Carbon family Questions in English

(A) The carbon-carbon $(C-C)$ bond length in diamond is $154 \, pm$.
In graphite, the $C-C$ bond length is $141.5 \, pm$ due to resonance.
In naphthalene and fullerene, the bond lengths are intermediate between single and double bonds.
Therefore, the $C-C$ bond length is maximum in diamond.

(B) The reaction is $\underline{S}O_2 + H_2O \longrightarrow H_2SO_3$.
In $H_2SO_3$,the oxidation state of sulfur is $+4$.
Sulfur in its $+4$ oxidation state forms sulfurous acid $(H_2SO_3)$,which carries the $-ous$ suffix.
Therefore,the product is an oxyacid with an $-ous$ suffix.

(D) Carbon monoxide $(CO)$ is a neutral oxide.
It does not react with water at room temperature $(R.T.)$ to form any oxy acid.
Therefore,the product is not an oxy acid,and it does not possess an $-ic$ or $-ous$ suffix.
Thus,the correct option is $D$.

(A) The reaction is: $\underline{S}O_3 + H_2O \longrightarrow H_2SO_4$.
In $SO_3$,the oxidation state of sulfur is $+6$.
In the product $H_2SO_4$ (sulfuric acid),the oxidation state of sulfur is also $+6$.
Since the oxidation state remains the same and the product is an oxy acid with the highest oxidation state for sulfur,it is named with the $-ic$ suffix (sulfuric acid).
Therefore,the correct option is $A$.

(A) The reaction is: $\underline{C}O_2 + H_2O \longrightarrow H_2CO_3$.
In $H_2CO_3$ (carbonic acid),the oxidation state of carbon is $+4$.
Since carbon is in its highest oxidation state $(+4)$,the corresponding oxy acid is named with the suffix $-ic$.
Therefore,$H_2CO_3$ is carbonic acid,which ends in $-ic$.

(D) The reaction $CCl_4 + H_2O \longrightarrow \text{No reaction}$ occurs because the carbon atom in $CCl_4$ is sterically hindered by four chlorine atoms and lacks vacant $d$-orbitals to accept a lone pair from the oxygen atom of water.
Since no reaction occurs,there is no product formed,and therefore,the product is neither an oxy acid with an $-ic$ suffix nor an $-ous$ suffix.
Thus,the correct option is $D$.

(D) The reaction $\underline{N}O + H_2O \longrightarrow$ shows no reaction at room temperature $(R.T.)$.
Since there is no product formed,the condition does not apply to any oxyacid classification.
Therefore,the correct description is that the product is not an oxyacid,neither with $-ic$ suffix nor with $-ous$ suffix.

(A) The reaction for the complete hydrolysis of $Cl_2O_7$ is:
$Cl_2O_7 + H_2O \to 2HClO_4$
In $HClO_4$ (perchloric acid),the oxidation state of chlorine is $+7$.
Since the chlorine atom is in its highest oxidation state $(+7)$,the corresponding oxyacid is named with the $-ic$ suffix (specifically,perchloric acid).
Therefore,the product is an oxyacid with an $-ic$ suffix.

(A) The reaction is $H_3PO_5 + H_2O \longrightarrow H_3PO_4 + H_2O_2$.
In the product $H_3PO_4$ (phosphoric acid),the phosphorus atom is in its $+5$ oxidation state.
The suffix $-ic$ is used for the acid where the central atom is in its higher oxidation state (e.g.,phosphoric acid).
Since $H_3PO_4$ is an oxy acid with an $-ic$ suffix,option $A$ is correct.

(A) The reaction given is: $\underline{Si}F_4 + 4H_2O \longrightarrow H_4SiO_4 + 4HF$.
Note: The balanced equation for complete hydrolysis is $\underline{Si}F_4 + 4H_2O \longrightarrow H_4SiO_4 + 4HF$.
The product $H_4SiO_4$ is orthosilicic acid,which can be written as $Si(OH)_4$ or $H_4SiO_4$.
In $H_4SiO_4$,the oxidation state of $Si$ is $+4$.
Since the highest oxidation state of $Si$ is $+4$,the corresponding oxyacid is named with the $-ic$ suffix (Silicic acid).
Therefore,the product is an oxyacid with an $-ic$ suffix.

(A) The reaction given is the hydrolysis of Caro's acid $(H_2SO_5)$:
$H_2SO_5 + H_2O \longrightarrow H_2SO_4 + H_2O_2$
In this reaction,the underlined atom is sulfur $(S)$.
The product $H_2SO_4$ is sulfuric acid,which is an oxyacid with an $-ic$ suffix.
The product $H_2O_2$ is hydrogen peroxide,which is not an oxyacid.
Since the question asks to assign the nature for the product of the underlined atom $(S)$,and $H_2SO_4$ is the primary sulfur-containing product,it is an oxyacid with an $-ic$ suffix.
Therefore,the correct description is that the product is an oxyacid with an $-ic$ suffix.

(A) The complete hydrolysis of $SiCl_4$ at room temperature $(R.T.)$ is given by the reaction:
$SiCl_4 + 4H_2O \longrightarrow H_4SiO_4 + 4HCl$
Here,$H_4SiO_4$ is orthosilicic acid,which can also be written as $Si(OH)_4$ or $H_2SiO_3 \cdot H_2O$.
In the nomenclature of oxy acids,$H_4SiO_4$ is referred to as silicic acid,which follows the $-ic$ suffix convention.
Therefore,the product is an oxy acid with an $-ic$ suffix.

(A) The reaction is: $\underline{Cr}O_2Cl_2 + 2H_2O \longrightarrow H_2CrO_4 + 2HCl$.
In $H_2CrO_4$,the oxidation state of $Cr$ is $+6$.
Since $Cr$ is in its highest oxidation state $(+6)$,the corresponding oxyacid is named chromic acid,which ends with the $-ic$ suffix.
Therefore,the product is an oxyacid with an $-ic$ suffix.

(D) The reaction $\underline{Si}O_2 + H_2O \longrightarrow$ No reaction indicates that silicon dioxide $(SiO_2)$ is insoluble in water and does not undergo hydrolysis at room temperature $(R.T.)$.
Since no reaction occurs,there is no formation of an oxyacid.
Therefore,the product is not an oxyacid,neither with $-ic$ suffix nor with $-ous$ suffix.
Thus,the correct option is $D$.

(B) The reaction for the complete hydrolysis of $P_4O_6$ is:
$P_4O_6 + 6H_2O \longrightarrow 4H_3PO_3$
In $P_4O_6$,the oxidation state of phosphorus is $+3$.
$H_3PO_3$ is phosphorous acid,which contains phosphorus in the $+3$ oxidation state.
Oxy acids with lower oxidation states are named with the suffix $-ous$,while those with higher oxidation states are named with the suffix $-ic$.
Since $H_3PO_3$ is an oxy acid with the $-ous$ suffix,the correct option is $B$.

(A) The given reaction is the hydrolysis of pyrosilicic acid $(H_6Si_2O_7)$:
$H_6Si_2O_7 + H_2O \longrightarrow 2H_4SiO_4$
$H_4SiO_4$ is orthosilicic acid.
In orthosilicic acid,the oxidation state of $Si$ is $+4$.
Since $Si$ is in its highest oxidation state $(+4)$,the corresponding oxy acid is named with the $-ic$ suffix (Silicic acid).
Therefore,the product is an oxy acid with an $-ic$ suffix.

(B) The hydrolysis of $SF_4$ at room temperature proceeds as follows:
$SF_4 + 3H_2O \longrightarrow H_2SO_3 + 4HF$
In the product $H_2SO_3$ (sulfurous acid),the oxidation state of sulfur is $+4$.
Since the suffix used is $-ous$ (sulfurous acid),the correct assignment is that the product is an oxyacid with an $-ous$ suffix.

(A) The reaction is the hydrolysis of oleum $(H_2S_2O_7)$:
$H_2S_2O_7 + H_2O \longrightarrow 2H_2SO_4$.
Here,the sulfur atom in $H_2S_2O_7$ has an oxidation state of $+6$.
The product formed is sulfuric acid $(H_2SO_4)$,where sulfur also has an oxidation state of $+6$.
Since the suffix for the acid with the higher oxidation state is $-ic$,the product is an oxy acid with an $-ic$ suffix.

(A) The reaction for the complete hydrolysis of phosphorus pentoxide $(P_4O_{10})$ is:
$P_4O_{10} + 6H_2O \longrightarrow 4H_3PO_4$
In the product $H_3PO_4$ (phosphoric acid),the oxidation state of phosphorus is $+5$.
Since the phosphorus atom is in its highest oxidation state $(+5)$,the corresponding oxyacid is named with the $-ic$ suffix (phosphoric acid).
Therefore,the correct description is that the product is an oxyacid with an $-ic$ suffix.

(A) The reaction is: $\underline{P}OCl_3 + 3H_2O \longrightarrow H_3PO_4 + 3HCl$.
In this reaction,the phosphorus atom in $\underline{P}OCl_3$ (where the oxidation state of $P$ is $+5$) undergoes complete hydrolysis to form phosphoric acid $(H_3PO_4)$.
$H_3PO_4$ is an oxy acid of phosphorus where phosphorus is in the $+5$ oxidation state,which corresponds to the $-ic$ suffix (phosphoric acid).
Therefore,the product is an oxy acid with an $-ic$ suffix.

(D) The reaction of $P_4$ with $H_2O$ at room temperature $(R.T.)$ shows no reaction.
Since no product is formed,it does not result in any oxy acid.
Therefore,the condition described in option $D$ is satisfied as the product is not an oxy acid.

(D) The reaction $\underline{S}_8 + H_2O \longrightarrow$ No reaction indicates that elemental sulfur $(S_8)$ does not undergo hydrolysis at room temperature $(R.T.)$.
Since no reaction occurs,no oxy acid is formed.
Therefore,the condition described in option $D$ is satisfied,as there is no product formed that could be classified as an oxy acid with either an $-ic$ or $-ous$ suffix.

(B) The reaction is: $\underline{S}OCl_2 + 2H_2O \longrightarrow H_2SO_3 + 2HCl$.
In this reaction,the sulfur atom in $SOCl_2$ (thionyl chloride) undergoes hydrolysis to form sulfurous acid $(H_2SO_3)$ and hydrochloric acid $(HCl)$.
The product $H_2SO_3$ is an oxy acid of sulfur where sulfur is in the $+4$ oxidation state.
Oxy acids of sulfur with the $+4$ oxidation state are named with the $-ous$ suffix (e.g.,sulfurous acid).
Therefore,the product is an oxy acid with an $-ous$ suffix.

(A) The reaction is: $\underline{Si}H_4 + 4H_2O \longrightarrow H_4SiO_4 + 4H_2$.
$H_4SiO_4$ is also written as $Si(OH)_4$ or $H_2SiO_3 \cdot H_2O$,which is silicic acid.
In silicic acid,the oxidation state of $Si$ is $+4$.
Since the maximum oxidation state of $Si$ is $+4$,the acid is named with the $-ic$ suffix (silicic acid).
Therefore,the product is an oxy acid with an $-ic$ suffix.

(A) The complete hydrolysis of $SOF_4$ is given by the reaction:
$SOF_4 + 3H_2O \longrightarrow H_2SO_4 + 4HF$
In the product $H_2SO_4$ (sulfuric acid),the sulfur atom is in the $+6$ oxidation state.
The suffix $-ic$ is used for the higher oxidation state of the central atom in an oxy acid.
Since $H_2SO_4$ is sulfuric acid,it ends with the $-ic$ suffix.
Therefore,the correct option is $A$.

(B) The thermal decomposition reactions are as follows:
$1. 2K_2Cr_2O_7 \xrightarrow{\Delta } 2K_2CrO_4 + Cr_2O_3 + \frac{3}{2}O_2$ (Produces $O_2$)
$2. Ag_2C_2O_4 \xrightarrow{\Delta } 2Ag + 2CO_2$ (Does not produce $O_2$)
$3. 2Pb(NO_3)_2 \xrightarrow{\Delta } 2PbO + 4NO_2 + O_2$ (Produces $O_2$)
$4. Ag_2CO_3 \xrightarrow{\Delta } 2Ag + CO_2 + \frac{1}{2}O_2$ (Produces $O_2$)
Thus,$Ag_2C_2O_4$ does not release oxygen gas upon heating.

(D) Amphoteric oxides are those that react with both acids and bases to form salts and water.
Common examples of elements forming amphoteric oxides include $Al$,$Zn$,$Pb$,$Sn$,$Be$,and $Ge$.
In option $D$,both $Ge$ (Germanium) and $Al$ (Aluminum) form amphoteric oxides ($GeO_2$ and $Al_2O_3$ respectively).
Therefore,the correct pair is $Ge$ and $Al$.

(D) Amphoteric oxides are those that react with both acids and bases to form salts and water.
Common examples of metals that form amphoteric oxides include $Al$,$Zn$,$Pb$,$Sn$,$Be$,and $Ga$.
In the given options,$Sn$ (Tin) and $Zn$ (Zinc) both form amphoteric oxides ($SnO_2$ and $ZnO$ respectively).
Therefore,the correct pair is $Sn$ and $Zn$.

(B) The tetrahalides of group $14$ elements are covalent and possess a tetrahedral geometry.
Except for carbon,all other group $14$ tetrahalides $(MX_4)$ have vacant $d$-orbitals in the central atom.
These vacant $d$-orbitals allow the central atom to accept electron pairs from donor molecules,thereby acting as $Lewis$ acids.
Carbon lacks vacant $d$-orbitals and cannot expand its octet,thus it does not act as a $Lewis$ acid.

(C) The reaction of $Al$ with $N_2$ at high temperature gives aluminum nitride:
$2Al + N_2 \rightarrow 2AlN \, (A)$
Hydrolysis of $AlN$ gives ammonia:
$AlN + 3H_2O \rightarrow Al(OH)_3 + NH_3 \uparrow$
The reaction of $Al$ with $C$ at high temperature gives aluminum carbide:
$4Al + 3C \rightarrow Al_4C_3 \, (B)$
Hydrolysis of $Al_4C_3$ gives methane:
$Al_4C_3 + 12H_2O \rightarrow 4Al(OH)_3 + 3CH_4 \uparrow$
Therefore,$A$ gives $NH_3$ and $B$ gives $CH_4$.

(D) The catenation power of an element is directly proportional to the strength of the bond formed between its own atoms (Element-Element bond energy).
Since the order of catenation power is $Ge < Si < C$,the order of bond energies must also be $Ge-Ge < Si-Si < C-C$.
Comparing the given values,the bond energies are approximately $350 \ kJ/mol$ for $C-C$,$180 \ kJ/mol$ for $Si-Si$,and $170 \ kJ/mol$ for $Ge-Ge$.
Therefore,the correct order is $350, 180, 170$.

(B) $1$. Both $CO_2$ and $SO_2$ react with lime water $(Ca(OH)_2)$ to form calcium carbonate $(CaCO_3)$ or calcium sulfite $(CaSO_3)$,which are white precipitates,thus turning lime water milky.
$2$. $SO_2$ acts as a reducing agent and turns acidified $K_2Cr_2O_7$ solution green due to the formation of $Cr_2(SO_4)_3$,but $CO_2$ does not.
$3$. $SO_2$ can act as both an oxidising and a reducing agent,while $CO_2$ is generally inert in redox reactions under standard conditions.
$4$. Since only statement $B$ is correct,the options provided are technically flawed. However,in many contexts,the question aims to highlight the common property of turning lime water milky.

(D) $1$. $SiC$ (Carborundum) is a covalent carbide. It is extremely hard and resistant to hydrolysis under normal conditions.
$2$. $Al_4C_3$ (Aluminum carbide) is an ionic carbide (methanide). On hydrolysis,it reacts with water to produce methane gas: $Al_4C_3 + 12H_2O \rightarrow 4Al(OH)_3 + 3CH_4$.
$3$. $Be_2C$ (Beryllium carbide) is also an ionic carbide (methanide). On hydrolysis,it produces methane gas: $Be_2C + 4H_2O \rightarrow 2Be(OH)_2 + CH_4$.
Therefore,both statements $B$ and $C$ are correct.

(C) Silicates are classified based on the number of oxygen atoms shared between $SiO_4^{4-}$ tetrahedral units.
$1$. Ortho silicates contain discrete $SiO_4^{4-}$ units.
$2$. Pyro silicates (also known as sorosilicates) are formed by sharing one oxygen atom between two $SiO_4^{4-}$ units,resulting in the $Si_2O_7^{6-}$ anion.
$3$. Cyclic silicates contain units where two oxygen atoms per tetrahedron are shared to form a ring structure.
$4$. Chain silicates involve sharing two oxygen atoms per tetrahedron to form a long chain.
Therefore,the silicates containing $Si_2O_7^{6-}$ units are called pyro silicates.

(B) True: $R_2SiO$ is the repeating unit of linear silicones.
$(b)$ False: $RSiCl_3$ on hydrolysis followed by dehydration gives cross-linked silicones. For linear silicones,$R_2SiCl_2$ is used.
$(c)$ True: Silicones are chemically inert and have high thermal stability,making them useful as heat insulators.
$(d)$ True: Silica $(SiO_2)$ reacts with hydrofluoric acid $(HF)$ to form hexafluorosilicic acid: $SiO_2 + 6HF \rightarrow H_2SiF_6 + 2H_2O$.
Therefore,the correct statements are $(a), (c),$ and $(d)$.

(A) Silicones are a group of organosilicon polymers,which have $(-R_2SiO-)$ as a repeating unit.
The starting materials for the manufacture of silicones are alkyl or aryl substituted silicon chlorides,$R_nSiCl_{(4-n)}$,where $R$ is an alkyl or aryl group.
Upon hydrolysis of these alkyl or aryl substituted silicon chlorides,silanols are formed,which undergo condensation polymerization to form silicones.
Among the given options,$(CH_3)_2SiCl_2$ is an alkyl-substituted silicon chloride,which on hydrolysis yields $(CH_3)_2Si(OH)_2$,followed by polymerization to form silicones. Therefore,option $A$ is the correct answer.

(B) $SiO_2$ (silica) is a covalent network solid where each $Si$ atom is $sp^3$ hybridized and bonded to four oxygen atoms in a tetrahedral geometry.
Due to the presence of $Si-O$ bonds and the nature of the oxide,$SiO_2$ is acidic in nature,as it reacts with strong bases to form silicates.

(D) The structure of $P_4O_{10}$ consists of a $P_4$ tetrahedron where each $P$ atom is bonded to three other $P$ atoms through oxygen bridges ($P-O-P$ bonds).
Additionally,each $P$ atom is bonded to a terminal oxygen atom via a double bond $(P=O)$.
These terminal $P=O$ bonds are shorter than the $P-O-P$ single bonds.
In $P_4O_{10}$,there are $4$ such terminal $P=O$ bonds.
Therefore,the number of short $P-O$ bonds is $4$.

(D) Zeolites are aluminosilicates with a three-dimensional network structure in which some silicon atoms are replaced by aluminum atoms,giving the framework a negative charge.
This framework is represented as $Al_xSi_{1-x}O_2$.
Therefore,the statement that $SiO_4^{4-}$ units are not replaced by $AlO_4^{5-}$ ions is incorrect,as this replacement is the fundamental reason for their ion-exchange properties.

(D) Aluminum is a highly reactive metal that forms a thin,protective oxide layer $(Al_2O_3)$ on its surface when exposed to air.
This oxide layer is transparent,hard,and prevents further oxidation or tarnishing of the underlying metal.
Because it does not tarnish or lose its reflectivity in air,it serves as an excellent material for making mirrors.

(C) $CCl_4$ does not undergo hydrolysis because carbon lacks $d$-orbitals in its valence shell,which prevents it from expanding its coordination number to accommodate incoming water molecules. In contrast,$SiCl_4$,$SiF_4$,and $PbCl_4$ can undergo hydrolysis due to the availability of vacant $d$-orbitals or the presence of lower-energy orbitals that facilitate the formation of intermediate complexes with water.

(B) In the carbon family (Group $14$),as we move down the group,the atomic number increases.
Due to the inert pair effect,the stability of the $+2$ oxidation state increases while the stability of the $+4$ oxidation state decreases.
Metallic character increases down the group.
Ionization energy decreases down the group.
Atomic size increases down the group due to the addition of new shells.

(C) Silicon is the second most abundant element in the Earth's crust,primarily found in the form of silica $(SiO_2)$ and silicates. These compounds are the primary components of rocks,sand,and soil. Therefore,silicon is an important constituent of rocks.

(D) Crystalline silicon has a diamond-like cubic structure where each silicon atom is tetrahedrally bonded to four other silicon atoms. Among the given allotropes of carbon,diamond possesses the same $sp^3$ hybridized tetrahedral structure as silicon. Therefore,diamond is isomorphous with crystalline silicon.

(D) In group $14$ elements,the stability of the $+2$ oxidation state increases down the group due to the inert pair effect.
$Pb$ and $Sn$ form stable monoxides ($PbO$ and $SnO$).
$Ge$ forms $GeO$,which is stable at high temperatures.
$Silicon$ $(Si)$ does not form a stable monoxide $(SiO)$ under standard conditions because $SiO$ is unstable and disproportionates into $Si$ and $SiO_2$.

(A) The compound $SnCl_4 \cdot 5H_2O$ is commonly known as 'Butter of Tin'.
It is a hydrated form of tin$(IV)$ chloride,which appears as a white,crystalline solid.

(C) The reaction between silica $(SiO_2)$ and sodium carbonate $(Na_2CO_3)$ is a fusion reaction used in glass manufacturing.
The chemical equation is: $SiO_2 + Na_2CO_3 \rightarrow Na_2SiO_3 + CO_2 \uparrow$.
As shown in the equation,$CO_2$ gas is evolved during this process.

(A) The stability of carbon tetrahalides $(CX_4)$ depends on the bond dissociation energy of the $C-X$ bond.
As the size of the halogen atom increases from $F$ to $I$,the bond length increases and the bond dissociation energy decreases.
Since the $C-F$ bond is the shortest and strongest among the carbon tetrahalides,$CF_4$ has the highest bond dissociation energy and is the most stable.

(D) Carbon $(C)$ and Silicon $(Si)$ belong to Group $14$ of the periodic table.
Carbon is a non-metal,while Silicon is a metalloid.
Due to the significant difference in their atomic size,ionization enthalpy,and electronegativity,they exhibit different physical and chemical properties.
For example,carbon forms stable $p\pi-p\pi$ multiple bonds,whereas silicon does not form such bonds effectively due to its larger atomic size.

(D) Carborundum is the common name for Silicon Carbide,which has the chemical formula $SiC$.
It is a very hard material used as an abrasive.