Carbon family Questions in English

(D) When heating silicon with methyl chloride at $573 \ K$ in the presence of $Cu$ powder (Direct Process),a mixture of methyl chlorosilanes is formed: $CH_3SiCl_3$,$(CH_3)_2SiCl_2$,$(CH_3)_3SiCl$,and $(CH_3)_4Si$.
Among these,the yield of $(CH_3)_4Si$ (tetramethylsilane) is the minimum.
This is because the substitution of all four chlorine atoms by bulky methyl groups is sterically hindered and thermodynamically less favorable compared to the formation of chlorosilanes.

(B) Silicones are a group of organosilicon polymers which have the general empirical formula $(R_2SiO)_n$.
They are synthesized by the hydrolysis of dialkyldichlorosilanes $(R_2SiCl_2)$,followed by condensation polymerization.
The structure consists of a repeating unit of $-(Si(R)_2-O)-$.
Thus,they are considered as polymers of $R_2SiO$.

(B) $SiO_2$ is an acidic oxide. It reacts with hydrofluoric acid $(HF)$ to form hexafluorosilicic acid and with strong bases like sodium hydroxide $(NaOH)$ to form sodium silicate.
$SiO_2 + 6HF \longrightarrow H_2[SiF_6] + 2H_2O$
$SiO_2 + 2NaOH \longrightarrow Na_2SiO_3 + H_2O$

(C) In three-dimensional silicates,all four oxygen atoms (corners) of each $SiO_4^{4-}$ tetrahedron are shared with other tetrahedra.
This results in a three-dimensional network structure.
Examples of such structures include various forms of silica like quartz,tridymite,and cristobalite.

(D) The hydrolysis of silicon tetrachloride $(SiCl_4)$ proceeds as follows:
$SiCl_4 + 4H_2O \longrightarrow H_4SiO_4 + 4HCl$
Here,'$X$' is silicic acid $(H_4SiO_4)$.
Upon heating at $1000^{\circ} C$,silicic acid undergoes dehydration:
$H_4SiO_4 \xrightarrow{\Delta, 1000^{\circ} C} SiO_2 + 2H_2O$
Here,'$Y$' is silicon dioxide $(SiO_2)$.
Therefore,'$X$' is $H_4SiO_4$ and '$Y$' is $SiO_2$.

(D) Orthosilicic acid $(H_4SiO_4)$,on heating at $1000^{\circ}C$,loses two water molecules to form silica $(SiO_2)$ as product $A$.
$H_4SiO_4 \xrightarrow[{-2H_2O}]{1000^{\circ}C} SiO_2 (A)$
Silica $(SiO_2)$ on reduction with carbon at high temperature gives carborundum $(SiC)$ as product $B$ and carbon monoxide $(CO)$.
$SiO_2 + 3C \xrightarrow{\Delta} SiC (B) + 2CO$
Therefore,$B$ is carborundum.

(A) Silica $(SiO_2)$ is used for making optical instruments due to its high transparency and thermal stability.

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

(C) The group $14$ elements have a valence shell electronic configuration of $ns^2 np^2$.
They commonly exhibit $+4$ and $+2$ oxidation states.
Due to the inert pair effect,the stability of the $+2$ oxidation state increases down the group $(C < Si < Ge < Sn < Pb)$.
Therefore,$Pb^{2+}$ is the most stable state for lead,and $Pb^{4+}$ is a strong oxidizing agent because it tends to reduce to $Pb^{2+}$.
Conversely,$Sn^{2+}$ is a reducing agent as it tends to oxidize to $Sn^{4+}$.
Since $Pb^{2+}$ is the most stable state for lead,it does not act as a reducing agent.
Thus,the statement that lead in $+2$ state acts as a good reducing agent is incorrect.

(B) In Group $14$,the acidic character of oxides decreases as we move down the group from $Si$ to $Pb$.
$SiO_2$ is a well-known acidic oxide.
$SnO_2$ and $PbO_2$ are amphoteric in nature.
$SnO$ is also amphoteric.
Therefore,$SiO_2$ is the correct acidic oxide.

(D) Catenation is the ability of an element to form stable bonds with other atoms of the same element to form long chains or rings.
This property decreases down the group in Group $14$ because the bond energy of $M-M$ bonds decreases as the atomic size increases.
$C$ shows the highest catenation.
$Pb$ (Lead) is a metal and does not exhibit catenation due to its large atomic size and weak $Pb-Pb$ bond energy.

(C) Most group $14$ elements do not react with water under ambient conditions.
However,tin $(Sn)$ reacts with steam at high temperatures to form tin dioxide $(SnO_2)$ and hydrogen gas $(H_2)$.
The chemical equation is: $Sn(s) + 2H_2O(g) \xrightarrow{\Delta} SnO_2(s) + 2H_2(g)$.
Carbon $(C)$,Germanium $(Ge)$,and Lead $(Pb)$ do not react with water under these conditions.

(D) Among group-$14$ elements,$Pb$ does not show catenation property.
The order of catenation is $C >> Si > Ge \simeq Sn$.

(D) ,$Si$,and $Ge$ do not react with water or steam.
$Sn$ reacts with steam at high temperatures to form tin$(IV)$ oxide and hydrogen gas.
The chemical equation is:
$Sn_{(s)} + 2H_2O_{(g)} \xrightarrow{\Delta} SnO_{2(s)} + 2H_{2(g)}$

(D) The balanced chemical equations are:
$P_2O_3 + 3H_2O \rightarrow 2H_3PO_3$ $(X = H_3PO_3)$
$P_4O_{10} + 6H_2O \rightarrow 4H_3PO_4$ $(Y = H_3PO_4)$
In $H_3PO_3$ (phosphorous acid),the structure contains $1$ $P=O$ bond,$2$ $P-OH$ bonds,and $1$ $P-H$ bond.
In $H_3PO_4$ (phosphoric acid),the structure contains $1$ $P=O$ bond and $3$ $P-OH$ bonds.
Thus,the number of $P=O$ bonds in $X$ and $Y$ are $1$ and $1$ respectively.

(B) Neutral oxides are those oxides that exhibit neither acidic nor basic properties when reacting with water.
Examples of neutral oxides include nitrous oxide $(N_2O)$,nitric oxide $(NO)$,and carbon monoxide $(CO)$.
$SO_2$ and $CO_2$ are acidic oxides because they react with water to form sulphurous acid $(H_2SO_3)$ and carbonic acid $(H_2CO_3)$,respectively.
$CaO$ is a metal oxide and is basic in nature,reacting with water to form calcium hydroxide $(Ca(OH)_2)$.

(A) The chemical reaction is: $C + 2H_2SO_4 \rightarrow CO_2 + 2SO_2 + 2H_2O$.
Both $CO_2$ and $SO_2$ are non-metallic oxides,which are acidic in nature.

(B) When concentrated $H_2SO_4$ reacts with non-metals,it acts as an oxidizing agent and produces the oxoacid of the non-metal along with $SO_2$ gas.
In the case of carbon,the reaction produces carbonic acid $(H_2CO_3)$ and sulfur dioxide $(SO_2)$.
Since $H_2CO_3$ is unstable,it immediately decomposes into carbon dioxide $(CO_2)$ and water $(H_2O)$.
The overall balanced chemical equation is:
$C + 2H_2SO_4 \xrightarrow{\Delta} CO_2 + 2SO_2 + 2H_2O$
Thus,the products formed are $CO_2, SO_2,$ and $H_2O$.
Therefore,option $(b)$ is the correct answer.

(D) mixture of helium and oxygen is used in the treatment of asthma.
Because helium has a low density,this mixture flows easily through restricted respiratory passages,making it easier for patients with asthma to breathe.

(C) In group $14$,the stability of the $+2$ oxidation state increases down the group due to the inert pair effect. Therefore,$Pb^{2+}$ is the most stable state for lead. Consequently,$Pb^{2+}$ acts as an oxidizing agent because it tends to gain electrons to reach the more stable $Pb^0$ state or remain in the $Pb^{2+}$ state,whereas $Sn^{2+}$ acts as a reducing agent because it tends to lose electrons to reach the more stable $Sn^{4+}$ state. Thus,the statement that lead in $+2$ oxidation state is a reducing agent is incorrect.

(A, C, D) $SiO_{2}$ is an acidic oxide and reacts with both strong bases and specific fluorinating agents.
$1$. $SiO_{2(s)} + 2F_{2(g)} \rightarrow SiF_{4(g)} + O_{2(g)}$
$2$. $SiO_{2(s)} + 6HF(aq.) \rightarrow H_{2}SiF_{6}(aq.) + 2H_{2}O(\ell)$
$3$. $SiO_{2(s)} + 2NaOH(aq.) \rightarrow Na_{2}SiO_{3}(aq.) + H_{2}O(\ell)$
Since $HF$,hot $NaOH$,and $F_{2}$ all react with $SiO_{2}$,the question implies identifying reagents that attack it. Given the standard multiple-choice format,this is a multiple-correct type question.

(D) The main reason that $SiCl_{4}$ is easily hydrolysed compared to $CCl_{4}$ is that silicon $(Si)$ has vacant $3d$-orbitals,which allow it to extend its coordination number beyond four.
Water molecules can coordinate with the vacant $d$-orbitals of the $Si$ atom in $SiCl_{4}$,facilitating the hydrolysis process.
In contrast,carbon $(C)$ lacks $d$-orbitals and cannot expand its coordination number beyond four,making $CCl_{4}$ resistant to hydrolysis by water.

(C) Silicon is the second most abundant element in the Earth's crust,but it does not occur in the free state in nature.
It is always found in the combined state,primarily as silica $(SiO_2)$ and silicates.
Therefore,the statement that silicon occurs in the free state in nature is incorrect.

(A) Silicones are synthetic polymers containing $R_2SiO$ repeating units.
Silicon oil is prepared by the hydrolysis and subsequent polymerisation of a mixture of $trimethylchlorosilane$ $( (CH_3)_3SiCl )$ and $dimethyldichlorosilane$ $( (CH_3)_2SiCl_2 )$.
The $trimethylchlorosilane$ acts as a chain terminator,controlling the chain length of the resulting silicone polymer,which determines its viscosity (forming oil).
These are widely used as lubricants and antifoaming agents.

(B) The correct answer is $HNO_2$.
In $HNO_2$ (nitrous acid),the oxidation state of nitrogen is $+3$.
Since the oxidation state of nitrogen ranges from $-3$ to $+5$,nitrogen in $HNO_2$ can be oxidized to $+5$ (acting as a reducing agent) or reduced to lower oxidation states (acting as an oxidizing agent).
Additionally,$NO_2^-$ (nitrite ion) acts as a ligand and is capable of forming complex compounds with transition metals.

(C) The stability of oxidation states in Group $14$ elements is governed by the inert pair effect.
For $Pb$,the $+2$ oxidation state is more stable than the $+4$ state due to the inert pair effect. Thus,the reaction $PbO_{2} + Pb \rightarrow 2PbO$ is spontaneous,meaning $\Delta_{r}G^{\circ}(1) < 0$.
For $Sn$,the $+4$ oxidation state is more stable than the $+2$ state. Thus,the reaction $SnO_{2} + Sn \rightarrow 2SnO$ is non-spontaneous,meaning $\Delta_{r}G^{\circ}(2) > 0$.
Therefore,the correct set is $\Delta_{r}G^{\circ}(1) < 0$ and $\Delta_{r}G^{\circ}(2) > 0$.

(A) The isotope ${}^{13}C$ is stable and not radioactive. The radioactive isotope of carbon is ${}^{14}C$. Therefore,statement $A$ is incorrect.

(A) Oxygen commonly exhibits an oxidation state of $-2$,but it also shows $-1$ (in peroxides),$-1/2$ (in superoxides),and even positive states when combined with fluorine (e.g.,$OF_2$). Therefore,the statement that oxygen exhibits only $-2$ oxidation state is incorrect.