Previous year Questions (2016-20) - The p-Block Elements Notes | EduRev

Q.1. The electron gain enthalpy (in kJ mol^–1) of fluorine, chlorine, bromine and iodine, respectively, are    (2020)
(1) −296, −325, −333 and −349
(2) −349, −333, −325 and −296
(3) −333, −349, −325 and −296
(4) −333, −325, −349 and −296
Ans. (3) 

The electron gain enthalpy of halogens is given below:

Q.2. Chlorine reacts with hot and concentrated NaOH and produces compounds (X) and (Y). Compound (X) gives white precipitate with silver nitrate solution. The average bond order between Cl and O atoms in (Y) is ______.    (2020)
Ans. (1.67) 

The reaction involved is


The structure of NaClO₃ is

Two Cl–O bonds are double bond and one is single bond, thus the average bond order between Cl and O is

Q.3. In the following reactions, products (A) and (B), respectively are:
NaOH (hot and conc.) + Cl₂ → (A) + side products
Ca(OH)₂ (dry) + Cl₂ → (B) + side products    (2020)
(1) NaClO₃ and Ca(OCl)₂
(2) NaClO₃ and Ca(ClO₃)₂
(3) NaOCl and Ca(OCl)₂
(4) NaOCI and Ca(CIO₃
 Ans. (1) 

The reaction involved is



Q.4. The number of bonds between sulphur and oxygen atoms in  and the number of bonds between sulphur and sulphur atoms in rhombic sulphur, respectively, are    (2020)
(1) 4 and 6
(2) 8 and 8
(3) 8 and 6
(4) 4 and 8
Ans. (2)

The structure of is

There is total of 8 bonds present in between S and O.
Struture of rhombic sulphur is:

There are a total of 8 S-S bonds.

Q.5. White phosphorus on reaction with concentrated NaOH solution in an inert atmosphere of CO₂ gives phosphine and compound (X). (X) on acidification with HCl gives compound (Y). The basicity of compound (Y) is    (2020)
(1) 2
(2) 1
(3) 4
(4) 3
Ans. (2) 

The reaction involved is


The structure of the hypophosphorous acid is:

Hypophosphorous acid have only one replaceable hydrogen atom, thus, its basicity is one.

Q.6. The reaction of H₃N₃B₃Cl₃(A) with LiBH₄ in tetrahydrofuran gives inorganic benzene (B). Further, the reaction of (A) with (C) leads to H₃N₃B₃(Me)₃. Compounds (B) and (C) respectively, are    (2020)
(1) borazine and MeBr
(2) diborane and MeMgBr
(3) boron nitride and MeBr
(4) borazine and MeMgBr
Ans. (4)

The reaction involved is



Q.7. The one that is extensively used as a piezoelectric material is:    (2019)
(1) tridymite
(2) amorphous silica
(3) quartz
(4) mica
Ans. (3)
Quartz exhibits piezo electricity and thus can be used as a piezoelectric material.

Q.8. Correct statements among a to d regarding silicones are:    (2019)
(a) They are polymers with hydrophobic character.
(b) They are biocompatible.
(c) In general, they have high thermal stability and low dielectric strength.
(d) Usually, they are resistant to oxidation and used as greases.
(1) (a), (b), (c) and (d)
(2) (a), (b) and (c) only
(3) (a) and (b) only
(4) (a), (b) and (d) only
Ans. (4)
Silicones are polymers containing Si—O—Si linkages with strong hydrophobic character.
Generally, they exhibit high thermal stability with high dielectric strength. Silicon greases are resistant to oxidation which are commonly used for greasing purposes.

Q.9. Good reducing nature of H₃PO₂ is attributed to the presence of:    (2019)
(1) Two P — OH bonds
(2) One P — H bond
(3) Two P — H bonds
(4) One P — OH bond
Ans. (3)
Structure of H₃PO₂:

Greater the number of P—H bonds present in the acid, greater will be its reducing property.

Q.10. Wilkinson catalyst is:    (2019)
(1) [(Ph₃P)₃ IrCl]
(2) [(Et₃P)₃ RhCl]
(3) [(Ph₃P)₃ RhCl]
(4) [(Et₃P)₃ IrCl]
Ans. (3)
Wilkinson’s catalyst is [Rh(PPh₃)₃Cl]

Q.11. The type of hybridisation and number of lone pair(s) of electrons of Xe in XeOF₄, respectively, are:    (2019)
(1) sp³d² and 1
(2) sp³d and 2
(3) sp³d² and 2
(4) sp³d and 1
Ans. (1)
, sp³d², no. of lone pair = 1

Q.12. Among the following reactions of hydrogen with halogens, the one that requires a catalyst is:    (2019)
(1) H₂ + I₂ → 2 HI
(2) H₂ + Cl₂ → 2 HCl
(3) H₂ + Br₂ → 2HBr
(4) H₂ + F₂ → 2HF
Ans. (1)
The reaction between I₂ and H₂ requires catalyst, whereas all other halogens react with H₂ without the requirement of a catalyst.

Q.13. The pair that contains two P-H bonds in each of the oxoacids is:    (2019)
(1) H₄P₂O₅ and H₄P₂O₆
(2) H₃PO₂ and H₄P₂O₅
(3) H₃PO₃ and H₃PO₂
(4) H₄P₂O₅ and H₃PO₃
Ans. (2)


Q.14. The number of 2-centre-2-electron and 3-centre-2-electron bonds in B₂H₆, respectively, are:    (2019)
(1) 2 and 1
(2) 4 and 2
(3) 2 and 2
(4) 2 and 4
Ans. (2)
Structure of B₂H₆:

∴ No. of 2-centre-2 electron bonds = 4,
No. of 3-centre-2 electron bonds = 2.

Q.15. The chloride that CANNOT get hydrolysed is:    (2019)
(1) PbCl₄
(2) CCl₄
(3) SnCl₄
(4) SiCl₄
Ans. (2)
CCl₄ cannot be hydrolysed due to absence of d orbitals at carbon atom.

Q.16. The hydride that is NOT electron deficient is:    (2019)
(1) SiH₄
(2) B₂H₆
(3) GaH₃
(4) AlH₃
Ans. (1)
SiH₄: Electron precise hydride
B₂H₆, GaH₃ and Al H₃ are electron deficient

Q.17. Iodine reacts with concentrated HNO₃ to yield Y along with other products. The oxidation state of iodine in Y, is:    (2019)
(1) 5 
(2) 7
(3) 3
(4) 1
Ans. (1)

In HIO₃ oxidation state of iodine is +5.

Q.18. The element that does NOT show catenation is:    (2019)
(1) Ge
(2) Si
(3) Sn
(4) Pb
Ans. (4)
Catenation power of the elements decreases as we move down in the group. Therefore, Pb does not show catenation property.

Q.19. Chlorine on reaction with hot and concentrated sodium hydroxide gives:    (2019)
(1) Cl^- and CIO^-₃ 
(2) Cl^- and ClO^-
(3) ClO^-₃ and ClO^-₂
(4) Cl^- and ClO^-₂
Ans. (1)
3Cl₂ + 6 NaOH → 5 NaCl + NaClO₃ + 3 H₂O

Q.20. The element that shows greater ability to form pπ - pπ multiple bonds, is:    (2019)
(1) Sn 
(2) C
(3) Ge
(4) Si
Ans. (2)
Carbon has unique ability to form pπ-pπ multiple bonds with itself and with other atoms of small size. Heavier elements do not form pπ-pπ bonds because their atomic orbitals are too large and diffuse to have effective overlapping. 

Q.21. Diborane (B₂H₆) reacts independently with O₂ and H₂O to produce, respectively:    (2019)
(1) B₂O₃ and H₃BO₃ 
(2) B₂O₃ and [BH₄]^-
(3) H₃BO₃ and B₂O₃ 
(4) HBO₂ and H₃BO₃
Ans. (1)
B₂H₆ + 3O₂ → B₂O₃ + 3H₂O
B₂H₆ + 6H₂O → 2H₃BO₃ + 6H₂

Q.22. The correct statement about ICI₅ and ICI^-₄ is:    (2019)
(1) both are is isostructural.
(2)  ICI₅ is trigonal bipyramidal and ICI^-₄ is tetrahedral.
(3) ICI₅ is square pyramidal and ICI^-₄ is tetrahedral.
(4) ICI₅ is square pyramidal and ICI^-₄ is square planar.
Ans. (4)
ICI₅ is sp³d² hybridised (5 bp, 1 lp)




Q.23. The ion that has sp³d² hydridization for the central atom, is:    (2019)
(1) [ICI₄]^- 
(2) [ICI₂]^- 
(3) [IF₆]^-    
(4) [BrF₂]^-
Ans. (1)


Q.24. The correct order of the oxidation states of nitrogen in NO, N₂O, NO₂ and N₂O₃ is:    (2019)
(1) NO₂ < NO < N₂O₃ < N₂O
(2) NO₂ < N₂O₃ < NO < N₂O
(3) N₂O < N₂O₃ < NO < NO₂
(4) N₂O < NO < N₂O₃ < NO₂
Ans. (4)
order of oxidation state –


Q.25. C₆₀, an allotrope of carbon cantains:    (2019)
(1) 12 hexagons and 20 pentagons.
(2) 18 hexagons and 14 pentagons.
(3) 16 hexagons and 16 pentagons.
(4) 20 hexagons and 12 pentagons.
Ans. (4)
Fullerene (C₆₀) contains 20 hexagons (six membered) rings and 12 pentagons (five membered rings)

Q.26. HF has highest boiling point among hydrogen halides, because it has:    (2019)
(1) strongest van der Waals’ interactions
(2) lowest ionic character
(3) strongest hydrogen bonding
(4) lowest dissociation enthalpy
Ans. (3)
Due to strong H-bonding between HF molecules. HF has highest boiling point among the hydrogen halides.

Q.27. The amorphous form of silica is:    (2019)
(1) Tridymite
(2) Kieselguhr
(3) Cristobalite
(4) Quartz
Ans. (2)
Quartz, tridymite and cristobalite are crystalline forms of silica, while kieselguhr is an amorphous form of silica.

Q.28. The oxoacid of Sulphur that does not contain bond between Sulphur atoms is:    (2019)
(1) H₂S₄O₆
(2) H₂S₂O₃
(3) H₂S₂O₇
(4) H₂S₂O₄
Ans. (3)


H₂S₂O₇ does not show bonding between sulphur atoms.

Q.29. The correct order of catenation is:    (2019)
(1) C > Sn > Si ≈ Ge
(2) C > Si > Ge ≈ Sn
(3) Si > Sn > C > Ge
(4) Ge > Sn > Si > C
Ans. (2)
The catenation property among 14^th group elements is based on bond enthalpy value of bond between the same element. The decreasing order of bond enthalpy values is
Bond enthalpy  in kJ/mol
∴ Decreasing order of catenation is
C > Si > Ge ≈ Sn

Q.30. The number of pentagons in C₆₀ and trigons (triangles) in white phosphorous, respectively, are:    (2019)
(1) 20 and 3
(2) 12 and 4
(3) 12 and 3
(4) 20 and 4
Ans. (2)
Number of pentagons in C₆₀ (Buckminsterfullerene) = 12
Number of triangles in P₄ (White phosphorous) = 4

Q.31. The noble gas that does NOT occur in the atmosphere is:    (2019)
(1) He
(2) Kr
(3) Ne
(4) Ra
Ans. (4)
Radon is radioactive element and not present in atmosphere.

Q.32. The basic structural unit of feldspar, zeolites, mica, and asbestos is:    (2019)
(1) (SiO₃)^2-
(2) SiO₂
(3) (SiO₄)^4-
(4)
Ans. (3)
These are examples of silicates, the basic unit of each of them is SiO.

Q.33. The C - C bond length is maximum in:    (2019)
(1) graphite
(2) C₇₀
(3) C₆₀
(4) diamond
Ans. (4)
Carbon-carbon bond length is maximum in diamond because diamond has all single bonds while graphite, C₇₀ and C₆₀ have single and double bonds.
 

Q.34. Which of the following compounds contain(s) no covalent bond(s)?    (2018)
KCl, PH₃, O₂, B₂H₆, H₂SO₄
(1) KCl, B₂H₆, PH₃
(2) KCl, H₂SO₄
(3) KCl
(4) KCl, B₂H₆
Ans. (3)
KCl is an ionic compound. It cannot from covalent bond. Elements of s - block & p - block combine to form ionic compounds.

Q.35. Total number of lone pair of electrons in I^-₃ ion is:    (2018)
(1) 3
(2) 6
(3) 9
(4) 12
Ans. (3)


Q.36. The compound that does not produce nitrogen gas by the thermal decomposition is:    (2018)
(1) Ba(N₃)₂ 
(2) (NH₄)₂Cr₂O₇ 
(3) NH₄NO₂ 
(4) (NH₄)2SO₄ 
Ans. (4)


Q.37. Xenon hexafluoride on partial hydrolysis produces compounds 'X' and 'Y'. Compounds 'X' and 'Y' and the oxidation state of Xe are respectively:    (2018)
(1) XeOF₄ (+ 6) and XeO₂F₂ (+6) 
(2) XeOF₄ (+ 6) and XeO₃ (+6) 
(3) XeO₂F₂ (+ 6) and XeO₂ (+4) 
(4) XeO₂(+ 4) and XeO₃ (+6) 
Ans. (1)


Q.38.
In hydrogen azide (above) the bond orders of bonds (I) and (II) are:    (2018)
(1) <2, <2  
(2) <2, >2 
(3) >2, >2 
(4) >2, <2 
Ans. (2)

Or


Q.39. The decreasing order of bond angles in BF₃, NH₃, PF₃ and I₃^- is:    (2018)
(1) BF₃ > NH₃ > PF₃ > I₃^– 
(2) I₃^- > BF₃ > NH₃ > PF₃ 
(3) I₃ > NH₃ > PF₃ > BF₃ 
(4) BF₃ > I₃ > PF₃ > NH₃ 
Ans. (2)



Q.40. Among the oxides of nitrogen:    (2018)
N₂O₃, N₂O₄ and N₂O₅ ; the molecule(s) having nitrogen-nitrogen bond is/are:  
(1) N₂O₃ and N₂O₄ 
(2) N₂O₄ and N₂O₅ 
(3) N₂O₃ and N₂O₅ 
(4) Only N₂O₅ 
Ans. (1)


Q.41. A group 13 element 'X' reacts with chlorine gas to produce a compound XCl₃. XCl₃ is electron deficient and easily reacts with NH₃ to form Cl₃X → NH₃ adduct; however, XCl₃ does not dimerize. X is:   (2018)
(1) B
(2) Al
(3) In
(4) Ga
Ans. (1)


Q.42. Which of the following reactions is an example of a redox reaction?    (2017) 
(1) XeF₄ + O₂F₂ → XeF₆ + O₂ 
(2) XeF₂ + PF₅ → [XeF]⁺ PF₆^- 
(3) XeF₆ + H₂O → XeOF₄ + 2HF 
(4) XeF₆ + 2H₂O → XeO₂F₂ + 4HF 
Ans. (1)
Xe is oxidised from +4(in XeF₄) to +6(in XeF₆) 
Oxygen is reduced from +1 (in O₂F₂) to zero (in O₂) 

Q.43. The products obtained when chlorine gas reacts with cold and dilute aqueous NaOH are    (2017)
(1) CIO^- and CIO₃^- 
(2) CIO₂^- and CIO^- 
(3) Cl^- and CIO^- 
(4) Cl^- and CIO₂^- 
Ans. (3)


Q.44. The number of S = O and S – OH bonds present in peroxodisulphuric acid and pyrosulphuric acid respectively are:    (2017)
(1) (2 and 4) and (2 and 4)
(2) (4 and 2) and (4 and 2)
(3) (4 and 2) and (2 and 4)
(4) (2 and 2) and (2 and 2)
Ans. (2)
Peroxodisulphuric acid
H₂S₂O₈ 
Pyrosulphuric acid
H₂S₂O₇ :  

Q.45. In which of the following reactions, hydrogen peroxide acts as an oxidizing agent?    (2017)
(1) HOCl + H₂O₂ → H₃O⁺ + Cl^- + O₂
(2) I₂ + H₂O₂ + 2OH^- → 2I^- + 2H₂O + O₂
(3) PbS + 4H₂O₂ → PbSO₄ + 4H₂O
(4) 2MnO₄^- + 3H₂O₂ → 2MnO₂ + 3O₂ + 2H₂O + 2OH^-
Ans. (3)


Q.46. The number of P–OH bonds and the oxidation state of phosphorus atom in pyrophosphoric acid (H₄P₂O₇) respectively are:    (2017)
(1) five and four
(2) four and five
(3) five and five
(4) four and four
Ans. (2)


Q.47. The correct sequence of decreasing number of π-bonds in the structure of H₂SO₃, H₂SO₄ and H₂S₂O₇ is:    (2017)
(1) H₂S₂O₇ > H₂SO₃ > H₂SO₄
(2) H₂S₂O₇ > H₂SO₄ > H₂SO₃
(3) H₂SO₄ > H₂S₂O₇ > H₂SO₃
(4) H₂SO₃ > H₂SO₄ > H₂S₂O₇
Ans. (2)




Q.48. XeF₆ on partial hydrolysis with water produces a compound ‘X’. The same compound ‘X’ is formed when XeF₆ reacts with silica. The compound ‘X’ is:    (2017)
(1) XeF₄
(2) XeF₂
(3) XeO₃
(4) XeOF₄
Ans. (4)

Q.49. The pair in which phosphorous atoms have a formal oxidation state of + 3 is:    (2016)
(1) Orthophosphorous and pyrophosphorous acids
(2) Pyrophosphorous and hypophosphoric acids
(3) Orthophosphorous and hypophosphoric acids
(4) Pyrophosphorous and pyrophosphoric acids
Ans. (1)
Phosphorus acid series contain phosphorus in the oxidation state (+ III).

Pyrophosphorus acid
      
orthophosphorus acid

Q.50. The non-metal that does not exhibit positive oxidation state is:    (2016)
(1) Fluorine
(2) Oxygen
(3) Chlorine
(4) Iodine
Ans. (1)
Fluorine is the most electronegative element in periodic table hence it shows –1 oxidation state in all its compounds.

Q.51. The group of molecules having identical shape is:    (2016)
(1)  PCl₅, IF₅, XeO₂F₂
(2) BF₃, PCl_3, XeO₃
(3)
(4) SF₄, XeF₄, CCl₄
Ans. (3)
are sp³d hybridized with 2 lone pair e's, hence all have (T-shape) identical shape.
  

Q.52. Which intermolecular force is most responsible in allowing xenon gas to liquefy?    (2016)
(1) Instantaneous dipole-induced dipole
(2) Ionic
(3) Ion-dipole
(4) Dipole-dipole
Ans. (1)
Instantaneous dipole-induced dipole forces are most responsible in allowing xenon gas to liquify.