Main Group Elements and Their Compounds MCQ Quiz - Objective Question with Answer for Main Group Elements and Their Compounds - Download Free PDF

The correct answer is Carbon.

Key Points

• Graphite, Diamond, and Buckminsterfullerene are allotropes of Carbon, meaning they are different structural forms of the same element.
• Graphite is composed of layers of carbon atoms arranged in a hexagonal lattice and is widely used as a lubricant and in pencils.
• Diamond is a crystalline form of carbon with a tetrahedral lattice structure, making it the hardest naturally occurring material.
• Buckminsterfullerene (C₆₀) is a molecule composed of 60 carbon atoms arranged in a spherical structure resembling a soccer ball, also known as a "buckyball."
• These allotropes exhibit distinct physical and chemical properties due to differences in their atomic arrangements and bonding.

Additional Information

• Allotropes:
  • Allotropes are different structural forms of the same element, where atoms are bonded in varying arrangements.
  • Examples include Carbon allotropes such as Graphite, Diamond, and Buckminsterfullerene.
• Graphite:
  • Graphite conducts electricity due to the free movement of electrons between its layers.
  • It has applications in batteries, electrodes, and as a lubricant.
• Diamond:
  • Diamond's extreme hardness is due to its strong covalent bonds in a tetrahedral lattice.
  • It is widely used in cutting tools, jewelry, and high-performance industrial applications.
• Buckminsterfullerene:
  • Discovered in 1985, Buckminsterfullerene is named after architect Buckminster Fuller.
  • It has potential applications in nanotechnology, medicine, and electronics.
• Carbon:
  • Carbon is a non-metal element with the atomic number 6 and symbol "C."
  • It is essential for life and forms the basis of organic chemistry.
  • Carbon exists in multiple forms, including amorphous carbon, graphene, and carbon nanotubes.

Concept:

Oxidation States of Sulfur in Dithionite and Dithionate Ions

• Dithionite (S₂O₄²⁻): In this ion, sulfur has an oxidation state of +3. This can be determined by considering the total charge on the ion and the oxidation state of oxygen, which is typically -2.
• Dithionate (S₂O₆²⁻): In this ion, sulfur has an oxidation state of +5. Similar reasoning applies, where the oxygen atoms contribute a charge of -2 each, and the overall charge of the ion is -2.

Explanation:

• In the dithionite ion (S₂O₄²⁻), the oxidation state of sulfur is +3. This is calculated by assigning oxygen an oxidation state of -2 (common for oxygen in compounds) and solving for sulfur. The sum of oxidation states must equal the charge of the ion, which is -2.
• In the dithionate ion (S₂O₆²⁻), the oxidation state of sulfur is +5. The same approach is used here, but with six oxygen atoms contributing a total of -12 to balance the charge of -2, the sulfur atoms must each have an oxidation state of +5.

Hence, the correct answer is +3 and +5.

Concept:

Conversion of Diamond to Graphite and Electrical Conductivity of Graphite

• Thermodynamics of Diamond to Graphite Conversion:
  • The conversion of diamond to graphite is thermodynamically spontaneous at room temperature and pressure because the Gibbs free energy change (ΔG) is negative (-2.90 kJ mol^-1).
  • However, the conversion is extremely slow due to the high activation energy required to break the strong covalent bonds in the diamond structure.
• Electrical Conductivity of Graphite:
  • Graphite consists of layers of hexagonally arranged carbon atoms held together by weak van der Waals forces.
  • Within each layer, carbon atoms are bonded via sp² hybridization, with delocalized π-electrons providing high conductivity.
  • Conductivity is high in the plane of the layers due to the presence of delocalized electrons.
  • Conductivity is low perpendicular to the plane because the layers are only weakly held by van der Waals forces, preventing effective electron movement.

Explanation:

• Statement I: "The conversion of diamond to graphite at room temperature and pressure is spontaneous (ΔG = -2.90 kJ mol^-1)."
  • Correct, because the negative Gibbs free energy indicates thermodynamic spontaneity.
  • However, the process is extremely slow due to kinetic constraints.
• Statement II: "The electrical conductivity of graphite perpendicular to the plane is low, whereas parallel to the plane is very high."
  • Correct, as conductivity is high along the layers due to delocalized electrons but low perpendicular to the layers due to weak interlayer interactions.

Therefore, the correct answer is Both Statement I and Statement II are correct.

Concept:

Reaction of Borax with Hydrochloric Acid

• Borax (sodium tetraborate, Na₂B₄O₇·10H₂O) reacts with aqueous hydrochloric acid (HCl) to produce boric acid (H₃BO₃) as the main product.
• In the reaction, the borax dissolves in the acidic solution and undergoes hydrolysis, releasing boric acid.
• The reaction is as follows:

  Na₂B₄O₇·10H₂O + 4HCl → 2NaCl + 4B(OH)₂ + 5H₂O

  Boric acid, H₃BO₃, is produced as the final product from the hydrolysis of B(OH)₂ and is found as a weak acid.

Explanation:

• The final product of the reaction between borax and aqueous hydrochloric acid is boric acid (H₃BO₃).
• Other products of the reaction include sodium chloride (NaCl) and water (H₂O).
• Thus, boric acid is the primary product formed from this reaction.

Hence, the correct answer is H₃BO₃ (boric acid).

Concept:

Color of Nitric Acid and its Photochemical Decomposition

• Concentrated nitric acid (HNO₃) is colorless when freshly prepared. However, upon keeping for a prolonged period, it often turns yellow due to the formation of nitrogen dioxide (NO₂) which dissolves in the acid, giving it a yellowish color.
• The yellow color in concentrated HNO₃ is due to the presence of nitrogen dioxide, which forms as a result of the photochemical decomposition of nitric acid.
• The reason provided (R) about photochemical decomposition of HNO₃ is correct, as exposure to light causes HNO₃ to decompose into nitrogen dioxide (NO₂) and oxygen (O₂).

Explanation:

• Both Assertion A and Reason R are correct. The yellow color of concentrated nitric acid on keeping is a result of nitrogen dioxide formation from the photochemical decomposition of HNO₃.
• Therefore, Reason R correctly explains Assertion A, as the yellow color is directly linked to NO₂ formation from the photochemical decomposition of HNO₃.

Hence, the correct answer is: Both A and R are correct and R is the correct explanation of A.

Key Points

• The s-block element that is a silvery- metal and is used in an alloy with copper or nickel to make various equipment is Beryllium.
• Beryllium is a lightweight and strong metal that is widely used in aerospace, defense, and nuclear industries.
• It has a high melting point, good thermal conductivity, and is also a good electrical conductor.
• Atomic number of Beryllium is - 4.

Additional Information

• Rubidium is a soft, silvery- metal that is highly reactive.
  • It is used in atomic clocks and as a catalyst in certain chemical reactions.
• Francium is a highly radioactive and unstable element.
  • It is extremely rare and has no practical applications.
• Caesium is a soft, silvery-golden metal that is also highly reactive.
  • It is used in atomic clocks, drilling fluids, and in the treatment of cancer.

The correct answer is NH₃ and HCl

Key Points

• Dalton's law is applicable only to a mixture of non-reacting gases.
• Ammonia (NH₃) reacts with hydrochloric acid at normal temperature and gives ammonium chloride(NH₄Cl).
• NH₃ + HCl    →    NH₄Cl
• So, this gaseous mixture does not follow Dalton's law.

Additional Information

Dalton's law:

• Dalton's law states that in a mixture of non-reacting gases, the total pressure exerted is equal to the sum of the partial pressures of the individual gases.
• This empirical law was observed by John Dalton in 1801.
• It was published in 1802.
• It is also called Dalton's law of partial pressures.
• Dalton's law is related to the ideal gas laws.
• Dalton's law is not strictly followed by real gases, with the deviation increasing with pressure.

Important Points

• Helium is a noble gas that will not react with hydrogen gas, so this gaseous mixture follows Dalton’s law.
• Nitrogen gas is not reactive in normal temperatures because the breakdown of nitrogen into atomic form is a highly endothermic process.
• Nitrogen and oxygen gas are not found in their atomic form in normal temperatures, so these gases will not react with each other. 

Concept:

Lewis acid and Base:

• Species that can accept a pair of the electron is a Lewis acid.
• Lewis acid has a generally positive charge.
• Lewis acid has a vacant unoccupied orbital called LUMO.
• Electron pairs from donor atoms are accepted in these orbitals.
• Lewis acids can be positive or neutral.
• Examples are H+, H₃O+ , CH₃+ , NO+ , AlCl₃ etc.
• Species that can donate a pair of electrons are called Lewis bases.
• Lewis bases have generally negative charges but can also be positive or neutral like NO⁺ and CO.
• They have the highest occupied molecular orbitals full.
• The interaction between the Lewis base and acid leads to the formation of a new coordinate bond between them as shown below-

Explanation:

• The strength of these Lewis acids will depend on the electron deficiency of the Boron atom.
• The more the electron deficiency over Boron, the higher will be its acidic strength.
• The substituents added to boron are halogen atoms Bromine, chlorine, and Fluorine.
• Boron and Fluorine belong to the same period of the periodic table and they have 2p-2p π orbital overlap.
• Due to this 2p-2p overlap, the electron deficiency over the boron atom is covered to a greater extent.
• Hence, the presence of fluorine decreases the acidity of Boron.
• The overlap between chlorine and boron is lesser in the case of Chlorine and least in the case of bromine.

• Bromine and Boron are not comparable in size and the weakest π overlap will take place between them.
• Hence, Bromine substituent will help in increasing the acidity of Boron.
• BFBr₂ due to the presence of two bromine atoms, the acidity will be highest.
• BFClBr, F tries to decrease the acidity whereas Cl and Br try to increase the acidity.
• In BF₂Br, two fluorine atoms tend to decrease the acidity of the compound and thus become less acidic than BFClBr.
• BF₂Cl is lesser acidic than BF2Br because, in BF₂Cl, Cl is comparative in size with boron than Bromine and decreases the acidity more than Chlorine.

Hence, the correct order of Lewis acidic strength is: BFBr₂ > BFClBr > BF₂Br > BF₂Cl.

CONCEPT:

Photolysis of SF₆ and Radical Formation

• In the upper atmosphere, SF₆ undergoes photolysis, where ultraviolet (UV) light breaks the sulfur-fluorine bond, resulting in the formation of species A (SF₅).
• Species A, being a radical, contains an unpaired electron on the sulfur atom and is paramagnetic.
• Species A can combine with O₂ to form radical B (SF₅O₂•), where the unpaired electron is now located on one of the oxygen atoms, making B a radical as well.

REACTIONS:

• Photolysis of SF₆ under UV light:
  • SF₆ → SF₅• + F• (Species A: SF₅ is a radical with an unpaired electron on sulfur)
• Formation of radical B:
  • SF₅• + O₂ → SF₄O• (Radical B has an unpaired electron on the oxygen atom)

EXPLANATION:

• Species A (SF₅•) is a paramagnetic radical with an unpaired electron on the sulfur atom.
• Species B (SF₄O•) has an unpaired electron on one of the oxygen atoms, indicating that the unpaired electrons in A and B are located on sulfur and oxygen atoms, respectively.

CONCLUSION:

The correct answer is Option 1: Unpaired electrons in A and B are located on sulfur and oxygen atoms, respectively.

Concept:

• Group 15 element includes nitrogen (N), phosphorus (P), arsenic (As), antimony (Sb), and bismuth (Bi). The p-block elements are also known as the Representative Elements which are placed on the right side of the main periodic table.
• The Group 15 elements have a particular name pnictogens.
• All the elements of group 15 form tri-halides of the types EX₃ and EX₅. Nitrogen does not form pentahalide due to its valence shell's non-availability of the d-orbitals. E= N, P, As, Sb and Bi. X=F, Cl, Br, and I.
• The halides of group 15 may exhibit Lewis acidity.

Explanation:

• AsCl3 can act as a lewis acid as As can use its vacant d orbitals to accommodate a Cl^- ion. Thus, AsCl3 can form complexes of the type [AsCl4]- in the presence of a chloride source.

​\({\rm{AsC}}{{\rm{l}}_{\rm{3}}}{\rm{ + C}}{{\rm{l}}^{\rm{ - }}} \to {\rm{ [AsC}}{{\rm{l}}_{\rm{4}}}{{\rm{]}}^{\rm{ - }}}\)

• PF₃ acts as a weak σ-donor and strong pi-acceptor ligand. This is because in the case of PF₃ due to the electron-withdrawing effect of fluoride ion the phosphorous atom becomes electron deficient and hence acts as a weak σ-donor ligand.
• In the solid-state SbF₅ forms a polymeric chain. The crystalline SbF5  material is a tetramer, meaning that it has the formula [SbF₄(μ-F)]₄. Where 2 Fluorine atoms are shared between three Sb centers.
• SbF₅ is a strong Lewis acid, exceptionally so toward sources of F⁻ ions to give the very stable anion [SbF₆]⁻ ion, called hexafluoroantimonate. [SbF₆]⁻. Due to the high affinity of SbF₅ towards fluoride ions, The reaction of SbF5 with anhydrous HF generates [H2F]+ ions.

​\({\rm{Sb}}{{\rm{F}}_{\rm{5}}}{\rm{ + 2 HF}} \to {{\rm{[Sb}}{{\rm{F}}_{\rm{6}}}{\rm{]}}^{\rm{ - }}}{\rm{ + }}{{\rm{H}}_{\rm{2}}}{{\rm{F}}^{\rm{ + }}} \).

Conclusion:

• Hence, the correct statements for the Group 15 halides are (i) and (iv).​

Concept:

• Zeolites are crystalline aluminosilicate materials that are commonly used as commercial adsorbents and catalysts.
• Zeolites mainly consist of the elements such as silicon, aluminum, and oxygen. It has a general formula of 

​\({\rm{M}}_{{{\rm{1}} \over {\rm{n}}}}^{{\rm{n + }}}\left( {{\rm{Al}}{{\rm{O}}_{\rm{2}}}^{\rm{ - }}} \right){\left( {{\rm{Si}}{{\rm{O}}_{\rm{2}}}} \right)_{\rm{x}}}{\rm{.y}}{{\rm{H}}_{\rm{2}}}{\rm{O}} \)

• Indium tin oxide or commonly known as ITO is a ternary composition of elements such as Indium (In), Tin (Sn), and Oxygen (O).
• LiCoO2 or Lithium cobaltate is a dark blue or bluish-gray crystalline solid. The oxidation state of Co in LiCoO2 is +3.
• Platinum (Pt) metal forms alloys with various types and amounts of transition metals such as Co, Ni, Fe, Cu, Pd, Au, Ag, Mo, Mn, and Al.

Explanation:

• Zeolites consist of metal ion \({\rm{M}}_{{{\rm{1}} \over {\rm{n}}}}^{{\rm{n + }}}\), which can be exchanged for others in a contacting electrolyte solution. H⁺ ion-exchanged zeolites are particularly useful as solid acid catalysts.
• Membrane separation using zeolites involves energy efficient way to capture CO₂.
• Solid LiCoO₂ consists of lithium-ion (Li⁺) layers between the extended anionic sheets of cobalt and oxygen atoms. It is used as a positive electrode of Li-ion batteries.
• Indium tin oxide or ITO has a very high melting point and it is used as transparent conducting oxide because of its electrical conductivity and optical transparency. It has applications on solar panel displays.
• Pt alloy is used in fuel cells as a platinum alloy–based alloy is the best electrocatalyst for the anodic and cathodic reactions in fuel cells.

Conclusion:

• Hence, the correct match is a – ii; b – i; c – iv; d – iii.

Concept:-

• ​The acidic or basic strength of any solution can be described on the basis of pH value.
• pH can be expressed as the decimal logarithm of the reciprocal of the hydrogen ion activity, aH+, in a solution.
• pH can be expressed as, 

pH=-log[aH+]

The pKa value is the negative base -10 logarithm of the acid dissociation constant (Ka) of a solution. The higher the value of pKa the lower will be the acid strength.

Usanovich proposed a very wide definition of acids and bases. According to usanovich's concept, an acid is any chemical species that would

• reacts with a base, or

• accepts an anion or electrons, or

• furnishes cations

Whereas a base is any chemical species that would

• reacts with an acid, or

• donatess anions or electrons, or

• combines with cations.

Explanation:-

For the reaction,

\({\rm{Xe + Pt}}{{\rm{F}}_{\rm{6}}}\buildrel {{\rm{S}}{{\rm{F}}_{\rm{6}}}} \over \longrightarrow {\left[ {{\rm{Xe}}} \right]^{\rm{ + }}}{\left[ {{\rm{Pt}}{{\rm{F}}_{\rm{6}}}} \right]^{\rm{ - }}}\)

​Xe donates an electron to form a mono-positive ion [Xe]⁺ and PtF₆ accepts an electron to form a mono-negative ion  [PtF6]^-.

• As per usanovich's concept for this reaction, Xe is a base as it donated an electron, and PtF₆ is an acid as it accepted an electron in the reaction.
• Now, for the reaction

\({\rm{Xe}}{{\rm{F}}_{\rm{4}}}{\rm{ + M}}{{\rm{e}}_{\rm{4}}}{\rm{NF}} \to {\left[ {{\rm{M}}{{\rm{e}}_{\rm{4}}}{\rm{N}}} \right]^{\rm{ + }}}{\left[ {{\rm{Xe}}{{\rm{F}}_{\rm{5}}}} \right]^{\rm{ - }}}\)

​Me₄NF donates a fluoride ion (F^-) to form a mono-positive ion [Me4NF]+ and XeF4 accepts a fluoride ion (F-) to form a mono-negative ion [XeF5]-.

• As per usanovich's concept for this reaction, Me4NF is a base as it donated a fluoride ion (F-), and XeF4 is an acid as it accepted a fluoride ion (F-).

Conclusion:-

• Hence, the correct statement for the reaction is Xe acts as a base and XeF4 acts as an acid.

Key Points

Boron Nitride:

• The elements of B and Indium form stable solid nitrides MN. BN has a layer lattice-like graphite while the others have a diamond-like structure.
• The common form of Boron nitride consists of a layer lattice similar to graphite.
• In each layer alternate Boron and Nitrogen, atoms form planar hexagons.
• The layers are stacked over one another so that the N atom of one layer is directly over the B atoms of the other layers.
• Difference from that of graphite where the hexagonal rings are stacked over alternate layers.
• In Boron nitride, the B-N distance in each layer is 145pm, which suggests that there is substantial π bonding within the layer.
• Boron Nitride is sometimes called 'inorganic graphite'. However, unlike graphite, it is colourless and thus an insulator.
• Hexagonal BN is converted to a cubic form comparable to diamond when heated at 1800° C under 85000 atm pressure, preferably in presence of an alkali or alkaline earth metal catalyst.
• This extremely hard variety, called borazon, is used in cutting a diamond.

Reactions of BN:

• Boron Nitrides react with nitrogen or ammonia to form (BN)_x.
• Boron nitride is a slippery white solid which melts under pressure at 3000° C.
• It is chemically inert to air, oxygen, hydrogen, chlorine, iodine etc. even on heating.
• It is decomposed by water, fluorine and HF.
• It is also decomposed on fusion with KOH or K₂CO₃.

2BN + 3F₂ → 2BF₃ + N₂.

Hence, the CORRECT statement about hexagonal boron nitride is it is reactive towards Fluorine.

CONCEPT:

Covalency in X-F Bonds

• The covalency in X-F bonds depends on the central atom's ability to accommodate multiple bonds with fluorine atoms. This generally increases as the central atom moves down the periodic table (from group 14 to group 17), as larger atoms have more available orbitals for bonding and can form higher coordination compounds.
• In general, the order of covalency increases with the size of the central atom because larger atoms can form more bonds. For example:
  • SiF₄ (Silicon tetrafluoride): Silicon belongs to group 14 and can form 4 covalent bonds with fluorine.
  • PF₅ (Phosphorus pentafluoride): Phosphorus belongs to group 15 and can form 5 covalent bonds with fluorine.
  • SF₆ (Sulfur hexafluoride): Sulfur belongs to group 16 and can form 6 covalent bonds with fluorine.
  • IF₇ (Iodine heptafluoride): Iodine belongs to group 17 and can form 7 covalent bonds with fluorine.
• As the size of the central atom increases, its ability to expand its octet and accommodate more fluorine atoms also increases, which leads to an increase in covalency in the X-F bonds.

EXPLANATION:

• The correct order of covalency in X-F bonds is based on the number of fluorine atoms attached to the central atom and its ability to form stable covalent bonds:
  • SiF₄ < PF₅ < SF₆ < IF₇
• This order reflects the increasing number of fluorine atoms and the corresponding increase in covalency as you move from SiF₄ (4 bonds) to IF₇ (7 bonds).

CONCLUSION:

The correct answer is Option 1: SiF₄ < PF₅ < SF₆ < IF₇.

Concept

• Back bonding occurs as electrons pass from one atom’s atomic orbital to another atom’s or ligand’s anti-bonding orbital. This form of bonding will occur between atoms in a compound when one atom has a lone pair of electrons and the other has a vacant orbital next to it.​

​

• Every ligand is a σ donor at first. 
• CO donates its electron to metal at first and forms a bond with it.
• The metal will donate electrons to carbon and shows back bonding. The electrons of metal will enter LUMO (lowest u occupied molecular orbital) of CO which is antibonding molecular orbital π *. 

​

• As we know whenever an electron enters in an antibonding orbital then the bond order decreases and hence the bond strength also decreases. So due to back bonding between metal and carbons, the CO bond strength decreases and there arises a slight double bond character between metal and carbon, and the bond order between the carbon and oxygen in carbonyl (CO) decreases from three to two.
• Hence M-C bond strength increase and the C-O bond strength decreases.
• M-C bond strength ∝ 1 / C-O bond strength
• Stretching frequency is given by \(ν= {{1 \over 2 \pi c} \sqrt{k \over μ} } \)   where k represents bond strength. According to this ν (frequency) ∝ k (bond strength)
• Thus, the greater the negative charge on the metal, the more will be the extent of back bonding with the carbonyl (CO) group, and hence lower will be its CO stretching frequency.

​Explanation:-

A. The CO stretching frequency in IR is less than 2143 cm-1.

• In the Lewis acid-base adduct (CF3)3B⋅CO, (CF3)3B acts as a Lewis acid and CO acts as a base (σ donor).
• Due to the donation of σ electrons from antibonding HOMO orbital will result in an increase in bond order and bond strength which results in a higher CO stretching frequency.
• Thus, the CO stretching frequency in IR will be more than 2143 cm-1 (in free CO).
• Thus, statement A is incorrect.

B. The 19F NMR spectrum shows one singlet resonance only.

• In the compound (CF3)3B⋅CO, the F atoms in each CF₃ is present in different chemical environments and thus show more than one singlet resonance.
• Thus, statement B is incorrect.

C. The point group of (CF3)3B⋅CO is C3v.

• The molecule (CF3)3B⋅CO, contains a C3 proper axis of symmetry and one σv plane.
•  So, the point group of (CF₃)₃B⋅CO is C_3v.

• Thus, statement C is correct.

D. (CF3)3B⋅CO reacts with KF to form K[(CF3)3BC(O)F].

• When (CF3)3B⋅CO reacts with KF it forms K[(CF3)3BC(O)F].

​(CF3)3B⋅CO + KF → K[(CF3)3BC(O)F]

• Thus, statement D is correct.

Conclusion:-

• Hence, the correct statements are C and D only.