Main Group Elements and Their Compounds MCQ Quiz in मल्याळम - Objective Question with Answer for Main Group Elements and Their Compounds - സൗജന്യ PDF ഡൗൺലോഡ് ചെയ്യുക

The correct answer is n-1 and 2n

Concept:-

• Polymer Structure: Polymers are large molecules comprised of repeating subunits called monomers. The properties of a polymer, such as its flexibility or hardness, are greatly influenced by how these monomers are connected and structured.
• Siloxane Chemistry: Siloxanes, a type of silicone, are polymers with a silicon-oxygen backbone. The number of Si-O-Si linkages corresponds to the number of repeating monomer units minus one in such structures.
• Hydrolysis Reactions: Hydrolysis of a bond involves the splitting of that bond upon the addition of water. In the context of this polymer, the methyl groups can be hydrolyzed under certain conditions, and the number of such hydrolyzable groups is directly proportional to the number of repeating units in the polymer.

Explanation:-

The formula given [(CH₃)₂SiO]n represents a type of polysiloxane, a polymer with a silicon-oxygen backbone and organic (in this case, methyl) side groups. The repeating unit (CH₃)₂SiO forms the "monomer" that repeats n times to form the polymer.

Si-O-Si Bonds: Every monomer unit shares an oxygen with another, forming an Si-O-Si bond between each pair of monomers.
So, for 'n' monomers, there will be 'n-1' Si-O-Si bonds because the chain ends are not linked to any other unit.

Hydrolyzable Methyl Groups: Each (CH₃)₂SiO unit has two methyl (CH₃) groups, which can be hydrolyzed in the presence of certain conditions.
Since we have 'n' such units in the polymer, we will have '2n' hydrolyzable methyl groups

The correct option is option 3

Concept:

• Aluminum fluoride (AlF3) exists as a solid under standard conditions and features a polymeric structure. Rather than existing as individual AlF3 molecules, it forms a three-dimensional network.
• The structure of aluminum fluoride is trigonal planar around each aluminum ion, and each aluminum ion is octahedrally coordinated by six fluorine atoms.
• Each fluorine anion is surrounded tetrahedrally by four aluminum cations. This means the crystal structure is cyclic and each unit cell includes many AlF3 formula units, which could be described as AlF₆ octahedra corner-sharing to form a 3D network.
• The polymeric structure also results in each individual Al and F atom being connected to more than one counterpart. Specifically, each Al atom is connected to 6 different F atoms and each F atom is connected to 4 different Al atoms.
• This arrangement is very different from the gaseous phase of aluminum fluoride, which is monomeric and consists of discrete AlF₃ molecules.
• The structure of aluminum fluoride (AlF₃) is similar to that of the mineral cryolite (Na3AlF6), a mineral that is important in the industrial manufacture of aluminum.

Explanation:

 Here in the structure, each Al atom is connected to 6 different F atoms, and each F atom is connected to 4 different Al atoms.

Explanation:

In BF₃ ( D_зh) due to stronger 2p(B)-2p(F) π back bonding; the bond order of each B-F bond is greater than one while in BF₄^- (T_d) bond order of B-F is one therefore conversion of BF₃ to BF4- accompanies with bond elongation. Point group T_d is a group of high symmetry than Dзh

Conclusion:-

• Hence, Conversion of boron trifluoride to tetrafluoroborate accompanies increase in symmetry and bond elongation

Concept:

inert pair effect -

• Inert pair effect is the inertness of the inner ns subshell towards a chemical reaction.
• Due to inert pair effect the two electrons of outermost ns subshell is remain unshared in compounds of post-transition elements.
• For ex - In group 4A the stability of lower oxidation state increases down the group due to inert pair effect.

Explanation:

 Inert pair effect is -

• Inert pair effect is the inertness of the inner ns subshell towards the chemical reaction.
• It is especially shown in the group 3A and 4A.

Consequence of inert pair effect,  on moving down the group stability of lower oxidation state increases and higher oxidation state decrease in group 4A and 3A.

→ Sn and Pb are the members of group 4A and hence only they show the inert pair effect.

Conclusion

Therefore, Sn and Pb shows the inert pair effect.

Hence, the correct answer is option 2.

Explanation:

The reaction of PCl5 with NH3, NH4Cl and (NH4)2SO4 can produce different products, as shown below:

→ PCl₅ + 4 NH₃ → 4 NH₄Cl + P(NH₂)₃Cl₂

In this reaction, PCl₅ reacts with ammonia (NH₃) to form ammonium chloride (NH₄Cl) and phosphoryl imidochloride P(NH2)4Cl . This is an intermediate product which can further react with NH3 to form the final product P(NH2)4Cl  as follows:

P(NH₂)₃Cl₂ + NH₃ → P(NH2)4Cl 

→ PCl₅ + 4 NH₄Cl → 4 NH₃ + P(NH₄)₃Cl₂

In this reaction, PCl₅ reacts with ammonium chloride (NH₄Cl) to form ammonia (NH₃) and phosphoryl trichloride (P(NH₄)₃Cl₂). This is an intermediate product which can further react with NH₃ to form the final product P(NH₄)₄Cl as follows:

PP(NH4)3Cl2 + NH₃ → P(NH₄)₄Cl

→ PCl₅ + (NH₄)₂SO₄ → 2 NH₄Cl + Cl₃P = NPOCl₂

In this reaction, PCl5 reacts with ammonium sulfate (NH4)2SO4 to form ammonium chloride (NH₄Cl) and phosphoryl nitrate Cl3P = NPOCl2

nPCl5 + nNH4Cl \(\stackrel{150^{\circ}\rm C}{\longrightarrow}\) (NPCl2)n + 4nHCl

Conclusion: Therefore, the major product obtained by the reaction of PCl5 with NH3, NH4Cl and (NH4)2SO4 respectively are (PNCl2)n ,P(NH2)4Cl and Cl3P = NPOCl2.

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:

Phosphorus heptasulfide, free from yellow and white phosphorus appears as light-yellow crystals, light-gray powder, or fused solid.

Explanation:

Therefore, the number of P-P bonds is 0.

Explanation:

• Statement A: GaBr₂ exist as Ga⁺[GaBr₄]^- or GaBr(ionic) and GaBr₃(covalent). Thus, Statement A is correct.​
• Statement B: TlCl₃ is covalent whereas TlI₃ is ionic. This is correct because TlCl₃ tends to form covalent bonds, whereas TlI₃ exists as an ionic compound (3Tl⁺ and \(I_3^-\)).
• Statement C: The coordination number of Gd in GdCl₃·6H₂O is 8. This is correct because in this hydrate, the gadolinium ion (Gd³⁺) is typically surrounded by eight water molecules in its coordination sphere.
• Statement D: (BN)_x exists in a diamond-like structure similar to carbon. This statement is correct because boron nitride can form a cubic structure (c-BN) that is isoelectronic and structurally similar to diamond.

Conclusion:

The correct option is: 4) A, B, C and D. All the statements A, B, C, and D are correct based on the bonding and structural properties of the given compounds.