Carbon family Questions in English

(A) The correct option is $A$.
When concentrated sulphuric acid reacts with $NaCl$,sodium bisulphate and $HCl$ gas $(X)$ are formed. The $HCl$ gas is acidic in nature and turns moist blue litmus paper red.
$2NaCl + H_2SO_4 \longrightarrow Na_2SO_4 + 2HCl(g) (X)$
Egg shell powder primarily consists of calcium carbonate $(CaCO_3)$. When $HCl$ gas $(X)$ is passed into a suspension of egg shell powder in water,it reacts with $CaCO_3$ to release $CO_2$ gas $(Y)$.
$CaCO_3 + 2HCl \longrightarrow CaCl_2 + H_2O + CO_2(g) (Y)$
$CO_2$ gas turns lime water milky due to the formation of insoluble calcium carbonate.
Thus,the gases $X$ and $Y$ are $HCl$ and $CO_2$ respectively.

(B) Assertion $A$ is correct: $CO$ is a neutral oxide,while $CO_2$ is an acidic oxide.
Reason $R$ is correct: $CO_2$ reacts with water to form carbonic acid $(H_2CO_3)$,which confirms its acidic nature $(CO_2 + H_2O \rightleftharpoons H_2CO_3)$. $CO$ is sparingly soluble in water and does not react to form an acid,which explains why it is neutral.
Since the acidic nature of $CO_2$ is directly linked to its ability to form an acid with water,$R$ is the correct explanation of $A$.

(D) Bond enthalpy is inversely proportional to the atomic size of the elements involved in the bond. \\ As we move down the group $14$ from $C$ to $Sn$,the atomic size increases,which leads to a decrease in the strength of the covalent bond formed between the atoms. \\ The atomic radii order is $C < Si < Ge < Sn$. \\ Therefore,the bond enthalpy order is $C - C > Si - Si > Ge - Ge > Sn - Sn$.

(B) $Li$ and $Cs$ are highly reactive metals that react vigorously with water.
$Br_2$ has a boiling point of $58^{\circ}C$,so it will change into a vapour state in boiling water $(100^{\circ}C)$.
$Ga$ has a melting point of $29^{\circ}C$ and a boiling point of $2400^{\circ}C$. Therefore,it remains in the liquid state within the temperature range of boiling water $(100^{\circ}C)$.

(A) For good quality cement,the ratio of silica $(SiO_2)$ to alumina $(Al_2O_3)$ should be between $2.5$ and $4.0$.

(D) Coke is largely used as a reducing agent in metallurgy.
In diamond,each carbon atom undergoes $sp^3$ hybridisation and is linked to four other carbon atoms by using hybridised orbitals in a tetrahedral fashion.
Buckminsterfullerene contains six-membered and five-membered rings and hence is a cage-like molecule.
Graphite is very soft and slippery. Hence,it is used as a dry lubricant in machines running at high temperature.

(C) The oxides of group $14$ elements show varying acidic and basic character.
$SiO_2$ is acidic.
$CO$ is neutral.
$GeO$ is basic.
$GeO_2$ is weakly acidic.
$SnO_2$ and $PbO_2$ are amphoteric in nature,meaning they react with both acids and bases.

(D) The nature of the given oxides is as follows:
Acidic oxides: $Cl_2O_7, SiO_2, N_2O_5$
Neutral oxides: $CO, NO, N_2O$
Amphoteric oxides: $Al_2O_3, SnO_2, PbO_2$
Therefore,the number of amphoteric oxides is $3$.

(C) $SiO_2$ and $GeO_2$ are acidic,whereas $SnO$ and $PbO$ are amphoteric in nature. This statement is true.
Carbon does not possess $d$-orbitals,so it cannot form $p \pi-d \pi$ bonds. Allotropic forms of carbon arise due to the property of catenation and $p \pi-p \pi$ bond formation. Thus,Statement $II$ is false.

(D) The reaction of a chloride salt with $MnO_2$ and concentrated $H_2SO_4$ produces chlorine gas,which is greenish-yellow in color.
$2NH_4Cl + MnO_2 + 2H_2SO_4 \xrightarrow{\Delta} MnSO_4 + (NH_4)_2SO_4 + 2H_2O + Cl_2 \uparrow$
Since $Cl_2$ is the greenish-yellow gas,the salt $(A)$ must be a chloride salt,which is $NH_4Cl$.

(A) Incorrect: Covalent radius increases down the group from $C$ to $Pb$ due to the addition of new shells.
$(B)$ Incorrect: Electronegativity does not decrease gradually; it remains almost constant from $Si$ to $Pb$ due to poor shielding by $d$ and $f$ orbitals.
$(C)$ Correct: Carbon has no $d$ orbitals,limiting its covalence to $4$,while others can expand it.
$(D)$ Correct: Heavier elements have large atomic sizes,making $p\pi-p\pi$ overlap ineffective.
$(E)$ Correct: Carbon can show negative oxidation states (e.g.,in $CH_4$,$C$ is $-4$).
Therefore,statements $(C), (D),$ and $(E)$ are correct.

(A) Fuming sulphuric acid,also known as oleum,is a solution of sulphur trioxide $(SO_3)$ in concentrated sulphuric acid $(H_2SO_4)$.
Its chemical formula is $H_2S_2O_7$ (pyrosulphuric acid).
In the formula $H_2S_2O_7$,the number of oxygen atoms is $7$.

(C) $STATEMENT-1$ is True: Due to the inert pair effect,the stability of the $+4$ oxidation state decreases down the group $14$ $(C > Si > Ge > Sn > Pb)$. Thus,$Pb^{4+}$ is highly unstable and acts as a strong oxidizing agent to reach the more stable $+2$ state.
$STATEMENT-2$ is False: The inert pair effect states that the lower oxidation state $(+2)$ becomes more stable than the higher oxidation state $(+4)$ for the heavier members of group $14$ (like $Pb$). Therefore,the higher oxidation state is less stable for heavier members.

(C) Amphoteric oxides are those that react with both acids and bases.
In option $[A]$,$Cr_2O_3$,$BeO$,$SnO$,and $SnO_2$ are all amphoteric.
In option $[B]$,$CrO$ is a basic oxide.
In option $[C]$,$NO$ is a neutral oxide and $B_2O_3$ is an acidic oxide.
In option $[D]$,$ZnO$,$Al_2O_3$,$PbO$,and $PbO_2$ are all amphoteric.
Therefore,both options $[A]$ and $[D]$ contain only amphoteric oxides.

(B) In the crystalline state of $AlCl_3$,the $Cl^{-}$ ions form a close-packed lattice structure.
$Al^{3+}$ ions occupy the octahedral voids within this lattice.
Since each $Al^{3+}$ ion is surrounded by $6$ $Cl^{-}$ ions in an octahedral geometry,the coordination number of $Al$ is $6$.

(B) Diamond is harder than graphite due to its three-dimensional network structure.
$(B)$ Graphite is a good conductor of electricity due to the presence of free electrons in its layers,whereas diamond is an insulator.
$(C)$ Diamond is a better conductor of heat than graphite due to its rigid lattice structure.
$(D)$ The $C-C$ bond order in graphite is $\sim 1.5$ (due to resonance),while in diamond it is $1$. Thus,statement $(D)$ is correct.
Therefore,statements $(B)$ and $(D)$ are correct.

(B) The hydrolysis of $(CH_3)_2SiCl_2$ leads to the formation of a linear silicone polymer because it has two reactive chlorine atoms that can form siloxane linkages in two directions.
To control the length of the polymer chain and prevent further growth,chain termination is performed using $(CH_3)_3SiCl$. This compound has only one reactive chlorine atom,which caps the end of the polymer chain,preventing further polymerization.
Therefore,$(CH_3)_2SiCl_2$ is used for the preparation of the linear polymer,and $(CH_3)_3SiCl$ is used for chain termination.

(D) The melting point of group $14$ elements decreases as we move down the group from $C$ to $Sn$,but $Pb$ has a slightly higher melting point than $Sn$ due to its metallic structure and bonding characteristics. The order of melting points is $C > Si > Ge > Pb > Sn$.
Atomic numbers and their corresponding elements:
$Z=6$ $(C)$: $3730 \ ^{\circ}C$
$Z=14$ $(Si)$: $1410 \ ^{\circ}C$
$Z=32$ $(Ge)$: $937 \ ^{\circ}C$
$Z=50$ $(Sn)$: $232 \ ^{\circ}C$
$Z=82$ $(Pb)$: $327 \ ^{\circ}C$
Comparing the given options $(Z=14, 6, 82, 50)$,the element with the lowest melting point is $Sn$ $(Z=50)$.

(C) In group $14$,the first ionisation enthalpy decreases down the group due to the increase in atomic size,but $Pb$ shows a higher value than $Sn$ due to the inert pair effect and poor shielding of $4f$ electrons.
Given the values,$A$ corresponds to $Sn$ $(708 \ kJ \ mol^{-1})$ and $B$ corresponds to $Pb$ $(715 \ kJ \ mol^{-1})$.
$Sn^{2+}$ acts as a reducing agent because it tends to get oxidised to $Sn^{4+}$,which is the more stable oxidation state for $Sn$.
$Pb^{4+}$ acts as an oxidising agent because it tends to get reduced to $Pb^{2+}$,which is the more stable oxidation state for $Pb$ due to the inert pair effect.
Therefore,$A^{2+}$ is reducing and $B^{4+}$ is oxidising.

(B) $1$. Density: The density of $Diamond$ $(3.51 \ g/cm^3)$ is greater than that of $Graphite$ $(2.26 \ g/cm^3)$. Thus,$Diamond > Graphite$ is correct.
$2$. $C-C$ bond strength: In $Diamond$,all $C-C$ bonds are single bonds ($sp^3$ hybridization). In $Graphite$,there is partial double bond character due to resonance ($sp^2$ hybridization). Therefore,$Graphite$ has higher bond strength than $Diamond$. The order $Graphite < Diamond$ is incorrect.
$3$. Thermodynamic stability: $Graphite$ is the most stable allotrope of carbon at room temperature and pressure. Thus,$Graphite > Diamond$ is correct.
$4$. Heat conductivity: $Diamond$ is a better conductor of heat than $Graphite$ at room temperature. Thus,$Diamond > Graphite$ is correct.
Conclusion: Option $B$ is the incorrect order.

(B) The stability of the species $[MX_6]^{2-}$ depends on the size of the central atom $M$ and the steric hindrance caused by the ligands $X$.
Silicon $(Si)$ is a small atom. When it is surrounded by six large chlorine atoms $(Cl)$,there is significant steric repulsion between the ligands,which makes the $[SiCl_6]^{2-}$ ion unstable.
In contrast,$[SiF_6]^{2-}$ is stable because the fluoride ion is small enough to fit around the silicon atom without excessive steric hindrance.
Therefore,$[SiCl_6]^{2-}$ is the species that is not stable.

(D) Lime water is a solution of calcium hydroxide,$Ca(OH)_2$.
When carbon dioxide $(CO_2)$ is passed through lime water,it forms an insoluble precipitate of calcium carbonate $(CaCO_3)$,which turns the solution milky: $Ca(OH)_2 + CO_2 \rightarrow CaCO_3 + H_2O$.
Similarly,sulfur dioxide $(SO_2)$ also reacts with lime water to form calcium sulfite $(CaSO_3)$,which is also an insoluble white precipitate,turning the solution milky: $Ca(OH)_2 + SO_2 \rightarrow CaSO_3 + H_2O$.
Therefore,both $CO_2$ and $SO_2$ turn lime water milky.

(C) In fullerene,each carbon atom is bonded to three other carbon atoms,forming a cage-like structure. Therefore,each carbon atom undergoes $sp^2$ hybridization.

(B) Pyrex glass is a type of borosilicate glass.
It is primarily composed of $60$ to $80 \%$ silica $(SiO_2)$,$10$ to $25 \%$ boron trioxide $(B_2O_3)$,and smaller amounts of aluminum oxide $(Al_2O_3)$ and other oxides.
Since $SiO_2$ constitutes the largest percentage,it is the chief constituent.

(C) In both silica $(SiO_{2})$ and silicates,the fundamental building block is the orthosilicate unit,which is the tetrahedral ($SiO_{4})^{4-}$ anion.
In this structure,one silicon atom is bonded to four oxygen atoms arranged at the corners of a tetrahedron.

(A) Carborundum is the common name for silicon carbide $(SiC)$.
It is produced by heating silica $(SiO_{2})$ with excess carbon in an electric furnace at high temperatures.
The chemical reaction is:
$SiO_{2} + 3C \longrightarrow SiC + 2CO$

(A) Monosilane $(SiH_{4})$ is highly reactive and spontaneously ignites in air to produce silica $(SiO_{2})$ and water. The combustion reaction is: $SiH_{4} + 2O_{2} \longrightarrow SiO_{2} + 2H_{2}O$. The white smoke produced consists of finely divided particles of $SiO_{2}$,which form characteristic vortex rings.

(D) $SO_2$,$CO_2$,and $N_2O_5$ are acidic oxides because they react with water to form acids ($H_2SO_3$,$H_2CO_3$,and $HNO_3$ respectively).
$N_2O$ (nitrous oxide) is a neutral oxide.
Therefore,$N_2O$ is not acidic in nature.

(D) $Cl_2O_7$ is an acidic oxide because it is a non-metallic oxide with a high oxidation state.
$CaO$ is a basic oxide because it is an oxide of an alkaline earth metal.
$B_2O_3$ is an acidic oxide as it is a non-metallic oxide.
$SnO$ is an amphoteric oxide because it can react with both acids and bases to form salts and water.

(A) The $S_8$ molecule has a puckered ring structure, often described as a crown-shaped conformation.
In this structure, each sulfur atom is bonded to two other sulfur atoms with a bond angle of $107^{\circ}$ and a bond length of $204 \text{ pm}$.

(A) Buckminster fullerene,also known as $C_{60}$,consists of $60$ carbon atoms arranged in a soccer ball-like structure.
It contains $20$ six-membered rings and $12$ five-membered rings.
Therefore,the number of six-membered and five-membered rings is $20$ and $12$ respectively.

(B) The property of low solubility of carbon dioxide in water makes it biologically and geo-chemically important.
Carbon dioxide on reaction with water forms carbonic acid,which dissociates to give $HCO_3^{-}$ ions.
The $H_2CO_3 / HCO_3^{-}$ buffer system helps to maintain the $pH$ of blood between $7.26$ and $7.42$.

(A) Among the given halides,only $CCl_4$ cannot be hydrolysed.
This is because carbon lacks vacant $d$-orbitals in its valence shell,which are required to accept the lone pair of electrons from water molecules during the hydrolysis process.
In contrast,$SiCl_4$,$GeCl_4$,and $SnCl_4$ possess vacant $d$-orbitals,allowing them to undergo hydrolysis.

(A) Silicones are a group of organosilicon polymers that have $-(R_2SiO)-$ as a repeating unit.
During the polymerization process,the chain length of the polymer can be controlled by adding $(CH_3)_3SiCl$,which acts as a chain terminator by blocking the end terminals of the polymer chain.

(C) $SnO_2$ (Tin$(IV)$ oxide) is amphoteric in nature,meaning it can react with both acids and bases.
$CO_2$ is an acidic oxide,and $Ag_2O$ is a basic oxide.

(C) Dry ice is the solid form of carbon dioxide.
It is produced by allowing liquified $CO_{2}$ to expand rapidly,which causes it to cool and solidify into a white,snow-like substance.

(D) The reactions involved are:
$Na_2SO_3(s) + 2HCl(aq) \rightarrow 2NaCl(aq) + SO_2(g) + H_2O(l)$
Here,$X = Na_2SO_3$ and $Y = SO_2$.
$SO_2$ decolourises acidified $KMnO_4$ solution:
$2KMnO_4 + 5SO_2 + 2H_2O \rightarrow K_2SO_4 + 2MnSO_4 + 2H_2SO_4$
When $H_2S$ (gas $Z$) is passed into an aqueous solution of $SO_2$ $(Y)$,colloidal sulphur is obtained:
$SO_2(g) + 2H_2S(g) \rightarrow 3S(s) + 2H_2O(l)$
Thus,$X = Na_2SO_3$ and $Z = H_2S$.

(B) $[SiCl_6]^{2-}$ does not exist.
The chlorine atom is significantly larger than the fluorine atom.
Due to the large size of $Cl$ atoms,there is significant interelectronic repulsion when six $Cl$ atoms attempt to surround the central $Si$ atom.
Consequently,the steric hindrance prevents the formation of the $[SiCl_6]^{2-}$ ion,making it unstable.

(D) In the solid state,$PCl_5$ exists as an ionic compound consisting of $[PCl_4]^+$ and $[PCl_6]^-$ ions.
The $[PCl_4]^+$ ion has a tetrahedral geometry due to $sp^3$ hybridization.
The $[PCl_6]^-$ ion has an octahedral geometry due to $sp^3d^2$ hybridization.
Therefore,solid $PCl_5$ contains both tetrahedral and octahedral ions.

(C) Dry ice is solid $CO_2$.
It is widely used as a refrigerant for ice-cream and frozen foods because it sublimes at $-78.5 \ ^{\circ}C$ at atmospheric pressure,providing intense cooling without leaving any liquid residue.

(B) $1$. Ionization enthalpy: For group $14$ elements,the order is $Ge > Sn > Pb$. This is correct due to the inert pair effect and shielding.
$2$. Melting point: The order is $Ge > Sn > Pb$. The given option $Ge > Pb > Sn$ is incorrect because $Sn$ has a lower melting point than $Pb$ due to its metallic structure.
$3$. Density: The order is $Pb > Sn > Ge$. This is correct as density increases down the group.
$4$. Electrical resistivity: The order is $Ge > Pb > Sn$. This is correct as $Ge$ is a metalloid (semiconductor) and $Sn, Pb$ are metals.

(C) Acidic oxides are typically non-metal oxides or oxides of metals in high oxidation states.
$CO_2$,$SiO_2$,and $GeO_2$ are acidic oxides.
$SnO$,$PbO$,and $PbO_2$ are amphoteric oxides.
$CO$ is a neutral oxide.
Therefore,in the pair $GeO_2$ and $SiO_2$,both oxides are acidic.

(D) In Group $14$ of the periodic table:
$C$ (Carbon) is a non-metal.
$Si$ (Silicon) is a metalloid.
$Ge$ (Germanium) is a metalloid.
$Sn$ (Tin) is a metal.
$Pb$ (Lead) is a metal.
Therefore,the elements with metallic nature are $Sn$ and $Pb$.

(D) Group-$14$ elements form halides of the formula $MX_2$ and $MX_4$ (where $X = F, Cl, Br, I$).
$PbI_4$ does not exist because lead $(Pb)$ is a heavy element and exhibits an inert pair effect.
The energy released during the formation of the $Pb-I$ bonds is not sufficient to promote an electron from the $6s$ orbital to the $6p$ orbital to achieve the $sp^3$ hybridization required for tetrahalide formation.
Consequently,$Pb$ prefers to exist in the $+2$ oxidation state rather than the $+4$ oxidation state when bonded to large,less electronegative atoms like iodine.

(C) $CCl_4$ does not undergo hydrolysis because $C$ lacks $d$-orbitals to accept electrons from water molecules,whereas $SiCl_4$ does undergo hydrolysis.
Therefore,statement $(I)$ is incorrect.
Group $14$ elements exhibit $(+4)$ and $(+2)$ oxidation states.
Due to the inert pair effect,the stability of the $(+4)$ oxidation state decreases down the group,while the stability of the $(+2)$ oxidation state increases.
Thus,$GeX_4$ is more stable than $GeX_2$,and $PbX_4$ is less stable than $PbX_2$.
Therefore,statements $(II)$ and $(III)$ are correct.
The stability of dihalides increases down the group,so statement $(IV)$ is incorrect.

(A) The atomic size increases down the group and decreases in a period from left to right.
As we move down the group $14$ $(C, Si, Ge, Sn, Pb)$,the atomic size increases.
Smaller atomic size leads to stronger overlapping of orbitals,resulting in a stronger bond.
Therefore,$C$ (carbon) has the smallest atomic size and the highest bond energy among the given elements.

(B) $Ge$ is an element of group $14$ (carbon family).
In group $14$,both $+2$ and $+4$ oxidation states are possible.
For $Ge$,the $+4$ oxidation state is more stable than the $+2$ oxidation state.
Due to the inert pair effect,which arises from the poor shielding of $s$-electrons by intervening $d$ and $f$-electrons,the stability of the $+4$ oxidation state decreases down the group from $Ge$ to $Pb$.
Consequently,for $Pb$,the $+2$ oxidation state is more stable than the $+4$ oxidation state.
Therefore,the statement "$GeX_2$ is more stable than $GeX_4$" is incorrect.

(C) Zinc oxide $(ZnO)$ is used in paints as a white pigment.
$B$ is zinc hydroxide,$Zn(OH)_2$,which is a white,water-insoluble compound.
When $Zn(OH)_2$ is heated,it decomposes to form zinc oxide $(A)$: $Zn(OH)_2 \xrightarrow{\Delta} ZnO + H_2O$.
$Zn(OH)_2$ dissolves in $NaOH$ to form sodium zincate $(C)$: $Zn(OH)_2 + 2NaOH \rightarrow Na_2ZnO_2 + 2H_2O$.
When $ZnO$ $(A)$ is heated with coke $(C)$,it produces zinc metal $(D)$ and carbon monoxide $(E)$: $ZnO + C \xrightarrow{\Delta} Zn + CO$.
$CO$ burns with a blue flame.
Thus,$D$ is $Zn$ and $C$ is $Na_2ZnO_2$.

(NONE) $SiO_2$ is an acidic oxide,so it acts as an acidic flux to remove basic impurities like $FeO$ by forming slag $(FeSiO_3)$.
$FeO + SiO_2 \rightarrow FeSiO_3$ (slag).
Option $A$ is correct.
The distance between layers in graphite is $335 \ pm$ or $3.35 \times 10^{-8} \ cm$. Option $B$ is correct.
$SiO_2$ reacts with $Na_2CO_3$ as follows:
$Na_2CO_3 + SiO_2 \rightarrow Na_2SiO_3 + CO_2 \uparrow$.
Option $C$ is correct.
In graphite,each carbon atom is $sp^2$ hybridized. Option $D$ is correct.
Since all given statements are correct,there is no incorrect statement among the options provided.