JEE Advanced (Subjective Type Questions): The p-Block Elements- 2 | Chapter-wise Tests for JEE Main & Advanced PDF Download

Q.22. A soluble compound of a poisonous element M, when heated with Zn/H₂SO₄ gives a colourless and extremely poisonous gaseous compound N, which on passing through a heated tube gives a silvery mirror of element M. Identify M and N.             (1997 - 2 Marks)

Ans. M = As,N = AsH₃

Solution. 

Q.23. Draw  the structure of a cyclic silicate, (Si₃O₉)^6– with proper labelling.         (1998 -  4 Marks)

Solution. In cyclic   three tetrahedra of    are joined together sharing two oxygen atoms per  tetrahedron.

Q.24. Thionyl chloride can be synthesized by chlorinating SO₂ using PCI₅. Thionyl chloride is used to prepare anhydrous ferric chloride starting from its hexahydrated salt.
 Alternatively, the anhydrous ferric chloride can also be prepared from its hexahydrated salt by treating with 2, 2 – dimethoxypropane. Discuss all this using balanced chemical equations.             (1998 -  6 Marks)

Solution.

Q.25. Reaction of phosphoric acid with Ca₅(PO₄)₃F yields a fertilizer “triple superphosphate”. Represent the same through balanced chemical equation.           (1998 -  2 Marks)

Solution. 

Q.26. In the following equation,           (1999 - 6 Marks)

A + 2B + H₂O → C + 2D
 (A = HNO₂, B = H₂SO₃,  C =  NH₂OH). Identify D. Draw the structures of A, B, C and D. 

Solution. The reaction is

The structures of A, B, C and D are as follows.

Q.27. In the contact process for industrial manufacture of sulphuric acid some amount of sulphuric acid is used as a starting material. Explain briefly. What is the catalyst used in the oxidation of SO₂?              (1999 - 4 Marks)

Solution. Sulphur trioxide produced in the  contact process is absorbed by sulphuric acid forming H₂S₂O₇. It is not dissolved in water as it gives a dense fog of sulphuric acid particles.
The catalyst used in the contact process is vanadium pentoxide. 

Q.28. The Haber process can be represented by the following scheme;

Identify A, B, C, D and E.                (1999 - 5 Marks)

Ans. 

Solution. In such a case

A = Ca(OH)₂, B = NH₄HCO₃, C = Na₂CO₃, D = NH₄Cl and E = CaCl₂

Q.29. Give an example of oxidation of one halide by another halogen. Explain the feasibility of the reaction           (2000 - 2 Marks).

Solution. More electronegative halogen displaces lesser electronegative halogen from its halide. Thus,

Cl₂ + 2KBr (or 2 KI) → 2KCl + Br₂ (or I₂)

Q.30. Draw the molecular structures of XeF₂, XeF₄ and XeO₂F₂ indicating the location of lone pair(s) of electrons.       (2000 - 3 Marks)

Solution. TIPS/Formulae :

Use the formula

H (hydridisation) 

V = number of electron in valence shell of central atom
M = number of monovalent atoms surrounding the central atom
C = Charge on cation
A = Charge on anion

XeF₂ :   Hence hybridisation is sp3d, and thus its structure is linear.

XeF, :  Hence hybridisation is sp³d². and thus its structure is square planar.

XeO₂F₂ :  Hence hybridisation is sp³d. and shape is see saw.

Q.31. Give reason(s) why elemental nitrogen exists as a diatomic molecule whereas elemental phosphorus as a tetraatomic molecule.           (2000 - 2 Marks)

Solution.  Elemental nitrogen exists as a diatomic molecule because nitrogen can form pπ - pπ multiple bonds which is not possible in case of phosphorus due to repulsion between non-bonded electrons of the inner core. There is no such repulsion in case of smaller nitrogen atoms as they have only 1s² electrons in their inner core.

Q.32. Compound (X) on reduction with LiAlH₄ gives a hydride (Y) containing 21.72% hydrogen along with other products. The compound (Y) reacts with air explosively resulting in boron trioxide. Identify (X) and (Y). Give balanced reactions involved in the formation of (Y) and its reaction with air.

Draw the structure of (Y).                         (2001 - 5 Marks)

Ans. BCl₃  or  BBr₃, B₂H₆

Solution. Since B₂O₃ is formed by reaction of (Y) with air, (Y) therefore should be B₂H₆ in which % of hydrogen is 21.72. The compound (X) on reduction with LiAlH₄ gives B₂H₆. Thus it is boron trihalide. The reactions are shown as:

Structure of B₂H₆ is as follows:

Thus the diborane molecule has four two-centre- two -electron bonds (2c – 2e bonds) also called usual bonds  and two three-centre-two -electron bonds (3c – 2e) also called banana bonds. Hydrogen attached to usual and banana bonds are called H_t (terminal H) and H_b (bridged H) respectively.

Q.33. Starting from SiCl₄, prepare the followin g in steps not exceeding the number given in parentheses (give reactions only):
 (i) Silicon (1)
 (ii) Linear silicone containing methyl groups only (4)
 (iii) Na₂SiO₃ (3)

Solution. (i) SiCl₄ + 2Mg(or Zn) → Si + 2MgCl₂ (or ZnCl₂)
(ii) SiCl₄ + 2CH₃MgCl → (CH₃)₂SiCl₂ + 2MgCl₂

Polymerisation continues on both ends to give linear silicone.

Q.34. Write balanced equations for the reactions of the following compounds with water : (2002 - 5 Marks) 

(i) Al₄C₃
 (ii) CaNCN
 (iii) BF₃
 (iv) NCl₃
 (v) XeF₄

Solution.

Ammonia formed dissolves in water to form NH₄OH

Q.35. How is boron obtained from borax? Give chemical equations with reaction conditions. Write the structure of B₂H₆ and its reaction with HCl.              (2002 - 5 Marks)

Solution. NOTE : When hot con centr ated HCl is added to borax (Na₂B₄O₇.10H₂O) the sparingly soluble H₃BO₃ is formed which on subsequent heating gives B₂O₃ which is reduced to boron on heating with Mg, Na or K

Na₂B₄O₇ (anhydrous) + 2HCl( hot, conc.)

→ 2NaCl+ H₂B₄O₇

B₂H₆ + HCl → B₂H₅Cl + H₂

[NOTE : Normally this reaction takes place in the presence of Lewis acid (AlCl₃)]

Q.36. Write down reactions involved in the extraction of Pb. What is the oxidation number of lead in litharge?       (2003 - 2 Marks)

Ans. O.N. of Pb in PbO is +2

Solution.

Q.37. Identify the following:

Also mention the oxidation state of  S in all the compounds.

Ans. 

Solution. 

Oxidation states of ‘S’ are :  + 4 in  (A) , (+ 6)  in B  and  + 2 in (C), + 2.5 in (D)

Q.38. AlF₃ is insoluble in anhydrous HF but it becomes soluble in presence of little amount of KF. Addition of boron trifluoride to the resulting solution causes reprecipitation of AlF₃.
 Explain with balanced chemical equations.         (2004 - 2 Marks)

Solution. HF is weakly dissociated, while KF is highly dissociated giving a high concentration of F– which leads to the formation of soluble AlF₆^3–.
AlF₃ + 3 KF → K₃[AlF₆]
Since BF₃ is more acidic than AlF₃, it pulls out F– from AlF₆^3– reprecipitating AlF₃.

Q.39. How many grams of CaO are required to neutralize 852 gm of P4O10 ? Draw structure of P₄O₁₀ molecule.             (2005 - 2 Marks)

Ans. 1008 g

Solution. 

Moles of CaO = 3 × 6 = 18; wt. of CaO = 18 × 56 = 1008 g

For structure of P₄O₁₀ : See question 20 of this section.

Q.40. Write the structures of (CH₃)₃ N and (Me₃Si)₃ N .  Are they isostructural? Justify your answer.              (2005 - 2 Marks)

Solution. (CH₃)₃ N and (Me₃Si)₃ N are not isostructural, the former is pyramidal while the latter is trigonal planar. Silicon has vacant d orbitals which can accommodate lone pair of electrons from N (back bonding) leading to planar shape.

Q.41.

Ans.  

Solution. 

A. Conc. H₂SO₄
B. Br₂

C. 

Reactions involved are