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Punjab State Board PSEB 12th Class Chemistry Book Solutions Chapter 6 General Principles and Processes of Isolation of Elements Textbook Exercise Questions and Answers.

PSEB Solutions for Class 12 Chemistry Chapter 6 General Principles and Processes of Isolation of Elements

PSEB 12th Class Chemistry Guide General Principles and Processes of Isolation of Elements InText Questions and Answers

Question 1.
Copper can be extracted by hydrometallurgy but not zinc% Explain.
Answer:
The E^⊖ value of zinc (Zn²⁺/Zn = – 0.76 V) is lower than that of copper (Cu2+/Cu = 0.34 V). This means that .zinc is a stronger reducing agent and can displace copper from solution of Cu²⁺ ions.
Zn(s) + Cu²⁺ (aq) → Zn²⁺(aq) + Cu(s)
In order to extract zinc by hydrometallurgy, we need stronger reducing agent like

etc. However, all these metals reduce water to hydrogen gas. Therefore, these metals cannot be used to displace Zn from solution of Zn²⁺ ions. Thus, copper can be extracted by hydrometallurgy but not zinc.

Question 2.
What is the role of depressant in froth floatation process?
Answer:
In the froth floatation process, the role of the depressants is to separate two sulphide ores by selectively preventing one ore from forming froth. For example, to separate two sulphide ores (ZnS and PbS), NaCN is used as a depressant which selectively allows PbS to come with froth, but prevents ZnS from coming to froth. This happens because NaCN reacts with ZnS to form Na₂[Zn(CN)₄].
4NaCN + ZnS → Na₂[Zn(CN)₄] + Na₂S

Question 3.
Why is the extraction of copper from pyrites more difficult than that from its oxide ore through reduction?
Answer:
The Gibbs free energy of formation (△_fG) of Cu₂S is less than that of H₂S and CS₂. Therefore, H₂ and C cannot reduce Cu₂S to Cu.

On the other hand, the Gibbs free energy of formation of Cu₂O is greater than that of CO. Hence, C can reduce Cu₂O to Cu.
C(s) + Cu₂O(s) → 2Cu(s) + CO(g)
Hence, the extraction of copper from its pyrite ore is difficult than from its oxide ore through reduction.

Question 4.
Explain:
(i) Zone refining
(ii) Column chromatography.
Answer:
(i) Zone refining : This method is based on the principle that impurities are more soluble in the molten state of metal (the melt) than in the solid state. In the process of zone refining, a circular mobile heater is fixed at one end of a rod of impure metal. As the heater moves, the molten zone of the rod also moves with it. As a result, pure metal crystallizes out of the melt and the impurities pass onto the adjacent molten zone. This process is repeated several times, which leads to the segregation of impurities at one end of the rod. Then, the end with the impurities is cut off. Silicon, boron, gallium, indium etc. can be purified by this process.

(ii) Column chromatography : Column chromatography is a technique used to separate different components of a mixture. It is a very useful technique used for the purification of elements available in minute quantities. It is also used to remove the impurities that are not very different in chemical properties from the element to be purified. Chromatography is based on the principle that different components of a mixture are differently adsorbed on an adsorbent. In chromatography, there are two phases: mobile phase and stationary phase. The stationary phase is immobile and immiscible. Al₂O₃ column is usually used as the stationary phase in column chromatography. The mobile phase may be a gas, liquid, or supercritical fluid in which the sample extract is dissolved. Then, the mobile phase is forced to move through the stationary phase. The component that is more strongly adsorbed on the column takes a longer time to travel through it than the component that is weakly adsorbed. The adsorbed components are then removed (eluted) using a suitable solvent (eluant).

Question 5.
Out of C and CO, which is a better reducing agent at 673 K?
Answer:
At 673 K, the value of △G_(CO,CO2) is less than that of △G_(C,CO).
Therefore, CO can be oxidised more easily to CO₂ than C to CO. Hence, CO is a better reducing agent than C at 673 K.

Question 6.
Name the common elements present in the anode mud in electrolytic refining of copper. Why are they so present ?
Answer:
In electrolytic refining of copper, the common elements present in anode mud are selenium, tellurium, silver, gold, platinum, and antimony.

These elements are very less reactive and are not affected during the purification process. Hence, they settle down below the anode as anode mud.

Question 7.
Write down the reactions taking place in different zones in the blast furnace during the extraction of iron.
Answer:
Reactions in blast furnace are as follows :
(a) Reactions at lower temperature range (500 to 800 K)
3Fe₂O₃ + CO → 2Fe₃O₄ + CO₂
Fe₃O₄ + 4CO → 3Fe + 4CO₂
Fe₂O₃ + C0 → 2FeO + CO₂

(b) Reactions at higher temperature range (900-1500 K)
C + CO₂ → 2CO
FeO + CO → Fe + CO₂

Question 8.
Write chemical reactions taking place in the extraction of zinc from zinc blende. .
Answer:
The different steps involved in the extraction of zinc from zinc blende (ZnS) are given below:
(i) Concentration of ore : First, the gangue from zinc blende is removed by the froth floatation method.
(ii) Conversion to oxide (Roasting) : Sulphide ore is converted into oxide by the process of roasting. In this process, ZnS is heated in a regular supply of air in a furnace at a temperature, which is below the melting point of Zn.
2ZnS + 3O₂2 → 2ZnO + 2SO₂

(iii) Extraction of zinc from zinc oxide (Reduction) : Zinc is extracted from zinc oxide by the process of reduction. The reduction of zinc oxide is carried out by mixing it with powdered coke and then, heating it at 1673 K.

(iv) Electrolytic refining: Zinc can be refined by the process of electrolytic refining. In this process, impure zinc is made the anode while a pure copper strip is made the cathode. The electrolyte used is an acidified solution of zinc sulphate (ZnSO₄). Electrolysis results in the transfer of zinc in pure form from the anode to the cathode.
Anode : Zn → Zn²⁺ + 2e^–
Cathode : Zn²⁺ + 2e^– → Zn

Question 9.
State the role of silica in the metallurgy of copper.
Answer:
During the roasting of pyrite ore, a mixture of FeO and Cu₂O is obtained.

The role of silica in the metallurgy of copper is to remove the iron oxide obtained during the process of rosting as ‘slag’. If the sulphide ore of copper contains iron, then silica (SiO₂) is added as flux before roasting. Then, FeO combines with silica to form iron silicate, FeSiO₃ (slag).

Question 10.
What is meant by the term “chromatography”?
Answer:
Chromatography is a collective term used for a family of laboratory techniques for the separation of mixtures. The term is derived from Greek words ‘chroma’ meaning ‘colour’ and ‘graphy5 meaning ‘writing’. Chromatographic techniques are based on the principle that different components are absorbed differently on an absorbent. There are several chromatographic techniques such as paper chromatography, column chromatography, gas chromatography, etc.

Question 11.
What criterion is followed for the selection of the stationary phase in chromatography?
Answer:
The stationary phase is selected in such a way that the components of the sample have different solubility’s in the phase. Hence, different components have different rates of movement through the stationary phase and as a result, can be separated from each other.

Question 12.
Describe a method for refining nickel.
Answer:
Nickel is refined by Mond’s process. In this process, nickel is heated in the presence of carbon monoxide to form nickel tetracarbonyl, which is a volatile complex.

Then, the obtained nickel tetracarbonyl is decomposed by subjecting it to a higher temperature (450-470 K) to obtain pure nickel metal.

Question 13.
How can you separate alumina from silica in a bauxite ore associated with silica? Give equations, if any.
Answer:
(i) To separate alumina from silica in a bauxite ore associated with silica, first the powdered ore is digested with a concentrated NaOH solution at 473-523 K and 35-36 bar pressure. This results in the leaching out of alumina (Al₂O₃) as sodium aluminate and silica (SiO₂) as sodium silicate leaving the impurities behind.

(ii) Then, CO₂gas is passed through the resulting solution to neutralise the aluminate in the solution, which results in the precipitation of hydrated alumina. To induce precipitation, the solution is seeded with freshly prepared samples of hydrated alumina.

(iii) During this process, sodium silicate remains in the solution. The obtained hydrated alumina is filtered, dried, and heated to get back pure alumina.

Question 14.
Giving examples, differentiate between ‘roasting’ and ‘calcination’.
Answer:

Roasting	Calcination
1.  Sulphur dioxide is produced along with metal oxide.	Carbon dioxide is produced along with metal oxide.
2.  Ore is heated in the presence of excess of air or oxygen.	Ore is heated in the absence or limited supply of air or O2.
3. Volatile impurities are removed as oxides, such as SO2, As2O3, etc.	Water and organic impurities are removed.

Question 15.
How is ‘cast iron’ different from ‘pig iron”?
Answer:
The iron obtained from blast furnaces is known as pig iron. It contains around 4% carbon and many impurities such as S, P, Si, Mn in smaller amounts.

Cast iron is obtained by melting pig iron and coke using a hot air blast. It contains a lower amount of carbon (3%) than pig iron. Unlike pig iron, cast iron is extremely hard and brittle.

Question 16.
Differentiate between “minerals” and “ores”.
Answer:

Mineral	Ore
Naturally occurring substances of metals present in the earth’s crust are called minerals.	Minerals which can be used to obtain the metal profitably are cahed ores.
All minerals are not ores.	All ores are essentially minerals too.
e.g,, bauxite (Al2O3-xH2O) and clay (A12O3 -2SiO2 -2H2O)	e.g., bauxite (A12O3 ∙xH2O)

Question 17.
Why copper matte is put in silica lined converter?
Answer:
Copper matte contains Cu₂S and FeS. Copper matte is put in a silica-lined converter to remove the remaining FeO and FeS present in the matte as slag (FeSiO₃). Also, some silica is added to the silica-lined converter. Then, a hot air blast is blown. As a result, the remaining FeS and FeO are converted to iron silicate (FeSiO₃) and Cu₂S is converted into metallic copper.
2FeS + 3O₂ → 2FeO + 2SO₂
FeO + SiO₂ → FeSiO₂
2Cu₂S + 3O₂ → 2Cu₂O + 2SO₂
2Cu₂O + Cu₂S → 6Cu + SO₂

Question 18.
What is the role of cryolite in the metallurgy of aluminium?
Answer:
Cryolite (Na₃AlF₆) has two roles in the metallurgy of aluminium :

1. To decrease the melting point of the mixture from 2323 K to 1140 K.
2. To increase the electrical conductivity of Al₂O₃.

Question 19.
How is leaching carried out in case of low grade copper ores?
Answer:
In case of low grade copper ores, leaching is carried out using acid or bacteria in the presence of air. In this process, copper goes into the solution as Cu²⁺ ions.
Cu(s) + 2H⁺(aq) + \(\frac{1}{2}\)O₂(g) → Cu²⁺(aq) + 2H₂O(l)
The resulting solution is treated with scrap iron or H₂ to get metallic copper.
Cu²⁺(aq) + H₂(g) → Cu(s) + 2H⁺(aq)

Question 20.
Why is zinc not extracted from zinc oxide through reduction using CO?
Answer:
The standard free energy of formation (△_fG^⊖) of CO₂ from CO is
higher than that of the formation of ZnO from Zn. Therefore, CO cannot be used to reduce ZnO to Zn.

Question 21.
The value of △_fG^⊖ for formation of Cr₂O₃ is – 540 kJmol^-1 and that of A₂O₃ is – 827 kJ mol^-1. Is the reduction of Cr₂O₃ possible with Al?
Answer:
The two thermochemial equations may be written as
(i) 2Al + \(\frac{1}{2}\)O₂ → Al₂O₃ △_fG^⊖ = -827kJmol^-1
(ii) 2Cr + \(\frac{1}{2}\)O₂ → Cr₂O₃ △_fG^⊖ = -540kJmol^-1
Subtracting equation (ii) from (i), we have
2Al + Cr₂O₃ → Al₂O₃ + 2Cr
△_fG^⊖ = -827-(-540)
= -287kJmol^-1
As △_fG^⊖ for the reduction reaction of Cr₂O₃ by Al is negative, this reaction is possible.

Question 22.
Out of C and CO, which is a better reducing agent for ZnO?
Answer:
The free energy of formation (△_fG^⊖) of CO from C becomes lower at temperatures above 1120 K whereas that of CO₂ from C becomes lower above 1323 K than △_fG^⊖ of ZnO. However, △_fG^⊖ of CO₂ from CO is always higher than that of ZnO. Therefore, C and reduce ZnO to Zn but not CO. Therefore, out of C can CO, C is a better reducing agent than CO for ZnO.

Question 23.
The choice of a reducing agent in a particular case depends on thermodynamic factor. How far do you agree with this statement? Support your opinion with two examples.
Answer:
For any spontaneous reaction, the Gibbs free energy change (△G) must be negative. △G = △H – T△S where △H is the enthalpy change during the reaction, T is the absolute temperature and △S is the change in entropy.

Consider the Ellingham diagram (given below) for some metal oxides. From the diagram, it is evident that metals for which the free energy of formation of their oxides is more negative can reduce those metal oxides for which the free energy of formation of their respective oxides is less negative. In other words, any metal will reduce the oxide of other metals which lie above it in the Ellingham diagram because the free energy will become more negative by an amount equal to the difference in the two graphs at that particular temperature. Thus, Al reduces FeO, Cr₂O₃ and NiO in Thermite reaction, but Al will not reduce MgO at a temperature below 1773 K.

It can be followed that:
(i) 2Al + Cr₂O₃→ Al₂O₃ + 2Cr
(Aluminothermic process)
(ii) 2Al + Fe₂O₃ → Al₂O₃ + 2Fe are spontaneous.
But Al can’t be used to reduce MgO below 1500°C. From the above it is clear that thermodynamic considerations help us in choosing a suitable reducing agent in metallurgy.

Question 24.
Name the processes from which chlorine is obtained as a by-product. What will happen if an aqueous solution of NaCl is subjected to electrolysis?
Answer:
(i) Down’s process for the manufacture of Na metal: When molten NaCl is subjected to electrolysis, chlorine is obtained as a by product at anode because in molten state only Na⁺ and Cl^– ions are present.
NaCl (melt) → Na⁺ (melt) + Cl^– (melt)
At cathode : Na⁺ (melt) + e^– → Na(s)
At anode : Cl^–(melt) → Cl(g) + e^–

(ii) Manufacture of NaOH : If an aqueous solution of NaCl is electrolysed, Cl₂ will be obtained at the anode but at the cathode, H₂ will be obtained instead of Na. This is because the standard reduction potential of Na (E^⊖ = – 2.71 V) is more negative than that of H₂O (E^⊖ = – 0.83 V). Hence, H₂O will get preference to get reduced at the cathode and as a result, H₂ is evolved.
NaCl(aq) → Na⁺(aq) + Cl^– (aq)
H₂O ⇌ H⁺(aq) + OH^–(aq)
At cathode : 2H₂O(l) + 2e^– → H₂(g) + 2OH^– (aq)
At anode : Cl^– (melt) → Cl(g) + e^–
2Cl (g) → Cl₂(g)
H₂ gas is obtained at cathode; chlorine gas at anode and NaOH is formed in the solution.
Na⁺(aq) + OH^–(aq) → NaOH (aq)

Question 25.
What is the role of graphite rod in the electrometallurgy of aluminium?
Answer:
In the electrometallurgy of aluminium, a fused mixture of purified alumina (Al₂O₃), cryolite (Na₃AlF₆) and fluorspar (CaF₂) is electrolysed. In this electrolysis, graphite is used as the anode and graphite-lined iron is used as the cathode. During the electrolysis, A1 is liberated at the cathode, while CO and CO₂ are liberated at the anode, according to the following equation :
At cathode: Al³⁺(melt) + 3e^– → Al(l)
At anode: C(s) + O^2- (melt) → CO(g) + 2e^–
C(s) + 2O^2- (melt) → CO₂(g) + 4e^–
If a metal is used instead of graphite as the anode, then 02will be liberated. This will not only oxidise the metal of the electrode, but also convert some of the A1 liberated at the cathode back into Al₂O₃. Hence, graphite is used for preventing the formation of O₂ at the anode.

Question 26.
Outline the principles of refining of metals by the following methods:
(i) Zone refining
(ii) Electrolytic refining
(iii) Vapour phase refining
Answer:
(i) Zone refining : This method is based on the principle that impurities are more soluble in the molten state of metal (the melt) than in the solid state. In the process of zone refining, a circular mobile heater is fixed at one end of a rod of impure metal. As the heater moves, the molten zone of the rod also moves with it. As a result, pure metal crystallizes out of the melt and the impurities pass onto the adjacent molten zone. This process is repeated several times, which leads to the segregation of impurities at one end of the rod. Then, the end with the impurities is cut off. Silicon, boron, gallium, indium etc. can be purified by this process.

(ii) Electrolytic refining : Electrolytic refining is the process of refining impure metals by using electricity. In this process, impure metal is made the anode and a strip of pure metal is made the cathode. A solution of a soluble salt of the same metal is taken as the electrolyte. When an electric current is passed, metal ions from the electrolyte are deposited at the cathode as pure metal and the impure metal from the anode dissolves into the electrolyte in the form of ions. The impurities present in the impure metal gets collected below the anode. This is known as anode mud.
At anode: M → Mⁿ⁺ + ne^–
At cathode: Mⁿ⁺ + ne^– → M

(iii) Vapour phase refining : Vapour phase refining is the process of refining metal by converting it into its volatile compound and then, decomposing it to obtain a pure metal. To carry out this process,

•  the metal should form a volatile compound with an available reagent, and
• the volatile compound should be easily decomposable so that the metal can be easily recovered.
  Nickel, zirconium, and titanium are refined using this method.

Question 27.
Predict conditions under which Al might be expected to reduce MgO. (Hint: See Intext Question 4)
Answer:
The equations for the formation of two oxides are :
\(\frac{4}{3}\)Al(s) + O₂(g) → \(\frac{2}{3}\)Al₂O₃(s)
2Mg(s) + O₂(g) → 2MgO(s)
If we observe the plots for the formation of the two oxides on the Ellingham diagram, we find the two curves intersect each other at a certain point. The corresponding value of △_fG^⊖ becomes zero for the reduction of MgO by aluminium metal.
2MgO(s) + \(\frac{4}{3}\)Al(s) ⇌ 2Mg(s) + \(\frac{2}{3}\)Al₂O₃(s)
This means that the reduction of MgO by A1 metal cannot occur below this temperature (1665 K). Instead, Mg can reduce Al₂O₃ to Al below 1665 K.
Aluminium metal (Al) can reduce MgO to Mg above 1665 K
because △_fG^⊖ for Al₂O₃ is less as compared to that of MgO.

Chemistry Guide for Class 12 PSEB General Principles and Processes of Isolation of Elements Textbook Questions and Answers

Question 1.
Which of the ores mentioned in Table 6.1 can be concentrated by magnetic separation method?
Answer:
If the ore or the gangue can be attracted by the magnetic field, then the ore can be concentrated by the process of magnetic separation. The ores of iron such as haematite (Fe₂O₃), magnetite (Fe₃O₄), siderite (FeCO₃) and iron pyrites (FeS₂) can be separated by the process of magnetic separation.

Question 2.
What is the significance of leaching in the extraction of aluminium?
Answer:
In the extraction of aluminium, the significance of leaching is to concentrate pure alumina (Al₂O₃) from bauxite ore. Bauxite usually contains silica, iron oxide, and titanium oxide as impurities. In the process of leaching, alumina is concentrated by digesting the powdered ore with a concentrated solution of NaOH at 473-523 K and 35-36 bar. Under these conditions, alumina (Al₂O₃) dissolves as sodium meta-aluminate and silica (SiO₂) dissolves as sodium silicate leaving the impurities behind.

The impurities are then filtered and the solution is neutralised by passing CO₂ gas. In this process, hydrated Al₂O₃ gets precipitated and sodium silicate remains in the solution. Precipitation is induced by seeding the solution with freshly prepared samples of hydrated Al₂O₃.
2Na[Al(OH)₄](aq) + CO₂(g) → Al₂O₃∙xH₂O(S) + 2NaHCO₃(aq)
Hydrated alumina
Hydrated alumina Al₂O₃∙xH₂O is filtered, dried, and heated to give back pure alumina (Al₂O₃).

Question 3.
The reaction,
Cr₂O₃ + 2Al > Al₂O₃ + 2Cr (△_fG^⊖ = -421kJ) is thermodynamically feasible as is apparent from the Gibbs energy value.
Why does it not take place at room temperature?
Answer:
The change in Gibbs energy is related to the equilibrium constant, K as
△G = – RT in K
At room temperature, all reactants and products of the given reaction are in the solid state. As a result, equilibrium does not exist between the reactants and th e prod ac ts lienee, the reaction does not take place at room temperature.
However, at a higher temperature, chromium melts and the reaction takes place.
We also know that according to the equation
△G = △H – T△S,
Increasing the temperature increases die value of T△S, making the value of △G more and more negative. Therefore, the reaction becomes more and more feasible as the temperature is increased.

Question 4.
Is it true that under certain conditions. Mg can reduce Al₂O₃ and Al can reduce MgO? What are those conditions?
Answer:
If we look at the Ellingbam diagram wo observe that the plots for Al and Mg cross each other at 1350°C (1623k) Below this temperature Mg can reduce Al₂O₃ and above this temperature., Al can reduce MgO.

Punjab State Board PSEB 12th Class Chemistry Book Solutions Chapter 7 The p-Block Elements Textbook Exercise Questions and Answers.

PSEB Solutions for Class 12 Chemistry Chapter 7 The p-Block Elements

PSEB 12th Class Chemistry Guide The p-Block Elements InText Questions and Answers

Question 1.
Discuss the general characteristics of group 15 elements with reference to their electronic configuration, oxidation state, atomic size, ionisation enthalpy and electronegativity.
Answer:
General trends in group 15 elements
(i) Electronic configuration : All the elements of group 15 have ns² np³ (5 valence electrons) electronic configuration in their valence shells.
The s-subshell is completely filled and p-subshell is exactly half-filled. This imparts extra stability to their electronic configuration.
Nitrogen (N₇) = [He] 2s² 2p³
Phosphorus (P₁₅) = [Ne] 3s² 3p³
Arsenic (As₃₃ ) = [Ar] 3d¹⁰ 4s² 4p³
Antimony (Sb₅₁) = [Kr] 4d¹⁰ 5s² 5p³
Bismuth (Bi₈₃) = [Xe] 4f¹⁴5d¹⁰ 6s² 6p³

(ii) Oxidation state : All these elements have 5 valence electrons and require three more electrons to complete their octets. However, gaining electrons is very difficult as the nucleus will have to attract three more electrons. This can take place only with nitrogen as it is the smallest in size and the distance between the nucleus and the valence shell is relatively small. The remaining elements of this group show a formal oxidation state of – 3 in their covalent compounds. In addition to the – 3 state, N and P also show – 1 and – 2 oxidation states.

All the elements present in this group show + 3 and + 5 oxidation states. However, the stability of + 5 oxidation state decreases down a group, whereas the stability of +3 oxidation state increases. This happens because of the inert pair effect.

(iii) Atomic size : On moving down a group, the atomic size increases. This increase, in the atomic size is attributed to an increase in the number of shells.

(iv) Ionisation enthalpy First ionisation enthalpy decreases on moving down a group. This is because of increasing atomic sizes. Ionisation enthalpy of group 15 elements is greater than that of group 14 elements and group 16 elements in the corresponding periods. The order of successive ionisation enthalpies as expected is . △_iH₁ < △_iH₂ < △_iH₃.
Electronegativity : The electronegativity value decreases down the group with increasing atomic size.

Question 2.
Why does the reactivity of nitrogen differ from phosphorus?
Answer:
Nitrogen is chemically less reactive. This is because of the high stability of its molecule, N₂. In N₂, the two nitrogen atoms form a triple bond. This triple bond has very high bond strength, which is very difficult to break. It is because of nitrogen’s small size that it is able to form pn-pn bonds with itself. This property is not exhibited by atoms such as phosphorus. Thus, phosphorus is more reactive than nitrogen.

Question 3.
Discuss the trends in chemical reactivity of group 15 elements.
Answer:
General trends in chemical reactivity of group 15 elements are as follows :
(i) Reactivity towards hydrogen : The elements of group 15 react with hydrogen to form hydrides of type EH₃, where E = N, P, As, Sb, or Bi. The stability of hydrides decreases on moving down from NH₃ to BiH₃.

(ii) Reactivity towards oxygen : The elements of group 15 form two types of oxides: E₂O₃ and E₂O₅ where E = N, P, As, Sb, or Bi. The oxide with the element in the higher oxidation state is more acidic than the other. However, the acidic character decreases on moving down a group.

(iii) Reactivity towards halogens : The group 15 elements react with halogens to form two series of salts: EX₃ and EX₅. However, nitrogen does not form NX₅ as it lacks the d-orbital. All trihalides (except NX₃) are stable.

(iv) Reactivity towards metals : The group 15 elements react with metals to form binary compounds in which metals exhibit – 3 oxidation states.

Question 4.
Why does NH3form hydrogen bond but PH₃ does not?
Answer:
Nitrogen is highly electronegative as compared to phosphorus. This causes a greater attraction of electrons towards nitrogen in NH₃ than towards phosphourus in PH₃. Hence, the extent of hydrogen bonding in PH₃ is very less as compared to NH₃.

Question 5.
How is nitrogen prepared in the laboratory? Write the chemical equations of the reactions involved.
Answer:
(i) In the laboratory, nitrogen is prepared by treating an aqueous solution of ammonium chloride with sodium nitrite.
NH₄Cl(aq) + NaNO₂(aq) → N₂(g) + 2H₂O(Z) + NaCl(aq)
NO and HNO₃ are produced in small amounts. These are impurities that can be removed on passing nitrogen gas through aqueous sulphuric acid, containing potassium dichromate.

(ii) Pure nitrogen is also obtained by thermal decomposition of sodium or barium azide.

Question 6.
How is ammonia manufactured industrially?
Answer:
Ammonia is prepared on a large-scale by the Haber’s process.

According to Le-Chatelier’s principle high pressure would favour the production of ammonia. Optimum conditions for production of NH₃ are
(i) Temperature—700 K
(ii) Pressure—200 × 10⁵ Pa
(iii) Catalyst—Fe₂O₃
(iv) Promotor—K₂O and Al₂O₃

Question 7.
Illustrate how copper metal can give different products on reaction with HNOs.
Answer:
Concentrated nitric acid is a strong oxidising agent. It is used for oxidising most metals. The products of oxidation depend on the concentration of the acid, temperature and also on the material undergoing oxidation.
3Cu + 8HNO_3(dilute) → 3CU(NO₃)₂ + 2NO + 4H₂O
Cu + 4HNO_3(conc.) → CU(NO₃)₂ + 2NO₂ + 2H₂O

Question 8.
Give the resonating structures of N02 and N205.
Answer:
1. Resonating structures of NO₂

2. Resonating structures of NO₂O₅

Question 9.
The HNH angle value is higher than HPH, HAsH and HSbH angles. Why?
[Hint : Can be explained on the basis of sp³ hybridisation in NH₃ and only s-p bonding between hydrogen and other elements of the group.]
Answer:
It can be explained on the basis of sp³ hybridisaton in NH₃ and only s-p bonding between hydrogen and other elements of the group.

As we move down the group, the size of the central atom goes on increasing and its electronegativity goes on decreasing. As a result, the bond pairs of electrons tend to lie away from the central atom as we move from NH₃ to SbH₃. In other words, the force of repulsion between the adjacent bond pairs is maximum in NH3 minimum in SbH3. Consequently the bond angle is maximum in NH₃ and minimum in SbH₃.

Question 10.
Why does R₃P = 0 exist but R₃N = 0 does not (R = alkyl group)?
Answer:
N due to the absence of d-orbitals, cannot form pn-dn multiple bonds. Thus, N cannot expand its covalency beyond four but in R₃N = O, N has a covalency of 5. So, the compound R₃N — O does not exist. On the other hand, P due to the presence of d-orbitals forms pπ-dπ multiple bonds and hence can expand its covalency beyond 4. Therefore, P forms R₃P = O in which the covalency of P is 5.

Question 11.
Explain why NH₃ is basic while BiH₃ is only feebly basic.
Answer:
Since, the atomic size of N (70 pm) is much smaller than that of Bi (148 pm), electron density on the N-atom is much higher than that on Bi-atom. As a result, the tendency of N in NH₃ to donate its lone pair of electrons is much higher than that of Bi in BiH₃. Thus, NH₃ is much more basic than BiH₃.

Question 12.
Nitrogen exists as diatomic molecule and phosphorus as P₄. Why?
Answer:
Nitrogen owing to its small size has a tendency to form pπ-pπ multiple bonds with itself. Nitrogen thus forms a very stable diatomic molecule, N₂. On moving down a group, the tendency to form pπ-pπ bonds decreases (because of the large size of heavier elements). Therefore, phosphorus (like other heavier metals) exists in the P₄ state.

Question 13.
Write main differences between the properties of white phosphorus and red phosphorus.
Answer:

White, Phosphorus	Red Phosphorus
1. It is a soft and waxy solid. It possesses a garlic smell.	It is a hard and crystalline solid, without any smell.
2. It is poisonous.	It is non-poisonous.
3. It is insoluble in water but soluble in carbon disulphide.	It is insoluble in both water and carbon disulphide.
4. It undergoes spontaneous combustion in air.	It is relatively less reactive.
5. P4 molecules are held by weak van der Waal’s forces.	P4 molecules are held by covalent bonds in polymeric structure.
6. Bums easily in Cl2 forming PCl3 and PCl5.	Combines with Cl2 only on heating.

Question 14.
Why does nitrogen show catenation properties less than phosphorus?
Answer:
Catenation is much more common in phosphorus compounds than in nitrogen compounds. This is because of the relative weakness of the N—N single bond as compared to the P—P single bond. Since nitrogen atom is smaller, there is greater repulsion of electron density of two nitrogen atoms, thereby weakening die N—N single bond.

Question 15.
Give the disproportionation reaction of H₃PO₃.
Answer:
On heating, orthophosphorus acid (H₃PO₃) disproportionates to give orthophosphoric acid (H₃PO₄) and phosphine (PH₃). The oxidation states of P in various species involved in the reaction are mentioned below :

Thus, H₃PO₃ (oxidation state +3) oxidises to give H₃PO₄ (oxidation state + 5) and reduce to produce phosphine (oxidation state – 3)

Question 16.
Can PCl₅ act as an oxidising as well as a reducing agent? Justify.
Answer:
PCl₅ can only act as an oxidising agent. The highest oxidation state that P can show is + 5. In PCl₅, phosphorus is in its highest oxidation state (+5). However, it can decrease its oxidation state and act as an oxidising agent. e.g.,

Question 17.
Justify the placement of O, S, Se, Te and Po in the same group of the periodic table in terms of electronic configuration, oxidation state and hydride formation.
Answer:
(i) Electronic configuration : The electronic configuration of the elements of group 16 is given below: .
O₈ = [He] 2s² 2p⁴
S₁₆ = [Ne] 3s² 3p⁴
Se₃₄ = [Ar]3d¹⁰ 4s² 4p⁴
Te₅₂ = [Kr] 4d¹⁰ 5s² 5p⁴
PO₈₄ = [Xe] 4f¹⁴ 5d¹⁰ 6s² 6p⁴
All these elements have similar valence shell configuration ns² np⁴, hence their position in group 16 with each other is justified.

(ii) Oxidation state : As these elements have six valence electrons (ns² np⁴), they should display an oxidation state of – 2. However, only oxygen predominantly shows the oxidation state of -2 owing to its high electronegativity. It also exhibits the oxidation state of -1 (H₂O₂), zero (O₂), and + 2(OF₂). However, the stability of the -2 oxidation state decreases on moving down a group due to a decrease in the electronegativity of the elements. The heavier elements of the group show an oxidation state of + 2, + 4, and + 6 due to the availability of d-orbitals.

(iii) Formation of hydrides : These elements form hydrides of formula H₂E, where E = O, S, Se, Te, Po. Oxygen and sulphur also form hydrides of type H₂E₂. These hydrides are quite volatile in nature.

Question 18.
Why is dioxygen a gas but sulphur a solid?
Answer:
Oxygen is smaller in size as compared to sulphur. Due to its smaller size, it can effectively form pπ—pπ bonds and form O₂(0=0) molecule. Also, the intermolecular forces in oxygen are weak van der Waal’s, which cause it to exist as gas. On the other hand, sulphur does not form M₂ molecule but exists as a puckered structure held together by strong covalent bonds. Hence, it is a solid.

Question 19.
Knowing the electron gain enthalpy values for O → O^– and O → O^2- as- 141 and 702 kJ mol^-1 respectively, how can you account for the formation of a large number of oxides having O ^2- species and not O^–?
[Hint : Consider lattice energy factor in the formation of compounds).
Answer:
Stability of an ionic compound depends on its lattice energy. More the lattice energy of a compound, more stable it will be.

Lattice energy is directly proportional to the charge carried by an ion. When a metal combines with oxygen, the lattice energy of the oxide involving O^2-ion is much more than the oxide involving O^– ion. Hence, the oxide having O^2- ions are more stable than oxides having O^–. Hence, we can say that formation of O^2- is energetically more favourable than formation of O^–.

Question 20.
Which aerosols deplete ozone?
Answer:
Freons or chlorofluorocarbons (CFCs) are aerosols that accelerate the depletion of ozone. In the presence of ultraviolet radiations, molecules of CFCs break down to form chlorine-free radicals that combine with ozone to form oxygen.

Question 21.
Describe the manufacture of H₂SO₄ by contact process?
Answer:
H₂SO₄ is prepared by contact process. The acid produced by this process is free from arsenic impurities and is of high purity. The process involves the following steps:
Step I: Preparation of sulphur dioxide : Preparation of SO₂ by burning of sulphur or roasting pyrites.
S₈(s) + 😯₂(g) → 8SO₂

Step II: Conversion of sulphur dioxide to sulphur trioxide : Sulphur dioxide convert into sulphur trioxide when SO₂ react with oxygen in presence of V₂O₅ at 720 K temperature.

Step III: Formation of oleum : Sulphur trioxide so formed is absorbed in sulphuric acid to form oleum.
SO₃ + H₂SO₄ → H₂S₂O₇

Step IV : Oleum change into H₂SO₄ :

Question 22.
How is SO₂ an air pollutant?
Answer:
Sulphur dioxide causes harm to the environment in many ways:
1. It combines with water vapour present in the atmosphere to form, sulphuric acid. This causes acid rain. Acid rain damages soil, plants, and buildings, especially those made of marble.

2. Even in very low concentrations, SO₂ causes irritation in the respiratory tract. It causes throat and eye irritation and can also affect the larynx to cause breathlessness.

3. It is extremely harmful to plants. Plants exposed to sulphur dioxide for a long time lose colour from their leaves. This condition is known as chlorosis. This happens because the formation of chlorophyll is affected by the presence of sulphur dioxide.

Question 23.
Why are halogens strong oxidising agents?
Answer:
The general electronic configuration of halogens is np⁵, where n = 2 – 6. Thus, halogens need only one more electron to complete their octet and to attain the stable noble gas configuration. Also, halogens are highly electronegative with low dissociation energies and high negative electron gain enthalpies. Therefore, they have a high tendency to gain an electron. Hence, they act as strong oxidising agents.

Question 24.
Explain why fluorine forms only one oxoacid, HOF.
Answer:
Fluorine is known to form only one oxoacid, HOF which is highly unstable. Other halogens form oxoacids of the type HOX, HXO₂, HXO₃ and HXO₄ (X = Cl, Br, I). Fluorine due to its small size, absence of d-orbital and high electronegativity cannot act as central atom in higher oxoacids and hence do not form higher oxoacids.

Question 25.
Explain why inspite of nearly the same electronegativity, nitrogen forms hydrogen bonding while chlorine does not.
Answer:
Electronegativity of both nitrogen (N) as well as chlorine (Cl) is 3.0. But only nitrogen forms hydrogen bonding not chlorine. The reason is that atomic size of N (atomic radius = 70 pm) is less as compared to chlorine (atomic radius =99 pm) therefore, N can cause greater polarisation of N—H bond than Cl in case of Cl—H bond. Hence, N atom is involved in hydrogen bonding and not chlorine.

Question 26.
Write two uses of ClO₂.
Answer:
Uses of ClO₂

1. It is used for purifying water.
2. It is used as a bleaching agent.

Question 27.
Why are halogens coloured?
Answer:
Almost all halogens are coloured. This is because halogens absorb radiations in the visible region. This results in the excitation of valence electrons to a higher energy region. Since, the amount of energy required for excitation differs for each halogen, each halogen displays a different colour.
Different colours of halogens are given below :
Fluorine — Yellow
Chlorine — Greenish yellow
Bromine — Red
Iodine — Violet

Question 28.
Write the reactions of F₂ and Cl₂ with water.
Answer:
(i) Fluorine reacts with water to produce oxygen and ozone.
2F₂(g) + 2H₂O(l) → O₂(g) + 4HF(aq)
3F₂(g) + 3H₂O(l) → 6HF (aq) + O₃(g)

(ii) Chlorine reacts with water in presence of sunlight to produce nascent oxygen.

Question 29.
How can you prepare Cl₂ from HCl and HCl from Cl₂? Write reactions only.
Answer:
HCl can be oxidised to Cl₂ by a number of oxidising agents like MnO₂, KMnO₄.
MnO₂ + 4HCl → MnCl₂ + Cl₂ + 2H₂O
C₂ can be reduced to HCl by reacting with H₂ in the presence of diffused sunlight.

Question 30.
What inspired N. Bartlett for carrying out reaction between Xe and PtF₆?
Answer:
Neil Bartlett observed that PtF₆ reacts with O₂ to yield an ionic solid, \(\mathrm{O}_{2}^{+} \mathrm{PtF}_{6}^{-}\).
O₂(g) + PtF₆(g) → \(\mathrm{O}_{2}^{+} \mathrm{PtF}_{6}^{-}\)
Here, O₂ gets oxidised to \(\mathrm{O}_{2}^{+}\) by PtF₆.
Since, the first ionisation enthalpy of Xe (1170 kJ mol^-1) is fairly close to that of O₂ molecules (1175 kJ mol^-1), Bartlett thought that PtF₆ should also oxidise Xe to Xe⁺. This inspired Bartlett to carry out the reaction between Xe and PtF₂. When Xe and PtF₆ were mixed, a rapid reaction occurred and a red solid with the formula, Xe⁺[PtF₆]^– was obtained.

Question 31.
What are the oxidation states of phosphorus in the following:
(a) H₃PO₃
(b) PCl₃
(c) Ca₃P₂
(d) Na₃PO₄
(e) POF₃.
Answer:
We know, the general valency of H = +1, O = – 2, Ca = + 2, Na = +1, F = -1, Cl = -1

Question 32.
Write balanced equations for the following:
(i) NaCl is heated with sulphuric acid in the presence of MnO₂.
(ii) Chlorine gas is passed into a solution of Nal in water.
Answer:
(i) 4NaCl + MnO₂ + 4H₂SO₄ → MnCl₂ + 4NaHSO₄ + 2H₂O + Cl₂
(ii) Cl₂ + Nal → 2NaCl + I₂

Question 33.
How are xenon fluorides XeF₂, XeF₄ and XeF₆ obtained?
Answer:
Preparation of XeF₂, XeF₄ and XeF₆ :

Question 34.
With what neutral molecule is CIO^– isoelectronic? Is that molecule a Lewis base?
Answer:
CIO^– has 26 electrons [ 17 (Cl) + 8 (0) + le^– (charge)]. The neutral molecule which is isoelectronic with it is C1F (17 + 9) = 26e^–. ClF is a Lewis base.

Question 35.
How are XeO₃ and XeOF₄ prepared?
Answer:
Hydrolysis of XeF₄ and XeF₆ with water gives XeO₃.
6XeF₄ + 12H₂O → 4Xe + 2XeO₃ + 24HF + 3O₂
XeF₆ + 3H₂O → XeO₃ + 6HF
In contrast, partial hydrolysis of XeF₆ gives XeOF₄.
XeF₆ + H₂O → XeOF₄ + 2HF

Question 36.
Arrange the following in the order of property indicated for each set:
(i) F₂, Cl₂, Br₂, I₂—increasing bond dissociation enthalpy.
(ii) HF, HCl, HBr, HI—increasing acid strength.
(iii) NH₃, PH₃, ASH₃, SbH₃, BiH₃—increasing base strength.
Answer:
(i) In the order of increasing bond dissociation enthalpy : Bond dissociation enthalpy decreases as the bond distance increases, so dissociation enthalpy increases as below :
I—I < F—F < Br—Br < Cl—Cl

(ii) In the order of increasing acid strength in water (i.e., aqueous solution) :
As the size of atom increases, then bond dissociation enthalpy of H—X bond decreases. So, acidic strength increases as below :
HF < HCl < HBr < HI

(iii) In the order of increasing base strength :
As we move from NH₃ to BiH₃, the size of the atom increases. Consequently, the electron density on the central atom decreases, so basic strength increases as below :

Question 37.
Which one of the following does not exist?
(i) XeOF₄
(ii) NeF₂
(iii) XeF₂
(iv) XeF₆
Answer:
The sum of first and second ionisation enthalpies of Ne are much higher than those of Xe. Thus, F₂ can oxidise Xe to Xe²⁺ but cannot oxidise Ne to Ne²⁺. In other words, NeF₂ does not exist and all the xenon fluorides (XeF₂ and XeF₆) and xenon oxyfluoride (XeOF₄) do exist.

Question 38.
Give the formula and describe the structure of a noble gas species which is isostructural with :
(i) \(\mathrm{ICl}_{4}^{-}\)
(ii) \(\mathrm{IBr}_{2}^{-}\)
(iii) \(\mathrm{BrO}_{3}^{-}\)
Answer:
(i) In \(\mathrm{ICl}_{4}^{-}\), the central atom “I” has 7 valence electrons and one negative charge. Four of these form single bonds with four Cl atoms (four bond pairs) while the remaining four constitute two lone pairs, so according to VSEPR theory, it should be square planar. \(\mathrm{ICl}_{4}^{-}\) has 7 + 4×7 + 1= 36 valence electrons. A noble gas species having 36 valence electrons is XeF4 (8 + 4 × 7 = 36). Thus, \(\mathrm{ICl}_{4}^{-}\) and XeF₄, both are square planar.

(ii) In \(\mathrm{IBr}_{2}^{-}\), the central atom “I” has 7 valence electrons and one negative charge. Two of these form two single bonds with two Br atoms, while the remaining six constitute three lone pair. Thus, I in \(\mathrm{IBr}_{2}^{-}\) has two bond pairs and three lone pairs, so according to VSEPR theory, it should be linear.

\(\mathrm{IBr}_{2}^{-}\) has 7 + 2×7 + 1=22 valence electrons. A noble gas species having 22 valence electrons is XeF2 (8 + 2 × 7 = 22). Thus, \(\mathrm{IBr}_{2}^{-}\) and XeF₂ both are linear.

(iii) In \(\mathrm{BrO}_{3}^{-}\), the central atom “Br” has seven electrons and one negative charge. Four of these electrons form double bonds with oxygen atoms while fifty electrons forms a single bond with O^– ion. The remaining two electrons form one lone pair, so according to VSEPR theory, it should be pyramidal.

\(\mathrm{BrO}_{3}^{-}\) has 7 + 3 × 6 + 1 = 26 valence electrons. A noble gas species having 26 valence electrons is XeO₃ (8 + 3 × 6 = 26). Thus, \(\mathrm{BrO}_{3}^{-}\) and XeO₃ both are pyramidal.

Question 39.
Why do noble gases have comparatively large atomic sizes?
Answer:
The atomic size, in the case of noble gases, is expressed in terms of van der radii whereas the atomic size of other members of the period is either metallic radii or covalent radii. As the van der radii is larger than both metallic as well as covalent radii, therefore the atomic size of noble gas is quite large. Among the noble gases, the atomic size increases down the group due to addition of new electronic shells.

Question 40.
List the uses of neon and argon gases.
Answer:
(i) Uses of Neon

• It is used in neon discharge lamps and signs which are used for advertising purposes.
• It is used in safety devices for protecting electrical instruments because it has a property of carrying exceedingly high currents under high voltage.

(ii) Uses of Argon

• It is widely used in filling incandescent metal filament electric bulbs.
• It is used for filling radio-valves, rectifiers and fluorescent tubes.

Chemistry Guide for Class 12 PSEB The p-Block Elements Textbook Questions and Answers

Question 1.
Why are pentahalides more covalent than trihalides?
Answer:
In pentahalides, the oxidation state is +5 and in trihalides, the oxidation state is +3. Since the metal ion with a high charge has more polarising power, pentahalides are more covalent than trihalides.

Question 2.
Why is BiH₃ the strongest reducing agent amongst all the hydrides of group 15 elements?
Answer:
As we move down a group, the atomic size increases and the stability of the hydrides of group 15 elements decreases. Since, the stability of hydrides decreases on moving from NH₃ to BiH₃, the reducing character of the hydrides increases on moving from NH₃ to BiH₃.

Question 3.
Why is N₂ less reactive at room temperature?
Answer:
The two N atoms in N₂ are bonded to each other by very strong triple covalent bonds. The bond dissociation energy of this bond is very high. As a result, N₂ is less reactive at room temperature.

Question 4.
Mention the conditions required to maximise the yield of ammonia.
Answer:
Ammonia is produced by Haber’s process as

Yield of ammonia is favoured by high pressure according to Le-Chatelier’s principle. Other conditions, that favour the production of ammonia are as follows:

1. High pressure (200 atm or 200 × 10⁵ Pa)
2. Temperature approximately 700 K
3. Use of a catalyst such as iron oxide mixed with small amounts of Mo or K₂O and Al₂O₃.

Question 5.
How does ammonia react with a solution of Cu²⁺ ?
Answer:
Ammonia reacts with a solution of Cu²⁺ by donating a lone pair of electrons.

Question 6.
What is the covalence of nitrogen in N₂O₅?
Answer:

From the structure of N₂O₅, it is evident that the covalence of nitrogen is 4.

Question 7.
Bond angle in \(\mathbf{P H}_{4}^{+}\) is higher than that in PH₃. Why?
Answer:
In PH₃, P is sp³ hybridised. Three orbitals are involved in bonding with three hydrogen atoms and the fourth one contains a lone pair. As lone pair-bond pair repulsion is stronger than bond pair-bond pair repulsion, the tetrahedral shape associated with sp³ bonding is changed to pyramidal. PH₃ combines with a proton to form \(\mathbf{P H}_{4}^{+}\) in which the lone pair is absent. Due to the absence of lone pair in \(\mathbf{P H}_{4}^{+}\), there is no lone pair-bond pair repulsion. Hence, the bond angle in \(\mathbf{P H}_{4}^{+}\) is higher than the bond angle in PH₃.

Question 8.
What happens when white phosphorus is heated with concentrated NaOH solution in an inert atmosphere of CO₂?
Answer:
White phosphorus dissolves in boiling NaOH in an inert atmosphere of CO₂ to give phosphine (PH₃) and sodium hypophosphite (NaH₂PO₂).

Question 9.
What happens when PCl₅ is heated?
Answer:
All the bonds that are present in PCl₅ are not similar. It has three equatorial and two axial bonds. The equatorial bonds are stronger than the axial ones. Therefore, when PCl₅ is heated strongly, it decomposes to form PCl₃.

Question 10.
Write a balanced-equation for the hydrolytic reaction of PCl₅ in heavy water.
Answer:
Hydrolytic reaction of PCl₅ in heavy water (D₂O)
PCl₅ + D₂O → POCl₃ + 2DCl
POCl₃ + 3D₂O → D₃PO₄ + 3DCl
Therefore, the net reaction can be written as
PCl₅ + 4D₂O → D₃PO₄ + 5DCl

Question 11.
What is the basicity of H₃PO₄?
Answer:
The structure of H₃PO₄ is as

Since there are three OH groups present in H₃PO₄ its basicity is three i.e., it is a tribasic acid.

Question 12.
What happens when H₃PO₃ is heated?
Answer:
H₃PO₃, on heating, undergoes disproportionation reaction to form PH₃ and H₃PO₄. The oxidation numbers of P in H₃PO₃,PH₃, and H₃PO₄ are +3, -3, and + 5 respectively. As the oxidation number of the same element is decreasing and increasing during a particular reaction, the reaction is a disproportionation reaction.

Question 13.
List the important sources of sulphur.
Answer:
Sulphur mainly exists in combined form in the fearth’s crust primarily as sulphates [gypsum (CaSO₄∙2H₂O), Epsom salt (MgSO₄∙7H₂O), baryte (BaSO₄)] and sulphides [galena (PbS), zinc blends (ZnS), copper pyrites (CuFeS₂)].

Traces of sulphur occur as H₂S in volcanoes. Organic materials such as eggs, garlic, onion, mustard, hair and wool also contain sulphur.

Question 14.
Write the order of thermal stability of the hydrides of group 16 elements.
Answer:
The thermal stability of hydrides decreases on moving down the group. This is due to a decrease in the bond dissociation enthalpy of hydrides on moving down the group.
Thus, the order of bond dissociation enthalpy is
H₂O > H₂S > H₂Se > H₂Te > H₂PO
This is also the order of thermal stability.

Question 15.
Why is H₂O a liquid and H₂S a gas?
Answer:
H₂O has oxygen as the central atom. Oxygen has smaller size and higher electronegativity as compared to sulphur. Therefore, there is extensive hydrogen bonding in H₂O, which is absent in H₂S. Molecules of H₂S are held together only by weak van der Waal’s forces of attraction.
Hence, H₂O exists as a liquid while H₂S as a gas.

Question 16.
Which of the following does not react with oxygen directly?
Zn, Ti, Pt, Fe
Answer:
Pt is a noble metal and does not react very easily. All other elements, Zn, Ti, Fe, are quite reactive. Hence, oxygen does not react with platinum (Pt) directly.

Question 17.
Complete the following reactions:
(i) C₂H₄ + O₂ →
(ii) 4Al + 3O₂ →
Answer:

Question 18.
Why does O₃ act as a powerful oxidising agent?
Answer:
Ozone is not a very stable compound under normal conditions and decomposes readily on heating to give a molecule of oxygen and nascent oxygen. Nascent oxygen, being a free redical, is very reactive.

Therefore, ozone acts as a powerful oxidising agent.

Question 19.
How is O₃ estimated quantitatively?
Answer:
Quantitatively, ozone can be estimated with the help of potassium iodide. When ozone is made to react with potassium iodide solution buffered with a borate buffer (pH 9.2), iodine is liberated. This liberated iodine can be titrated against a standard solution of sodium thiosulphate using starch as an indicator. The reactions involved in the process are given below :

Question 20.
What happens when sulphur dioxide is passed through an aqueous solution of Fe(Ill) salt?
Answer:
When SO₂ is passed through an aqueous solution of Fe(III) i.e., ferric salt, it is reduced to Fe(II) i.e. ferrous salt. Here, SO₂ acts as a reducing agent.
2Fe³⁺ + SO₂ + 2H₂O → 2Fe²⁺ + \(\mathrm{SO}_{4}^{2-}\) + 4H⁺

Question 21.
Comment on the nature of two S-O bonds formed in SO₂ molecule. Are the two S—O bonds in this molecule equal?
Answer:
Both the S—O bonds in SO₂ are covalent and have equal strength due to resonating/canonical structure. These are equal with bond length = 143 pm. The resonating structures of SO₂ are as follows :

Question 22.
How is the presence of SO₂ detected?
Answer:
SO₂ is a colourless and pungent smelling gas. Two tests to detect the presence of SO₂ are as follows:
(i) SO₂ decolourises acidified KMnO₄ solution.

(ii) SO₂ changes the colour of acidified potassium dichromate solution from orange to green

Question 23.
Mention three areas in which H₂SO₄ plays an important role.
Answer:
Sulphuric acid is an important industrial chemical and is used for a lot of purposes. Some important uses of sulphuric acid are given below:

1. It is used in fertiliser industry. It is used to make various fertilisers such as ammonium sulphate and calcium super phosphate.
2. It is used in the manufacture of pigments, paints, and detergents.
3. It is used in the manufacture of storage batteries.

Question 24.
Write the conditions to maximise the yield of H₂SO₄ by contact process.
Answer:
The key step in the manufacture of H₂SO₄ is catalytic oxidation of SO₂ to produce SO₃ in presence of V₂O₅.

The reaction is exothermic, reversible and the forward reaction results in the decrease in volume. Thus, according to Le-Chatelier’s principle, the forward reaction should be favoured by low temperature and high pressure. But the temperature should not be very low otherwise the rate of reaction will become very slow.

Question 25.
Why is K_a2 << K_a1 for H₂SO₄ in water?
Answer:
H₂SO₄ is a strong dibasic acid. It ionises in two steps and has two dissociation constants.
H₂SO₄(aq) + H₂O(l) → H₃O⁺(aq) + \(\mathrm{HSO}_{4}^{-}\)(aq); K_a1 >10
\(\mathrm{HSO}_{4}^{-}\)(aq) + H₂O(l) → H₃O⁺(aq) + \(\mathrm{SO}_{4}^{-}\)(aq); K_a2 = 1.2 × 10^-2
K_a1 >> K₁₂
Because the negatively charged HSO₄ ions have much less tendency to donate a proton to H₂O as compared to neutral H₂SO₄.

Question 26.
Considering the parameters such as bond dissociation enthalpy, electron gain enthalpy and hydration enthalpy, compare the oxidising power of F₂ and C₂.
Answer:
The electrode potential depends upon the parameters indicated below :

(Values of kJ mole-1)	△dissH	△egH	△hydH
Fluorine	158.8	-333	515
Chlorine	242.6	-349	381

The two factors, high hydration enthalpy of F^-1 ion (515 kJ mol^-1) and low F—F bond dissociation enthalpy more than compensate the less negative electron gain enthalpy of fluorine. Due to this, electrode potential of F₂ (+2.87 V) is much higher than that of Cl₂ (+1.36 V) and hence F₂ is a stronger oxidising agent than Cl₂.

Question 27.
Give two examples to show the anomalous behaviour of fluorine.
Answer:
Anomalous behaviour of fluorine

1. It forms only one oxoacid as compared to other halogens that form a number of oxoacids.
2. Ionisation enthalpy, electronegativity, and electrode potential of fluorine are much higher than expected.

Question 28.
Sea is the greatest source of some halogens. Comment.
Answer:
Sea water contains chlorides, bromides, and iodides of Na, K, Mg, and Ca. However, it primarily contains NaCl. The deposits of dried up sea beds contain sodium chloride and camallite, KCl ∙ MgCl₂ ∙ 6H₂O. Marine life also contains iodine in their systems. For example, sea weeds contain upto 0.5% iodine as sodium iodide. Thus, sea is the greatest source of some halogens.

Question 29.
Give the reason for bleaching action of Cl₂.
Answer:
When chlorine reacts with water, it produces nascent oxygen. This nascent oxygen then combines with the coloured substances present in the organic matter to oxide them into colourless substances.
Cl₂ + H₂ → 2HCl + [O]
Coloured substance + [O] → Colourless substance
Bleaching action of chlorine creates permanent effect. It bleaches the vegetable or organic matter in the presence of moisture.

Question 30.
Name two poisonous gases which can be prepared from chlorine gas.
Answer:
Two poisonous gases that can be prepared from chlorine gas are

1. Phosgene (COCl₂)
2. Mustard gas (ClCH₂CH₂SCH₂CH₂Cl)

Question 31.
Why is ICl more reactive than I₂?
Answer:
ICl is more reactive than I₂ because I—Cl bond in IC1 is weaker than I—I bond in I₂ due to less bond dissociation energy consequently I-Cl break easily to form halogen atoms which readily bring about the reactions.

Question 32.
Why is helium used in diving apparatus?
Answer:
Air contains a large amount of nitrogen and the solubility of gases in liquids increases with increase in pressure. When sea divers dive deep into the sea, large amount of nitrogen dissolves in their blood. When they come back to the surface, solubility of nitrogen decreases and it separates from the blood and forms small air bubbles. This leads to a dangerous medical condition called bends. Therefore, air in oxygen cylinders used for diving is diluted with helium gas. This is done as He is sparingly less soluble in blood.

Question 33.
Balance the following equation: XeF₆ + H₂O → XeO₂F₂ + HF
Answer:
Balanced equation
XeF₆ + 2H₂O → XeO₂F₂ + 4HF

Question 34.
Why has it been difficult to study the chemistry of radon?
Answer:
It is difficult to study the chemistry of radon because it is a radioactive substance having a half-life of only 3.82 days. Also, compounds of radon such as RnF₂ have not been isolated. They have only been identified.

Punjab State Board PSEB 12th Class Chemistry Important Questions Chapter 5 Surface Chemistry Important Questions and Answers.

PSEB 12th Class Chemistry Important Questions Chapter 5 Surface Chemistry

Very Short Answer Type Questions

Question 1.
Define desorption.
Answer:
The process of removal of an adsorbed substance from a surface on which it is adsorbed is called desorption.

Question 2.
What is the effect of temperature on chemisorption?
Answer:
Chemisorption initially increases then decreases with rise in temperature. The initial increase is due to the fact that heat supplied acts as activation energy. The decrease afterwards is due to the exothermic nature of adsorption equilibrium.

Question 3.
What is the role of diffusion in heterogeneous catalysis?
Answer:
The gaseous molecules diffuses on to the surface of the solid catalyst and get adsorbed. After the required chemical changes, the products diffuse away from the surface of the catalyst leaving the surface free for more reactant molecules to get adsorbed and undergo reaction.

Question 4.
What is the type of charge on Agl colloidal sol formed when AgNO₃ solution is added to KI solution?
Answer:
Negatively charged sol, Agl/I^– is formed when AgNO₃ solution is added to KI solution.

Question 5.
What causes Brownian movement in a colloidal solution?
Answer:
Unbalanced bombardment of the particles of dispersed phase by molecules of dispersion medium causes Brownian motion. This stabilises the sol.

Question 6.
Based on the type of dispersed phase, what type of colloid is micelles?
Answer:
Associated colloids

Question 7.
Name the temperature above which the formation of micelles takes place.
Answer:
Kraft temperature.

Question 8.
How do emulsifying agents stabilise the emulsion?
Answer:
The emulsifying agent forms an interfacial layer between suspended particles and the dispersion medium thereby stabilising the emulsion.

Question 9.
Write the dispersed phase and dispersion medium of butter.
Answer:
Dispersed phase — Liquid
Dispersion medium — Solid.

Question 10.
Write the main reason for the stability of colloidal sols.
Answer:
All the particles of colloidal sol carry the same charge so they keep on repelling each other and do not aggregate together to form bigger particles.

Question 11.
How is Brownian movement responsible for the stability of sols?
Answer:
The Brownian movement has a stirring effect, which does not allow the particles to settle down.

Short Answer Type Questions

Question 1.
Differentiate among a homogeneous solution, a suspension and a colloidal solution, giving a suitable example of each.
Answer:

Property	Homogeneous solution	Colloidal solution	Suspension
(i) Particle size	Less than 1 nm	Between 1 nm to 1000 nm	More than 1000 nm
(ii) Separation by
ordinary filtration	Not possible	Not possible	Not possible
ultra filtration	Not possible	Possible	Possible
(iii)   Settling of particles	Do not settle	Settle only on coagulation	Settle under gravity
(iv) Appearance	Transparent	Opaque	Translucent
(v) Example	Glucose dissolved in water	Smoke, milk, gold sol	Sand in water

Question 2.
Classify colloids where the dispersion medium is water. State their characteristics and write an example of each of these classes.
Answer:
These are of two types
(i) Hydrophilic
Stability: More stable as the stability is due to charge and water envelope surrounding the sol particles.
Nature: Reversible
Examples: Starch, gum etc.

(ii) Hydrophobic
Stability: Less stable as the stability is due to charge only.
Nature: Irreversible
Examples: Metal hydroxide like Fe(OH)₃ and metal sulphide like As₂S₃.

Question 3.
Explain the cleansing action of soap. Why do soaps not work in hard water?
Answer:
The cleansing action of soap such as sodium stearate is due to the fact that soap molecules form micelle around the oil droplet in such a way that hydrophobic part of the stearate ions is in the oil droplet and hydrophilic part projects out of the grease droplet like the bristles. Since the polar groups can interact with water, the oil droplet surrounded by stearate ions is now pulled in water and removed from the dirty surface. Thus, soap helps in emulsification and washing away of oils and fats.

Hard water contains calcium and magnesium salts. In hard water, soap gets precipitated as calcium and magnesium soap which being insoluble stick to the clothes as gummy mass. Therefore, soaps do not work in hard water.

Question 4.
Adsorption of a gas on the surface of solid is generally accompanied by a decrease in entropy still it is a spontaneous process. Why?
Answer:
According to the equation
△G = △H – T△S
For a process to be spontaneous, △G should be negative. Even though △S is negative here, △G is negative because reaction is highly exothermic, i.e., △H is negative.

Question 5.
Define the following terms:
(i) Brownian movement,
(ii) Peptization.
Answer:
(i) Brownian movement : The motion of the colloidal particles in a zig zag path due to unbalanced bombardment by the particles of dispersion medium is called Brownian movement.

(ii) Peptization : The process of converting a precipitate into colloidal sol by shaking it with dispersion medium in the presence of a small amount of suitable electrolyte is called peptization. During peptization, the precipitate absorbs one of the ions of the electrolyte on its surface. This causes development of positive or negative charge on precipitates, which ultimately break up into particles of colloidal dimension.

Question 6.
(i) Write the expression for Freundlich’s equation to describe the behaviour of adsorption from solution.
(ii) What causes charge on sol particles?
(iii) Name the promoter used in the haber’s process for the manufacture of ammonia.
Answer:
(i) \(\frac{x}{m}\) = KC^{\(\frac{1}{n}\)}
(ii) The charge on the sol particles is due to :

•  electron capture by sol particles during electro dispersion.
• preferential anolsorption of ions from solution.
• formulation of electrical double layer.

(iii) Molybdenum acts in a promoter for iron.

Long Answer Type Questions

Question 1.
Consider the adsorption isotherms given alongside and interpret the variation in the extent of adsorption (xlm) when

(a) (i) temperature increases at constant pressure.
(ii) pressure increases at constant temperature.
(b) Name the catalyst and the promoter used in Haber’s process for manufacture of ammonia.
Answer:
(a) (i) At constant pressure, extent of adsorption \(\left(\frac{x}{m}\right)\) decreases with increase in temperature as adsorption is an exothermic process.

(ii) At constant temperature, first adsorption \(\left(\frac{x}{m}\right)\) increases with increase in pressure up to a particular pressure and then it
At low pressure, \(\frac{x}{m}\) = kp m
At intermediate range of pressure, \(\frac{x}{m}\) = kp^1/n (n > 1)
At high pressure, \(\frac{x}{m}\) = k (independent of pressure)

(b) Finely divided iron is used as a catalyst and molybdenum is used as promoter.

Question 2.
Explain the following observations:
(i) Sun looks red at the time of setting.
(ii) Cottrell’s smoke precipitator is fitted at the mouth of the chimney used in factories.
(iii) Physical adsorption is multilayered while chemical adsorption is monolayered.
Answer:
(i) At the time of setting, the sun is at horizon. The light emitted by the sun has to travel a relatively longer distance through the atmosphere. As a result, blue part of light is scattered away by the particulate in the atmosphere causing red part to be visible.

(ii) Cottrell’s smoke precipitator, neutralises the charge on unburnt carbon particles, coming out of chimney and they get precipitated and settle down at the floor of the chamber.

(iii) Physical adsorption involves van der Waals’ forces, so any number of layers may be formed one over the other on the surface of the adsorbent. Chemical adsorption takes place as a result of the reaction between adsorbent and adsorbate. When the surface of adsorbent is covered with one layer, no further reaction can take place.

Punjab State Board PSEB 6th Class Hindi Book Solutions Hindi Grammar मुहावरे Exercise Questions and Answers, Notes.

PSEB 6th Class Hindi Grammar मुहावरे (2nd Language)

मुहावरे अर्थ एवं वाक्य प्रयोग सहित :

1. अटल रहना = टिके रहना।
प्रयोग – हमें अपने धर्म पर अटल रहना चाहिए।

2. अपनी जान तक लड़ा देना = मरने – मारने से न डरना
प्रयोग – वीर पुरुष लक्ष्य प्राप्त करने के लिए अपनी जान तक लड़ाते हैं।

3. आनन्द के समुद्र में डूबना = बहुत खुश होना
प्रयोग – शब्द कीर्तन सुनकर सभी आनन्द के समुद्र में डूब गए।

4. आड़े समय में पल्ला पकड़ना = मुसीबत के समय सहारा ढूँढ़ना
प्रयोग – सुरेन्द्र ने आड़े समय में पल्ला पकड़ा था, मैं उसे नहीं भूल सकता।

5. आँसुओं से हाथ भिगोना = किसी को बिना कारण दुःख देकर खुश होना
प्रयोग – दुष्ट लोग आँसुओं से हाथ भिगोना मामूली – सी बात समझते हैं।

6. आँख की किरकिरी = दिल में चुभना।
प्रयोग – दूसरों की आँख की किरकिरी मत करो।

7. कोई कसर न छोड़ना = कोई कमी न रहने देना
प्रयोग – वह पास नहीं हो सका पर मेहनत में उसने कोई कसर नहीं छोड़ी।

8. चिन्ता में डूब जाना = बहुत फिक्र करना
प्रयोग – बेटे के न आने पर माता जी चिन्ता में डूब गईं।

9. चेहरा खिल उठना = खुश हो जाना।
प्रयोग – अपने मित्र को देख कर उसका चेहरा खिल उठा।

10. जाल में फँसी मछली = मुसीबत में पड़ा आदमी
प्रयोग – बीमारी में बूढ़ा, जाल में फँसी मछली की तरह तड़प रहा था।

11. जिगर का टुकड़ा = प्यारा बेटा
प्रयोग – राम अपने जिगर के टुकड़े को पल भर के लिए भी दूर नहीं करता।

12. जी ललचाना = इच्छा होना
प्रयोग – जलेबियाँ बनती देख मोहन का जी ललचाने लगा।

13. जीवन फूंक देना = नई जान डालना, उत्साह पैदा करना
प्रयोग – स्वामी विवेकानन्द ने लोगों में जीवन फूंक दिया था।

14. जीवन लीला समाप्त होना = मर जाना
प्रयोग – कुछ दिन की बीमारी के बाद लाला जी की जीवन लीला समाप्त हो गई।

15. दंग रह जाना = हैरान होना
प्रयोग – सर्कस के करतब देख कर बच्चे दंग रह जाते हैं।

16. दाल न गलना = मनचाहा काम न होना
प्रयोग – चले जाओ, यहाँ तुम्हारी दाल न गलेगी।

17. तलवार संभालना = लड़ने के लिए तैयार होना।
प्रयोग – देश की रक्षा के लिए सभी ने तलवार संभाल ली।

18. दिमाग दौड़ाना = गहराई से सोचना
प्रयोग – मोहन ने बहुत दिमाग दौड़ाया पर प्रश्न समझ में नहीं आया।

19. ठाठ का जीवन व्यतीत करना = शान – शौकत से रहना
प्रयोग – पिता की मृत्यु पर देव ठाठ का जीवन व्यतीत करने लगा।

20. हाथ से दस हाथ हो जाना = बहत से सहायता करने वाले मिल जाने
प्रयोग – दोनों लड़के विदेश से लौटने पर लाला जी के हाथ से दस हाथ हो गए।

21. धोखा – धड़ी करना = हेराफेरी करना
प्रयोग – किसी के साथ भी धोखा – धड़ी करना पाप है।

22. धूम मचाना = प्रसिद्ध होना
प्रयोग – जादूगर ने शहर में धूम मचा दी।

23. डट जाना = अड़े रहना
प्रयोग – उपकार परीक्षा के दिनों पढ़ाई में डट जाती है।

24. नयनों से गंगा – यमुना बहाना = खूब आँसू बहाना
प्रयोग – बेटे की मौत पर बुढ़िया के नयनों से गंगा – यमुना बहने लगी।

25. नाक होना = स्वाभिमानी होना
प्रयोग – नेता जी ने अपने काम से नाक होना सिद्ध कर दिया।

26. नौ दो ग्यारह होना = भाग जाना
प्रयोग – सिपाही को देखते ही चोर नौ दो ग्यारह हो गया।

27. नौ नकद न तेरह उधार = तुरन्त उचित कीमत ले लेना
प्रयोग – अच्छे दुकानदार का नियम है – नौ नकद न तेरह उधार।

28. पलक झपकते ही = एक दम थोड़ी – सी देरी में
प्रयोग – पलक झपकते ही घोड़ा दीवार कूद गया।

29. पसीना बहाना = कड़ी मेहनत करना
प्रयोग – आजकल निर्वाह के लिए हर आदमी को पसीना बहाना पड़ता है।

30. परमात्मा में लीन होना = परलोक सिधार जाना।
प्रयोग – भक्त हमेशा परमात्मा में लीन होना चाहता है।

31. प्राण पखेरू उड़ना = मर जाना
प्रयोग – सीढ़ियों से नीचे गिरते ही बुढ़िया के प्राण पखेरू उड़ गए।

32. प्रसन्नता की लहर दौड़ना = बहुत खुश होना
प्रयोग – भारत की जीत पर सब लोगों के चेहरों पर प्रसन्नता की लहर दौड़ गई।

33. फूला न समाना = बहुत खुश होना
प्रयोग – पास होने पर मोहन फूला न समा रहा था।

34. मन जीत लेना = अपना बना लेना
प्रयोग – छोटी बहू ने अपने अच्छे व्यवहार से सब का मन जीत लिया।

35. मिट्टी में मिलाना = नष्ट करना
प्रयोग – शराबी बेटे ने पिता का सारा धन मिट्टी में मिला दिया।

36. मैदान साफ़ नज़र आना = कोई बाधा या मुश्किल न दिखना
प्रयोग – आगे बढ़ो हर क्षेत्र में मैदान साफ़ नज़र आता है।

37. मोल – भाव करना = कोई वस्तु खरीदते समय कीमत घटाना – बढ़ाना।
प्रयोग – हर चीज़ मोल – भाव करके ही लेनी चाहिए।

38. मौत के घाट उतारना = मार डालना
प्रयोग – गुरु जी ने दोनों पठानों को मौत के घाट उतार दिया।

39. व्यवसाय चमकना = व्यापार में वृद्धि होना
प्रयोग – मेहनत करने से व्यवसाय चमकने लगता है।

40. विचित्र भूमिका निभाना = अद्भुत कार्य करना
प्रयोग – युद्ध में सेनापति ने शत्रुओं पर विजय दिलाने में विचित्र भूमिका निभाई।

41. विदा लेना = चले जाना।
प्रयोग – दस दिन के बाद स्वामी जी ने विदा ले ली।

42. सन्नाटा छाना = चुप्पी होना
प्रयोग – कर्गों से शहर में सन्नाटा छा जाता है।

43. सुगन्धि आना = मन को अच्छा लगना
प्रयोग – चन्दन की सुगन्धि आना स्वाभाविक है।

44. हाथ बंटाना = मदद करना
प्रयोग – विद्यार्थियों को घर के कामों में भी हाथ बंटाना चाहिए।

45. हाथ – पाँव मारना = कोशिश करना
प्रयोग – हाथ – पाँव मारने से ही काम बन सकता है।

46. हालत बहुत पतली होना = पैसे की कमी होना
प्रयोग – आय का साधन न रहने से उसके घर की हालत बहुत पतली हो गई।

47. हिरण की तरह चौकड़ी भरना = खूब उछलते हुए आगे बढ़ना
प्रयोग – कई धावक हिरण की तरह चौकड़ी भरते हैं।

Punjab State Board PSEB 12th Class Chemistry Book Solutions Chapter 5 Surface Chemistry Textbook Exercise Questions and Answers.

PSEB Solutions for Class 12 Chemistry Chapter 5 Surface Chemistry

PSEB 12th Class Chemistry Guide Surface Chemistry InText Questions and Answers

Question 1.
Distinguish between the meaning of the terms adsorption and absorption.
Give one example of each.
Answer:

Adsorption	Absorption
1. It is the surface phenomenon.	It is the bulk phenomenon.
2. It is the phenomenon as a result of which the species of one substance gets concentrated mainly on the surface of another substance.	It is the phenomenon as a result of which one substance gets distributed uniformly throughout the total volume of another substance.
3. Adsorption is fast in the beginning then slows down due to non­availability of the surface.	Absorption proceeds at uniform rate.
4. The concentration on the surface of the adsorbent is different from that in the bulk. e.g., Water vapours on silica gel.	The concentration is same throughout the material. e.g., Water vapours are absorbed by anhydrous CaCl2.

Question 2.
What is the difference between physisorption and chemisorption?
Answer:

Physisorption	Chemisorption
1. In this type of adsorption, the adsorbate is attached to the surface of the adsorbent with weak van der Waal’s forces of attraction.	In this type of adsorption, strong chemical bonds are formed between the adsorbate and the surface of the adsorbent.
2. No new compound is formed in the process.	New compounds are formed at the surface of the adsorbent.
3. It is generally found to be reversible in nature.	It is usually irreversible in nature.
4. Enthalpy of adsorption is low as weak van der Waal’s forces of attraction are involved. The values lie in the range of 20-40 kJ mol-1.	Enthalpy of adsorption is high as chemical bonds are formed. The values lie in the range of 40-400 kJ mol-1.
5. It is favoured by low temperature conditions.	It is favoured by high temperature conditions.
6. It is an example of multi-layer adsorption	It is an example of mono-layer adsorption.

Question 3.
Give reason why a finely divided substance is more effective as an adsorbent.
Answer:
Adsorption is a surface phenomenon. Therefore, adsorption is directly proportional to the surface area. A finely divided substance has a large surface area. Both physisorption and chemisorption increase with an increase in the surface area. Hence, a finely divided substance behaves as a good adsorbent.

Question 4.
What are the factors which influence the adsorption of a gas on a solid?
Answer:
There are various factors that affect the rate of adsorption of a gas on a solid surface.

1. Nature of the gas : Easily liquefiable gases such as NH₃, HCl etc. are adsorbed to a great extent in comparison to gases such as H₂, O₂ etc. This is because van der Waal’s forces are stronger in easily liquefiable gases.
2. Surface area of the solid : The greater the surface area of the adsorbent, the greater is the adsorption of a gas on the solid surface.
3. Effect of pressure : Adsorption is a reversible process and is accompanied by a decrease in pressure. Therefore, adsorption increases with an increase in pressure.
4. Effect of temperature : Adsorption is an exothermic process. Thus, in accordance with Le-Chatelier’s principle, the magnitude of adsorption decreases with an increase in temperature.

Question 5.
What is an adsorption isotherm? Describe Freundlich adsorption isotherm.
Answer:

The plot between the extent of absorption \(\left(\frac{x}{m}\right)\) against the pressure of gas (P) at constant temperature (T) is called the adsorption isotherm.

Freundlich adsorption isotherm : Freundlich adsorption isotherm gives an empirical relationship between the quantity of gas adsorbed by the unit mass of solid adsorbent and pressure at a specific temperature.
From the given plot it is clear that at pressure P_s, \(\frac{x}{m}\) reaches the maximum value. Ps is called the saturation pressure. Three cases arise from the graph now :
Case I-At low pressure
The plot is straight and sloping, indicating that the pressure is directly proportional to \(\frac{x}{m}\) i.e., \(\frac{x}{m}\) ∝ P.
\(\frac{x}{m}\) = kP (k is a constant)

Case II-At high pressure
When pressure exceeds, the saturated pressure, \(\frac{x}{m}\) becomes independent of P values.
\(\frac{x}{m}\) ∝ P^o
\(\frac{x}{m}\) = kP^o

Case III-At intermediate pressure
At intermediate pressure, \(\frac{x}{m}\) depends on P raised to the powers between 0 and 1. This relationship is known as the Freundlich adsorption isotherm.
\(\frac{x}{m}\) ∝ P^{\(\frac{1}{n}\)}
\(\frac{x}{m}\) = kP^1/n n > 1
Now, taking log
log\(\frac{x}{m}\) = log k + \(\frac{1}{n}\)logP
On plotting the graph between log \(\left(\frac{x}{m}\right)\) and log P, a straight line is obtained with the slope equal to \(\frac{1}{n}\) and intercept equal to log k.

Question 6.
What do you understand by activation of adsorbent? How is it achieved?
Answer:
By activating an adsorbent, we tend to increase the adsorbing power of the adsorbent. Some ways to activate an adsorbent are:

1. By increasing the surface area of the adsorbent. This can be done by breaking it into smaller pieces or powdering it.
2. Some specific treatments can also lead to the activation of the adsorbent. For example, wood charcoal is activated by heating it between 650 K and 1330 K in vacuum or air. It expels all the gases absorbed or adsorbed and thus, creates a space for adsorption of gases.

Question 7.
What role does adsorption play in heterogeneous catalysis?
Answer:
Heterogeneous catalysis : A catalytic process in which the catalyst and the reactants are present in different phases is known as a heterogeneous catalysis. This heterogeneous catalytic action can be explained in terms of the adsorption theory. The mechanism of catalysis involves the following steps:

1. Adsorption of reactant molecules on the catalyst surface.
2. Occurrence of a chemical reaction through the formation of an intermediate.
3. Desorption of products from the catalyst surface.
4. Diffusion of products away from the catalyst surface.

In this process, the reactants are usually present in the gaseous state and the catalyst is present in the solid state. Gaseous molecules are then adsorbed on the surface of the catalyst. As the concentration of reactants on the surface of the catalyst increases, the rate of reaction also increases. In such reactions, the products have very less affinity for the catalyst and are quickly desorbed, thereby making the surface free for other reactants.

Question 8.
Why is adsorption always exothermic?
Answer:
Adsorption is always exothermic. This statement can be explained in two ways:
(i) Adsorption leads to a decrease in the residual forces on the surface of the adsorbent. This causes a decrease in the surface energy of the adsorbent. Therefore, adsorption is always exothermic.

(ii) AH of adsorption is always negative. When a gas is adsorbed on a solid surface, its movement is restricted leading to a decrease in the entropy of the gas i.e., ∆S is negative. Now for a process to be spontaneous, ∆G should be negative.
∴ ∆G – ∆H – T∆S
Since, ∆S is negative, ∆H has to be negative to make ∆G negative. Hence, adsorption is always exothermic.

Question 9.
How are the colloidal solutions classified on the basis of physical stjates of the dispersed phase and dispersion medium?
Answer:
One criterion for classifying colloids is the physical state of the dispersed phase and dispersion medium. Depending upon the type of the dispersed phase and dispersion medium (solid, liquid, or gas), there can be eight types of colloidal systems.

Dispersed phase	Dispersion medium	Type of colloid	Example
Solid	Solid	Solid Sol	Gemstones, glasses
Solid	Liquid	Sol	Paints, cell fluids
Solid	Gas	Aerosol	Smoke, dust
Liquid	Solid	Gel	Cheese, butter
Liquid	Liquid	Emulsion	Milk, hair cream
Liquid	Gas	Aerosol	Fog, mist, cloud
Gas	Solid	Solid Sol	Pumice stone, foam rubber
Gas	Liquid	Foam	Froth, soap lather

Question 10.
Discuss the effect of pressure and temperature on the adsorption of gases on solids.
Answer:
Effect of pressure on adsorption : At constant temperature, the extent of adsorption of a gas (x / m) on a solid increases with pressure. A graph between x / m and the pressure p of a gas at constant temperature is called adsorption isotherm.

(i) At lower range of pressure, x / m is directly proportional tothe applied pressure.
\(\frac{x}{m}\) ∝ p¹ or \(\frac{x}{m}\) = kp

(ii) At high pressure range, the extent of adsorption of a gas (x / m) is independent of the applied pressure, i.e.,
\(\frac{x}{m}\) ∝ p^o or \(\frac{x}{m}\) = k

(iii) At intermediate pressure range, the value of x / m is proportional to a fractional power of pressure, i. e.,
\(\frac{x}{m}\) ∝ p^1/n or \(\frac{x}{m}\) = kp^1/n
where 1 / n is a fraction. Its value may be between 0 and 1.
log\(\left(\frac{x}{m}\right)\) = log k + \(\frac{1}{n}\) log p

Effect of temperature on adsorption : Adsorption is generally temperature dependent. Mostly adsorption processes are exothermic and hence adsorption decreases with increasing temperature. However, for an endothermic adsorption process, adsorption increases with increase in temperature.

Question 11.
What are lyophilic and lyophobic sols? Give one example of each type. Why are hydrophobic sols easily coagulated?
Answer:
(i) Lyophilic sols : Colloidal sols directly formed by mixing substances in a suitable dispersion medium are called lyophilic sols. These sols are reversible in nature i.e., if two constituents of the sol are separated by any means (such as evaporation), then the sol can be prepared again by simply mixing the dispersion medium with the dispersion phase and shaking the mixture, e.g., gum, gelatin, starch, rubber etc.

(ii) Lyophobic sols : When substances such as metals and their sulphides etc. are mixed with the dispersion medium, they do not form colloidal sols. Their colloidal sols can be prepared only by special methods. Such sols are called lyophobic sols. These sols are irreversible in nature, e.g., gold sol, AS₂O₃ etc.

Now, the stability of hydrophilic sols depends on two things—the presence of a charge and the salvation of colloidal particles. On the other hand, the stability of hydrophobic sols is only because of the presence of a charge. Therefore, the latter are much less stable than the former. If the charge of hydrophobic sols is removed (by addition of electrolytes), then the particles present in them come closer and form aggregates, leading to precipitation.

Question 12.
What is the difference between multimolecular and macromolecular colloids? Give one example of each. How are associated colloids different from these two types of colloids?
Answer:
(i) In multi-molecular colloids, the colloidal particles are an aggregate of atoms or small molecules with a diameter of less than 1 nm. The molecules in the aggregate are held together by van der Waal’s forces of attraction. Examples of such colloids include gold sol and sulphur sol.

(ii) In macro-molecular colloids, the colloidal particles are large molecules . having colloidal dimensions. These particles have a high molecular mass. When these particles are dissolved in a liquid, sol is obtained. For example: starch, nylon, cellulose, etc.

(iii) Certain substances tend to behave like normal electrolytes at lower concentrations. However, at higher concentrations, these substances behave as colloidal solutions due to the formation of aggregated particles. Such colloids are called associated colloids.

Question 13.
What are enzymes? Write in brief the mechanism of enzyme catalysis.
Answer:
Enzymes are basically protein molecules of high molecular masses. These form colloidal solutions when dissolved in water. These are complex, nitrogenous organic compounds produced by living plants and animals. Enzymes are also called “biochemical catalysts’.

On the surface of the enzymes, various cavities are present with characteristic shapes. These cavities possess active groups such as—NH₂, —COOH, etc. The reactant molecules having a complementary shape fit into the cavities just like a key fits into a lock. This leads to the formation of an activated complex. This complex then decomposes to give the product.

1. Binding of enzyme to substrate (reactant) to form activated complex.
   E + S → ES*
2. Decomposition of the activated complex to form product.
   ES* → E + P

Question 14.
How are colloids classified on the basis of
(i) physical states of components
(ii) nature of dispersion medium and
(iii) interaction between dispersed phase and dispersion medium?
Answer:
(i) One criterion for classifying colloids is the physical state of the dispersed phase and dispersion medium. Depending upon the type of the dispersed phase and dispersion medium (solid, liquid, or gas), there can be eight types of colloidal systems.

Dispersed phase	Dispersion medium	Type of colloid	Example
Solid	Solid	Solid Sol	Gemstones, glasses
Solid	Liquid	Sol	Paints, cell fluids
Solid	Gas	Aerosol	Smoke, dust
Liquid	Solid	Gel	Cheese, butter
Liquid	Liquid	Emulsion	Milk, hair cream
Liquid	Gas	Aerosol	Fog, mist, cloud
Gas	Solid	Solid Sol	Pumice stone, foam rubber
Gas	Liquid	Foam	Froth, soap lather

(ii) On the basis of the nature of dispersion medium, colloids can be divided as:

Dispersion medium	Name of sol
Water	Aquasol or hydrosol
Alcohol	Alcosol
Benzene	Benzosol
Gases	Aerosol

(iii) On the basis of the nature of the interaction between the dispersed phase and dispersion medium, the colloids can be classified as lyophilic (solvent attracting) and lyophobic (solvent repelling).

Question 15.
Explain what is observed
(i) when a beam of light is passed through a colloidal sol.
(ii) an electrolyte, NaCl is added to hydrated ferric oxide sol.
(iii) electric current is passed through a colloidal sol?
Answer:
(i) When a beam of light is passed through a colloidal solution, then scattering of light is observed. This is known as the Tyndall effect. This scattering of light illuminates the path of the beam in the colloidal solution.

(ii) When NaCl is added to hydrated ferric oxide sol, it dissociates to give Na⁺ and Cl^– ions. Particles of ferric oxide sol are positively charged. Thus, they get coagulated in the presence of negatively charged Cl^– ions.

(iii) The colloidal particles are charged and carry either a positive or negative charge. The dispersion medium carries an equal and opposite charge. This makes the whole system neutral. Under the influence of an electric current, the colloidal particles move towards the oppositely charged electrode. When they come in contact with the electrode, they lose their charge and coagulate.

Question 16.
What are emulsions? What are their different types? Give example of each type.
Answer:
The colloidal solution in which both the dispersed phase and dispersion medium are liquids is called an emulsion. There are two types of emulsions:
(a) Oil in water type : Here, oil is the dispersed phase while water is the dispersion medium. For example: milk, vanishing cream, etc.

(b) Water in oil type : Here, water is the dispersed phase while oil is the dispersion medium. For example: cold cream, butter, etc.

Question 17.
What is demulsification? Name two demulsifiers.
Answer:
The process of decomposition of an emulsion into its constituent liquids is called demulsification. Examples of demulsifiers are surfactants, ethylene oxide, etc.

Question 18.
Action of soap is due to emulsification and micelle formation. Comment.
Answer:
The cleansing action of soap is due to emulsification and micelle formation. Soaps are basically sodium and potassium salts of long chain fatty acids, R-COO^– Na⁺. The end of the molecule to which the sodium is attached is polar in nature, while the alkyl-end is non-polar. Thus, a soap molecule contains a hydrophilic (polar) and a hydrophobic (non-polar) part.

When soap is added to water containing dirt, the soap molecules surround the dirt particles in such a manner that their hydrophobic parts get attached to the dirt molecule and the hydrophilic parts point away from the dirt molecule. This is known as micelle formation. Thus, we can say that the polar group dissolves in water while the non-polar group dissolves in the dirt particle. Now, as these micelles are negatively charged, they do not coalesce and a stable emulsion is formed.

Question 19.
Give four examples of heterogeneous catalysis.
Answer:
Examples of heterogeneous catalysis
(i) Oxidation of sulphur dioxide to form sulphur trioxide. In this reaction, Pt acts as a catalyst.

(ii) Formation of ammonia by the combination of dinitrogen and dihydrogen in the presence of finely divided iron.

This process is called the Haber’s process.

(iii) Oswald’s process: Oxidation of ammonia to nitric oxide in the presence of platinum.

(iv) Hydrogenation of vegetable oils in the presence of Ni.

Question 20.
What do you mean by activity and selectivity of catalysts?
Answer:
(a) Activity of a catalyst : The activity of a catalyst is its ability to increase the rate of a particular reaction. Chemisorption is the main factor in deciding the activity of a catalyst. The adsorption of reactants on the catalyst surface should be neither too strong nor too weak. It should just be strong enough to make the catalyst active.

(b) Selectivity of the catalyst : The ability of the catalyst to direct a reaction to yield a particular product is referred to as the selectivity of the catalyst. For example, by using different catalysts, we can get different products for the reaction between H₂ and CO.

Question 21.
Describe some features of catalysis by zeolites.
Answer:
1. Zeolites are widely used as catalysts in petrochemical industries for cracking of hydrocarbons and isomerisation. They are also used for removing permanent hardness of water,
e.g., ZSM-5 is a catalyst used in petroleum industry

2. Zeolites are shape selective catalysts having honey comb like structure.
3. They are microporous aluminosilicates with Al—O—Si framework and general formula M x / n [(AlO₂)_x (SiO₂)_y] ∙ mH₂O
4. The reactions taking place in zeolites depend upon the size and shape of the reactant and product molecules as well as upon the pores and cavities of the zeolites.

Question 22.
What is shape selective catalysis?
Answer:
A catalytic reaction which depends upon the pore structure of the catalyst and on the size of the reactant and the product molecules is called shape-selective catalysis. For example, catalysis by zeolites is a shape-selective catalysis. The pore size present in the zeolites ranges from 260-740 pm. Thus, molecules having a pore size more than this cannot enter the zeolite and undergo the reaction.

Question 23.
Explain the following terms:
(i) Electrophoresis
(ii) Coagulation
(iii) Dialysis
(iv) Tyndall effect.
Answer:
(i) Electrophoresis : The movement of colloidal particles under the influence of an applied electric field is known as electrophoresis. Positively charged particles move to the cathode, while negatively charged particles move towards the anode. As the particles reach oppositely charged electrodes, they become neutral and get coagulated.

(ii) Coagulation : The process of settling down of colloidal particles i.e., conversion of a colloid into a precipitate is called coagulation.

(iii) Dialysis : The process of removing a dissolved substance from a colloidal solution by the means of diffusion through a membrane is known as dialysis. This process is based on the principle that ions and small molecules can pass through animal membranes unlike colloidal particles.

(iv) Tyndall effect : When a beam of light is allowed to pass through a colloidal solution, it becomes visible like a column of light. This is known as the Tyndall effect. This phenomenon takes place as particles of colloidal dimensions scatter light in all directions.

Question 24.
Give four uses of emulsions.
Answer:
Uses of emulsions

1. Cleansing action of soaps is based on the formation of emulsions.
2. Digestion of fats in intestines takes place by the process of emulsification.
3. Antiseptics and disinfectants when added to water form emulsions.
4. The process of emulsification is used to make medicines.

Question 25.
What are micelles? Give an example of a micellers system.
Answer:
The aggregate of colloidal particles which have both hydrophobic and hydrophilic parts are called micelles. These are formed above a particular temperature called Krafts temperature (T_k)and above certain concentrations, called Critical Miceller Concentration (CMC).

These molecules are arranged radially with the hydrocarbon or non-polar part towards the centre and the polar part towards the periphery, e.g., soap solution in water is an example of micelles system.

Question 26.
Explain the terms with suitable examples:
(i) Alcosol
(ii) Aerosol
(iii) Hydrosol
Answer:
(i) Alcosol : A colloidal solution having alcohol as the dispersion medium and a solid substance as the dispersed phase is called an alcosol.
For example: colloidal sol of cellulose nitrate in ethyl alcohol is an alcosol.

(ii) Aerosol : A colloidal solution having a gas as the dispersion medium and a solid as the dispersed phase is called an aerosol. For example: fog, mist, cloud, etc.

(iii) Hydrosol: A colloidal solution having water as the dispersion medium and a solid as the dispersed phase is called a hydrosol. For example: starch sol or gold sol etc.

Question 27.
Comment on the statement that “colloid is not a substance but a state of substance”.
Answer:
Common salt (a typical crystalloid in an aqueous medium) behaves as a colloid in a benzene medium. Hence, we can say that a colloidal substance does not represent a separate class of substances. When the size of the solute particle lies between 1 nm and 1000 nm, it behaves as a colloid.

Hence, we can say that colloid is not a substance but a state of the substance which is dependent on the size of the particle. A colloidal state is intermediate between a true solution and a suspension.

Chemistry Guide for Class 12 PSEB Surface Chemistry Textbook Questions and Answers

Question 1.
Write any two characteristics of chemisorption.
Answer:

1. Chemisorption is highly specific in nature. It occurs only if there is a possibility of chemical bonding between the adsorbent and the adsorbate.
2. Like physisorption, chemisorption also increases with an increase in the surface area of the adsorbent.

Question 2.
Why does physisorption decrease with the increase of temperature?
Answer:
Physisorption is exothermic in nature. Therefore, in accordance with Le-Chateliere’s principle, it decreases with an increase in temperature. This means that physisorption occurs more readily at a lower temperature.

Question 3.
Why are powdered substances more effective adsorbents than their crystalline forms?
Answer:
Powdered substances are more effective adsorbents than their crystalline forms because when a substance is powdered, its surface area increases and physisorption is directly proportional to the surface area of the adsorbent.

Question 4.
In Haber’s process, hydrogen is obtained by reacting methane with steam in presence of NiO as catalyst. The process is known as steam reforming.
Why is it necessary to remove CO when ammonia is obtained by Haber’s process?
Answer:
Carbon monoxide acts as a poison for catalyst iron and promoter molybdenum in Haber’s process, i.e., the efficiency of catalyst and promoter is decreased. It also combines with Fe to form iron carbonyl, Fe(CO)₅ which interfere in the production of ammonia. Hence, CO must be removed from the reaction mixture.

Question 5.
Why is the ester hydrolysis slow in the beginning and becomes faster after sometime?
Answer:
The chemical equation for ester hydrolysis can be represented as:
Ester + Water → Acid + Alcohol
The acid produced in the reaction acts as a catalyst and makes the reaction faster. Substances that act as catalysts in the same reaction in which they are obtained as products are known as autocatalysts.

Question 6.
What is the role of desorption in the process of catalysis?
Answer:
The role of desorption in the process of catalysis is to make the surface of the solid catalyst free for the fresh adsorption of the reactants on the surface.

Question 7.
What modification can you suggest in the Hardy-Schulze law?
Answer:
Hardy-Schulze law states that ‘the greater the valence of the flocculating ion added, the greater is its power to cause precipitation.’ This law takes into consideration only the charge carried by an ion, not its size. The smaller the size of an ion, the more will be its polarising power. Thus, Hardy-Schulze law can be modified in terms of the polarising power of the flocculating ion. Thus, the modified Hardy-Schulze law can be stated as ‘the greater the polarising power of the flocculating ion added, the greater is its power to cause precipitation.’

Question 8.
Why is it essential to wash the precipitate with water before estimating it quantitatively?
Answer:
When a substance gets precipitated, some ions that combine to form the precipitate get adsorbed on the surface of the precipitate. Therefore, it becomes important to wash the precipitate before estimating it quantitatively in order to remove these adsorbed ions or other such impurities.

Punjab State Board PSEB 12th Class Chemistry Book Solutions Chapter 13 Amines Textbook Exercise Questions and Answers.

PSEB Solutions for Class 12 Chemistry Chapter 13 Amines

PSEB 12th Class Chemistry Guide Amines InText Questions and Answers

Question 1.
Write IUPAC names of the following compounds and classify them into primary, secondary, and tertiary amines.
(i) (CH₃)₂CHNH₂
(ii) CH₃(CH₂)₂NH₂
(iii) CH₃NHCH(CH₃)₂
(iv) (CH₃)₃CNH₂
(v) C₆H₅NHCH₃
(vi) (CH₃CH₂)₂NCH₃
(vii) m-BrC₆H₄NH₂
Answer:
(i) Propan-2-amine (1° amine)
(ii) Propan-1-amine (1° amine)
(iii) N-Methylpropan-2-amine (2° amine)
(iv) 2-Methylpropan-2-amine (1° amine)
(v) AT-Methylbenzenamine or N-methylaniline (2° amine)
(vi) N-Ethyl-N-methyl ethan amine (3° amine)
(vii) 3-Bromobenzenamine or 3-bromoaniline (1° amine)

Question 2.
Give one chemical test to distinguish between the following pairs of compounds.
(i) Methylamine and dimethylamine
(ii) Secondary and tertiary amines
(iii) Ethylsunine and aniline
(iv) Aniline and benzylrnnlne
(v) Aniline and N-inethylaniline.
Answer:
(i) Methylamlne is 1° amine, therefore, it gives carbylamine test, i.e., when heated with an alcoholic solution of KOH and CHCl₃, it gives an offensive smell of methyl carbylamine. In contrast, dimethylamine is a secondary amine and hence does not give this test.

(ii) Secondary amines (both aliphatic and aromatic) react with nitrous acid to form nitrosamines (yellow oily liquid)

When the yellow oily liquid is warmed with a crystal of phenol and a few drops of conc. H₂SO₄, a greenish solution is formed. It changes to red on dilution with water but changes to deep blue on addition of aqueous NaOH solution. Tertiary amines do not give this test.

(iii) Ethylamine (primary aliphatic amine) and aniline (primary aromatic amine) can be distinguished by azo dye test.
Aniline responds to this test whereas ethyl amine does not.

(iv) Benzylantine reacts with nitrous acid to form a diazonium salt which being unstable even at low temperature, decomposes with evolution of N₂ gas.

Aniline reacts with HNO₂ to form benzene diazonium chloride which is stable at 273-278 K and hence does not decompose to evolve N₂ gas.

(v) Aniline being a primary amine gives carbylamine test, i.e., when heated with an alcoholic solution of KOH and CHCl₃, it gives an offensive smell of phenyl isocyanide. In contrast, N-methyl aniline, being secondary amine does not give this test.

Question 3.
Account for the following:
(i) pK_b of aniline is more than that of methylamine.
(ii) Ethylamme is soluble in water whereas aniline is not.
(iii) Methylamine in water reacts with ferric chloride to precipitate hydrated ferric oxide.
(iv) Although the amino group is o-and p-directing in aromatic electrophilic substitution reactions, aniline on nitration gives a substantial amount of m -nitroaniline.
(v) Aniline does not undergo Friedel Crafts reaction.
(vi) Diazonium salts of aromatic amines are more stable than those of aliphatic amines.
(vii) Gabriel phthalimide synthesis is preferred for synthesizing primary amines.
Answer:
(i) In aniline due to resonance, the lone pair of electrons on the N-atom are delocalized over the benzene ring. Due to this, electron density on the nitrogen decreases. On the other hand, in CH₃NH₂, +I effect of CH₃ increases the electron density on the N-atom. Consequently, aniline is a weaker base than methylamine and hence its pK_b value is higher than that of methylamine.

(ii) Ethylamine dissolves in water because it forms H-bonds with water molecules as shown below :

In aniline, due to the large hydrocarbon part the extent of H-bonding decreases considerably and hence aniline is insoluble in water.
(iii) Methylamine in water reacts with ferric chloride to precipitate hydrated fem oxide:

Due to the +I effect of – CH₃ group, methylamine is more basic than water. Therefore, in water, methylamine produces OH^– ions by accepting H⁺ ions from water.

Ferric chloride (FeCl₃) dissociates in water to form Fe³⁺ and Cl^– ions.
FeCl₃ → Fe³⁺+ +3Cl^–
Then, OH^– ion reacts with Fe³⁺ ion to form a precipitate of hydrated ferric oxide.
2Fe³⁺ +6OH^– → Fe₂O₃ -3H₂O Hydrated ferric oxide

(iv) Nitration is usually carried out with a mixture of cone. HNO₃ and cone. H₂SO₄.
In presence of these acids, most of aniline gets protonated to form anilinium ion. Thus, in presence of acids, the reaction mixture consists of aniline and anilinium ion. The -NH₂ group in aniline is o, p-directing and activating while the -NH₃ group in anilinium ion is m-directing and deactivating.
Nitration of aniline mainly gives p-nitroaniline. On the other hand, the nitration of anilinium ion gives m-nitroaniline.

Thus, nitration of aniline gives a substantial amount of m-nitroaniline due to protonation of the amino group.

(v) Aniline being a Lewis base, reacts with Lewis acid AlCl₃ to form a salt.

As a result, N of aniline acquires positive charge and hence, it acts as a strong deactivating group for electrophilic substitution reaction. Consequently, aniline does not undergo Friedel Crafts’ reaction.
(vi) The diazonium ion undergoes resonance as shown below:

This resonance accounts for the stability of the diazonium ion. Hence, diazonium salts of aromatic amines are more stable than those of aliphatic amines.

(vii) Gabriel phthalimide synthesis results in the formation of 1° amine only. 2° or 3° amines are not formed in this synthesis. Thus, a pure 1° amine can be obtained. Therefore, Gabriel phthalimide synthesis is preferred for synthesizing primary amines.

Question 4.
Arrange the following:
(i) In decreasing order of the pK_b values :
C₂H₅NH₂, C₆H₅NHCH₃, (C₂H₅)₂NH and C₆H₅NH₂
(ii) In increasing order of basic strength :
C₆H₅NH₂, C₆H₅N(CH₃)₂, (C₂H₅)₂NH and CH₃NH₂
(iii) In increasing order of basic strength :
(a) Aniline, p-nitroaniline andp-toluidine
(b) C₆H₅NH₂, C₆H5NHCH₃, C₆H₅CH₂NH₂.
(iv) In decreasing order of basic strength in gas phase:
C₂H₅NH₂,(C₂H₅)₂NH,(C₂H₅)₃N and NH₃
(v) In increasing order of boiling point:
C₂H₅OH, (CH₃)₂NH, C₂H₅NH₂
(vi) In increasing order of solubility in water :
C₆H₅NH₂, (C₂H₅)₂NH, C₂H₅NH₂.
Answer:
(i) C₆H₅NH₂ > C₆H₅NHCH₃ > C₂H₅NH₂ > (C₂H₅)₂NH
(ii) C₆H₅NH₂C₆H₅N(CH₃)₂ < CH₃NH₂ < (C₂H₅)₂NH

(b) C₆H₅NH₂ < C₆H₅NHCH₃ < C₆H₅CH₂NH₂.
(iv) (C₂H₅)₃N > (C₂H₅)₂NH > C₂H₅NH₂ > NH₃.
(v) (CH₃)₂NH < C₂H₅NH₂ < C₂H₅OH
(vi) C₆H₅NH₂ < (C₂H₅)₂NH < C₂H₅NH₂.

Question 5.
How will you convert:
(i) Ethanoic acid into methanamine
(ii) Hexanenitrile into 1-amino pentane
(iii) Methanol to ethanoic acid
(iv) Ethanamine into methanamine
(v) Ethanoic acid into propanoic acid
(vi) Methanamine into ethanamine
(vii) Nitromethane into dimethylamine
(viii) Propanoic acid into ethanoic acid
Answer:

Question 6.
Describe a method for the identification of primary, secondary, and tertiary times. Also, write chemical equations of the reactions involved.
Answer:
Primary, secondary and tertiary amines can be identified and distinguished by Hmsberg’s test. In this test, the amines are allowed to react with Hinsberg’s reagent, benzene sulphonyl chloride (C₆H₅SO₂Cl). The three types of amines react differently with Hinsbergs reagent. Therefore, they can be easily identified using Hinsberg’s reagent. Primary amines react with benzene sulphonyl chloride to form N-allyl benzene sulphonyl amide which is soluble in alkali.

Due to the presence of a strong electron-withdrawing sulphonyl group in the sulphonamide, the H-atom attached to nitrogen can be easily released as proton. So, it is acidic and dissolves in alkali. Secondary amines react with Hinsberg’s reagent to give a sulphonamide which is insoluble in alkali.

There is no H-atom attached to the N-atom in the sulphonamide. Therefore, it is not acidic and insoluble in alkali.
On the other hand, tertiary amines do not react with Hinsberg’s reagent at all.

Question 7.
Write short notes on the following:
(i) Carbylamine reaction
(ii) Diazotisation
(iii) Hofmann’s bromamide reaction
(iv) Coupling reaction
(v) Ammonolysis
(vi) Acetylation
(vii) Gabriel phthalimide synthesis.
Answer:
(i) Carbylamine reaction: Carbylamine reaction is used as a test for the identification of primary amines. When aliphatic and aromatic primary amines are heated with chloroform and ethanolic potassium hydroxide, carbylamines (or isocyanides) are formed. These carbylamines have very unpleasant odours. Secondary and tertiary amines do not respond to this test.

(ii) Diazotisation: Aromatic primary amines react with nitrous acid (prepared in situ from NaNO₂ and a mineral acid such as HCl) at low temperatures (273-278 K) to form diazonium salts. This conversion of aromatic primary amines into diazonium salts is known as diazotization. For example, on treatment with NaNO₂ and HCl at 273 – 278 K, aniline produces benzene diazonium chloride, with NaCl and H₂O as by-products.

(iii) Hofmann’s bromamide reaction: When an amide is treated with bromine in an aqueous or ethanolic solution of sodium hydroxide, a primary amine with one carbon atom less than the original amide is produced. This degradation reaction is known as Hofmann bromamide reaction. This reaction involves the migration of an alkyl or aryl group from the carbonyl carbon atom of the amide to the nitrogen atom.

(iv) Coupling reaction: The reaction of joining two aromatic rings through the -N=N- bond is known as a coupling reaction. Arenediazonium salts such as benzene diazonium salts react with phenol or aromatic amines to form colored azo compounds.

It can be observed that the para-positions of phenol and aniline are coupled with the diazonium salt. This reaction proceeds through electrophilic substitution.

(v) Ammonolysis: When an alkyl or benzyl halide is allowed to react with an ethanolic solution of ammonia, it undergoes nucleophilic substitution reaction in which the halogen atom is replaced by an amino (- NH₂) group. This process of cleavage of the carbon-halogen bond is known as ammonolysis.

When this substituted ammonium salt is treated with a strong base such as sodium hydroxide, amine is obtained.

Though primary amine is produced as the major product, this process produces a mixture of primary, secondary and tertiary amines, and also a quaternary ammonium salt as shown.

(vi) Acetylation: The process in which acetyl (CH₃CO -) group is introduced in a molecule, is called acetylation, reagents used for this purpose are acetyl chloride or acetic anhydride.
(vii) Gabriel phthalimide synthesis: Gabriel phthalimide synthesis is a very useful method for the preparation of aliphatic primary amines.

It involves the treatment of phthalimide with ethanolic potassium hydroxide to form potassium salt of phthalimide. This salt is further heated with alkyl halide, followed by alkaline hydrolysis to yield the corresponding primary amine.

(vii) Gabriel phthalimide synthesis: Gabriel phthalimide synthesis is a very useful method for the preparation of aliphatic primary amines. It involves the treatment of phthalimide with ethanolic potassium hydroxide to form potassium salt of phthalimide. This salt is further heated with an alkyl halide, followed by alkaline hydrolysis to yield the corresponding primary amine.

Question 8.
Accomplish the following conversions:
(i) Nitrobenzene to benzolc acid
(ii) Benzene to m-bromophenol
(iii) Benzoic acid to aniline
(iv) Aniline to 2,4,6-tribromofluorobenzene
(v) Benzyl chloride to 2-phenylethanolamine
(vi) Chiorobenzene top-chloroaniline
(vii) Aniline top-bromoaniline
(viii) Benzainide to toluene
(ix) Aniline to benzyl alcohol.
Answer:

Question 9.
Give the structures of A, B and C In the following reactions:

Answer:

Question 10.
An aromatic compound ‘A’ on treatment with aqueous ammonia and heating forms compound ‘B’ which on heating with Br₂ and KOH forms a compound ‘C’ of molecular formula C₆H₇N. Write the structures and IUPAC names of compounds A, B, and C.
Answer:
It is given that compound ‘C’ having the molecular formula, C₆H₇ N is formed by heating compound ‘B’ with Br₂ and KOH. This is a Hofmann bromamide degradation reaction. Therefore, compound ‘B’ is an amide, and compound ‘C’ is an amine. The only amine having the molecular formula, C₆H₇N is aniline, (C₆H₅NH₂).

Therefore, compound ‘B’ (from which ‘C’ is formed) must be benzamide, (C₆H₅CONH₂)

Further, benzamide is formed by heating compound ‘A’ with aqueous ammonia. Therefore, compound ‘A’ must be benzoic acid.

The given reactions can be explained with the help of the following equations:

Question 11.
Complete the following reactions:
(i) C₆H₅ NH₂ + CHCl₃ + alc.KOH →
(ii) C₆H₅N₂Cl+H₃PO₂+H₂O →
(iii) C₆H₅NH₂ + H₂SO₄(conc.) →
(iv) C₆H₅N₂Cl+C₂H₅OH →
(v) C₆H₅NH₂+Br₂(aq) ) →
(vi) C₆H₅NH₂ + (CH₃CO)₂O →

Answer:

Question 12.
Why cannot aromatic primary amines be prepared by Gabriel phthalimide synthesis?
Answer:
Gabriel phthalimide synthesis is used for the preparation of aliphatic primary amines. It involves nucleophilic substitution (S_N2) of alkyl halides by the anion formed by the phthalimide.

But aryl halides do not undergo nucleophilic substitution with the anion formed by the phthalimide.

Hence, aromatic primary amines cannot be prepared by this process.

Question 13.
Write the reactions of (i) aromatic and (ii) aliphatic primary amines with nitrous acid.
Answer:
(i) Aromatic amines react with nitrous acid (prepared in situ from NaNO₂ and a mineral acid such as HCl) at 273 – 278 K to form stable aromatic diazonium salts i.e., NaCl and H₂O.

(ii) Aliphatic primary amines react with nitrous acid (prepared in situ from NaNO₂ and a mineral acid such as HCl) to form unstable aliphatic diazonium salts, which further produce alcohol and HCl with the evolution of N₂ gas.

Question 14.
Give plausible explanation for each of the following:
(i) Why are amines less acidic than alcohols of comparable molecular masses?
(ii) Why do primary amines have higher boiling point than tertiary amines?
(iii) Why are aliphatic amines stronger bases than aromatic amines?
Answer:
(i) Amines are less acidic than alcohols of comparable molecular masses because N – H bond is less polar than O – H bond. Hence, amines release H⁺ ion with more difficulty as compared to alcohol.

(ii) Intermolecular hydrogen bonding is present in primary amines but not in tertiary amines (H-atom absent in amino group) so primary amines have higher boiling point than tertiary amines.

(iii) Aliphatic amines are stronger bases than aromatic amines due to following reasons :
(a) Electron pair on the nitrogen atom of aromatic amines is involved in conjugation with the π-electron pairs of the ring as follows :

(b) Anilinium ion obtained by accepting a proton is less stabilized by resonance.

So, aniline is a weaker base than alkyl amines, in which the +1 effect increases the electron density on the nitrogen atom.

Chemistry Guide for Class 12 PSEB Amines Textbook Questions and Answers

Question 1.
Classify the following amines as primary, secondary or tertiary:

(iii) (C₂H₅)₂CHNH₂
(iv) (C₂H₅)₂NH
Answer:
(i) Primary
(ii) Tertiary
(iii) Primary
(iv) Secondary.

Question 2.
(i) Write structures of different isomeric amines corresponding to the molecular formula, C₄H₁₁N.
(ii) Write IUPAC names of all the isomers.
(iii) What type of isomerism is exhibited by different pairs of amines?
Answer:
(i) and (ii)
Primary amines :

Secondary amines:

Tertiary amine:

(iii) Chain isomers : (a) and (d), (b) and (c)
Position isomers : (a) and (b), (f) and (g)
Metamers : (e) and (f)
Functional isomers: All primary amines are functional isomers of secondary and tertiary amines and vice-versa.

Question 3.
How will you convert:
(i) Benzene into Aniline ?
(ii) Benzene into N, N-Dimethylaniline
(iii) Cl-(CH₂)₄ -Cl into hexane-1, 6-diamine?
Answer:

(ii) Benzene is converted into aniline which can be subsequently heated with excess of methyl iodide under pressure to obtain N, N-dimethylaniline.

Question 4.
Arrange the following in increasing order of their basic strength:
(i) C₂H₅NH₂,C₆H₆NH₂,NH₃,C₆H₅CH₂NH₂,(C₂H₅)₂NH
(ii) C₂H₅NH₂, (C₂H₆)₂NH, (C₂H₅)₃N, C₆H₅NH₂
(iii) CH₃NH₂, (CH₃)₂NH, (CH₃)₃N, C₆H₅NH₂, C₆H₅CH₂NH₂
Answer:
(i) C₆H₅NH₂ < NH₃ < C₆H₅CH₂NH₂ < C₂H₅NH₂< (C₂H₅)₂NH
(ii) C₆H₅NH₂ < C₂H₅NH₂ < (C₂H₅)₃N < (C₂H₅)₂NH
(iii) C₆H₅NH₂ < C₆H₅CH₂NH₂ < (CH₃)₃N < CH₃NH₂ < (CH₃)₂NH

Question 5.
Complete the following acid-base reactions and name the products:
(i) CH₃CH₂CH₂NH₂ + HCl →
(ii) (C₂H₅)₃N + HCl →
Answer:

Question 6.
Write reactions of the final alkylation product of aniline with excess of methyl iodide in the presence of sodium carbonate solution.
Answer:

Question 7.
Write the chemical reaction of aniline with benzoyl chloride and write the name of the product obtained.
Answer:

Question 8.
Write structures of different isomers corresponding to the molecular formula, C₃H₉N. Write IUPAC names of the isomers which will liberate nitrogen gas on treatment with nitrous acid.
Answer:
In all, four structural isomers are possible. These are as follows:
Primary amines :

Secondary amInes :

Tertiary amines :

Only primary amines react with HNO₂ to liberate N₂ gas.

Question 9.
Convert:
(i) 3-Methylaniline into 3-nitrotoluene
(ii) Aniline into 1,3,5-tribromobenzene.
Answer:

Punjab State Board PSEB 12th Class Chemistry Book Solutions Chapter 2 Solutions Textbook Exercise Questions and Answers.

PSEB Solutions for Class 12 Chemistry Chapter 2 Solutions

PSEB 12th Class Chemistry Guide Solutions InText Questions and Answers

Question 1.
Define the term solution. How many types of solutions are formed? Write briefly about each type with an example.
Answer:
Homogeneous mixtures of two or more than two components are known as solutions. Solute and solvent are two components of a solution.
There are three types of solutions.
(i) Gaseous solution: The solution in which the solvent is a gas is known as a gaseous solution. In these solutions, the solute may be liquid, solid, or gas. For example, a mixture of oxygen and nitrogen gas is a gaseous solution.

(ii) Liquid solution : The solution in which the solvent is a liquid is known as a liquid solution. In these solutions, the solute may be gas, liquid, or solid. For example, a solution of ethanol in water is a liquid solution.

(iii) Solid solution : The solution in which the solvent is a solid is known as a solid solution. In these solution, the solute may be gas, liquid or solid. For example, a solution of copper in gold is a solid solution.

Question 2.
Give an example of a solid solution in which the solute is a gas.
Answer:
In case a solid solution is formed between two substances (one having very large particles and the other having very small particles), an interstitial solid Solution will be formed. For example, a solution of hydrogen in palladium is a solid solution in which the solute is a gas.

Question 3.
Define the following terms:
(i) Mole fraction
(ii) Molality
(iii) Molarity
(iv) Mass percentage.
Answer:
(i) Mole fraction: The mole fraction of a component in a mixture is defined as the ratio of the number of moles of the component (solute or solvent) to the total number of moles of all the components in the mixture.
i.e., Mole fraction of a component

Mole fraction is denoted by ‘χ’
If in a binary solution, the number of moles of the solute and the solvent are nA and nB respectively, then the mole fraction of the solute in the solution is given by,
χ_A = \(\frac{n_{A}}{n_{A}+n_{B}}\)
Similarly, the mole fraction of the solvent in the solution is given as:
χ_A = \(\frac{n_{B}}{n_{A}+n_{B}}\)

(ii) Molality: Molality (m) is defined as the number of moles of the solute per kilogram of the solvent. It is expressed as:

Unit of molality is mol kg^-1 or m (molal).

(iii) Molarity: Molarity (M) is defined as the number of moles of the solute dissolved in one litre of the solution.

(iv) Mass percentage: The mass percentage of a component of a solution is defined as the mass of the solute in grams present in 100 g of the solution.

Question 4.
Concentrated nitric acid used in laboratory work is 68% nitric acid by mass in aqueous solution. What should be the molarity of such a sample of the acid if the density of the solution is 1.504gmL ^-1?
Solution:
68% HNO₃ by mass means
Mass of HNO₃ (w) = 68 g
Mass of solution = 100 g
Molar mass of nitric acid (HN0),
(M) = 1 × 1 + 1 × 14 + 3 × l6 = 63g mol^-1
Then, number of moles of HNO₃ = \(\frac{W}{M}\) = \(\frac{68}{63}\) mol
= 1.079 mol
Density of solution = 1.504 g mL^-1
∴ Volume of solution = \(\frac{\text { Mass }}{\text { Density }}\) = \(\frac{100}{1.504}\) mL
66.49 mL
66.49 × 10^-3 L

= \(\frac{1.079 \mathrm{~mol}}{66.49 \times 10^{-3} \mathrm{~L}}\) = 16.23 mol L^-1

Question 5.
A solution of glucose in water is labelled as 10% w/w, what would be the molality and mole fraction of each component in the solution? If the density of solution is 1.2 gmL^-1, then what shall be the molarity of the solution?
Solution:
10% w/w solution of glucose means 10 g of glucose is present in 100 g of the solution i.e., 90 g of water.
Molar mass of glucose
(C₆H₁₂O₆) = 6 × 12 + 12 × 1 + 6 × 16 = 180 g mol^-1
Then, number of moles of glucose = \(\frac{10}{180}\)mol = 0.056 mol

Question 6.
How many mL of 0.1 M HCl are required to react completely with lg mixture of Na₂CO₃ and NaHCO₃ containing equimolar amounts of both?
Solution:
Let the mass of Na₂CO₃ = x g
Mass of mixture = 1.0 g
Then, the mass of NaHCO₃ = (1 – x) g
Molar mass of Na₂CO₃ = 2 × 23 + 1 × 12 + 3 × 16 =106 g mol^-1
∴ Number of moles of Na₂CO₃ = \(\frac{\text { Mass }}{\text { Molar mass }}\) = \(\frac{x}{106}\) = mol
Molar mass of NaHCO₃ = 1 × 23 + 1 × 1 + 1 × 12 + 3 × 16 = 84g mol^-1
∴ Number of moles of NaHCO₃ = \(\frac{\text { Mass }}{\text { Molar mass }}\) = \(\frac{1-x}{84}\) mol
According to the question, the mixture contains equimolar amounts of Na₂CO3and NaHCO₃
Moles of Na₂CO₃ = Moles of NaHCO₃
\(\frac{x}{106}=\frac{1-x}{84}\)
⇒ 84x =106 -106x
⇒ 190x =106
⇒ x = 0.5579 g
Thus, number of moles of Na₂CO₃ = \(\frac{0.5579}{106}\) mol = 0.0053 mol
Number of moles of NaHCO₃ = \(\frac{1-0.5579}{84}\)mol = 0.0053 mol

To calculate the moles of HCl required
HCl reacts with Na₂CO₃ and NaHCO₃ according to the following equations
2HCl + Na₂CO₃ → 2NaCl + H₂O + CO₂
HCl + NaHCO₃ → NaCl + H₂O + CO₂;
1 mol of Na₂CO₃ reacts with 2 mol of HCl.
Therefore, 0.0053 mol of Na₂CO₃ reacts with 2 × 0.0053 mol = 0.0106 mol
Similarly, 1 mol of NaHCO₃ reacts with 1 mol of HCl.
Therefore, 0.0053 mol of NaHCO₃ reacts with 0.0053 mol of HCl.
Total moles of HCl required = (0.0106 + 0.0053) mol
= 0.0159 mol

To calculate the volume of 0.1 MHC1
0.1 mol of 0.1 M HCl present in 1000 mL of HCl
Therefore, 0.0159 mol of HCl will be present in HCl
= \(\frac{1000 \times 0.0159}{0.1}\) mol
= 159 mL

Question 7.
A solution is obtained by mixing 300 g of 25% solution and 400 g of 40% solution by mass. Calculate the mass percentage of the resulting solution.
Solution:
Mass of solute in I solution = \(\frac{25}{100}\) × 300 g = 75 g
Mass of solute in II solution = \(\frac{40}{100}\) × 400 g = 160 g
After mixing both solutions
Total mass of solute = 75 + 160 = 235 g
Total mass of solution = 300 + 400 = 700 g
Therefore, mass percentage (w/w) of the solute in the resulting solution
= \(\frac{235}{700}\) × 100%
= 33.57%
And, mass percentage (w/w) of the solvent in the resulting solution.
= (100 – 33.57)%
= 66.43%

Question 8.
An antifreeze solution is prepared from 222.6 g of ethylene glycol (C₂H₆O₂ ) and 200 g of water. Calculate the molality of the solution. If the density of the solution is 1.072 g mL ‘1, then what shall be the molarity of the solution?
Solution:
To calculate the molality of the solution
Molar mass of ethylene glycol (C₂H₆O₂)
= 2 x 12 + 6 x 1 + 2 x 16 = 62 g mol^-1
Mass of ethylene glycol = 222.6 g
Number of moles of ethylene glycol = \(\frac{\text { Mass of ethylene glycol }}{\text { Molar mass of ethylene glycol }}\)
= \(\frac{222.6 \mathrm{~g}}{62 \mathrm{~g} \mathrm{~mol}^{-1}}\) = 3.59 mol
Mass of water = 200 g = 0.2 kg
Therefore, molality of the solution \(=\frac{\text { No. of moles of ethylene glycol }}{\text { Mass of solvent in kg }}\)
= \(\frac{3.59 \mathrm{~mol}}{0.200 \mathrm{~kg}}\) = 17.95m

To calculate the molarity of the solution
Density of the solution = 1.072 g mL^-1
Mass of solution = Mass of solute + Mass of solvent
222.6 g + 200 g = 422.6 g
∴ Volume of the solution = \(\frac{\text { Mass }}{\text { Density }}\) = \(\frac{422.6 \mathrm{~g}}{1.072 \mathrm{~g} \mathrm{~mL}^{-1}}\)
= 394.22 mL = 0.3942 x 10^-3 L
⇒ Molarity of the solution = \(\frac{\text { Moles of ethylene glycol }}{\text { Volume of solution in litre }}\)
= \(\frac{3.59 \mathrm{~mol}}{0.3942 \times 10^{-3} \mathrm{~L}}\) = 9.1 M

Question 9.
A sample of drinking water was found to be severely contaminated with chloroform (CHCl₃) supposed to be a carcinogen. The level of contamination was 15 ppm (by mass):
(i) express this in percent by mass
(ii) determine the molality of chloroform in the water sample.
Solution:
Let the mass of solution be 10⁶ g.
Mass of solute, CHCl₃ = 15 g

(i) Therefore, percent by mass of CHCl₃ = \(\frac{\text { Mass of } \mathrm{CHCl}_{3}}{\text { Mass of solution }}\) × 100%
\(\frac{15}{10^{6}}\) × 100%
= 1.5 × 10^-3%

(ii) Molar mass of chloroform (CHCl₃ ) = 1 × 12 + 1 × 1 + 3 × 35.5
= 119.5g mol^-1

= \(\frac{\frac{15}{119.5} \mathrm{~mol}}{10^{6} \times 10^{-3} \mathrm{~kg}}\) = 1.26 × 10 4 mol^-4 kg^-1
1.26 × 10^-4 m

Question 10.
What role does the molecular interaction play in a solution of alcohol and water?
Answer:
In pure alcohol and water, the molecules are held tightly by a strong hydrogen bonding. The interaction between the molecules of alcohol and water is weaker than alcohol-alcohol and water-water interactions. As a result, when alcohol and water are mixed, the intermolecular interactions become weaker and the molecules can easily escape. This increases the vapour pressure of the solution, which in turn lowers the boiling point of the resulting solution.

Question 11.
Why do gases always tend to be less soluble in liquids as the temperature is raised?
Answer:
Solubility of gases in liquids decreases with an increase in temperature. This is because dissolution of gases in liquids is an exothermic process.
Gas + Liquid → Solution + Heat
Therefore, when the temperature is increased, heat is supplied and the equilibrium shifts backwards, thereby decreasing the solubility of gases.

Question 12.
State Henry’s law and mention some important applications?
Answer:
Henry’s law states that partial pressure of a gas in the vapour phase is proportional to the mole fraction of the gas in the solution. If p is the partial pressure of the gas in the vapour phase and % is the mole fraction of the gas, then Henry’s law can be expressed as:
P = K_Hχ
Where, K_H is Henry’s law constant

Some important applications of Henry’s law are mentioned below:
(i) Bottles are sealed under high pressure to increase the solubility of C02 in soft drinks and soda water.

(ii) Henry’s law states that the solubility of gases increases with an increase in pressure. Therefore, when a scuba diver dives deep into the sea, the increased sea pressure causes the nitrogen present in air to dissolve in his blood in great amounts. As a result, when he comes back to the surface, the solubility of nitrogen again decreases and the dissolved gas is released, leading to the formation of nitrogen bubbles in the blood. This results in the blockage of capillaries and leads to a medical condition known as “bends’ or ‘decompression sickness’. Hence, the oxygen tanks used by scuba divers are filled with air and diluted with helium to avoid bends.

(iii) The concentration of oxygen is low in the blood and tissues of people living at high altitudes such as climbers. This is because at high altitudes, partial pressure of oxygen is less than that at ground level. Low-blood oxygen causes climbers to become weak and disables them from thinking clearly. These are symptoms of anoxia.

Question 13.
The partial pressure of ethane over a solution containing 6.56 × 10^-3g of ethane is 1 bar. If the solution contains 5.00 × 10^-2 g of ethane, then what shall be the partial pressure of the gas?
Solution:
According to Henry’s law the mass of the gas dissolved in solution x Partial pressure (p) (At constant temperature)
(6.56 × 10^-3g) ∝ 1 bar
(5.00 × 10^-2g) ∝ p
or p = \(\frac{5.0 \times 10^{-2} \mathrm{~g}}{6.56 \times 10^{-3} \mathrm{~g}}\) × 1 bar = 7.62 bar

Question 14.
What is meant by positive and negative deviations from Raoult’s law and how is the sign of Δ_mix H related to positive and negative deviations from Raoult’s law?
Answer:
Positive deviation: ‘When the vapour pressure of a solution is higher than the predicted value by Raoult’s law, it is called positive deviation’. In such cases intermolecular interactions between solute and solvent particles (A and B) are weaker than those between solute-solute (A – A) and solvent-solvent (B – B). Hence, the molecules of (A or B) will escape more easily from the surface of solution than in their pure state. Therefore, the vapour pressure of the solution will be higher. Characteristics of a solution showing positive deviation :
(i) P_A > \(p_{A}^{0}\) χ_A; P_B > \(p_{B}^{0}\) χ_B
(ii) ΔH_mix >0;i.e., + ve
(iii) ΔV_mix > 0, i.e., + ve

Examples of solutions showing positive deviation:
(i) Ethyl alcohol and water
(ii) Acetone and carbon disulphide
(iii) Carbon tetrachloride and benzene
(iv) Acetone and benzene

Negative deviation: “When the vapour pressure of a solution is lower than the predicted value by Raoult’s law, it is called negative deviation.’ In case of negative deviation the intermolecular attractive forces between A – A and B – B are weaker than those between A – B. It leads to decrease in vapour pressure resulting in negative deviation.

Characteristics of a solution showing negative deviation:
(i) P_A < \(p_{A}^{0}\) χ_B; P_B < \(p_{B}^{0}\) χ_B
(ii) ΔV_mix < 0; i. e., – ve; because weak A – A and B – B bonds are broken
and strong A – B bonds are formed. Heat is consequently released.
(iii) _mix<0;i.e.,-ve

Examples of solutions showing negative deviation:
(i) HNO₃ and water
(ii) Chloroform and acetone
(iii) Acetic acid and pyridine
(iv) Hydrochloric acid and water

Question 15.
An aqueous solution of 2% non-volatile solute exerts a pressure of 1.004 bar at the normal boiling point of the solvent. What is the molar mass of the solute?
Solution:
Here,
Vapour pressure of the solution at normal boiling point (p₁) = 1.004 bar
Vapour pressure of pure water at normal boiling point (\(p_{1}^{0}\)) = 1.013 bar
Mass of solute, (w₂) = 2 g
Mass of solvent (water), (w₁ ) = 100 – 2 = 98g
Molar mass of solvent (water), (M₁) = 18 g mol^-1
Molar mass of solute (M₂) = ?
According to Raoult’s law,

Hence, the molar mass of the solute is 41.35 g mol^-1.

Question 16.
Heptane and octane form an ideal solution. At 373 K, the vapour pressures of the two liquid components are 105.2 kPa and 46.8 kPa respectively. What will be the vapour pressure of a mixture of 26.0 g of heptane and 35 g of octane?
Solution:
Vapour pressure of heptane (\(p_{1}^{0}\)) = 105.2 kPa
Vapour pressure of octane (p\(p_{2}^{0}\)) = 46.8 kPa
Mass of heptane = 26.0 g
Mass of octane = 35 g
Molar mass of heptane (C₇H₁₆) = 7 × 12 + 16 × 1 = 100 g mol^-1;
∴ Number of moles of heptane = \(\frac{26}{100}\) mol = 0.26 mol
Molar mass of octane (C₈H₁₈) = 8 × 12 + 18 × 1 = 114g mol^-1
∴ Number of moles of octane = \(\frac{35}{114}\) mol = 0.31 mol
Mole fraction of heptane, χ₁ = \(\frac{0.26}{0.26+0.31}\) = 0.456
Mole fraction of octane, χ₂ = 1 – 0.456 = 0.544
Now, partial pressure of heptane, p₁ = χ₁ \(p_{1}^{0}\)
= 0.456 × 105.2 = 47.97 kPa
Partial pressure of octane, p₂ = χ₂\(p_{2}^{0}\)
= 0.544 × 46.8 = 25.46 kPa
Hence, vapour pressure of solution, p_total = P₁ + p₂
= 47.97 + 25.46 = 73.43 kPa

Question 17.
The vapour pressure of water is 12.3 kPa at 300 K. Calculate vapour pressure of 1 molal solution of a non-volatile solute in it.
Solution:
1 molal solution means 1 mol of the solute is present in 1000 g of the solvent (water).
Molar mass of water = 18 g mol^-1
> Number of moles present in 1000 g of water = \(\frac{1000}{18}\) = 55.56 mol
Therefore, mole fraction of the solute
χ₂ = \(\frac{1}{1+55.56}\) = 0.0177
It is given that,
Vapour pressure of water, \(p_{1}^{0}\) = 12.3 kPa
∴ \(\frac{p_{1}^{0}-p_{1}}{p_{1}^{0}}\) = χ₂
Applying the relation,
⇒ \(\frac{12.3-p_{1}}{12.3}\) = 0.0177
⇒ 12.3 – p₁ = 0.0177 × 12.3
⇒ 12.3 -P₁ = 0.2177
⇒ P₁ = 12.3 – 0.2177
⇒ p₁ = 12.0823
= 12.08 kPa
Hence, the vapour pressure of the solution is 12.08 kPa.

Question 18.
Calculate the mass of a non-volatile solute (molar mass 40 g mol^-1) which should be dissolved in 114 g octane to reduce its vapour pressure to 80%.
Solution:

Question 19.
A solution containing 30 g of non-volatile solute exactly in 90 g of water has a vapour pressure of 2.8 kPa at 298 K. Further, 18 g of water is then added to the solution and the new vapour pressure becomes 2.9 kPa at 298 K. Calculate :
(i) molar mass of the solute
(ii) vapour pressure of water at 298 K.
Solution:
(i) Let, the molar mass of the solute be M g mol^-1
Now, the no. of moles of solvent (water), = n₁ = \(\frac{\text { Mass }}{\text { Molar mass }}\)
= \(\frac{90 \mathrm{~g}}{18 \mathrm{~g} \mathrm{~mol}^{-1}}\) = 5mol
And, the no. of moles of solute, n₂ = \(\frac{\text { Mass }}{\text { Molar mass }}\)
= \(\frac{30 \mathrm{~g}}{\mathrm{Mg} \mathrm{mol}^{-1}}=\frac{30}{M}\) mol
Vapour pressure of I solution
p₁ = 2.8 kPa
According to Raoult’s law

Question 20.
A 5% solution (by mass) of cane sugar in water has freezing point of 271 K. Calculate the freezing point of 5% glucose in water if freezing point of pure water is 273.15 K.
Solution:
Let the mass of solution = 100 g
∴ Mass of cane sugar (w₂) = 5 g
ΔT_f = (273.15 – 271) K = 2.15 K
Molar mass of cane sugar (C₁₂H₂₂O₁₁), (M₂) = 12 x 12 + 22 x 1 +11 x 16
= 342 g mol^-1
Mass of solvent (water), (w₁) = 100 – 5 = 95 g

(Molar mass of glucose = 180 g mol^-1)
ΔT_f = 4.08 K
Freezing point of glucose solution
T_f = \(T_{f}^{0}\) – ΔT_f = 273.15 – 4.08 = 269.07 K

Question 21.
Two elements A and B form compounds having formula AB₂ and AB₄. When dissolved in 20 g of benzene (C₆H₆, 1 g of AB₂ lowers the freezing point by 2.3 K whereas 1.0 g of AB₄ lowers it by 1.3 K. The molar depression constant for benzene is 5.1 K kg mol^-1. Calculate atomic masses of A and B.
Solution:
We know,
M₂ = \(\frac{1000 \times w_{2} \times K_{f}}{\Delta T_{f} \times w_{1}}\)
Then M_AB2 = \(\frac{1000 \times 1 \times 5.1}{2.3 \times 20}\) = 110.87 g mol^-1
Then M_AB2 = \(\frac{1000 \times 1 \times 5.1}{1.3 \times 20}\) = 196.15 g mol^-1
Let the atomic masses of A and B be x and y respectively.
Molar mass of AB2 = x + 2 y = 110.87 …………… (i)
Molar mass of AB4 = x + 4y = 196.15 ………… (ii)
Subtracting equation (i) from (ii) , we get
2y=85.28 y = 42.64
Putting the value of ‘/ in equation (i), we get
x + 2 x 42.64 = 110.87
⇒ x = 25.59
Hence, the atomic masses of A and B are 25.59 u and 42.64 u respectively.

Question 22.
At 300 K, 36 g of glucose present in a litre of its solution has an osmotic pressure of 4.98 bar. If the osmotic pressure of the solution is 1.52 bars at the same temperature, what would be its concentration?
Solution:
Here, T = 300 K, π = 1.52 bar, R = 0.083 bar L K^-1mol^-1
Applying the relation, π = CRT
⇒ C = \(\frac{\pi}{R T}\)
= \(\frac{1.52 \mathrm{bar}}{0.083 \mathrm{bar} \mathrm{L} \mathrm{K}^{-1} \mathrm{~mol}}\)
= 0.061 mol
Since, the volume of the solution is 1 L, the concentration of the solution would be 0.061 M.

Question 23.
Suggest the most important type of intermolecular attractive interaction in the following pairs. .
(i) n-hexane and n-octane
(ii) I₂ and CCl₄
(iii) NaClO₄ and water
(iv) methanol and acetone
(v) acetonitrile (CH₃CN) and acetone (C₃H₆O).
Answer:
(i) Van der Wall’s forces of attraction. (London forces)
(ii) Van der Wall’s forces of attraction. (London forces) .
(iii) Ion-dipole interaction. ‘
(iv) Dipole-dipole interaction.
(v) Dipole-dipole interaction.

Question 24.
Based on solute-solvent interactions, arrange the following in order of increasing solubility in n-octane and explain. Cyclohexane, KCl, CH₃OH, CH₃CN.
Answer:
(i) Cyclohexane and n-octane both are non-polar.
So, they will mix completely in all proportions.
(ii) KCl is an ionic compound, but n-octane is non-polar.
So, KCl will not dissolve in n-octane.
(iii) CH₃OH and CH₃CN both are polar but CH₃CN is less polar than CH3OH. As the solvent is non-polar CH3CN will dissolve more than CH₃OH in n-octane.
Therefore, the order of solubility in n-octane will be KCl < CH₃OH < CH₃CN < Cyclohexane

Question 25.
Amongst the following compounds, identify which are insoluble, partially soluble and highly soluble in water?
(i) phenol
(ii) toluene
(iii) formic acid
(iv) ethylene glycol
(v) chloroform
(vi) pentanol.
Answer:
(i) Phenol (C₆H₅OH) has the polar group -OH and non-polar group -C₆H₅. Thus, phenol is partially soluble in water.

(ii) Toluene (C₆H₅ – CH₃) has no polar groups. Thus, toluene is insoluble in water.

(iii) Formic acid (HCOOH) has the polar group -OH and can form H-bond with water. Thus, formic acid is highly soluble in water.

(iv) Ethylene glycol has polar -OH group and can form H-bond. Thus, it is highly soluble in water.
(v) Chloroform is insoluble in water because it cannot form hydrogen bonds with water.

(vi) Pentanol (C₅H₁₁OH) has polar -OH group, but it also contains a very bulky non-polar -C₅H₁₁ group. Thus, pentanol is partially soluble in water.

Question 26.
If the density of some lake water is 1.25 g mL^-1 and contains 92 g of Na⁺ ions per kg of water, calculate the molality of Na+ ions in the lake.
Solution:
Number of moles present in 92 g of Na⁺ ions
= \(\frac{92 \mathrm{~g}}{23 \mathrm{~g} \mathrm{~mol}^{-1}}\) = 4 mol

= \(\frac{4 \mathrm{~mol}}{1 \mathrm{~kg}}\) = m

Question 27.
If the solubility product of CuS is 6 × 10¹⁶, calculate the maximum molarity of CuS in aqueous solution.
Solution:
Solubility product of CuS, K_sp = 6 × 10^-16 .
Let s be the solubility of CuS in mol L^-1. Then,

= s × s
= s²
Then we have, K _sp s² = 6 × 10^-16
⇒ s = \(\sqrt{6 \times 10^{-16}}\)
= 2.45 x 10^-8 mol^-1
Hence, the maximum molarity of CuS in an aqueous solution is 2.45 × ^-8 mol^-1.

Question 28.
Calculate the mass percentage of aspirin (C₉H₈O₄) in
acetonitrile (CH₃CN) when 6.5 g of C₉H₈O₄ is dissolved in 450 g of CH₃CN.
Solution:
Mass of aspirin = 6.5 g
Mass of acetonitrile = 450 g
Then, total mass of the solution = (6.5 + 450) g = 456.5 g
Therefore, mass percentage of C₉H₈O₄ = \(\frac{6.5}{456.5}\) × 100%
= 1.424%

Question 29.
Nalorphene (C₁₉H₂₁NO₃), similar to morphine, is used to combat withdrawal symptoms in narcotic users. Dose of nalorphene generally given is 1.5 mg. Calculate the mass of 1.5 × 10^-3 m aqueous solution required for the above dose.
Solution:
1.5 × 10^-3 m solution means that 1.5 × 10^-3 mole of nalorphene is dissolved in 1 kg of water.
Molar mass of C₁₉H₂₁N0₃
= 19 × 12 + 21 + 14 + 48
= 311 g mol^-1
∴ 1.5 × 10^-3 mole of nalorphene
= 1.5 × 10^-3 × 311 g = 0.467 g
= 467 mg
Mass of solution
= 1000 g + 0.467 g
= 1000.467 g
Thus, for 467 mg of nalorphene solution required 1000.467.
For 1.5 mg nalorphene = \(\frac{1000.467 \times 1.5}{467}\) = 3.21 g

Question 30.
Calculate the amount of benzoic acid (C₆H₅COOH) required for preparing 250 mL of 0.15 M solution in methanol.
Solution:
Molarity = 0.15 M or 0.15 mol L^-1
Volume of solution = 250 mL = 0.25 L
Molar mass of benzoic acid (C₆H₅COOH) = 7 × 12 + 6 × 1 + 2 × 16
= 122 g mol^-1
Molality \(\frac{\text { Mass }}{\text { Molar mass }}\) × \(\frac{1}{\text { Volume (L) }}\)
0.15 mol L^-1 = \(\frac{w}{122 \mathrm{~g} \mathrm{~mol}^{-1}}\) × \(\frac{1}{0.25 \mathrm{~L}}\)
Mass of solute = (0.15 × 122 × 0.25) g = 4.575 g

Question 31.
The depression in freezing point of water observed for the same amount of acetic acid, trichloroacetic acid and trifluoroacetic acid increases in the order given above. Explain briefly.
Answer:

Among H, Cl, and F, H is least electronegative while F is most electronegative. Then, F can withdraw electrons towards itself more than Cl and H. Thus, trifluoroacetic acid can easily lose H⁺ ions i.e., trifluoroacetic acid ionises to the largest extent. Now, the more ions produced, the greater is the depression of the freezing point. Hence, the depression in the freezing point increases in the order:
Acetic acid < trichloroacetic acid < trifluoroacetic acid

Question 32.
Calculate the depression in the freezing point of water when 10 g of CH₃CH₂CHClCOOH is added to 250 g of water, K_a = 1.4 × 10^-3, K_f = 1.86 K kg mol^-1.
Solution:
Mass of solute (CH₃CH₂CHClCOOH) = 10 g
Molar mass of
CH₃CH₂CHClC00H = 4 × 12 + 7 × 1 + 1 × 35.5 + 2 × 16 = 48 + 7 + 35.5 + 32
= 122.5 g mol^-1
\frac{\text { Mass / Molar mass }}{\text { Mass of solvent (Kg) }} = \(\frac{\text { Mass / Molar mass }}{\text { Mass of solvent (Kg) }}[latex/latex]
= [latex]\frac{10 \mathrm{~g}}{\left(122.5 \mathrm{~g} \mathrm{~mol}^{-1}\right) \times(0.25 \mathrm{Kg})}\)
= 0.326 m
Let α be the degree of dissociation of CH₃CH₂CHClCOOH then


Total no. of moles after dissociation = 1 – α + α + α
= 1 + α
Van’t Hoff factor
Total no. of moles after dissociation

∴ i = \(\frac{1+\alpha}{1}\)
= 1 + α
= 1 + 0.0655
= 1.0655
Hence, the depression in the freezing point of water is given as:
ΔT_f = i.K_fm
= 1.0655 × 1.86 kg mol^-1 × 0.326 mol kg^-1
= 0.65K

Question 33.
19.5 g of CH₂FCOOH is dissolved in 500 g of water. The depression in the freezing point of water observed is 1.0°C. Calculate the van’t Hoff factor and dissociation constant of fluoroacetic acid.
Solution:
Calculation of Van’t Hoff factor (i)
Given, w₁ = 500 g = 0.5 kg, w₂ = 19.5 g, K_f = 1.86 K kg mol^-1, ΔT_f = 1 K
Molar mass of CH₂FCOOH (M₂)
= 2 × 12 + 3 × 1 + 1 × 19 + 2 × 16
= 24 + 3 + 19 + 32
= 78 g mol^-1
ΔT_f = i K_f m

Question 34.
Vapour pressure of water at 293 K is 17.535 mm Hg. Calculate the vapour pressure of water at 293 K when 25 g of glucose is dissolved in 450 g of water.
Solution:
Vapour pressure of water, \(\) = 17.535 mm Hg
Mass of glucose, w₂ = 25 g
Mass of water, w₁ = 450 g
Molar mass of glucose (C₆H₁₂O₆), M₂ = 6 × 12 + 12 × 1 + 6 × 16
= 180 g mol^-1

we know that
\(\frac{p_{1}^{0}-p_{1}}{p_{1}^{0}}\) = \(\frac{n_{2}}{n_{2}+n_{1}}\)
⇒ \(\frac{17.535-p_{1}}{17.535}\) = \(\frac{0.139}{0.139+25}\)
⇒ 17.535 – p₁ = \(\frac{0.139 \times 17.535}{25.139}\)
⇒ 17.535 – p₁ = 0.097
⇒ P₁ = 17.44 mm Hg
Hence, the vapour pressure of water is 17.44 mm Hg.

Question 35.
Henry’s law constant for the molality of methane in benzene at 298 K is 4.27 x 10⁵ mm Hg. Calculate the solubility of methane in benzene at 298 K under 760 mm Hg.
Solution:
Here, p = 760 mm Hg, K_H = 4.27 × 10⁵ mm Hg (at 298 K)
According to Henry’s law, p = K_Hχ
χ = \(\frac{p}{k_{\mathrm{H}}}\)
= \(\frac{760 \mathrm{~mm} \mathrm{Hg}}{4.27 \times 10^{5} \mathrm{~mm} \mathrm{Hg}}\)
= 177.99 × 10^-5
= 178 × 10^-5
Hence, the mole fraction of methane in benzene is 178 × 10^-5.

Question 36.
100 g of liquid A (molar mass 140 g mol^-1) was dissolved in 1000 g of liquid B (molar mass 180 g mol^-1). The vapour pressure of pure liquid B was found to be 500 torr. Calculate the vapour pressure of pure liquid A and its vapour pressure in the solution if the total vapour pressure of the solution is 475 Torr.
Solution:
Number of moles of liquid A, n_A = \(\frac{w_{1}}{M_{1}}\) = \(\frac{100}{140}\) mol = 0.714 mol
Number of moles of liquid B,n_B = \(\frac{w_{2}}{M_{2}}\) = \(\frac{1000}{180}\) mol = 5.556 mol
Then, mole fraction of A, χ_A = \(\frac{n_{A}}{n_{A}+n_{B}}\)
= \(\frac{0.714 \mathrm{~mol}}{(0.714+5.556) \mathrm{mol}}\) = 0.114
Mole fraction of B, χ_B = 1 – 0.114 = 0.886
Vapour pressure of pure liquid B, \(p_{B}^{0}\) = 500 torr
Therefore, vapour pressure of liquid B in the solution,
P_B = \(p_{B}^{0}\)χ_B
= 500 × 0.886
= 443 torr
Total vapour pressure of the solution, p_total = 475 torr
∴ Vapour pressure of liquid A in the solution,
P_A = P_total – P_B
= 475 – 443 = 32 torr
Now, P_A = \(p_{A}^{0}\)χ_A
⇒ \(\frac{p_{A}}{\chi_{A}}\) = \(\frac{32}{0.114}\)
= 280.7 torr
Hence, the vapour pressure of pure liquid A is 280.7 torr.

Question 37.
Vapour pressures of pure acetone and chloroform at 328 K are 741.8 mm Hg and 632.8 nun Hg respectively. Assuming that they form ideal solution over the entire range of composition, plot
P_total’ P_chloroform and P_acetone as a function of x_acetone experimental data observed for different compositions of mixture is

Plot this data also on the same graph paper. Indicate whether it has positive deviation or negative deviation from the ideal
solution.

It can be observed from the graph that the plot for the ptotai of the solution curves downwards. Therefore, the solution shows negative deviation from the ideal behaviour.

Question 38.
Benzene and toluene form ideal solution over the entire range of composition. The vapour pressure of pure benzene and toluene at 300 K are 50.71 mm Hg and 32.06 mm Hg respectively.
Calculate the mole fraction of benzene in vapour phase if 80 g of benzene is mixed with 100 g of toluene.
Solution:
Molar mass of benzene (C₆H₆) = 6 × 12 + 6 × 1 = 78g mol^-1
Molar mass of toluene (C₆H₅CH₃ ) = 7 × 12 + 8 × 1 = 92 g mol^-1
No. of moles present in 80 g of benzene = \(\frac{80}{78}\) mol = 1.026 mol
No. of moles present in 100 g of toluene = \(\frac{100}{92}\) mol = 1.087 mol
Mole fraction of benzene, χ_C6H6, = \(\frac{1.026}{1.026+1.087}\) = 0.486
∴ Mole fraction of toluene,χ_C6H5CH35013 = 1 – 0.486 = 0.514
It is given that vapour pressure of pure benzene, \(p_{\mathrm{C}_{6} \mathrm{H}_{6}}^{0}\) = 50.71 mm Hg
Vapour pressure of pure toluene, \(p_{\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{CH}_{3}}^{0}\) = 32.06 mm Hg
Therefore, partial vapour pressure of benzene,
P_total = χ_C6H6 × \(p_{\mathrm{C}_{6} \mathrm{H}_{6}}^{0}\)
= 0.486 × 50.71
= 24.645 mm Hg
Partial vapour pressure of toluene, P_C6H5CH3 = χ_C6H5CH3 × \(P_{\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{CH}_{3}}^{0}\)
= 0.514 × 32.06
= 16.479 mm Hg
Total vapour pressure of solution (p) = 24.645 + 16.479
= 41.124 mm Hg
Mole fraction of benzene in vapour phase
= \(\frac{\chi_{\mathrm{C}_{6} \mathrm{H}_{6}} \times p_{\mathrm{C}_{6} \mathrm{H}_{6}}^{0}}{p_{\text {total }}}\)
= \(\frac{0.486 \times(50.71) \mathrm{mm}}{(41.124) \mathrm{mm}}\)
= 0.599 ≅ 0.6

Question 39.
The air is a mixture of a number of gases. The mqjor components are oxygen and nitrogen with approximate proportion of 20% is to 79% by volume at 298 K. The water is in equilibrium with air at a pressure of 10 atm. At 298 K if the Henry’s law constants for oxygen \md nitrogen are 330 × 10⁷ mm and 6.51 × 10⁷ mm respectively, calculate the composition of these gases in water.
Solution:
Percentage of oxygen (O₂) in air = 20%
Percentage of nitrogen (N₂) in air = 79%
Also, it is given that water is in equilibrium with air at a total pressure of 10 atm that is, (10 × 760) mm = 7600 mm
Therefore, partial pressure of oxygen,
PO₂ = \(\frac{20}{100}\) x 7600 mm
= 1520 mm Hg
Partial pressure of nitrogen, pN₂ = \(\frac{79}{100}\) x 7600 mm
= 6004 mm Hg
Now, according to Henry’s law,
p = K_H.χ
For oxygen:

\(\frac{6004 \mathrm{~mm}}{6.51 \times 10^{7} \mathrm{~mm}}\)
(Given K_H = 6.51 × 10⁷ mm)
= 9.22 × 10^-5
Hence, the mole fractions of oxygen and nitrogen in water are 4.61 × 10^-5 and 9.22 × 10^-5 respectively.

Question 40.
Determine the amount of CaCl₂ (i = 2.47) dissolved in 2.5 litre of water such that its osmotic pressure is 0.75 atm at 27°C.
Solutio:
We know that,
π = i \(\)RT
⇒ π = i \(\)RT
⇒ w = \(\)
Given,
π = 0.75 atm
V = 2.5L
i = 2.47
T = (27 + 273)K = 300K
R = 0.0821 L atm K^-1mol^–
Molar mass of CaCl₂(M) = 1 × 40 + 2 × 35.5 = 111 g mol ^-1
Therefore, w = \(\frac{0.75 \times 111 \times 2.5}{2.47 \times 0.0821 \times 300}\) = 3.42g
Hence the required amount of CaCl₂ is3.42g

Question 41.
Determine the osmotic pressure of a solution prepared by dissolving 25 mg of K₂SO₄ in 2 litre of water at 25° C, assuming that it is completely dissociated.
Solution:
When K₂SO₄ is dissolved in water, K⁺ and \(\mathrm{SO}_{4}^{2-}\) ions are produced.

Total number of ions produced = 3
∴ i = 3
Given, w = 25 mg = 0.025 g, V = 2 L
T = 25°C = (25 + 273) K = 298 K
Also, we know that R = 0.0821 L atm K^-1 mol^-1
Molar Mass of K₂SO₄ (M) = (2 × 39) + (1 × 32) + (4 × 16) = 174 g mol^-1
Applying the following relation,
π = i \(\frac{n}{V}\) RT = i \(\frac{w}{M} \frac{1}{V}\) RT
= 3 × \(\frac{0.025}{174}\) × 1 × 0.0821 × 298 = 5.27 × 10^-3 atm

Chemistry Guide for Class 12 PSEB Solutions Textbook Questions and Answers

Question 1.
Calculate the mass percentage of benzene (C₆H₆) and carbon tetrachloride (CCl₄) if 22 g of benzene is dissolved in 122 g of carbon tetrachloride.
Solution:

Alternatively,
Mass percentage of CCl₄ = (100 – 15.28)% = 84.72%

Question 2.
Calculate the mole fraction of benzene in solution containing 30% by mass in carbon tetrachloride.
Solution:
Let the total mass of the solution be 100 g and the mass of benzene be 30 g.
∴ Mass of carbon tetrachloride = (100 – 30) g = 70 g
Molar mass of benzene (C₆H₆) = (6 × 12 + 6 × 1) g
∴ mol^-1 = 78 g mol^-1

Question 3.
Calculate the molarity of each of the following solutions:
(a) 30 g of CO(NO₃) ₂.6H₂O in 4.3 L of solution
(b) 30 mL of 0.5 M H₂SO₄diluted to 500 mL.
Solution:
Molarity is given by

(a) Molar mass of CO (NO₃)₂.6H₂O = 59 + 2 (14 + 3 × 16) + 6 × 18
= 291 g mol^-1
∴ Moles of Co (NO₃)₂.6H₂O = \(\frac{30}{291}\) mol = 0.103 mol
Volume of solution = 4.3 L
Therefore, molarity = \(\frac{0.103 \mathrm{~mol}}{4.3 \mathrm{~L}}\) = 0.023 M

(b) Number of moles present in 1000 mL of 0.5 M H₂SO₄ = 0.5 mol
∴ Number of moles present in 30 mL of 0.5 M H₂SO₄
= \(\frac{0.5 \times 30}{1000}\) mol
= 0.015 mol
Volume of solution = 500 mL = 0.5 L
Therefore, molarity = \(\frac{0.015}{0.5 \mathrm{~L}}\) mol = 0.03M

Question 4.
Calculate the mass of urea (NH₂CONH₂) required in making 2.5 kg of 0.25 molal aqueous solution.
Solution:
Mass of required aqueous solution = 2.5 kg = 2500 g
Molar mass of urea (NH₂CONH₂) = 2 (1 × 14 + 2 × 1) + 1 × 12 +1 × 16
= 60 g mol^-1
0.25 molal aqueous solution of urea means 0.25 mole of urea is dissolved in 1000 g of water.
Mass of water = 1000 g
Moles of urea = 0.25 mol
Mass of urea = No. of moles of urea × Molar mass of urea
∴ Mass of 0.25 moles of urea = 0.25 × 60 = 15 g
Mass of solution = 1000 + 15 = 1015 g
1015 g of aqueous solution contains urea = 15 g
∴ 2500 g of aqueous solution will require urea
= \(\frac{15 g}{1015 g}\) × 2500 g = 36.95 g

Question 5.
Calculate (a) molality (b) molarity and (c) mole fraction of KI. If the density of 20% (mass/mass) aqueous KI is 1.202 g mL^-1.
Solution:
(a) Molar mass of KI = 39 +127 = 166 g mol^-1
20% (mass/mass) aqueous solution of KI means 20 g of KI is present in 100 g of solution.
∴ 20 g of KI is present in (100 – 20) g of water = 80 g of water

Question 6.
H₂S, a toxic gas with rotten egg like smell, is used for the qualitative analysis. If the solubility of H₂S in water at STP is 0.195 m, calculate Henry’s law constant.
Solution:
It is given that the solubility of H₂S in water at STP is 0.195 m, i.e., 0.195 mol of H₂S is dissolved in 1000 g of water.

Question 7.
Henry’s law constant for CO₂ in water is 1.67 × 10⁸ Pa at 298 K.
Calculate the quantity of CO₂ in 500 mL of soda water when packed under 2.5 atm COa pressure at 298 K.
Solution:
Given, K_H = 1.67 × 10⁸ Pa, p_CO2 = 2.5 atm = 2.5 × 10⁵ Pa
According to Henry’s law
p_CO2 = K_HχCO2 = \(\frac{P_{\mathrm{CO}_{2}}}{K_{\mathrm{H}}}\)
\(\frac{2.5 \times 10^{5}}{1.67 \times 10^{8}}\) = 1.5 × 10^-3 ……….. (i)
Mass of water = Density of water × Volume of water
= 1 g/mL × 500 mL = 500 g
Number of moles of water, (n_H2O)
= \(\frac{\text { Mass of water }}{\text { Molar mass }}=\frac{500 \mathrm{~g}}{18 \mathrm{~g} / \mathrm{mol}}\)
= 27.78 mol
χ_CO2 = \(\frac{n_{\mathrm{CO}_{2}}}{n_{\mathrm{H}_{2} \mathrm{O}}+n_{\mathrm{CO}_{2}}}=\frac{n_{\mathrm{CO}_{2}}}{n_{\mathrm{H}_{2} \mathrm{O}}}\)
⇒ n_CO2 = χ_CO2n_H2O
= 1.5 × 10^-3 × 27.78 mol
= 41.67 × 10^-3 mol
Mass of CO₂ = No. of moles of CO₂ × Molar mass
= 41.67 × 10^-3 × 44 =1.834 g

Question 8.
The vapour pressure of pure liquids A and B are 450 and 700 mm Hg respectively, at 350 K. Find out the composition of the liquid mixture if total vapour pressure is 600 mm Hg. Also find the composition of the vapour phase.
Solution:
Given, p\(p_{A}^{0}\) = 450 mm Hg, \(p_{B}^{0}\) = 700 mm Hg,
P_total = 600 mm Hg
According to Raoult’s law
P_A = \(p_{A}^{0}\)χ_A
P_B = \(p_{B}^{0}\)χ_B – \(p_{A}^{0}\) (1 – χ_A )

Therefore, total pressure, p_total = p_A + p_B
⇒ p_total = \(p_{A}^{0}\)χ_A + \(p_{B}^{0}\) (1 – χ_A)
⇒ p_total = \(p_{A}^{0}\)χ_A + \(p_{B}^{0}\) – \(p_{B}^{0}\)χ_A
⇒ p_total = \(p_{A}^{0}\) – \(p_{B}^{0}\) χ_A + \(p_{B}^{0}\)
⇒ 600 = (450 – 700)χ_A + 700
⇒ -100 = -250χ_A
⇒ χ_A = 0.4
Mole fraction of A (χ_A) = 0.4
Mole fraction of B (χ_B) = 1 – 0.4 = 0.6
Now, P_A = P_Aχ_A = 450 × 0.4
= 180 mm Hg
P_B = \(p_{B}^{0}\)χ_B
= 700 × 0.6
= 420 mm Hg
Now, in the vapour phase:
Mole fraction of liquid A = \(\frac{p_{A}}{p_{A}+p_{B}}\)
\(\frac{180}{180+420}=\frac{180}{600}\) = 0.30
And, mole fraction of liquid B = 1 – 0.30 = 0.70

Question 9.
Vapour pressure of pure water at 298 K is 23.8 mm Hg. 50 g of urea (NH₂CONH₂) dissolved in 850 g of water. Calculate the vapour pressure of water for this solution and its relative lowering.
Solution:
Given, vapour pressure of water, \(p_{1}^{0}\) = 23.8 mm Hg
Weight of water w₁, = 850 g
Weight of urea, w₂ = 50 g
Molecular weight of water, M₁ = 18 g mol^-1
Molecular weight of urea, M₂ = 60 g mol^-1

Now, we have to calculate vapour pressure of water in the solution. We take vapour pressure as p₁.
Now, from Raoult’s law, we have

Question 10.
Boiling point of water at 750 mm Hg is 99.63°C. How much sucrose is to be added to 500 g of water such that it boils at 100°C.
Solution:
Here, elevation in boiling point ΔT_b = (100 + 273) – (99.63 + 273)
= 0.37 K
Mass of water, w₁ = 500 g
Molar mass of sucrose (C₁₂H₂₂O₁₁), M₂ = 12 × 12 + 22 × 1 + 11 × 16
= 342 g mol^-1
Molal elevation constant, K_b = 0.52 K kg mol^-1
We know that,
ΔT_b = \(\frac{K_{b} \times 1000 \times w_{2}}{M_{2} \times w_{1}}\)
⇒ w₂ = \(\frac{\Delta T_{b} \times M_{2} \times w_{1}}{K_{b} \times 1000}\)
= \(\frac{0.37 \times 342 \times 500}{0.52 \times 1000}\)
= 121.67 g
Hence, 121.67 g of sucrose is to be added. ’

Question 11.
Calculate the mass of ascorbic acid (Vitamin C, C₆H₈O₆) to be dissolved in 75 g of acetic acid to lower its melting point by 1.5°C. K_f = 3.9 Kkg mol^-1.
Solution:
Mass of acetic acid, w₁ = 75 g
Molar mass of ascorbic acid (C₆H₈O₆), M = 6 × 12 + 8 × 1 + 6 × 16
= 176 g mol^-1
Depression in melting point (AT_f ) = 1.5 K
Molal depression constant (K_f ) = 3.9 K kg mol^-1
We know that,
ΔT_b = \(\frac{K_{f} \times w_{2} \times 1000}{M_{2} \times w_{1}}\)
⇒ w₂ = \(\frac{\Delta T_{f} \times M_{2} \times w_{1}}{K_{f} \times 1000}\)
= \(\frac{1.5 \times 176 \times 75}{3.9 \times 1000}\)
= 5.08 g
Hence, 5.08 g of ascorbic acid is needed to be dissolved.

Question 12.
Calculate the osmotic pressure in pascals exerted by a solution prepared by dissolving 1.0 g of polymer of molar mass 185,000 in 450 mL of water at 37°C. ,
Solution:
It is given that,
Volume of water, V = 450 mL = 0.45 L
Temperature, T = (37 + 273) K = 310 K
R = 8.314 K Pa L K^-1 mol^-1
= 8.314 × 10³ Pa LK^-1 mol^-1
Number of moles of the polymer, n = \(\frac{1}{185000}\) mol
We know that,
Osmotic pressure, n = \(\frac{n}{V}\)RT
= \(\frac{1}{185000}\) mol × \(\frac{1}{0.45 \mathrm{~L}}\) × 8.314 × 10³ Pa LK^-1 mol^-1 × 310 K
= 30.98 Pa

Punjab State Board PSEB 12th Class Chemistry Book Solutions Chapter 12 Aldehydes, Ketones, and Carboxylic Acids Textbook Exercise Questions and Answers.

PSEB Solutions for Class 12 Chemistry Chapter 12 Aldehydes, Ketones, and Carboxylic Acids

PSEB 12th Class Chemistry Guide Aldehydes, Ketones, and Carboxylic Acids InText Questions and Answers

Question 1.
What is meant by the following terms? Give an example of the reaction in each case.
(i) Cyanohydrin
(ii) Acetal
(iii) Semicarbazone
(iv) Aldol
(v) Hemiaeetal
(vi) Oxime
(vii) Ketal
(viii) Imine
(ix) 2, 4-DNP derivative
(x) Schiffs base.
Answer:
(i) Aldehydes and ketones react with hydrogen cyanide (HCN) to yield cyanohydrins.
It is catalyzed by a base and the generated cyanide ion (CN^–) being a stronger nucleophile readily adds to carbonyl compounds to yield corresponding cyanohydrin.

(ii) gem-Dialkoxy compounds in which the two alkoxy groups are present on the terminal carbon atom are called acetals. These are produced by the action of an aldehyde with two equivalents of a monohydric alcohol in the presence of dry HCl gas.

These are easily hydrolyzed by dilute mineral acids to regenerate the original aldehydes. Therefore, these are used for the protection of aldehydic group in organic synthesis.

(iii) Semicarbazones are derivatives of aldehydes and ketones and are produced by the action of semicarbazide on them in weak acidic medium.

These are used for identification and characterization of aldehydes and ketones.
(iv) Aldehydes and ketones having at least one α -hydrogen undergo a reaction in the presence of dilute alkali as catalyst to form f3-hydroxy aldehydes (aldol) or β-hydroxy ketones (ketol), respectively. This is known as aldol condensation.

(v) gem-Alkoxyalcohols are called hemiacetals. These are produced by addition of one molecule of a monohydric alcohol to an aldehyde in presence of dry HCl gas.

(vi) Oximes are produced when aldehydes or ketones react with hydroxylamine in weak acidic medium.

(vii) gem-Dialkoxyalkanes are called ketals. In ketals, the two alkoxy groups are present on the same carbon within the chain. These are produced when a ketone is heated with ethylene glycol in presence of diy HCl gas or p-toluene sulphonic acid (PTS).

These are easily hydrolyzed by dilute mineral acids to regenerate the original ketones. Therefore, ketals are used for protection of keto groups in organic synthesis.

(viii) Compounds containing group are called imines. These are produced when aldehydes and ketones react with ammonia derivatives.

Z = Alkyl, aryl – NH₂,-OH,-C₆H₅NH,-NHCONH₂ etc.

(ix) 2, 4-Dinitrophenythydrazones (i.e., 2, 4-DNP derivatives) are produced when aldehydes or ketones react with 2, 4-dinitrophenyl hydrazine in weakly acidic medium.

2, 4-DNP derivatives are used for identification and characterization of aldehydes and ketones.
(x) Aldehydes and ketones react with primary aliphatic or aromatic amines to form azomethines or Schiffs bases.

Question 2.
Name the following compounds according to IUPAC system of nomenclature:
(i) CH₃CH(CH₃)CH₂CH₂CHO
(ii) CH₃CH₂COCH(C₂H₅)CH₂CH₂Cl
(iii) CH₃CH=CHCHO
(iv) CH₃COCH₂COCH₃
(v) CH₃CH(CH₃)CH₂C(CH₃)₂COCH₃
(vi) (CH₃)₃CCH₂COOH
(vii) OHCC₆H₄CHO-p
Answer:
(i) 4-Methylpentanal
(ii) 6-Chloro-4-ethylhexan-3-one
(iii) But-2-en-l-al
(iv) Pentane-2,4-dione
(v) 3,3,5-Trimethylhexan-2-one
(vi) 3,3-Dimethylbutanoic acid
(vii) Benzene-1,4-dicarbaldehyde

Question 3.
Draw the structures of the following compounds.
(i) 3-Methylbutanal
(ii) p-Nitropropiophenone
(iii) p-Methylbenz aldehyde
(iv) 4-Methylpent-3-en-2-one
(v) 4-Chloropentan-2-one
(vi) 3-Bromo-4-phenylpentanoic acid
(vii) p,p’-Dihydroxybenzophenone
(viii) Hex-2-en-4-ynoic acid
Answer:

Question 4.
Write the IUPAC names of the following ketones and aldehydes. Wherever possible, give also common names.
(i) CH₃CO(CH₂)₄CH₃
(ii) CH₃CH₂CH Br CH₂CH(CH₃) CHO
(iii) CH₃(CH₂)₅CHO
(vi) PhCOPh

Answer:

IUPAC Name	Common Name
(i)Heptane-2-one	Methyl n-propyl ketone
(ii) 4-Bromo-2-methyl hexanal	γ -Bromo- α –methyl caproaldehyde
(iii) Heptanal	n-heptyl aldehyde
(iv) 3-Phenyprop-2-enal	Β- Phenylacrole in
(v) Cyclopentane-carbaldehyde	Cyclopentane carbaldehyde
(vi) Diphenyl methanone	Benzophenone

Question 5.
Draw structures of the following derivatives.
(i) The 2,4-dinitrophenylhydrazone of benzaldehyde
(ii) Cyclopropanone oxime
(iii) Acetaldehydedimethylacetal
(iv) The semicarbazone of cyclobutanone
(v) The ethylene ketal of hexan-3-one
(vi) The methyl hemiacetal of formaldehyde
Answer:

Question 6.
Predict the products formed when cyclohexane carbaldehyde reacts with following reagents.
(i) PhMgBr and then H₃O⁺
(ii) Tollens’ reagent
(iii) Semicarbazide and weak acid
(iv) Excess ethanol and acid
(v) Zinc amalgam and dilute hydrochloric acid
Answer:

Question 7.
Which of the following compounds would undergo aldol condensation, which the Cannizzaro reaction and which neither? Write the structures of the expected products of aldol condensation and Cannizzaro reaction.
(i) Methanal
(ii) 2-Methylpentanal
(iii) Benzaldehyde
(iv) Benzophenone
(v) Cyclohexanone
(vi) 1-Phenyipropanone
(vii) Phenylacetaldehyde
(viii) Butan.1-oI
(ix) 2, 2-Dimethylbutanal
Answer:
Aldol condensation

Cannizzaro reaction

(iv) Benzophenone: It is a ketone, so it does not undergo Cannizzaro’s reaction. Without a-hydrogen, it cannot participate in aldol condensation.
(viii) Butan-1-ol: It is an alcohol. So, it cannot participate in any of the above 2 reactions.

Question 8.
How will you convert ethanal into the following compounds?
(i) Butane-1, 3-diol
(ii) But-2-enal
(iii) But-2-enoic acid
Answer:
(i) On treatment with dilute alkali, ethanal produces 3-hydroxybutanal gives butane-1, 3-diol on reduction.

(ii) On treatment with dilute alkali, ethanal gives 3-hydroxybutanal which on heating produces but-2-enal.

(iii) When treated with Tollen’s reagent, but-2-ena produced in the above reaction produces but-2-enoic acid.

Question 9.
Write structural formulas and names of four possible aldol condensation products from propanal and butanal.
In each case, indicate which aldehyde acts as nucleophile and which as electrophile.
(i) Taking two molecules of propanal, one which acts as a nucleophile and the other as an electrophile.

(ii) Taking two molecules of butanal, one which acts as a nucleophile and the other as an electrophile.

(iii) Taking one molecule each of propanal and butanal in which propanal acts as a nucleophile and butanal acts as an electrophile.

(iv) Taking one molecule each of propanal and butanal in which propanal acts as an electrophile and butanal acts as a nucleophile.

Question 10.
An organic compound with the molecular formula C₉H₁₀O forms 2, 4-DNP derivative, reduces Tollen’s reagent, and undergoes Cannizzaro reaction. On vigorous oxidation, it gives 1,2-benzene dicarboxylic acid. Identify the compound.
Answer:
It is given that the compound (with molecular formula C₉H₁₀O) forms 2,4-DNP derivative and reduces Tollen’s reagent. Therefore, the given compound must be an aldehyde. Again, the compound undergoes Cannizzaro reaction and on oxidation gives 1, 2-benzene dicarboxylic acid. Therefore, the -CHO group is directly attached to a benzene ring and this benzaldehyde is ortho-substituted. Hence, the compound is 2-ethyl benzaldehyde.

The given reactions can be explained by the following equations:

Question 11.
An organic compound (A) (molecular formula C₈H₁₆O₂) was hydrolysed with dilute sulphuric acid to give a carboxylic acid (B) and an alcohol (C). Oxidation of (C) with chromic acid produced (B). (C) on dehydration gives but-l-ene. Write equations for the reactions involved.
Answer:
As the ester (A) with molecular formula C₈H₁₆O₂ upon hydrolysis gives carboxylic acid (B) and the alcohol (C) and oxidation of (C) with chromic acid produces the acid (B), therefore, both the carboxylic acid (B) and alcohol (C) must contain the same number of carbon atoms. Now ester (A) contains eight carbon atoms, therefore, both the carboxylic acid (B) and the alcohol (C) must contain four carbon atoms each. As the alcohol (C) on dehydration gives but-l-ene, therefore, (C) must be a straight chain alcohol, i.e., butan-l-ol.

If C is butane-lol, then the acid (B) which it gives on oxidation must be butanoic acid and the ester (A) must be butyl butanoate. The relevant equations for all the reactions involved may be explained as follow :

Question 12.
Arrange the following compounds in increasing order of their property as indicated:
(i) Acetaldehyde, Acetone, Di-tert-butyl ketone, Methyl tert-butyl ketone (reactivity towards HCN)
(ii) CH₃CH₂CH(Br)COOH, CH₃CH(Br)CH₂COOH,
(CH₃)₂CHCOOH, CH₃CH₂CH₂COOH (acid strength)
(iii) Benzoic acid, 4-Nitrobenzoic acid, 3, 4-Dinitrobenzoic acid, 4-Methoxybenzoic acid (acid strength)
Answer:
(i) When HCN reacts with a compound, the attacking species is a nucleophile, CN^–. Therefore, as the negative charge on the compound increases, its reactivity with HCN decreases. In the given compounds, the +I effect increases as shown below. It can be observed that steric hindrance also increases in the same.

Hence, the given compounds can be arranged according to their increasing reactivities toward HCN as:
Di-tert-butyl ketone < Methyl tert-butyl ketone < Acetone < Acetaldehyde

(ii) After losing a proton, carboxylic acids gain a negative charge as shown: R – COOH →R – COO^– + H⁺ Now, any group that will help stabilise the negative charge will increase the stability of the carboxyl ion and as a result, will increase the strength of the acid. Thus, groups having +I effect will decrease the strength of the acids, and groups having – I effect will increase the strength of the acids.

In the given compounds, -CH₃ group has +I effect and Br^– group has – I effect. Thus, acids containing Br^– are stronger. Now, the +I effect of isopropyl group is more than that of n-propyl group. Hence, (CH₃)₂CHCOOH is a weaker acid than CH₃CH₂CH₂COOH.

Also, the -I .effect grows weaker as distance increases. Hence, CH₃CH(Br)CH₂COOH is a weaker acid than CH₃CH₂CH(Br)COOH.
Hence, the strengths of the given acids increase as:
(CH₃)₂CHCOOH < CH₃CH₂CH₂COOH < CH₃CH(Br) CH₂COOH < CH₃CH₂CH(Br) COOH

(iii) As we have seen in the previous case, electron-donating groups decrease the strengths of acids, while electron-withdrawing groups increase the strengths of acids. As methoxy group is an electron-donating group, 4- methoxy benzoic acid is a weaker acid than benzoic acid.

Nitro group is an electron-withdrawing group and will increase the strengths of acids. As 3,4-di nitrobenzoic acid i.ontains two nitro groups, it is a slightly stronger acid than 4-nitrobenzoic acid. Hence, the strengths of the given acids increase as:
4-Methoxybenzoic acid <Benzoic acid <4-Nitrobenzoic acid <3,4-Dinitrobenzoic acid.

Question 13.
Give simple chemical tests to distinguish between the following pairs of compounds:
(i) Propanal and Propanone
(ii) Acetophenone and Benzophenone
(iii) Phenol and Benzoic acid
(iv) Benzoic acid and Ethyl benzoate
(v) Pentan-2-one and Pentan-3-one
(vi) Benzaldehyde and Acetophenone
(vil) Ethanal and Propanal
Answer:
(i) Propanal and propanone:
lodoform test: This test is given by propanone and not by propanal.
Propanone on reacting with hot NaOH/I₂ gives a yellow precipitate of CHI₃ while propanal does not.
2NaOH + I₂ → NaI + NaOI + H₂O

(ii) Acetophenone and benzophenone: Acetophenone responds to iodoform test, but benzophenone does not.

(iii) Phenol and benzoic acid: Benzoic acid reacts with NaHCO₃ giving CO₂ gas with effervescence, whereas phenol does not.

(iv) Benzoic acid and ethyl benzoate: Benzoic acid on reaction with sodium hydrogen carbonate give out CO₂ gas with effervescence, while ethyl benzoate does not.

(v) Pentan-2-one and pentane-3-one: Pentan-2-one, when treated with NaOI(I2/NaOH), gives yellow precipitate of iodoform but pentane-3-one does not give this test.

(vi) Benzaldehyde and acetophenone: Benzaldehyde “being an aldehyde gives silver mirror with Tollen’s reagent but acetophenone being a ketone does not give this test.

(vii) Ethanal and propanal: Ethanal responds to iodoform test, while propanal does not.

Question 14.
How will you prepare the following compounds from, benzene? You may use any inorganic reagent and any organic reagent having not more than one carbon atom.
(i) Methyl benzoate
(ii) m -Nitrobenzoic acid
(iii) p-Nitrobenzoic acid
(iv) Phenylacetic acid
(v) p-Nitrobenz aldehyde.
Answer:

Question 15.
How will you bring about the following conversions in not more than two steps?
(i) Propanone to Propene
(ii) Benzoic acid to Benzaldehyde
(iii) Ethanol to 3-Hydroxybutanal
(iv) Benzene to m-Nitroacetophenone
(y) Benzaldehyde to Benzophenone
(vi) Bromobenzene to 1-Phenylethanol
(vii) Benzaldehyde to 3-Phenylpropan-1-oI
(viii) Benzaldebyde to a -Hydroxy phenylacetic acid
(ix) Benzoic acid to m-Nitro benzyl alcohol
Answer:

Question 16.
Describe the following:
(i) Acetylation
(ii) Cannizzaro reaction
(iii) Cross aldol condensation
(iv) Decarboxylation
Answer:
(i) Acetylation: The introduction of an acetyl functional group into an organic compound is known as acetylation. It is usually carried out in the presence of a base such as pyridine, dimethylaniline, etc. This process involves the substitution of an acetyl group for an active hydrogen atom. Acetyl chloride and acetic anhydride are commonly used as acetylating agents.
For example, acetylation of ethanol produces ethyl acetate.

(ii) Cannizzaro reaction: The self oxidation-reduction (disproportionation) reaction of aldehydes having no a-hydrogens on treatment with concentrated alkalis is known as the Cannizzaro reaction. In this reaction, two molecules of aldehydes participate where one is reduced to alcohol and the other is oxidized to carboxylic acid.
For example, when ethanol is treated with concentrated potassium hydroxide, ethanol and potassium ethanoate are produced.

(iii) Cross-aldol condensation: When aldol condensation is carried out between two different aldehydes, or two different ketones, or an aldehyde and a ketone, then the reaction is called a cross-aldol condensation. If both the reactants contain a-hydrogens, four compounds are obtained as products. For example, ethanol and propanal react to give four products. CH₃CHO + CH₃CH₂CHO

(iv) Decarboxylation: Decarboxylation refers to the reaction in which carboxylic acids lose carbon dioxide to form hydrocarbons when their sodium salts are heated with soda-lime (a mixture of NaOH and CaO in ratio 3:1).

Decarboxylation also takes place when aqueous solutions of alkali metal salts of carboxylic acids are electrolyzed. This electrolytic process is known as Kolbe’s electrolysis.

Question 17.
Complete each synthesis by giving missing starting material, reagent or products


Answer:

Question 18.
Give plausible explanation for each of the following:
(i) Cyclohexanone forms cyanohydrin in good yield but 2, 2, 6 trimethyl cyclohexanone does not.
(ii) There are two -NH₂ groups in semicarbazide. However, only one is involved in the formation of semicarbazones.
(iii) During the preparation of esters from a carboxylic acid and an alcohol in the presence of an acid catalyst, the water or the ester should be removed as soon as it is formed.
Answer:
(i) Cyclohexanones form cyanohydrins according to the following equation :

In this case, the nucleophile CN^– can easily attack without any steric hindrance. However, in the case of 2, 2, 6- trimethyl cyclohexanone, methyl groups at α-positions offer steric hindrances and as a result, CN^– cannot attack effectively.

For this reason, it does not form a cyanohydrin.
(ii) Semicarbazide undergoes resonance involving only one of the two -NH₂ groups, which is attached directly to the carbonyl carbon atom.

Therefore, the electron density on -NH₂ group involved in the resonance also decreases. As a result, it cannot act as a nucleophile.
Since the other -NH₂ group is not involved in resonance; it can act as a nucleophile and can attack carbonyl-carbon atoms of aldehydes and ketones to produce semicarbazones.
(iii) Ester along with water is formed reversibly from a carboxylic acid and an alcohol in presence of an acid.

If either water or ester is not removed as soon as it is formed, then it reacts to give back the reactants as the reaction is reversible. Therefore, to shift the equilibrium in the forward direction i.e., to produce more ester, either of the two should be removed.

Question 19.
An organic compound contain 69.77% carbon, 11.63% hydrogen and rest oxygen. The molecular mass of the compound Is 86. It does not reduce Tollens’ reagent but forms an additional compound with sodium hydrogen sulphite and give positive iodoform test. On vigorous oxidation, it gives ethanoic and propanoic acid. Write the possible structure of the compound.
Solution:

Empirical formula of the compound A = C₅H₁₀O
Molecular formula of the compound A = n (Empirical formula)
n = \(\frac{\text { Molecular mass of compound } A}{\text { Empirical formula mass of compound } A} \)
Molecular mass of compound A =86

Empirical formula mass of compound
A = 5 x 12 + 1 x 10+1 x 16
= 60+10+16 = 86
n = \(\frac{86}{86}\) = 1

Molecular formula of the compound
A = 1 (C₅H₁₀O) = C₅H₁₀O
As the compound A forms addition compound with NaHSO₃ therefore it must be either an aldehyde or ketone. As it does not reduce Tollen’s reagent and give positive iodoform test therefore it must be a methyl ketone. As on oxidation, the compound A gives a mixture of ethanoic acid and propane acid, therefore compound A is

Question 20.
Although phenoxide ion has more number of resonating structures than carboxylate ion, carboxylic acid is a stronger
acid than phenol. Why?
Answer:
(i) Resonance structures of phenoxide ion are:

It can be observed from the resonance structures of phenoxide ion that in II, III and IV, less electronegative carbon atoms carry a negative charge. Therefore, these three structures contribute negligibly towards the resonance stability of the phenoxide ion. Hence, these structures can be eliminated. Only structures I and V carry a negative charge on the more electronegative oxygen atom.

(ii) Resonance structures of carboxylate ion are :

In the case of carboxylate ions, resonating structures I and II contain a charge carried by a more electronegative oxygen atom. Further, in resonating structures I and II, the negative charge is delocalized over two oxygen atoms. But in resonating structures I and V of the phenoxide ion, the negative charge is localized on the same oxygen atom. Therefore, the resonating structures of carboxylate ion contribute more towards its stability than those of phenoxide ion. As a result, carboxylate ion is more resonance-stabilized than phenoxide ion. Hence, carboxylic acid is a stronger acid than phenol.

Chemistry Guide for Class 12 PSEB Aldehydes, Ketones, and Carboxylic Acids Textbook Questions and Answers

Question 1.
Write the structures of the following compounds :
(i) α-Methoxypropionaldehyde
(ii) 3-Hydroxybutanal
(iii) 2-Hydroxycyclopentane carbaldehyde
(iv) 4-Oxopentanal
(v) Di-sec. butyl ketone
(vi) 4-Fluoroacetophenone
Answer:

Question 2.
Write the structures of products of the following reactions;


Answer:

Question 3.
Arrange the following compounds in increasing order of their boiling points.
CH₃CHO, CH₃CH₂OH, CH₃OCH₃, CH₃CH₂CH₃
Answer:
Hydrocarbons are non-polar having weakest attractive forces; Ethers are polar (dipolar forces); Aldehydes have strong dipolar interactions;
Alcohols have maximum intermolecular forces due to hydrogen bonding.
CH₃CH₂CH₃ < CH₃OCH₃ < CH₃CHO< CH₃CH₂OH.

Question 4.
Arrange the following compounds in increasing order of their reactivity in nucleophilic addition reactions.
(i) Ethanal, Propanal, Propanone, Butanone.
(ii) Benzaldehyde, p-Tolualdehyde, p -Nitrobenz aldehyde, Acetophenone.
(Hint: Consider steric effect and electronic effect)
Answer:

The +I effect of the alkyl group increases in the order:
Ethanal < Propanal < Propanone < Butanone
The electron density at the carbonyl carbon increases with the increase in the +I effect. As a result, the chances of attack by a nucleophile decrease. Hence, the increasing order of the reactivities of the given carbonyl compounds in nucleophilic addition reactions is:
Butanone < Propanone < Propanal < Ethanal

The + I effect is more in ketone than in aldehyde. Hence, acetophenone is the least reactive in nucleophilic addition reactions. Among aldehydes, the +1 effect is the highest in p-total dehyde because of the presence of the electron-donating -CH₃ group and the lowest in p-nitrobenzaldehyde because of the presence of the electron-withdrawing 2 groups. Hence, the increasing order of the reactivities of the given compounds is :
Acetophenone < p – tolualdehyde <Benzaldehyde <p – Nitrobenzaldehyde

Question 5.
Predict the products of the following reactions:

Answer:

Question 6.
Give the IUPAC names of the following compounds :
(i) PhCH₂CH₂COOH
(ii) (CH₃)₂C = CHCOOH

Answer:
(i) 3-Phenylpropanoic acid
(ii) 3-Methylbut-2-enoic acid
(iii) 2-Methylcyclopentanecarboxylic acid
(iv) 2,4,6-Trinitrobenzoic acid

Question 7.
Show how each of the following compounds can be converted to benzoic acid.
(i) Ethylbenzene
(ii) Acetophenone
(iii) Bromobenzene
(iv) Phenylethene (Styrene)
Answer:

Question 8.
Which acid of each pair shown here would you expect to be stronger?
(i) CH₃CO₂H or CH₂FCO₂H
(ii) CH₂FCO₂H or CH₂CICO₂H
(iii) CH₂FCH₂CH₂CO₂H or CH₃CHFCH₂CO₂H

Answer:

The +I effect of the -CH₃ group increases the electron density on the O-H bond. Therefore, the release of protons becomes difficult. On the other hand, the – I effect of F decreases the electron density on the O-H bond. Therefore, protons can be released easily. Hence, CH₂FCO₂H is a stronger acid than CH₃CO₂H.

F has a stronger – I effect than Cl. Therefore, CH₂FCO₂H can release protons more easily than CH₂ClCO₂H.
Hence, CH₂FCO₂H is a stronger acid than CH₂ClCO₂H.

The inductive effect decreases with an increase in distance. Hence, the +I effect of F in CH₃CHFCH₂CO₂H is more than it is in CH₂FCH₂CH₂CO₂H.
Hence, CH₃CHFCH₂CO₂H is a stronger acid than CH₂FCH₂CH₂CO₂H.

Due to the – I effect of F, it is easier to release proton in the case of compound (A). However, in the case of compound (B), release of proton is difficult due to the +1 effect of -CH₃ group. Hence, (A) is a stronger acid than (B).