Carbon family Questions in English

(N/A) The hydrolysis of $RSiCl_3$ leads to the formation of a silanetriol,$RSi(OH)_3$,which undergoes polymerization to form a cross-linked silicone polymer.
Step $1$: Hydrolysis
$RSiCl_3 + 3H_2O \rightarrow RSi(OH)_3 + 3HCl$
Step $2$: Polymerization
The silanetriol units undergo condensation polymerization by eliminating water molecules to form a three-dimensional cross-linked structure:
$n RSi(OH)_3 \rightarrow (RSiO_{1.5})_n + 1.5n H_2O$
The resulting structure is a cross-linked polymer where each silicon atom is bonded to an $R$ group and three oxygen atoms,which are shared with other silicon atoms.

(D) Silicones are used as sealants,greases,and electrical insulators. They are also used for the waterproofing of fabrics. Due to their biocompatible nature,they are widely used in surgical and cosmetic implants.

(N/A) Feldspar,zeolites,mica,and asbestos are silicate minerals. The basic structural unit of silicates is $SiO_{4}^{4-}$. In this unit,the silicon atom is bonded to four oxygen atoms in a tetrahedral fashion.
In silicate compounds,these units are joined together via corners by sharing $1, 2, 3,$ or $4$ oxygen atoms. When silicate units are linked together,they form chain,ring,sheet,or three-dimensional structures. The negative charge on the silicate structure is neutralized by positively charged metal ions. Two important man-made silicates are glass and cement.

(N/A) If aluminium atoms replace a few silicon atoms in the three-dimensional network of silicon dioxide,the overall structure,known as aluminosilicate,acquires a negative charge. Cations such as $Na^{+}$,$K^{+}$,and $Ca^{2+}$ balance this negative charge. Examples include feldspar and zeolites.
Uses: Zeolites are widely used as catalysts in petrochemical industries for the cracking of hydrocarbons and isomerisation. For example,$ZSM-5$ (a type of zeolite) is used to convert alcohols directly into gasoline. Hydrated zeolites are used as ion exchangers in the softening of "hard" water.

(N/A) Carbon atoms have the unique tendency to link with one another through covalent bonds to form long chains and rings. This property of self-linking is called catenation.
Carbon also exhibits the ability to form $p\pi - p\pi$ multiple bonds with itself and with other atoms of small size and high electronegativity,such as $C=C, C \equiv C, C=O, C \equiv N$.

(B) On a small scale,pure $CO$ is prepared by the dehydration of formic acid $(HCOOH)$ using concentrated $H_2SO_4$ at $373 \ K$.
$HCOOH \xrightarrow[conc. \ H_2SO_4]{373 \ K} CO + H_2O$

(A) Aluminosilicates are formed when some of the $Si^{4+}$ ions in the three-dimensional network of $SiO_2$ are replaced by $Al^{3+}$ ions.
This substitution creates a negative charge on the framework,which is balanced by the presence of additional cations like $Na^+$,$K^+$,or $Ca^{2+}$.

(A) The general formula of silicone is $(R_{2}SiO)_{n}$,where $R$ represents an alkyl or aryl group (e.g.,methyl or phenyl).
The empirical formula of silicone is $R_{2}SiO$.
These compounds are structurally similar to ketones $(R_{2}C=O)$,which is why they are named silicones.

(N/A) $(1)$ The half-life of $^{14}C$ is $5770 \ years$.
$(2)$ $^{14}C$ isotope of carbon is used in radiocarbon dating.
$(3)$ The main component of ceramic,cement,and glass is $SiO_2$ (silica).

(N/A) Pure forms of germanium and silicon are primarily used as semiconductors in the manufacturing of electronic devices such as transistors,diodes,and integrated circuits.

(A) The hydrolysis of silicon tetrachloride $(SiCl_4)$ proceeds as follows:
$SiCl_4 + 4H_2O \rightarrow Si(OH)_4 + 4HCl$.
Thus,$Si(OH)_4$ (orthosilicic acid) is indeed the product formed at the end of the hydrolysis of $SiCl_4$.
Therefore,the statement is $True$.

(C) Catenation is the ability of an element to form bonds with other atoms of the same element to form long chains or rings.
In group $14$,the tendency to show catenation decreases down the group due to the decrease in bond dissociation enthalpy of the $M-M$ bond as the atomic size increases.
The order of catenation is $C \gg Si > Ge = Sn$.
$Pb$ does not show catenation due to its large atomic size and weak $Pb-Pb$ bond.

(B) Carbon has a small atomic size and high electronegativity,which allows its $2p$ orbitals to overlap effectively with the $2p$ orbitals of other small,highly electronegative atoms like $N$,$O$,and itself to form $p\pi-p\pi$ multiple bonds.
Therefore,carbon forms $p\pi-p\pi$ bonds with elements that have small size and high electronegativity.

(A) Fullerenes were discovered by $H.W. \ Kroto$,$R.E. \ Smalley$,and $R.F. \ Curl$ in $1985$.

(D) Diamond is used as an abrasive for sharpening hard tools.
It is used in making dies for wire drawing.
It is also used in the manufacture of tungsten filaments for electric light bulbs.

(A) Fullerene $(C_{60})$ consists of pentagons and hexagons.
Due to the delocalization of electrons, the bond lengths are intermediate between single and double bonds.
The $C-C$ single bond length is $143.5 \ pm$ and the $C=C$ double bond length is $138.3 \ pm$.

(D) $(i)$ $CO_{2}$ is mainly used to prepare urea.
$(ii)$ $CO_{2}$ is used to prepare carbonated water.
$(iii)$ It is used as a fire extinguisher.

(N/A) Dry ice is produced by allowing liquefied $CO_{2}$ to expand rapidly,which causes it to solidify into a white,snow-like solid.
Its uses include:
$1$. It is used as a refrigerant for ice-cream and frozen foods.
$2$. It is used in the entertainment industry to create fog effects.

(A) Silica $(SiO_2)$ occurs in several crystallographic forms,which are interconvertible at suitable temperatures. The most common forms are $Quartz$,$cristobalite$,and $tridymite$.

(B) The starting materials for the manufacture of silicones are alkyl or aryl substituted silicon chlorides,$R_nSiCl_{(4-n)}$,where $R$ is an alkyl or aryl group.

(N/A) The basic structural unit of silicates is the $SiO_4^{4-}$ tetrahedron. Examples include $Feldspar$,$Zeolites$,$Mica$,and $Asbestos$.

(N/A) The manmade silicate is $Glass$ or $Cement$.
$ZSM-5$ is used to convert alcohols directly into gasoline.
Hydrated zeolites are used as ion exchangers in softening of hard water.

(N/A) If aluminium atoms replace a few silicon atoms in the three-dimensional network of silicon dioxide,the overall structure,known as aluminosilicate,acquires a negative charge.
Cations such as $Na^{+}$,$K^{+}$,and $Ca^{2+}$ are used to balance the electric charge.

(D) The basic structural unit of silicates is the $SiO_4^{4-}$ tetrahedron. When these units are linked together by sharing oxygen atoms,they can form various structures such as chains,rings,sheets,or three-dimensional networks.

(B) $Kieselghur$ is a naturally occurring,soft,siliceous sedimentary rock that is easily crumbled into a fine white to off-white powder.
It is an amorphous form of silica $(SiO_2)$ derived from the remains of diatoms.
Due to its porous nature,it is widely used in filtration plants and as an abrasive.

(A) Statement $(1)$ is True (Fullerene $C_{60}$ consists of $20$ hexagons and $12$ pentagons).
Statement $(2)$ is True ($CO$ is a neutral oxide,whereas $CO_2$ is an acidic oxide).
Statement $(3)$ is True (Boron fibres are used in high-strength,lightweight composite materials).
Statement $(4)$ is False (The bead formed with $CoO$ in the borax bead test is blue in colour,not yellow).

(A) $(1)$ True: As we move down the group $13$,the metallic character increases,and the oxides become more basic. Indium and thallium oxides are basic.
$(2)$ False: While $C$ and $Si$ are non-metals and $Ge$ is a metalloid,tin $(Sn)$ and lead $(Pb)$ are metals with relatively low melting points compared to other transition metals.

(N/A) $(i)$ $CCl_4$ is a non-polar covalent compound and cannot form hydrogen bonds with polar $H_2O$ molecules. Furthermore,$C$ does not have vacant $d$-orbitals to accommodate the lone pair of electrons from the oxygen atom of the water molecule,making it resistant to hydrolysis.
In contrast,$SiCl_4$ is easily hydrolyzed by water because the central $Si$ atom has vacant $d$-orbitals that can accommodate the lone pair of electrons from the oxygen atom of the water molecule. The hydrolysis reaction is as follows:
$SiCl_4 + 4H_2O \rightarrow Si(OH)_4 + 4HCl$
$(ii)$ Carbon atoms have a strong tendency to link with one another through covalent bonds to form long chains and rings; this property is called catenation. This is primarily because the $C-C$ bond enthalpy is very high,making the bond exceptionally strong. As we move down the group,the atomic size increases and electronegativity decreases,which leads to a significant reduction in the bond strength and,consequently,the tendency for catenation decreases. The order of catenation is $C \gg Si > Ge > Sn$. Lead $(Pb)$ does not show catenation.

(N/A) In $CO_2$, the carbon atom undergoes $sp$ hybridization. Two $sp$ hybridized orbitals of the carbon atom overlap with two $p$-orbitals of oxygen atoms to form two sigma bonds, while the other two electrons of the carbon atom are involved in $p\pi-p\pi$ bonding with oxygen atoms. This results in a linear shape with both $C-O$ bonds of equal length $(115 \ pm)$ and no dipole moment. $O=C=O \leftrightarrow -O-C \equiv O^{+} \leftrightarrow O^{+} \equiv C-O^{-}$
Silicon dioxide is a covalent, three-dimensional network solid in which each silicon atom is covalently bonded in a tetrahedral manner to four oxygen atoms. Each oxygen atom is in turn covalently bonded to another silicon atom. Each corner is shared with another tetrahedron. The entire crystal may be considered as a giant molecule in which eight-membered rings are formed with alternate silicon and oxygen atoms.
$(b)$ Silicon has vacant $d$-orbitals in its valence shell, due to which it can accommodate $6$ electrons from $6$ fluorine atoms, whereas carbon does not have $d$-orbitals and cannot expand its covalence beyond four.

(N/A) In $CO_{2}$, the carbon atom undergoes $sp$ hybridization. Two $sp$ hybridized orbitals of the carbon atom overlap with two $p$ orbitals of oxygen atoms to form two sigma bonds, while the other two electrons of the carbon atom are involved in $p\pi-p\pi$ bonding with oxygen atoms.
This results in a linear shape [with both $C-O$ bonds of equal length $(115 \ pm)$] with no dipole moment. Carbon can form a double bond with oxygen due to its small size.
Silicon dioxide is a covalent, three-dimensional network solid in which each silicon atom is covalently bonded in a tetrahedral manner to four oxygen atoms. Each oxygen atom is, in turn, covalently bonded to another silicon atom.
Each corner is shared with another tetrahedron. The entire crystal may be considered a giant molecule in which eight-membered rings are formed with alternate silicon and oxygen atoms. Silicon cannot form a stable $p\pi-p\pi$ double bond with oxygen due to its large size and lower electronegativity compared to carbon.

(N/A) Due to the inert pair effect,$Tl$ is more stable in the $+1$ oxidation state than in the $+3$ oxidation state. Therefore,$Tl(NO_3)_3$ acts as a strong oxidizing agent.
Carbon atoms have the tendency to link with one another through covalent bonds to form chains and rings. This property is called catenation. This is because $C-C$ bonds are very strong. Down the group,the atomic size increases and electronegativity decreases,and thereby,the tendency to show catenation decreases. This can be clearly seen from bond enthalpy values. The order of catenation is $C \gg Si > Ge \approx Sn$. Lead does not show catenation.

(A-5, B-3, C-4, D-1, 2) $A-5, B-3, C-4, D-1, 2$
$A$. In $BF_4^-$,$B$ is $sp^3$ hybridized,surrounded by $4$ bond pairs,and has no lone pair,resulting in a tetrahedral shape.
$B$. In $AlCl_3$,the octet of $Al$ is incomplete,making it an electron-deficient compound and a Lewis acid.
$C$. In $SnO$,the oxidation state of $Sn$ is $+2$,which can be further oxidized to $+4$.
$D$. In $PbO_2$,the oxidation state of $Pb$ is $+4$. Due to the inert pair effect,$Pb^{2+}$ is more stable than $Pb^{4+}$. Thus,$PbO_2$ acts as a strong oxidizing agent by reducing to $Pb^{2+}$.

The element is carbon $(C)$,which is tetravalent and forms monoxide $(CO)$ and dioxide $(CO_2)$.
Producer gas is a mixture of $CO_{(g)}$ and $N_{2_{(g)}}$.
Formation of producer gas:
$2C_{(s)} + O_{2_{(g)}} + 4N_{2_{(g)}} \xrightarrow{1273 \ K} 2CO_{(g)} + 4N_{2_{(g)}}$
Carbon monoxide acts as a strong reducing agent and reduces ferric oxide $(Fe_2O_3)$ to iron $(Fe)$.
Reduction of ferric oxide:
$Fe_2O_{3_{(s)}} + 3CO_{(g)} \xrightarrow{\Delta} 2Fe_{(s)} + 3CO_{2_{(g)}}$

(N/A) The non-metals with high melting and boiling points are $Boron$ and $Carbon$ (in the form of diamond).

(B) The given reactions are as follows:
$1$. Reaction of inorganic sulphite $X$ $(Na_{2}SO_{3})$ with dilute $H_{2}SO_{4}$:
$Na_{2}SO_{3} + H_{2}SO_{4} \rightarrow Na_{2}SO_{4} + H_{2}O + SO_{2} (Y)$
$2$. Reaction of $Y$ $(SO_{2})$ with $NaOH$:
$SO_{2} + 2NaOH \rightarrow Na_{2}SO_{3} (X) + H_{2}O$
$3$. Reaction of $X$ $(Na_{2}SO_{3})$ with $Y$ $(SO_{2})$ and water:
$Na_{2}SO_{3} + H_{2}O + SO_{2} \rightarrow 2NaHSO_{3} (Z)$
Thus,$Y$ is $SO_{2}$ and $Z$ is $NaHSO_{3}$.

(A) Let us analyze each statement:
$(a)$ Incorrect: Solid $CO_{2(s)}$ (dry ice) is used as a refrigerant,not gaseous $CO_{2(g)}$.
$(b)$ Incorrect: The structure of $C_{60}$ (Buckminsterfullerene) contains $20$ six-membered rings and $12$ five-membered rings,not $12$ six-membered and $20$ five-membered.
$(c)$ Correct: $ZSM-5$ is a shape-selective catalyst used in the petrochemical industry to convert alcohols directly into gasoline.
$(d)$ Correct: Carbon monoxide $(CO)$ is a colorless,odorless,and neutral gas.
Therefore,statements $(c)$ and $(d)$ are correct.

(B) $CO_{2}$ is an acidic oxide.
$SnO_{2}$ is an amphoteric oxide,meaning it reacts with both acids and bases.
$SiO_{2}$ is an acidic oxide.
$GeO_{2}$ is predominantly acidic in nature.

(A) Methanides are carbides that react with water to produce methane $(CH_{4})$ gas.
$Be_{2}C + 4 H_{2}O \rightarrow 2 Be(OH)_{2} + CH_{4}$
$Al_{4}C_{3} + 12 H_{2}O \rightarrow 4 Al(OH)_{3} + 3 CH_{4}$
$CaC_{2}$ is an acetylide,producing ethyne $(C_{2}H_{2})$.
$Mg_{2}C_{3}$ is an allylide,producing propyne $(C_{3}H_{4})$.
Therefore,$Be_{2}C$ and $Al_{4}C_{3}$ are methanides.

(A) The structure of $P_4O_{10}$ consists of a tetrahedral arrangement of four phosphorus atoms.
Each phosphorus atom is bonded to three oxygen atoms via $P-O-P$ bridges,and one terminal oxygen atom via a $P=O$ double bond.
Counting the bonds in the structure:
$1$. Total $P-O-P$ bonds = $6$.
$2$. Total $P=O$ bonds = $4$.
$3$. Total $P-P$ bonds = $0$.
Therefore,the order is $P-O-P (6) > P=O (4) > P-P (0)$.

(C) The reaction between phosphorus trichloride $(PCl_3)$ and phosphonic acid $(H_3PO_3)$ is given by:
$PCl_3 + 3H_3PO_3 \rightarrow 3H_3PO_2Cl + H_3PO_3$ (Note: The standard reaction for the formation of pyrophosphorous acid is $PCl_3 + 3H_3PO_3 \rightarrow 3H_3PO_2Cl + H_3PO_3$ is not the intended path here).
Based on the provided image,the reaction is $PCl_3 + H_3PO_3 \rightarrow H_4P_2O_5 + 3HCl$.
The product $H_4P_2O_5$ is pyrophosphorous acid.
In the structure of pyrophosphorous acid $(H_4P_2O_5)$,there are two $P-OH$ groups.
Since only the hydrogen atoms attached to oxygen atoms (in $P-OH$ groups) are ionisable,there are $2$ ionisable hydrogens in the molecule.

(D) The structure of $C_{60}$ (Buckminsterfullerene) consists of $60$ carbon atoms arranged in a truncated icosahedron shape.
It contains $20$ six-membered rings (hexagons) and $12$ five-membered rings (pentagons).
Each carbon atom is $sp^2$ hybridized and forms three sigma bonds.
The five-membered rings are fused only to six-membered rings,while the six-membered rings are fused to both six and five-membered rings.
Therefore,the statement that it contains $12$ six-membered rings and $24$ five-membered rings is incorrect.

(D) The given reactions are part of the Solvay process for the production of sodium carbonate.
$1$. $2NH_3 + H_2O(A) + CO_2 \rightarrow (NH_4)_2CO_3$
$2$. $(NH_4)_2CO_3 + H_2O + CO_2(B) \rightarrow 2NH_4HCO_3$
$3$. $NH_4HCO_3 + NaCl \rightarrow NaHCO_3(C) + NH_4Cl$
Comparing these with the given equations,we identify $A = H_2O$,$B = CO_2$,and $C = NaHCO_3$.

(B) The compound $[SiCl_6]^{2-}$ does not exist.
This is because the six large $Cl^-$ ions cannot be accommodated around the small $Si^{4+}$ atom due to steric hindrance and the inability of the silicon atom to effectively coordinate six large chloride ions.

(B) The enamel of human teeth is primarily composed of hydroxyapatite,which is a crystalline form of calcium phosphate,$Ca_{10}(PO_4)_6(OH)_2$.
In this structure,calcium exists in the $Ca^{2+}$ oxidation state.
Phosphorus exists in the phosphate ion $(PO_4)^{3-}$,where the oxidation state of phosphorus is $+5$ $(P^{5+})$.
Fluoride ions $(F^{-})$ are also often incorporated into the enamel structure to form fluorapatite,which strengthens the teeth.
Therefore,$P^{3+}$ is not present in the enamel of the teeth.

(B) The hydrolysis of $PCl_{3}$ proceeds as follows:
$PCl_{3} + 3H_{2}O \longrightarrow H_{3}PO_{3} (A) + 3HCl$
However,the question implies a stepwise hydrolysis where $A$ is an intermediate like $PCl_{2}(OH)$ or $PCl(OH)_{2}$.
Upon further hydrolysis,$A$ converts to $B$,which is phosphorous acid $(H_{3}PO_{3})$.
The structure of $H_{3}PO_{3}$ is $H-P(=O)(OH)_{2}$.
In $H_{3}PO_{3}$,only the two hydrogen atoms attached to oxygen atoms are ionisable as protons $(H^{+})$.
The hydrogen atom directly attached to the phosphorus atom is not ionisable.
Therefore,the number of ionisable protons in $B$ is $2$.

(A) $(CH_3)_4 Si$ is a simple silane.
$(CH_3) Si(OH)_3$ undergoes polymerization to form $2D$-silicone (cross-linked).
$(CH_3)_2 Si(OH)_2$ undergoes polymerization to form linear chain silicone.
$(CH_3)_3 Si(OH)$ undergoes condensation to form a dimer,$(CH_3)_3 Si-O-Si(CH_3)_3$.
Thus,the correct matching is $A-III, B-IV, C-I, D-II$.

(B) The reaction of ethanol with $PCl_{3}$ produces ethyl chloride and phosphorous acid $(H_{3}PO_{3})$,which is compound $A$.
$3C_{2}H_{5}OH + PCl_{3} \rightarrow 3C_{2}H_{5}Cl + H_{3}PO_{3} (A)$
Further reaction of $H_{3}PO_{3}$ with $PCl_{3}$ yields pyrophosphorous acid $(H_{4}P_{2}O_{5})$,which is compound $B$.
$2H_{3}PO_{3} + PCl_{3} \rightarrow H_{4}P_{2}O_{5} (B) + PCl_{3} + ...$ (The reaction stoichiometry leads to $H_{4}P_{2}O_{5}$).
In the structure of pyrophosphorous acid $(H_{4}P_{2}O_{5})$,there are two $P-H$ bonds.
Protons attached directly to the phosphorus atom $(P-H)$ are non-ionisable.
Therefore,there are $2$ non-ionisable protons in $H_{4}P_{2}O_{5}$.

(D) In fullerene, each carbon atom is bonded to three other carbon atoms. Two types of these bonds exist: those with greater single bond character and longer length $(143.5 \text{ pm})$ and those with greater double bond character and shorter length $(138.3 \text{ pm})$.

(B) In cyclic silicates,two oxygen atoms of each $SiO_4^{4-}$ tetrahedron are shared to form a ring with the general formula $[SiO_3]_n^{2n-}$.
The figure shows a ring consisting of $6$ silicon atoms.
Substituting $n = 6$ into the general formula,we get $[SiO_3]_6^{2(6)-} = [Si_6O_{18}]^{12-}$.
Thus,the cyclic silicate ion is $[Si_6O_{18}]^{12-}$.

(D)
As element $X$ forms a stable product of the type $XCl_4$,it must be tetravalent.
Among the given elements,$Al$ is trivalent ($+3$ oxidation state),$Na$ is monovalent ($+1$ oxidation state),$Ca$ is divalent ($+2$ oxidation state),and $Si$ is tetravalent ($+4$ oxidation state).
Therefore,$Si$ forms $SiCl_4$,which is a stable compound.
Thus,option $(d)$ is correct.