Carbon family Questions in English

(B) The reaction of methyl chloride with silicon in the presence of copper catalyst at $570 \ K$ is known as the Direct Process,which produces methylchlorosilanes.
$(P)$ is dimethyl dichlorosilane,$(CH_3)_2 SiCl_2$.
Hydrolysis of $(CH_3)_2 SiCl_2$ yields dimethyl silanediol,$(CH_3)_2 Si(OH)_2$,which is $(Q)$.
$(CH_3)_2 Si(OH)_2$ undergoes condensation polymerization to form straight-chain silicone polymers.
Reaction:
$2 CH_3 Cl + Si \xrightarrow[570 \ K]{\text{Cu powder}} (CH_3)_2 SiCl_2 (P)$
$(CH_3)_2 SiCl_2 + 2 H_2 O \longrightarrow (CH_3)_2 Si(OH)_2 (Q) + 2 HCl$
$(n) (CH_3)_2 Si(OH)_2 \longrightarrow \text{Straight chain polymer} + (n) H_2 O$

(C) Buckminsterfullerene $(C_{60})$ is an allotrope of carbon consisting of $60$ carbon atoms arranged in a soccer ball-like structure.
It contains $20$ six-membered rings $(x = 20)$ and $12$ five-membered rings $(y = 12)$.
Therefore,the value of $(x+y) = 20 + 12 = 32$.

(B) Electron precise hydrides are formed by elements of group $14$ of the periodic table.
These elements have exactly the number of electrons required to form the necessary covalent bonds (e.g.,$CH_4$,$SiH_4$,$GeH_4$).
Among the given options,$C, Si, Ge$ belong to group $14$ and thus form electron precise hydrides.

(C) The acidity is directly proportional to the tendency of the hydrogen donor.
It depends upon the bond dissociation enthalpy of the $H-X$ bond,where $X$ is $O, S, Se, Te$.
As we move down the group,the bond dissociation energy of hydrides decreases due to an increase in bond length,which makes the release of $H^+$ ions easier.
Therefore,the acidic strength increases down the group.
Hence,the correct order of acidic character of the given hydrides is $H_2Te > H_2Se > H_2S > H_2O$.

(B) $I)$ $CO$ binds to hemoglobin to form carboxyhemoglobin,which is about $300$ times more stable than oxyhemoglobin,thus reducing the oxygen-carrying capacity of blood. This statement is correct.
$II)$ Producer gas is a mixture of carbon monoxide $(CO)$ and nitrogen $(N_2)$,typically produced by the partial combustion of carbon in air. This statement is correct.
$III)$ The $C-O$ bond length in $CO_2$ is $115 \ pm$. This statement is correct.

(C) Buckminster fullerene $(C_{60})$ is an allotropic form of carbon that exhibits aromatic character due to the delocalization of $\pi$-electrons within its cage-like structure.
$(I)$ The number of five-membered rings in all fullerenes is $12$.
$(II)$ The number of six-membered rings in $C_{60}$ is $20$.

(D) graphite crucible is a container used for melting and casting non-ferrous metals such as gold,silver,and aluminium.
Activated charcoal is an ideal water filter because it removes toxins from water without stripping it of essential salts and minerals.
Carbon black is primarily used as a pigment and reinforcing filler in tires and rubber products,not as a fuel.
Therefore,only statements $(i)$ and $(ii)$ are correct.

(B) $CO$ is a neutral oxide.
$GeO_2$ is an acidic oxide (as $Ge$ is a metalloid and higher oxidation state oxides of group $14$ are acidic).
$SnO$ and $PbO$ are amphoteric oxides.
Therefore,the correct option is $B$.

(B) Statement $I$ is correct as $Ge$ and $Sn$ exist in the earth's crust in trace amounts $(1-2 \ ppm)$.
Statement $II$ is correct as $Pb$ belongs to group $14$ and has four valence electrons,all of which are used to form four $Pb-F$ bonds,resulting in a tetrahedral geometry.
Statement $III$ is incorrect because the stability of the $+2$ oxidation state increases down the group due to the inert pair effect,making it more stable for heavier elements like $Pb$. For $Ge$,the $+4$ oxidation state is more stable than the $+2$ state,so $GeX_4$ is more stable than $GeX_2$.

(A) The thermodynamic stability of carbon allotropes is determined by their standard enthalpy of formation.
Graphite is the standard state of carbon,meaning its $\Delta H_f^{\circ} = 0 \ kJ \ mol^{-1}$.
Other allotropes like diamond have a positive enthalpy of formation $(\Delta H_f^{\circ} = +1.9 \ kJ \ mol^{-1})$,making them less stable than graphite.
The order of thermodynamic stability is: $\text{Graphite} > \text{Diamond} > \text{Fullerene} > \text{Coke}$.

(A) The correct matches are as follows:
$A$. Carbon black is used as a filler in automobile tyres $(IV)$.
$B$. Graphite is used as electrodes in batteries $(I)$.
$C$. Diamond is used as an abrasive due to its extreme hardness $(III)$.
$D$. Activated charcoal is used in air conditioning systems to adsorb poisonous gases $(V)$.
Therefore,the correct sequence is $A-IV, B-I, C-III, D-V$.

(A) Buckminsterfullerene $(C_{60})$ consists of $20$ six-membered rings and $12$ five-membered rings.
Therefore,$x = 20$ and $y = 12$.
The sum $x + y = 20 + 12 = 32$.

(B) Fullerene and graphite are crystalline allotropes of carbon.
In fullerene $(C_{60})$,each carbon atom is bonded to three other carbon atoms,resulting in $sp^2$ hybridisation.
Similarly,in graphite,each carbon atom is bonded to three other carbon atoms in a hexagonal planar structure,which also results in $sp^2$ hybridisation.

(C) The coordination number of an element depends on the availability of vacant orbitals in the valence shell.
Carbon is in the $2^{nd}$ period with the electronic configuration $1s^2 2s^2 2p^2$. It lacks vacant $d$-orbitals,limiting its maximum covalency to $4$.
Germanium is in the $4^{th}$ period with the electronic configuration $[Ar] 3d^{10} 4s^2 4p^2$.
Due to the presence of vacant $d$-orbitals,germanium can expand its coordination number beyond $4$.

(B) Fullerenes are allotropes of carbon consisting of clusters of carbon atoms,such as $C_{60}$,$C_{70}$,$C_{76}$,and $C_{84}$.
Fullerenes are synthesized by heating graphite in an electric arc in the presence of an inert gas like helium or argon at low pressure.
During this process,the carbon vaporizes and condenses to form a soot containing various fullerenes.
$C_{60}$ (Buckminsterfullerene) is the most abundant product,while $C_{70}$ is obtained in smaller quantities.

(A) When carbon combines with oxygen,it can form four distinct oxides or oxocarbons:
$1$. Carbon monoxide: $CO$
$2$. Carbon dioxide: $CO_2$
$3$. Carbon suboxide: $C_3O_2$
$4$. Mellitic anhydride: $C_{12}O_9$
Thus,there are $4$ such compounds.

(A) $CCl_4$ cannot undergo hydrolysis because the $C$ atom,being a member of the second period,does not have vacant $d$-orbitals in its valence shell.
Therefore,it cannot accommodate the lone pair donated by the oxygen atom of the water molecule.
In contrast,$SiCl_4$,$VCl_4$,and $TiCl_4$ possess vacant $d$-orbitals,allowing them to undergo hydrolysis.
Hence,the correct option is $A$.

(B) In group $14$,as we move down the group,the atomic size increases and the bond energy of the element-element bond decreases.
This leads to a decrease in the tendency to show catenation.
The order of catenation is: $C >> Si > Ge \approx Sn$.
Lead $(Pb)$ does not show catenation due to its large size and weak bond strength.
Thus,the correct option is $B$.

(A) $(I)$ In diamond,each carbon atom is $sp^3$ hybridised. This statement is correct.
$(II)$ Graphite consists of planar hexagonal layers of carbon atoms where each carbon is $sp^2$ hybridised. This statement is correct.
$(III)$ Silicones are organosilicon polymers with the general formula $(R_2SiO)_n$. Due to the presence of non-polar alkyl groups,they are hydrophobic (water-repelling) in nature. This statement is correct.
$(IV)$ The tendency for catenation in group $14$ elements decreases down the group due to the decrease in bond dissociation energy of the element-element bond. The correct order is $C >> Si > Ge \approx Sn$. Thus,statement $(IV)$ is incorrect.
Therefore,the correct statements are $(I)$,$(II)$,and $(III)$.

(C) Statement $I$ is correct: $Sn^{2+}$ is a reducing agent and $Pb^{4+}$ is a strong oxidising agent due to the inert pair effect.
Statement $II$ is incorrect: Silicon exists as $[SiF_6]^{2-}$ but not as $[SiCl_6]^{2-}$ because the large size of $Cl^-$ ions makes the $[SiCl_6]^{2-}$ ion sterically unstable.
Statement $III$ is incorrect: The hybridisation of carbon in fullerene is $sp^2$,as each carbon atom is bonded to three other carbon atoms.
Statement $IV$ is incorrect: Among $Ge$,$Sn$,and $Pb$,the melting point order is $Ge > Sn > Pb$,so the lowest melting point is for $Pb$.
Therefore,statements $II$,$III$,and $IV$ are incorrect. However,based on the provided options,the most appropriate choice is $II$ and $III$.

(A) The statement '$CO$ is used in the manufacture of Urea' is incorrect.
In the industrial production of urea,$NH_3$ and $CO_2$ are used as raw materials,not $CO$.
The reaction is: $2NH_3 + CO_2 \rightarrow NH_2CONH_2 + H_2O$.
Quartz is indeed a piezoelectric material.
Silicones are synthetic polymers used as electrical insulators due to their high thermal stability and water-repellent nature.
$ZSM-5$ is a type of zeolite catalyst used in the petrochemical industry to convert alcohols directly into gasoline.

(C) $CO$ (carbon monoxide) is a strong reducing agent and is used in the extraction of many metals from their oxides. However,it is not strong enough to reduce alkali metal oxides (like $Na_2O$ or $K_2O$) to their respective alkali metals because alkali metals are highly electropositive and have a very high affinity for oxygen. Thus,statement $C$ is incorrect.

(C) Statement $a$ is incorrect because $C_{60}$ (Buckminsterfullerene) consists of $20$ six-membered rings and $12$ five-membered rings,not the other way around.
Statement $b$ is correct; the $H_2CO_3 / HCO_3^-$ buffer system is essential for maintaining the blood $pH$ in the range of $7.26$ to $7.42$.
Statement $c$ is correct; graphite has a layered structure with weak van der Waals forces between layers,allowing them to slide over each other,making it an excellent dry lubricant at high temperatures.
Therefore,statements $b$ and $c$ are correct.

(C) $[SiF_6]^{2-}$ is known to exist,but due to the large size of the chlorine $(Cl)$ atom and steric hindrance,$[SiCl_6]^{2-}$ ions are not known to exist. Therefore,the statement in option $C$ is incorrect.

(D) Carbon monoxide is estimated using iodine pentoxide $(I_2O_5)$ as per the following reaction:
$I_2O_5 + 5 CO \longrightarrow I_2 + 5 CO_2$
The liberated iodine is then titrated against a standard sodium thiosulfate solution to determine the amount of $CO$ present.

(D) The correct matches are as follows:
$A$. Kieselghur is an amorphous form of silica used in $III$. Filtration plants.
$B$. Silica gel is used as $I$. Chromatographic material.
$C$. $ZSM-5$ is a type of zeolite catalyst used to $IV$. Convert alcohol directly into gasoline.
$D$. Hydrated zeolites are used for $II$. Softening of hard water.
Therefore,the correct matching is $A-III, B-I, C-IV, D-II$.

(C) Silicones are organosilicon polymers with the general formula $(R_2SiO)_n$.
They possess several unique properties:
$1$. They are bio-compatible and are used in surgical and dental implants.
$2$. They exhibit high thermal stability due to the strong $Si-O$ bond.
$3$. They are hydrophobic (water-repelling) in nature due to the organic side groups.
$4$. They have high dielectric strength,making them excellent electrical insulators.
Therefore,the statement that they have low dielectric strength is incorrect.

(D) Silica $(SiO_2)$ is a covalent network solid with a three-dimensional tetrahedral structure,where each silicon atom is bonded to four oxygen atoms.
Option $A$ is correct because $SiO_2$ is acidic and reacts with bases.
Option $B$ is correct because $SiO_2$ is inert to most acids but reacts with hydrofluoric acid $(HF)$ to form silicon tetrafluoride $(SiF_4)$.
Option $C$ is correct because $SiO_2$ reacts with $NaOH$ to form sodium silicate $(Na_2SiO_3)$.
Option $D$ is incorrect because,unlike graphite which has a two-dimensional layered structure,silica has a three-dimensional giant covalent network structure.

(A) Statement-$I$ is correct because Carbon $(C)$ does not have vacant $d$-orbitals to accommodate the lone pair of electrons from water,whereas Silicon $(Si)$ has vacant $3d$-orbitals to accept the lone pair from water molecules,facilitating hydrolysis.
Statement-$II$ is correct because for Germanium $(Ge)$,the $+4$ oxidation state is more stable than the $+2$ oxidation state due to the relatively weaker inert pair effect compared to heavier elements like $Pb$.

(C) The correct answer is $SiCl_4$.
$1$. Hydrolysis of $P_4 O_{10}$: $P_4 O_{10} + 6 H_2 O \rightarrow 4 H_3 PO_4$. Both products are not solid under standard conditions.
$2$. Hydrolysis of $F_2$: $2 F_2 + 2 H_2 O \rightarrow 4 HF + O_2$. Both $HF$ and $O_2$ are not solids.
$3$. Hydrolysis of $SiCl_4$: $SiCl_4 + 2 H_2 O \rightarrow SiO_2 + 4 HCl$. Here,$SiO_2$ (silica) is a solid product.
$4$. $N_2$ does not undergo hydrolysis under normal conditions.

(A) $i$. The basic structural unit of all silicates is the $SiO_4^{4-}$ tetrahedron,which is correct.
$ii$. Silicones are synthetic polymers that are chemically inert,water-repellent,and biocompatible,which is correct.
$iii$. Producer gas is a mixture of carbon monoxide $(CO)$ and nitrogen $(N_2)$,which is correct.
Therefore,all statements $i, ii,$ and $iii$ are correct.

(C) The melting and boiling points of group $13$ elements are lower than those of corresponding group $14$ elements due to the stronger metallic bonding in group $14$ elements. Thus,statement $(i)$ is incorrect.
$SiO$ is less stable than $SiO_2$ and exists only at high temperatures. Thus,statement $(ii)$ is correct.
$PbI_4$ does not exist because $Pb$ prefers the $+2$ oxidation state over the $+4$ state due to the inert pair effect. Thus,statement $(iii)$ is correct.
Buckminster fullerene $(C_{60})$ contains $20$ six-membered rings and $12$ five-membered rings. Thus,statement $(iv)$ is incorrect.
Therefore,the correct statements are $(ii)$ and $(iii)$.

(D) Silicon $(IV)$ chloride $(SiCl_4)$ undergoes hydrolysis to form silicic acid,which is $Si(OH)_4$ or $H_4SiO_4$.
The chemical reaction is as follows:
$SiCl_4 + 4H_2O \rightarrow Si(OH)_4 + 4HCl$
In the structure of silicic acid $(Si(OH)_4)$,the silicon atom is bonded to four $-OH$ groups.
Therefore,the number of $-OH$ groups present in $X$ is $4$.

(B) When silicon tetrachloride $(SiCl_4)$ reacts with an excess of water,it undergoes complete hydrolysis.
Silicon has vacant $d$-orbitals,which allow the water molecule to attack the silicon atom,leading to the substitution of chlorine atoms by hydroxyl groups.
The reaction proceeds as follows:
$SiCl_4 + 4H_2O \rightarrow Si(OH)_4 + 4HCl$
Therefore,the major product is silicic acid,$Si(OH)_4$.

(A) $SiO_2$ is a covalent three-dimensional network solid.
Silica (commonly known as quartz) is $SiO_2$,which is usually used to manufacture glass,ceramics,and piezoelectric materials.
The most common use of silica is in water filtration; because of its uniform shape and size,it is an effective filtration bed that removes contaminants from water.
$SiO_2$ is less reactive due to high $Si-O$ bond enthalpy. Therefore,the statement that it is highly reactive is incorrect.

(B) The two most abundant elements in the Earth's crust are oxygen and silicon. The compound $(A)$ formed by these elements is silicon dioxide $(SiO_2)$,which is widely used in building construction (e.g.,in sand and concrete).
When silicon dioxide $(SiO_2)$ reacts with carbon at high temperatures,it undergoes a reduction reaction:
$SiO_2 + 3C \xrightarrow{\Delta} SiC + 2CO$
Here,$(B)$ is carbon monoxide $(CO)$,which is a poisonous,stable diatomic gas.

(A) Silicones are used as greases (lubricants) and in surgical implants due to their inert nature and biocompatibility.
Zeolites are used as catalysts (e.g.,$ZSM-5$) and as ion exchangers in water softening.
Therefore,the correct matches are:
$A$ (Silicones) $\rightarrow$ $(ii)$ Greases,$(iii)$ In surgical implants.
$B$ (Zeolites) $\rightarrow$ $(i)$ As a catalyst,$(iv)$ As ion exchangers.
The correct code is $A-(ii, iii), B-(i, iv)$.

(C) In pyrosilicate $(Si_2O_7)^{6-}$,two silicate units $(SiO_4)^{4-}$ share one oxygen atom. The structure is represented as $(Si_2O_7)^{6-} = O_3Si-O-SiO_3$.

(A) $(CH_3)_3SiCl$ is used during the polymerisation of organosilicones because it acts as a chain terminator.
When added to the reaction mixture,it reacts with the terminal $-OH$ groups of the growing siloxane chain,forming a stable $(CH_3)_3Si-O-$ group at the end.
This prevents further condensation and thus allows the control of the chain length of the organosilicone polymers.
Hence,the correct option is $A$.

(D) Silicones are synthetic organosilicon polymers with the general formula $(R_2SiO)_n$.
They possess a hydrophobic (water-repelling) nature due to the presence of non-polar organic groups attached to the silicon atoms.
They are widely used in medical applications because they are biocompatible.
However,silicones are excellent electrical insulators,not conductors,due to the strong $Si-O$ bonds and the absence of free electrons or ions for charge transport.

(C) Statement $(I)$ is correct: Germanium exists only in traces $(1.5 \ ppm)$ and is mainly recovered from flue dust arising from the roasting of zinc ores.
Statement $(II)$ is incorrect: The electronegativity of group $14$ elements decreases from carbon to silicon and remains approximately constant from silicon to $Sn$ $(Si \approx Ge \approx Sn)$.
Statement $(III)$ is correct: All the group $14$ elements (Carbon,Silicon,Germanium,Tin,Lead) are available in the solid state at room temperature.
Therefore,statements $(I)$ and $(III)$ are correct.

(B) $[SiF_6]^{2-}$ is formed but $[SiCl_6]^{2-}$ is not because the size of the $Cl^-$ ion is significantly larger than that of the $F^-$ ion.
Due to the larger size of $Cl^-$,there is significant steric hindrance around the central $Si$ atom,which prevents the formation of the $[SiCl_6]^{2-}$ complex.
While the electronegativity of $F$ is indeed higher than that of $Cl$,this property is not the primary reason for the stability of $[SiF_6]^{2-}$ over $[SiCl_6]^{2-}$.
Therefore,both $(A)$ and $(R)$ are correct statements,but $(R)$ is not the correct explanation for $(A)$.

(C) Methyl chloride reacts with silicon in the presence of copper powder as a catalyst at $570 \ K$ to form dimethyldichlorosilane as the major product. This is the first step in the synthesis of silicones.
The reaction is: $2 CH_3Cl + Si \xrightarrow[570 \ K]{Cu} (CH_3)_2SiCl_2$
Comparing this with the given reaction,we get:
$X = Si$,$Y = Cu$,and $Z = (CH_3)_2SiCl_2$.

(B) Photovoltaic cells are devices that convert sunlight directly into electricity. $Amorphous$ $silicon$ is widely used in the manufacturing of these solar cells because it is an effective semiconductor material that can be deposited in thin films,making it cost-effective for large-scale energy production.

(A) Carbon belongs to the second period and has only $s$ and $p$ orbitals in its valence shell $(n=2)$. It lacks $d$-orbitals,so it cannot expand its octet,limiting its maximum covalency to $4$.
Silicon $(Si)$ and Germanium $(Ge)$ belong to the third and fourth periods respectively. They possess vacant $d$-orbitals in their valence shells ($n=3$ and $n=4$).
Due to the availability of these $d$-orbitals,$Si$ and $Ge$ can undergo $sp^3d^2$ hybridisation,allowing them to form $6$ bonds,thus achieving a maximum covalency of $6$.
Therefore,both the assertion and the reason are correct,and the reason correctly explains the assertion.

(D) The correct answer is $D$.
Water gas is a mixture of $CO$ and $H_2$.
It does not contain $CO_2$ as a major component.
$SiO_2$ reacts with $HF$ to form $SiF_4$ and $H_2O$,making it soluble.
$(CH_3)_2SiCl_2$ undergoes hydrolysis followed by condensation to form linear silicones.
Zeolites are used for water softening by ion exchange.

(A) The tendency for catenation is directly proportional to the bond dissociation energy of the element-element bond.
As we move down the group from $C$ to $Ge$,the atomic size increases,which leads to a decrease in the overlap of orbitals and consequently a decrease in bond strength.
The bond dissociation energies for $C-C$,$Si-Si$,and $Ge-Ge$ are approximately $348 \ kJ \ mol^{-1}$,$180 \ kJ \ mol^{-1}$,and $167 \ kJ \ mol^{-1}$ respectively.
Therefore,the correct order of bond energies matches the order of catenation tendency.

(A) The species $[SiCl_6]^{2-}$ does not exist.
Silicon $(Si)$ has a small atomic size and cannot accommodate six large $Cl^-$ ions around it due to steric hindrance.
Additionally,the $Si-Cl$ bond is not strong enough to stabilize the hexacoordinated structure.
In contrast,$[SiF_6]^{2-}$ exists because the $F^-$ ion is small enough to fit around the $Si$ atom.

(D) $I)$ $SnF_4$ is an ionic compound due to the high electronegativity difference between $Sn$ and $F$. This statement is correct.
$II)$ Due to the inert pair effect,the stability of the $+2$ oxidation state increases down the group for group $14$ elements. Thus,the stability of dihalides increases down the group. This statement is correct.
$III)$ $Ge$ is in the $+4$ oxidation state in $GeCl_4$ and $+2$ in $GeCl_2$. Since $+4$ is more stable for $Ge$ than $+2$,$GeCl_4$ is more stable than $GeCl_2$. This statement is incorrect.