Carbon family Questions in English

(D) Aluminum forms a thin,protective layer of aluminum oxide $(Al_2O_3)$ on its surface when exposed to air. This layer is transparent,hard,and non-porous,which prevents further oxidation or tarnishing of the underlying metal. Because it does not tarnish,it maintains its reflective properties,making it an excellent material for mirrors.

(B) In the carbon family $(Group \ 14)$,as we move down the group,the atomic number increases.
Due to the inert pair effect,the stability of the $+2$ oxidation state increases while the stability of the $+4$ oxidation state decreases.
Therefore,the correct statement is that the stability of their $+2$ oxidation state increases.

(C) Silicon is found in most rocks as silica $(SiO_2)$ and silicates.

(D) Silica $(SiO_2)$ is soluble in hydrofluoric acid $(HF)$ because it reacts to form hexafluorosilicic acid.
The chemical reaction is:
$SiO_2 + 6HF \to H_2SiF_6 + 2H_2O$

(A) Volatility is inversely proportional to the molecular mass of the compound.
As we move down the group from $C$ to $Sn$,the molecular mass increases,which leads to stronger van der Waals forces of attraction.
Since $CH_4$ has the lowest molecular mass among the given hydrides,it has the weakest intermolecular forces and is therefore the most volatile.

(D) Diamond has a three-dimensional network structure similar to that of crystalline silicon. Both crystallize in the cubic system,making them isomorphous.

(C) Pyro silicates are also known as sorosilicates.
They contain the $Si_2O_7^{6-}$ unit,which is formed by joining two $SiO_4^{4-}$ tetrahedra by sharing one oxygen atom at one corner.
Therefore,the correct pyro silicate ion is $Si_2O_7^{6-}$.

(A) In silicates,the sharing of oxygen atoms determines the structure.
When $3$ oxygen atoms of each $[SiO_4]^{4-}$ tetrahedron are shared,it results in a two-dimensional sheet structure.
This is characteristic of sheet silicates (phyllosilicates).

(B) $C_{60}$ is a cage-like molecule known as Buckminsterfullerene.
It consists of $60$ carbon atoms arranged in a structure containing $12$ pentagons and $20$ hexagons.

(D) In group $14$ elements,the stability of the $+2$ oxidation state increases down the group due to the inert pair effect.
$Ge$ is at the top of the group,so $Ge(IV)$ is more stable than $Ge(II)$. Thus,$Ge(II)$ tends to lose two more electrons to reach the stable $+4$ state,acting as a strong reducing agent.
Conversely,$Pb$ is at the bottom of the group,so $Pb(II)$ is more stable than $Pb(IV)$ due to the inert pair effect. Thus,$Pb(IV)$ tends to gain two electrons to reach the stable $+2$ state,acting as a strong oxidizing agent.

(A) $CO$ (carbon monoxide) burns with a blue flame in the presence of oxygen according to the reaction: $2CO(g) + O_2(g) \rightarrow 2CO_2(g)$.

(D) The stability of dihalides of group $14$ elements increases as we move down the group.
This is due to the inert pair effect,which makes the $+2$ oxidation state more stable for heavier elements.
The stability order is: $SiX_2 < GeX_2 < SnX_2 < PbX_2$.

(A) Carborundum $(SiC)$ is produced by heating silica $(SiO_2)$ with excess carbon (coke) in an electric furnace at high temperatures.
The chemical reaction is:
$SiO_2 + 3C \xrightarrow{\text{Heat}} SiC + 2CO$
Thus,silica is heated with carbon to obtain carborundum.

(C) The hydrolysis of dialkyldichlorosilanes followed by polymerization leads to the formation of silicones.
$2H_2O + (Me)_2SiCl_2 \rightarrow (Me)_2Si(OH)_2 + 2HCl$
$(Me)_2Si(OH)_2$ undergoes polymerization to form linear silicones: $n(Me)_2Si(OH)_2 \rightarrow [-(Me)_2Si-O-]_n + nH_2O$.

(B) The red solid is $HgI_2$ (Mercury$(II)$ iodide).
$1$. $HgI_2$ is insoluble in water but dissolves in $KI$ solution due to the formation of a soluble complex: $HgI_2 + 2KI \rightarrow K_2[HgI_4]$.
$2$. Upon heating,$HgI_2$ decomposes to form mercury metal and iodine vapor: $HgI_2 \xrightarrow{\Delta} Hg + I_2(g)$.
$3$. The iodine vapor appears violet,and the mercury metal forms droplets on the cooler parts of the test tube.

(C) The acidic strength of hydrides of non-metals increases down the group as the bond dissociation energy decreases.
$HF$ is a weak acid due to strong hydrogen bonding.
$HBr$ is a strong acid because the $H-Br$ bond is relatively weak compared to $H-F$.
$H_2Te$ is a stronger acid than $H_2Se$ due to the increase in atomic size down the group.
$H_3P$ (Phosphine) is a very weak base.
Therefore,the statement '$HBr$ (Strong acid)' is correct.

(D) Zeolites are aluminosilicates in which some $SiO_4^{4-}$ units are replaced by $AlO_4^{5-}$ and $AlO_6^{9-}$ ions,creating a framework with negative charge that is balanced by cations.

(C) Polyphosphates,such as sodium hexametaphosphate (commonly known as Calgon),are used to soften water. They work by forming soluble complexes with cationic components like $Ca^{2+}$ and $Mg^{2+}$ ions present in hard water,thereby sequestering them.

(A) At $25\,^oC$ and atmospheric pressure,$P$ (Phosphorus) exists in the solid state. $Hg$ (Mercury) is a liquid,$Cl$ (Chlorine) is a gas,and $Br$ (Bromine) is a liquid.

(A) $ZnO$ is an amphoteric oxide because it reacts with both acids and bases to form salt and water.
Reaction with acid: $ZnO + H_2SO_4 \rightarrow ZnSO_4 + H_2O$
Reaction with base: $ZnO + 2NaOH \rightarrow Na_2ZnO_2 + H_2O$
$CaO$,$BaO$,and $SrO$ are basic oxides.

(B) Zeolites are aluminosilicates with a three-dimensional network structure where some $Si^{4+}$ ions are replaced by $Al^{3+}$ ions,resulting in $AlO_4^{5-}$ units instead of $SiO_4^{4-}$ units.
Option $B$ is incorrect because it states the charge on the aluminosilicate unit as $AlO_4^{6-}$,whereas the correct charge is $AlO_4^{5-}$ due to the substitution of $Si^{4+}$ by $Al^{3+}$.

(A) The reaction between $SnO_2$ and excess $HCl$ is as follows:
$SnO_2 + 4HCl \rightarrow SnCl_4 + 2H_2O$
Since $HCl$ is in excess,$SnCl_4$ reacts further with $HCl$ to form a complex anion:
$SnCl_4 + 2HCl \rightarrow H_2[SnCl_6]$
This dissociates to give $[SnCl_6]^{2-}$ ions in the solution.

(A) Mercury$(II)$ chloride,$HgCl_2$,is a covalent compound that exhibits significant volatility and undergoes sublimation upon heating.
It is commonly known as corrosive sublimate.

(C) The hydrolysis of $(Me)_2SiCl_2$ involves the replacement of chlorine atoms with hydroxyl groups,followed by condensation polymerization.
$n(Me)_2SiCl_2 + 2nH_2O \to n(Me)_2Si(OH)_2 + 2nHCl$
$n(Me)_2Si(OH)_2 \to [-O-Si(Me)_2-O-]_n + nH_2O$
Thus,the final product is a linear silicone polymer: $[-O-Si(Me)_2-O-]_n$.

(C) Carbon $(C)$ and tin $(Sn)$ both belong to Group $14$ of the periodic table.
Elements in the same group exhibit similar chemical properties due to having the same valence shell electronic configuration.

(D) The hydrides of the given elements are:
$NaH$ (Sodium hydride)
$H_2O$ (Water)
$BH_3$ (Borane)
$SiH_4$ (Silane)
Comparing the number of hydrogen atoms per atom of the element:
For $NaH$,ratio is $1:1$.
For $H_2O$,ratio is $2:1$.
For $BH_3$,ratio is $3:1$.
For $SiH_4$,ratio is $4:1$.
Thus,$Si$ forms a hydride with the maximum number of hydrogen atoms per atom of the element.

(A) The correct option is $A$.
In group $14$ elements,as we move down the group,the stability of the $+2$ oxidation state increases due to the inert pair effect.
The inert pair effect is the reluctance of the $ns^2$ electron pair to participate in bonding.
Therefore,the stability order for the $+2$ oxidation state is $Si^{2+} < Ge^{2+} < Sn^{2+} < Pb^{2+}$.
Thus,$Pb^{2+}$ is the most stable.

(D) Lead $(Pb)$ is soluble in dilute nitric acid $(HNO_3)$ because it forms soluble lead nitrate $(Pb(NO_3)_2)$.
However,lead becomes passive towards concentrated $HNO_3$ due to the formation of a protective layer of lead oxide $(PbO_2)$ on its surface.

(B) $[CCl_6]^{2-}$ does not exist because carbon has a maximum covalency of $4$ due to the absence of $d$-orbitals in its valence shell.
It cannot expand its octet to accommodate six chlorine atoms.

(A) The reaction between aluminum and metal oxides (like $Fe_2O_3$) is highly exothermic,leading to the formation of $Al_2O_3$.
This large amount of heat released is utilized in the process known as thermite welding to join broken iron rails or machine parts.

(D) Both $CO_2$ and $SO_2$ turn lime water milky due to the formation of insoluble metal carbonates or sulfites.
$Ca(OH)_2(aq) + CO_2(g) \to CaCO_3(s) \downarrow + H_2O(l)$
$Ca(OH)_2(aq) + SO_2(g) \to CaSO_3(s) \downarrow + H_2O(l)$
Therefore,the correct option is $(D)$.

(C) The correct answer is $(C)$.
In Haber's process,iron $(Fe)$ is used as a catalyst.
In Deacon's process,copper$(II)$ chloride $(CuCl_2)$ is used as a catalyst.
In the lead chamber process,nitrogen oxides $(NO_x)$ are used as catalysts.
In the Solvay process,no catalyst is used; it involves the reaction of brine with ammonia and carbon dioxide to produce sodium bicarbonate.

(B) Nitrolim is a mixture of calcium cyanamide $(CaCN_{2})$ and carbon $(C)$.
It is formed by the reaction of calcium carbide $(CaC_{2})$ with nitrogen $(N_{2})$ at high temperatures:
$CaC_{2} + N_{2} \rightarrow CaCN_{2} + C$.

(A) The atomic number of the element is $Z = 114$.
The electronic configuration is determined by filling orbitals in the order of increasing energy: $[Rn] \ 5f^{14} \ 6d^{10} \ 7s^2 \ 7p^2$.
Since the valence shell configuration is $ns^2 np^2$ (where $n=7$),the element belongs to the $14^{th}$ group,which is the carbon family.

(A) $Me_3SiCl$ acts as a chain terminator.
Upon hydrolysis,it produces $Me_3SiOH$,which contains only one reactive site,thereby limiting the growth of the polymer chain.
On the other hand,$Me_2SiCl_2$ leads to linear polymers,and $RSiCl_3$ (like $MeSiCl_3$ or $PhSiCl_3$) leads to cross-linked high molecular mass polymers.

(D) The basic structural unit of all silicates is the $SiO_4^{4-}$ tetrahedron,in which one silicon atom is bonded to four oxygen atoms in a tetrahedral arrangement.
This unit is derived from silicic acid,$H_4SiO_4$,by the removal of four protons ($H^+$ ions) as shown in the reaction:
$H_4SiO_4 \rightarrow SiO_4^{4-} + 4H^+$

(D) $1$. Beryl $(Be_3Al_2Si_6O_{18})$ is a well-known example of a cyclic silicate (metasilicate) where six $SiO_4$ tetrahedra are linked in a ring.
$2$. $Mg_2SiO_4$ (Forsterite) is an orthosilicate (nesosilicate) containing discrete $SiO_4^{4-}$ units.
$3$. The basic structural unit of all silicates is the $SiO_4^{4-}$ tetrahedron.
$4$. Feldspars are three-dimensional tectosilicates where some silicon atoms are replaced by aluminum atoms,making them aluminosilicates. Therefore,the statement that 'Feldspars are not aluminosilicates' is incorrect.

(C) In a pyrosilicate structure,two $[SiO_4]^{4-}$ tetrahedral units are joined by sharing one oxygen atom at one corner.
This results in the formula $[Si_2O_7]^{6-}$.
Therefore,the correct answer is pyrosilicate.

(A) $SnO_2$ is an amphoteric oxide because it reacts with both acids and bases.
Reaction with acid:
$SnO_2 + 4 HCl \longrightarrow SnCl_4 + 2 H_2O$
Reaction with base:
$SnO_2 + 2 NaOH \longrightarrow Na_2SnO_3 + H_2O$

(A) The $ns^2$ electrons of the outermost shell in group $14$ elements may not participate in bonding due to the inert pair effect.
This effect increases down the group,making the lower oxidation state $(+2)$ more stable for heavier elements.
For $Sn$ (tin),the $+4$ oxidation state is more stable.
For $Pb$ (lead),the $+2$ oxidation state is more stable due to the significant inert pair effect.
Therefore,the most characteristic oxidation states for lead and tin are $+2$ and $+4$ respectively.

(B) In chain silicates,each $SiO_4^{4-}$ tetrahedron shares two oxygen atoms with other tetrahedra to form a long chain.
This results in the general formula $(SiO_3^{2-})_n$ for pyroxenes.
Another type of chain silicate is the double chain silicate (amphiboles),which has the formula $(Si_4O_{11}^{6-})_n$.

(A) The reluctance of valence shell $ns^2$ electrons to participate in bonding due to poor shielding of $d$ and $f$ orbitals is known as the inert pair effect.
As we move down group $14$,the stability of the $+2$ oxidation state increases relative to the $+4$ oxidation state.
Therefore,the stability of dihalides $(MX_2)$ increases in the order: $SiX_2 < GeX_2 < SnX_2 < PbX_2$.

(B) The hydrolysis of substituted chlorosilanes leads to the formation of silicones.
$1$. $R_3SiCl$ gives a dimer on hydrolysis,which terminates the chain.
$2$. $R_2SiCl_2$ gives linear silicone polymers.
$3$. $RSiCl_3$ has three chlorine atoms attached to the silicon atom. Upon hydrolysis,it forms $RSi(OH)_3$,which has three hydroxyl groups available for condensation. This allows the polymer to grow in three dimensions,resulting in a cross-linked silicone polymer.
Therefore,$RSiCl_3$ is the correct answer.

(C) The hydrolysis of carbides depends on the nature of the bond between the metal/non-metal and carbon.
$Al_4C_3$ reacts with water to produce methane $(CH_4)$.
$Mg_2C_3$ reacts with water to produce propyne $(CH_3C \equiv CH)$.
$CaC_2$ reacts with water to produce ethyne $(HC \equiv CH)$.
$SiC$ (Silicon carbide or Carborundum) is a covalent network solid with a very stable structure. Due to its extremely high lattice energy and covalent nature,it is chemically inert and does not undergo hydrolysis with water.

(B) Two-dimensional sheet silicates: In such silicates,three oxygen atoms of each tetrahedral unit are shared with adjacent $[SiO_4]^{4-}$ tetrahedral units.
Such sharing forms a two-dimensional sheet structure with the general formula $(Si_2O_5)_n^{2n-}$.
Example: Talc $Mg_3(Si_2O_5)_2(OH)_2$.
Hence,option $B$ is correct.

(B) Holme signals are used as a distress signal at sea. They consist of a mixture of $CaC_2$ (calcium carbide) and $Ca_3P_2$ (calcium phosphide). When this mixture is thrown into water,it produces $C_2H_2$ (acetylene) and $PH_3$ (phosphine). The $PH_3$ gas catches fire spontaneously on contact with air,igniting the $C_2H_2$ gas and producing a bright light and smoke. The reaction is: $CaC_2 + Ca_3P_2 + H_2O \rightarrow Ca(OH)_2 + C_2H_2 + PH_3$.

(A) In $diamond$,each carbon atom is $sp^3$ hybridized and bonded to four other carbon atoms in a tetrahedral geometry,forming a rigid $3D$ network structure. It does not contain any hexagonal rings.
In $graphite$,carbon atoms are arranged in layers of hexagonal rings.
In $fullerenes$ (like $C_{60}$),the structure consists of both hexagonal and pentagonal rings.
Therefore,$diamond$ is the correct answer.