increasing atomic number. Explain.
Sol: In group 13 and 14, as we move down the group, the tendency of s-electrons of the valence shell to participate in bond formation decreases. This is due to ineffective shielding of s-electrons of the valence shell by the intervening d- and f-electrons. This is called inert pair effect.
Due to this, s-electrons of the valence shell of group 13 and 14 are unable to participate in bonding. Hence, +1 and +2 oxidation states, in group 13 and 14 respectively, become -more stable with increasing atomic number.

Q29. Carbon and silicon both belong to the group 14, but in spite of the stoichiometric similarity, the dioxides (i.e., carbon dioxide and silicon dioxide) differ in their structures. Comment.

Q30. If a trivalent atom replaces a few silicon atoms in three dimensional network of silicon dioxide, what would be the type of charge on overall structure?
Sol: If a few tetrahedral Si atoms in a three dimensional network structure of Si0₂ are replaced by an equal number of trivalent atoms, then one valence electron of each Si atom will become free. As a result, each substitution of Si atom by a trivalent atom introduces one unit negative charge into the three dimensional network structure of Si0₂. Hence, Si0₂ becomes negatively charged.

Q31. When BC1₃ is treated with water, it hydrolyses and forms [B(OH)₄]” only whereas A1C1₃ in acidified aqueous solution forms [A1(H₂0)₆]³⁺ Explain what is the hybridisation of boron and aluminium in these species?
Sol: BC1₃ + 3H₂0→ B(OH)₃ + 3HC1
B(0H)₃ + H₂0→[B(OH)4]^– + H⁺
B(OH)₃ due to its incomplete octet accepts an electron pair (as OH^–) to give [B(OH)₄]^–. Boron in this ion involves one 2s orbital and three 2p orbitals. Thus, hybridisation of B in [B(OH)₄]^– is sp³.

Q32. Aluminium dissolves in mineral acids and aqueous alkalies and thus shows amphoteric character, A piece of aluminium foil is treated with dilute hydrochloric acid or dilute sodium hydroxide solution in a test tube and on bringing a burning matchstick near the mouth of the test tube, a pop sound indicates the evolution of hydrogen gas. The same activity when performed with concentrated nitric acid, reaction doesn’t proceed. Explain the reason.
Sol: Al being amphoteric dissolves both in acids and alkalies evolving H₂ gas which bums with a pop sound.

Q31. Explain the following:

• Gallium has higher ionisation enthalpy than aluminium.
• Boron does not exist as B³⁺
• Aluminium forms [A1F₆]^3- ion but boron does not form [BF₆]^3-
• PbX₂ is more stable than PbX₄.
• Pb⁴⁺ acts as an oxidising agent but Sn²⁺ acts as a reducing agent.
• Electron gain enthalpy of chlorine is more negative as compared to fluorine.
• TI(N0₃)₃ acts as an oxidising agent.
• Carbon shows catenation property but lead does not.
• BF₃ does not hydrolyse.
• Why does the element silicon, not form a graphite like structure whereas carbon does.

Sol: (i) Due to ineffective shielding of valence electrons by the intervening 3d electrons, the effective nuclear charge on Ga is slightly higher than that on A1 and hence the ∆H_i of gallium is slightly higher than that of Al. Boron has three electrons in the valence shell. Because of its small size and high sum of the first three ionisation enthalpies (i.e., ∆

(ii) Boron has three electrons in the valence shell. Because of its small size and high sum of the first three ionisation enthalpies (i.e., ∆_iH₁ + ∆_iH₂ + ∆_iH₃), boron does not lose all its valence electrons to form B³⁺ ions.

(iii) Al has vacant d-orbitals and hence can expand its coordination number from 4 to 6 and hence forms octahedral [A1F₆]^3- ion in which Al undergoes sp³d² hybridisation. In contrast, B does not have d-orbitals. Therefore, it can have a maximum coordination number of 4. Therefore, B forms [BF₄]^– (in which B is sp³-hybridised) but not [BF₆]^3-.

(iv) Due to inert pair effect, +2 oxidation state of Pb is more stable than its +4 oxidation state. Consequently, PbX₂ in which the oxidation state of Pb is +2 is more stable than PbX₄ in which the oxidation state of Pb is +4.

(v) Inert pair effect is less prominent in Sn than in Pb. Therefore, +2 oxidation of Sn is less stable than its +4 oxidation state. In other words, Sn²⁺ can easily lose two electrons to form Sn⁴⁺ and hence Sn²⁺ acts as a reducing agent.
Sn²⁺→Sn⁴⁺ + 2e

In contrast, the inert pair effect is’ more prominent in Pb than in Sn. Therefore, +2 oxidation state of Pb is more stable than its +4 oxidation state. In other words, Pb⁴⁺ can easily lose two electrons to form Pb²⁺ and hence Pb⁴⁺ acts as an oxidising agent.

Pb⁴⁺ + 2e^–→-Pb²⁺

(vi) Due to small size, the electron-electron repulsions in the relatively compact 2p-subshell of F are quite strong and hence the incoming electron is not accepted with the same ease as in case of bigger Cl atom where repulsions are comparatively weak. Thus, electron gain enthalpy of chlorine is more negative as compared to that of fluorine.

(vii) Due ta strong inert pair effect, the +3 oxidation state of T1 is less stable than its +1 oxidation state. Since in T1(N0₃)₃, oxidation state of T1 is +3, therefore, it can easily gain two electrons to form T1N0₃ in which the oxidation state of T1 is +1. Consequently, T1(N0₃)₃ acts as an oxidising agent.

(viii) Property of catenation depends upon the strength of element-element bond which, in turn, depends upon the size of the element. Since the atomic size of carbon is much smaller than that of lead, therefore, carbon-carbon bond strength is much higher than that of lead-lead bond. Due to stronger C-C than Pb-Pb bonds, carbon has a much higher tendency for catenation than lead.

(ix) Unlike other boron halides, BF₃ does not hydrolyse completely. Instead, it hydrolyses incompletely to form boric acid and fluoroboric acid. This is because the HF first formed reacts with H₃B0₃.

(x) In graphite, carbon is sp²-hybridised and each carbon is linked to three other carbon atoms by forming hexagonal rings. Each carbon is now left with one unhybridised p-orbital which undergoes sideways overlap to form three p-p double bonds. Thus, graphite has two-dimensional sheet like (layered) structure consisting of a number of benzene rings fused together. Silicon, on the other hand, does not form an analogue of carbon because of the following reason:
Due to bigger size and smaller electronegativity of Si than C, it does not undergo sp²-hybridisation and hence it does not form p-p double bonds needed for graphite like structure. Instead, it prefers to undergo only sp³-hybridisation and hence silicon has diamond like three­dimensional network

Q34. Identify the compound A, X and Z in the following reactions:

Q35. Complete the following chemical equations:

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.
Q36. Match the species given in Column I with the properties mentioned in Column II.

Column I	Column II
(i) BF4–	(a) Oxidation state of central atom is +4
(ii) A1C13	(b) Strong oxidising agent
(iii) SnO	(c) Lewis acid
(iv) Pb02	(d) Can be further oxidised
	(e) Tetrahedral shape

Sol: (i→e); (ii → c); (iii —> d); (iv →a, b)
(i) BF₄^– : Tetrahedral shape, sp³ hybridisation, regular geometry.
(ii) AlCl₃: Octet of A1 not complete, acts as Lewis acid.
(iii) SnO: Sn²⁺ can show +4 oxidation state.
(iv)Pb0₂: Oxidation.state of Pb in Pb0₂ is +4. Due to inert pair effect, Pb⁴⁺ is less stable than Pb²⁺ and hence acts as strong oxidising agent.

Q37. Match the species given in Column I with properties given in Column II.

Column I	Column II
(i) Diborane	(a) Used as a flux for soldering metals
(ii) Gallium                     ‘	(b) Crystalline form of silica
(iii) Borax	(c) Banana bonds
(iv) Aluminosilicate	(d) Low melting, high boiling, useful for measuring high temperatures
(v) Quartz	(e) Used as catalyst in petrochemical industries

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

• BH₃ is unstable, forms diborane B₂H₆ by 3 centre -2 electron bonds, shows banana bonds.
• Gallium with low melting point and high boiling point makes it useful to measure high temperatures.
• Borax is used as a flux for soldering metals.
• Aluminosilicate is used as catalyst in petrochemical industries.
• Quartz, is a crystalline form of silica.

 

_Q38.  Match the species given in Column I with the hybridisation given in Column II.

Column I	Column II
(i) Boron in [B(OH)4]“	(a) sp2
(ii) Aluminium in [A1(H20)6]3+	(b) sp3
(iii) Boron in B2H6	(c) sp3d2
(iv) Carbon in Buckminsterfullerene
(v) Silicon in SiO44-
(vi) Germanium in [GeCl6]2-

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

• Boron in [B(OH)₄]^– is sp³
• Aluminium in [A1(H₂0)₆]³⁺ is sp³d² hybridised
• Boron in B₂H₆ is sp³
• Carbon in Buckminsterfullerene sp² is hybridised.
• Silicon in SiO₄^4- is sp³
• Germanium in [GeCl₆]^2- is sp³ d²

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): If aluminium atoms replace a few silicon atoms in three dimensional network of silicon dioxide, the overall structure acquires a negative charge.
Reason (R): Aluminium is trivalent while silicon is tetravalent.
(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) In aluminosilicates (anion), some of the silicon atoms are replaced by aluminium. Since, aluminium is trivalent while silicon is tetravalent, hence we get negatively charged ion.

Q40. Assertion (A): Silicones are water repelling in nature.
Reason (R): Silicones are organosilicon polymers, which have (-R₂SiO-) as repeating unit.
(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) Silicones are organosilicon polymers and they are hydrophobic in nature. Silicones neither react nor absorb water molecules.

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NCERT Exemplar Class 11 Chemistry Chapter 9 Hydrogen are part of NCERT Exemplar Class 11 Chemistry. Here we have given NCERT Exemplar Class 11 Chemistry Chapter 9 Hydrogen.

NCERT Exemplar Class 11 Chemistry Chapter 9 Hydrogen

Multiple Choice Questions
Single Correct Answer Type

Q1. Hydrogen resembles halogens in many respects for which several factors are responsible. Of the following factors which one is most important in this respect?
(a) Its tendency to lose an electron to form a cation.
(b) Its tendency to gain a single electron in its valence shell to attain stable electronic configuration.
(c) Its low negative electron gain enthalpy value.
(d) Its small size.
Sol: (b) Halogens have the tendency to gain one electron and acquire inert gas configuration. Hydrogen also accepts one electron and acquires helium configuration.

Q2. Why does H⁺ ion always get associated with other atoms or molecules?
(a) Ionisation enthalpy of hydrogen resembles that of alkali metals.
(b) Its reactivity is similar to halogens.
(c) It resembles both alkali metals and halogens.
(d) Loss of an electron from hydrogen atom results in a nucleus of very small size as compared to other atoms or ions. Due to small size it can not exist freely.
Sol: (d) H→H⁺ +e^–
H⁺ has a very small size (~1.5 x 10^-3 pm) compared to normal atomic and ionic sizes of 50 to 220 pm. It does not exist freely and is always associated with other atoms or molecules.

Q3. Metal hydrides are ionic, covalent or molecular in nature. Among LiH, NaH, KH, RbH, CsH, the correct order of increasing ionic character is
(a) LiH > NaH > CsH > KH > RbH
(b) LiH < NaH < KH < RbH < CsH
(c) RbH > CsH > NaH > KH > LiH
(d) NaH > CsH > RbH > LiH > KH
Sol: (b) Ionic character increases as the size of the atom increases.
LiH < NaH < KH < RbH < CsH

Q4. Which of the following hydrides is electron-precise hydride?
(a) B₂H₆
(b) NH₃                    
(c) H₂0                    
(d) CH₄
Sol: (d) CH₄ is an electron precise hydride since there are exact number of electrons to form normal covalent bonds.

Q5. Radioactive elements emit a, p and y rays and are characterized by their half lives. The radioactive isotope of hydrogen is
(a) Protium (b) Deuterium (c) Tritium (d) Hydronium
Sol: (c) Nucleides with n/p (neutron-proton) ratio > 1.5 are usually radioactive. For example, tritium (n = 2,p = 1).

Q6. Consider the reactions
(A) H₂0₂ + 2HI → I₂ + 2H₂0
(B) HOCl + H₂O₂ → H₃0⁺ + Cl^‑ + 0₂
Which of the following statements is correct about H₂0₂ with reference to these reactions? Hydrogen peroxide is   _______      
(a) an oxidizing agent in both (A) and (B)
(b) an oxidizing agent in (A) and reducing agent in (B)
(c) a reducing agent in (A) and oxidizing agent in (B)
(d) a reducing agent in both (A) and (B)

O.N. of oxygen is decreased from -1 (H₂0₂) to -2 (H₂0), therefore, it is reduced and acts as an oxidizing agent.

O.N. of oxygen is increased from -1 (H₂0₂) to 0 (0₂), therefore, it is oxidized and acts as a reducing agent.

Q7. The oxide that gives H₂0₂ on treatment with dilute H₂S0₄ is
(a) Pb0₂
(b) Ba0₂ -8H₂0
(c) Mn0₂
(d) Ti0₂

Q8. Which of the following equatibns depicts the oxidizing nature of H₂0₂?
(a) 2Mn0^–₄ + 6H⁺ + 5H₂0₂ → 2Mn²⁺ + 8H₂0 + 50₂
(b) 2Fe³⁺ + 2H⁺ + H₂0₂ → 2Fe²⁺ + 2H₂0 + 0₂
(c) 2I^– + 2H⁺ + H₂0₂ → I₂ + 2H₂0
(d) KI0₄ + H₂0₂ → KI0₃ + H₂0 + 0₂

Q9. Which of the following equations depicts reducing nature of H₂0₂?
(a) 2[Fe(CN)₆]^4- + 2H⁺ + H₂0₂ → 2[Fe(CN)₆]^3- + 2H₂0
(b) I₂ + H₂0₂ + 2OH^–→ 2I^– + 2H₂0 + 0₂
(c) Mn²⁺ + H₂0₂ → Mn⁴⁺ + 20H
(d) PbS + 4H₂0₂ → PbS0₄ + 4H₂0
Sol: (b) I₂ + H₂0₂ + 2OH^–→ 2I^– + 2H₂0 + 0₂
I₂ is reduced to I^–. Thus, H₂0₂ acts as a reducing agent.

Q10. Hydrogen peroxide is ._________ .
(a) an oxidizing agent
(b) a reducing agent
(c) both an oxidizing and a reducing agent
(d) neither oxidizing nor reducing agent.
Sol: (c) H₂0₂ acts both as oxidizing and reducing agent.

Q11. Which of the following reactions increases production of dihydrogen from synthesis gas?

Sol:(c) To increase the production of H2 from synthesis gas, CO is oxidized to C0₂ by passing it over steam at 673 K in presence of a catalyst.

Q12. When sodium peroxide is treated with dilute sulphuric acid, we get .
(a) sodium sulphate and water
(b) sodium sulphate and oxygen
(c) sodium sulphate, hydrogen and oxygen
(d) sodium sulphate and hydrogen peroxide.

Q13. Hydrogen peroxide is obtained by‘the electrolysis of _______.
(a) water
(b) sulphuric acid
(c) hydrochloric acid
(d) fused sodium peroxide

Q14. Which of the following reactions is an example of use of water gas in the synthesis of other compounds?

Sol: (d) Water gas is used in synthesis of compounds like methanol.

Q15. Which of the following ions will cause hardness in water sample?

Q16. Which of the following compounds is used for water softening?
(a) Ca₃(P0₄)₂
(b) Na₃P0₄           
(c)  Na₆P₆0₁₈                             
(d) Na₂HP0₄
Sol: (c)Na₆P₆0₁₈
(Sodium hexametaphosphate) commercially known as calgon is used for water softening.
2CaCl₂ + Na₂[Na₄(P0₃)₆] →Na₂[Ca₂(P0₃)₆] + 4NaCl

Q17. Elements of which of the following group(s) of periodic table do not form hydrides?
(a) Groups 7,8, 9
(b) Group 13
(c) Group 15,16, 17
(d) Group 14
Sol: (a) Group 7, 8, 9 elements do not form hydrides.

Q18. Only one element of forms hydride.
(a) group 6
(b) group 7
(c) group 8
(d) group 9
Sol: (a) Only one element chromium from group 6 forms hydride, (CrH).

More than One Correct Answer Type
Q19. Which of the following statements are not true for hydrogen?
(a) It exists as diatomic molecule.
(b) It has one electron in the outermost shell.
(c) It can lose an electron to form a cation which can freely exist.
(d) It forms a large number of ionic compounds by losing an electron.
Sol. (c, d) It can lose an electron to form a cation which cannot freely exist.
Generally, it does not form ionic compounds by losing an electron but forms a large number of covalent compounds by sharing electron.

Q20. Dihydrogen can be prepared on commercial scale by different methods. In its preparation by the action of steam on hydrocarbons, a mixture of CO and H2 gas is formed. It is known as _____ .
(a)water gas
(b) syn gas
(c) producer gas
(d) industrial gas
Sol: (a, b) A mixture of CO + H₂ is known as water gas or syn gas (synthesis gas).

Q21. Which of the following statement(s) is/are correct in the case of heavy water?
(a) Heavy water is used as a moderator in nuclear reactor.
(b) Heavy water is more effective as solvent than ordinary water.
(c) Heavy water is more associated than ordinary water.
(d) Heavy water has lower boiling point than ordinary water.
Sol: (a, c) Heavy water is used as a moderator in nuclear reactor and is more associated than ordinary water.

Q22.Which of the following statements about hydrogen are correct?
(a) Hydrogen has three isotopes of which protium is the most common.
(b) Hydrogen never acts as cation in ionic salts.
(c) Hydrogen ion, H⁺, exists freely in solution.
(d) Dihydrogen does not act as a reducing agent.

Q23. Some of the properties of water are described below. Which of them is/are not correct?
(a) Water is known to be universal solvent.
(b) Hydrogen bonding is present to a large extent in liquid water.
(c) There is no hydrogen bonding in the frozen state of water.
(d) Frozen water is heavier than liquid water.
Sol:(c, d) There is extensive hydrogen bonding in ice. Ice is lighter than water due to empty spaces present in tetrahedrons formed by hydrogen bonds.

Q24. Hardness of water may be temporary or permanent. Permanent hardness is due to the presence of
(a) Chlorides of Ca and Mg in water
(b) Sulphate of Ca and Mg in water
(c) Hydrogen carbonates of Ca and Mg in water
(d) Carbonates of alkali metals in water.
Sol: (a, b) Permanent hardness of water is due to presence of following salts: CaCl₂; MgCl₂; CaS0₄; MgS0₄

Q25. Which of the following statements are correct?
(a) Elements of group 15 form electron deficient hydrides.
(b) All elements of group 14 form electron precise hydrides.
(c) Electron precise hydrides have tetrahedral geometries.
(d) Electron rich hydrides can act as Lewis acids.
Sol: (b, c) All elements of group 14 form electron precise hydrides like CH4 which are tetrahedral in geometry.

Q26. Which of the following statements are correct?
(a) Hydrides of group 13 act as Lewis acids.
(b) Hydrides of group 14 are electron deficient hydrides.
(c) Hydrides of group 14 act as Lewis acids.
(d) Hydrides of group 15 act as Lewis bases.
Sol: (a, d) All elements of group 13 will form electron deficient compounds which acts as Lewis acids.
All elements of group 14 will form electron precise compounds.
Electron rich hydrides have excess electrons which are present as lone pairs. Elements of group 15-17 form such compounds. NH3 has one lone pair, H20 has two lone pairs and HF has three lone pairs, and so these compounds act as Lewis bases.

Q27. Which of the following statements are correct?
(a) Metallic hydrides are deficient of hydrogen.
(b) Metallic hydrides conduct heat and electricity.
(c) Ionic hydrides do not conduct electricity in solid state.
(d) Ionic hydrides are very good conductors of electricity in solid state.
Sol: (a, b, c) Metallic hydrides are non-stoichiometric hydrides. They conduct heat and electricity. Ionic hydrides conduct electricity only in molten or aqueous state.

Short Answer Type Questions
Q28. How can production of hydrogen from water gas be increased by using water gas shift reaction?

Q29. What are metallic/interstitial hydrides? How do they differ from molecular hydrides?
Sol: Metallic hydrides are formed by d- and f-^:block elements. Their hydrides conduct heat and electricity. They are non-stoichiometric, being deficient in hydrogen. For example, LaH_2.87, ybH_2.55etc.

Metallic hydrides		Molecular hydrides
(1)	These are formed by el­and f-block elements.	(1)	These are formed by p-block elements and some s-block elements (Be and Mg).
(2)	They conduct electricity.	(2)	They do not conduct electricity.
(3)	They are hard and have metallic luster.	(3)	They are volatile compounds having low melting and boiling points.

Q30. Name the classes of hydrides to which H₂0, B₂H₆ and NaH belong.
Sol: H₂0 – Electron rich covalent hydride/molecular hydride
B₂H₆ – Electron deficient molecular hydride
NaH – Ionic hydride

Q31. If same mass of liquid water and a piece of ice are taken, then why is the density of ice less than that of liquid water?
Sol: The mass per unit volume (i.e., mass/volume) is called density. Since water expands on freezing, therefore, volume of ice for the same mass of water is more than liquid water. In other words, density of ice is lower than liquid water and hence ice floats on water.

Q33. Give reasons:
(i) Lakes freeze from top towards bottom.
(ii) Ice floats on water.
Sol: (i) During severe winter, the temperature of water in the lake keeps on decreasing. Since cold water is heavier, it keeps on going into the interior of the lake while warm water keeps on coming to the surface of the lake. This process continues till the temperature of entire water of lakes becomes 4°C. Since density of water is maximum at 277 K, any further decrease in the temperature will decrease its density. As a result, the temperature of the surface water keeps on decreasing and it ultimately freezes. Now, any further decrease in the temperature will decrease the temperature of water below 4°C. This process continues and as a result, the lakes keep on freezing from top to bottom.
(ii) Density of ice is less than water due to presence of empty spaces created because of H-bonding between H₂0 molecules. Hence, ice floats on water.

Q34. What do you understand by the term ‘auto-protolysis of water? What is its significance?
Sol: Autoprotolysis is a reaction in which two same molecules react to give ions with proton transfer. Water undergoes autoproteolysis, i.e., a proton from one molecule is transferred to another molecule.

On account of autoprotolysis, water is amphoteric in nature.

Q35. Discuss briefly de-mineralisation of water by ion exchange resin.
Sol: Demineralised water free from all soluble mineral salts is obtained by passing water successively through a cation exchange and an anion exchange resin.
In cation exchange process, H⁺ exchanges for Na⁺, Ca²⁺, Mg²⁺ and in the anion exchange process OH exchanges for anions like CH, HCO3, S0^2-₄, etc. H⁺ and OH^– released combine to form water.
H⁺ + OH^– → H₂0

Q36. Molecular hydrides are classified as electron deficient, electron precise and electron rich compounds. Explain each type with two examples.
Sol: Covalent or molecular hydrides are classified into three categories:
(i) Electron deficient hydrides: These hydrides do not have sufficient number of electrons to form normal covalent bonds. Examples are the hydrides of group 13 such as B₂H₆, (AlH₃)_n etc.
(ii) Electron precise hydrides: These have exact number of electrons to form normal covalent bonds. Examples are the hydrides of group 14 such as CH₄, SiH₄, etc.
(iii) Electron rich hydrides: These have more number of electrons than normal covalent bonds. The excess electrons are present in the form of lone pairs. Examples are the hydrides of group 15, 16 and 17 such as

Q37. How is heavy water prepared? Compare its physical properties with those of ordinary water.
Sol: Heavy water can be prepared by exhaustive electrolysis of water. Comparison of physical properties of H₂0 and D₂0.

	Property	H2o	D2o
(i)	Molecular mass (g mol-1)	18.015	20.027
(ii)	Melting point (K)	273.0	276.8
(iii)	Boiling point (K)	373.0	374.4
(iv)	Density (298 K) g cm-3	1.0000	1.1059
(v)	Enthalpy of vapourisation (kJ mol-1)	40.66	41.61

 

Q38. Write one chemical reaction for the preparation of D₂0₂.
Sol: D₂0₂ can be prepared by the reaction of D₂S0₄ dissolved in water over Ba0₂.
Ba0₂ + D₂S0₄→ BaS0₄ + D₂0₂

Q39. Calculate the strength of 5 volume H₂0₂
Sol: 5 volume H₂0₂ solution means that 1 L of 5 volume H₂0₂ on decomposition gives 5L of 0₂ at NTP.

Q40. (i)Draw the gas phase and solid phase structure of H₂0₂.
(ii) H₂0₂ is a better oxidizing agent than water. Explain.
Sol: (i) Structure of H₂0₂ is slightly different in gas phase and solid phase.

Q41. Melting point, enthalpy of vapourisation and viscosity data of H20 and D20 are given below:

	H2o	D2o
Boiling point / K	373.0	374.4
Enthalpy of vapourisation at (373 K)/ kJ mol-1	40.66	41.61
Viscosity/centipoise	0.8903	1.107

On the basis of this data, explain in which of these liquids intermolecular forces are stronger?

Sol: The melting point, enthalpy of vapourisation and viscosity values of all these items depend upon the intermolecular forces

of attraction. Since their values are higher for D₂0 as compared to those of H₂0, therefore, intermolecular forces of attraction

are stronger in D₂0 than in H₂0.

Q42. Dihydrogen reacts with dioxygen (0₂) to form water. Write the name and formula of the product when the isotope of hydrogen which has one proton and one neutron in its nucleus is treated with oxygen. Will the reactivity of both the isotopes be the same towards oxygen? Justify your answer.
Sol:The isotope of hydrogen which contains one proton and one neutron is deuterium (D). Thus, when dideuterium reacts with dioxygen, deuterium oxide, i.e., heavy water is produced.

The reactivity of H₂ and D₂ towards oxygen will be different. Since the D – D bond is stronger than H – H bond, therefore, H₂ is more reactive than D₂ towards reaction with oxygen.

Q43. Explain why HC1 is a gas and HF is a liquid.
Sol: Due to greater electronegativity of F over Cl, F forms stronger H-bonds as compared to Cl. As a result, more energy is needed to break the H-bonds in HF than in HC1 and hence the b.p. of HF is higher than that of HCl. Consequently, H-F is liquid while HC1 is a gas at room temperature.

Q44. When the first element of the periodic table is treated with dioxygen, it gives a compound whose solid state floats on its liquid state. This compound has an ability to act as an acid as well as a base. What products will be formed when this compound undergoes autoionisation?
Sol: The first element is hydrogen and its molecular form is dihydrogen (H2). It reacts with oxygen to form water whose solid state is ice which is lighter than water and floats over water.
Water is amphoteric in nature, i.e., it acts as an acid in the presence of strong bases and acts as a base in the presence of strong acids.

Q45. Rohan heard that instructions were given to the laboratory attendant to store a particular chemical, i.e., keep it in the dark room, add some urea in it, and keep it away from dust. This chemical acts as an oxidizing as well as a reducing agent in both acidic and alkaline media. This chemical is important for use in the pollution control treatment of domestic and industrial effluents.
(i) Write the name of this compound.
(ii) Explain why such precautions are taken for storing this chemical.
Sol: (i) The name of the compound is H₂0₂. It acts as an oxidizing as well as reducing agent in both acidic and basic media.
(ii) H₂0₂ is decomposed by light and dust particles. Urea is added as a negative catalyst, i.e., to check its decomposition.

2H₂0₂(1)→2H₂0(1) + 0₂(g)

Because of the oxidizing properties, H₂0₂ is widely used to control pollution by oxidation of harmful cyanides and obnoxious smelling sulphides present in domestic and industrial effluents. It also helps in sewage disposal by supplying 0₂ for oxidation of organic matter present – in sewage waters.

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NCERT Exemplar Class 11 Chemistry Chapter 8 Redox Reactions are part of NCERT Exemplar Class 11 Chemistry. Here we have given NCERT Exemplar Class 11 Chemistry Chapter 8 Redox Reactions. https://www.cbselabs.com/ncert-exemplar-problems-class-11-chemistry-chapter-8-redox-reactions/

NCERT Exemplar Class 11 Chemistry Chapter 8 Redox Reactions

Multiple Choice Questions
Single Correct Answer Type

Q1. Which of the following is not an example of redox reaction? 
(a) CuO + H₂ → Cu + H₂0                    
(b) Fe₂0₂ + 3CO → 2Fe + 3C0₂
(c).2K + F₂ →2KF                                   
(d) BaCl₂ + H₂S0₄ →BaS0₄ + 2FIC1
Sol: (d) BaCl₂ + H₂S0₄ —> BaS0₄ + 2HC1 is not a redox reaction. It is an example of double displacement reactions.

Q2. The more positive the value of E°, the greater is the tendency of the species to get reduced. Using the standard electrode potential of redox couples given below, find out which of the following is the strongest oxidizing agent. E° values: Fe³⁺/Fe²⁺ = +0.77; I₂(g)/I = +0.54;
Cu²⁺/Cu = +0.34; Ag⁺/Ag = +0.80 V
(a) Fe³⁺
(b) I₂(g)                    
(c) Cu²⁺                    
(d) Ag⁺
Sol: (d) Since Ag⁺/Ag has highest positive value of E°, therefore, Ag⁺ is the strongest oxidizing agent with highest tendency to get reduced

Q3. E° values of some redox complexes are given below. On the basis of these values choose the correct option.
E° values: Br₂/Br^– = +1.90; Ag⁺/Ag(s) = +0.80 Cu²⁺/Cu(s) = +0.34; I₂(s)/I^– = +0.54 V
(a) Cu will reduce Br^–
(b) Cu will reduce Ag
(c) Cu will reduce I^–                               
(d) Cu will reduce Br₂

Sol: (d) Copper will reduce Br₂, if the E° of the redox reaction, 2Cu + Br₂
CuBr₂ is +ve.

Since E° of this reaction is +ve, therefore, Cu can reduce Br₂ and hence option (d) is correct.

Q4. Using the standard electrode potential, find out the pair between which redox reaction is not feasible.
E° values: Fe³⁺/Fe²⁺ = +0.77; I₂/I^– = +0.54;
Cu²⁺/Cu = +0.34; Ag⁺/Ag = +0.80 V
(a) Fe³⁺ and I
(b) Ag⁺ and Cu
(c) Fe³⁺ and Cu
(d) Ag and Fe³⁺
Sol: (d)

Q5. Thiosulphate reacts differently with iodine and bromine in the reactions given below:
2S₂0₃^2_ + I₂→S₄0₆^2- + 2I^–
S₂0₃^2- + 2Br₂ + 5H₂0 →2S0₄^2- + 4Br^– + 10H⁺
Which of the following statements justifies the above dual behaviour of thiosulphate?
(a) Bromine is a stronger oxidant than iodine.
(b) Bromine is a weaker oxidant than iodine.
(c) Thiosulphate undergoes oxidation by bromine and reduction by iodine in these reactions.
(d) Bromine undergoes oxidation and iodine undergoes reduction in these reactions.

Bromine being stronger oxidizing agent than I₂, it oxidises S of S₂O^2-₃ to SO₄^2- whereas I₂ oxidises it only into S₄O₆^2- ion.

Q6. The oxidation number of an element in a compound is evaluated on the basis of certain rules. Which of the following rules is not correct in this respect?
(a) The oxidation number of hydrogen is always +1.
(b) The algebraic sum of all the oxidation numbers in a compound is zero.
(c) An element in the free or the uncombined state bears oxidation number zero.
(d) In all its compounds, the oxidation number of fluorine is -1.
Sol: (a) Oxidation number of hydrogen is -1 in metal hydrides like NaH.

Q7. In which of the following compounds, an element exhibits two different oxidation states.
(a) NH₂OH
(b) NH₄NO₃
(c) N₂H₄
(d) N₃H

Sol: (b) NH₄NO₃ is an ionic compound consisting of NH₄⁺ and NO₃^– ion.
The oxidation number of N in two species is different as shown below:
In NH;,
Let the oxidation state of N in NH₄⁺ be x.
x + 4x(+l) = +l
x = -3
In NO₃^–
Let the oxidation state of N in NO₃^–  be y,
y + 3 x (-2) = -1
y = +5

Only in arrangement (a), the O.N. of central atom increases from left to right. Therefore, option (a) is correct.

Q9. The largest oxidation number exhibited by an element depends on its outer electronic configuration. With which of the following outer electronic configurations the element will exhibit largest oxidation number?
(a) 3d¹4s²               
(b) 3d² 4s²               
(c) 3d⁵4s¹                 
(d) 3d⁵4s²
Sol: (d) Highest O.N. of any transition element = (n – 1)d electrons +ns electrons. Therefore, larger the number of electrons in the 3d orbitals, higher is the maximum O.N.
(a) 3d¹4s²= 3; ‘
(b) 3d² 4s² = 3 + 2 = 5;
(c) 3d⁵4s¹=5 + 1=6
(d) 3d⁵4s² = 5+2 = 7

Q10. Identify disproportionation reaction
(a) CH₄ + 20₂ → C0₂ + 2H₂0
(b) CH₄ + 4C1₂ → CC1₄ + 4HCl
(c) 2F₂ + 20H^–→2F^– + OF₂ + H₂0
(d) 2N0₂ + 20H^– → N0₂ + NO^–₃ + H₂0
Sol: (d) Reactions in which the same substance is oxidized as well as reduced are called disproportionation reactions. Writing the O.N. of each element above its symbol in the given reactions,

Thus, in reaction (d), N is both oxidized as well as reduced since the O.N. of N increases from +4 in N0₂ to +5 in N0^–₃ and decreases from +4 in N0₂ to +3 in N0^–₂.

Q11. Which of the following elements does not show disproportionation tendency?
(a) Cl
(b) Br  
(c) F  
(d) I
Sol: (c) Being the most electronegative element, F can only be reduced and hence it always shows an oxidation number of-1. Further, due to the absence of d-orbitals, it cannot be oxidized and hence it does not show +ve oxidation numbers. In other words, F cannot be simultaneously oxidized as well as reduced and hence does not show disproportionation reactions. Thus, option (c) is correct.

More than One Correct Answer Type

Q12. Which of the following statement(s) is/are not true about the following decomposition reaction?
2KCIO3 →2KC1 + 30₂
(a) Potassium is undergoing oxidation.
(b) Chlorine is undergoing oxidation.
(c) Oxygen is reduced.
(d) None of the species are undergoing oxidation or reduction.
Sol: (a, b, c, d) Writing the oxidation number of each element above its symbol,

(a) The O.N. of K does not change, K. undergoes neither reduction nor oxidation. Thus, option (a) is not correct.
(b) The O.N. of chlorine decreases from +5 in KCl0₃ to -l in KCl, hence, Cl undergoes reduction.
(c) Since, O.N. of oxygen increases from -2 in KC10₃ to 0 in 0₂, oxygen is oxidized.
(d) This statement is not correct because Cl is undergoing reduction and O is undergoing oxidation.

Q13. Identify the correct statement(s) in relation to the following reaction:
Zn + 2HCl → ZnCl₂ + H₂
(a) Zinc is acting as an oxidant.
(b) Chlorine is acting as a reductant.
(c) Hydrogen ion is acting as an oxidant.
(d) Zinc is acting as a reductant.

(a) The O.N. of Zn increases from 0 to +2 (in ZnCl₂) and therefore, Zn acts as a reductant and not as an oxidant. Hence, option (a) is not correct.
(b) The O.N. of Cl does not change and therefore, it neither acts as a reductant nor an oxidant. Hence, option (b) is not correct.
(c) The O.N. of H decreases from +1 in H⁺ to 0 in H₂. Therefore, H⁺ acts an oxidant. This option is correct.
(d) Zinc acts as reductant because its O.N. changes from 0 to +2. This option is correct.

Q14. The exhibition of various oxidation states by an element is also related to the outer orbital electronic configuration of its atom. Atom(s) having which of the following outermost electronic configurations will exhibit more than one oxidation state in its/their compounds.
(a) 3s¹
(b) 3d^l4s²                 
(c)  3d²4s²
(d) 3s²3p³

Sol:(b, c, d) Elements which have only s-electron in the valence shell do not show more than one oxidation state. Thus, element with 3s¹ as outer electronic configuration shows only one oxidation state of +1.
Transition elements, i.e., elements (b, c) having incompletely filled orbitals in the penultimate shell show variable oxidation states. Thus, element with outer electronic configuration as 3d¹ 4s² shows variable oxidation states of +2 and +3 and the element with outer electronic configuration as 3d²4s² shows variable oxidation states of +2, +3 and +4.
p-Block elements also show variable oxidation states due to a number of reasons such as involvement of J-orbitals and inert pair effect. For example, element (d) with 3s² 3p³ as (i.e., P) as the outer electronic configuration shows variable oxidation states of +3 and +5 due to involvement of d-orbitals.

Q15. Identify the correct statements with reference to the given reaction.
P₄ + 30H^– + 3H₂0→ PH₃ + 3H₂ P0^–₂
(a) Phosphorus is undergoing reduction only.
(b) Phosphorus is undergoing oxidation only.
(c) Phosphorus is undergoing oxidation as well as reduction.
(d) Hydrogen is undergoing neither oxidation nor reduction
Sol:

Because O.N. of P increases from 0 (P₄) to +1 (H₂P0^–₂) and decreases from 0 (P₄) to -3 (PH₃), therefore, P has undergone both oxidation as well as reduction. So, option (c) is correct. Option (d) is also correct because O.N. of H remains +1 in all the compounds and hence hydrogen is undergoing neither oxidation nor reduction.

Q16. Which of the following electrodes will act as anodes, when connected to Standard Hydrogen Electrode?
(a)   A1^3-/A1;  E°= -1.66 V
(b)    Fe²⁺ /Fe;  E°= -0.44 V
(c) Cu²⁺/ Cu E°=34 V
(d) F₂(g)/2F^–(aq) E°= 2.87 V
Sol:(a, b) The electrodes having negative electrode potentials are stronger reducing agents than H2 gas and therefore, will act as anodes.

Short Answer Type Questions
Q17. The reaction Cl₂(g) + 20H^–(aq)→ Cl0^–(aq) + Cl^–(aq) + H₂0(l) represents the process of bleaching. Identify and name the species that bleaches the substances due to its oxidizing action.
Sol:

In this reaction, O.N. of Cl increases from 0 (in Cl₂) to +1 (in CIO^–) and decreases to -1 (in Cl^–). Therefore, Cl₂ is both oxidized to CIO^– and reduced to Cl^–. Since Cl^– ion cannot act as an oxidizing agent (because it cannot decrease its O.N. lower than -1), therefore, Cl₂ bleaches substances due to oxidizing action of hypochlorite, ClO ion.

Q18. Mn0^2-₄ undergoes disproportionation reaction in acidic medium but Mn0^–₄ does not. Give reason.

Q19. PbO and Pb0₂ react with HC1 according to following chemical equations:
2PbO + 4HCl → 2PbCl₂ + 2H₂0
Pb0₂ + 4HC1 → PbCl₂ + Cl₂ + 2H₂0
Why do these compounds differ in their  reactivity?

In reaction (i), O.N. of none of the atoms undergo a change. Therefore, it is not a redox reaction. It is an acid-base reaction, because PbO is a basic oxide which reacts with HCl acid.
The reaction (ii) is a redox reaction in which Pb0₂   gets reduced and acts as an oxidizing agent.

Q20. Nitric acid is an oxidizing agent and reacts with PbO but it does not react with Pb0₂. Explain why?
Sol: Nitric acid is an oxidizing agent and reacts with PbO to give a simple acid- base reaction without any change in oxidation state. In Pb0₂, Pb is in +4 oxidation state and cannot be oxidized further, hence no reaction takes place between Pb0₂ and HN0₃.

Q21. Write balanced chemical equations for the following reactions:
(i) Permanganate ion (Mn0^–₄) reacts with sulphur dioxide gas in acidic medium to produce Mn²⁺ and hydrogensulphate ion. (Balance by ion- electron method)
(ii) Reaction of liquid hydrazine (N₂H₄) with chlorate ion (C10^–₃) in basic medium produces nitric oxide gas and chloride ion in gaseous state. (Balance by oxidation number method)
(iii) Dichlorine heptaoxide (C1₂0₇) in gaseous state combines with an aqueous solution of hydrogen peroxide in acidic medium to give chlorite ion (C10^–₂) and oxygen gas. (Balance by ion-electron method)
Sol: (i) 2Mn0^–₄ + 5S0₂ + 2H₂0 + H⁺→5HS0^–₄ + 2Mn²⁺
Balancing by ion-electron method:

Q22. Calculate the oxidation number of phosphorus in the following species.
(a) HPO₃^2- and (b) P0₄^3-
Sol: (a) Let the oxidation number of P inHPO₃^2- be x. So,
+1 + x + (—6) = —2 x = +3
(b) Let the oxidation number of P in P0³₄^– be x. So, x + (-8) = -3
x = +5

Q23. Calculate the oxidation number of each sulphur atom in the following compounds:
(a) Na₂S₂0₃               
(b) Na₂S₄0₆             
(c) Na₂S0₃
(d) Na₂S0₄
Sol: (a) Na₂S₂0₃

Thus, the oxidation states of two S atoms in Na₂S₂0₃ are -2 and +6.
(b) Na₂S₄0₆

Q24. Balance the following equations by the oxidation number method.
(i) Fe²⁺ + H⁺ + Cr₂0₇^2- →Cr³⁺ + Fe³⁺ + H₂0
(ii) I₂ + N0^–₃→ N0₂ +I0₃
(iii) I₂ + S₂0₃^2- →I^– + S₄0₆^2- ‘
(iv) MnO, + C₂0₄^2-→ Mn²⁺ + CO₂

Total increase in O.N. = 5×2= 10
Total decrease in O.N. = 1
To equalize O.N. multiply NO₃^–, by 10
I₂ + 10no₃^–→ 10no₂ + IO₃^–
Balancing atoms other than O and H
I₂ + 10no₃^–→ 10NO₂ + 2 IO₃^–
Balancing O and H
I₂ + lO no₃^– + 8H⁺→ 10NO₂ + 2 IO₃^– + 4H₂0

To equalize O.N. multiply C0₂ by 2. –
Mn0₂ + C₂0^2-₄ →Mn²⁺ + 2CO₂
Balance H and O by adding 2H₂0 on right side, and 4H⁺ on left side of equation.
MnO₂ + C₂0^2-₄ + 4H⁺ →Mn²⁺ + 2CO₂ + 2H₂0

Q25. Identify the redox reactions out of the following reactions and identify the oxidizing and reducing agents in them

Sol: (i) Writing the O.N. of on each atom,

Here, O.N. of Cl increases from -1 (in HCl) to 0 (in Cl₂). Therefore, Cl^– is oxidized and hence HCl acts as a reducing agent.
The O.N. of N decreases from +5 (in HN0₃) to +3 (in NOC1) and therefore, HN0₃ acts as an oxidizing agent.
Thus, reaction (i) is a redox reaction.

Here O.N. of none of the atoms undergo a change and therefore, this is not a redox reaction.

Here, O.N. of Fe decreases from +3 (in Fe₂0₃) to 0 (in Fe) and therefore, Fe₂0₃ acts as an oxidizing agent.
O.N. of C increases from +2 (in CO) to +4 (in C0₂) and therefore, CO acts as a reducing agent. Thus, this is a redox reaction.

Here O.N. of none of the atoms undergo a change and therefore, it is not a redox reaction.

Here. O.N. of N increases from -3 (in NH₃) to 0 in (N₂) and therefore, NH₃ acts as a reducing agent. O.N. of O decreases from 0 (in 0₂) to -2 (in H,0) and therefore, 0₂ acts as an oxidizing agent.
Thus, reaction (v) is a redox reaction.

Dividing the equation into two half reactions:
Oxidation half reaction: I^–→ I₂
Reduction half reaction: Cr₂0₇^2-→ Cr³⁺
Balancing oxidation and reduction half reactions separately as:

(ii) Step-1: Separate the equation into two half reactions.
The oxidation number of various atoms are shown below:

(i) Balance all atoms other than H and O. This step is not needed, because, it is already balanced.
(ii) The oxidation number on left is +2 and on right is +3. To account for the difference, the electron is added to the right as:
Fe²⁺→Fe³⁺ + e^–
(iii) Charge is already balanced.
(iv) No need to add H or O.
The balanced half equation is:
Fe²⁺→ Fe³⁺ + e^–         …(i)
Consider the second half equation
Cr₂0₇^2-→ Cr³⁺

(i) Balance the atoms other than H and O.

Cr₂0₇^2-→2Cr³⁺
(ii) The oxidation number of chromium on the left is +6 and on the right is +3. Each chromium atom must gain three electrons. Since there are two Cr atoms, add 6e^– on the left.
Cr₂0₇^2- + 6e ^–→ 2Cr³⁺
(iii) Since the reaction takes place in acidic medium add 14H⁺ on the left to equate the net charge on both sides.
Cr₂0₇^2-+6e^–+14H⁺→2Cr³⁺

(iv) To balance FI atoms, add 7H₂O molecules on the right.

Cr₂0₇^2- + 6e^– + 14H⁺→ 2Cr³⁺ + 7H₂0          .. .(ii)

This is the balanced half equation.

Step-3: Now add up the two half equations. Multiply eq. (i) by 6 so that electrons are balanced.

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NCERT Exemplar Class 11 Chemistry Chapter 6 Thermodynamics

Multiple Choice Questions
Single Correct Answer Type

Q1. Thermodynamics is not concerned about .
(a) energy changes involved in a chemical reaction.
(b) the extent to which a chemical reaction proceeds.
(C) the rate at which a reaction proceeds.
(d) the feasibility of chemical reaction.
Sol: (c) Thermodynamics is not concerned with rate at which a reaction proceeds. The rate of reaction is dealt by kinetics.

Q2. Which of the following statements is correct?
(a) The presence of reacting species in a covered beaker is an example of open system.
(b) There is an exchange of energy as well as matter between the system and the surroundings in a closed system.
(c) The presence of reactants in a closed vessel made up of copper is an example of a closed system.
(d) The presence of reactants in a thermos flask or any other closed insulated vessel is an example of a closed system.
Sol: (c) For a closed vessel made up of copper, no matter can be exchanged between the system and the surroundings but energy exchange can occur through its walls.

Q3. The state of gas can be described by quoting the relationship between
(a) pressure, volume, temperature
(b) temperature, amount, pressure
(c) amount, volume, temperature
(d) pressure, volume, temperature, amount
Sol: (d) State of a system can be described by state functions or state variables which are pressure, volume, temperature and amount of the gas (PV= nRT).

Q4. The volume of gas is reduced to half from its original volume. The specific
heat will .
(a) reduce to half (b) be doubled
(c) remain constant (d) increase four times
Sol: (c) The specific heat of a substance is the heat required to raise the temperature of 1 gram of a substance by one degree (1 K or 1 °C). It is an intensive property and is independent of the volume of the substance.

Q5. During complete combustion of one mole of butane, 2658 kJ of heat is released. The thermochemical reaction for above change is

Sol: (c) Exothermic reaction for combustion of one mole of butane is represented as:

Q7. In an adiabatic process, no transfer-of heat takes place between system and surroundings. Choose the correct option for free expansion of an ideal gas under adiabatic condition from the following.

Sol: (c) For free expansion w = 0 For adiabatic process q = 0 From first law of thermodynamics,
∆U=q + w
= 0 + 0 = 0
Since there is no change of internal energy, hence temperature will also remain constant, i.e., ∆T = 0

Q8. The pressure-volume work for an ideal gas can be calculated by using the expression

The work can also be calculated from the pV

plot by using the area under curve within the specified limits. When an ideal gas is compressed (a) reversibly or (b) irreversibly from V_i to V_f, choose the correct option.
(a) w (reversible) = w (irreversible)
(b) w (reversible) < w (irreversible)
(c) w (reversible) > w (irreversible)
(d) w (reversible) = w (irreversible) + p_ex. ∆V
Sol: (b) w (reversible) < w (irreversible)

Area under the curve is greater in irreversible compression than that of reversible compression.

Q9. The entropy change can be calculated by using the expression ∆S = q _rev / T.  When water freezes in a glass beaker, choose the correct statement amongst the following:

When water freezes in a glass beaker, choose the correct statement amongst the following:

(a) ∆S_(system) decreases but ∆S_{(surroundings)} remains the same.
(b) ∆S_(system) increases but ∆S_{(surroundings)} decreases.
(C) ∆S_(system) decreases but ∆S_{(surroundmgs)} increases.
(d) ∆S_(system) decreases but ∆S_{(surroundings)} also decreases.

Q10. On the basis of thermochemical equations (i), (ii) and (iii), find out which of the algebraic relationships given in options (a) to (d) is correct.

Q11. Consider the reactions given below. On the basis of these reactions find out which of the algebraic relations given in options (a) to (d) is correct?
(i) C(g) + 4H(g) → CH₄(g); ∆_rH= kJ mol^-1
(ii) C(graphite, s) + 2H₂(g) → CH₄(g); ∆_rH = y kJ mol ¹
(a) x = y                   (b) x = 2y           (c)x >y     (d)x< y
Sol: (c) x > y because same bonds are formed  in reactions (i) and (ii)  but bonds
between reactant molecules are broken only in reaction (ii). As energy is absorbed when bonds are broken, energy released in reaction (i) is greater than that in reaction (ii).

Q12. The enthalpies of elements in their standard states are taken as zero. The enthalpy of formation of a compound
(a) is always negative
(b) is always positive
(c) may be positive or negative
(d) is never negative
Sol:(c) Heat of formation of a compound may be positive or negative, e.g.,

Q13. Enthalpy of sublimation of a substance is equal to
(a) enthalpy of fusion + enthalpy of vapourisation
(b) enthalpy of fusion
(c) enthalpy of vapourisation
(d) twice the enthalpy of vapourisation.
Sol: (a) Enthalpy of sublimation of a substance is equal to enthalpy of fusion + enthalpy of vapourisation.
Sublimation is direct conversion of solid to vapour, i.e., solid → vapour
Writing in two steps, we have solid → liquid → vapour.
Solid → liquid requires enthalpy of fusion.
Liquid →vapour requires enthalpy of vapourisation

Q14. Which of the following is not correct?
(a) ∆G is zero for a reversible reaction.
(b) ∆G is positive for a spontaneous reaction
(c) ∆G is negative tor a spontaneous reaction
(d) ∆G is positive for a non-spontaneous reaction.
Sol:(b) ∆G gives a criterion for spontaneity at constant pressure and temperature.
(i) If ∆G is negative (< 0). the process is spontaneous.
(ii) If ∆G is positiv e (> 0). the process is non-spontaneous.
(iii) If ∆G is zero then reaction is at equilibrium.

More than One Correct Answer Type
Q15. Thermodynamics mainly deals with
(a) interrelation of various forms of energy and their transformation front one from to another.
(b) energy changes in the processes which depend only on initial and final states of the microscopic system containing a few molecules.
(c) how and at what rate these energy transformations are carried out.
(d) the system in equilibrium state or moving from one equilibrium state to another equilibrium state.
Sol: (a, d) Thermodynamics deals with interrelation of various forms of energy and their transformation into each other. It also deals with thermal or mechanical equilibrium. However, if does not tell anything about the rate of reaction.

Q16. In an exothermic reaction, heat is evolved, and system loses heat to the surroundings. For such system
(a) q_P will be negative                               
(b) ∆_γHwill be negative
(c) q_p will be positive                                
(d) ∆_γHwill be positive.
Sol:(a, b) For an exothermic reaction, q_p = -ve, ∆_γH = -ve

Q17. The spontaneity means, having the potential to proceed without the assistance of external agency. The processes which occur spontaneously are
(a) flow of heat from colder to warmer body.
(b) gas in a container contracting into one comer.
(c) gas expanding to fill the available volume.
(d) burning carbon in oxygen to give carbon dioxide.
Sol:(c, d) Gas expands or diffuses in available space spontaneously, e.g., leakage of cooking gas gives smell of ethyl mercaptan spontaneously. Moreover, burning of carbon to C0₂ is also spontaneous.

Q18. For an ideal gas. the work of reversible expansion under isothermal condition 1.0 mol of an ideal gas is expanded isothermally and reversibly to ten times of its original volume, in two separate experiments. The expansion is carried out at 300 K and at 600 K respectively. Choose the correct option.
can be calculated by using expression w = -nRT In V_f / V_i A sample containing
(a) Work done at 600 K is 20 times the work done at 300 K.
(b) Work done at 300 K is twice the work done at 600 K
(c) Work done at 600 K is twice the work done at 300 K.
(d) ∆U= 0 in both cases.

i.e., work done at 600 K is twice the work done at 300 K. Since each case involves isothermal expansion of an ideal gas, there is no change in internal energy, i.e., ∆U = 0.

Q19. Consider the following reaction between zinc and oxygen and choose the correct options out of the options given below:
2Zn(s) + 0₂(g) → 2ZnO(s); ∆H=-693.8 kJ mol^-1
(i) The enthalpy of two moles ZnO is less than the total enthalpy of two moles of Zn and one mole of oxygen by 693.8 kJ.
(ii) The enthalpy of two moles of ZnO is more than the total enthalpy of two moles of Zn and one mole of oxygen by 693.8 kJ.
(iii) 8 kJ mol ^-1 energy is evolved in the reaction.
(iv) 693.8 kJ mol^-1 energy is absorbed in the reaction.

Short Answer Type Questions

Q20. 18.0 g of water completely vapourises at 100°C and 1 bar pressure and the enthalpy change in the process is
40.79 kJ mol^-1. What will be the enthalpy change for vapourising two moles of water under the same conditions? What is the standard enthalpy of vapourisation for water?
Sol: Enthalpy of a reaction is the energy change per mole for the process.
18 g of H₂0 = 1 mole (∆H_vap = 40.79 kJ moE¹)
Enthalpy change for vapourising 2 moles of H₂0 = 2 x 40.79 = 81.58 kJ ∆H°_vap = 40.79 kJ mol ^-1

Q21. One mole of acetone requires less heat to vapourise than 1 mol of water. Which of the two liquids has higher enthalpy of vapourisation?
Sol: Water has higher enthalpy of vapourisation. (∆H_r)_water > (∆H_r)_acetone

Q24. Enthalpy is an extensive property. In general, if enthalpy of an overall reaction A→B along one route is ∆_rH and ∆_rH₁, ∆_rH₂, ∆_rH₃ …. represent enthalpies of intermediate reactions leading to product B. What will be the relation between ∆_rH for overall reaction and ∆_rH₁, ∆_rH₂….. etc. for intermediate reactions.
Sol: By Hess’s law ∆_rH = ∆_rH₁+ ∆_rH₂ + ∆_rH₃……………………

Q25. The enthalpy of atomisation for the reaction CH₄(g) → C(g) + 4H(g) is 1665 kJ mol^-1. What is the bond energy of C – H bond?
Sol: CH₄ → C + 4H, ∆H= + 1665 kJ mol^-1          ,
Bond energy of (C – H) bond = 1665/4 =416.2 kJ mol

Q27. Given that ΔH= 0 for mixing of two gases. Explain whether the diffusion of these gases into each other in a closed container is a spontaneous process or not?
Sol: It is a spontaneous process. Although enthalpy change is zero but randomness or disorder (ΔS) increases and ΔS is positive. Therefore, in equation, ΔG = ΔH – TΔS, the term TΔS will be negative. Hence ΔG will be negative.

Q28. Heat has randomising influence on a system and temperature is the measure of average chaotic motion of particles in the system. Write the mathematical relation which relates these three parameters.
Sol: Heat has randomising influence on a system and temperature is the measure of average chaotic motion of particles in the system. The mathematicalrelation which relates these three parameters is ΔS = q _rev/ T
Here, ΔS = change in entropy                                                  ^
q_rcv = heat of reversible reaction ‘
T = temperature

Q29. Increase in enthalpy of the surroundings is equal to decrease in enthalpy of the system. Will the temperature of system and surroundings be the same when they are in thermal equilibrium?
Sol: Yes, when the system and the surroundings are in thermal equilibrium, their temperatures are same.

Q30. At 298 K, K_p for the reaction N₂0₄(g)⇌ 2N0₂(g) is 0.98. Predict whether the reaction is spontaneous or not.
Sol: Δ_rG° = -RT ln K_p
= -RT ln (0.98)
Since In (0.98) is negative
.’. Δ_rG° is positive
=> the reaction is non spontaneous

Q31. A sample of 1.0 mol of a monoatomic ideal gas is taken through a cyclic process of expansion and compression as shown in the figure. What will be the value of ΔHfor the cycle as a whole?

Sol: For a cyclic process, ΔH = 0

Q32. The standard molar entropy of H₂O(l) is 70 J K^-1 mol^-1. Will the standard molar entropy H₂0(s) be more, or less than 70 J K ^-1 mol^-1?
Sol: The standard molar entropy of H₂0 (1) is 70 J K^-1 mol^-1. The solid form of H₂0 is ice. In ice, molecules of H₂0 are less random than in liquid water. Thus, molar entropy of H₂0 (s) < molar entropy of H₂0 (1). The standard molar entropy of H₂0 (s) is less than 70 J K ¹ mol^-1.

Q33. Identify the state functions and path functions out of the following: enthalpy, entropy, heat, temperature, work, free energy.
Sol: State functions: Enthalpy, entropy, temperature, free energy Path functions: Heat, work

Q34. The molar enthalpy of vapourisation of acetone is less than that of water. Why?
Sol: Molar enthalpy of vapourisation is more for water due to hydrogen bonding between water molecules.

Q35. Which quantity out of Δ_rG and Δ_rG° will be zero at equilibrium?
Sol: Gibbs energy for a reaction in which all reactants and products are in standard state. Δ_rG° is related to the equilibrium constant of the reaction as follows
Δ_rG = A_rG° + RT In K
At equilibrium, 0 = Δ_rG° + RT In A^– ({Δ_rG = 0) or    Δ_rG° =-RT lnK
Δ_rG° = 0 when K= 1
For all other values of K, A_rG° will be non-zero.

Q36. Predict the change in internal energy for an isolated system at constant volume.
Sol: For an isolated system w = 0, q = 0
Since ΔU= q + w = 0 + 0 = 0, ΔU= 0

Q37. Although heat is a path function but heat absorbed by the system under certain specific conditions is independent of path. What are those conditions? Explain.
Sol: At constant volume
q = ΔU + (-w)
-w = pΔ q = AU + pΔV
ΔV = 0 (at constant volume)
Hence, q_v = ΔU + 0 = ΔU= change in internal energy At constant pressure, q_p = AU + pΔV
Since ΔU + pΔV=ΔH
=> q_p = ΔH change in enthalpy
Hence, at constant volume and at constant pressure, heat change is a state function because it is equal to ΔU and ΔH respectively which are state functions.

Q38. Expansion of a gas in vacuum is called free expansion. Calculate the work done and the change in internal energy when 1 litre of ideal gas expands isothermally into vacuum until its total volume is 5 litre.
Sol: During free expansion, external pressure is zero, so Work done, w = -p_extΔV
= -0(5 – 1) = 0
Since the gas is expanding isothermally, therefore, q = 0
ΔU = q + w =0+0=0

Q39. Heat capacity (C_P) is an extensive property but specific heat (c) is an intensive property. What will be the relation between C_p and c for 1 mol of water?
Sol: For water, molar heat capacity = 18 x Specific heat or
C_p = 18 x c
But, specific heat,
C = 4.18 J g^-1 K^-1 Heat capacity,
C_p = 18 x 4.18 JK ¹ = 75.24 JK^-1

Q40. The difference between C_p and C_v can be derived using the empirical relation H = U + pV. Calculate the difference between C_p and C_v for 10 moles of an ideal gas.
Sol: Given that, C_v = heat capacity at constant volume,
C_p = heat capacity at constant pressure Difference between C_p and C_v is equal to gas constant (R).
.’. C_p – C_v = nR                                (where, n = no. of moles)
= 10 x 8.314 = 83.14J

Q41. If the combustion of 1 g of graphite produces 20.7 kJ of heat, what will be molar enthalpy change? Give the significance of sign also.
Sol: Molar enthalpy change for graphite (ΔH)
= enthalpy change for 1 g x molar mass of C = -20.7×12 = -2.48 x 10² kJ mol^-1
Since the sign of ΔH = -ve, it is an exothermic reaction.

Q42. The net enthalpy change of a reaction is the amount of energy required to break all the bonds in reactant molecules minus amount of energy required to form all the bonds in the product molecules. What will be the enthalpy change for the following reaction?
H₂(g) + Br₂(g)→2HBr(g)
Given that bond energy of H₂, Br₂ and HBr is 4.35 kJ mol^-1,192 kJ mol^-1 and 368 kJ mol ^-1 respectively.

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We hope the NCERT Exemplar Class 11 Chemistry Chapter 6 Thermodynamics help you. If you have any query regarding NCERT Exemplar Class 11 Chemistry Chapter 6 Thermodynamics, drop a comment below and we will get back to you at the earliest.

NCERT Exemplar Class 11 Chemistry Chapter 5 States of Matter are part of NCERT Exemplar Class 11 Chemistry. Here we have given NCERT Exemplar Class 11 Chemistry Chapter 5 States of Matter. https://www.cbselabs.com/ncert-exemplar-problems-class-11-chemistry-chapter-5-states-matter/

NCERT Exemplar Class 11 Chemistry Chapter 5 States of Matter

Multiple Choice Questions
Single Correct Answer Type

NCERT Exemplar Class 11 Chemistry States Of Matter 

Q1. A person living in Shimla observed that cooking food without using pressure cooker takes more time. The reason for this observation is that at high altitude
(a) pressure increases.
(b) temperature decreases.
(c) pressure decreases.
(d) temperature increases.
Sol: (c) At high altitudes, pressure is low. Hence, boiling takes place at lower temperature and therefore, cooking takes more time. In pressure cooker, pressure is high and hence boiling point increases.

Q2. Which of the following properties of water can be used to explain the spherical shape of rain droplets?
(a)Viscosity
(b)Surface tension
(c)Critical phenomena
(d)Pressure
Sol: (b) Due to surface tension, the surface of the water drops is under tension and tends to take spherical shape to reduce the tension.

States Of Matter Class 11 NCERT Exemplar

Q3. A plot of volume (V) versus temperature (T) for a gas at constant pressure is a straight line passing through the origin.
The plots at different values of pressure are shown in the figure. Which , of the following order of pressure is correct for this gas?

(a)P₁ >P2 >P₃ >P₄
(b) P₁ =p₂ =p₃ =p₄
(c) P₁ <P₂<P₃<P4
(d) p_x <p₂=p₃<p₄

Sol: (c) At a particular temperature pV= constant.
Thus,
P₁V₁ =P₂V₂=P₃V₃=P₄V₄
As V₁> V₂ > V₃ > V ₄, therefore, P₁ <P₂ < P₃ < P₄.

Q4. The interaction energy of London force is inversely proportional to sixth power of the distance between two interacting particles but their magnitude depends upon
(a) charge of interacting particles.
(b) mass of interacting particles.
(c) polarisability of interacting particles.
(d) strength of permanent dipoles in the particles.
Sol: (c) London dispersion forces operate only over very short distance. The energy of interaction varies as
1/(distance between two interacting particles)⁶ Large or more complex are the molecules, greater is the magnitude of London forces. This is obviously due to the fact that the large electron clouds are easily distorted or polarised.
Hence, greater the polarisability of the interacting particles, greater is the magnitude of the interaction energy.

NCERT Exemplar Class 11 Chemistry Chapter 5

Q5. Dipole-dipole forces act between the molecules possessing permanent dipole. Ends of dipoles possess ‘partial charges’. The partial charge is
(a) more than unit electronic charge.
(b) equal to unit electronic charge.
(c) less than unit electronic charge.
(d) double the unit electronic charge.
Sol: (c) Partial charge is a small charge developed by displacement of electrons. It is less than unit electronic charge and is represented as δ⁺ or δ^–

Q6. The pressure of a 1:4 mixture of dihydrogen and dioxygen enclosed in a vessel is one atmosphere. What would be the partial pressure of dioxygen?
(a) 0.8 x 10⁵atm                                     
(b) 0.008 Nm^-2
(c) 8 x 10⁴ Nm ^-2                                     
(d) 0.25 atm
Sol: (c) Let the number of moles of dihydrogen and dioxygen be 1 and 4.

NCERT Exemplar States Of Matter Class 11

Q7. As the temperature increases, average kinetic energy of molecules increases. What would be the effect of increase of temperature on pressure provided the volume is constant?
(a) Increases
(b) Decreases
(c) Remains same
(d) Becomes half
Sol: (a) At constant volume, as the temperature is increased, pressure also increases.

Q8. Gases possess characteristic critical temperature which depends upon the magnitude of intermolecular forces between the particles.
Following are the critical temperatures of some gases.

Gases	h2	He	02	n2
Critical temperature in Kelvin	33.2	5.3	154.3	126

From the above data what would be the order of liquefaction of these gases? Start writing the order from the gas liquefying first.
(a) H₂, He, 0₂, N₂
(b) He,0₂,H₂,N₂
(c) N₂,0₂,He,H₂                                       
(d) 0₂,N₂,H₂,He
Sol: (d) Higher the critical temperature, more easily is the gas liquefied. Hence, order of liquefaction starting with the gas liquefying first will be: 0₂, N₂, H₂, He.

States Of Matter NCERT Exemplar Class 11 

Q9. What is SI unit of viscosity coefficient (η)?
(a) pascal
(b) N s m^-2              
(c) km ^-2 s                
(d) N m^-2

Solu:

Q10. Atmospheric pressures recorded in different cities are as follows:

Cities	Shimla	Bengaluru	Delhi	Mumbai
p in N/m2	1.01 x 105	1.2 x 105	1.02 x 105	1.21 x 105

Consider the above data and mark the place at which liquid will boil first.
(a) Shimla
(b) Bengaluru
(c) Delhi
(d) Mumbai
Sol: (a) Shimla has lowest atmospheric pressure, hence liquid will boil first in this city. Boiling of a liquid takes place when the vapour pressure becomes equal to the atmospheric pressure.

States Of Matter Class 11 Exemplar NCERT 

Q11. Which curve in figure represents the curve of ideal gas?

(a) only
(b) C and D only
(c)E and F only                                    
(d)A and B only

Sol: (a) For curve B, value of PV is constant and for an ideal gas plot of PV vs P is a straight line.

Q12. Increase in kinetic energy can overcome intermolecular forces of attraction. How will the viscosity of liquid be affected by the increase in temperature?
(a)Increase
(b)No effect
(c) Decrease
(d) No regular pattern will be followed
Sol: (c) Intermolecular force of liquid decreases with increase in temperature, hence viscosity of liquid also decreases. However, some exceptions are there like liquid proteins and liquid sulphur.

Q13.How does the surface tension of a liquid vary with increase in temperature?
(a) Remains same
(b) Decrease
(c) Increase
(d) No regular pattern is followed
Sol: (b) Surface tension of a liquid decreases with increase in temperature due to less forces of attraction between the molecules.

More than One Correct Answer Type
Q14.With regard to the gaseous state of matter which of the following statements are correct?
(a) Complete order of molecules (b) Complete disorder of molecules
(c) Random motion of molecules (d) Fixed position of molecules
Sol: (b, c) In gaseous state, molecules are in a state of random motion, i.e., it is the state in which molecules are disorderly arranged. Gaseous state has higher entropy than the liquid as well as solid.

Q15. Which of the following figures does not represent 1 mole of dioxygen gas at STP?
(a) 16 grams of gas                                
(b) 22.7 litres of gas
(c) 6.022 x 10²³ dioxygen molecules
(d) 11.2 litres of gas
Sol: (a, d) 1 mole of dioxygen represents 32 g of 0₂, 22.7 L of 0₂ or 6.022 x 10²³ molecules of o₂ gas.

Q16. Under which of the following two conditions applied together, a gas deviates most from the ideal behaviour?
(a) Low pressure (b) High pressure
(c) Low temperature (d) High temperature
Sol: (b, c) A gas which obeys the ideal gas equation, p V = nRTunder all conditions of temperature and pressure is called an ‘ideal gas’.
However, there is no gas which obeys the ideal gas equation under all conditions of temperature and pressure. Hence, the concept of ideal gas is only theoretical or hypothetical. The gases are found to obey the gas laws fairly well when the pressure is low or the temperature is high.
Such gases are, therefore, known as ‘real gases’. All gases are real gases. Hence, at high pressure and low temperature, a real gas deviates most from ideal behaviour.

Q17. Which of the following changes decrease the vapour pressure of water kept in a sealed vessel?
(a) Decreasing the quantity of water
(b) Adding salt to water
(c) Decreasing the volume of the vessel to one-half
(d) Decreasing the temperature of water
Sol: (b, d) Vapour pressure does not depend upon the quantity of water or size of the vessel. It decreases on adding salt to water or decreasing the temperature of water.

Short Answer Type Questions

Q18. If 1 gram of each of the following gases are taken at STP, which of the gases will occupy (a) greatest volume and (b) smallest volume? CO, H₂0, CH₄, NO

Q19. Physical properties of ice, water and steam are very different. What is the chemical composition of water in all the three states?
Sol: The chemical composition of water remains the same in all the physical states, i.e., solid, liquid and gas.

Q20. The behaviour of matter in different states is governed by various physical laws. According to you what are the factors that determine the state of matter?
Sol: Pressure, temperature, mass and volume are the factors that determine the

Q21. Use the information and data given below to answer the questions (a) to (c):
• Stronger intermolecular forces result in higher boiling point.
• Strength of London forces increases with the number of electrons in the molecule.
• Boiling point of HF, HC1, HBr and HI are 293 K, 189 K, 206 K and 238 K respectively.
(a) Which type of intermolecular forces are present in the molecules HF, HC1, HBr and HI?
(b) Looking at the trend of boiling points of HC1, HBr and HI, explain out of dipole-dipole interaction and London interaction, which one is predominant here?
(c) Why is boiling point of hydrogen fluoride highest while that of hydrogen chloride lowest?
Sol: (a) All the given molecules viz. HF, HC1, HBr and HI have permanent di-poles. Hence, all of them possess dipole-dipole and London forces. HF in addition to dipole-dipole and London forces also has hydrogen bonding.
(b) Electronegativity of Cl, Br and I is in the order: Cl > Br > I. Therefore, polar character and hence dipole-dipole interactions should be in the order HCl > HBr > HI. But boiling points are found to be in the order HCl < HBr < HI. This shows that London forces are predominant. This is because London forces increase as the number of electrons in the molecule increases. In this case, the number of electrons increases from HC1 to HI.
(c) Due to very high electronegativity of F, HF is most polar and also there is hydrogen bonding present in it. Hence, it has the highest boiling point.

Q22. What will be the molar volume of nitrogen and argon at 273.15 K and 1 atm?
Sol: When temperature and pressure of a gas are 273.15 K (or 0°C) and 1 atm
(or 1 bar or 10⁵ pascal), such conditions are called standard temperature and pressure conditions (STP). Under these conditions, the volume occupied by 1 mole of each and every gas is 22.4 L. Hence, the molar volume of N₂ and Ar at 273.15 K and 1 atm is 22,4 L.

Q23. A gas that follows Boyle’s law, Charles’ law and Avogadro’s law is called an ideal gas. Under what conditions a real gas would behave ideally?
Sol:  At high temperature and low pressure, the gases behave ideally since the two postulates of kinetic theory of gases are true under these conditions.
(i) The volume of a molecule of a gas is negligible as compared to its complete volume.
(ii) There is negligible force of attraction between the molecules of a gas.

 

Q24. Two different gases ‘A’ and ‘9’ are filled in separate containers of equal capacity under the same conditions of temperature and pressure. On increasing the pressure slightly, the gas ‘A’ liquefies but gas ‘B’ does not liquefy even on applying high pressure until it is cooled. Explain this phenomenon.
Sol: Gas ‘A’ is at critical temperature and therefore liquefies. Gas ‘B’ is at a temperature higher than critical temperature and therefore, does not liquefy even on applying high pressure.

Q25. Value of universal gas constant (R) is same for all gases. What is its physical significance?

Q26. One of the assumptions of kinetic theory of gases states that “there is no force of attraction between the molecules of a gas.” How far is this statement correct? Is it possible to liquefy an ideal gas? Explain.
Sol: The statement that there is no force of attraction between the molecules of a gas is true at low pressure and high temperature only. If the statement is true under all conditions of temperature and pressure, it will not be possible to liquefy an ideal gas.

Q27. The magnitude of surface tension of liquid depends on the attractive forces between the molecules. Arrange the following in increasing order of surface tension:

water, alcohol (C₂H₅OH) and hexane [CH₃(CH₂)₄CH₃)].

Sol: In the above given molecules, only hexane [CH₃(CH₂)₄CH₃] is a non-polar molecule in which only London dispersion forces exist. These forces are very weak while both water and ethanol are polar molecules in which dipole- dipole interactions as well as H-bonding exists.
However, since H-bonding interactions are much stronger in water than ethanol, therefore, it possesses stronger intermolecular forces than alcohol and hexane. Hence, the increasing order of surface tension is
Hexane < Alcohol < Water

Q28. Pressure exerted by saturated water vapour is called aqueous tension. What correction term will you apply to the total pressure to obtain pressure of dry gas?
Sol: Whenever a gas is collected over water, it is moist and saturated with water vapour which exerts their own pressure. The pressure due to water vapour is called aqueous tension. Thus, the total pressure of the gas (p_{moist gas}) is

P_{moist gas} = P _drygas

Thus P _drygas  is given as

P _drygas = P_{moist gas} – aqueous tension. Hence, the correction term applied to the total pressure of the gas in order to obtain pressure of dry gas is P_{moist gas} – aqueous tension.

Q29. Name the energy which arises due to motion of atoms or molecules in a body. How is this energy affected when the temperature is increased?
Sol: The energy generated due to motion of atoms or molecules in a body is thermal energy. It is measured as average kinetic energy of molecules. It increases with increase in temperature.
K. E ∝ T

Q30. Name two intermolecular forces that exist between HF molecules in liquid state.
Sol: (i) Dipole-dipole interactions
(ii) Hydrogen bonding

Q31. One of the assumptions of kinetic theory of gases is that there is no force of attraction between the molecules of a gas.
State and explain the evidence that shows that the assumption is not applicable for real gases.
Sol: Real gases can be liquefied on cooling and compressing which shows that there are forces of attractions between gaseous molecules.

Q32. Compressibility factor, Z, of a gas is given as Z = PV/nRT
(i) What is the value of Z for an ideal gas?
(ii) For real gas what will be the effect on value of Z above Boyle’s temperature?
Sol: (i) For ideal gas, Z = 1.
(ii) For a real gas, above Boyle’s temperature, gas shows positive deviation and hence Z > 1

Q33. The critical temperature (T_c) and critical pressure (P_c) of C0₂ are 30.98°C and 73 atm respectively. Can C0₂(g) be liquefied at 32°C and 80 atm pressure?
Sol: C0₂ gas cannot be liquefied at 32°C by applying a pressure of 80 atm. This is because the temperature is higher than critical temperature of C0₂.

Q34. For real gases the relation between P, V and T is given by van der Waals equation:

where ‘a’ and ‘b’ are van der Waals constants, ‘nb’ is approximately equal to the total volume of the molecules of a gas.
‘a’ is the measure of magnitude of intermolecular attraction.
(i) Arrange the following gases in the increasing order of ‘b’. Give reason. 0₂, C0₂, H₂, He
(ii) Arrange the following gases in the decreasing order of magnitude of ‘a’. Give reason.CH₄, O₂, H₂
Sol: (i) As ‘b’ represents molar volume occupied by the gas molecules, greater the size of the molecules, greater is the volume occupied by 1 mole of molecules. The size and hence the value of ‘b’ increase in the order: H₂ < He < 0₂ < C0₂.
As all the given molecules are non-polar, the magnitude of intermolecular attractions and hence the value of ‘a’ increases with the increase in number of electrons in the molecule, i.e., in the order: CH₄ > 0₂ > H₂. (Greater the number of electrons, greater is the size of electron cloud, greater is the polarization of the molecule, greater is the attraction).

Q35. The relation between pressure exerted by an ideal gas (P_ideal) and observed pressure (P_real) is given by the equation

If pressure is taken in N m^-2, number of moles in mol and volume in m³,Calculate the unit of ‘a’. What will be the unit of ‘a’ when pressure is in atmosphere and volume in dm³?

Q36. Name two phenomena that can be explained on the basis of surface tension.
Sol:(i) Spherical shape of liquid drops.
(ii) Cleansing action of soaps and detergents. ‘
(iii) Capillary action —» Rise and dip of liquid in the column of capillary.

Q37. Viscosity of a liquid arises due to strong intermolecular forces existing between the molecules. Stronger the intermolecular forces, greater is the viscosity. Name the intermolecular forces existing in the following liquids and arrange them in the increasing order of their viscosities. Also give reason for the assigned order in one line.Water, Hexane (CH₃CH₂CH₂CH₂CH₂CH₃), Glycerine (CH₂OHCH(OH)CH₂OH)

Sol: The increasing order of viscosity is Hexane < Water < Glycerine
In hexane, there are only London forces, whereas in water and glycerine there is hydrogen bonding which is stronger in case of glycerine (because of the presence of three -OH groups) than that in water.

Q38. Explain the effect of increasing the temperature of a liquid on intermolecular forces operating between its particles. What will happen to the viscosity of a liquid if its temperature is increased?
Sol: If the temperature is increased, the intermolecular forces become weak and kinetic energy increases. As the temperature increases the kinetic energy can overcome intermolecular forces, hence viscosity decreases and liquids can flow easily.

Q39. The variation of pressure with volume of the gas at different temperatures can be graphically represented as shown in figure.

On the basis of this graph answer the following questions.
(i) How will the volume of a gas change if its pressure is increased at constant temperature?
(ii) At a constant pressure, how will the volume of a gas change if the temperature is increased from 200 K to 400 K?
Sol: (i) The volume of a gas will decrease if the pressure on the gas is increased keeping the temperature constant.
(ii) The volume of a gas increases if the temperature is increased keeping the pressure constant.

Q40. Pressure versus volume graphs for a real gas and an ideal gas are shown in the figure.

Answer the following questions on the basis of this graph.
(i) Interpret the behaviour of real gas with respect to ideal gas at low pressure.
(ii) Interpret the behaviour of real gas with respect to ideal gas at high pressure.
(iii) Mark the pressure and volume by drawing a line at the point where real gas behaves as an ideal gas.

Sol: (i) At low pressure, the two* curves almost coincide. This shows that at low pressure, the real gases show very small deviation from ideal behaviour.
(ii) At high pressure, the curves are far apart. This shows that real gases show large deviations at high pressure.
(iii) At point A, where the two curves intersect each other, the real gas behaves exactly like ideal gas and the pressure and volume corresponding to this point are P₁ and V₁

Matching Column Type Questions
Q41. Match the graph between the following variables with their names.

Column I (Graphs)	Column II (Names)
(i) Pressure vs temperature graph at constant molar volume.	(a) Isotherms
(ii) Pressure vs volume graph at constant temperature.	(b) Constant temperature curve
(iii) Volume vs temperature graph at constant pressure.	(c) Isochores
	(d) Isobars

Sol: (i) →(c); (ii) → (a); (iii) → (d)
(i) Pressure vs temperature graph at constant volume – Isochores
(ii) Pressure vs volume graph at constant temperature – Isotherms
(iii) Volume vs temperature graph at constant pressure – Isobars

Q42. Match the following gas laws with the equation representing them.

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): Three states of matter are the result of balance between intermolecular forces and thermal energy of the molecules. .
Reason (R): Intermolecular forces tend to keep the molecules together but thermal energy of molecules tends to keep them apart.
(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) A is true but R is false.
(d) A is false but R is true.
Sol: (a) States of matter of a substance depend on the intermolecular forces.

Intermolecular forces keep the molecules together while thermal energy of molecules tends to keep them apart.

Q45. Assertion (A): At constant temperature, PV vs V  plot for real gases is not a straight line.
Reason (R): At high pressure all gases have Z> 1 but at intermediate pressure most gases have Z < 1.
(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) A is true but R is false.
(d) A is false but R is true.
Sol: (b) PV versus Fplot is straight line for ideal gases but not for real gases.
At high pressure, there is positive deviation from ideal behaviour (Z < 1) but at intermediate or moderate pressure, the real gases pass negative deviation from ideal behavior (Z < 1).

Q46. Assertion (A): The temperature at which vapour pressure of a liquid is equal to the external pressure is called boiling temperature.
Reason (R): At high altitude atmospheric pressure is high.
(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) A is true but R is false.
(d) A is false but R is true.
Sol: (c) The temperature at which the vapour pressure of a liquid becomes equal to the external pressure is called its boiling point. Atmospheric pressure decreases with increase in altitude.

Q47. Assertion (A): Gases do not liquefy above their critical temperature, even on applying high pressure.
Reason (R): Above critical temperature, the molecular speed is high and intermolecular attractions cannot hold the molecules together because they escape because of high speed.
(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) A is true but R is false.
(d) A is false but R is true.
Sol:(a) According to Andrews, a real gas cannot be liquefied‘above critical ‘ temperature, whatever pressure is applied. Intermolecular force of gas molecules is low above critical temperature.

Q48. Assertion (A): At critical temperature liquid passes into gaseous state imperceptibly and continuously.
Reason (R): The density of liqtiid and gaseous phase is equal to critical temperature.
(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) A is true but R is false.
(d) A is false but R is true.
Sol:(a) At critical state, surface tension of liquid is zero hence liquid passes into gaseous state imperceptibility and continously . At this stage , the liquid and vapour phase becomes equal.

Q49. Assertion (A): Liquids tend to have maximum number of molecules at their surface.
Reason (R): Small liquid drops have spherical shape.
(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) A is true but R is false.
(d) A is false but R is true.
Sol: (d) Liquid tends to acquire minimum surface area due to surface tension. Thus, small liquid drops are spherical. In spherical shape, surface area is minimum.
Long Answer Type Questions
Q50. Isotherms of carbon dioxide at various temperatures are represented in the following figure. Answer the following questions based on this figure.

(i) In which state will C0₂ exist between the points a and b at temperature T₁
(ii) At what point will Co₂ start liquefying when temperature is T₁?
(iii) At what point will C0₂ be completely liquefied when temperature is T₂?
(iv) Will condensation take place when the temperature is T₃
(v) What portion of the isotherm at T₁ represent liquid and gaseous C0₂ at equilibrium?

Sol: (i) At T₁ (between points ‘a’ and ‘b’) C0₂ exists in gaseous state.
(ii) At point b, C0₂ starts liquefying at T₁
(iii) At temperature T₂, at point ‘g’ C0₂ will be completely liquefied.
(iv) No condensation at T₃ since T₃ > T_c.
(iv) At T₁ liquid and gaseous C0₂ are at equilibrium between b and c.

Q51. The variation of vapour pressure of different liquids with temperature is shown in figure

(i) Calculate graphically boiling points of liquids A and B.
(ii) If we take liquid C in a closed vessel and heat it continuously, at what temperature will it boil?
(iii) At high altitude, atmospheric pressure is low (say 60 mm Hg). At what temperature liquid D boils?
(iv) Pressure cooker is used for cooking food at hill station. Explain in terms of vapour pressure why is it so?

Sol: (i) Boiling point of liquid A ≃315 K, B ≃ 345 K
(ii) In a closed vessel, liquid C will not boil because pressure inside keeps on increasing.
(iii) Temperature corresponding to 60 mm ≃ 313 K.
(iv) A liquid boils when its vapour pressure becomes equal to atmospheric pressure. At hill station, atmospheric pressure is low. Therefore, liquid boils at a lower temperature and cooking is not perfect. In a pressure cooker the pressure inside increases and the liquid boils at a higher temperature.

Q52. Why does the boundary between liquid phase and gaseous phase disappear on heating a liquid up to critical temperature in a closed vessel? In this situation what will be the state of the substance?
Sol: When a liquid is heated up to its critical temperature in a closed vessel, it does not pass through a two phase region and substances remain in one phase. There is a continuity between a gaseous and liquid state. The term fluid is used for either a liquid or a gas to recognize this continuity. Liquid and gas can be distinguished only when the fluid is below its critical temperature and the surfaces separating them can be seen. At critical temperature, liquid passes into gaseous state continuously and the surface separating the two phases disappears. A gas below the critical temperature can be liquefied by applying pressure.

Q53. Why does sharp glass edge become smooth on heating it up to its melting point in a flame? Explain which property of liquids is responsible for this phenomenon.
Sol: On heating the glass, it melts and takes up rounded shape at the edges which has minimum surface area. This is due to the property of surface tension of liquids.

Q54. Explain the term ‘laminar flow’. Is the velocity of molecules same in all the layers in laminar flow? Explain your answer.
Sol: When a liquid flows over a fixed surface, the layer of molecules in the immediate contact of surface is stationary. The velocity of the upper layers increases as the distance of layers from the fixed layer increases.
This type of flow in which there is a regular gradation of velocity on passing from one layer to the next is called laminar flow.

In laminar flow, the velocity of molecules is not same in all the layers because every layer offers some resistance or friction to the layer immediately below it.

Q55. Isotherms of carbon dioxide gas are shown in figure. Mark a path for changing . gas into liquid such that only one phase (i.e. either a gas or liquid) exists at any time during the change. Explain how the temperature, volume and pressure should be changed to carry out the change.

Sol: In isotherm of carbon dioxide, it is possible to change a gas into liquid or a liquid into gas by a process in which always a single phase is present.

If we move vertically from point A to F by increasing the temperature, then we can reach the point G by compressing the gas at constant temperature along this (isotherm at 31.1°C). Now we can move vertically downwards to D by lowering the temperature. As soon as we cross point H on critical isotherm, we get liquid. If process is carried out at critical temperature, substance always remains in one phase. Hence the path for the change is A → F → G → H → D

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