Carbon family Questions in English

(D) The black sulphide precipitate is lead$(II)$ sulphide $(PbS)$.
When $PbS$ is treated with ozone $(O_3)$,it undergoes oxidation to form lead$(II)$ sulphate $(PbSO_4)$,which is a white compound.
The balanced chemical equation is:
$PbS + 4O_3 \rightarrow PbSO_4 + 4O_2$
(Black) $\rightarrow$ (White)
Therefore,the correct option is $D$.

(D) To determine if an $O$ atom is present between two central atoms,we examine the structures of the given oxyacids:
$1$. $H_2S_2O_8$ (Peroxodisulfuric acid): Contains a peroxide linkage,$S-O-O-S$.
$2$. $H_4P_2O_7$ (Pyrophosphoric acid): Contains a $P-O-P$ linkage.
$3$. $H_2S_2O_5$ (Disulfurous acid): Contains an $S-O-S$ linkage.
$4$. $H_2S_2O_4$ (Dithionous acid): The structure is $HO-S-S-OH$,where the two sulfur atoms are directly bonded to each other without an intervening oxygen atom.
Therefore,the correct option is $D$.

(D) Carbon and Nitrogen possess a unique ability to form $p\pi-p\pi$ multiple bonds with themselves and with other small,highly electronegative atoms like oxygen.
This is due to their small atomic size,which allows for effective side-on overlap of $2p$ orbitals.
Examples include $C=C, C=O, N=N, N=O$,etc.
Therefore,the correct option is $D$.

(C) The stability of hydrides of group $14$ elements decreases down the group as the bond dissociation enthalpy decreases due to an increase in the size of the central atom.
The order of stability is: $CH_4 > SiH_4 > GeH_4 > SnH_4 > PbH_4$.
Therefore,$PbH_4$ (Plumbane) is the least stable hydride.

(C) According to the inert pair effect,as we move down the group,the stability of the lower oxidation state increases.
For group $14$ elements,the stability of the $+2$ oxidation state increases in the order: $Ge^{2+} < Sn^{2+} < Pb^{2+}$.
This is because the $ns^2$ electrons become more reluctant to participate in bonding due to poor shielding by $d$ and $f$ orbitals.

(A) The extent of hydrolysis depends on the availability of vacant $d$-orbitals and the electrophilicity of the central atom.
$CCl_4$ does not undergo hydrolysis because carbon lacks vacant $d$-orbitals.
For the others,the hydrolysis rate increases with the increase in the positive charge density on the central atom,which makes it more susceptible to nucleophilic attack by $H_2O$.
Thus,the correct increasing order is $CCl_4 < MgCl_2 < AlCl_3 < SiCl_4 < PCl_5$.

(B) At high temperature,nitrogen reacts with calcium carbide to form calcium cyanamide $(CaCN_2)$.
The reaction is as follows:
$CaC_2 + N_2 \rightarrow CaCN_2 + C$

(A) In a $SiO_4^{4-}$ tetrahedron,there are $4$ oxygen atoms.
- In sheet silicates,$3$ oxygen atoms are shared,leaving $1$ oxygen atom unshared (monovalent).
- In cyclic silicates,$2$ oxygen atoms are shared,leaving $2$ oxygen atoms unshared.
- In single chain silicates,$2$ oxygen atoms are shared,leaving $2$ oxygen atoms unshared.
- In double chain silicates,some tetrahedra share $2$ and some share $3$ oxygen atoms.
Since sheet silicates have $3$ shared oxygen atoms per tetrahedron,they possess $1$ monovalent (unshared) oxygen atom per unit.
Therefore,the correct option is $A$.

(B) $PbI_4$ does not exist because $Pb(IV)$ is a strong oxidizing agent and $I^-$ is a strong reducing agent.
$Pb(IV)$ oxidizes $I^-$ to $I_2$ and itself gets reduced to $Pb(II)$.
Therefore,the stable compound formed is $PbI_2$.

(B) The silicate anion consists of three $SiO_4$ tetrahedra linked in a chain.
Each $SiO_4$ unit has a charge of $-4$.
In a chain of $n$ tetrahedra,the number of shared oxygen atoms is $(n-1)$.
For $n = 3$,the number of shared oxygen atoms is $3 - 1 = 2$.
Each shared oxygen atom reduces the total negative charge by $1$ unit (since it is shared between two $Si$ atoms,effectively contributing $-1$ charge to each $SiO_4$ unit instead of $-2$).
Alternatively,the formula for a linear chain of $n$ tetrahedra is $[Si_nO_{3n+1}]^{(2n+2)-}$.
For $n = 3$,the formula is $[Si_3O_{3(3)+1}]^{(2(3)+2)-} = [Si_3O_{10}]^{8-}$.
Thus,the charge of the silicate anion is $-8$.

(C) Silicon reacts with $HF$ to form silicon tetrafluoride $(SiF_4)$,which then reacts with excess $HF$ to form hexafluorosilicic acid $(H_2SiF_6)$.
The chemical reactions are:
$Si + 4HF \rightarrow SiF_4 + 2H_2$
$SiF_4 + 2HF \rightarrow H_2SiF_6$

(C) Carbon does not have vacant $d$-orbitals in its valence shell.
Therefore,it cannot expand its coordination number beyond $4$ to accommodate the incoming water molecule.
As a result,$CCl_4$ does not undergo hydrolysis at room temperature,whereas $SiCl_4$,$SnCl_4$,and $PbCl_4$ can undergo hydrolysis due to the presence of vacant $d$-orbitals.

(B) $SiCl_4$ on hydrolysis gives silicic acid.
The reaction is as follows:
$SiCl_4 + 4H_2O \rightarrow Si(OH)_4 + 4HCl$
$Si(OH)_4$ is also known as orthosilicic acid or simply silicic acid. Hence,the correct option is $B$.

(C) Cyclic silicates contain $(SiO_3)_n^{2n-}$ ions,which are formed by linking three or more tetrahedral $SiO_4^{4-}$ units in a cyclic manner.
In these structures,each $SiO_4^{4-}$ unit shares two oxygen atoms with other units,resulting in the general formula $(SiO_3)_n^{2n-}$.

(C) $SnCl_2$ acts as a reducing agent because there is a high tendency for $Sn(II)$ ions to oxidize to $Sn(IV)$ ions.
This occurs because $Sn^{4+}$ is more stable than $Sn^{2+}$ due to the inert pair effect in the $p$-block elements.
Therefore,option $C$ is the correct answer.

(B) The plague of tin refers to the allotropic transformation of white tin ($\beta$-tin) to grey tin ($\alpha$-tin).
This process occurs at temperatures below $13.2 \ ^{\circ}C$.
It is an autocatalytic process that leads to the crumbling or disintegration of tin objects,often described as the 'tin plague'.

(D) Tin$(IV)$ chloride pentahydrate is known as butter of tin.
It is represented as $SnCl_4 \cdot 5H_2O$.

(B) The $Si_2O_7^{6-}$ anion is known as a pyrosilicate or disilicate anion.
It is formed by the condensation of two $SiO_4^{4-}$ tetrahedra.
In this structure,one oxygen atom is shared between the two silicon atoms,while the other three oxygen atoms on each silicon remain unshared.
Therefore,the correct condition is that one oxygen of a $SiO_4$ tetrahedron is shared with another $SiO_4$ tetrahedron.

(B) The hydrolysis of $R_2SiCl_2$ occurs as follows:
$R_2SiCl_2 + 2H_2O \to R_2Si(OH)_2 + 2HCl$
Compound $(A)$ is a dihydroxy silane $(R_2Si(OH)_2)$.
Upon polymerization,these dihydroxy silane units undergo condensation to form linear silicone chains:
$n R_2Si(OH)_2 \to -[Si(R)_2-O]_n- + nH_2O$
Thus,compound $(B)$ is a linear silicone.

(D) In group $14$,the metallic character increases down the group as the atomic size increases.
$CO_2$ and $SiO_2$ are acidic in nature.
$GeO_2$ is weakly acidic,$SnO_2$ is amphoteric,and $PbO_2$ is amphoteric.
Among the monoxides,$CO$ is neutral,$GeO$ is basic,while $SnO$ and $PbO$ are amphoteric.
However,comparing the basicity of oxides of group $14$ elements,$PbO$ is considered the most basic among the given options because $Pb$ is the most metallic element in the group,leading to the highest ionic character in its oxide.

(D) In a $SiO_4^{4-}$ tetrahedral unit,there are $4$ oxygen atoms.
If $3$ oxygen atoms are shared with other tetrahedra,each shared oxygen contributes $1/2$ to the unit.
The number of oxygen atoms per silicon atom is calculated as: $1$ (unshared oxygen) $+ 3 \times (1/2)$ (shared oxygens) $= 1 + 1.5 = 2.5$.
Thus,the ratio of $Si:O$ is $1:2.5$,which is $2:5$.
The formula for the sheet silicate is $(Si_2O_5)_n^{2n-}$,which corresponds to the sharing of $3$ oxygen atoms per tetrahedron.

(B) In an amphibole silicate structure,the tetrahedra are linked in double chains.
In this structure,some tetrahedra share $2$ corners and some share $3$ corners.
The average number of corners shared per tetrahedron is calculated as $\frac{2+3}{2} = 2.5$ or $2\frac{1}{2}$.

(C) chain of three $SiO_4$ tetrahedra sharing corners has the formula $[Si_3O_{10}]^{8-}$.
To balance this charge,we need $8$ units of positive charge.
The mineral contains $Ca^{2+}$,$Cu^{2+}$,and $H_2O$ in a $1:1:1$ ratio.
Let the number of formula units be $x$. The total charge is $x(Ca^{2+} + Cu^{2+}) = x(2+2) = 4x$.
For $4x = 8$,we get $x = 2$.
Thus,the formula is $Ca_2Cu_2Si_3O_{10} \cdot 2H_2O$.

(D) $I$. Pyro silicates: ($1$ corner $O$ atom per tetrahedron is shared).
$II$. Cyclic silicates: ($2$ corners per tetrahedron are shared).
$III$. Double chain silicates: (Some units share $2$ corners,others share $3$ corners).
$IV$. Single chain silicates: ($2$ corners per tetrahedron are shared).
$V$. $3D$ silicates: ($4$ corners per tetrahedron are shared).
$VI$. Sheet silicates: ($3$ corners per tetrahedron are shared).
Therefore,in both cyclic silicates $(II)$ and single chain silicates $(IV)$,only $2$ corners per tetrahedron are shared.

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

(D) Carbon and Nitrogen have a unique ability to form $p\pi-p\pi$ multiple bonds with themselves and with other atoms of small size and high electronegativity.
This is due to their small atomic size and high electronegativity,which allows for effective lateral overlap of $p$-orbitals.
Examples of such multiple bonding include $C=C$,$C=O$,$N=N$,and $N=O$.

(A) $1$. Identify the elements: $X$ has $6$ electrons,so it is Carbon $(C)$. $Y$ has $11$ electrons,so it is Sodium $(Na)$. $Z$ has $35$ electrons,so it is Bromine $(Br)$.
$2$. Determine the hydrides: $X$ forms $CH_4$ (methane),$Y$ forms $NaH$ (sodium hydride),and $Z$ forms $HBr$ (hydrogen bromide).
$3$. Analyze properties: $CH_4$ is a colourless,non-polar gas insoluble in water. $NaH$ is a silver/grey ionic solid that reacts with water to form $NaOH$ (an alkali) and $H_2$ gas. $HBr$ is a colourless gas that dissolves in water to form hydrobromic acid,which is a strong acid.
$4$. Conclusion: Option $A$ correctly describes these properties.

(D) The compound zinc oxide $(ZnO)$ is known as philosopher's wool.
When $ZnO$ is heated with $BaO$ at $1100 \ ^oC$,it forms barium zincate $(BaZnO_2)$.
The chemical reaction is:
$ZnO + BaO \rightarrow BaZnO_2$

(D) The reaction of zinc with very dilute nitric acid $(HNO_3)$ is a redox reaction where zinc is oxidized to zinc nitrate and the nitrate ion is reduced to ammonium nitrate $(NH_4NO_3)$.
The balanced chemical equation is:
$4Zn + 10HNO_3 (\text{very dilute}) \rightarrow 4Zn(NO_3)_2 + NH_4NO_3 + 3H_2O$
Thus,the product formed is ammonium nitrate,which contains the $NH_4^+$ ion.

(B) Anhydrous mercurous chloride $(Hg_2Cl_2)$ is prepared by the reaction of mercuric chloride $(HgCl_2)$ with metallic mercury $(Hg)$.
$HgCl_2 + Hg \rightarrow Hg_2Cl_2$

(D) $PH_3$ is a covalent compound.

(C) Calcium phosphide $(Ca_3P_2)$ reacts with water to produce phosphine $(PH_3)$ gas. When $PH_3$ comes in contact with air,it undergoes spontaneous combustion to produce dense white clouds of phosphoric acid $(H_3PO_4)$,which are used in smoke screens.

(A) $NO$ (Nitric oxide) is a neutral oxide because it does not react with either acids or bases and does not affect litmus paper.

(C) $P_4O_{10}$ is an acidic oxide,whereas $NH_3$ is a basic gas.
Because of their opposite chemical nature,they react with each other to form ammonium phosphate.
The chemical reaction is:
$6NH_3 + P_4O_{10} + 6H_2O \to 4(NH_4)_3PO_4$
Therefore,$P_4O_{10}$ cannot be used as a drying agent for $NH_3$.

(C) The atomicity of sulfur $(S_8)$ is $8$.
To have half the atomicity of sulfur,the element must have an atomicity of $8 / 2 = 4$.
Phosphorus exists as $P_4$,which has an atomicity of $4$.

(A) $SO_3$ is acidic in nature.
$N_2O$ is a neutral oxide.
$BeO$ is an amphoteric oxide.
$HgO$ is a basic oxide because it is a metal oxide.

(A) $NO$ is a neutral oxide.
$CO_2$ is an acidic oxide.
$CaO$ is a basic oxide.
$ZnO$ is an amphoteric oxide.

(D) The reaction of potassium ferrocyanide with concentrated $H_2SO_4$ is as follows:
$K_4[Fe(CN)_6] + 6H_2SO_4 + 6H_2O \to 2K_2SO_4 + FeSO_4 + 3(NH_4)_2SO_4 + 6CO$
As shown in the chemical equation,carbon monoxide $(CO)$ gas is evolved during this process.

(A) $SO_2$ acts as a reducing agent and reduces bromine water,causing it to become colorless.
The chemical reaction is: $SO_2 + Br_2 + 2H_2O \to 2HBr + H_2SO_4$.

(A) $ZnO$ is an amphoteric oxide because it reacts with both acids and bases to form salts and water.

(C) Bleaching powder $(CaOCl_2)$ is prepared by the action of chlorine gas on dry slaked lime $(Ca(OH)_2)$.
The chemical reaction is:
$Ca(OH)_2 + Cl_2 \to CaOCl_2 + H_2O$
Therefore,the correct option is $C$.

(D) Ions that possess properties similar to halide ions and consist of two or more electronegative atoms (at least one of which is nitrogen) are known as $Pseudohalides$. Examples include $CN^-$,$OCN^-$,and $SCN^-$.

(D) Concentrated hydrochloric acid kept in open air releases $HCl$ gas vapors.
$HCl$ gas has a strong affinity for moisture present in the air.
As a result,it forms tiny droplets of liquid solution that appear as a white smoky cloud.

(D) Zeolites are microporous,crystalline solids with a well-defined structure.
They consist of $Si$,$Al$,and $O$ atoms.
In their structure,some $Si^{4+}$ ions are replaced by $Al^{3+}$ ions,which creates a negative charge on the framework.
This negative charge is balanced by cations like $Na^+$,$K^+$,or $Ca^{2+}$.
The statement that $SiO_4^{4-}$ is replaced by $AlO_4^{5-}$ and $AlO_6^{9-}$ is incorrect because the framework is based on $AlO_4$ tetrahedra,not $AlO_6$ octahedra.

(C) In pyrosilicates,two $SiO_4^{4-}$ tetrahedral units are joined by sharing one oxygen atom at one corner.
This results in the formation of the pyrosilicate ion,which has the formula $[Si_2O_7]^{6-}$.
The structure is represented as:
$O_3Si-O-SiO_3^{6-}$.

(A) An amphoteric oxide is one that reacts with both acids and bases to form salts and water.
$SnO_2$ is an amphoteric oxide.
It reacts with acids: $SnO_2 + 2H_2SO_4 \to Sn(SO_4)_2 + 2H_2O$
It reacts with bases: $SnO_2 + 2NaOH \to Na_2SnO_3 + H_2O$

(A) Graphite consists of layers of carbon atoms arranged in a hexagonal network.
These layers are held together by weak van der Waals forces.
Due to this layer-type structure,the layers can slide over one another,which makes graphite soft and slippery,allowing it to be used as a dry lubricant.

(A) Hydrogen sulphide gas reacts with lead acetate to form lead sulphide,which is black in color.
$H_2S + (CH_3COO)_2Pb \rightarrow PbS \downarrow (\text{black}) + 2 CH_3COOH$

(B) Pseudohalides are polyatomic anions that resemble halide ions in their chemical properties. They typically contain nitrogen,such as $CN^{\Theta}$,$OCN^{\Theta}$,$SCN^{\Theta}$,and $N_3^{\Theta}$ $(NNN^{\Theta})$.
$RCOO^{\Theta}$ (carboxylate ion) is not a pseudohalide because it does not exhibit the characteristic properties of halide ions.

(B) The reaction between lead$(IV)$ oxide $(PbO_2)$ and concentrated nitric acid $(HNO_3)$ is as follows:
$2PbO_2 + 4HNO_3 \rightarrow 2Pb(NO_3)_2 + 2H_2O + O_2$
From the balanced chemical equation,it is clear that oxygen gas $(O_2)$ is evolved.