Carbon family Questions in English

(A) The thermal decomposition of lead nitrate is represented by the following chemical equation:
$2Pb(NO_3)_2(s) \xrightarrow{\Delta} 2PbO(s) + 4NO_2(g) + O_2(g)$
As shown in the reaction,the heating of lead nitrate produces lead oxide $(PbO)$,nitrogen dioxide $(NO_2)$,and oxygen $(O_2)$.
Therefore,the correct option is $A$.

(C) When sulfur $(S)$ is burned in the presence of oxygen,it reacts to form sulfur dioxide $(SO_2)$,which is a gas at room temperature.
$S(s) + O_2(g) \to SO_2(g)$

(A) $NO$ is an example of a neutral oxide as it neither shows acidic nor basic character.

(C) When tin $(Sn)$ is boiled with a concentrated alkali solution (such as $NaOH$),it undergoes an oxidation reaction to form stannate ions.
The chemical reaction is: $Sn + 2OH^- + 2H_2O \rightarrow [Sn(OH)_6]^{2-} + 2H_2$ or more commonly represented as the formation of the stannate ion,$SnO_3^{2-}$.
Thus,the correct product formed is $SnO_3^{2-}$.

(C) suboxide is an oxide in which the element is in a lower oxidation state than the normal oxide.
Carbon suboxide,also known as tricarbon dioxide,has the chemical formula $C_3O_2$ or $O=C=C=C=O$.
It is a stable member of the series of linear oxocarbons and is characterized by its cumulene structure.

(C) $SO_2$ is highly soluble in water,forming sulphurous acid. Therefore,it cannot be collected over water.
$H_2O + SO_2 \to H_2SO_3$ (Sulphurous acid)

(A) The hydrolysis of $R_3SiCl$ produces $R_3SiOH$ as an intermediate.
$R_3SiCl + H_2O \to R_3SiOH + HCl$
Since $R_3SiOH$ has only one reactive $-OH$ group,it can only undergo condensation to form a dimer,$R_3Si-O-SiR_3$,and cannot form a long-chain polymer.
$R_3SiOH + HOSiR_3 \to R_3Si-O-SiR_3 + H_2O$.

(D) $CCl_4 + H_2O \to$ No reaction.
Carbon atom does not have vacant $d$-orbitals to accept the lone pair of electrons from water molecules,hence it cannot undergo hydrolysis.

(C) The bleaching action of bleaching powder $(CaOCl_2)$ is due to the release of nascent oxygen.
The chemical reaction is: $CaOCl_2 \to CaCl_2 + [O]$.
The nascent oxygen $([O])$ acts as a strong oxidizing agent,which oxidizes colored substances to colorless substances.

(A) The chemical formula of bleaching powder is $CaOCl_2$ or $Ca(OCl)_2 \cdot CaCl_2 \cdot Ca(OH)_2 \cdot H_2O$.
It is essentially a mixture of calcium hypochlorite $Ca(OCl)_2$ and calcium chloride $CaCl_2$.
The effective component responsible for the bleaching action is the hypochlorite ion $(OCl^-)$,which releases chlorine gas upon reaction with acids or atmospheric $CO_2$.
Therefore,the effective component is chlorine.

(C) Chlorine with cold and dilute sodium hydroxide gives a mixture of chloride salt,hypochlorite,and water.
$Cl_{2(g)} + 2NaOH_{(aq)} \rightarrow NaCl_{(aq)} + NaOCl_{(aq)} + H_2O_{(l)}$
Chlorine with hot and concentrated sodium hydroxide gives a mixture of chloride salt,chlorate,and water.
$3Cl_{2(g)} + 6NaOH_{(aq)} \rightarrow 5NaCl_{(aq)} + NaClO_{3(aq)} + 3H_2O_{(l)}$

(A) Bromine vapour is a non-polar substance.
When $CS_2$ (Carbon disulphide) or $CCl_4$ (Carbon tetrachloride) is added,bromine dissolves in these solvents,reducing the vapour concentration and thus the intensity of the brown colour.
Animal charcoal powder adsorbs bromine gas,which also reduces the intensity of the brown colour.
Pieces of marble $(CaCO_3)$ do not react with bromine vapour,so the intensity of the brown colour does not decrease appreciably.

(A) The reaction between $HgCl_2$ and $Hg(CN)_2$ results in the formation of an addition compound.
The chemical equation is: $HgCl_2 + Hg(CN)_2 \to HgCl_2 \cdot Hg(CN)_2$.

(D) $1$. $Ni(CO)_4$ involves $sp^3$ hybridization with no unpaired electrons,making it diamagnetic. This statement is true.
$2$. $CO_2$ is an acidic oxide,whereas $Sn(OH)_2$ is amphoteric. $CO_2$ acts as a stronger Lewis acid. This statement is true.
$3$. Graphite has free electrons due to its layered structure,allowing it to conduct electricity,while diamond has a rigid 3D covalent structure with no free electrons. This statement is true.
$4$. $CCl_4$ is inert towards hydrolysis because carbon cannot expand its octet,whereas $BCl_3$ is readily hydrolysed due to the presence of an empty $p$-orbital on the boron atom. The statement '$CCl_4$ is hydrolysed whereas $BCl_3$ is inert' is false.

(D) Bleaching powder,$CaOCl_2$,is unstable and reacts with atmospheric $CO_2$ or undergoes auto-oxidation over time.
When kept for a long time,it decomposes into calcium chloride $(CaCl_2)$ and calcium chlorate $(Ca(ClO_3)_2)$.
The reaction is: $6CaOCl_2 \rightarrow 5CaCl_2 + Ca(ClO_3)_2$.

(D) $PbI_4$ is the least stable halide among the given options due to the following reasons:
$(I)$ The size of the iodine atom is very large,which makes the $Pb-I$ bond weak.
$(II)$ Due to the inert pair effect,the $+2$ oxidation state of $Pb$ is significantly more stable than the $+4$ oxidation state. Consequently,$PbI_4$ tends to decompose into $PbI_2$ and $I_2$.

(D) Flint glass is a type of optical glass that contains a high percentage of lead$(II)$ oxide $(PbO)$,which increases its refractive index and dispersion. Therefore,lead is present in maximum quantity in flint glass.

(C) Concentrated $HCl$ is a highly volatile liquid that releases $HCl$ gas.
$HCl$ gas has a very strong affinity for water vapor present in the atmosphere.
When $HCl$ gas molecules come in contact with atmospheric moisture,they absorb the water vapor to form tiny droplets of concentrated hydrochloric acid solution.
These microscopic droplets scatter light and appear as a white cloud or fumes in the air.

(C) An amphoteric oxide is one that can react with both acids and bases.
$SnO_2$ is amphoteric in nature because it reacts with both strong bases and strong acids:
$1. \text{Reaction with base: } SnO_2 + 2NaOH \to Na_2SnO_3 + H_2O$
$2. \text{Reaction with acid: } SnO_2 + 4HCl \to SnCl_4 + 2H_2O$
$CO_2$ and $SiO_2$ are acidic oxides,while $CaO$ is a basic oxide.

(D) The composition of common glass,also known as soda-lime glass,is represented by the formula $Na_2O \cdot CaO \cdot 6SiO_2$.
Thus,the correct option is $D$.

(A) Philosopher's wool is $ZnO$.
When $ZnO$ is heated with $BaO$ at $1100\,^{\circ}C$,it forms barium zincate.
The reaction is: $ZnO + BaO \xrightarrow{1100\,^{\circ}C} BaZnO_2$.
Therefore,the correct compound is $BaZnO_2$.

(C) Carbon has the electronic configuration $1s^2 2s^2 2p^2$.
It does not have $d-$orbitals in its valence or penultimate shell.
Therefore,the statement that carbon differs due to the presence of $d-$orbitals in the penultimate shell is false.

(A) Silicon chloroform $(SiHCl_3)$ is prepared by the reaction of silicon with hydrogen chloride gas at high temperatures $(300 \ ^\circ C)$.
The chemical equation is:
$Si + 3HCl \to SiHCl_3 + H_2$

(A) The correct answer is $(A)$.
Helium is used as a diluent for oxygen in deep-sea diving because it is much less soluble in blood than nitrogen,even under high pressure.
This prevents the condition known as 'the bends' or decompression sickness.
$A$ mixture of $80\% \text{ helium}$ and $20\% \text{ oxygen}$ is commonly used by sea divers for respiration.

(D) The reaction of gas $A$ $(CO_2)$ with slaked lime $(Ca(OH)_2)$ forms a white precipitate of calcium carbonate $(CaCO_3)$:
$Ca(OH)_2(aq) + CO_2(g) \rightarrow CaCO_3(s) + H_2O(l)$
On prolonged bubbling of $CO_2$,the precipitate dissolves due to the formation of soluble calcium bicarbonate $(Ca(HCO_3)_2)$:
$CaCO_3(s) + H_2O(l) + CO_2(g) \rightarrow Ca(HCO_3)_2(aq)$
On heating the resultant solution,the calcium bicarbonate decomposes to reform the white precipitate of $CaCO_3$ and releases gas $B$ $(CO_2)$:
$Ca(HCO_3)_2(aq) \xrightarrow{\Delta} CaCO_3(s) + H_2O(l) + CO_2(g)$
Thus,both gases $A$ and $B$ are $CO_2$.

(B) The red solid is $HgI_2$ (mercuric iodide).
$HgI_2$ is insoluble in water but dissolves in a solution of $KI$ due to the formation of a soluble complex: $HgI_2 + 2KI \to K_2[HgI_4]$.
Upon heating,$HgI_2$ undergoes thermal decomposition: $HgI_2 \to Hg + I_2$.
The $I_2$ vapours appear as violet-coloured fumes,and metallic mercury $(Hg)$ droplets condense on the cooler parts of the test tube.

(C) In graphite,carbon atoms are $sp^2$ hybridized and arranged in planar hexagonal layers held together by weak van der Waals forces,which allows the layers to slide over each other,making it soft.
In diamond,each carbon atom is $sp^3$ hybridized and bonded to four other carbon atoms in a rigid,three-dimensional tetrahedral network,making it extremely hard.

(A) Producer gas is a mixture of carbon monoxide $(CO)$ and nitrogen $(N_2)$.
It is typically produced by passing air over red-hot coke.
The correct option is $A$.

(B) $Graphite$ is an allotrope of carbon where each carbon atom is $sp^2$ hybridized and bonded to three other carbon atoms in a hexagonal planar structure. The fourth valence electron of each carbon atom remains free,which allows it to conduct electricity and heat. Therefore,$(b)$ is the correct answer.

(D) Diamond is a covalent network solid where each carbon atom is $sp^3$ hybridized and bonded to four other carbon atoms in a tetrahedral geometry.
It is the hardest known natural material.
It is a poor conductor of electricity because there are no free electrons.
However,it is an excellent conductor of heat due to the strong covalent bonding network.
Therefore,the statement that diamond is a bad conductor of heat is false,and the only correct description among typical properties is that it is an allotrope of carbon.

(B) In the given options,$Graphite$ is a good conductor of electricity.
This is because each carbon atom in $Graphite$ is $sp^2$ hybridized,leaving one valence electron free to move,which allows for the conduction of electricity.
$Diamond$ is an insulator,$Silicon$ is a semiconductor,and $Amorphous \ carbon$ is a poor conductor.

(A) Humphry Davy and Michael Faraday demonstrated through combustion experiments that diamond is an allotrope of carbon,as it produces only $CO_2$ upon burning in oxygen.

(B) The basic structural unit of all silicates is the $[SiO_4]^{4-}$ tetrahedron,in which one silicon atom is bonded to four oxygen atoms at the corners of a regular tetrahedron.

(C) Zeolite is a three dimensional silicate. In these silicates,all four oxygen atoms of the $(SiO_4)^{4-}$ tetrahedra are shared with other tetrahedra,resulting in a three dimensional network structure.

(B) Graphite is used as a moderator to slow down the speed of fast-moving neutrons in atomic reactors.

(B) Silicon is the second most abundant element in the Earth's crust,after oxygen. It is primarily found in the form of silica $(SiO_2)$ and silicates,which are the major components of rocks and minerals.

(A) Feldspar is a common group of rock-forming minerals. The chemical formula for Potassium Feldspar (Orthoclase) is $KAlSi_3O_8$,which is equivalent to $K_2O \cdot Al_2O_3 \cdot 6SiO_2$.

(B) Anhydrous $FeCl_3$ is prepared by the action of dry chlorine gas on heated iron filings.
The reaction is: $2Fe + 3Cl_2 \to 2FeCl_3$.
Heating hydrated ferric chloride $(FeCl_3 \cdot 6H_2O)$ leads to hydrolysis,producing $Fe_2O_3$ instead of anhydrous $FeCl_3$.

(D) The correct answer is $D$.
Cuprous chloride $(CuCl)$ slowly oxidizes in the presence of air and moisture to form a green-colored basic cupric chloride compound.
The reaction is:
$CuCl \xrightarrow{\text{air}} 3CuO \cdot CuCl_2 \cdot 3H_2O$ (green colored).

(B) The thermal decomposition of copper$(II)$ nitrate occurs as follows:
$2Cu(NO_3)_2 \xrightarrow{\Delta} 2CuO + 4NO_2 + O_2$
Upon strong heating,copper$(II)$ nitrate decomposes to form copper$(II)$ oxide $(CuO)$,nitrogen dioxide $(NO_2)$,and oxygen $(O_2)$.
Therefore,the final product obtained is copper oxide.

(C) Ferric oxide $(Fe_2O_3)$ is a basic oxide. It does not react with bases like $NaOH$ to form a salt and water. Instead,it remains insoluble in $NaOH$ solution. However,in the context of chemical reactions involving iron oxides,$Fe(OH)_3$ is the precipitate formed when iron$(III)$ salts react with bases. Given the options provided,$Fe(OH)_3$ is the relevant species associated with iron$(III)$ in basic conditions. The reaction is: $Fe_2O_3 + 3H_2O \to 2Fe(OH)_3$.

(A) . $ZnO$ is an amphoteric oxide.
$ZnO + H_2SO_4 \to ZnSO_4 + H_2O$
$ZnO + 2NaOH \to Na_2ZnO_2 + H_2O$
$ZnO$ reacts with both acids and bases to form salt and water,which is the characteristic property of amphoteric oxides.

(B) In the laboratory,hydrogen sulfide $(H_2S)$ is prepared by the action of dilute sulfuric acid $(H_2SO_4)$ on ferrous sulfide $(FeS)$.
The chemical equation for the reaction is: $FeS + H_2SO_4 (\text{dil.}) \to FeSO_4 + H_2S \uparrow$.

(A) The presence of $Si$ (silicon) in steel imparts a fibrous structure to it,which improves its mechanical properties.

(B) Pyrite $(FeS_2)$ is commonly known as fool's gold due to its metallic luster and pale brass-yellow hue,which resembles gold.

(C) The correct method for the preparation of silver nitrate $(AgNO_3)$ is the reaction of silver metal with hot dilute nitric acid $(HNO_3)$.
The balanced chemical equation is:
$3Ag + 4HNO_3 (\text{dilute}) \xrightarrow{\Delta} 3AgNO_3 + NO + 2H_2O$

(B) The hydrolysis of organosilicon chlorides leads to the formation of silicones.
$R_3SiCl$ yields a dimer,$(R_3Si)_2O$.
$R_2SiCl_2$ yields linear silicones.
$RSiCl_3$ has three chlorine atoms attached to the silicon atom,which upon hydrolysis leads to the formation of a cross-linked or three-dimensional silicone polymer structure.

(C) The hydrolysis of dimethyldichlorosilane,$(Me)_2SiCl_2$,proceeds as follows:
$1$. First,the chlorine atoms are replaced by hydroxyl groups to form the unstable intermediate $(Me)_2Si(OH)_2$.
$2$. This intermediate undergoes rapid condensation polymerization by eliminating water molecules to form a linear silicone polymer.
$3$. The reaction is represented as: $n(Me)_2SiCl_2 + 2nH_2O \rightarrow [-O-Si(Me)_2-O-]_n + 2nHCl$.
$4$. Thus,the final product is a linear silicone polymer,represented by option $(C)$.

(A) The reaction between carbon monoxide $(CO)$ and sodium hydroxide $(NaOH)$ occurs under high pressure and temperature conditions to form sodium formate $(HCOONa)$.
$CO + NaOH \xrightarrow{\Delta, \text{pressure}} HCOONa$
Therefore,the correct option is $(A)$.

(D) The reaction of metal oxides with aluminum is highly exothermic,releasing a large amount of energy. This process is known as the thermite reaction. $Fe_{2}O_{3} + 2Al \rightarrow 2Fe + Al_{2}O_{3} + \text{Heat}$. This heat is sufficient to melt the metal,making it useful for joining railway tracks,which is known as thermite welding.