Carbon family Questions in English

(D) The catenation property is directly related to the bond dissociation energy of the element-element bond.
As the atomic size increases down the group $(C < Si < Ge)$,the bond length increases and the bond dissociation energy decreases.
The bond dissociation energies for these bonds are approximately:
$C-C$: $348 \ kJ \ mol^{-1}$
$Si-Si$: $180 \ kJ \ mol^{-1}$
$Ge-Ge$: $167 \ kJ \ mol^{-1}$
Therefore,the correct order of bond energies is $348, 180, 167$ for $C-C, Si-Si,$ and $Ge-Ge$ respectively.

(B) Lead $(Pb)$ is a dense metal with a high atomic number,which makes it highly effective at absorbing and blocking high-energy radiation such as $X$-rays and gamma rays. Therefore,it is commonly used in radiation shields.

(D) Silicones are a group of organosilicon polymers which have the general empirical formula $(R_2SiO)_n$,where $R$ is an alkyl or aryl group.
These compounds have a repeating unit of $R_2Si-O-SiR_2$ linkage.
The backbone of silicones consists of alternating silicon and oxygen atoms,i.e.,$-Si-O-Si-O-Si-$ linkage.

(C) The central atom $C$ (Carbon) belongs to the second period and has a maximum covalency of $4$ because it lacks $d$-orbitals in its valence shell.
Therefore,it cannot expand its octet to accommodate six ligands.
In contrast,$Si$,$Ge$,and $Sn$ belong to the third,fourth,and fifth periods respectively,and possess vacant $d$-orbitals,allowing them to form hexacoordinated species like $[SiCl_6]^{2-}$,$[GeF_6]^{2-}$,and $[SnCl_6]^{2-}$.

(A) Volatility is inversely proportional to the boiling point of a substance.
For the hydrides of group $14$ elements $(CH_4, SiH_4, GeH_4, SnH_4)$,the molecular mass increases as we move down the group from $C$ to $Sn$.
As the molecular mass increases,the magnitude of van der Waals forces of attraction increases,leading to an increase in the boiling point.
Therefore,the boiling point order is $CH_4 < SiH_4 < GeH_4 < SnH_4$.
Since $CH_4$ has the lowest boiling point,it is the most volatile compound among the given options.

(C) In group $14$ elements,the stability of the $+2$ oxidation state increases down the group due to the inert pair effect.
Carbon and silicon show $+4$ as the most stable oxidation state.
Germanium shows both $+2$ and $+4$,but $+4$ is more stable.
Tin $(Sn)$ shows both $+2$ and $+4$,with $+4$ being more stable.
Lead $(Pb)$ shows both $+2$ and $+4$,but due to the inert pair effect,the $+2$ oxidation state is more stable than the $+4$ oxidation state.

(D) Carbon is a non-metal. While it shows catenation and forms multiple bonds,it is a poor conductor of electricity (except for graphite,which is a conductor,not a semiconductor). Silicon and germanium are the elements in Group $14$ that exhibit semiconductor properties. Therefore,the statement that carbon exhibits semiconductor properties is incorrect.

(C) Diamond is a crystalline allotrope of $Carbon$.
Emerald is a variety of the mineral beryl,which is a cyclosilicate of beryllium and aluminum with the chemical formula $Be_3Al_2(SiO_3)_6$.
Therefore,diamond is a form of $Carbon$ and emerald is a form of $Silica$ (as it is a silicate mineral).

(D) Diamond has a three-dimensional network structure where each carbon atom is $sp^3$ hybridized and bonded to four other carbon atoms in a tetrahedral geometry.
This structure is identical to the crystal lattice of crystalline silicon.
Therefore,diamond is isomorphous with crystalline silicon.

(C) Pyrosilicates are a class of silicates that contain the $Si_2O_7^{6-}$ ion.
This ion is formed by the sharing of one oxygen atom between two $SiO_4^{4-}$ tetrahedral units.
Therefore,the correct formula for the pyrosilicate ion is $Si_2O_7^{6-}$.

(C) Flint glass is a type of optical glass that has a high refractive index and low Abbe number. Due to these properties,it is widely used in the manufacturing of lenses and prisms.

(C) Asbestos is a naturally occurring fibrous silicate mineral. It is well-known for its high thermal resistance and ability to withstand very high temperatures without being damaged,making it useful for fireproof materials.

(A) In $[SiO_4]^{4-}$ units,if $3$ oxygen atoms are shared per tetrahedron,it leads to the formation of a two-dimensional sheet structure (also known as phyllosilicates).
In a pyrosilicate,only $1$ oxygen atom is shared.
In a linear chain silicate,$2$ oxygen atoms are shared.
In a three-dimensional silicate,all $4$ oxygen atoms are shared.

(B) $C_{60}$ is a form of fullerene,also known as buckminsterfullerene.
It consists of $60$ carbon atoms arranged in a soccer ball-like structure.
This structure is composed of $12$ pentagonal faces and $20$ hexagonal faces.

(B) Carbon has a small atomic size,which allows its $p$-orbitals to undergo effective side-to-side overlap,forming stable $p\pi-p\pi$ multiple bonds.
In the case of heavier elements in the carbon family (like $Si, Ge, Sn, Pb$),the atomic size is much larger.
Due to the large atomic radii,the $p$-orbitals are more diffuse and the distance between the nuclei is greater,which prevents effective side-to-side overlap.
Therefore,they do not form stable $p\pi-p\pi$ bonds.

(A) Thermodynamically,graphite is the most stable allotrope of carbon at standard temperature and pressure ($298 \ K$,$1 \ atm$).
Although diamond is kinetically stable,it is thermodynamically unstable relative to graphite.
Therefore,the conversion of diamond to graphite is spontaneous,but the rate of this reaction is extremely slow at room temperature,making it appear stable.

(A) $C_{60}$ is a fullerene known as buckminsterfullerene.
It consists of $60$ carbon atoms arranged in a structure that resembles a soccer ball.
Each carbon atom in $C_{60}$ is $sp^2$ hybridized.
The structure is a truncated icosahedron,which contains $20$ hexagonal rings and $12$ pentagonal rings.

(D) Lead$(II)$ acetate,with the chemical formula $(CH_3COO)_2Pb$,is commonly known as lead sugar because of its sweet taste.
It is a white crystalline substance that is soluble in water and glycerol.

(D) Allotropy is the property of an element to exist in two or more different physical forms in the same physical state.
In Group $14$ elements,Carbon $(C)$,Silicon $(Si)$,and Tin $(Sn)$ exhibit allotropy.
Carbon exists as diamond,graphite,and fullerenes.
Silicon exists in crystalline and amorphous forms.
Tin exists as grey tin $(\alpha-Sn)$ and white tin $(\beta-Sn)$.
Lead $(Pb)$ does not exhibit allotropy due to its metallic nature and the inert pair effect,which makes it stable in its standard metallic form.

(D) White lead is chemically $2PbCO_3 \cdot Pb(OH)_2$.
When exposed to air containing hydrogen sulfide $(H_2S)$,it reacts to form lead sulfide $(PbS)$,which is black in color.
The chemical reaction is: $Pb(OH)_2 \cdot 2PbCO_3 + 3H_2S \rightarrow 3PbS + 2CO_2 + 4H_2O$.
Thus,$H_2S$ is the gas responsible for blackening the surface.

(B) The phenomenon of $Tin \ Plague$ refers to the allotropic transformation of white $Tin$ $(\beta-Sn)$ to grey $Tin$ $(\alpha-Sn)$ at low temperatures.
This transformation is accompanied by a change in density,which causes the metal to crumble into a grey powder,resembling a plague.

(D) The phenomenon known as $Tin \ Cry$ refers to the crackling sound produced when a rod of tin is bent. This sound is caused by the friction between the crystals of tin as they slide over each other during the bending process.

(B) The stability of hydrides of group $14$ elements decreases down the group as the bond dissociation energy decreases due to the increase in atomic size and decrease in the strength of the $M-H$ bond.
Among the given options,$PbH_4$ (Plumbane) is the hydride of the element at the bottom of group $14$.
Therefore,$PbH_4$ is the least stable hydride.

(D) Litharge $(PbO)$ is used in the manufacturing of special glass,glazing of pottery,and as a pigment in paints.
However,lead storage batteries primarily use lead dioxide $(PbO_2)$ at the cathode and spongy lead $(Pb)$ at the anode,not litharge $(PbO)$.
Therefore,the correct answer is $D$.

(B) The element $Y$ forms two chlorides $YCl_n$ and $YCl_{n-2}$,which indicates that the element exhibits variable valency with a difference of $2$ in oxidation states. This is a characteristic property of the inert pair effect observed in $p$-block elements.
Specifically,for group $14$ elements like $Sn$ (Tin),it forms $SnCl_4$ (a volatile liquid) and $SnCl_2$ (a solid).
Thus,element $Y$ belongs to group $14$.

(D) In group $14$ elements,as we move down the group from $Ge$ to $Pb$,the stability of the $+2$ oxidation state increases and the stability of the $+4$ oxidation state decreases due to the inert pair effect.
$Ge(II)$ tends to lose two more electrons to reach the more stable $+4$ state,making it a strong reducing agent.
Conversely,$Pb(IV)$ tends to gain two electrons to reach the more stable $+2$ state,making it a strong oxidizing agent.
Therefore,the observed behavior is due to the more pronounced inert pair effect in $Pb$ compared to $Ge$.

(A) Carbon monoxide $(CO)$ burns with a characteristic blue flame in the presence of oxygen to form carbon dioxide $(CO_2)$. The reaction is: $2CO(g) + O_2(g) \rightarrow 2CO_2(g)$. Hydrogen $(H_2)$ also burns with a pale blue flame,but in the context of $p$-block elements chemistry,$CO$ is the classic example associated with this property.

(D) In group $14$ elements,the stability of the $+2$ oxidation state increases down the group due to the inert pair effect.
As we move from $Si$ to $Pb$,the tendency of the $ns^2$ electrons to participate in bonding decreases.
Therefore,the stability of dihalides ($+2$ oxidation state) follows the order: $SiX_2 < GeX_2 < SnX_2 < PbX_2$.

(A) Carborundum is silicon carbide $(SiC)$. It is prepared by heating silica $(SiO_2)$ or silicon with carbon (coke) in an electric furnace at high temperatures.
The chemical reaction is: $SiO_2 + 3C \rightarrow SiC + 2CO$ or $Si + C \rightarrow SiC$.
Therefore,heating silicon with carbon at high temperature produces carborundum.

(B) Graphite is a form of carbon where each carbon atom is $sp^2$ hybridized and bonded to three other carbon atoms in a hexagonal planar structure.
This leaves one electron per carbon atom free,which is delocalized over the layers.
Due to these delocalized electrons,graphite acts as a good conductor of electricity.
However,compared to metals,its conductivity is often described as moderate or semi-metallic in nature,making it the correct choice among the given options.

(B) Phosphorus is a non-metal belonging to group $15$.
$PH_3$ (phosphine) is formed by the sharing of electrons between phosphorus and hydrogen atoms.
Since both phosphorus and hydrogen are non-metals with relatively small differences in electronegativity,the bond formed is covalent in nature.
Therefore,$PH_3$ is a covalent hydride.

(A) Neutral oxides are those which do not react with either acids or bases.
$NO$ (Nitric oxide),$N_2O$ (Nitrous oxide),and $CO$ (Carbon monoxide) are well-known examples of neutral oxides.
$CO_2$ is an acidic oxide.
$CaO$ and $Na_2O$ are basic oxides.
Therefore,$NO$ is the correct answer.

(C) The reaction of $N_2$ with calcium carbide $(CaC_2)$ at high temperatures (approx. $1000 \ ^\circ C$) produces calcium cyanamide $(CaCN_2)$ and carbon $(C)$.
The chemical equation is: $CaC_2 + N_2 \rightarrow CaCN_2 + C$.
This mixture of $CaCN_2$ and $C$ is known as nitrolim,which is used as a fertilizer.

(B) When silver nitrate $(AgNO_3)$ is strongly heated,it undergoes thermal decomposition to form metallic silver $(Ag)$,nitrogen dioxide $(NO_2)$,and oxygen gas $(O_2)$.
The balanced chemical equation for this reaction is:
$2AgNO_3(s) \xrightarrow{\Delta} 2Ag(s) + 2NO_2(g) + O_2(g)$

(C) The atomicity of sulfur $(S_8)$ is $8$.
Half of the atomicity of sulfur is $8 / 2 = 4$.
Among the given elements,phosphorus $(P_4)$ exists as a tetratomic molecule,meaning its atomicity is $4$.
Therefore,phosphorus has half the atomicity of sulfur.

(C) When $Na_2SO_3$ (sodium sulfite) reacts with dilute $H_2SO_4$,it undergoes a double displacement reaction to form sodium sulfate,water,and sulfur dioxide gas.
The chemical equation is: $Na_2SO_3 + H_2SO_4 \rightarrow Na_2SO_4 + H_2O + SO_2 \uparrow$.
$Na_2SO_4$ and $NaHSO_4$ do not react with dilute $H_2SO_4$ to produce $SO_2$.
$Na_2S$ reacts with dilute $H_2SO_4$ to produce $H_2S$ gas,not $SO_2$.

(B) An oxide that reacts with both acids $(HCl)$ and bases $(NaOH)$ is known as an amphoteric oxide.
$ZnO$ is an amphoteric oxide.
It reacts with $HCl$ as follows: $ZnO + 2HCl \rightarrow ZnCl_2 + H_2O$.
It reacts with $NaOH$ as follows: $ZnO + 2NaOH \rightarrow Na_2ZnO_2 + H_2O$ (sodium zincate).
Therefore,$ZnO$ is the correct answer.

(A) The cyclic trimer of sulfur trioxide $(SO_3)$ is known as $\gamma-SO_3$ or trimeric sulfur trioxide,which has the formula $(SO_3)_3$ or $S_3O_9$.
In this structure,three $SO_4$ tetrahedra are linked together by sharing oxygen atoms at the corners.
The structure consists of a six-membered ring of alternating sulfur and oxygen atoms $(-S-O-S-O-S-O-)$.
There are no direct $S-S$ bonds in the cyclic trimer of sulfur trioxide.
Therefore,the number of $S-S$ bonds is $0$.

(B) The reaction between calcium hydroxide $(Ca(OH)_2)$ and sulfur dioxide $(SO_2)$ is a neutralization reaction where a base reacts with an acidic oxide.
The balanced chemical equation is:
$Ca(OH)_2 + SO_2 \rightarrow CaSO_3 + H_2O$
Here,$CaSO_3$ (calcium sulfite) is formed as the product.

(D) $1$. Lead nitrate $(2Pb(NO_3)_2 \xrightarrow{\Delta} 2PbO + 4NO_2 + O_2)$ releases oxygen on heating.
$2$. Potassium chlorate $(2KClO_3 \xrightarrow{\Delta} 2KCl + 3O_2)$ releases oxygen on heating.
$3$. Mercuric oxide $(2HgO \xrightarrow{\Delta} 2Hg + O_2)$ releases oxygen on heating.
$4$. Manganese dioxide $(MnO_2)$ is a stable oxide and does not decompose to release oxygen upon simple heating; it is often used as a catalyst for the decomposition of other substances.

(D) Ions that consist of two or more electronegative atoms,which have properties similar to those of halide ions $(X^-)$,are known as pseudohalides. Examples include $CN^-$,$OCN^-$,$SCN^-$,and $N_3^-$. These ions contain at least one nitrogen atom and exhibit chemical behavior analogous to halides.

(A) In an alkaline medium,$Na_2CO_3$ reacts with $SO_2$ to form $Na_2SO_3$ (sodium sulphite).
The chemical reaction is as follows:
$Na_2CO_3 + SO_2 \rightarrow Na_2SO_3 + CO_2$
When $SO_2$ reacts with caustic alkalies or carbonates in an alkaline medium,it forms sulphites $(Na_2SO_3)$. In an acidic medium,it would form bisulphites $(NaHSO_3)$.

(B) Silicones are organosilicon polymers containing $Si-O-Si$ linkages.
They are formed by the hydrolysis of alkyl or aryl substituted chlorosilanes (e.g.,$R_2SiCl_2$).
The general formula for linear silicones is $(R_2SiO)_n$,where the repeating unit is $-(R_2Si-O)-$

(D) Ceramics are inorganic,non-metallic,solid materials. They are classified into various types,including oxides and non-oxides.
Non-oxide ceramics do not contain oxygen in their chemical structure.
$1$. $B_4C$ (Boron carbide) is a carbide.
$2$. $SiC$ (Silicon carbide) is a carbide.
$3$. $Si_3N_4$ (Silicon nitride) is a nitride.
Since none of these compounds contain oxygen,they are all classified as non-oxide ceramics. Therefore,the correct option is $D$.

(B) Graphite is thermodynamically the most stable allotrope of carbon. That is why $\Delta_fH^o$ (graphite) is taken as $0 \ kJ \ mol^{-1}$.
$\Delta_fH^o$ (diamond) $= +1.90 \ kJ \ mol^{-1}$
$\Delta_fH^o$ (fullerene) $= +38.1 \ kJ \ mol^{-1}$
Since the standard enthalpy of formation of graphite is the lowest (zero),it is the most stable form.

(B) When $SiF_4$ reacts with water,it undergoes hydrolysis to form silicic acid $(H_4SiO_4)$ and hydrofluoric acid $(HF)$.
The balanced chemical equation is:
$SiF_4 + 4H_2O \to H_4SiO_4 + 4HF$

(A) Silica $(SiO_2)$ on heating with carbon $(C)$ at an elevated temperature produces carborundum,which is silicon carbide $(SiC)$.
$SiO_2 + 3C \xrightarrow{\Delta} SiC + 2CO$
Carborundum is an extremely hard substance used as an abrasive.

(C) $Sn^{2+}$ is a strong reducing agent and it gets oxidized to $Sn^{4+}$.
The reaction with mercuric chloride $(HgCl_2)$ proceeds as follows:
$2HgCl_2 + SnCl_2 \to Hg_2Cl_2 + SnCl_4$ (white precipitate)
$Hg_2Cl_2 + SnCl_2 \to 2Hg + SnCl_4$ (grey precipitate of metallic mercury)
Thus,the Assertion is correct because $SnCl_2$ acts as a reducing agent,not an oxidizing agent. Therefore,the Reason is incorrect.

(B) Diamond is a bad conductor because all $4$ valence electrons of carbon are involved in $sp^3$ covalent bonding,leaving no free electrons.
Graphite is a good conductor of electricity because each carbon atom is $sp^2$ hybridized,leaving one free electron per carbon atom in its lattice.
Both the Assertion and the Reason are correct statements,but the fact that graphite is a good conductor does not explain why diamond is a bad conductor. Therefore,the Reason is not the correct explanation of the Assertion.