Carbon family Questions in English

(C) Amphoteric oxides are those that react with both acids and bases.
Among the given oxides:
$1$. $CO_2$ is acidic.
$2$. $GeO_2$ is amphoteric.
$3$. $SnO_2$ is amphoteric.
$4$. $PbO_2$ is amphoteric.
$5$. $CO$ is neutral.
$6$. $GeO$ is amphoteric.
$7$. $SnO$ is amphoteric.
$8$. $PbO$ is amphoteric.
Thus,the amphoteric oxides are $GeO_2, SnO_2, PbO_2, GeO, SnO, PbO$.
The total number of amphoteric oxides is $6$.

(A) $GeO_2$ is acidic in nature.
$CO$ is a neutral oxide.
$PbO_2$ is amphoteric in nature.
$SnO$ is amphoteric in nature.

(C) In group $14$,most $MX_4$ halides are covalent. However,$SnF_4$ and $PbF_4$ are ionic due to the large electronegativity difference between the metal and fluorine. Thus,Statement $I$ is correct.
For 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 the $+4$ oxidation state is more stable than the $+2$ oxidation state. Therefore,$GeCl_4$ is more stable than $GeCl_2$. Thus,Statement $II$ is incorrect.

(B) Most of the $MX_4$ $(X = F, Cl, Br, I)$ compounds of group $14$ elements are covalent in nature.
However,$SnF_4$ and $PbF_4$ are exceptions as they exhibit ionic character due to the high electronegativity of fluorine and the size of the central metal atoms.

(D) In group $14$,the elements are $C, Si, Ge, Sn, Pb$.
$C$ and $Si$ are non-metals.
$Ge$ is a metalloid.
$Sn$ and $Pb$ are metals.
Thus,there is only $1$ metalloid $(Ge)$ in group $14$.
Therefore,the statement in option $(D)$ is incorrect.

(D) Among the group $14$ elements,$Sn$ (tin) reacts with steam at high temperatures to produce tin dioxide and hydrogen gas.
The chemical reaction is: $Sn_{(s)} + 2 H_2O_{(g)} \rightarrow SnO_{2(s)} + 2 H_{2(g)}$.

(B) The presence of vacant $d$-orbitals in $Si$,$Ge$,and $Sn$ allows for the formation of octahedral complexes.
However,$[SiCl_6]^{2-}$ does not exist because the six large $Cl^{-}$ ions cannot be accommodated around the small $Si^{4+}$ cation due to steric hindrance.

(D) Among the dioxides of group-$14$ elements ($CO_2$,$SiO_2$,$GeO_2$,$SnO_2$,and $PbO_2$),$PbO_2$ is weakly basic in nature.
Due to the inert pair effect,the $+2$ oxidation state of $Pb$ is more stable than the $+4$ oxidation state.
Therefore,$PbO_2$ (where $Pb$ is in $+4$ state) has a strong tendency to get reduced to $Pb^{2+}$ $(PbO)$,which makes it a powerful oxidizing agent.

(B) Carbon $(C)$ reacts with oxygen to form carbon monoxide $(CO)$,which is a neutral oxide.
Carbon $(C)$ also reacts with oxygen to form carbon dioxide $(CO_2)$,which is an acidic oxide.
The reactions are:
$2C(s) + O_2(g) \rightarrow 2CO(g)$ (Neutral oxide)
$C(s) + O_2(g) \rightarrow CO_2(g)$ (Acidic oxide)
Therefore,the element $(X)$ is carbon $(C)$.

(B) Carbon $(C)$ forms two common oxides: carbon monoxide $(CO)$ and carbon dioxide $(CO_2)$.
$CO$ is a neutral oxide,while $CO_2$ is acidic in nature.
However,in the context of Group $14$ elements,$CO_2$ is acidic,and $SiO_2$ is also acidic.
Among the given options,Carbon $(C)$ is the only element whose dioxide $(CO_2)$ is acidic,and while $CO$ is neutral,it is the most appropriate answer in the context of non-metallic character compared to the other metallic elements ($Sn$,$Ge$,$Pb$) which form amphoteric oxides.

(A) The inert pair effect refers to the reluctance of the $ns^2$ electron pair to participate in bonding due to poor shielding by $d$ and $f$ orbitals.
For lead $(Pb)$,which belongs to group $14$,the $+2$ oxidation state becomes more stable than the $+4$ oxidation state as we move down the group.
Therefore,lead forms compounds in the $+2$ oxidation state due to the inert pair effect.

(D) smoke screen is a dense cloud of smoke used to conceal military positions.
Calcium phosphide $(Ca_3P_2)$ reacts with water to produce phosphine $(PH_3)$ gas:
$Ca_3P_2 + 6H_2O \rightarrow 3Ca(OH)_2 + 2PH_3$
Phosphine $(PH_3)$ is spontaneously flammable in air and burns to form phosphorus pentoxide $(P_2O_5)$:
$2PH_3 + 4O_2 \rightarrow P_2O_5 + 3H_2O$
The resulting $P_2O_5$ forms a dense white smoke screen.

(A) Carbon $(C)$ reacts with conc. $HNO_3$ to form $CO_2$ (linear) and $NO_2$ (bent). Both are acidic oxides.
Carbon $(C)$ reacts with conc. $H_2SO_4$ to form $CO_2$ (linear) and $SO_2$ (bent). Both are acidic oxides.
Thus,$X = C$ and $Z = C$.

(A) Carbon $(C)$ is the element that is oxidized by hot concentrated $H_2SO_4$ to produce two gaseous products.
The chemical reaction is:
$C + 2 H_2SO_4 \rightarrow CO_2(g) + 2 SO_2(g) + 2 H_2O(l)$
In this reaction,both $CO_2$ and $SO_2$ are gaseous products.

(B) When gaseous $HCl$ is passed into an aqueous solution of sodium sulfite $(Na_2SO_3)$,a double displacement reaction occurs.
The reaction is: $Na_2SO_3(aq) + 2HCl(g) \rightarrow 2NaCl(aq) + H_2O(l) + SO_2(g)$.
The products formed are sodium chloride $(NaCl)$,sulfur dioxide $(SO_2)$,and water $(H_2O)$.

(D) Density is defined as mass per unit volume. In a group,as we move down,the atomic mass increases significantly more than the atomic volume. Therefore,the density increases down the group. Among the given elements $(C, Si, Sn, Pb)$,$Pb$ (Lead) is at the bottom of the group and has the highest atomic mass,resulting in the highest density.

(B) Germanium $(Ge)$ belongs to group $14$ of the periodic table and exhibits semiconductor properties.
$Fe$ (Iron) and $Cu$ (Copper) are metallic conductors.
Diamond is a covalent solid and acts as an insulator.

(B) $Si$ $(Silicon)$ and $Ge$ $(Germanium)$ are well-known semiconductors.
$As$ $(Arsenic)$ is a metalloid often used as a dopant in semiconductors.
$Pb$ $(Lead)$ is a metal and acts as a conductor,not a semiconductor.

(B) The stability of the $+2$ oxidation state in Group $14$ elements increases down the group due to the inert pair effect.
As we move from $Si$ to $Pb$,the $ns^2$ electrons become more reluctant to participate in bonding.
Therefore,the stability of dihalides $(MX_2)$ follows the order: $SiX_2 < GeX_2 < SnX_2 < PbX_2$.

(C) $[SiCl_6]^{2-}$ does not exist because six large $Cl^{-}$ ions cannot be accommodated around the small $Si^{4+}$ ion due to steric hindrance and inter-electronic repulsion between the chloride ions.
In contrast,$[SiF_6]^{2-}$ exists because the $F^{-}$ ion is small enough to fit around the $Si^{4+}$ ion.
$[GeCl_6]^{2-}$ and $[Sn(OH)_6]^{2-}$ are stable due to the larger size of the central $Ge^{4+}$ and $Sn^{4+}$ ions,which can accommodate six ligands.

(D) Graphite has a two-dimensional sheet-like structure.
In graphite,each carbon atom is bonded to three other carbon atoms in the same plane,forming hexagonal rings.
Therefore,each carbon atom in graphite is $sp^2$ hybridized.
In contrast,in diamond,each carbon atom is bonded to four other carbon atoms in a tetrahedral arrangement,making it $sp^3$ hybridized.

(B) Non-metallic oxides are generally acidic in nature. Carbon is a non-metal belonging to group $14$ ($IVA$ group).
$C + O_2 \longrightarrow CO_2$
$CO_2 + H_2O \longrightarrow H_2CO_3$ (Carbonic acid)
Since $CO_2$ is a gas and forms an acidic solution,the element belongs to group $IV$ $(IVA)$.

(A) The acidic oxides are $B_2O_3, SiO_2,$ and $CO_2$.
$CO$ is a neutral oxide.
$Al_2O_3$ and $PbO_2$ are amphoteric oxides.
$Tl_2O_3$ is a basic oxide.
Therefore,the total number of acidic oxides is $3$.

(B) The melting point order of group $14$ elements is $C > Si > Ge > Sn > Pb$.
Among the given options,$Pb$ $(Lead)$ has the lowest melting point.

(A) The stability of oxidation states in Group $14$ elements is governed by the inert pair effect.
As we move down the group from $Si$ to $Pb$,the stability of the $+2$ oxidation state increases,while the stability of the $+4$ oxidation state decreases.
For Lead $(Pb)$,which is at the bottom of the group,the $+2$ oxidation state is significantly more stable than the $+4$ oxidation state due to the inert pair effect.
Therefore,$Pb^{2+} > Pb^{4+}$ is the correct order of stability.

(C) The inert pair effect results in the stabilization of an oxidation state that is two units less than the common group oxidation state.
The stability of the lower oxidation state increases down the group.
Thus,$Pb^{2+}$ is much more stable than $Pb^{4+}$,which makes $Pb^{4+}$ a strong oxidizing agent as it readily gets reduced to $Pb^{2+}$.
Conversely,$Sn^{2+}$ is less stable than $Sn^{4+}$,so $Sn^{2+}$ acts as a reducing agent and gets oxidized to $Sn^{4+}$.

(D) The enamel of teeth is composed of hydroxyapatite,which has the formula $Ca_{10}(PO_4)_6(OH)_2$.
When it reacts with $F^{-}$ ions,the $OH^{-}$ ions are replaced by $F^{-}$ ions to form fluorapatite:
$Ca_{10}(PO_4)_6(OH)_2 + 2 F^{-} \longrightarrow Ca_{10}(PO_4)_6 F_2 + 2 OH^{-}$
This product,$Ca_{10}(PO_4)_6 F_2$,can be represented as $3[Ca_3(PO_4)_2] \cdot CaF_2$.
Therefore,option $(D)$ is the correct answer.

(C) Silica $(SiO_2)$ is an acidic oxide and acts as an acid flux to remove basic impurities like $CaO$ or $FeO$ by forming slag.
In the reaction between $SiO_2$ and $Na_2CO_3$,$CO_2$ gas is evolved,not $CO$.
The reaction is: $Na_2CO_3 + SiO_2 \rightarrow Na_2SiO_3 + CO_2 \uparrow$.
Therefore,the statement in option $C$ is incorrect as it mentions the liberation of $CO$ instead of $CO_2$.

(C) The elements belonging to group $14$ are $C$,$Si$,$Ge$,$Sn$,and $Pb$.
$C$ and $Si$ are non-metals,$Ge$ is a metalloid,while $Sn$ and $Pb$ are metals.
These elements form covalent networks in their solid state.
The melting point depends on the strength of the interatomic forces (covalent or metallic bonds).
As we move down the group,the atomic size increases,which leads to a decrease in the strength of the interatomic forces.
Therefore,the melting point decreases from $C$ to $Pb$.
Thus,the correct order of melting temperature is $C > Si > Ge$.

(B) $(i)$ Quartz is a crystalline form of $SiO_2$ and exhibits piezoelectric properties, making it useful in electronic devices. This statement is correct.
$(ii)$ Group $14$ elements (except $C$) have vacant $d$-orbitals, allowing them to accept lone pairs from water molecules, leading to hydrolysis. $CCl_4$ cannot be hydrolyzed due to the absence of $d$-orbitals. This statement is correct.
$(iii)$ In graphite, the $C-C$ bond distance within the layers is $141.5 \ pm$, not $154 \ pm$ (which is the $C-C$ bond length in diamond). This statement is incorrect.
$(iv)$ $SiO_2$ is acidic in nature and is insoluble in $HCl$, but it reacts with $HF$ to form $SiF_4$. This statement is incorrect.
Therefore, statements $(i)$ and $(ii)$ are correct.

(A) $I$. Graphite consists of layers of carbon atoms arranged in hexagonal rings,which is correct.
$II$. Buckminsterfullerene $(C_{60})$ is aromatic in nature due to the delocalization of $\pi$-electrons in the spherical cage,so this statement is incorrect.
$III$. The distance between two adjacent layers in graphite is $340 \ pm$,not $141.5 \ pm$. The $C-C$ bond length within the layer is $141.5 \ pm$. Thus,this statement is incorrect.
$IV$. In both graphite and Buckminsterfullerene,each carbon atom is $sp^2$ hybridized. Thus,this statement is correct.
Therefore,statements $I$ and $IV$ are correct.

(B) $(i)$ $CCl_4$ does not undergo hydrolysis because it lacks vacant $d$-orbitals to accept lone pairs from water molecules.
$(ii)$ The structure of diamond consists of a $3-D$ network of carbon atoms with directional covalent bonds throughout the lattice.
$(iii)$ The thermodynamic stability order of carbon allotropes is $\text{Graphite} > \text{Diamond} > \text{Fullerene}$.
$(iv)$ Glass is a man-made amorphous silicate.
Therefore,statements $(ii)$ and $(iv)$ are correct.

(C) Graphite is an allotrope of carbon that has a layered structure,making it soft and slippery,which allows it to be used as a dry lubricant at high temperatures.
Thus,Assertion $(A)$ is true.
The reason for this soft and slippery nature is that the hexagonal layers of carbon atoms are held together by weak van der Waals forces,allowing them to slip over one another.
Thus,Reason $(R)$ is true and it correctly explains why graphite acts as a lubricant.

(C) The oxides of $Sn$ and $Pb$ (Group $14$) exhibit amphoteric character.
Specifically,$SnO_2, SnO, PbO,$ and $PbO_2$ are amphoteric because they can react with both acids and bases.
This property arises due to the stability of different oxidation states ($+2$ and $+4$) influenced by the inert pair effect.

(A) Carbon $(C)$ has the smallest atomic size and high bond dissociation energy,which allows it to form stable $C-C$ bonds,leading to the property of maximum catenation.
Germanium $(Ge)$ is the least abundant element among the group $14$ elements on the earth's crust.
Therefore,$X = C$ and $Y = Ge$.

(D) The isotope of carbon containing $7$ neutrons is $_{6}^{13}C$,which has a natural abundance of $1.1 \%$. This statement is correct.
$(b)$ Among the $IV$ $A$ (Group $14$) elements,the melting point generally decreases down the group. $Sn$ (Tin) has a lower melting point than $Pb$ (Lead) due to its metallic structure and bonding characteristics. This statement is correct.
$(c)$ Silicon is the $2^{nd}$ most abundant element by mass in the earth's crust,accounting for approximately $27.7 \%$. This statement is correct.
$(d)$ Elemental carbon (specifically in the form of diamond) shows the highest electrical resistivity among the Group $14$ elements because diamond is an electrical insulator. This statement is correct.
Therefore,all the statements are correct.

(D) Statement $1$ is correct: Charcoal is produced by the destructive distillation of wood (heating in the absence of air).
Statement $2$ is correct: Charcoal is considered an impure form of graphite.
Statement $3$ is incorrect: Graphite is thermodynamically the most stable allotrope of carbon at room temperature and pressure,not diamond.
Statement $4$ is incorrect: Coke is used as a reducing agent,not an oxidising agent,during the extraction of iron from its ores (e.g.,$Fe_2O_3 + 3C \rightarrow 2Fe + 3CO$).
Therefore,statements $1$ and $2$ are correct.

(C) In the carbon family (Group $14$),the density of elements increases as we move down the group due to the increase in atomic mass being more significant than the increase in atomic volume. Thus,$Pb$ has the highest density among the given options.
Conversely,the boiling point decreases as we move down the group because the metallic character increases and the strength of interatomic forces (covalent bonding) decreases with increasing atomic size. Therefore,$Pb$ has the lowest boiling point among the given options.
Hence,$Pb$ is the correct answer.

(D) . Graphite is an electrical conductor,but the conductivity is direction dependent. This is correct.
$B$. Diamond is more dense than graphite. This is correct due to the compact structure of diamond compared to the layered structure of graphite.
$C$. Diamond is metastable. This is correct because it is kinetically stable but thermodynamically unstable relative to graphite.
$D$. Graphite is thermodynamically less stable allotrope of carbon. This is incorrect. Graphite is the most thermodynamically stable allotrope of carbon at standard temperature and pressure. Hence,option $D$ is the incorrect statement.

(C) Graphite reacts with potassium vapour at $300^{\circ} C$ to form an intercalation compound $C_8K$.
In this compound,the potassium atoms donate their valence electrons to the graphite layers.
The presence of these delocalized electrons makes the compound electrically conducting.
Furthermore,the presence of unpaired electrons or the specific electronic structure of the interlayer state imparts paramagnetic character to the compound.

(C) Buckminsterfullerene is a type of fullerene with the molecular formula $C_{60}$.
It has a cage-like structure of fused rings that resembles a soccer ball.
Buckminsterfullerene contains $20$ six-membered rings and $12$ five-membered rings.
Therefore,$X=20$ and $Y=12$.

(B) $(I)$ The catenation property of group $14$ elements decreases from carbon to tin. Carbon has a small atomic size,resulting in strong $C-C$ bonds,which leads to maximum catenation. As we move down the group,atomic size increases,bond strength decreases,and the tendency for catenation decreases.
$(II)$ Fullerene $(C_{60})$ consists of $20$ six-membered rings and $12$ five-membered rings. The statement in the question incorrectly swaps these values.
$(III)$ $SiO_2$ is an acidic oxide and is soluble in strong alkaline solutions like $NaOH$ to form sodium silicate: $2NaOH + SiO_2 \longrightarrow Na_2SiO_3 + H_2O$.
Therefore,statements $I$ and $III$ are correct.

(C) Carbon has an electronic configuration of $2, 4$.
In diamond,each carbon atom is $sp^3$ hybridized and forms four strong covalent bonds with four other carbon atoms in a tetrahedral arrangement.
In graphite,each carbon atom is $sp^2$ hybridized and forms three covalent bonds with other carbon atoms,leaving one electron delocalized.
In $C_{60}$ (fullerene),each carbon atom is also $sp^2$ hybridized and forms three bonds.

(C) The gas produced by the passage of air over hot coke is called producer gas.
The reaction proceeds as:
$2 C (s) + (O_2 + 3.76 N_2) (g) \xrightarrow{\Delta} 2 CO (g) + 3.76 N_2 (g)$
This mixture of $CO$ and $N_2$ is known as producer gas.

(B) Silicones are organosilicon polymers which have the general empirical formula $(R_2SiO)_n$.
They are hydrophobic in nature due to the presence of organic groups $(R)$ attached to the silicon atoms.
Because of this hydrophobic nature,they are widely used for water proofing of fabrics.
Thus,both Assertion $(A)$ and Reason $(R)$ are correct,and $(R)$ is the correct explanation of $(A)$.

(C) In Group $14$,the melting point generally decreases as we move down the group due to the weakening of metallic bonding.
However,there is an anomaly where the melting point of $Pb$ is slightly higher than that of $Sn$.
The order of melting points for the elements is $C > Si > Ge > Pb > Sn$.
Therefore,$Sn$ has the lowest melting point.