PSEB 8th Class Agriculture Book Solutions Guide in Punjabi English Medium

Punjab State Board Syllabus PSEB 8th Class Agriculture Book Solutions Guide Pdf in English Medium and Punjabi Medium are part of PSEB Solutions for Class 8.

PSEB 8th Class Agriculture Guide | Agriculture Guide for Class 8 PSEB in English Medium

Agriculture Guide for Class 8 PSEB | PSEB 8th Class Agriculture Book Solutions

PSEB 8th Class Agriculture Book Solutions Guide in English Medium

PSEB 8th Class Agriculture Book Solutions Guide in Hindi Medium

PSEB 8th Class Agriculture Book Solutions Guide in Punjabi Medium

  • Chapter 1 ਭੂਮੀ ਅਤੇ ਭੂਮੀ ਸੁਧਾਰ
  • Chapter 2 ਪਨੀਰੀਆਂ ਤਿਆਰ ਕਰਨਾ
  • Chapter 3 ਜ਼ਮੀਨ ਦੀ ਪੈਮਾਇਸ਼ ਤੇ ਰੀਕਾਰਡ
  • Chapter 4 ਸੂਰਜੀ ਊਰਜਾ
  • Chapter 5 ਖੁੰਬਾਂ ਦੀ ਕਾਸ਼ਤ
  • Chapter 6 ਮਧੂ ਮੱਖੀ ਪਾਲਣ
  • Chapter 7 ਬਹੁ-ਭਾਂਤੀ ਖੇਤੀ
  • Chapter 8 ਜੈਵਿਕ ਖੇਤੀ
  • Chapter 9 ਖੇਤੀ ਮਸ਼ੀਨਰੀ ਅਤੇ ਉਸਦੀ ਸਾਂਭ-ਸੰਭਾਲ
  • Chapter 10 ਫ਼ਲਾਂ ਅਤੇ ਸਬਜ਼ੀਆਂ ਦੀ ਸਾਂਭ-ਸੰਭਾਲ
  • Chapter 11 ਫ਼ਲ ਅਤੇ ਸਬਜ਼ੀਆਂ ਤੋਂ ਅਚਾਰ, ਮੁਰੱਬੇ ਅਤੇ ਸ਼ਰਬਤ ਬਨਾਉਣਾ

PSEB 10th Class Agriculture Book Solutions Guide in Punjabi English Medium

PSEB 10th Class Agriculture Book Solutions

Punjab State Board Syllabus PSEB 10th Class Agriculture Book Solutions Guide Pdf in English Medium and Punjabi Medium are part of PSEB Solutions for Class 10.

PSEB 10th Class Agriculture Guide | Agriculture Guide for Class 10 PSEB in English Medium

Agriculture Guide for Class 10 PSEB | PSEB 10th Class Agriculture Book Solutions

PSEB 10th Class Agriculture Book Solutions in English Medium

PSEB 10th Class Agriculture Book Solutions in Hindi Medium

PSEB 10th Class Agriculture Book Solutions Guide in Punjabi Medium

  • Chapter 1 ਖੇਤੀਬੜੀ ਸਹਿਯੋਗੀ ਸੰਸਥਾਵਾਂ
  • Chapter 2 ਖੇਤੀ ਗਿਆਨ-ਵਿਗਿਆਨ ਦਾ ਸੋਮਾ : ਪੰਜਾਬ ਐਗਰੀਕਲਚਰਲ ਯੂਨੀਵਰਸਿਟੀ
  • Chapter 3 ਹਾੜ੍ਹੀ ਦੀਆਂ ਫ਼ਸਲਾਂ
  • Chapter 4 ਸਰਦੀ ਦੀਆਂ ਸਬਜ਼ੀਆਂ ਦੀ ਕਾਸ਼ਤ
  • Chapter 5 ਫ਼ਲਦਾਰ ਬੂਟਿਆਂ ਦੀ ਕਾਸ਼ਤ
  • Chapter 6 ਖੇਤੀ ਜੰਗਲਾਤ
  • Chapter 7 ਆਰਥਿਕ ਵਿਕਾਸ ਵਿਚ ਖੇਤੀ ਦਾ ਯੋਗਦਾਨ
  • Chapter 8 ਖੇਤੀ ਆਧਾਰਿਤ ਉਦਯੋਗਿਕ ਧੰਦੇ
  • Chapter 9 ਤਸਦੀਕਸ਼ੁਦਾ ਬੀਜ ਉਤਪਾਦਨ
  • Chapter 10 ਫ਼ਸਲਾਂ ਲਈ ਲਾਭਦਾਇਕ ਅਤੇ ਹਾਨੀਕਾਰਕ ਜੀਵ
  • Chapter 11 ਪੌਦਾ ਰੋਗ ਨਿਵਾਰਨ ਕਲੀਨਿਕ

PSEB 9th Class Agriculture Book Solutions Guide in Punjabi English Medium

PSEB 9th Class Agriculture Book Solutions

Punjab State Board Syllabus PSEB 9th Class Agriculture Book Solutions Guide Pdf in English Medium and Punjabi Medium are part of PSEB Solutions for Class 9.

PSEB 9th Class Agriculture Guide | Agriculture Guide for Class 9 PSEB in English Medium

Agriculture Guide for Class 9 PSEB | PSEB 9th Class Agriculture Book Solutions

PSEB 9th Class Agriculture Book Solutions Guide in Punjabi Medium

  • Chapter 1 ਸਾਉਣੀ ਦੀਆਂ ਫ਼ਸਲਾਂ
  • Chapter 2 ਸਾਉਣੀ ਦੀਆਂ ਸਬਜ਼ੀਆਂ
  • Chapter 3 ਫੁੱਲਾਂ ਦੀ ਕਾਸ਼ਤ
  • Chapter 4 ਖੇਤੀ ਉਤਪਾਦਾਂ ਦਾ ਮੰਡੀਕਰਨ
  • Chapter 5 ਬੀਜ, ਖਾਦ ਅਤੇ ਕੀੜੇਮਾਰ ਜ਼ਹਿਰਾਂ ਦਾ ਕੁਆਲਟੀ ਕੰਟਰੋਲ
  • Chapter 6 ਪਸ਼ੂ ਪਾਲਣ
  • Chapter 7 ਦੁੱਧ ਅਤੇ ਦੁੱਧ ਤੋਂ ਬਣਨ ਵਾਲੇ ਪਦਾਰਥ
  • Chapter 8 ਮੁਰਗੀ ਪਾਲਣ
  • Chapter 9 ਸੂਰ, ਭੇਡਾਂ/ਬੱਕਰੀਆਂ ਅਤੇ ਖ਼ਰਗੋਸ਼ ਪਾਲਣਾ
  • Chapter 10 ਮੱਛੀ ਪਾਲਣ
  • Chapter 11 ਕੁਝ ਨਵੇਂ ਖੇਤੀ ਵਿਸ਼ੇ

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

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
PSEB 12th Class Chemistry Important Questions Chapter 7 The p-Block Elements 1
Answer:
(i) PCl3 + 3H2O → H3PO3 + 3HCl
(ii) XeF2 + PF5 → [XeF]+ [PF6]
PSEB 12th Class Chemistry Important Questions Chapter 7 The p-Block Elements 2

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

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

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.

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

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 :
PSEB 12th Class Chemistry Important Questions Chapter 7 The p-Block Elements 3
While N2O3 is the anhydride of nitrous acid, HNO2. It dissolves in water to form the unstable acid as follows :
PSEB 12th Class Chemistry Important Questions Chapter 7 The p-Block Elements 4
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.

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

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 :
PSEB 12th Class Chemistry Important Questions Chapter 7 The p-Block Elements 5
As a result of resonance, the two O—O bond lengths in O3 are equal.

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

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.
PSEB 12th Class Chemistry Important Questions Chapter 7 The p-Block Elements 6

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.

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

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

(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)
PSEB 12th Class Chemistry Important Questions Chapter 7 The p-Block Elements 8

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

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.

PSEB 12th Class Chemistry Important Questions Chapter 7 The p-Block Elements 9

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.

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

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
PSEB 12th Class Chemistry Important Questions Chapter 7 The p-Block Elements 10
(c) N2O4
(d) Bleaching action of chlorine is permanent due to oxidation.
Cl + H2O → 2HCl + [O]
(d) 3HNO2 → HNO3 + H2O + 2NO

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

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.

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

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.
PSEB 12th Class Chemistry Solutions Chapter 7 The p-Block Elements 1

Question 6.
How is ammonia manufactured industrially?
Answer:
Ammonia is prepared on a large-scale by the Haber’s process.
PSEB 12th Class Chemistry Solutions Chapter 7 The p-Block Elements 2
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
PSEB 12th Class Chemistry Solutions Chapter 7 The p-Block Elements 3

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

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 + 8HNO3(dilute) → 3CU(NO3)2 + 2NO + 4H2O
Cu + 4HNO3(conc.) → CU(NO3)2 + 2NO2 + 2H2O

Question 8.
Give the resonating structures of N02 and N205.
Answer:
1. Resonating structures of NO2
PSEB 12th Class Chemistry Solutions Chapter 7 The p-Block Elements 4
2. Resonating structures of NO2O5
PSEB 12th Class Chemistry Solutions Chapter 7 The p-Block Elements 5

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.
PSEB 12th Class Chemistry Solutions Chapter 7 The p-Block Elements 6
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.

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

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.

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

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 :
PSEB 12th Class Chemistry Solutions Chapter 7 The p-Block Elements 7
Thus, H3PO3 (oxidation state +3) oxidises to give H3PO4 (oxidation state + 5) and reduce to produce phosphine (oxidation state – 3)

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

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.,
PSEB 12th Class Chemistry Solutions Chapter 7 The p-Block Elements 8

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.

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

Question 19.
Knowing the electron gain enthalpy values for O → O and O → O2- 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 O2-ion is much more than the oxide involving O ion. Hence, the oxide having O2- ions are more stable than oxides having O. Hence, we can say that formation of O2- 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.
PSEB 12th Class Chemistry Solutions Chapter 7 The p-Block Elements 9

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 :
PSEB 12th Class Chemistry Solutions Chapter 7 The p-Block Elements 10

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

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.

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

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

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

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.
PSEB 12th Class Chemistry Solutions Chapter 7 The p-Block Elements 11

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

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.

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

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
PSEB 12th Class Chemistry Solutions Chapter 7 The p-Block Elements 13

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 :
PSEB 12th Class Chemistry Solutions Chapter 7 The p-Block Elements 14

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

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 :
PSEB 12th Class Chemistry Solutions Chapter 7 The p-Block Elements 15

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

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.
PSEB 12th Class Chemistry Solutions Chapter 7 The p-Block Elements 16

(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.
PSEB 12th Class Chemistry Solutions Chapter 7 The p-Block Elements 17

(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.
PSEB 12th Class Chemistry Solutions Chapter 7 The p-Block Elements 18

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.

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

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.

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

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
PSEB 12th Class Chemistry Solutions Chapter 7 The p-Block Elements 19
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.
PSEB 12th Class Chemistry Solutions Chapter 7 The p-Block Elements 20

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

Question 6.
What is the covalence of nitrogen in N2O5?
Answer:
PSEB 12th Class Chemistry Solutions Chapter 7 The p-Block Elements 21
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.
PSEB 12th Class Chemistry Solutions Chapter 7 The p-Block Elements 22

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).
PSEB 12th Class Chemistry Solutions Chapter 7 The p-Block Elements 23

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

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.
PSEB 12th Class Chemistry Solutions Chapter 7 The p-Block Elements 24

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
PSEB 12th Class Chemistry Solutions Chapter 7 The p-Block Elements 25
Since there are three OH groups present in H3PO4 its basicity is three i.e., it is a tribasic acid.
PSEB 12th Class Chemistry Solutions Chapter 7 The p-Block Elements 26

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

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.
PSEB 12th Class Chemistry Solutions Chapter 7 The p-Block Elements 27

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.

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

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:
PSEB 12th Class Chemistry Solutions Chapter 7 The p-Block Elements 28

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

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.
PSEB 12th Class Chemistry Solutions Chapter 7 The p-Block Elements 29
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 :
PSEB 12th Class Chemistry Solutions Chapter 7 The p-Block Elements 30

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+

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

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 :
PSEB 12th Class Chemistry Solutions Chapter 7 The p-Block Elements 31

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.
PSEB 12th Class Chemistry Solutions Chapter 7 The p-Block Elements 32

(ii) SO2 changes the colour of acidified potassium dichromate solution from orange to green
PSEB 12th Class Chemistry Solutions Chapter 7 The p-Block Elements 33

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.

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

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.
PSEB 12th Class Chemistry Solutions Chapter 7 The p-Block Elements 34
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 Ka2 << Ka1 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); Ka1 >10
\(\mathrm{HSO}_{4}^{-}\)(aq) + H2O(l) → H3O+(aq) + \(\mathrm{SO}_{4}^{-}\)(aq); Ka2 = 1.2 × 10-2
Ka1 >> 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 :
PSEB 12th Class Chemistry Solutions Chapter 7 The p-Block Elements 35

(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.

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

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.

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

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.

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

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.

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

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

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

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.

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

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.

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

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.
PSEB 12th Class Chemistry Important Questions Chapter 6 General Principles and Processes of Isolation of Elements 1

(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
PSEB 12th Class Chemistry Important Questions Chapter 6 General Principles and Processes of Isolation of Elements 2

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

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.
PSEB 12th Class Chemistry Important Questions Chapter 6 General Principles and Processes of Isolation of Elements 3

(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.
PSEB 12th Class Chemistry Important Questions Chapter 6 General Principles and Processes of Isolation of Elements 4

(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.
PSEB 12th Class Chemistry Important Questions Chapter 6 General Principles and Processes of Isolation of Elements 5
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)

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

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

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

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.

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

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.

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

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
PSEB 12th Class Chemistry Solutions Chapter 6 General Principles and Processes of Isolation of Elements 1
PSEB 12th Class Chemistry Solutions Chapter 6 General Principles and Processes of Isolation of Elements 2 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.

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

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.
PSEB Solutions for Class 12 Chemistry Chapter 6 General Principles and Processes of Isolation of Elements 3

(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).
PSEB Solutions for Class 12 Chemistry Chapter 6 General Principles and Processes of Isolation of Elements 4

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.

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

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 + CO → 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.
PSEB 12th Class Chemistry Solutions Chapter 6 General Principles and Processes of Isolation of Elements 5

(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.
PSEB Solutions for Class 12 Chemistry Chapter 6 General Principles and Processes of Isolation of Elements 6
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).
PSEB 12th Class Chemistry Solutions Chapter 6 General Principles and Processes of Isolation of Elements 7

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

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.
PSEB 12th Class Chemistry Solutions Chapter 6 General Principles and Processes of Isolation of Elements 8
Then, the obtained nickel tetracarbonyl is decomposed by subjecting it to a higher temperature (450-470 K) to obtain pure nickel metal.
PSEB Solutions for Class 12 Chemistry Chapter 6 General Principles and Processes of Isolation of Elements 9

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

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.
PSEB 12th Class Chemistry Solutions Chapter 6 General Principles and Processes of Isolation of Elements 10

(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.
PSEB Solutions for Class 12 Chemistry Chapter 6 General Principles and Processes of Isolation of Elements 11

(iii) During this process, sodium silicate remains in the solution. The obtained hydrated alumina is filtered, dried, and heated to get back pure alumina.
PSEB 12th Class Chemistry Solutions Chapter 6 General Principles and Processes of Isolation of Elements 12

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.
PSEB 12th Class Chemistry Solutions Chapter 6 General Principles and Processes of Isolation of Elements 13
Water and organic impurities are removed.
PSEB 12th Class Chemistry Solutions Chapter 6 General Principles and Processes of Isolation of Elements 14

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.

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

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.

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

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.
PSEB Solutions for Class 12 Chemistry Chapter 6 General Principles and Processes of Isolation of Elements 15

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 → Al2O3fG = -827kJmol-1
(ii) 2Cr + \(\frac{1}{2}\)O2 → Cr2O3fG = -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.

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

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.
PSEB Solutions for Class 12 Chemistry Chapter 6 General Principles and Processes of Isolation of Elements 16
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.

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

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) + O2- (melt) → CO(g) + 2e
C(s) + 2O2- (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.

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

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.
PSEB Solutions for Class 12 Chemistry Chapter 6 General Principles and Processes of Isolation of Elements 3

(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
PSEB Solutions for Class 12 Chemistry Chapter 6 General Principles and Processes of Isolation of Elements 17

(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.
PSEB 12th Class Chemistry Solutions Chapter 6 General Principles and Processes of Isolation of Elements 18

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

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.
PSEB Solutions for Class 12 Chemistry Chapter 6 General Principles and Processes of Isolation of Elements 19
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).
PSEB Solutions for Class 12 Chemistry Chapter 6 General Principles and Processes of Isolation of Elements 20

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

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.

PSEB 12th Class Chemistry Important Questions Chapter 5 Surface Chemistry

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.

PSEB 12th Class Chemistry Important Questions Chapter 5 Surface Chemistry

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

PSEB 12th Class Chemistry Important Questions Chapter 5 Surface Chemistry

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.

PSEB 12th Class Chemistry Important Questions Chapter 5 Surface Chemistry

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

PSEB 12th Class Chemistry Important Questions Chapter 5 Surface Chemistry

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.

PSEB 12th Class Chemistry Important Questions Chapter 5 Surface Chemistry

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.

PSEB 12th Class Chemistry Important Questions Chapter 5 Surface Chemistry

Long Answer Type Questions

Question 1.
Consider the adsorption isotherms given alongside and interpret the variation in the extent of adsorption (xlm) when
PSEB 12th Class Chemistry Important Questions Chapter 5 Surface Chemistry 1
(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, \(\frac{x}{m}\) = kp1/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.

PSEB 12th Class Chemistry Important Questions Chapter 5 Surface Chemistry

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.

PSEB 12th Class Chemistry Solutions Chapter 5 Surface Chemistry

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.

PSEB 12th Class Chemistry Solutions Chapter 5 Surface Chemistry

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:
PSEB 12th Class Chemistry Solutions Chapter 5 Surface Chemistry 1
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 Ps, \(\frac{x}{m}\) 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.
\(\frac{x}{m}\) ∝ Po
\(\frac{x}{m}\) = kPo

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\(\frac{1}{n}\)
\(\frac{x}{m}\) = kP1/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.
PSEB 12th Class Chemistry Solutions Chapter 5 Surface Chemistry 2

PSEB 12th Class Chemistry Solutions Chapter 5 Surface Chemistry

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.

PSEB 12th Class Chemistry Solutions Chapter 5 Surface Chemistry

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.
PSEB 12th Class Chemistry Solutions Chapter 5 Surface Chemistry 3
(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.,
\(\frac{x}{m}\) ∝ po 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.,
\(\frac{x}{m}\) ∝ p1/n or \(\frac{x}{m}\) = kp1/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.

PSEB 12th Class Chemistry Solutions Chapter 5 Surface Chemistry

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’.
PSEB 12th Class Chemistry Solutions Chapter 5 Surface Chemistry 4
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).

PSEB 12th Class Chemistry Solutions Chapter 5 Surface Chemistry

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.

PSEB 12th Class Chemistry Solutions Chapter 5 Surface Chemistry

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.
PSEB 12th Class Chemistry Solutions Chapter 5 Surface Chemistry 5

(ii) Formation of ammonia by the combination of dinitrogen and dihydrogen in the presence of finely divided iron.
PSEB 12th Class Chemistry Solutions Chapter 5 Surface Chemistry 6
This process is called the Haber’s process.

(iii) Oswald’s process: Oxidation of ammonia to nitric oxide in the presence of platinum.
PSEB 12th Class Chemistry Solutions Chapter 5 Surface Chemistry 7

(iv) Hydrogenation of vegetable oils in the presence of Ni.
PSEB 12th Class Chemistry Solutions Chapter 5 Surface Chemistry 8

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.
PSEB 12th Class Chemistry Solutions Chapter 5 Surface Chemistry 9

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
PSEB 12th Class Chemistry Solutions Chapter 5 Surface Chemistry 10
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.

PSEB 12th Class Chemistry Solutions Chapter 5 Surface Chemistry

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 (Tk)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.

PSEB 12th Class Chemistry Solutions Chapter 5 Surface Chemistry

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.

PSEB 12th Class Chemistry Solutions Chapter 5 Surface Chemistry

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.

PSEB 12th Class Chemistry Solutions Chapter 5 Surface Chemistry

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.

PSEB 12th Class Chemistry Solutions Chapter 11 Alcohols, Phenols and Ethers

Punjab State Board PSEB 12th Class Chemistry Book Solutions Chapter 11 Alcohols, Phenols, and Ethers Textbook Exercise Questions and Answers.

PSEB Solutions for Class 12 Chemistry Chapter 11 Alcohols, Phenols, and Ethers

PSEB 12th Class Chemistry Guide Alcohols, Phenols and Ethers Intext Questions and Answers

Question 1.
Write IUPAC names of the following compounds:
PSEB 12th Class Chemistry Solutions Chapter 11 Alcohols, Phenols and Ethers 1
Answer:
(i) 2, 2, 4-Trimethylpentan-3-ol
(ii) 5-Ethylheptane-2,4-diol
(iii) Butane-2, 3-diol
(iv) Propane-1, 2, 3-triol
(v) 2-Methylphenol
(vi) 4-Methylphenol
(vii) 2, 5-Dimethylphenol
(viii) 2, 6-Dimethylphenol
(ix) l-Methoxy-2-methylpropane
(x) Ethoxybenzene
(xi) 1-Phenoxyheptane
(xii) 2-Ethoxybutane

Question 2.
Write structures of the compounds whose IUPAC names are as follows:
(i) 2-Methylbutan-2-ol
(ii) l-Phenylpropan-2-ol
(iii) 3,5-Dimethylhexane -1, 3, 5-triol
(iv) 2,3 – Diethylphenol
(v) 1 -Ethoxypropane
(vi) 2-Ethoxy-3-methylpentane
(vii) Cyclohexylmethanol (viii) 3-Cyclohexylpentan-3-ol
(ix) Cyclopent-3-en-l-ol
(x) 3-Chloromethylpentan-l-ol.
Answer:
PSEB 12th Class Chemistry Solutions Chapter 11 Alcohols, Phenols and Ethers 3

PSEB 12th Class Chemistry Solutions Chapter 11 Alcohols, Phenols and Ethers 4

Question 3.
(i) Draw the structures of all isomeric alcohols of molecular formula C5H12O and give their IUPAC names.
(ii) Classify the isomers of alcohols in question 3(i) as primary, secondary, and tertiary alcohols.
Answer:
(i) Eight isomers are possible. These are :
PSEB 12th Class Chemistry Solutions Chapter 11 Alcohols, Phenols and Ethers 5
(ii) Primary : (a), (d), (e) arid (g); Secondary : (b), (c), (h); Tertiary : (f).

Question 4.
Explain why propanol has higher boiling point than that of the hydrocarbon, butane?
Answer:
Propanol undergoes intermolecular H-bonding because of the presence of -OH group. On the other hand, butane does not.
PSEB 12th Class Chemistry Solutions Chapter 11 Alcohols, Phenols and Ethers 6
Therefore, extra energy is required to break hydrogen bonds. For this reason, propanol has a higher boiling point than hydrocarbon butane.

Question 5.
Alcohols are comparatively more soluble in water than hydrocarbons of comparable molecular masses. Explain this fact.
Answer:
Alcohols form H-bonds with water due to the presence of -OH group. However, hydrocarbons cannot form H-bonds with water.
PSEB 12th Class Chemistry Solutions Chapter 11 Alcohols, Phenols and Ethers 7
As a result, alcohols are comparatively more soluble in water than hydrocarbons of comparable molecular masses.

Question 6.
What is meant by the hydroboration-oxidation reaction? Illustrate it with an example.
Answer:
The addition of diborane to alkenes to form trialkyl boranes followed by their oxidation with alkaline hydrogen peroxide to form alcohols is called hydroboration-oxidation reaction.
PSEB 12th Class Chemistry Solutions Chapter 11 Alcohols, Phenols and Ethers 8
The alcohol obtained by this process is contrary to the Markovnikov’s rule.

Question 7.
Give the structures and IUPAC names of monohydric phenols of molecular formula, C7H8O.
Answer:
The three isomers are given as follows :
PSEB 12th Class Chemistry Solutions Chapter 11 Alcohols, Phenols and Ethers 9

Question 8.
While separating a mixture of ortho and para-nitrophenols by steam distillation, name the isomer which will be steam volatile. Give reason.
Answer:
o-Nitrophenol is steam volatile as it exists as discrete molecules due to intramolecular H-bonding and hence can be separated by steam distillation from p-nitrophenol which is less volatile as it exists as associated molecules because of intermolecular H-bonding.
PSEB 12th Class Chemistry Solutions Chapter 11 Alcohols, Phenols and Ethers 10

Question 9.
Give the equations of reactions for the preparation of phenol from cumene.
Answer:
PSEB 12th Class Chemistry Solutions Chapter 11 Alcohols, Phenols and Ethers 11

Question 10.
Write chemical reaction for the preparation of phenol from chlorobenzene.
Answer:
Chlorobenzene is fused with NaOH (at 623 K and 300 atm pressure) to produce sodium phenoxide, which gives phenol on acidification.
PSEB 12th Class Chemistry Solutions Chapter 11 Alcohols, Phenols and Ethers 12

Question 11.
Write the mechanism of hydration of ethene to yield ethanol.
Answer:
The mechanism of hydration of ethene to yield ethanol involves three steps.
Step 1:
Protonation of ethene to form carbocation by electrophilic attack of H3O+:
H2O+H+ → H3O+
PSEB 12th Class Chemistry Solutions Chapter 11 Alcohols, Phenols and Ethers 13

Step 2 :
Nucleophilic attack of water on carbocation:
PSEB 12th Class Chemistry Solutions Chapter 11 Alcohols, Phenols and Ethers 14

Step 3:
Deprotonation to form ethanol:
PSEB 12th Class Chemistry Solutions Chapter 11 Alcohols, Phenols and Ethers 15

Question 12.
You are given benzene, cone. H2SO4 and NaOH. Write the equations for the preparation of phenol using these reagents.
Answer:
PSEB 12th Class Chemistry Solutions Chapter 11 Alcohols, Phenols and Ethers 16

Question 13.
Show how will you synthesize:
(i) 1-phenyl ethanol from a suitable alkene.
(ii) cyclohexyl methanol using an alkyl halide by an SN2 reaction.
(iii) pentane-l-ol using a suitable alkyl halide?
Answer:
(i) By acid-catalyzed hydration of ethylbenzene (styrene), 1-phenyl ethanol can be synthesized.
PSEB 12th Class Chemistry Solutions Chapter 11 Alcohols, Phenols and Ethers 17
(ii) When chloro methylcyclohexane is treated with sodium hydroxide, cyclohexyl methanol is obtained.
PSEB 12th Class Chemistry Solutions Chapter 11 Alcohols, Phenols and Ethers 18
(iii) When 1-chloromethane is treated with NaOH, pentane-l-ol is produced.
PSEB 12th Class Chemistry Solutions Chapter 11 Alcohols, Phenols and Ethers 19

Question 14.
Give two reactions that show the acidic nature of phenol. Compare acidity of phenol with that of ethanol.
Answer:
The reactions showing acidic nature of phenol are :
(i) Reaction with sodium: Phenol reacts with active metals like sodium to liberate H2 gas.
PSEB 12th Class Chemistry Solutions Chapter 11 Alcohols, Phenols and Ethers 20
(ii) Reaction with NaOH: Phenol dissolves in NaOH to form sodium phenoxide and water.
PSEB 12th Class Chemistry Solutions Chapter 11 Alcohols, Phenols and Ethers 21
Comparison of acidic character of phenol and ethanol:
Phenol is more acidic than ethanol. This is due to the reason that phenoxide ion left after the loss of a proton from phenol is stabilized by resonance while ethoxide ion left after loss of a proton from ethanol is not stabilized by resonance.
PSEB 12th Class Chemistry Solutions Chapter 11 Alcohols, Phenols and Ethers 22

Question 15.
Explain why is ortho nitrophenol more acidic than ortho methoxy phenol?
Answer:
PSEB 12th Class Chemistry Solutions Chapter 11 Alcohols, Phenols and Ethers 23
The nitro-group is an electron-withdrawing group. The presence of this group in the ortho position decreases the electron density in the O-H bond. As a result, it is easier to lose a proton. Also, the o-nitro phenoxide ion formed after the loss of protons is stabilized by resonance. Hence, ortho nitrophenol is a stronger acid. On the other hand, methoxy group is an electron-releasing group. Thus, it increases the electron density in the O-H bond and hence, the proton cannot be given out easily. For this reason,ortho-nitrophenol is more acidic than ortho- methoxy phenol.

Question 16.
Explain how does the -OH group attached to a carbon of benzene ring activates it towards electrophilic substitution?
Answer:
Phenol may be regarded as a resonance hybrid of following structures:
PSEB 12th Class Chemistry Solutions Chapter 11 Alcohols, Phenols and Ethers 24
Thus, due to +R effect of the -OH group, the electron density in the benzene ring increases thereby facilitating the attack of an electrophile. In other words, presence of -OH group, activates the benzene ring towards electrophilic substitution reactions. Now, since the electron density is relatively higher at the two o-and one p-position, electrophilic substitution occurs mainly at o-and p-positions.

Question 17.
Give equations of the following reactions:
(i) Oxidation of propane-1-ol with alkaline KMnO4 solution.
(ii) Bromine in CS2 with phenol.
(iii) Dilute HNO3 with phenol.
(iv) Treating phenol with chloroform in presence of aqueous NaOH.
Answer:
PSEB 12th Class Chemistry Solutions Chapter 11 Alcohols, Phenols and Ethers 25

Question 18.
Explain the following with an example.
(i) Kolbe’s reaction.
(ii) Reimer-Tlemimnn reaction.
(iii) WiIHamon ether synthesis.
(iv) Unsymmetrical ether.
Answer:
(i) Kolbe’s reaction: When phenol is treated with sodium hydroxide, sodium phenoxide is produced. This sodium phenoxide when treated with carbon dioxide, followed by acidification, undergoes electrophilic substitution to give ortho-hydroxybenzoic acid as the main product. This reaction is known as Kolbe’s reaction.
PSEB 12th Class Chemistry Solutions Chapter 11 Alcohols, Phenols and Ethers 26

(ii) Reimer-Telemann reaction: When phenol is treated with chloroform (CHCl3 ) in the presence of sodium hydroxide, a -CHO group is introduced at the ortho position of the benzene ring.
PSEB 12th Class Chemistry Solutions Chapter 11 Alcohols, Phenols and Ethers 27
This reaction is known as the Reimer-Tiemann reaction. The intermediate is hydrolyzed in the presence of alkalis to produce salicylaldehyde.
PSEB 12th Class Chemistry Solutions Chapter 11 Alcohols, Phenols and Ethers 28

(iii) Williamson ether synthesis: Williamson ether synthesis is a laboratory method to prepare symmetrical and unsymmetrical ethers by allowing alkyl halides to react with sodium alkoxides.
PSEB 12th Class Chemistry Solutions Chapter 11 Alcohols, Phenols and Ethers 29
This reaction involves SN2 attack of the alkoxide ion on the alkyl halide. Better results are obtained in case of primary alkyl halides.
PSEB 12th Class Chemistry Solutions Chapter 11 Alcohols, Phenols and Ethers 30
If the alkyl halide is secondary or tertiary, then elimination competes over substitution.

(iv) Unsymmetrical ether: An unsymmetrical ether is an ether where two groups on the two sides of an oxygen atom differ (i.e., have an unequal number of carbon atoms). For example: ethyl methyl ether (CH3-O-CH2CH3), methyl phenyl ether (CH3 -0 – C6H5), etc.

Question 19.
Write the mechanism of acid-catalyzed dehydration of ethanol to yield ethene.
Answer:
The mechanism of acid-catalyzed dehydration of ethanol to yield ethene involves the following three steps:
Step 1:
Protonaüon of ethanol to form ethyl oxonium ion:
PSEB 12th Class Chemistry Solutions Chapter 11 Alcohols, Phenols and Ethers 31

Step 2 :
Formation of carbocation (rate-determining step):
PSEB 12th Class Chemistry Solutions Chapter 11 Alcohols, Phenols and Ethers 32

Step 3 :
Elimination of a proton to form ethene:
PSEB 12th Class Chemistry Solutions Chapter 11 Alcohols, Phenols and Ethers 33
The acid consumed in Step 1 is released in Step 3. After the formation of ethene, it is removed to shift the equilibrium in a forward direction.

Question 20.
How are the following conversions carried out?
(i) Propene → Propan-2-ol
(ii) Benzyl chloride → Benzyl alcohol
(iii) Ethyl magnesium chloride → Propan-1-ol.
(iv) Methyl magnesium bromide → 2-Methylpropan-2-ol.
Answer:
(i) If propene is allowed to react with water in the presence of an acid as a catalyst, then propane-2-ol is obtained.
PSEB 12th Class Chemistry Solutions Chapter 11 Alcohols, Phenols and Ethers 34

(ii) If benzyl chloride is treated with NaOH (followed by acidification) then benzyl alcohol is produced.
PSEB 12th Class Chemistry Solutions Chapter 11 Alcohols, Phenols and Ethers 35

(iii) When ethyl magnesium chloride is treated with methanal, an adduct is produced which gives propane-1-ol on hydrolysis.
PSEB 12th Class Chemistry Solutions Chapter 11 Alcohols, Phenols and Ethers 36

(iv) When methyl magnesium bromide is treated with propanone, an adduct is produced which gives 2- methylpropane-2-ol on hydrolysis.
PSEB 12th Class Chemistry Solutions Chapter 11 Alcohols, Phenols and Ethers 37

Question 21.
Name the reagents used in the following reactions:
(i) Oxidation of a primary alcohol to carboxylic acid.
(ii) Oxidation of a primary alcohol to aldehyde.
(iii) Bromination of phenol to 2,4,6-tribromophenol.
(iv) Benzyl alcohol to benzoic acid.
(v) Dehydration of propane-2-ol to propene.
(vi) Butan-2-one to butan-2ol.
Answer:
(i) Acidified potassium dichromate (K2Cr2O7/H2SO4)
(ii) Pyridinium chlorochromate (PCC)
(iii) Bromine water (Br2/H2O)
(iv) Acidified potassium permanganate (KMnO4/H2SO4)
(v) 85% phosphoric acid (H3PO4)
(vi) NaBH4 or LiAlH4.

Question 22.
Give reason for the higher boiling point of ethanol in comparison to methoxymethane.
Answer:
Ethanol undergoes intermolecular H-bonding due to the presence of -OH group, resulting in the association of molecules. Extra energy is required to break these hydrogen bonds. On the other hand, methoxymethane does not undergo H-bonding. Hence, the boiling point of ethanol is higher than that of methoxymethane.
PSEB 12th Class Chemistry Solutions Chapter 11 Alcohols, Phenols and Ethers 38

Question 23.
Give IUPAC mimes of the following ethers:
PSEB 12th Class Chemistry Solutions Chapter 11 Alcohols, Phenols and Ethers 39
(ii) CH3OCH2CH2Cl
(iii) O2N- C6H4 – OCH3(p)
(iv) CH3CH2CH2OCH3
PSEB 12th Class Chemistry Solutions Chapter 11 Alcohols, Phenols and Ethers 40
PSEB 12th Class Chemistry Solutions Chapter 11 Alcohols, Phenols and Ethers 41
Answer:
(i) 1-Ethoxy-2-methylpropane
(ii) 2-Chioro- 1 -methoxyethane
(iii) 4-Nitroanisole
(iv) 1 -Methoxypropane
(v) 4-Ethoxy- 1, 1 -dime thylcyclohexane
(vi) Ethoxybenzene.

Question 24.
Write the names of reagents and equations for the preparation of the following ethers by WilÌianlÑon’s synthesis:
(i) 1-Propoxypropane
(ii) Ethoxybenzene
(iii) 2.Methoxy-2-methylpropane
(iv) 1 -Methoxyethane
Answer:
PSEB 12th Class Chemistry Solutions Chapter 11 Alcohols, Phenols and Ethers 42

Question 25.
Illustrate with examples the limitations of Willi9mon synthesis for the preparation of certain types of ethers.
Answer:
The reaction of Williamson synthesis involves SN2 attack of an alkoxide ion on a primary alkyl halide.
PSEB 12th Class Chemistry Solutions Chapter 11 Alcohols, Phenols and Ethers 43
But if secondary or tertiary alkyl halides are taken in place of primary alkyl halides, then elimination would compete over substitution. As a result, alkenes would be produced. This is because alkoxides are nucleophiles as well as strong bases. Hence, they react with alkyl halides, which results in an elimination reaction.
PSEB 12th Class Chemistry Solutions Chapter 11 Alcohols, Phenols and Ethers 44

Question 26.
How is 1-propoxy propane synthesized from propane-l-ol? Write mechanism of this reaction.
Answer:
1-propoxy propane can be synthesized from propane-1-ol by dehydration Propan- 1 -ol undergoes dehydration in the presence of protic acids (such as H2SO4, H3PO4) to give 1-propoxy propane.
PSEB 12th Class Chemistry Solutions Chapter 11 Alcohols, Phenols and Ethers 45
The mechanism of this reaction involves the following three steps:
Step 1: Protonation
PSEB 12th Class Chemistry Solutions Chapter 11 Alcohols, Phenols and Ethers 46
Step 2: Nucleophilic attack
PSEB 12th Class Chemistry Solutions Chapter 11 Alcohols, Phenols and Ethers 47
Step 3: Deprotonanon
PSEB 12th Class Chemistry Solutions Chapter 11 Alcohols, Phenols and Ethers 48

Question 27.
Preparation of ethers by acid-catalyzed dehydration of secondary or tertiary alcohols is not a suitable method. Give
reason.
Answer:
Acid-catalyzed dehydration of primary alcohols to ethers occurs by SN2 reaction involving nucleophilic attack of the alcohol molecule on the protonated alcohol molecule.
PSEB 12th Class Chemistry Solutions Chapter 11 Alcohols, Phenols and Ethers 49
Under these conditions, secondary and tertiary alcohols, however, give alkenes rather than ethers. This is because due to steric hindrance, nucleophilic attack of the alcohol molecule on the protonated alcohol molecule does not occur. Instead, protonated secondary and tertiary alcohols lose a molecule of water to form stable secondary and tertiary carbocations. These carbocations prefer to lose a proton to form alkenes rather than undergoing nucleophilic attack by alcohol molecules to form ethers.
PSEB 12th Class Chemistry Solutions Chapter 11 Alcohols, Phenols and Ethers 50
Similarly, tertiary alcohols give alkenes rather than ethers.
PSEB 12th Class Chemistry Solutions Chapter 11 Alcohols, Phenols and Ethers 51

Question 28.
Write the equation of the reaction of hydrogen iodide with:
(i) 1-propoxy propane
(ii) Methoxybenzene and
(iii) Benzyl ethyl ether
Answer:
PSEB 12th Class Chemistry Solutions Chapter 11 Alcohols, Phenols and Ethers 52
PSEB 12th Class Chemistry Solutions Chapter 11 Alcohols, Phenols and Ethers 53

Question 29.
ExplaIn the fact that in aryl alkyl ethers
(i) the alkoxy group activates the benzene ring towards electrophilic substitution and
(ii) it directs the incoming substituents to ortho and para positions in benzene ring.
Answer:
(i) In aryl alkyl ethers, the +Reffect of the alkoxy group (OR) increases the electron density in the benzene ring thereby activating the benzene ring towards electrophilic substitution reactions.
PSEB 12th Class Chemistry Solutions Chapter 11 Alcohols, Phenols and Ethers 54

(ii) Since the electron density increases more at the two orthos and one para position as compared to meta-positions, electrophilic substitution reactions mainly occur at ortho-and para-positions. For example,
PSEB 12th Class Chemistry Solutions Chapter 11 Alcohols, Phenols and Ethers 55

Question 30.
Write the mechanism of the reaction of III with methoxymethane.
Answer:
When equimolar amounts of HI and methoxymethane are taken; a mixture of methyl alcohol and iodomethane are formed.
Mechanism
Step I:
PSEB 12th Class Chemistry Solutions Chapter 11 Alcohols, Phenols and Ethers 56

Step II :
PSEB 12th Class Chemistry Solutions Chapter 11 Alcohols, Phenols and Ethers 57
If HI is present in excess, CH3OH formed in step II is further converted into CH3I.

Step III:
PSEB 12th Class Chemistry Solutions Chapter 11 Alcohols, Phenols and Ethers 58
Step IV:
PSEB 12th Class Chemistry Solutions Chapter 11 Alcohols, Phenols and Ethers 59

Question 31.
Write equations of the following reactions:
(i) Friedel-Crafts reaction-alkylation of anisole.
(ii) Nitration of anisole.
(iii) Bromlnation of anisole in ethanoic acid medium.
(iv) Friedel-Craft’s acetylation of anisole.
Answer:
(i) Friedel-Crafts reaction (Alkylation of anisole)
PSEB 12th Class Chemistry Solutions Chapter 11 Alcohols, Phenols and Ethers 60

(ii) Nitration of anisole
PSEB 12th Class Chemistry Solutions Chapter 11 Alcohols, Phenols and Ethers 61

(iii) Bromination of anisole in ethanoic acid medium
PSEB 12th Class Chemistry Solutions Chapter 11 Alcohols, Phenols and Ethers 62

(iv) Friedel-Crafts acetylation of anisole
PSEB 12th Class Chemistry Solutions Chapter 11 Alcohols, Phenols and Ethers 63

Question 32.
Show ho would you synthesize the following alcohols from appropriate alkenes?
PSEB 12th Class Chemistry Solutions Chapter 11 Alcohols, Phenols and Ethers 64
Answer:
The given alcohols can be synthesized by applying Markovnikov’s rule of acid-catalyzed hydration of appropriate alkenes. –
PSEB 12th Class Chemistry Solutions Chapter 11 Alcohols, Phenols and Ethers 65

Question 33.
When 3-methyl butan-2-ol is treated with HBr, the following reaction takes place:
PSEB 12th Class Chemistry Solutions Chapter 11 Alcohols, Phenols and Ethers 66
Give a mechanism for this reaction.
(Hint: The secondary carbocation formed in step It rearranges to a more stable tertiary carbocation by a hydride ion shift from 3rd carbon atom.)
Answer:
Mechanism: The reaction takes place through the following mechanism :
Step I: Formation of protonated alcohol.
PSEB 12th Class Chemistry Solutions Chapter 11 Alcohols, Phenols and Ethers 67
Step II: Formation of carbocatlon.
PSEB 12th Class Chemistry Solutions Chapter 11 Alcohols, Phenols and Ethers 68
2° carbocation being less stable undergoes hydride shift to form more stable 3° carbocation.
PSEB 12th Class Chemistry Solutions Chapter 11 Alcohols, Phenols and Ethers 69
Step III: Attack of nucleophile
PSEB 12th Class Chemistry Solutions Chapter 11 Alcohols, Phenols and Ethers 70

Chemistry Guide for Class 12 PSEB Alcohols, Phenols, and Ethers Textbook Questions and Answers

Question 1.
Classify the following as primary, secondary and tertiary alcohols:
PSEB 12th Class Chemistry Solutions Chapter 11 Alcohols, Phenols and Ethers 71
Answer:
Primary alcohols (i), (ii), (iii)
Secondary alcohols (iv) and (v)
Tertiary alcohols (vi)

Question 2.
Identify allylic alcohols in the above examples.
Answer:
Allylic alcohols (ii) and (vi).

Question 3.
Name the following compounds according to IUPAC system :
PSEB 12th Class Chemistry Solutions Chapter 11 Alcohols, Phenols and Ethers 72
PSEB 12th Class Chemistry Solutions Chapter 11 Alcohols, Phenols and Ethers 73
Answer:
(i) 3-Chloromethyl-2-isopropylpentan-l-ol _
(ii) 2, 5-Dimethylhexane-l,3-diol .
(iii) 3-Bromocyclohexanol
(iv) Hex-l-en-3-ol
(v) 2-Bromo-3-methylbut-2-en-l-ol.

Question 4.
Show how are the following alcohols prepared by the reaction of a suitable Grignard reagent on methanal.
PSEB 12th Class Chemistry Solutions Chapter 11 Alcohols, Phenols and Ethers 74
Answer:
PSEB 12th Class Chemistry Solutions Chapter 11 Alcohols, Phenols and Ethers 75
PSEB 12th Class Chemistry Solutions Chapter 11 Alcohols, Phenols and Ethers 76

Question 5.
Write structures of the products of the following reactions:
PSEB 12th Class Chemistry Solutions Chapter 11 Alcohols, Phenols and Ethers 77
Answer:
PSEB 12th Class Chemistry Solutions Chapter 11 Alcohols, Phenols and Ethers 78

Question 6.
Give structures of the products you would expect when each of the following alcohol reacts with
(a) HCl-ZnCl2
(b) HBr and
(c) SOCl2.
Answer:
PSEB 12th Class Chemistry Solutions Chapter 11 Alcohols, Phenols and Ethers 79
PSEB 12th Class Chemistry Solutions Chapter 11 Alcohols, Phenols and Ethers 80
PSEB 12th Class Chemistry Solutions Chapter 11 Alcohols, Phenols and Ethers 81
PSEB 12th Class Chemistry Solutions Chapter 11 Alcohols, Phenols and Ethers 82

Question 7.
Predict the major product of acid-catalyzed dehydration of
(i) 1-methyl cyclohexanol and
(ii) butane-l-ol.
Answer:
(i) Acid-catalyzed dehydration of 1-methyl cyclohexanol can give two products, I and II. Since product (I) is more substituted, according to Saytzeff rule, it is the major product.
PSEB 12th Class Chemistry Solutions Chapter 11 Alcohols, Phenols and Ethers 83

Question 8.
Ortho and para-nitrophenols are more acidic than phenol. Draw the resonance structures of the corresponding phenoxide ions.
Answer:
The resonance structures of o – and p -nitrophenoxide ions and phenoxide ion are given as follows :
PSEB 12th Class Chemistry Solutions Chapter 11 Alcohols, Phenols and Ethers 85

PSEB 12th Class Chemistry Solutions Chapter 11 Alcohols, Phenols and Ethers 86
It is clear from the above structures that due to -R effect of NO2 group, o-and p-nitro phenoxide ions are more stable than phenoxide ions. Consequently, o-and p-nitrophenols are more acidic than phenol.

Question 9.
Write the equations involved in the following reactions :
(i) Reimer-Tiemann reaction
(ii) Kolbe’s reaction –
Answer:
PSEB 12th Class Chemistry Solutions Chapter 11 Alcohols, Phenols and Ethers 87

Question 10.
Write the reactions of Williamson synthesis of 2- ethoxy-3-methyl pentane starting from ethanol and 3- methyl pentane-2-ol.
Answer:
PSEB 12th Class Chemistry Solutions Chapter 11 Alcohols, Phenols and Ethers 88

Question 11.
Which of the following is an appropriate set of reactants for the preparation of l-methoxy-4- nitrobenzene and why ?
PSEB 12th Class Chemistry Solutions Chapter 11 Alcohols, Phenols and Ethers 89
Answer:
(ii) because inset (i), sodium methoxide (CH3ONa) is a strong nucleophile as well as a strong base. So, elimination reaction predominates over substitution reaction.

Question 12.
Predict the products of the following reactions :
(i) CH3 – CH2 – CH2 – O – CH3 + HBr →
PSEB 12th Class Chemistry Solutions Chapter 11 Alcohols, Phenols and Ethers 90
Answer:
PSEB 12th Class Chemistry Solutions Chapter 11 Alcohols, Phenols and Ethers 91