JEE Advanced (Subjective Type Questions): The s-Block Elements | Chapter-wise Tests for JEE Main & Advanced PDF Download

Q.1. Give reasons for the following :

(i) Sodium carbonate is made by Solvay process but the same process is not extended to the manufacture of potassium carbonate.
 (ii) Hydrogen peroxide is a better oxidising agent than water.
 (iii) Magnesium oxide is used for the lining of steel making furnace.
 (iv) Why is sodium chloride added during electrolysis of fused anhydrous magnesium chloride? 
 (v) Hydrogen peroxide acts as an oxidising as well as a reducing agent.
 (vi) The crystalline salts of alkaline earth metals contain more water of crystallisation than the corresponding alkali metal salts.
 (vii) BeCl₂ can be easily hydrolysed.

Solution. (i) Potassium ca r bon ate can n ot be man u factur ed by Solvay process, since, unlike sodium hydrogen carbonate, potassium hydrogen carbonate is rather too soluble in water to be precipitated like NaHCO₃.
(ii) H₂O₂ is a better oxidising agent than H₂O because oxidation number of oxygen in H₂O₂ is –1 and that in water it is –2. So H₂O₂ easily reduces to –2 oxidation number.
(iii) MgO is used for the lining of steel making furnace because it acts as basic flux and facilitates the removal of acidic impurities of Si, P and S from steel through slag formation.
(iv) The anhydrous magnesium chloride is fused with NaCl to provide conductivity to the electrolyte and to lower the fusion temperature of anhydrous MgCl₂.
NOTE : NaCl prevents hydrolysis of MgCl₂.
(v) The oxidation state of oxygen in H₂O₂ (i.e. –1) can be changed to 0 or –2 i.e oxygen in H₂O₂ exists in an intermediate oxidation state with respect to O2 and O2–.
Hence it acts both as an oxidising and reducing agent.
(vi) NOTE : Smaller the size of cation, higher will be hydration tendency because hydration energy of cation is inversely proportional to size of cation.
The size of alkaline earth metal ions are smaller than the size of alkali metal ions. So in crystalline form the salts of alkaline earth metals have more water molecules than those of alkali metals.
(vii) BeCl₂ is hydrolysed due to high polarising power and presence of vacant p-orbitals in Be-atom.

Q.2. How will you prepare bleaching powder from slaked lime

Solution. Bleaching powder, Ca(OCl)₂, can be prepared by passing chlorine through Ca(OH)₂ solution.

Q.3. Write down the balanced equations for the reactions when:

(i) Calcum phosphate is heated with a mixture of sand and carbon;
 (ii) An alkalin e solution of potassium fer ricyan ide is reacted with hydrogen peroxide.
 (iii) Car bon dioxide is passed th rough a concen tr ated aqueous solution of sodium chloride saturated with ammonia.
 (iv) Potassium ferricyanide reacts with hydrogen peroxide in basic solution.
 (v) Carbon dioxide is passed through a suspension of lime stone in water.

Solution. (i) 

This is the electrothermal process to extract phosphorus from phosphorite or bone ash [Ca₃(PO₄)₂].
(ii) Ferricyanide is oxidised to ferrocyanide on treatment with alkali

2K₃[Fe(CN)₆] + H₂O₂ + 2KOH → 2K₄[Fe(CN)₆] + 2H₂O + O₂

(iii) NaCl + NH₄OH + CO₂ → NH₄Cl + NaHCO₃
(iv) 2K₃[Fe(CN)₆] + H₂O₂ + 2KOH → 2K₄[Fe(CN)₆] + 2H₂O + O₂

NOTE : Suspension oif lime stone is CaCO₃.

Q.4. Give briefly the isolation of magnesium from sea water by the Dow process. Give equations for the steps involved.

Solution. In sea water Mg exists as MgCl₂.

On treating sea water with slaked lime Mg(OH)₂ is obtained.

On reacting Mg(OH)₂ with HCl, MgCl₂ is obtained.
Mg(OH)₂ + 2HCl → MgCl₂ + H2O

From MgCl₂, Mg is obtained by reduction of MgCl₂ with CaC₂.

MgCl₂ + CaC₂ → Mg + CaCl₂ + 2C

Q.5. Complete and balance the following reactions :

Ca₅ (PO₄)₃ F + H₂SO₄+ H₂O   + 5CaSO₄ .2H₂O+ ......... .

Solution. Ca₅(PO₄)₃F + 5H₂SO₄ + 10H₂O   3H₃PO₄ + 5CaSO₄.2H₂O + HF

Q.6. A 5.0 cm³ solution of H₂O₂ liberates 0.508 g of iodine from an acidified KI solution. Calculate the strength of H₂O₂ solution in terms of volume strength at STP.

Ans. 4.48 volumes

Solution.

i.e. 254 gm of I₂ is released by 34 gm H₂O₂
∴ 0.508 gm of I₂ will be released by

5 ml of H₂O₂ sol. contains 0.068 gm of H₂O₂.

NOTE : The strength of H₂O₂ is generally calculated in terms of volume strength. According to which 10 volume of H₂O₂ means that 1 ml of H₂O₂ sol gives 10 ml of O₂ at STP.

i.e., 68 gm of H₂O₂ gives 22,400 ml of O₂ at STP or 1 ml of H₂O₂ sol

or   1 ml of H₂O₂ sol gives 4.48 ml of O₂ i.e. strength of H₂O₂ sol is 4.48 volumes

Q.7. Explain the difference in the nature of bonding in LiF and LiI.

Solution. LiF has more ionic character while LiI has more covalent character. The latter is due to the greater  polarizability of larger iodide ion than the fluoride ion.

Q.8. To a 25ml H₂O₂ solution, excess of acidified solution of potassium iodide was added. The iodine liberated required 20 ml of 0.3 N sodium thiosulphate Solution. Calculate the volume strength of H₂O₂ solution.

Ans. 1.344

Solution. Meq. of H₂O₂ = Meq.of  Na₂S₂O₃

Q.9. Give reactions for the oxidation of hydrogen peroxide with potassium permanganate in acidic medium.

Solution. 2KMnO₄ + 3H₂SO₄ → K₂SO₄ + 2MnSO₄ + 3H₂O + 5 [O]

Q.10. Element A burns in nitrogen to give an ionic compound B. Compound B reacts with water to give C and D. A solution of C becomes ‘milky’ on bubbling carbon dioxide. Identify A, B, C and D.

Ans. Ba, Ba₃N₂, Ba(OH)₂, BaCO₃

Solution. 

M may be either Ca or Ba.

NOTE : It is not magnesium because Mg(OH)₂ has very low solubility in water.

If we consider Ba as M then A is Ba, B is Ba₃N₂, C is Ba(OH)₂, D is BaCO₃.

Q.11. Arrange the following sulphates of alkaline earth metals in or der of decreasing th ermal stability : BeSO₄, MgSO₄, CaSO₄, SrSO₄

Ans. SrSO₄ > CaSO₄ > MgSO₄ > BeSO₄

Solution. SrSO₄ > CaSO₄ > MgSO₄ > BeSO₄ (Based upon size of cation or ionic character)

Q.12. Work out the following using chemical equation: Chlorination of calcium hydroxide produces bleaching powder.

Solution. 

Q.13. Hydr ogen per oxide acts both as an oxidising and as a reducing agent in alkaline solution towards certain first row transition metal ions. Illustrate both these properties of H₂O₂ using chemical equations.

Solution. When H₂O₂ acts as oxidising agent, following reaction takes place:

H₂O₂ + 2e^– → 2OH^–

While regarding its action as reducing agent, the following reaction takes place:

H₂O₂ + 2OH^– → O₂ + 2H₂O + 2e^–

Examples of oxidising character of H2O2 in alkaline medium

2Cr(OH)₃ + 4NaOH + 3H₂O₂ → 2Na₂CrO₄ + 8H₂O

Here Cr3+ (Cr is a first row transition metal) is oxidised to Cr6+.

Example of reducing character of H₂O₂ in alkaline medium

2K₃[Fe(CN)₆]+2KOH+H₂O₂ → 2K₄[Fe(CN)₆]+2H₂O+O₂

Here Fe³⁺ (Fe is a first row transition metal) is reduced to Fe²⁺.