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

PSEB 12th Class Chemistry Important Questions Chapter 7 The p-Block Elements

Very Short Answer Type Questions

Question 1.
What is the difference between the nature of n-bonds present in H₃PO₃ and HNO₃ molecules?
Answer:
In H₃PO₃, there is pπ-dπ bond whereas in HNOs there is pπ-pπ bond.

Question 2.
Complete the following equations:
(i) PCl₃ + H₂O →
(ii) XeF₂ + PF₅ →

Answer:
(i) PCl₃ + 3H₂O → H₃PO₃ + 3HCl
(ii) XeF₂ + PF₅ → [XeF]⁺ [PF₆]^–

Question 3.
Which allotrope of sulphur is thermally stable at room temperature?
Answer:
Rhombic sulphur

Question 4.
O—O bond has lower bond dissociation enthalpy than S—S bond. Why?
Answer:
Due to smaller size the lone pairs of electrons on the O atom repel the bond pair of O—O bond to a greater extent as compared to the lone pairs of electrons on S atom in S—S bond. Consequently O—O bond has lower bond dissociation enthaltpy than S—S bond.

Question 5.
How would you account for the following:
(i) H₂S is more acidic than H₂O.
(ii) Both O₂ and F₂ stabilise higher oxidation states hut the ability of oxygen to stabilise the higher oxidation state exceeds that of fluorine.
Answer:
(i) This is because bond dissociation enthalpy of H—S bond is lower than that of H—O bond.
(ii) This is due to tendency of oxygen to form multiple bonds with metal atom.

Question 6.
Why solid PCl₅ is ionic in nature?
Answer:
Because in solid state, PCl₅ exists as [PCl₄]⁺[PCl₆]^– and conducts electricity on melting.

Question 7.
On adding NaOH to ammonium sulphate, a colourless gas with pungent odour is evolved which form a blue coloured complex with Cu²⁺ ion. Identify the gas.
Answer:
Ammonia (NH₃).

Question 8.
N₂O₅ is more acidic thanNaO₃. Why?
Answer:
N₂O₅ is the anhydride of nitric acid, forms the stable acid with water as follows :

While N₂O₃ is the anhydride of nitrous acid, HNO₂. It dissolves in water to form the unstable acid as follows :

Hence, N₂O₅ is more acidic thanN₂O₃.

Question 9.
Why is nitric oxide paramagnetic in gaseous state but the solid obtained on cooling is diamagnetic?
Answer:
In gaseous state, NO₂ exists as monomer which has one unpaired electron but in solid state, it dimerises to NO₂ so no unpaired electron is left hence, the solid formed is diamagnetic.

Question 10.
In the preparation of H₂SO₄ by contact process, why is SO₃ not absorbed directly in water to form H₂SO₄ ?
Answer:
Acid fog is formed, which is difficult to condense.

Short Answer Type Questions

Question 1.
Account for the following:
(i) Bi(V) is a stronger oxidising agent than Sb(V).
(ii) N—N single bond is weaker than P—P single bond.
Answer:
(i) Due to inert pair effect +3 oxidation state of Bi is more stable than its +5 oxidation state while +5 oxidation state of Sb is more stable than its +3 oxidation state. Therefore, Bi (V) can accept a pair of electrons to form more stable Bi (III) more easily than Sb (V). Hence, Bi (V) is a stronger oxidising agent than Sb (V).

(ii) N—N single bond is weaker than P—P single bond due to large interelectronic repulsion between the lone pairs of electrons present on the N atoms of N—N bond having small bond length.

Question 2.
Account for the following:
(i) PCl₅ is more covalent than PCl₃.
(ii) Iron on reaction with HCl forms FeCl₂ and not FeCl₃.
(iii) The two O—O bond lengths in the ozone molecule are equal.
Answer:
(i) The oxidation state of central atom, i.e., phosphorus is +5 in PCl₅ whereas it is +3 in PCl₃. Higher the positive oxidation of central atom, more will be its polarising power which, in turn, increases the covalent character of bond formed between the central atom and the atoms surrounding it.

(ii) Iron reacts with HCl to form FeCl₂ and H₂.
Fe + 2HCl → FeCl₂ + H₂
H₂ thus produced prevents the oxidation of FeCl₂ to FeCl₃.

(iii) Ozone is a resonance hybrid of the following two main structures :

As a result of resonance, the two O—O bond lengths in O₃ are equal.

Question 3.
How would you account for the following:
(i) The electron gain enthalpy with negative sign is less for oxygen than that for sulphur.
(ii) Fluorine never acts as the central atom in polyatomic interhalogen compounds.
Answer:
(i) This is due to smaller size of oxygen the electron cloud is distributed over a small region of space, making electron density high which repels the incoming electrons.

(ii) Fluorine never acts as the central atom in polyatomic interhalogen compounds since it is the most electronegative element of the group.

Question 4.
PCl₅ reacts with finely divided silver on heating and a white silver salt is obtained, which dissolves on adding excess aqueous NH₃ solution. Write the reactions involved to explain what happens.
Answer:
PCl₅ on reaction with finely divided silver produced silver halide.
PCl₅ + 2Ag → 2AgCl + PCl₃
AgCl on further reaction with aqueous ammonia solution produces a soluble complex of [Ag(NH₃)₂]⁺Cl^–.

Question 5.
Account for the following:
(i) Sulphur in vapour state exhibits paramagnetism.
(ii) H₃PO₂ is a stronger reducing agent than H₃PO₂.
Answer:
(i) In vapour form, sulphur partly exists as S₂ molecules which have two unpaired electrons in the antibonding n molecular orbitals like 02 molecule and hence, exhibits paramagnetism.

(ii) Acids which contain P—H bonds, have reducing character. Since, H₃PO₂ contains two P—H bonds while H₃PO₃ contains only one P—H bond, therefore H₃PO₂ is a stronger reducing agent than H₃PO₃.

Question 6.
What happens when:
(i) ortho phosphorus acid is heated?
(ii) XeF₆ undergoes complete hydrolysis?
Answer:
(i) On heating, ortho phosphorus acid disproportionates to give orthophosphoric acid and phosphine gas.

(ii) When XeF₆ undergoes complete hydrolysis, it forms XeO₃.
XeF₆ + 3H₂O → 6HF + XeO₃

Long Answer Type Questions

Question 1.
(a) Account for the following:
(i) Acidic character increases from HF to HI.
(ii) There is large difference between the melting and boiling points of oxygen and sulphur.
(iii) Nitrogen does not form pentahalide.
(b) Draw the structures of the following:
(i) ClF₃
(ii) XeF₄
Answer:
(a) (i) As the size of halogen atom increases from F to I, the bond dissociation enthalpy of H—X bond decreases from H—F to H—I. Due to this, acidic character increases from HF to HI.

(ii) Because of small size and high electronegativity oxygen forms pπ-pπ multiple bonds and exists as a diatomic, O₂ molecule. The molecules are held together by weak van der Waal forces. Sulphur on the other hand due to its higher tendency for catenation and lower tendency for pπ-pπ multiple bond formation, forms octa-atomic, S₈ molecule. Because of bigger size of S₈ molecule than O₂ molecule the force of attraction holding the S₈ molecules together are much stronger than O₂ molecules. Hence, there is large difference between the melting and boiling points of oxygen and sulphur.

(iii) Nitrogen with n = 2, has s and p-orbitals only. It does not have d-orbitals to expand its covalency beyond four. Due to this, it does not form pentahalide.

(b)

Question 2.
(a) Give reasons for the following:
(i) Bond enthalpy of F₂ is lower than that of Cl₂.
(ii) PH₃ has lower boiling point than NH₃.
(b) Draw the structures of the following molecules :
(i) BrF₃
(ii)BrF₅
(iii) (HPO₃)₃
Answer:
(a) (i) Bond dissociation enthalpy decreases as the bond distance increases from F₂ to I₂ because of the corresponding increase in the size of the atom as we move from F to I. The F—F bond dissociation enthalpy is, however, smaller than that of Cl—Cl and even smaller than that of Br—Br. This is because F atom is very small and hence the three lone pairs of electrons on each F atom repel the bond pair holding the F-atoms in F2 molecule resulting lower bond enthalpy than Cl₂.

(ii) Unlike NH₃, PH₃ molecules are not associated through hydrogen bonding in liquid state. That is why the boiling point of PH₃ is lower than NH₃.

Question 3.
(a) Complete the following chemical reaction equations:
(i) AgCl(s) + NH₃ (aqr) →
(ii) P₄(s) + NaOH(aq) + H₂O(l) →
(b) Explain the following observations :
(i) H₂S is less acidic than H₂Te.
(ii) Fluorine is a stronger oxidising agent than chlorine.
(iii) Noble gases are the least reactive elements.
Answer:
(a) (i) AgCl + 2NH₃ → [Ag(NH₃)₂]⁺ Cl^–
(ii) P₄ + 3NaOH + 3H₂O → PH₃ + 3NaH₂PO₂

(b) (i) This is because bond dissociation enthalpy of H—Te bond is less than H—S as the size of Te is larger than S.
(ii) Fluorine is a stronger oxidising agent than chlorine due to low dissociation enthalpy of F—F bond and high hydration enthalpy of F^– ions.
(iii) Noble gases are the least reactive elements due to fully filled outermost shells, high ionisation enthalpy and positive electron gain enthalpy.

Question 4.
(a) Arrange the following in the order of property indicated against each set:
(i) F₂, Cl₂, Br₂, I₂ (increasing bond dissociation enthalpy)
(ii) H₂O, H₂S, H₂Se, H₂Te (increasing acidic character)
(b) A colourless gas ‘A’ with a pungent odour is highly soluble in water and its aqueous solution is weakly basic. As a weak base it precipitates the hydroxides of many metals from their salt solution. Gas ‘A* finds application in detection of metal ions. It gives a deep blue colouration with copper ions. Identify the gas A’ and write the chemical equations involved in the following :
(i) Gas ‘A’ with copper ions
(ii) Solution of gas ‘A’ with ZnSO₄ solution.
Answer:
(a) (i) I₂ < F₂ < Br₂ < Cl₂
(ii) H₂O < H₂S < H₂Se < H₂Te

(b) The gas ‘A’ is ammonia (NH₃).
(i) Cu²⁺(aq) + 4 NH₃(aq) ⇌ [Cu(NH₃)₄]²⁺(aq)
(ii) ZnSO₄(aq) + 2 NH₄OH(aq) → Zn(OH)₂(s) + (NH₄) ₂ SO₄(aq)

Question 5.
Answer the following questions
(a) Write the formula of the neutral molecule which is isoelectronic with ClO^–.
(b) Draw the shape of H₂S₂O₇.
(c) Nitric acid forms an oxide of nitrogen on reaction with P₄. Write the formula of the stable molecule formed when this oxide undergoes dimerisation.
(d) Bleaching action of chlorine is permanent. Justify.
(e) Write the disproportionation reaction of that oxoacid of nitrogen in which nitrogen is in + 3 oxidation state.
Answer:
(a) ClF

(c) N₂O₄
(d) Bleaching action of chlorine is permanent due to oxidation.
Cl + H₂O → 2HCl + [O]
(d) 3HNO₂ → HNO₃ + H₂O + 2NO

Punjab State Board PSEB 12th Class Chemistry Book Solutions Chapter 8 The d-and f-Block Elements Textbook Exercise Questions and Answers.

PSEB Solutions for Class 12 Chemistry Chapter 8 The d-and f-Block Elements

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

Question 1.
Write down the electronic configuration of:
(i) Cr³⁺
(ii) Cu⁺
(iii) Co²⁺
(iv) Mn²⁺
(v) Pm³⁺
(vi) Ce⁴⁺
(vii) Lu²⁺
(viii) Th⁴⁺
Answer:
(i)Cr³⁺ = [Ar] 3d³
(ii) Cu⁺ = [Ar] 3d¹⁰
(iii) Co²⁺ = [Ar] 3d⁷
(iv) Mn²⁺ = [Ar] 3d⁵
(v) Pm³⁺ = [Xe] 4f⁴
(vi) Ce⁴⁺ = [Xe]
(vii) Lu²⁺ = [Xe] 4f¹⁴5d¹
(viii) Th⁴⁺ = [Rn]

Question 2.
Why are Mn²⁺ compounds more stable than Fe²⁺ towards oxidation to their + 3 state?
Answer:
Electronic configuration of Mn²⁺ is [Ar]¹⁸ 3d⁵
Electronic configuration of Fe²⁺ is [Ar]¹⁸ 3d⁶
It is known that half-filled and fully-filled orbitals are more stable. Therefore, Mn in (+ 2) state has a stable d⁵ configuration. This is the reason Mn²⁺ shows resistance to oxidation to Mn³⁺. Also, Fe²⁺ has 3d⁶ configuration and by losing one electron, its configuration changes to a more stable 3d⁵ configuration. Therefore, Fe²⁺ easily gets oxidised to Fe³⁺ oxidation state.

Question 3.
Explain briefly how +2 state becomes more and more stable in the first half of the first row transition elements with increasing atomic number?
Answer:
As the atomic number increases from 21 to 25, the number of electrons in the 3d-orbital also increases from 1 to 5. +2 oxidation state is attained by the loss of the two 4s electrons by these metals. Sc does not exhibit +2 oxidation state. As the number of d-electrons in +2 state increases from Ti to Mn, the stability of +2 state increases (d-orbital gradually becoming half filled). Mn(+2) has d⁵ electrons which is highly stable.

Question 4.
To what extent do the electronic configurations decide the stability of oxidation states in the first series of the transition elements? Illustrate your answer with examples.
Answer:
The stability of oxidation states in the first series of the first transition elements are related to their electronic configurations.

The first five elements of the first transition series upto Mn in which the 3d-subshell is not more than half-filled, the minimum oxidation state is given by fife number of electrons in the outer s-subshell and the maximum oxidation state is given by the sum of the outer s and d-electrons. For example, Sc does not show +2 oxidation state. Its electronic configuration is 4s² 3d¹. It loses all the three electrons to form SC³⁺.+3 oxidation state is very stable as by losing all three electrons, it attains the stable configuration of Argon. For Mn, +2 oxidation state is very stable, as after losing two 4s electrons, the d-orbitals become half-filled.

Question 5.
What may be the stable oxidation state of the transition elements with the following d-electron configurations in the ground state of their atoms?
3d³, 3d⁵, 3d⁸ and 3d⁴
Answer:
Stable oxidation states:
3d³ (vanadium): +2, +3, +4, +5
3d⁵ (chromium): +3, +4, +6
3d⁵ (manganese): +2, +4, +6, +7
3d⁸ (cobalt) : +2, +3 (in complexes)
3d⁴ : There is no d4 configuration in the ground state.

Question 6.
Name the oxometal anions of the first series of the transition metals in which the metal exhibits the oxidation state equal to its group number.
Answer:
(i) Vanadate, \(\mathrm{VO}_{3}^{-}\)
Oxidation state of V is + 5.

(ii) Chromate, \(\mathrm{CrO}_{4}^{2-}\)
Oxidation state of Cr is + 6.

(iii) Permanganate, \(\mathrm{MnO}_{4}^{-}\)
Oxidation state of Mn is + 7.

Question 7.
What is lanthanoid contraction? What are the consequences of lanthanoid contraction?
Answer:
As we move along the lanthanoid series, the atomic number increases gradually by one. This means that the number of electrons and protons present in an atom also increases by one. As electrons are being added to the same shell, the effective nuclear charge increases. This happens because the increase in nuclear attraction due to the addition of proton is more pronounced than the increase in the interelectronic repulsions due to the addition of electron. Also, with the increase in atomic number, the number of electrons in the 4f orbital also increases. The 4f electrons have poor shielding effect. Therefore, the effective nuclear charge experienced by the outer electrons increases. Consequently, the attraction of the nucleus for the outermost electrons increases. This results in a steady decrease in the size of lanthanoids with the increase in the atomic number. This is termed as lanthanoid contraction.
Consequences of lanthanoid contraction

1. There is similarity in the properties of second and third transition series.
2. Separation of lanthanoids is possible due to lanthanoid contraction.
3. It is due to lanthanoid contraction that there is variation in the basic strength of lanthanoid hydroxides. (Basic strength decreases from La(OH)₃ to Lu(OH)₃.)

Question 8.
What are the characteristics of the transition elements and why are they called transition elements? Which of the d-block elements may not he regarded as the transition eledaents?
Answer:
Characteristics of the Transition Elements (d-Block)
1. Electronic configuration : General electronic configuration of these elements is (n -1) d^1-10 ns^1-2.

2. Physical properties : These elements have metallic properties such as metallic lustre, high tensile strength, ductility, malleability, high thermal and electrical conductivity, low volatility (except Zn, Cd, Hg), hardness, etc. Their melting points are high.

3. Atomic and ionic size : In a given series, there is a progressive decrease in radius with increasing atomic number.

4. Ionisation enthalpies : Due to an increase in nuclear charge which accompanies the filling of inner d-orbitals, there is an increase in ionisation enthalpy along each series of the transition elements from left to right.

5. Oxidation states : These elements exhibit variable oxidation states.
e.g.,1 Transition series:

6. Trends in M²⁺/M^⊖ values: The general trend towards less negative E⁰ values across the series is related to the general increase in the sum of the first and second ionisation enthalpies.

7. Trends in M³⁺/M²⁺E^⊖ values : Low value of Ee shows the stability of ion (either d⁵ or d¹⁰ configuration).

8. Magnetic properties : These elements show diamagnetism and paramagnetism.
9. Formation of coloured salts : The compounds of transition elements form coloured ions, e.g., Mn³⁺ violet; Fe²⁺ green, etc.

10. Complex formation : These elements form complex compounds due to their small size and high charge density, e.g., [PtCl₄]^2-

11. Catalyst: Many of these elements are used as catalyst e.g., V₂O₅ is used as a catalyst in contact process for the manufacture of H₂SO₄.

12. Interstitial compound formation : Transition elements form interstitial compounds. It means the compounds in which H, C or N etc. are trapped inside the crystal lattices of metals.

13. Alloy formation : Because of similar radii and other characteristics alloys are readily formed by these metals.
The d-block elements are called transition elements because these elements represent change or transition in properties from s-block to p-block elements.
The electronic configuration of Zn, Cd and Hg is represented by the general formula (n – 1)d¹⁰ns². These elements have completely filled d-orbitals in ground state as well as in their common oxidation states. Therefore, they may not be regarded as the transition elements.

Question 9.
In what way is the electronic configuration of the transition elements different from that of the non-transition elements?
Answer:
Transition elements contain incompletely filled d-subshell, Le., their electronic configuration is (n – 1)d^1-10ns^1-2 whereas non-transition elements have no d-subshell or their subshell is completely filled and have ns^1-2 or ns²np^1-6 in their outermost shell.

Question 10.
What are the different oxidation states exhibited by the lanthanoids?
Answer:
In the lanthanoid series, + 3 oxidation state is most common i.e., Ln(III) compounds are predominant. However, + 2 and + 4 oxidation states can also be found in the solution or in solid compounds.

Question 11.
Explain giving reasons:
(i) Transition metals and many of their compounds show paramagnetic behaviour.
(ii) The enthalpies of atomisation of the transition metals are high.
(iii) The transition metals generally form coloured compounds.
(iv) Transition metals and their many compounds act as good catalyst.
Answer:
(i) Transition metals and many of their compounds show paramagnetic behaviour. Paramagnetism arises due to the presence of unpaired electrons with each electron having a magnetic moment associated with its spin angular momentum and orbital angular momentum. However, in the first transition series, the orbital angular momentum is quenched. Therefore, the resulting paramagnetism is only because of the unpaired electron.

(ii) Transition metals have high effective nuclear charge and a large number of valence electrons. Therefore, they form very strong metallic bonds. As a result, the enthalpy of atomisation of transition metals is high.

(iii) Formation of coloured compounds by transition metals is due to partial adsorption of visible light. The electron absorbs the radiation of a particular frequency (of visible region) and jumps into next orbital.

(iv) Catalysts, at the solid surface, involve the formation of bonds between reactants molecules and atoms of the surface of the catalyst (I row transition metals utilised 3d and 4s-electrons for bonding). This has the effect of increasing the concentration of the reactants at the catalyst surface and also lowering of the activation energy.
Transition metal ions show variable oxidation states so they are effective catalysts, e.g.

Mechanism of catalysing action of Fe³⁺ in the above reaction
(a) 2Fe³⁺ + 2I^– → 2Fe²⁺ + I₂
(b) 2Fe²⁺ + \(\mathrm{S}_{2} \mathrm{O}_{8}^{2-}\) → 2Fe³⁺ + \(2 \mathrm{SO}_{4}^{2-}\)

Question 12.
What are interstitial compounds? Why are such compounds well known for transition metals?
Answer:
Interstitial compounds are those in which small atoms occupy the interstitial sites in the crystal lattice. Interstitial compounds are well known for transition metals because small-sized atoms of H, B, C, N, etc., can easily occupy position in the voids present in the crystal lattices of transition metals.

Question 13.
How is the variability in oxidation states of transition metals different from that of the non-transition metals? Illustrate with examples.
Answer:
The oxidation states of transition elements differ from each other by unity e.g., Fe²⁺ and Fe³⁺, Cu+ and Cu²⁺ (due to incomplete filling of d-orbitals) whereas oxidation states of non-transition elements normally differ by two units e.g., Pb²⁺ and Pb⁴⁺, Sn²⁺ and Sn⁴⁺ etc.

Question 14.
Describe the preparation of potassium dichromate from iron chromite ore. What is the effect of increasing pH of a solution of potassium dichromate?
Answer:
Potassium dichromate is prepared from chromite ore FeCr₂O₄ as follows :
Step I: Conversion of chromite ore to sodium chromate.

Effect of pH : The chromates and dichromates can interconverted to each other in aqueous solution by increasing the pH of the solution.
\(\mathrm{CrO}_{4}^{2-}\) + 2OH^– ⇌ \(2 \mathrm{CrO}_{4}^{2-}\) + H₂O

Question 15.
Describe the oxidising action of potassium dichromate and write the ionic equations for its reaction with
(i) iodide
(ii) iron (H) solution and
(iii) H₂S.
Answer:
(i) \(\mathrm{Cr}_{2} \mathrm{O}_{7}^{2-}\) + 14H⁺ + 6I^– → 2Cr³⁺ + 7H₂O + 3I₂
(ii) \(\mathrm{Cr}_{2} \mathrm{O}_{7}^{2-}\) + 14H⁺ + 6Fe²⁺ → 2Cr³⁺ + 7H₂O + 6Fe³⁺
(iii) \(\mathrm{Cr}_{2} \mathrm{O}_{7}^{2-}\) + 8H⁺ + 3H₂S → Cr³⁺ + 7H₂O + 3S

Question 16.
Describe the preparation of potassium permanganate. How does the acidified permanganate solution reacts with
(i) iron
(ii) ions
(iii) SO₂ and
(iii) oxalic acid? Write the ionic equations for the reactions.
Answer:
Preparation of potassium permanganate, KMnO₄: Potassium permanganate is prepared by the fusion of MnO₂ (pyrolusite) with potassium hydroxide and an oxidising agent such as KNO₃ to form potassium manganate which disproportionate in a neutral or acidic solution to form permanganate.
2MnO₂ + 4KOH + O₂ → 2K₂MnO₄ + H₂O
\(3 \mathrm{MnO}_{4}^{2-}\) + 4H₊ → \(\mathrm{MnO}_{4}^{-}\) + MnO₂ + H₂O
On large scale : It is prepared by the alkaline oxidative fusion of MnO₂ to form potassium manganate. The electrolytic oxidation of potassium manganate in alkaline solution produces KMnO₄, at the anode.

Reaction of KMnO₄ in acidic medium
(a) Oxidises iron (II) ions to iron (III) ions.
\(\mathrm{MnO}_{4}^{-}\) + 5Fe²⁺ + 8H²⁺ → Mn²⁺ + 5Fe³⁺ + 4H₂O
(b) Oxidises SO₂ to sulphuric acid.
\(2 \mathrm{MnO}_{4}^{-}\) + 5SO₂ + 2H₂O → 2Mn²⁺ + 5SO₂ + 4H₂O
(c) Oxidises oxalic acid to carbon dioxide.
\(2 \mathrm{MnO}_{4}^{-}\) + \(5 \mathrm{C}_{2} \mathrm{O}_{4}^{2-}\) + 16H⁺ → 2Mn²⁺ + 10CO₂ + 8H₂O

Question 17.
For M²⁺/M and M³⁺/M²⁺ systems, the E^⊖ values for some metals are as follows:

Use this data to comment upon :
(i) the stability of Fe³⁺ in acid solution as compared to that of Cr³⁺ or Mn³⁺ and
(ii) the ease with which iron can be oxidised as compared to a similar process for either chromium or manganese metal.
Answer:
(i) Higher the reduction potential of a species, greater is the tendency for its reduction to take place. Therefore, Mn³⁺ with highest reduction potential would be readily reduced to Mn²⁺ and hence is the least stable.
Thus, from the value of reduction potential, it is clear that the stability of Fe³⁺ in acidic solution is more than Mn³⁺ but less than that of Cr³⁺.

(ii) Lower the reduction potential or higher the oxidation potential of a species, greater the ease with which its oxidation will take place. Thus, order of tendency to undergo oxidation is Fe < Cr < Mn.

Question 18.
Predict which of the following will be coloured in aqueous solution?
Ti³⁺, V³⁺, Cu⁺, Sc³⁺, Mn²⁺, Fe³⁺ and Co²⁺. Give reasons for each.
Answer:
Only those ions will be coloured which have incompletely filled d-orbitals. Those with fully-filled or empty d-orbitals are colourless. Due to d-d transition, Ti³⁺, V³⁺, Mn²⁺, Fe³⁺ and Co²⁺ are coloured.

Question 19.
Compare the stability of + 2 oxidation state for the elements of the first transition series.
Answer:
The decreasing negative electrode potentials of M²⁺/M⁺¹ in the first transition series shows that in general, the stability of +2 oxidation state decrease from left-to right (exception being Mn and Zn). The decrease in the negative electrode potentials is due to increase in the sum IE₂ + IE₂.

Question 20.
Compare the chemistry of actinoids with that of the lanthanoids with special reference to :
(i) electronic configuration
(ii) oxidation states
(iii) atomic and ionic sizes and
(iv) chemical reactivity.
Answer:
(i) Electronic configuration : The general electronic configuration of lanthanoids is [Xe]₅₄ 4f_1-14 5d_0-16s₂ whereas that of actinoids is [Rn]₈₆5f_1-146d_0-17s₂. Thus, lanthanoids belong to 4/-series whereas actinoids belong to 5/-series.

(ii) Oxidation states : Lanthanoids show limited oxidation states (+2, +3, +4), out of which, +3 is most common. This is because of a large energy gap between 4/, 5d and 6s subshells. On the other hand, actinoids show a large number of oxidation states because of small energy gap between 5/, 6d and 7s subshells.

(iii) Atomic and ionic sizes : Both show decrease in size of their atoms or ions in +3 oxidation state. In lanthanoids, the decrease is called lanthanoid contraction whereas in actinoids, it is called actinoid contraction. However, the contraction is greater from element to element in actinoids due to poorer shielding by 5/-electrons.

(iv) Chemical reactivity : In general, the earlier members of the lanthanoid series are quite reactive (similar to calcium) but with increasing atomic number, they behave more like aluminium.
Values for E^⊖ for the half-reaction :
Ln³⁺ (aq) + 3e^– → Ln(s)
are in the range of -2.2 to -2.4 V except for Eu for which the value is -2.0 V. This is of course, a small variation.

The metals combine with hydrogen when gently heated in the gas. The carbides, Ln₃C , Ln₂C₃ and LnC₂ are formed when the metals are heated with carbon. They liberate hydrogen from dilute acids and bum in halogens to form halides. They form oxides and hydroxides —M₂O₃ and M(OH)₃. The hydroxides are definite compounds, not just hydrated oxides, basic like alkaline earth metal oxides and hydroxides.

The actinoids are highly reactive metals, especially when finely divided. For example, the action of boiling water on them, gives a mixture of oxide and hydride and combination with most non-metals takes place at moderate temperatures. Hydrochloric acid attacks all metals but most are slightly affected by nitric acid owing to the formation of protective oxide layers; alkalis have no action.

Question 21.
How would you account for the following:
(i) Of the d⁴ species, Cr²⁺ is strongly reducing while manganese (III) is strongly oxidising.
(ii) Cobalt (II) is stable in aqueous solution but in the presence of complexing reagents, it is easily oxidised.
(iii) The d¹ configuration is very unstable in ions.
Answer:
(i) E^⊖ value for Cr³⁺ /Cr²⁺ is negative (-0.41 V) whereas E^⊖ value for Mn³⁺/Mn²⁺ is positive (+1.57 V). Thus, Cr²⁺ ions can easily undergo oxidation to give Cr³⁺ ions and, therefore, act as strong reducing agent. On the other hand, Mn³⁺ can easily undergo reduction to give Mn²⁺ and hence act as oxidising agent.

(ii) Co(III) has greater tendency to form coordination complexes than Co (II). Thus, in the presence of ligands, Co (II) charges to Co (III), i.e., it easily oxidised.

(iii) The ions with d1 configuration have the tendency to lose the only electron present in d-subshell to acquire stable d⁰ configuration. Therefore, they are unstable and undergo oxidation or disproportionation.

Question 22.
What is meant by ‘disproportionation’? Give two examples of disproportionation reaction in aqueous solution.
Answer:
Disproportionation reactions are those reactions in which the same substance undergoes oxidation as well as reduction. In disproportion reaction, oxidation number of an element increases as well as decreases to form two different products, e.g.,

Question 23.
Which metal in the first series of transition metals exhibits +1 oxidation state most frequently and why?
Answer:
Copper has electronic configuration 3d¹⁰ 4s¹ will attain a completely filled d-orbital and a stable configuration on losing 4s¹ electron i.e., [Ar] 3d¹⁰.

Question 24.
Calculate the number of unpaired electrons in the following gaseous ions: Mn³⁺, Cr³⁺, V³⁺ and Ti³⁺. Which one of these is the most stable in aqueous solution?
Answer:
Mn³⁺ = 3d⁴ = 4 unpaired electrons
Cr³⁺ = 3d³ = 3 unpaired electrons
V³⁺ = 3d² = 2 unpaired electrons
Ti³⁺ = 3d¹ =1 unpaired electron
Out of these species Cr³⁺ is the most stable in aqueous solution due to its tendency of complex formation.

Question 25.
Give examples and suggest reasons for the following features of the transition metal chemistry :
(i) The lowest oxide of transition metal is basic, the highest is amphoteric/acidic.
(ii) A transition metal exhibits higher oxidation state in oxides and fluorides.
(iii) The highest oxidation state is exhibited in oxo-anions of a metal.
Answer:
(i) The lowest oxide of transition metal is basic because the metal atom has low oxidation state. This means that it can donate valence electrons which are not involved in bonding to act like a base. Whereas the highest oxide is amphoteric/acidic due to the highest oxidation state as the valence electrons are involved in bonding and are unavailable. For example, MnO is basic whereas Mn₂O₇ is acidic.

(ii) A transition metal exhibits higher oxidation states in oxides and fluorides because oxygen and fluorine are highly electronegative elements, small in size (and strongest oxidising agents). For example, osmium shows an oxidation state of +6 in O₂F₆ and vanadium shows an oxidation state of +5 in V₂O₅.

(iii) Oxometal anions have the highest oxidation state, e.g., Cr in \(\mathrm{Cr}_{2} \mathrm{O}_{7}^{2-}\) has an oxidation state of +6 whereas Mn in \(\mathrm{MnO}_{4}^{-}\) has an oxidation state of +7. This is again due to the combination of the metal with oxygen, which is highly electronegative and oxidising element.

Question 26.
Indicate the steps in the preparation of:
(i) K₂Cr₂O₇ from chromite ore.
(ii) KMnO₄ from pyrolusite ore.
Answer:
(i) Potassium dichromate is prepared from chromite ore FeCr₂O₄ as follows :
Step I: Conversion of chromite ore to sodium chromate.

Effect of pH : The chromates and dichromates can interconverted to each other in aqueous solution by increasing the pH of the solution.
\(\mathrm{CrO}_{4}^{2-}\) + 2OH^– ⇌ \(2 \mathrm{CrO}_{4}^{2-}\) + H₂O
(ii) Preparation of potassium permanganate, KMnO₄: Potassium permanganate is prepared by the fusion of MnO₂ (pyrolusite) with potassium hydroxide and an oxidising agent such as KNO₃ to form potassium manganate which disproportionate in a neutral or acidic solution to form permanganate.
2MnO₂ + 4KOH + O₂ → 2K₂MnO₄ + H₂O
\(3 \mathrm{MnO}_{4}^{2-}\) + 4H₊ → \(\mathrm{MnO}_{4}^{-}\) + MnO₂ + H₂O
On large scale : It is prepared by the alkaline oxidative fusion of MnO₂ to form potassium manganate. The electrolytic oxidation of potassium manganate in alkaline solution produces KMnO₄, at the anode.

Reaction of KMnO₄ in acidic medium
(a) Oxidises iron (II) ions to iron (III) ions.
\(\mathrm{MnO}_{4}^{-}\) + 5Fe²⁺ + 8H²⁺ → Mn²⁺ + 5Fe³⁺ + 4H₂O
(b) Oxidises SO₂ to sulphuric acid.
\(2 \mathrm{MnO}_{4}^{-}\) + 5SO₂ + 2H₂O → 2Mn²⁺ + 5SO₂ + 4H₂O
(c) Oxidises oxalic acid to carbon dioxide.
\(2 \mathrm{MnO}_{4}^{-}\) + \(5 \mathrm{C}_{2} \mathrm{O}_{4}^{2-}\) + 16H⁺ → 2Mn²⁺ + 10CO₂ + 8H₂O

Question 27.
What are alloys? Name an important alloy which contains some of the lanthanoid metals. Mention its uses.
Answer:
An alloy is a homogeneous mixture of two or more metals, or metals and non-metals. An important alloy containing lanthanoid metals is misch metal which contains 95% lanthanoid metals and 5% iron along with traces of S, C, Ca and Al. It is used in Mg-based alloy to produce bullets, shells and lighter flints.

Question 28.
What are inner-transition elements? Decide which of the following atomic numbers are the atomic numbers of the inner-transition elements:
29, 59, 74, 95, 102, 104.
Answer:
The f-block elements, i.e., in which the last electron enters into /-subshell are called inner-transition elements. These include lanthanoids (58-71) and actinoids (90-103). Thus, elements with atomic numbers 59, 95 and 102 are inner-transition elements.

Question 29.
The chemistry of the actinoid elements is not so smooth as that of the lanthanoids. Justify this statement by giving some examples from the oxidation state of these elements.
Answer:
Lanthanoids show a limited number of oxidation state, viz., +2, +3 and +4 (out of which +3 is the most common). This is because of a large energy gap between 4/, 5d and 6s subshells. The dominant oxidation state of actinoids is also +3 but they show a number of other oxidation states also, e.g., uranium(Z = 92) and plutonium (Z = 94), show +3, +4, +5 and +6, neptunium (Z = 94) shows +3, +4, +5 and +7, etc. This is due to small energy difference between 5/, 6 d and 7s subshells of the actinoids.

Question 30.
Which is the last element in the series of the actinoids? Write the electronic configuration of this element. Comment on the possible oxidation state of this element.
Answer:
The last element in the actinoid series is lawrencium (Lr). Its atomic number is 103 and its electronic configuration is [Rn]5f¹⁴ 6d¹ 7s². The most common oxidation state displayed by it is + 3; because after losing 3 electrons it attains stable f¹⁴ configuration.

Question 31.
Use Hund’s rule to derive the electronic configuration of Ce³⁺ ion and calculate its magnetic moment on the basis of spin only formula.
Answer:
₅₈Ce =[Xe]⁵⁴f¹5d¹6s²
Ce³⁺ = [Xe]⁵⁴4f¹, i.e., there is only one unpaired electron, i.e., n = 1. Hence, p = \(\sqrt{n(n+2)}\) = \(\sqrt{1(1+2)}\) = \(\sqrt{3}\) = 1.73 BM

Question 32.
Name the members of the lanthanoid series which exhibit +4 oxidation states and those which exhibit +2 oxidation states. Try to correlate this type of behaviour with the electronic configurations of these elements.
Answer:
+4 = ₅₈Ce , ₅₉Pr, ₆₀Nd, ₆₅Tb, ₆₆Dy
+2 = ₆₀Nd, ₆₂Su ₆₃Eu, ₆₉Tm, ₇₀Yb
+2 oxidation state is exhibited when the lanthanoid has the configuration 5d⁰6s², so that 2 electrons are easily lost. +4 oxidation state is exhibited when the configuration left is close to 4f⁰ (e.g., 4f⁰, 4f¹, 4f²) or close to 4f⁷ (e.g., 4f⁷or 4f⁸)

Question 33.
Compare the chemistry of the actinoids with that of lanthanoids with reference to:
(i) electronic configuration
(ii) oxidation states and
(iii) chemical reactivity.
Answer:

Characteristics	Lanthanoids	Actinoids
(i) Electronic configuration	[Xe] 4/1-145d0-16s2	[Rn] 5/1“14 6d0_1 7s2
(ii) Oxidation states	Besides +3 oxidation state lanthanoids show +2 and +4 oxidation state only in a few cases.	Besides +3 oxidation state, actinoids show higher oxidation state of +4, +5, +6, +7 also because of smaller energy gap between 5/, 6d and 7s subshell.
(iv) General chemical reactivity elements	These are less reactive metals. Lesser tendency towards complex formation. Do not form oxacation. Compounds are less basic.	These are highly reactive metals Greater tendency towards complex formation. Form oxocation.Compounds are more basic.

Question 34.
Write the electronic configurations of the elements with the atomic numbers 61, 91,101 and 109.
Answer:
Z = 61 (Promethium, Pm), electronic configuration [Xe] 4f⁵5d⁰6s²
Z = 91 (Protactium, Pa), electronic configuration = [Rn] 5f²6d¹s²
Z = 101 (Mendelevium, Md), electronic configuration = [Rn] 5f¹³6d⁰7s²
Z =109 (Meitnerium, Mt), electronic configuration = [Rn] 5f¹⁴6d⁷ 7s²

Question 35.
Compare the general characteristics of the first series of the transition metals with those of the second and third series metals in the respective vertical columns. Give special emphasis on the following points :
(i) Electronic configurations,
(ii) Oxidation states,
(iii) Ionisation enthalpies and
(iv) Atomic sizes
Answer:
(i) Electronic configurations : The elements in the same vertical column generally have similar electronic configurations. Although the first series shows only two exceptions, i.e., Cr = 3d⁵4s¹ and Cu = 3d¹⁰4s¹ but the second series shows more exceptions, e.g., Mo(42) = 4ds⁵s¹,Tc(43) = 4d⁶5s¹,Ru(44) = 4d⁷5s¹ Rh(45) = 4d⁸5s¹, Pb(46) = 4d¹⁰5s⁰, Ag(47) = 4d¹⁰5s¹. Similarly, in the third series, W(74) = 5d⁴6s², Pt(78) = 5d⁹6s¹ and Au(79) = 5d¹⁰ 6d¹. Hence, in the same vertical column, in a number of cases, the electronic configuration of the three series are not similar.

(ii) Oxidation states : The elements in the same vertical column generally show similar oxidation states. The number of oxidation states shown by the elements in the middle of each series is maximum and minimum at the extreme ends.

(iii) Ionisation enthalpies : The first ionisation enthalpies in each series generally increase gradually as we move from left to right though some exceptions are observed in each series. The first ionisation enthalpies of some elements in the second (4d) series are higher while some of them have lower value than the elements of 3d series in the same vertical column. However, the first ionisation enthalpies of third (5d) series are higher than those of 3d and 4d series. This is because of weak shielding of nucleus of 4/-electrons in the 5d-series.

(iv) Atomic sizes : Generally, ions of the same charge or atoms in a given series show progressively decrease in radius with increasing atomic number though the decrease is quite small. But the size of the atoms of the 4d-series is larger than the corresponding elements of the 3d-series whereas those of corresponding elements of the 5d-series are nearly the same as those of 4d-series due to lanthanoid contraction.

Question 36.
Write down the number of 3d electrons in each of the following ions:
Ti²⁺, V²⁺, Cr³⁺, Mn²⁺, Fe²⁺, Fe³⁺, Co²⁺, Ni²⁺ and Cu²⁺. Indicate how would you expect the five 3d-orbitals to be occupied for these hydrated ions (octahedral).
Answer:

Question 37.
Comment on the statement that elements of the first transition series possess many properties different from those of heavier transition elements.
Answer:
The given statement is true. Some evidences in support of this statement are given below :

1. Atomic radii of the heavier transition elements (4 d and 5d series) are larger than those of the corresponding elements of the first transition series though those of 4d and 5d series are very close to each other.
2. Ionisation enthalpies of 5d series are higher than the corresponding elements of 3d and 4d series.
3. Enthalpies of atomisation of 4d and 5d series are higher than the corresponding elements of first series.
4. Melting and boiling points of heavier transition elements are greater than those of the first transition series due to stronger intermetallic bonding.

Question 38.
What can be inferred from the magnetic moment values of the following complex species?

Example	Magnetic Moment (BM)
K4[Mn(CN)6]	2.2
[Fe(H2O)6]2+	5.3
K2[Mncl4]	5.9

Answer:
Magnetic moment (μ) = \(\sqrt{n(n+2)}\) BM
For n = 1, μ = \(\sqrt{1(1+2)}\) = \(\sqrt{3}\) = 1.73;
For n = 2, μ = \(\sqrt{2(2+2)}\) = \(\sqrt{8}\) = 2.83;
For n = 3, μ = \(\sqrt{3(3+2)}\) = \(\sqrt{15}\) = 3.87;
For n = 4, μ = \(\sqrt{4(4+2)}\) = \(\sqrt{24}\) = 4.90;
For n = 5, μ = \(\sqrt{5(5+2)}\) = \(\sqrt{35}\) = 5.92
K₄[Mn(CN)₆]
Here, Mn is in +2 oxidation, state, i.e., as Mn²⁺. μ = 2.2 BM shows it has only one unpaired electron.
Hence, when CN^– ligands approach Mn2+ion, the electrons in 3d pair up
Hence, CN^– is a strong ligand. The hybridisation involved is d²sp³ forming inner orbital octahedral complex.

[Fe(H₂O)₆]²⁺
Here, Fe is in +2 oxidation state, i.e., as Fe²⁺. μ = 5.3 BM shows that there are four unpaired electrons. This means that the electrons in 3d do not pair up when the ligand H₂O molecules approach. Hence, H₂O is a weak ligand. To accommodate the electrons donated by six H20 molecules, the hybridisation will be sp³d².
Hence, it will be an outer orbital octahedral complex.

K₂[MnCl₄]
Here, Mn is in +2 oxidation state, i.e., as Mn²⁺. μ = 5.9 BM shows that there are five unpaired electrons. Hence, the hybridisation involved will be sp³ and the complex will be tetrahedral.

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

Question 1.
Silver atom has completely filled d-orbitals (4d¹⁰) in its ground state. How can you say that it is a transition element?
Answer:
Silver (Z = 47) can show +2 oxidation state in which it has incompletely filled d-subshell (4d⁹ configuration). So, silver is a transition element.

Question 2.
In the series Sc (Z = 21) to Zn (Z = 30), the enthalpy of atomisation of zinc is the lowest i.e., 126 kJ mol^-1. Why?
Answer:
In the first series of transition elements Sc to Zn, all elements have one or more unpaired electrons except zinc which has no unpaired electron as its outer electronic configuration is 3d¹⁰4s². Hence, interatomic metallic bonding (M-M bonding) is weaker in zinc. Therefore, enthalpy of atomisation is lowest.

Question 3.
Which of the 3d series of the transition metals exhibits the largest number of oxidation states and why?
Answer:
Mn (atomic number = 25) has electronic configuration[Ar] 3d⁵ 4s².
Mn has the maximum number of unpaffed electrons present in the d-subshell (5 electrons). Hence, Mn exhibits the largest number of oxidation states, i.e., + 2 to + 7 in its compounds.

Question 4.
The E^⊖(M²⁺/M) value for copper is positive (+0.34 V). What is possible reason for this? (Hint: Consider its high △_aH^⊖ and low △_hydH^⊖)
Answer:
E (M²⁺/M) for any metal is related to the sum of the enthalpy change taking place in the following steps:
M(s) + △_aH → M(g), (△_aH = Enthalpy of atomisation)
M(g) + △_iH → M²⁺(g) (△_iH = Ionisation enthalpy)
M²⁺(g) + aq → M²⁺(aq) + △_hydH
(△_hydH = Hydration enthalpy)
Copper has high enthalpy of ionisation and relatively low enthalpy of hydration. So, E^⊖(Cu²⁺/Cu) is positive. The high energy to transform Cu(s) to Cu²⁺(aq) is not balanced by its hydration enthalpy.

Question 5.
How would you account for the irregular variation of ionisation enthalpies (first and second) in the first series of the transition elements?
Answer:
Irregular variation of ionisation enthalpies is mainly attributed to varying degree of stability of different 3d configuration (e.g., d⁰, d⁵, d¹⁰ are exceptionally stable).

Question 6.
Why is the highest oxidation state of a metal exhibited in its oxide or fluoride only?
Answer:
Oxygen and fluorine have small size and high electronegativity. Hence, they can oxidise the metal to the highest oxidation state.

Question 7.
Which is a stronger reducing agent Cr²⁺ or Fe²⁺ and why?
Answer:
The following reactions are involved when Cr²⁺ and Fe²⁺ act as reducing agents.

The value is -0.41 V and value is +0.77 V. This means that Cr²⁺ can be easily oxidised to Cr³⁺, but Fe²⁺ does not get oxidised to Fe³⁺ easily. Therefore, Cr²⁺ is a better reducing agent that Fe²⁺.

Question 8.
Calculate the spin only magnetic moment of M²⁺ (aq) ion (Z = 27).
Answer:
Electronic configuration of M atom with Z = 27 is [Ar] 3d⁷ 4s².
∴ Electronic configuration of M²⁺ = [Ar]3d⁷, i.e.,

Hence, it has three unpaired electrons.
∴ Spin only magnetic moment (μ) = \(\sqrt{n(n+2)}\)
= \(\sqrt{3(3+2)}\)
= \(\sqrt{15}\)
= 3.87 BM

Question 9.
Explain why Cu⁺ ion is not stable in aqueous solutions.
Answer:
In aqueous solutions Cu⁺ undergoes disproportionation to form a more stable Cu²⁺ ion.
2Cu⁺(aq) > Cu²⁺(aq) + Cu(s)
The higher stability of Cu²⁺ in aqueous solutions may be attributed to its greater negative △_hyd.H⁰ than that of Cu⁺. It compensates the second ionisation enthalpy of Cu⁺ involved in the formation of Cu²⁺ ions.

Question 10.
Actinoid contraction is greater from element to element than lanthanoid contraction. Why?
Answer:
In actinoids, 5f orbitals are filled. These 5f orbitals have a poorer shielding effect than 4f orbitals (in lanthanoids). Thus, the effective nuclear charge experienced by electrons in valence shells in case of actinoids is much more than that experienced by lanthanoids. Hence, the size contraction in actinoids is greater as compared to that in lanthanoids.

Punjab State Board PSEB 6th Class Hindi Book Solutions Chapter 4 इंद्रधनुष Textbook Exercise Questions and Answers.

PSEB Solutions for Class 6 Hindi Chapter 4 इंद्रधनुष (2nd Language)

Hindi Guide for Class 6 PSEB इंद्रधनुष Textbook Questions and Answers

इंद्रधनुष अभ्यास

1. नीचे गुरुमुखी और देवनागरी लिपि में दिये गये शब्दों को पढ़ें और हिंदी शब्दों को लिखने का अभ्यास करें:

• ਰੰਗ = रंग
• ਕਿਰਨਾਂ = किरणें
• ਖੁਸ਼ੀ = खुशी

उत्तर :
विद्यार्थी इन हिन्दी शब्दों को अपनी अभ्यास पुस्तिका (कॉपी) में लिखने का अभ्यास करें।

2. नीचे एक ही अर्थ के लिए पंजाबी और हिंदी भाषा में शब्द दिये गये हैं। इन्हें ध्यान से पढ़ें और हिंदी शब्दों को लिखें :

• ਬੱਦਲ = मेघ
• ਸਤਰੰਗੀ ਪੀਂਘ = इन्द्रधनुष
• ਅਕਾਸ਼ = नभ
• ਧਰਤੀ = धरा
• ਛੇਕ = छेद
• ਵਿਹੜਾ = आँगन
• ਵਰਖਾ, ਮੀਂਹਿ = वर्षा
• ਕੁਦਰਤ = प्रकृति
• ਅਚਾਨਕ = सहसा
• ਝੂਲਾ = झूला

उत्तर :
विद्यार्थी दिए गए हिन्दी भाषा के शब्दों को अपनी अभ्यास – पुस्तिका (कापी) में लिखने का अभ्यास करें।

3. निम्नलिखित प्रश्नों के उत्तर एक या दो वाक्यों में लिखें :

(क) कविता में बादलों का रंग कैसा बताया गया है?
उत्तर :
कविता में बादलों का रंग काला बताया गया है।

(ख) वर्षा के बाद प्रकृति कैसी दिखाई देती है?
उत्तर :
वर्षा के बाद प्रकृति हरी – भरी दिखाई देती है।

(ग) वर्षा के बाद सूर्य दिखाई देने पर नभ पर क्या दिखाई देता है?
उत्तर :
वर्षा के बाद सूर्य दिखाई देने पर नभ पर इन्द्रधनुष दिखाई देता है।

(घ) इन्द्रधनुष का आकार कैसा होता है?
उत्तर :
इन्द्रधनुष का आकार झूले जैसा होता है।

(ङ) कवि ने इन्द्रधनुष के लिए अन्य कौन-सा शब्द प्रयोग किया है?
उत्तर :
कवि ने इन्द्रधनुष के लिए ‘सतरंगा’ शब्द प्रयोग किया है।

4. निम्नलिखित प्रश्नों के उत्तर चार या पाँच वाक्यों में लिखें :

(क) सावन के महीने में प्रकृति हरी-भरी क्यों दिखाई देती है?
उत्तर :
सावन के महीने में आकाश में बादल उमड़ – घुमड़ कर आते हैं और खूब वर्षा करते हैं जिससे सारी प्रकृति हरी – भरी दिखने लगती है।

(ख) इस माह की अन्य क्या विशेषताएँ हैं?
उत्तर :
सावन के महीने में आकाश घने काले बादलों से ढक जाता है। काले बादलों को देखकर बच्चे, बूढ़े तथा युवा सभी खुशियों से भर कर नाचने लगते हैं। उनके चेहरे खुशियों से खिल उठते हैं। गर्मी से राहत मिलती है। मोर नाचने लगता है, मेंढक टर्राने लगते हैं। सारी धरती हरी भरी हो उठती है और आकाश पर इन्द्रधनुष दिखने लगता है।

(ग) वर्षा ऋतु से हमें मुस्कराते रहने का क्या संदेश मिलता है?
उत्तर :
वर्षा ऋतु में उमड़ घुमड़ कर आए काले बादल हमें संदेश देते हैं कि हमेशा मुसकराते रहो और अपने चारों ओर खुशियाँ बाँटते चलो।

5. इन्द्रधनुष के चित्र को देखो और नीचे दिए गये शब्दों के पर्याय ढूँढ़कर लिखें :

1. प्रभु = ____________, ____________
2. मेघ = ____________, ____________
3. नभ = ____________, ____________
4. किरण = ____________, ____________

उत्तर :

1. प्रभु, परमात्मा, ईश्वर।
2. मेघ, जलद, बादल।
3. नभ, आकाश, गगन।
4. किरण, कर, मयूख।

6. कविता की पंक्तियाँ पूरी करो

(क) वर्षा थमी धरा महकी,
________________________।
(ख) नभ पर रंगों का मेला,
________________________।
(ग) हृदय हार नभ रानी का,
________________________।
उत्तर :
(क) वर्षा थमी धरा महकी,
प्रकृति दिखती हरी भरी।
(ख) नभ पर रंगों का मेला,
अर्ध वृत्त जैसा फैला।
(ग) हृदय हार नभ रानी का,
मोहित मन हर प्राणी का।

7. पढ़ो, समझो और दो-दो नये शब्द लिखो :

• र् + य = र्य = सूर्य,
• र् + म = में = ______________
• इ + च = र्च = ______________
• र् + षा = र्षा = ______________
• प + र = प्र = प्रकृति,
• क् + र = क्र = ______________
• द् + र = द्र = ______________
• ब् + र = ब्र = ______________

उत्तर :

• र् + य = र्य = सूर्य, धैर्य।
• र + म = र्म = चर्म, गर्म।
• र + च = र्च = चर्च, खर्च।
• र् + षा = र्षा = वर्षा, हर्षा।
• प् + र = प्र = प्रकृति, प्रकाश।
• क + र = क्र = क्रय, विक्रय।
• द् + र = द्र = दरिद्र, द्रविड़।
• ब् + र = ब्र = ब्राज़ील, सब्र।

अध्यापन निर्देश :
1. अध्यापक विद्यार्थियों को बताये कि हलन्त ‘र’ अर्थात् स्वर रहित ‘र’ अपने से अगले व्यंजन के ऊपर (‘) लगाया जाता है। इसे रेफ (‘) कहते हैं। जैसे – र् + य – र्य (सूर्य, कार्य आदि)। इसी तरह अध्यापक यह भी बताये कि ‘र’ से पहले हलन्त व्यंजन (अ रहित व्यंजन) हो तो ‘र’ उसके नीचे लिखा जाता है और उसका हलन्त हट जाता है जैसे – प् + र – प्र (प्रकृति, प्रभु आदि) ‘र’ के इस रूप को पदेन कहते हैं।

2. यदि ‘र’ के बाद मात्रा सहित व्यंजन आता है तो ‘र’ मात्रा के बाद लगता है जैसे – वर्षा में ‘र’ आ की मात्रा पर लगा है। इसी तरह ‘ई’ की मात्रा तथा ‘ओ’ की मात्रा के बाद ‘र’ का प्रयोग होता है। जैसे – ‘पूर्वी’ में तथा ‘धर्मों’ में क्रमशः ‘ई’ और ‘ओ’ की मात्रा के बाद ‘र’ का प्रयोग हुआ है।

8. सोचिए और लिखिए :

(i) इन्द्रधनुष का चित्र बनायें और उसमें अध्यापक की मदद से या स्वयं रंग भरें।
उत्तर :
विद्यार्थी स्वयं प्रयास करें।

(ii) इन्द्रधनुष में कौन-कौन से रंग होते हैं?
उत्तर :
लाल, नीला, हरा, पीला, नारंगी, बैंगनी, जामुनी।

(iii) सावन के महीने का आनन्द लें। सावन के महीने की रोचक बातें अपनी अभ्यास-पुस्तिका में लिखें।
उत्तर :
विद्यार्थी स्वयं प्रयास करें।

(iv) यदि वर्षा न हो तो क्या होगा?
उत्तर :
यदि वर्षा न हो तो सारी धरती प्यासी हो जाएगी। पेड़ – पौधे सब सड़ और मर जाएंगे। नदियों में पानी भी नहीं होगा। मनुष्य और प्रकृति सब त्राहि – त्राहि कर उठेंगे।

(v) यदि वर्षा अधिक होगी तो क्या होगा?
उत्तर :
यदि वर्षा अधिक होगी तो नदी – नाले जल से लबालब भर जाएंगे। चारों ओर पानी ही पानी दिखाई देगा। बाढ़ आ जाएगी और लोग अपने सामान उठा कर भागते – छिपते दिखाई देंगे।

9. चित्र देखकर अपनी कल्पना से पाँच वाक्य लिखें :

1. ____________________
2. ____________________
3. ____________________
4. ____________________
5. ____________________
6. ____________________
7. ____________________

उत्तर :

1. वर्षा हो रही है।
2. बच्चे वर्षा में भीग रहे हैं।
3. लड़कियाँ सावन के झूले झूल रही हैं।
4. पक्षी वर्षा से भीगने से बचने के लिए छिप रहे हैं।
5. बच्चा वर्षा में नाच रहा है।
6. चारों ओर पानी ही पानी है।
7. आकाश से वर्षा की बूंदें धरती पर गिर रही हैं।

अध्यापन निर्देश :
इन्द्रधनुष को अंग्रेजी में Rainbow कहते हैं।
इन्द्रधनुष के रंग (हिंदी में)
बैंगनी, जामुनी (गहरा नीला), नीला, हरा,
पीला, नारंगी (संतरी), लाल
इन्द्रधनुष के रंग (हिंदी में ) याद रखने
का संक्षिप्त रूप :
बैंजानीहपीनाला
इन्द्रधनुष के रंग (अंग्रेज़ी में)
Violet, Indigo, Blue, Green Yellow, Orange, Red
संक्षिप्त रूप : VIBGYOR

योग्यता विस्तार
वर्षा के दिनों में इन्द्रधनुष ऐसे समय बनता है जब सूरज पश्चिम में डूब रहा होता है। तब यह पूर्व की ओर बनता है। वैसे इसके बनने के लिए वर्षा के साथ-साथ साफ आकाश और सूरज की रोशनी का होना भी जरूरी है। हवा में मौजूद पानी की नन्ही-नन्ही बूंदों पर पड़ती सूर्य किरणों से हमें सात रंगों की छटा का बहुत ही सुंदर कुदरती घेरा दिखायी देता है।

सहायक क्रिया
1. रोशनी के आगे पारदर्शी पेन को सामने रखकर इन्द्रधनुष के सभी रंगों का अवलोकन करें।
2. विज्ञान की प्रयोगशाला से प्रिज्म लेकर (रोशनी में) इन्द्रधनुषी रंगों का अवलोकन करें।

बहुविकल्पीय प्रश्न

प्रश्न 1.
इन्द्रधनुष का आकार कैसा होता है ?
(क) झूले जैसा
(ख) फूले जैसा
(ग) फूल जैसा
(घ) धूल जैसा।
उत्तर :
(क) झूले जैसा

प्रश्न 2.
कवि ने इन्द्रधनुष के लिए किस शब्द का प्रयोग किया है ?
(क) इन्द्रा
(ख) इन्द्री
(ग) सतरंगा
(घ) सतरंगी।
उत्तर :
(ग) सतरंगा

प्रश्न 3.
इन्द्रधनुष में कितने रंग होते हैं ?
(क) पाँच
(ख) छह
(ग) सात
(घ) आठ।
उत्तर :
(ग) सात

प्रश्न 4.
इन्द्रधनुष में कौन-कौन से रंग होते हैं ?
(क) लाल-नीला
(ख) हरा-पीला
(ग) नारंगी, बैंगनी, जामुनी
(घ) ये सभी।
उत्तर :
(घ) ये सभी।

इंद्रधनुष Summary in Hindi

इंद्रधनुष कविता का सार

आकाश में घने काले बादल उमड़ आए। वे बरसे और ऐसा लगा जैसे आकाश में लाखों छेद हो गए हों। वर्षा रुकते ही हरी – भरी प्रकृति शोभा देने लगी। सूर्य के प्रकट होते ही सुनहरी किरणें फैली। सात रंगों का प्यारा इन्द्रधनुष प्रकट हो गया जिसने अपनी सुन्दरता से सभी का दिल जीत लिया। यह तो परमात्मा का झूला है। जब वर्षा ऋतु मुस्काती है तो वह सतरंगी हो जाती है।

पद्यांशों के सरलार्थ

1. उमड़े बरसे काले मेघ,
नभ में जैसे लाखों छेद।
वर्षा थमी धरा महकी,
प्रकृति दिखती हरी भरी।

कठिन शब्दों के अर्थ – उमड़े – घिर कर आना, फैल जाना। बरसे – बरसना। मेघ बादल। थमी – रुकी। महकी – सुगन्ध से भर गई। प्रकृति – कुदरत।

प्रसंग – प्रस्तुत पंक्तियाँ हमारी हिन्दी की पाठ्यपुस्तक ‘आओ हिन्दी सीखें’ की कविता ‘इन्द्रधनुष’ से ली गई हैं। इसमें कवि ने इन्द्रधनुष की सुन्दरता का वर्णन किया है।

सरलार्थ – कवि कहता है कि उमड – घुमड कर काले बादल आकाश पर छा गए हैं। उनसे वर्षा ऐसे होने लगी जैसे आकाश में लाखों छेद एक साथ हो गए हों। वर्षा के रुक जाने पर सारी धरती से सौंधी – सी महक आने लगी और सारी कुदरत हरी – भरी दिखाई देने लगी।

भावार्थ – कवि ने वर्षा के बाद धरती की शोभा का वर्णन किया है।

2. सहसा सूर्यदेव आये,
सोने – सी किरणें लाये।
नभ पर रंगों का मेला,
अर्ध – वृत्त जैसा फैला।

कठिन शब्दों के अर्थ – सहसा – अचानक। सूर्यदेव – सूरज देवता। नभ – आकाश। अर्ध – आधा। वृत्त – गोला।

प्रसंग – प्रस्तुत पंक्तियाँ हमारी हिन्दी पाठ्यपुस्तक ‘आओ हिन्दी सीखें’ की कविता ‘इन्द्रधनुष’ से ली गई हैं। इसमें कवि ने इन्द्रधनुष की सुन्दरता का वर्णन किया है। इन पंक्तियों में कवि वर्षा के बाद का दृश्य बताते हुए कहता है

सरलार्थ – कवि कहता है कि वर्षा रुकते ही अचानक आकाश पर सूरज देवता आ गए और वह अपने साथ सोने जैसी सुनहरे रंग की किरणें लाए। चारों तरफ आकाश में सुन्दर किरणें, फैल गई। किरणों के फैलते ही आकाश में आधे गोलाकार में इन्द्रधनुष दिखने लगा और ऐसा लगने लगा जैसे आकाश में रंगों का मेला लग गया हो।

भावार्थ – बरसात के बाद सूर्य निकलते ही इन्द्र धनुष की सुन्दरता आकाश में बिखर गई।

3. यह है इन्द्रधनुष प्यारा,
सतरंगा न्यारा – न्यारा।
हृदय – हार नभ रानी का,
मोहित मन हर प्राणी का।

कठिन शब्दों के अर्थ – सतरंगा – सात रंगों का। न्यारा – निराला, अद्भुत। हृदय हार – गले का हार। नभ – रानी – आकाश की रानी। मोहित – आकर्षित। हर – सभी। प्राणी – जीव, लोग।

प्रसंग – प्रस्तुत पंक्तियाँ हमारी हिन्दी पाठ्यपुस्तक ‘आओ हिन्दी सीखें’ की कविता ‘इन्द्रधनुष’ से ली गई हैं। इसमें कवि ने इन्द्रधनुष की सुन्दरता का वर्णन किया है।

सरलार्थ – कवि कहता है कि आकाश में सात रंगों से रंगा हुआ अद्भुत – सा दिखाई देने वाला रंग – बिरंगा यह इन्द्रधनुष है। यह इन्द्रधनुष इतना सुन्दर है कि लगता है जैसे आकाश की रानी के गले का सुन्दर हार हो। कवि कहता है कि यह इतना सुन्दर दृश्य है कि सभी लोगों के मन को मोह लेता है।

भावार्थ – इन्द्रधनुष की सुन्दरता मनभावन है।

4. सुन्दर यह प्रभु का झूला,
देख – देख कर मन फूला।
झूला प्रभु के आंगन में,
किरणें झूलें सावन में।

कठिन शब्दों के अर्थ – फूला – प्रसन्न हुआ। प्रभु – ईश्वर।

प्रसंग – प्रस्तुत पंक्तियाँ हमारी हिन्दी पाठ्यपुस्तक ‘आओ हिन्दी सीखें’ की कविता ‘इन्द्रधनुष’ से ली गई हैं। इसमें कवि ने वर्षा के पश्चात् आकाश में दिखने वाले इन्द्रधनुष की सुन्दरता का वर्णन किया है।

सरलार्थ – कवि कहता है कि इन्द्रधनुष की सुन्दरता को देखकर मन प्रसन्न हो जाता है। ऐसा लगता है जैसे ईश्वर ने इस सुन्दर झूले को अपने आंगन में डाल रखा है और सावन के मौसम में सूर्य की सुन्दर किरणें इस पर झूला झूलती हैं।

भावार्थ – इन्द्रधनुष तो परमात्मा का झूला प्रतीत होता है।

5. वर्षा ऋतु मुस्काती है,
सतरंगी हो जाती है।
आओ हम भी मुस्काएँ,
रंग खुशी से बिखराएँ।

प्रसंग – यह पद्यांश हिन्दी की पाठ्यपुस्तक आओ हिन्दी सीखें’ में संकलित ‘इन्द्रधनुष’। कविता से लिया गया है। इसमें कवि ने वर्षा के पश्चात् आकाश में दिखने वाले इन्द्रधनुष : की सुन्दरता का वर्णन किया है। कवि कहता है

सरलार्थ – कवि कहता है कि इन्द्रधनुष को देखकर वर्षा ऋतु भी मानो मुसकाने लगती है और उसके रंगों से यह भी सतरंगी हो जाती है। कवि कहता है कि आओ हम भी इन्द्रधनुष के समान जीवन में मुसकाएँ और खुशियों के रंग बिखराएँ।

भावार्थ – कवि ने जीवन में मुसकान बिखेरने का आह्वान किया हैं।

Punjab State Board PSEB 6th Class Hindi Book Solutions Chapter 3 जय जवान! जय किसान! Textbook Exercise Questions and Answers.

PSEB Solutions for Class 6 Hindi Chapter 3 जय जवान! जय किसान! (2nd Language)

Hindi Guide for Class 6 PSEB जय जवान! जय किसान! Textbook Questions and Answers

अभ्यास 

1. नीचे गुरुमुखी और देवनागरी लिपि में दिये गये शब्दों को पढ़ें और हिंदी शब्दों को लिखने का अभ्यास करें:

• ਜਨਮ = जन्म
• ਯਾਤਰਾ = यात्रा
• ਵਿਅਕਤੀ = व्यक्ति
• ਪੁੱਤਰ = पुत्र
• ਸ਼ਕਤੀ = शक्ति
• ਪੁੱਤਰੀ = पुत्री
• ਗੰਗਾ = गंगा
• ਸਮਾਪਤ = समाप्त
• ਸਮਾਂ = समय
• ਲਿਖਤੀ = लिखित
• ਅੱਖਾਂ = आँखों
• ਪਰੀਖਿਆ = परीक्षा

उत्तर :
विद्यार्थी हिन्दी में दिए गए शब्दों को समझें और उन्हें अपनी उत्तर पुस्तिका (कॉपी) पर लिखने का अभ्यास करें।

2. नीचे एक ही अर्थ के लिए पंजाबी और हिन्दी भाषा में शब्द दिये गये हैं। इन्हें ध्यान से पढ़ें और हिंदी शब्दों को लिखें :

• ਇੱਜ਼ਤ = सम्मान
• ਪੱਕਾ ਇਰਾਦਾ = दृढ़ निश्चय ग्ला
• ਇੱਕ ਕਲਾਸ ਵਿੱਚ ਪੜ੍ਹਨ ਵਾਲੇ = सहपाठी
• ਹਾਲਾਤਾਂ = परिस्थितियाँ
• ਉੱਚੀ ਪਦਵੀ = उच्च पद
• ਧੰਨ ਦੀ ਕਮੀ = आर्थिक तंगी
• ਤਰਲੇ ਪਾਉਣਾ = गुहार लगाना
• ਮੰਤਵ, ਉਦੇਸ਼ = उद्देश्य
• ਮੁਸ਼ਕਿਲ = विकट
• ਜਲ = कारावास

उत्तर :
विद्यार्थी हिन्दी में दिए गए शब्दों को ध्यान से पढ़ें और इन्हें अपनी अभ्यास पुस्तिका (कॉपी) पर लिखने का अभ्यास करें।

3. निम्नलिखित प्रश्नों के उत्तर एक वाक्य में दें :

(क) ‘जय जवान ! जय किसान!’ का नारा किसने दिया?
उत्तर :
‘जय जवान ! जय किसान !’ का नारा श्री लाल बहादुर शास्त्री जी ने दिया था।

(ख) लाल बहादुर शास्त्री जी का जन्म कब और कहाँ हुआ?
उत्तर :
श्री लाल बहादुर शास्त्री जी का जन्म 2 अक्तूबर, सन् 1904 ई० को बनारस के कस्बे मुगलसराय में हुआ।

(ग) जेल से आने के बाद लाल बहादुर जी ने कौन-सी पढ़ाई पूरी की?
उत्तर :
जेल से आने के बाद लाल बहादुर जी ने शास्त्री की पढ़ाई पूरी की।

(घ) पुत्री के बीमार होने पर उन्हें कितने दिन के लिए रिहा किया गया?
उत्तर :
पुत्री के बीमार होने पर उन्हें पन्द्रह दिनों के लिए रिहा किया गया।

(ङ) शास्त्री जी का देहान्त कब हुआ?
उत्तर :
11 जनवरी, सन् 1966 ई० को शास्त्री जी का देहान्त हुआ।

(च) शास्त्री जी अपने सिद्धांतों से समझौता करने को क्या समझते थे?
उत्तर :
शास्त्री जी अपने सिद्धान्तों से समझौता करना आत्मघातक समझते थे।

4. निम्नलिखित प्रश्नों के उत्तर तीन-चार वाक्यों में लिखें :

(क) शास्त्री जी अपने मित्रों से उधार क्यों नहीं माँगना चाहते थे?
उत्तर :
बचपन से ही अपने बनाए सिद्धान्तों पर चलने वाले शास्त्री जी अपने साथी से उधार मांगना अपने गौरव के विरुद्ध समझते थे इसीलिए वह अपने मित्रों से उधार नहीं मांगना चाहते थे।

(ख) लाल बहादुर शास्त्री जी ने लिखित शर्त पर जेल से छूटने से क्यों इन्कार किया?
उत्तर :
लाल बहादुर शास्त्री जी जेल में थे जब उन्हें अपनी बेटी के बीमार होने का समाचार मिला। सरकार ने उन्हें रिहा करने के लिए राजनैतिक आन्दोलन में भाग न लेने की लिखित शर्त रखी। इस पर इस देश के वीर सपूत ने लिखित शर्तों पर रिहा होने से इन्कार कर दिया।

(ग) मौत से जूझते पुत्र को छोड़ कर शास्त्री जी वापिस जेल क्यों चले गए?
उत्तर :
अंग्रेज़ सरकार ने शास्त्री जी को केवल एक सप्ताह के लिए ही जेल से रिहा किया था। टायफायड से ग्रस्त बेटे की सेवा करते हुए सप्ताह बीत गया। अब उन्हें जेल वापिस लौटना था अत: बेटे के लाख मना करने पर भी आज़ादी के मतवाले शास्त्री जी मौत से जूझते अपने पुत्र को छोड़कर वापिस जेल चले गए।

(घ) लाल बहादुर शास्त्री जी में कौन-से ऐसे गुण थे जिससे वे उच्च पद को प्राप्त कर सके?
उत्तर :
साहसी तथा स्वभाव से विनम्र, दृढ़ निश्चयी, लगन के पक्के, सिद्धान्तवादी जैसे गुणों के बल पर ही लाल बहादुर शास्त्री प्रधानमन्त्री जैसे उच्च पद को प्राप्त कर सके।

(ङ) लाल बहादुर शास्त्री जी के जीवन से हमें क्या प्रेरणा मिलती है?
उत्तर :
लाल बहादुर शास्त्री जी के जीवन से हमें प्रेरणा मिलती है कि अगर हमारा निश्चय पक्का है, संकल्प दृढ़ है, तो हम जीवन में कोई भी उच्च पद प्राप्त कर सकते हैं। जीवन में सफलता हमारे कदम चूमेगी।

5. नीचे दिये गये शब्दों में से उपयुक्त शब्द चुनकर रिक्त स्थान भरें :

निश्चय दूसरे एक पैसा टाइफाइड 11 जनवरी 1966 कारावास
(क) लाल बहादुर शास्त्री भारत के _______________ प्रधानमंत्री थे।
(ख) वे अपने _______________ पर दृढ़ रहने वाले व्यक्ति थे।
(ग) नाविक गंगा पार ले जाने का _______________ किराया लेता था।
(घ) नैनी _______________ के दौरान उन्हें अपनी पुत्री के बीमार होने का समाचार मिला।
(ङ) उनके पुत्र को _______________ हो गया था।
(च) उनका देहान्त _______________ को हुआ।
उत्तर :
(क) दूसरे,
(ख) निश्चय,
(ग) एक पैसा,
(घ) कारावास,
(ङ) टाइफाइड,
(च) 11 जनवरी, सन् 1966

7. (क) विपरीत शब्दों का मिलान करें :

उत्तर :
विपरीत शब्दों का सही मिलान

• साधारण = विशेष।
• बीमार = तन्दरुस्त।
• सम्मान = अपमान।
• परिश्रम = आलस्य।
• साहसी = कायर।
• गौरव = लाघव।
• आग्रह = दुराग्रह।
• जन्म = मृत्यु

(ख) नये शब्द बनाओ

1. प्रधान + मंत्री = प्रधानमंत्री
2. चिर + निद्रा = ___________________
3. अन्न + दाता = ___________________
4. सह + पाठी = ___________________
5. राष्ट्र + प्रेम = ___________________
6. आत्म + घातक = ___________________

उत्तर :

1. प्रधान + मंत्री = प्रधानमंत्री।
2. चिर + निद्रा = चिरनिद्रा।
3. अन्न + दाता = अन्नदाता।
4. सह + पाठी = सहपाठी।
5. राष्ट्र + प्रेम = राष्ट्रप्रेम।
6. आत्म + घातक = आत्मघातक।

(ग) लिंग बदलो

1. माता = _____________________
2. अध्यापक = _____________________
3. पुत्री = _____________________
4. बहन = _____________________
5. बाबू = _____________________

उत्तर :
लिंग बदलो

1. माता = पिता।
2. अध्यापक = अध्यापिका।
3. पुत्री = पुत्र।
4. बहन = भाई।
5. बाबू = बबुआइन।

8. संयुक्त अक्षरों से नये शब्द बनाओ

1. दृष्टि = कष्ट = _____________
2. शास्त्री = स्त्र = _____________
3. निश्चय = श्च = _____________
4. आत्म = त्म = _____________
5. आर्थिक = र्थ (र + थ) = _____________
6. इच्छा = च्छ = _____________

उत्तर :
संयुक्त अक्षरों से नये शब्द

1. दृष्टि = कष्ट, नष्ट।
2. शास्त्री = स्त्र = स्त्री, मिस्त्री।
3. निश्चय = श्च पश्चात्, पश्चाताप।
4. आत्म त्म = खत्म, आत्मा।
5. आर्थिक = र्थ (र् + थ) = दर्शनार्थ, अर्थ।
6. आन्दोलन = नन्द, आनन्द।

9. निम्नलिखित वाक्यों को ध्यानपूर्वक पढ़ें :

1. (क) पिता की मृत्यु के बाद लाल बहादुर शास्त्री पर भारी कष्ट आ पड़े। (साधारण तरीके से कही गई बात)
(ख) पिता की मृत्यु के बाद लाल बहादुर शास्त्री पर तो जैसे पहाड़ ही टूट पड़ा। (विशेष तरीके से कही गई बात)

2. (क) भारतीय जवान अपनी जान की परवाह न करके सीमा पर देश की रक्षा करते हैं। (साधारण तरीके से कही गई बात)
(ख) भारतीय जवान अपनी जान हथेली पर रखकर सीमा पर देश की रक्षा करते हैं। (विशेष तरीके से कही गई बात)

उपर्युक्त उदाहरणों में ‘क’ वाक्य साधारण तरीके से तथा ‘ख’ वाक्य विशेष तरीके से कहे गये हैं। इसी कारण ‘ख’ वाक्य ‘क’ वाक्यों की अपेक्षा अधिक सशक्त व प्रभावशाली हैं। इस प्रकार विशेष शब्द प्रयोग को मुहावरा कहते हैं। मुहावरे सीधी-साधी बात को अनोखे ढंग से प्रकट करते हैं। ये वाक्य के अंश होते हैं।

निम्नलिखित मुहावरों के अर्थ दिए गए हैं, इनके वाक्य बनायें :

मुहावरा – अर्थ = वाक्य = _______________

1. स्वर्ग सिधारना – मर जाना = _______________
2. गुदड़ी का लाल – निर्धन परिवार में जन्मा गुणी व्यक्ति = _______________
3. जिगर का टुकड़ा – बहुत प्यारा = _______________
4. जीवन लीला समाप्त होना – मर जाना = _______________
5. भरे मन से विदा लेना – दु:खी मन से जाना = _______________
6. चिर निद्रा में सोना – मृत्यु को प्राप्त होना = _______________

उत्तर :

1. स्वर्ग सिधारना – मर जाना – मोहन के पिता जी कल स्वर्ग सिधार गए।
2. गुदड़ी का लाल – निर्धन परिवार में जन्मा गुणी व्यक्ति – लाल बहादुर शास्त्री वास्तव में गुदड़ी के लाल थे।
3. जिगर का टुकड़ा – बहुत प्यारा – दिनेश ने अपने जिगर के टुकड़े को गले से लगा लिया।
4. जीवन लीला समाप्त होना – मर जाना – गरीबी से तंग आकर भिखारिन ने अपनी जीवन लीला समाप्त कर ली।
5. भरे मन से विदा लेना – दुखी मन से जाना – हमने भारी मन से विदा ली और चल पड़े।
6. चिर निद्रा में सोना – मृत्यु को प्राप्त होना – 11 जनवरी, सन् 1966 को शास्त्री जी चिर निद्रा में सो गए।

प्रयोगात्मक व्याकरण
(क) (1) लाल बहादुर शास्त्री भारत के दूसरे प्रधानमंत्री थे।
(2) उनका जन्म बनारस में हुआ।
(3) वे मेला देखने गंगा पार गये।
उत्तर :
व्यक्तिवाचक संज्ञा –
(1) लाल बहादुर शास्त्री।
(2) लाल बहादुर शास्त्री, राम दुलारी, शारदा प्रसाद।
(3) नैनी।

ऊपर लिखे पहले वाक्य में ‘लाल बहादुर शास्त्री’ किसी विशेष व्यक्ति के नाम का बोध कराता है। ‘भारत’ शब्द से देश विशेष का ज्ञान होता है। दूसरे वाक्य में ‘बनारस’ कहने से एक विशेष स्थान का ही विचार मन में आता है। इसी प्रकार तीसरे वाक्य में ‘गंगा’ शब्द से एक विशेष नदी अर्थात् गंगा नदी का बोध होता है।

अतः जिस शब्द से किसी विशेष व्यक्ति, स्थान या वस्तु का बोध हो, उसे व्यक्तिवाचक संज्ञा कहते हैं।

नीचे लिखे वाक्यों में से व्यक्तिवाचक संज्ञा छाँटिये :

(1) लाल बहादुर शास्त्री एक महान नेता थे।
(2) लाल बहादुर शास्त्री की माता का नाम राम दुलारी व पिता का नाम शारदा प्रसाद था।
(3) वे एक बार नैनी जेल गये।

(ख) (1) लाल बहादुर शास्त्री का जन्म साधारण परिवार में हुआ था।
(2) शास्त्री जी की पुत्री बीमार थी।
(3) उनका स्कूल गंगा पार था।
(4) शास्त्री जी अनेक बार जेल गये।

ऊपर के प्रथम वाक्य में परिवार’, दूसरे में ‘पुत्री’ और तीसरे में स्कूल’ और चौथे में ‘जेल’ ऐसे शब्द हैं जो किसी विशेष व्यक्ति, वस्तु या स्थान को सूचित नहीं करते।

• ‘परिवार’ किसी भी परिवार के लिए कहा जा सकता है।
• हरेक ‘पुत्री’ के लिए ‘पुत्री’ शब्द का प्रयोग किया जाता है।
• प्रत्येक ‘स्कूल’ को ‘स्कूल’ ही कहा जाता है।
• हरेक ‘जेल’ के लिए ‘जेल’ शब्द का ही प्रयोग होता है।

अर्थात ‘परिवार’ कहने से सभी परिवारों, ‘पुत्री’ कहने से सभी पुत्रियों, ‘स्कूल’ कहने से सभी स्कूलों तथा जेल कहने से सभी जेलों का बोध होता है, किसी एक का नहीं।

अतः ये शब्द किसी विशेष व्यक्ति, वस्तु या स्थान का बोध नहीं कराते अपितु ये शब्द पूरी जाति

(वर्ग) का ज्ञान कराते हैं। इसीलिए ये शब्द जातिवाचक संज्ञाएँ हैं।
अतः जिस शब्द से किसी व्यक्ति, वस्तु या स्थान की सम्पूर्ण जाति या वर्ग का बोध हो, उसे जातिवाचक संज्ञा कहते हैं।

निम्नलिखित में से जातिवाचक संज्ञाएँ छाँटिये :
(1) लाल बहादुर शास्त्री निश्चय पर दृढ़ रहने वाले व्यक्ति थे।
(2) भारतीय सेना ने दुश्मन को धूल चाटने पर मजबूर कर दिया।
(3) अंत में सरकार को स्वयं ही झुकना पड़ा।
उत्तर :
जातिवाचक संज्ञा शब्द
(1) व्यक्ति।
(2) भारतीय सेना, दुश्मन।
(3) सरकार।

नीचे दिये गए शब्दों में बने बनाए शब्द मिलेंगे। शब्द में से कम से कम दो शब्द ढूंढ़िए और लिखिए :

उत्तर :

• प्रधानमंत्री – प्रधान, धान, मंत्र, मंत्री।
• बनारस – बना, नार, रस।
• सरकार – सर, सरक, सरका, कार।
• लगातार – लग, लगा, लगाता, गात, तार।
• उदाहरण – दाह, हर, हरण।
• मतवाले – मत, तवा, वाले।
• देखकर – दे, देख, कर।
• विश्वास – विश्व, श्वास, वास।

बहुविकल्पीय प्रश्न

प्रश्न 1.
‘जय जवान! जय किसान !’ का नारा किसने दिया ?
(क) लाल बहादुर शास्त्री ने
(ख) लाला लाजपतराय
(ग) बालगंगाधर तिलक
(घ) नेहरू।
उत्तर :
(क) लाल बहादुर शास्त्री

प्रश्न 2.
शास्त्री जी का जन्म कब हुआ ?
(क) 2 अक्तूबर, 1903 ई० को
(ख) 2 अक्तूबर, 1904 ई० को
(ग) 2 अक्तूबर, 1905 ई० को
(घ) 2 अक्तूबर, 1908 ई० को।
उत्तर :
(ख) 2 अक्तूबर, 1904 ई० को

प्रश्न 3.
शास्त्री जी को पढ़ने के लिए क्या पार करना पड़ता था ?
(क) गंगा
(ख) यमुना
(ग) सरस्वती
(घ) गोदावरी।
उत्तर :
(क) गंगा

प्रश्न 4.
शास्त्री जी को किस आंदोलन में जेल जाना पड़ा ?
(क) असहयोग आंदोलन
(ख) सविनय अवज्ञा आंदोलन
(ग) भारत छोड़ो
(घ) देश छोड़ो।
उत्तर :
(क) असहयोग आंदोलन

प्रश्न 5.
शास्त्री जी की मृत्यु कब हुई ?
(क) 11 जनवरी, 1965 को
(ख) 11 जनवरी, 1966 को
(ग) 11 जनवरी, 1967 को
(घ) 11 जनवरी, 1968 को।
उत्तर :
(ख) 11 जनवरी 1966 को।

जय जवान ! जय किसान ! Summary in Hindi

जय जवान ! जय किसान ! पाठ का सार

जय जवान! जय किसान! का नारा देने वाले लाल बहादुर शास्त्री का जन्म 2 अक्तूबर, सन् 1904 ई० को बनारस के एक कस्बे मुग़लसराय में एक साधारण परिवार में हुआ। इनकी माता का नाम राम दुलारी था और इनके पिता शारदा प्रसाद जी एक अध्यापक थे। बचपन में ही इनके पिता स्वर्ग सिधार गए।

इतनी छोटी आयु में ही इन पर मुसीबतों का पहाड़ टूट पड़ा पर दृढ़ संकल्पी और साहसी इस वीर बालक ने मुश्किलों से हार न मानी। वे जीवन – पथ पर निरन्तर मुसीबतों से जूझते हुए आगे बढ़ते गए।

इनको पढ़ने के लिए गंगा – पार करके जाना पड़ता था। नाविक गंगा नदी को पार करवाने का एक पैसा किराया लेता था। इनके पास पैसा नहीं होता था तो वे रोज़ाना नदी को तैर कर ही पार कर लिया करते थे। कई बार इनके दोस्तों ने इनका किराया देना भी चाहा तो ये विनम्रता से उन्हें मना कर देते। उन्होंने अपने जीवन के कुछ सिद्धान्त बना लिए थे और उन्हीं सिद्धान्तों पर वे आजीवन चलते रहे।

सन् 1921 में जब असहयोग आन्दोलन शुरू हुआ तो इन्होंने भी अपनी पढ़ाई छोड़ कर इस आन्दोलन में भाग लिया। इस कारण इन्हें जेल भी जाना पड़ा। जेल से रिहा होकर इन्होंने अपनी ‘शास्त्री’ की पढ़ाई पूरी की और तभी से इनके नाम के साथ शास्त्री शब्द भी जुड़ गया।

देश की स्वतन्त्रता के लिए अपने परिवार का भी बलिदान कर देने वाले यह एकमात्र देशभक्त थे। एक बार यह कारावास में थे कि इन्हें अपनी पुत्री के सख्त बीमार होने की खबर मिली। सभी ने इन्हें पैरोल पर रिहा होकर पुत्री की देखभाल करने की सलाह दी। सरकार इन्हें राजनैतिक आन्दोलन में भाग न लेने की लिखित शर्त पर रिहा करने को तैयार थी। इन्होंने यह पेशकश ठुकरा दी।

तत्पश्चात् सरकार ने इन्हें बिना शर्त 15 दिनों के लिए रिहा तो कर दिया लेकिन तब तक बहुत देरी हो चुकी थी। जब यह अपनी बेटी के पास पहुँचे तो वह मृत्यु को प्राप्त हो चुकी थी। ऐसा एक बार फिर हुआ। यह जेल में ही थे कि इनका बेटा टायफायड का शिकार हो गया। अंग्रेज़ सरकार ने रिहा होने के लिए शर्त लगा दी। इन्होंने कोई शर्त न मानी तो आखिर अंग्रेज़ सरकार ने एक सप्ताह के लिए इन्हें रिहा कर दिया।

जब यह बेटे के पास पहुंचे तो वह 106 डिग्री बुखार से तड़प रहा था, उसके होंठ भी सूज गए थे। बेटे की देखभाल करते – करते सप्ताह बीत गया। इन्हें अब वापिस जाना था। बेटे ने रो रो कर कहा कि बाबू जी अभी मत जाइए। लेकिन शास्त्री जी ने भरे मन से बेटे से हाथ जोड़कर विदा ली और मातृभूमि की रक्षा के लिए जेल की तरफ चल पड़े।

सच में, ऐसे थे वे दृढ़ संकल्पी, भारत देश के वीर सपूत, जिन्होंने कभी भी अपने सिद्धान्तों से समझौता नहीं किया। भारत का यह वीर जवान 11 जनवरी, सन् 1966 ई० को चिरनिद्रा में सो गया। समस्त भारतीयों ने अश्रुपूरित नेत्रों से उन्हें भावभीनी विदाई दी। उनका नाम भारतीय इतिहास में सदा अमर रहेगा।

जय जवान ! जय किसान ! कठिन शब्दों के अर्थ –

• जान हथेली पर रखना = बलिदान को तैयार रहना, मरने की परवाह न करना।
• परिश्रम = मेहनत।
• अन्नदाता = अन्न देने वाले।
• विकास = उन्नति, तरक्की, प्रगति।
• सम्मान = इज्जत।
• स्वर्ग सिधारना = मर जाना।
• प्रबल = तेज़।
• पहाड़ टूटना = मुश्किलें आना।
• विनम्र = कोमल।
• नाविक = नाव चलाने वाला, मल्लाह।
• सहपाठियों = साथ पढ़ने वाले।
• पाट = किनारा।
• गौरव = बड़प्पन।
• अडिग = स्थिर।
• निराले = अद्भुत, सबसे अलग।
• कारावास – जेल।
• सिद्धान्तों = नियमों।
• जीवन लीला समाप्त होना = मारे जाना।
• डगमगाए = लड़खड़ाए।
• विकट = जटिल, बड़ी मुश्किल।
• जिगर का टुकड़ा = बेटा।
• मौत से जूझना = मृत्यु से लड़ना।
• आत्मघातक = अपनी हत्या आप करने वाला।
• दृढ़ = पक्का।
• निश्चय = इरादा।
• चिरनिद्रा में सोना = मर जाना।
• प्रेरणा स्रोत = प्रेरणा देने वाले।
• सदैव = हमेशा।

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 12th Class Chemistry Important Questions Chapter 6 General Principles and Processes of Isolation of Elements

Very Short Answer Type Questions

Question 1.
Zinc acts as a reducing agent in the extraction of silver. Comment.
Answer:
Zinc acts as a reducing agent in the extraction of silver. It reduces Ag⁺ to Ag and itself get oxidised to Zn²⁺.
2Na[Ag(CN)₂] + Zn → Na₂[Zn(CN)₄] + 2Ag↓

Question 2.
Winch reducing agent is employed to get copper from the leached low grade copper ore?
Answer:
Scrap iron, Cu²⁺(aq) + Fe(s) → Cu(s) + Fe²⁺(aq)
or H₂ gas, Cu²⁺(aq) + H₂(g) → Cu(s) + 2H⁺(aq)

Question 3.
Name the method used for refining of zirconium.
Answer:
Van Arkel method

Question 4.
Name the method that is used for refining of nickel.
Answer:
Mond process (Vapour phase refining)

Question 5.
Name the method used for refining of copper metal.
Answer:
Electrolytic refining

Question 6.
Although carbon and hydrogen are better reducing agents but they are not used to reduce metallic oxides at high temperatures. Why?
Answer:
At high temperature carbon and hydrogen react with metals to form carbides and hydrides respectively.

Question 7.
What is the function of collectors in the froth floatation process for the concentration of ores?
Answer:
Collectors (e.g., pine oil, xanthates etc.) enhance non-wettability of the ore particles.

Question 8.
Why is it that only sulphide ores are concentrated by froth floatation process?
Answer:
This is because the sulphide ore particles are preferentially wetted by oil and gangue particles are preferentially wetted by water.

Question 9.
At temperatures above 1073 K, coke can be used to reduce FeO to Fe. How can you justify this reduction with Ellingham diagram?
Answer:
Using Ellingham diagram, we observe that at temperature greater than 1073 K; △G(C, CO) < △G (Fe, FeO).
Hence, coke can reduce FeO to Fe.

Question 10.
The mixture of compounds A and B is passed through a column of Al₂O₃ by using alcohol as eluant. Compound A is eluted in preference to compound B. Which of the compounds A or B, is more readily adsorbed on the column?
Answer:
Since, compound ‘A’ comes out before compound ‘B’ the compound ‘B’ is more readily adsorbed on the column.

Short Answer Type Questions

Question 1.
Write the role of:
(i) I₂ in the van Arkel method of refining.
(ii) Dilute NaCN in the extraction of silver.
Answer:
(i) Impure titanium is heated with iodine to form volatile TiI₄, which decomposes on tungsten filament at high temperature to give pure titanium.

(ii) Dilute NaCN forms a soluble complex with Ag or Ag₂S while the impurities remain unaffected which are filtered off.
4Ag + 8NaCN + O₂ + 2H₂O → 4Na[Ag(CN)₂] + 4NaOH
or

Question 2.
Describe the role of
(i) Iodine in the refining of zirconium.
(ii) NaCN in the extraction of gold from gold ore.
Write chemical equations for the involved reactions.
Answer:
(i) Impure zirconium is heated with iodine to form volatile compound ZrI₄ which on further heating over tungsten filament decomposes to give pure zirconium.

(ii) Gold ore is leached with dilute solution of NaCN in the presence of air from which the metal is obtained later by replacement.
4Au + 8NaCN + O₂ + 2H₂O → 4Na[Au(CN)₂] + 4NaOH

Question 3.
Explain the role of each of the following in the extraction of metals from their ores:
(i) CO in the extraction of nickel.
(ii) Zinc in the extraction of silver.
Answer:
(i) CO in the extraction of nickel: Impure nickel is heated in a stream of carbon monoxide when volatile nickel tetracarbonyl is formed and the impurities are left behind in the solid state. The vapour of nickel tetracarbonyl is taken to a decomposer chamber maintained at 450-470 K where it decomposes to give pure nickel metal and carbon monoxide.

(ii) Zinc in the extraction of silver : Silver present in the ore is leached with dilute solution of NaCN in the presence of air or oxygen to form a soluble complex.

Silver is then recovered from the complex by displacement method using more electropositive zinc metal.
2[Ag(CN)₂]^– (aq) + Zn(s) → 2Ag(s) + [Zn(CN)₂]^2- (aq)

Question 4.
Wrought iron is the purest form of iron. Write a reaction used for the preparation of wrought iron from cast iron. How can the impurities of sulphur, silicon and phosphorus be removed from cast iron?
Answer:

This reaction takes place in reverberatory furnace lined with haematite.

(b) Limestone is added as flux. Impurities of S, Si and P oxidise and pass into slag. The metal is removed and freed from slag by passing through rollers.

Question 5.
Write the chemical reactions involved in the extraction of gold by cyanide process. Also give the role of zinc in the reaction.
Answer:
(i) 4Au(s) + 8CN^– (aq) + 2H₂O(aq) + O₂(g) → 4[Au(CN)₂]^– (aq) + 4OH^–(aq)
(ii) 2[Au(CN)₂]^– (aq) + Zn(s) → 2Au(s) + [Zn(CN)₄]^2- (aq)
Zinc acts as a reducing agent in this reaction.

Question 6.
Describe the role of
(i) NaCN in the extraction of gold from its ore.
(ii) Cryolite in the extraction of aluminium from pure alumina.
(iii) CO in the purification of nickel.
Answer:
(i) Gold is leached with a dilute solution of NaCN in the presence of air.
(ii) Cryolite lowers the high melting point of alumina and makes it a good conductor of electricity.
(iii) CO forms a volatile complex with metal nickel which is further decomposed to give pure Ni metal.

Long Answer Type Questions

Question 1.
(a) Explain how an element can be extracted using an oxidation reaction?
(b) What do you mean by refining? Mention some of the methods used for refining of metals.
Answer:
(a) Some of the extractions, particularly of non-metals are based upon oxidation.
A very common example of extraction based on oxidation is the extraction of chlorine from brine (Chlorine is abundant in sea water as common salt).
2Cl^–(aq) + 2H₂O(l) → 2OH^–(aq) + H₂(g) + Cl₂(g)
The △G⁰ for this reaction is + 422 kJ. When it is converted to E⁰ (using △G⁰ = -nE⁰F), we get E⁰ = -2.2 V. Naturally, it will require an external e.m.f. that is greater than 2.2 V. But the electrolysis requires an excess potential to overcome some other hindering reactions. Thus, Cl₂ is obtained by electrolysis giving out H₂ and aqueous NaOH as by products. Electrolysis of molten NaCl is also carried out. But in that case, Na metal is produced and not NaOH.

The extraction of gold and silver involves leaching the metal with CN^–. This is also an oxidation reaction (Ag → Ag⁺ or Au → Au⁺). The metal is later recovered by displacement method.
4Au(s) + 8CN^–(aq) + 2H₂O(aq) + O₂(g) → 4[Au(CN₂)]^–(aq) + 4OH(aq)
2[Au(CN)₂]^–(aq) + Zn(s) → 2Au(s) + [Zn(CN)₄]^2- (aq)
In this reaction zinc acts as a reducing agent.

(b) A metal extracted by any method is usually contaiminated with some impurity. For obtaining metals of high purity, several techniques are used depending upon the difference in properties of the metal and the impurity. The process is called refining. Some of them are listed below :

1. Distillation,
2. Liquation,
3. Electrolysis,
4. Zone-refining,
5. Vapour phase refining,
6. Chromatographic methods.

Question 2.
How is the concept of coupling reactions useful in explaining the occurrence of non-spontaneous thermochemical reactions? Explain giving an example?
Answer:
Coupled reactions : Many reactions which are non-spontaneous (△G^⊖ is positive) can be made to occur spontaneously if these are coupled with reactions having larger negative free energy. By coupling means carrying out simultaneously both non- spontaneous and spontaneous reactions. For example, decomposition of Fe₂O₃into iron is a non-spontaneous reaction (△G^⊖ = +1487 kJ mol^-1). However, this decomposition can take place spontaneously if carbon monoxide is simultaneously burnt in oxygen (△G^⊖ = – 514.4 kJ mol^-1).
2Fe₂O₃(s) → 4Fe(s) + 3O₂(g); …(i);
△G^⊖ = + 1487.0 kJmol^-1
2CO(g) + O₂(g) → 2CO₂(g); … (ii);
△G^⊖ = -514.4 kJmol^-1
Multiplying equation (ii) by 3 and then adding to equation (i), we get
6CO(g) + 3O₂(g) → 6CO₂(g) △G^⊖ = -1543.2 kJ mol^-1
2Fe₂O₃ (s) → 4Fe(s) + 3O₂(s) △G^⊖ = +1487.0 kJ mol^-1
2Fe₂O₃(s) + 6CO(g) → 4Fe(s) + 6CO₂(g) △G^⊖ = – 56.2 kJ mol^-1
Since, △G^⊖ in the reduction of Fe₂O₃ with CO is negative, therefore, the reaction is feasible and spontaneous.

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.