TS EAMCET 2021 Chemistry Question Paper with Answer and Solution

(C) The molecular formula $C_6H_{10}$ corresponds to an alkyne with $6$ carbon atoms. The branched isomers are those that do not have a straight chain of $6$ carbons. The possible branched isomers are:
$1$. $3-\text{methyl}-1-\text{pentyne}$ $(CH_3-CH_2-CH(CH_3)-C \equiv CH)$
$2$. $4-\text{methyl}-1-\text{pentyne}$ $(CH_3-CH(CH_3)-CH_2-C \equiv CH)$
$3$. $3-\text{methyl}-2-\text{pentyne}$ $(CH_3-CH_2-C(CH_3)-C \equiv C-CH_3)$
$4$. $3,3-\text{dimethyl}-1-\text{butyne}$ $((CH_3)_2C(C \equiv CH)-CH_3)$
Thus,there are $4$ possible branched isomers.

(C) The reaction sequence is as follows:
$1$. Chlorination of propane $(CH_3CH_2CH_3)$ with $Cl_2$ in the presence of light $(hv)$ gives $1$-chloropropane as the major product: $CH_3CH_2CH_3 + Cl_2 \xrightarrow{hv} CH_3CH_2CH_2Cl + HCl$.
$2$. Dehydrohalogenation of $1$-chloropropane with alcoholic $KOH$ gives propene: $CH_3CH_2CH_2Cl \xrightarrow{\text{alc. } KOH} CH_3CH=CH_2 + KCl + H_2O$.
$3$. High-temperature chlorination of propene $(773 \ K)$ leads to allylic substitution,forming allyl chloride: $CH_3CH=CH_2 + Cl_2 \xrightarrow{773 \ K} ClCH_2CH=CH_2 + HCl$.
$4$. Nucleophilic substitution of the allylic chloride with aqueous $AgOH$ (or $NaOH$) replaces the $Cl$ atom with an $OH$ group to form allyl alcohol: $ClCH_2CH=CH_2 + AgOH_{(aq)} \longrightarrow HOCH_2CH=CH_2 + AgCl$.

(D) The eclipsed conformation of ethane is the least stable due to torsional strain caused by the repulsion between the hydrogen atoms on adjacent carbon atoms.
Conversely,the staggered conformation is the most stable as the hydrogen atoms are as far apart as possible,minimizing repulsion.
The energy barrier for rotation about the $C-C$ single bond in ethane is approximately $12.5 \ kJ / mol$ $(3 \ kcal / mol)$.
Therefore,the energy difference between the staggered and eclipsed conformations is $12.5 \ kJ / mol$.

(D) The reaction in option $(a)$ is the catalytic hydrogenation of an alkyne,which produces an alkane $(CH_3-CH_2-CH_2-CH_3)$.
The reaction in option $(b)$ is Kolbe's electrolytic decarboxylation,which produces an alkane ($CH_3-CH_2-CH_2-CH_2-CH_2-CH_2-CH_2-CH_3$ from the radical coupling of the pentanoate ion).
The reaction in option $(c)$ is the Wurtz reaction,which produces an alkane $(CH_3-CH_2-CH_2-CH_3)$.
The reaction in option $(d)$ involves an alkyl fluoride $(CH_3-CH_2-CH_2-CH_2-F)$ with $Zn/H^{\oplus}$. Alkyl fluorides are generally unreactive towards reduction by $Zn/H^{\oplus}$ due to the high strength of the $C-F$ bond. Therefore,this reaction does not produce an alkane.

(C) The reaction $CH_2=CH_2 + H_2 \xrightarrow{Ni, 300^{\circ}C} CH_3-CH_3$ is the catalytic hydrogenation of ethene to ethane.
This specific process,where alkenes are reduced to alkanes using a metal catalyst like $Ni$,$Pt$,or $Pd$,is known as the Sabatier-Senderens reaction.

(B) When propyne $(CH_3-C \equiv CH)$ is passed through a red hot iron tube at $873 \ K$,it undergoes cyclic trimerization to form an aromatic compound called mesitylene ($1,3,5$-trimethylbenzene).
The reaction is as follows:
$3CH_3-C \equiv CH \xrightarrow{\text{Red hot Fe, } 873 \ K} C_9H_{12}$ (Mesitylene).
The molecular formula of mesitylene is $C_9H_{12}$.

(D) The reaction of an alkyne with excess $Br_2$ in $CCl_4$ is an electrophilic addition reaction.
During this process,both $\pi$-bonds of the triple bond are broken,and four new $\sigma$-bonds are formed with bromine atoms.
For hex$-1-$yne $(CH_3(CH_2)_3C \equiv CH)$,the addition occurs at the $C-1$ and $C-2$ positions.
The final product is $1,1,2,2-$tetrabromohexane.
$CH_3(CH_2)_3C \equiv CH + 2Br_2 \xrightarrow{CCl_4} CH_3(CH_2)_3C(Br)_2-CH(Br)_2$

(D) The stability of alkenes in fused ring systems is governed by Bredt's rule and the degree of substitution at the double bond.
Structure $III$ is a tetrasubstituted alkene (specifically,a tetrasubstituted double bond at the bridgehead position),which is highly stable due to hyperconjugation and inductive effects.
Structure $I$ is a trisubstituted alkene.
Structure $II$ is a disubstituted alkene.
Greater substitution leads to higher stability.
Therefore,the correct order of stability is $III > I > II$.

(C) Step $1$: Dehydrohalogenation of $C_6H_5-CHBr-CH_2Br$ with $alc. KOH$ followed by $NaNH_2$ leads to the formation of phenylacetylene $(C_6H_5-C \equiv CH)$.
Step $2$: The cyclic trimerization of phenylacetylene occurs when passed through a red hot iron tube at $873 \ K$.
Step $3$: The trimerization of $C_6H_5-C \equiv CH$ yields $1,3,5-triphenylbenzene$ as the major product due to steric hindrance,which favors the meta-substituted product.

(B) This reaction is known as Friedel-Crafts alkylation. In the presence of anhydrous $AlCl_3$,$n-$propyl chloride acts as an alkylating agent. It is an electrophilic aromatic substitution reaction.
Step $1$: Generation of electrophile:
$CH_3-CH_2-CH_2-Cl + AlCl_3 \rightarrow CH_3-CH_2-CH_2^+ + AlCl_4^-$
Step $2$: Carbocation rearrangement:
The primary carbocation $(CH_3-CH_2-CH_2^+)$ formed in step $1$ rearranges to form a more stable secondary carbocation $(CH_3-CH^+-CH_3)$ via a $1,2-$hydride shift.
Step $3$: Electrophilic attack:
This secondary carbocation attacks the benzene ring to form the intermediate.
Step $4$: Deprotonation:
The final step is the abstraction of a proton from the intermediate to restore aromaticity,yielding isopropyl benzene (cumene) as the major product.

(A) The given reactions are analyzed as follows:
$1.$ Friedel-Crafts alkylation of benzene with $t$-butyl chloride in the presence of $AlCl_3$ yields $t$-butyl benzene.
$2.$ Benzene reacts with $2$-methylpropene (isobutylene) in the presence of $BF_3/HF$ to form $t$-butyl benzene via a carbocation intermediate.
$3.$ Benzene reacts with $t$-butyl alcohol in the presence of $H_2SO_4$ to form $t$-butyl benzene through the formation of a $t$-butyl carbocation.
$4.$ Benzene reacts with butanoyl chloride via Friedel-Crafts acylation to form $1$-phenylbutan-$1$-one. Subsequent Clemmensen reduction $(Zn/Hg, HCl)$ yields $n$-butylbenzene,not $t$-butyl benzene.
Thus,reactions $1, 2,$ and $3$ produce $t$-butyl benzene.

(B) The reaction of benzene with acetyl chloride in the presence of anhydrous $AlCl_3$ is a Friedel-Crafts acylation reaction,which yields acetophenone $(C_6H_5COCH_3)$ as the major product.
In the structure of acetophenone:
$1$. The benzene ring contains $3$ $\pi$-bonds.
$2$. The carbonyl group $(C=O)$ contains $1$ $\pi$-bond.
Total number of $\pi$-bonds = $3 + 1 = 4$.
$3$. The oxygen atom in the carbonyl group has $2$ lone pairs of electrons.
Total number of lone pairs = $2$.
Thus,the number of $\pi$-bonds and lone pairs are $4$ and $2$ respectively.

(C) Temporary hardness of water is removed by Clark's process,where slaked lime,$Ca(OH)_2$,is added to hard water.
The chemical reactions involved are:
$Ca(HCO_3)_{2(aq)} + Ca(OH)_{2(aq)} \rightarrow 2CaCO_{3(s)} + 2H_2O_{(l)}$
$Mg(HCO_3)_{2(aq)} + 2Ca(OH)_{2(aq)} \rightarrow 2CaCO_{3(s)} + Mg(OH)_{2(s)} + 2H_2O_{(l)}$
Note: In the case of magnesium bicarbonate,the product is magnesium hydroxide,$Mg(OH)_2$,due to its lower solubility product compared to $MgCO_3$.
Thus,the insoluble precipitates formed are $Mg(OH)_2$ and $CaCO_3$.

(C) $SO_3 + D_2O \rightarrow D_2SO_4$ (Similar to $SO_3 + H_2O \rightarrow H_2SO_4$)
$CaC_2 + 2D_2O \rightarrow Ca(OD)_2 + C_2D_2$ (Similar to $CaC_2 + 2H_2O \rightarrow Ca(OH)_2 + C_2H_2$)
$Al_4C_3 + 12D_2O \rightarrow 4Al(OD)_3 + 3CD_4$ (Similar to $Al_4C_3 + 12H_2O \rightarrow 4Al(OH)_3 + 3CH_4$)
Thus,the correct match is $A-V, B-I, C-II$.

(C) moderator is a material used in a nuclear reactor to slow down the neutrons produced from fission.
Heavy water $(D_2O)$ is used as a moderator in nuclear reactors because it slows down neutrons effectively and also has a low probability of neutron absorption.

(C) The degree of dissociation $\alpha$ for a weak acid is related to the dissociation constant $K_a$ and concentration $C$ by the formula: $\alpha = \sqrt{\frac{K_a}{C}}$.
Given: $C = 0.1 \ M$ and $\alpha = 5 \% = 0.05$.
Rearranging the formula for $K_a$ (or $K_c$ for the acid dissociation equilibrium): $K_a = \alpha^2 \times C$.
Substituting the values: $K_a = (0.05)^2 \times 0.1$.
$K_a = 0.0025 \times 0.1 = 0.00025 = 2.5 \times 10^{-4}$.
Thus,the value of $K_c$ is $2.5 \times 10^{-4}$.

(A) For a basic buffer solution,the $pOH$ is given by the Henderson-Hasselbalch equation:
$pOH = pK_{b} + \log \frac{[Salt]}{[Base]}$
Given that the molar concentrations of the base and its conjugate acid (salt) are the same,we have $[Salt] = [Base]$.
Substituting this into the equation:
$pOH = pK_{b} + \log(1)$
Since $\log(1) = 0$,we get:
$pOH = pK_{b}$
Therefore,the $pOH$ of the buffer solution is the same as the $pK_{b}$ of the base.

(D) The acid dissociation constants $(K_a)$ for the given acids are as follows:
$HCN$: $K_a = 4.9 \times 10^{-10}$ $(III)$
$H_2C_2O_4$ (Oxalic acid): $K_a = 5.6 \times 10^{-2}$ $(IV)$
$H_2S$: $K_a = 8.9 \times 10^{-8}$ $(II)$
Niacin (Vitamin $B_3$): $K_a = 1.5 \times 10^{-5}$ $(V)$
Therefore,the correct matching is $A-III, B-IV, C-II, D-V$.

(D) For a tribasic acid like $H_3PO_4$,there are three ionisation constants. The overall ionisation constant $(K)$ is the product of the ionisation constants of the three individual steps.
Step $1$: $H_3PO_4 \rightleftharpoons H^+ + H_2PO_4^-$; $K_1 = \frac{[H^+][H_2PO_4^-]}{[H_3PO_4]}$ $(i)$
Step $2$: $H_2PO_4^- \rightleftharpoons H^+ + HPO_4^{2-}$; $K_2 = \frac{[H^+][HPO_4^{2-}]}{[H_2PO_4^-]}$ $(ii)$
Step $3$: $HPO_4^{2-} \rightleftharpoons H^+ + PO_4^{3-}$; $K_3 = \frac{[H^+][PO_4^{3-}]}{[HPO_4^{2-}]}$ $(iii)$
Adding the three equations gives the overall reaction: $H_3PO_4 \rightleftharpoons 3H^+ + PO_4^{3-}$.
The overall equilibrium constant $K$ is given by $K = \frac{[H^+]^3[PO_4^{3-}]}{[H_3PO_4]}$.
Multiplying the three stepwise equilibrium constants:
$K_1 \times K_2 \times K_3 = \frac{[H^+][H_2PO_4^-]}{[H_3PO_4]} \times \frac{[H^+][HPO_4^{2-}]}{[H_2PO_4^-]} \times \frac{[H^+][PO_4^{3-}]}{[HPO_4^{2-}]}$
$K_1 \times K_2 \times K_3 = \frac{[H^+]^3[PO_4^{3-}]}{[H_3PO_4]} = K$
Therefore,$K = K_1 \times K_2 \times K_3$.

(A) Let the direction cosines be $l, m, n$. Given the angles $\alpha = \frac{\pi}{4}$,$\beta = \frac{\pi}{3}$,and $\gamma = \theta$.
Then $l = \cos \frac{\pi}{4} = \frac{1}{\sqrt{2}}$ and $m = \cos \frac{\pi}{3} = \frac{1}{2}$.
We know that $l^2 + m^2 + n^2 = 1$.
Substituting the values,we get $\left(\frac{1}{\sqrt{2}}\right)^2 + \left(\frac{1}{2}\right)^2 + \cos^2 \theta = 1$.
$\frac{1}{2} + \frac{1}{4} + \cos^2 \theta = 1$.
$\frac{3}{4} + \cos^2 \theta = 1$.
$\cos^2 \theta = 1 - \frac{3}{4} = \frac{1}{4}$.
Since $0 < \theta < \frac{\pi}{2}$,$\cos \theta$ must be positive,so $\cos \theta = \frac{1}{2}$.
Thus,the direction cosines are $\frac{1}{\sqrt{2}}, \frac{1}{2}, \frac{1}{2}$.

(A) Boron has the atomic number $5$ and electronic configuration $1s^2 2s^2 2p^1$.
Due to the absence of $d$-orbitals in the valence shell,boron cannot expand its coordination number beyond $4$.
Therefore,it cannot form a hexacoordinated complex like $[B(H_2O)_6]^{3+}$.
Thus,the complex $[B(H_2O)_6]^{3+}$ does not exist.

(D) Amphoteric oxides are those that exhibit both acidic and basic properties.
$Ga_2O_3$ is a well-known amphoteric oxide.
$As_4O_{10}$ and $Sb_4O_{10}$ are also amphoteric in nature as they show acidic character with strong bases and basic character with strong acids.
$B_2O_3$ is acidic,and $Tl_2O$ is basic.
Therefore,the set of amphoteric oxides is $Ga_2O_3, As_4O_{10}, Sb_4O_{10}$.

(D) $1$. Diborane $(B_2H_6)$ is an electron-deficient species that acts as a Lewis acid. It reacts with carbon monoxide $(CO)$,a Lewis base,to form a coordinate bond adduct:
$B_2H_6 + 2CO \longrightarrow 2BH_3 \cdot CO$ $(A-II)$
$2$. Diborane burns in oxygen to form boron trioxide $(B_2O_3)$:
$B_2H_6 + 3O_2 \longrightarrow B_2O_3 + 3H_2O$ $(B-I)$
$3$. Diborane undergoes hydrolysis with water to produce boric acid $(H_3BO_3)$:
$B_2H_6 + 6H_2O \longrightarrow 2H_3BO_3 + 6H_2$ $(C-III)$
Thus,the correct match is $A-II, B-I, C-III$.

(A) $Al(OH)_3$ acts as a base and reacts with acid $HCl$ to give salt and water:
$Al(OH)_3 + 3HCl \longrightarrow AlCl_3 + 3H_2O$
$Al(OH)_3$ also acts as an acid and reacts with base $NaOH$ to give salt and water:
$Al(OH)_3 + NaOH \longrightarrow Na[Al(OH)_4]$
Since $Al(OH)_3$ reacts with both acids and bases,it is amphoteric in nature.

(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.

(C) In $diamond$,each carbon atom is $sp^3$ hybridized and bonded to $4$ other carbon atoms in a rigid,three-dimensional tetrahedral network. This structure prevents the movement of atoms,making it extremely hard.
In $graphite$,each carbon atom is $sp^2$ hybridized,forming planar hexagonal layers. These layers are held together by weak $van \ der \ Waals$ forces,which allow the layers to slide over each other,making $graphite$ soft.

(A) Carboxyhaemoglobin is a stable complex of carbon monoxide and haemoglobin that forms in red blood cells upon contact with carbon monoxide.
When the concentration of carboxyhaemoglobin in the blood reaches $3-4 \%$,it reduces the oxygen-carrying capacity of the blood.
This condition leads to symptoms such as headache,increased stress on the heart,and cardiovascular problems.
Therefore,statements $(I)$ and $(II)$ 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.

(B) The first ionisation energy is the energy required to remove the most loosely bound electron from an isolated gaseous atom.
Nitrogen $(Z = 7)$ has the electronic configuration $1s^2 2s^2 2p^3$.
Due to the exactly half-filled $2p$-subshell,nitrogen exhibits extra stability,resulting in a higher first ionisation energy compared to its immediate neighbors,carbon $(1086 \ kJ \ mol^{-1})$ and oxygen $(1314 \ kJ \ mol^{-1})$.
Among the given values,$1402 \ kJ \ mol^{-1}$ corresponds to nitrogen,while $1681 \ kJ \ mol^{-1}$ corresponds to fluorine,which has a higher effective nuclear charge.
Thus,the correct value for nitrogen is $1402 \ kJ \ mol^{-1}$.

(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 $(Ca(OH)_2)$.

(B) Amphoteric oxides are those that react with both acids and bases to form salt and water.
$BeO$ (Beryllium oxide),$ZnO$ (Zinc oxide),and $Sb_2O_3$ (Antimony$(III)$ oxide) are amphoteric.
$CO$ is neutral.
$CaO$ is basic.
$SO_2$ and $SO_3$ are acidic.
Therefore,there are $3$ amphoteric oxides.

(A) The reaction of sulphur $(S_8)$ with hot aqueous sodium hydroxide $(NaOH)$ is a disproportionation reaction.
In this reaction,sulphur is both oxidized and reduced to form sodium sulfide $(Na_2S)$ and sodium thiosulphate $(Na_2S_2O_3)$.
The balanced chemical equation is:
$S_8 + 12 NaOH \longrightarrow 4 Na_2S + 2 Na_2S_2O_3 + 6 H_2O$.
Thus,the products $P$ and $Q$ are $Na_2S$ and $Na_2S_2O_3$ respectively.

(B) In the given matching:
$A$. $CH_3-C^{+}(CH_3)-CH_3$ is a tertiary $(3^\circ)$ carbocation because the positive charge is on a carbon bonded to three other carbons. So,$A-IV$.
$B$. $CH_3-C^{+}H-CH_3$ is a secondary $(2^\circ)$ carbocation because the positive charge is on a carbon bonded to two other carbons. So,$B-I$.
$C$. $CH_3-CH_2^+$ is a primary $(1^\circ)$ carbocation because the positive charge is on a carbon bonded to one other carbon. So,$C-III$.
$D$. $CH_3^+$ is a methyl carbocation. So,$D-II$.
The correct sequence is $A-IV, B-I, C-III, D-II$.

(B) The electron gain enthalpy of noble gases is positive due to their stable electronic configuration $(ns^2 np^6)$.
Among the given noble gases,$Ne$ has the highest electron gain enthalpy (most positive value).
As we move down the group,the atomic size increases,and the effective nuclear charge experienced by the incoming electron decreases,making it progressively easier to add an electron compared to the smaller atoms,although the values remain positive.
Therefore,$Ne$ has the highest electron gain enthalpy among the options.

(B) disproportionation reaction is a redox reaction in which the same element in a single reactant is both oxidized and reduced.
$(i) \ \stackrel{0}{Cl}_{2} + 2KI \longrightarrow 2KCl + \stackrel{0}{I}_{2}$: This is a displacement reaction where $Cl$ is reduced and $I$ is oxidized. Not disproportionation.
$(ii) \ \stackrel{0}{Cl}_{2} + 2OH^{-} \longrightarrow \stackrel{+1}{ClO}^{-} + \stackrel{-1}{Cl}^{-} + H_2O$: Here,$Cl$ is oxidized from $0$ to $+1$ and reduced from $0$ to $-1$. This is a disproportionation reaction.
$(iii) \ \stackrel{0}{Mg} + 2\stackrel{+1}{H}Cl \longrightarrow \stackrel{+2}{Mg}Cl_{2} + \stackrel{0}{H}_{2}$: This is a displacement reaction where $Mg$ is oxidized and $H$ is reduced. Not disproportionation.
$(iv) \ 2H_2\stackrel{-1}{O}_{2} \longrightarrow 2H_2\stackrel{-2}{O} + \stackrel{0}{O}_{2}$: Here,$O$ is reduced from $-1$ to $-2$ and oxidized from $-1$ to $0$. This is a disproportionation reaction.
Therefore,reactions $(ii)$ and $(iv)$ are disproportionation reactions.

(B) disproportionation reaction is a redox reaction in which the same element is simultaneously oxidized and reduced,resulting in products with higher and lower oxidation states than the reactant.
In option $A$,$Cl_2$ $(0)$ goes to $ClO^-$ $(+1)$ and $Cl^-$ $(-1)$,which is a disproportionation reaction.
In option $B$,$F_2$ $(0)$ reacts with $OH^-$ to form $OF_2$ and $F^-$. In $OF_2$,the oxidation state of $F$ is $-1$ and in $F^-$,it is also $-1$. The oxidation state of $O$ changes from $-2$ in $OH^-$ to $+2$ in $OF_2$ and remains $-2$ in $H_2O$. Since the oxidation state of fluorine only decreases,this is not a disproportionation reaction.
In option $C$,$ClO_3^-$ $(+5)$ goes to $Cl^-$ $(-1)$ and $ClO_4^-$ $(+7)$,which is a disproportionation reaction.
In option $D$,$ClO^-$ $(+1)$ goes to $Cl^-$ $(-1)$ and $ClO_3^-$ $(+5)$,which is a disproportionation reaction.

(C) $Pb_3O_4$ is a mixed oxide,which can be represented as $2PbO \cdot PbO_2$.
In $PbO$,the oxidation state of $Pb$ is $+2$.
In $PbO_2$,the oxidation state of $Pb$ is $+4$.
Calculation for $PbO$: $x + (-2) = 0 \Rightarrow x = +2$.
Calculation for $PbO_2$: $x + 2(-2) = 0$ $\Rightarrow x - 4 = 0$ $\Rightarrow x = +4$.
Thus,the oxidation states of $Pb$ in $Pb_3O_4$ are $+2$ and $+4$.

(B) The given chemical equation is: $aCaCO_3 + bH_3PO_4 \rightarrow pCa_3(PO_4)_2 + qCO_2 + rH_2O$
To balance the equation,first balance the $Ca$ and $PO_4$ groups: $3CaCO_3 + 2H_3PO_4 \rightarrow 1Ca_3(PO_4)_2 + qCO_2 + rH_2O$
Next,balance the carbon atoms: $3CaCO_3 + 2H_3PO_4 \rightarrow 1Ca_3(PO_4)_2 + 3CO_2 + rH_2O$
Finally,balance the hydrogen and oxygen atoms: $3CaCO_3 + 2H_3PO_4 \rightarrow 1Ca_3(PO_4)_2 + 3CO_2 + 3H_2O$
Comparing this with the given equation,we get $a = 3$ and $b = 2$.
Therefore,the ratio $a / b = 3 / 2$.

(A) $KMnO_4$ oxidises oxalic acid in acidic medium to carbon dioxide. The balanced chemical equation is as follows:
$2 KMnO_4 + 5 H_2 C_2 O_4 + 3 H_2 SO_4 \rightarrow 2 MnSO_4 + 10 CO_2 + 8 H_2 O + K_2 SO_4$
According to the balanced equation,$2 \text{ moles}$ of $KMnO_4$ produce $10 \text{ moles}$ of $CO_2$.
Therefore,the number of $CO_2$ moles produced per mole of $KMnO_4$ is $\frac{10}{2} = 5$.

(A) The electrical conductivity of ions in an aqueous solution depends on their ionic mobility.
Ionic mobility is inversely proportional to the size of the hydrated ion.
Smaller ions have a higher charge density,which causes them to be more extensively hydrated.
Therefore,the order of hydrated ionic size is $Li^{+} > Na^{+} > K^{+} > Cs^{+}$.
Consequently,the order of ionic mobility and electrical conductivity is $Cs^{+} > K^{+} > Na^{+} > Li^{+}$.

(C) When $Na$ dissolves in liquid ammonia $(NH_{3(l)})$,it forms a deep blue coloured solution due to the presence of ammoniated electrons $(e^-(NH_3)_x)$.
The reaction is: $Na + (x+y)NH_3 \longrightarrow [Na(NH_3)_x]^+ + [e(NH_3)_y]^-$.
On standing,this solution reacts to form sodium amide $(NaNH_2)$ and hydrogen gas $(H_2)$:
$Na + NH_3 \longrightarrow NaNH_2 + \frac{1}{2} H_2$.
Thus,the solution is blue,$X$ is $NaNH_2$,and $Y$ is $\frac{1}{2} H_2$.

(A) Beryllium is a member of Group $2$ (alkaline earth metals). When powdered Beryllium is heated in air,it reacts with both oxygen $(O_2)$ and nitrogen $(N_2)$ present in the air.
The reactions are as follows:
$2Be(s) + O_2(g) \rightarrow 2BeO(s)$
$3Be(s) + N_2(g) \rightarrow Be_3N_2(s)$
Thus,the products formed are Beryllium oxide $(BeO)$ and Beryllium nitride $(Be_3N_2)$.

(B) Assertion $(A)$ is true because $LiCl$ and $MgCl_2$ exhibit significant covalent character due to the high polarizing power of small $Li^+$ and $Mg^{2+}$ ions (Fajans' rule),making them soluble in organic solvents like ethanol.
Reason $(R)$ is true because lithium and magnesium are indeed harder than other elements in their respective groups due to their small atomic size and strong metallic bonding.
However,the hardness of the metals is not the reason for the solubility of their chlorides in ethanol; the solubility is due to the covalent nature of the compounds.
Therefore,both statements are true,but the reason is not the correct explanation for the assertion.

(D) $I$. $LiF$ has a very high lattice enthalpy due to the small size of both $Li^+$ and $F^-$ ions,which makes it sparingly soluble in water. This is a correct statement.
$II$. $LiBr$ is soluble in acetone due to its covalent character (Fajans' rule),which makes this statement incorrect as it claims it is not soluble.
$III$. $LiCl$ is soluble in pyridine due to its covalent character,which makes this statement correct.
$IV$. The melting point of alkali metal halides follows the order $MF > MCl > MBr > MI$ because the lattice energy decreases as the size of the halide ion increases. This is a correct statement.
Therefore,statements $I$,$III$,and $IV$ are correct.

(A) The properties described are characteristic of Beryllium $(Be)$.
$1$. $BeSO_4$ is soluble in water.
$2$. $BeO$ is amphoteric in nature (reacts with both acids and bases).
$3$. $Be(OH)_2$ is insoluble in water but dissolves in $NaOH$ solution to form beryllate,$Na_2[Be(OH)_4]$,due to its amphoteric nature.
Therefore,the metal $M$ is $Be$.

(A) The solubility of alkaline earth metal sulphates decreases down the group from $Be$ to $Ba$.
This is because the hydration enthalpy decreases more rapidly than the lattice energy as the size of the cation increases.
For $BeSO_4$,the hydration enthalpy is significantly higher than its lattice energy,making it highly soluble in water.
Therefore,the correct option is $BeSO_4$.

(C) An oxide that reacts with both acids and bases is known as an amphoteric oxide.
$Al_2O_3$ is an amphoteric oxide.
It reacts with hydrochloric acid $(HCl)$ to form aluminum chloride:
$Al_2O_3 + 6HCl \rightarrow 2AlCl_3 + 3H_2O$.
It also reacts with sodium hydroxide $(NaOH)$ to form sodium aluminate:
$Al_2O_3 + 2NaOH \rightarrow 2NaAlO_2 + H_2O$.

(A) The correct matches are as follows:
$A$. $H_3BO_3$ (Boric acid) exhibits intermolecular hydrogen bonding in its solid state. Thus,$A-3$.
$B$. $AlCl_3$ exists as a dimer $(Al_2Cl_6)$ in the vapor phase or non-polar solvents. Thus,$B-1$.
$C$. $B_3N_3H_6$ (Borazine) is known as inorganic benzene due to its structural similarity to benzene. Thus,$C-4$.
$D$. $BF_3$ involves back bonding between the lone pair of fluorine and the vacant p-orbital of boron. Thus,$D-2$.
Therefore,the correct sequence is $A-3, B-1, C-4, D-2$.

(B) The atomic masses of elements are: $Na = 23 \ g/mol$,$S = 32 \ g/mol$,$O = 16 \ g/mol$,$H = 1 \ g/mol$.
The molecular mass of $Na_2SO_4 \cdot 10H_2O = (2 \times 23) + 32 + (4 \times 16) + (10 \times 18) = 46 + 32 + 64 + 180 = 322 \ g/mol$.
Given mass of $Na_2SO_4 \cdot 10H_2O = 32.2 \ g$.
Moles of $Na_2SO_4 \cdot 10H_2O = \frac{32.2 \ g}{322 \ g/mol} = 0.1 \ mol$.
In one mole of $Na_2SO_4 \cdot 10H_2O$,there are $4$ oxygen atoms in $Na_2SO_4$ and $10$ oxygen atoms in $10H_2O$,totaling $14$ oxygen atoms.
Moles of oxygen atoms $= 0.1 \ mol \times 14 = 1.4 \ mol$.
Mass of oxygen $= 1.4 \ mol \times 16 \ g/mol = 22.4 \ g$.

(B) The given sequence of reaction is $A$ $\xrightarrow{k_1} X$ $\xrightarrow{k_2} Y$ $\xrightarrow{k_3} Z$.
In a sequence of consecutive reactions,the slowest step is the rate-determining step.
The rate of a step is directly proportional to its rate constant $k$.
Given $k_3 > k_2 > k_1$,the step with the smallest rate constant $k_1$ is the slowest step.
Therefore,the conversion $A \longrightarrow X$ is the rate-determining step.

(D) The rate law expression is given by $\text{Rate} = k[A]^\alpha[B]^\beta$.
From entry $1$ and $2$,$[A]$ is constant. Taking the ratio:
$\frac{4 \times 10^{-2}}{2 \times 10^{-2}} = \frac{k[0.02]^\alpha[0.04]^\beta}{k[0.02]^\alpha[0.02]^\beta}$
$2 = [2]^\beta \Rightarrow \beta = 1$.
From entry $2$ and $3$,$[B]$ is constant. Taking the ratio:
$\frac{8 \times 10^{-2}}{4 \times 10^{-2}} = \frac{k[0.04]^\alpha[0.04]^\beta}{k[0.02]^\alpha[0.04]^\beta}$
$2 = [2]^\alpha \Rightarrow \alpha = 1$.
Substituting $\alpha = 1$ and $\beta = 1$ into entry $1$:
$2 \times 10^{-2} = k[0.02]^1[0.02]^1$
$k = \frac{2 \times 10^{-2}}{4 \times 10^{-4}} = \frac{200}{4} = 50 \text{ } M^{-1}s^{-1}$.

(A) For a first order reaction,the relationship between the rate constant $k$ and the half-life $t_{1/2}$ is given by the formula:
$k = \frac{0.693}{t_{1/2}}$
Given $t_{1/2} = 1200 \ s$.
Substituting the value:
$k = \frac{0.693}{1200} \ s^{-1}$
$k = 0.0005775 \ s^{-1}$
Rounding to two significant figures,we get:
$k \approx 5.8 \times 10^{-4} \ s^{-1}$

(A) The relation between temperature and activation energy is given by the Arrhenius equation: $\ln \left(\frac{k_2}{k_1}\right) = \frac{E_a}{R} \left(\frac{1}{T_1} - \frac{1}{T_2}\right)$.
Given: $k_2 = 2k_1$,$T_1 = 300 \ K$,$T_2 = 600 \ K$,$R = 8.314 \ J \ K^{-1} \ mol^{-1}$.
Substituting the values: $\log(2) = \frac{E_a}{2.303 \times 8.314} \left(\frac{1}{300} - \frac{1}{600}\right)$.
$0.3010 = \frac{E_a}{19.147} \left(\frac{2-1}{600}\right)$.
$0.3010 = \frac{E_a}{19.147 \times 600}$.
$E_a = 0.3010 \times 19.147 \times 600 \approx 3458 \ J / mol = 3.46 \ kJ / mol$.

(B) Tetrachloroethene $(Cl_2C=CCl_2)$ is widely used as a solvent for dry cleaning.
Liquid carbon dioxide $(CO_2)$ is also used as an environmentally friendly solvent for dry cleaning.
Hydrogen peroxide $(H_2O_2)$ is used in the process of laundry for bleaching clothes to achieve better whiteness.
Acetaldehyde $(CH_3CHO)$ is not used as a dry cleaning agent.
Therefore,$a, b$ and $c$ are the chemicals used in the context of dry cleaning and laundry processes.

(C) For $[Pt(NH_3)_2Cl(NO_2)]$:
Let the oxidation number of $Pt = x$.
$NH_3$ is a neutral ligand $(0)$,$Cl^-$ is $-1$,and $NO_2^-$ is $-1$.
$x + 2(0) + (-1) + (-1) = 0$
$x - 2 = 0 \implies x = +2$.
For $[CoCl_2(en)_2]^{\oplus}$:
Let the oxidation number of $Co = y$.
$en$ (ethylenediamine) is a neutral ligand $(0)$ and $Cl^-$ is $-1$.
$y + 2(-1) + 2(0) = +1$
$y - 2 = +1 \implies y = +3$.
Thus,the oxidation states are $+2$ and $+3$ respectively.

(D) In $Fe_2(CO)_9$,there are $3$ bridging $CO$ ligands and $6$ terminal $CO$ ligands.
In $Co_2(CO)_8$,there are $2$ bridging $CO$ ligands and $6$ terminal $CO$ ligands.
Therefore,the number of bridged $CO$ ligands in $Fe_2(CO)_9$ and $Co_2(CO)_8$ are $3$ and $2$,respectively.

(B) The Wilkinson's catalyst is a well-known organometallic coordination complex of rhodium.
Its chemical formula is $[RhCl(PPh_3)_3]$,where $PPh_3$ represents triphenylphosphine.
It is widely used as a catalyst for the hydrogenation of alkenes.
Therefore,the correct option is $B$.

(C) $[Ni(CN)_4]^{2-}$: $Ni^{2+}$ is $3d^8$. $CN^-$ is a strong field ligand,causing pairing of electrons. Hybridization is $dsp^2$,resulting in a square planar,diamagnetic complex. Matches $III$.
$[Ni(CO)_4]$: $Ni$ is $3d^8 4s^2$. $CO$ is a strong field ligand,causing pairing of electrons. Hybridization is $sp^3$,resulting in a tetrahedral,diamagnetic complex. Matches $II$.
$[NiCl_4]^{2-}$: $Ni^{2+}$ is $3d^8$. $Cl^-$ is a weak field ligand,no pairing occurs. Hybridization is $sp^3$,resulting in a tetrahedral,paramagnetic complex. Matches $I$.
Therefore,the correct match is $A-III, B-II, C-I$.

(C) In $[MnCl_6]^{3-}$,the oxidation state of $Mn$ is $+3$. The electronic configuration of $Mn^{3+}$ is $[Ar] 3d^4$. Since $Cl^-$ is a weak field ligand,no pairing occurs,resulting in $4$ unpaired electrons.
In $[Fe(CN)_6]^{3-}$,the oxidation state of $Fe$ is $+3$. The electronic configuration of $Fe^{3+}$ is $[Ar] 3d^5$. Since $CN^-$ is a strong field ligand,pairing occurs,leaving $1$ unpaired electron.
In $[Co(C_2O_4)_3]^{3-}$,the oxidation state of $Co$ is $+3$. The electronic configuration of $Co^{3+}$ is $[Ar] 3d^6$. Since $C_2O_4^{2-}$ acts as a strong field ligand with $Co^{3+}$,all electrons are paired,resulting in $0$ unpaired electrons.

(C) According to the spectrochemical series,ligands are arranged in increasing order of their crystal field splitting strength.
The order of field strength for the given ligands is $CO > NH_3 > C_2O_4^{2-} > N^{3-}$.
$CO$ is a strong field ligand (pi-acceptor),while $N^{3-}$ is a weak field ligand (pi-donor).

(C) The magnetic moment is given by the formula $\mu = \sqrt{n(n+2)} \ BM$,where $n$ is the number of unpaired electrons.
Given $\mu = 5.92 \ BM$,we have $\sqrt{n(n+2)} = 5.92$.
Solving for $n$,we get $n(n+2) \approx 35$,which implies $n = 5$.
For a high spin octahedral complex with $5$ unpaired electrons,the electrons occupy the $t_{2g}$ and $e_g$ orbitals according to Hund's rule.
The configuration is $t_{2g}^3 e_g^2$.

(A) The reaction of $[Ni(H_2O)_6]^{2+}$ with $en$ proceeds in steps by replacing water ligands:
$1$. For $1:1$ ratio: $[Ni(H_2O)_6]^{2+} + en \rightarrow [Ni(H_2O)_4(en)]^{2+} + 2H_2O$. The product is pale blue $(I)$.
$2$. For $1:2$ ratio: $[Ni(H_2O)_6]^{2+} + 2en \rightarrow [Ni(H_2O)_2(en)_2]^{2+} + 4H_2O$. The product is blue/purple $(II)$.
$3$. For $1:3$ ratio: $[Ni(H_2O)_6]^{2+} + 3en \rightarrow [Ni(en)_3]^{2+} + 6H_2O$. The product is violet $(III)$.
Thus,the correct match is $A-I, B-II, C-III$.

(B) The reaction $4FeCl_3 + 3K_4[Fe(CN)_6] \rightarrow Fe_4[Fe(CN)_6]_3 + 12KCl$ produces Prussian blue,which is $Fe_4[Fe(CN)_6]_3$ $(I)$.
Zinc chloride reacts with excess sodium hydroxide to form sodium zincate: $ZnCl_2 + 4NaOH \rightarrow Na_2ZnO_2 + 2NaCl + 2H_2O$ $(V)$.
Ferric chloride is reduced by hydrogen sulfide to ferrous chloride: $2FeCl_3 + H_2S \rightarrow 2FeCl_2 + 2HCl + S$ $(II)$.
Thus,the correct matching is $A-I, B-V, C-II$.

(B) The atomic number of $Co$ is $27$.
Electronic configuration of $Co$ is $[Ar] 3d^7 4s^2$.
When $Co$ forms $Co^{2+}$,it loses two electrons from the $4s$ orbital.
Electronic configuration of $Co^{2+}$ is $[Ar] 3d^7 4s^0$.
The $3d^7$ configuration has $5$ orbitals,where electrons are filled according to Hund's rule: two orbitals are fully filled (paired) and three orbitals are singly occupied (unpaired).
Therefore,the number of unpaired electrons is $3$.

(C) Copper is a transition metal. Concentrated nitric acid $(HNO_3)$ acts as a strong oxidizing agent. When copper reacts with concentrated $HNO_3$,it gets oxidized to copper$(II)$ nitrate,and the nitric acid is reduced to nitrogen dioxide $(NO_2)$ gas.
The balanced chemical equation for the reaction is:
$Cu(s) + 4HNO_3(conc.) \rightarrow Cu(NO_3)_2(aq) + 2NO_2(g) + 2H_2O(l)$
Thus,the main products formed are copper$(II)$ nitrate $(Cu(NO_3)_2)$ and nitrogen dioxide $(NO_2)$.

(B) The basicity of lanthanide hydroxides decreases as the ionic radius decreases with an increase in atomic number (lanthanide contraction).
Therefore,the basicity follows the order: $La(OH)_3 > Ce(OH)_3 > ... > Lu(OH)_3$.
Comparing $Ce(OH)_3$ and $Ce(OH)_4$,the hydroxide with the lower oxidation state of the metal $(Ce^{3+})$ is more basic than the one with the higher oxidation state $(Ce^{4+})$ because the higher charge density of $Ce^{4+}$ increases the covalent character of the $Ce-OH$ bond.
Thus,$Ce(OH)_3$ is the most basic among the given options.

(C) According to Faraday's law of electrolysis,the mass of a substance deposited is given by $m = \frac{M \times Q}{n \times F}$,where $M$ is the molar mass,$n$ is the valency factor,and $Q$ is the charge passed.
Since $Q$ is constant,the mass deposited is proportional to the equivalent mass,which is $\frac{M}{n}$.
For $Zn^{2+}$ $(ZnSO_4)$: Equivalent mass = $\frac{65.4}{2} = 32.7 \ g$.
For $Fe^{3+}$ $(FeCl_3)$: Equivalent mass = $\frac{56}{3} \approx 18.67 \ g$.
For $Ag^+$ $(AgNO_3)$: Equivalent mass = $\frac{108}{1} = 108 \ g$.
For $Ni^{2+}$ $(NiCl_2)$: Equivalent mass = $\frac{58.7}{2} = 29.35 \ g$.
Comparing the equivalent masses,$Ag$ has the highest value $(108 \ g)$.
Therefore,$Ag$ will be deposited in the maximum amount.

(D) In the electrochemical series,$Cu$ is placed below hydrogen. Hence,it has a positive value of standard reduction potential $(E^{\circ}_{M^{2+}/M})$.
The standard reduction potentials for the given elements are:
$Ti (Z=21) \Rightarrow -1.63 \ V$
$Cr (Z=24) \Rightarrow -0.90 \ V$
$Ni (Z=28) \Rightarrow -0.257 \ V$
$Cu (Z=29) \Rightarrow +0.337 \ V$
Comparing these values,the highest reduction potential is of $Cu$ because it is the only one with a positive value,while $Ni$,$Cr$,and $Ti$ have negative reduction potential values.

(D) The cell reaction is a concentration cell: $Cu(s) | Cu^{2+}(0.1 \ M) || Cu^{2+}(1.0 \ M) | Cu(s)$.
For a concentration cell,the standard cell potential $E^{\circ}_{cell} = E^{\circ}_{cathode} - E^{\circ}_{anode} = 0.34 \ V - 0.34 \ V = 0 \ V$.
The Nernst equation is $E_{cell} = E^{\circ}_{cell} - \frac{0.0591}{n} \log \frac{[Cu^{2+}]_{anode}}{[Cu^{2+}]_{cathode}}$.
Here,$n = 2$,$[Cu^{2+}]_{anode} = 0.1 \ M$,and $[Cu^{2+}]_{cathode} = 1.0 \ M$.
Substituting the values: $E_{cell} = 0 - \frac{0.0591}{2} \log \left( \frac{0.1}{1.0} \right)$.
$E_{cell} = -0.02955 \times \log(10^{-1}) = -0.02955 \times (-1) = 0.02955 \ V \approx 0.029 \ V$.

(C) According to Faraday's second law of electrolysis,the mass of different substances deposited by the same quantity of electricity is proportional to their equivalent masses $(E)$.
$W \propto E$
$W = \frac{E \times Q}{F}$
Since $Q$ and $F$ are constant,$W \propto E = \frac{\text{Atomic mass}}{\text{Valency}}$
Let the valencies of $A, B, C$ be $n_A, n_B, n_C$ respectively.
$W_A : W_B : W_C = \frac{8}{n_A} : \frac{18}{n_B} : \frac{50}{n_C}$
$2.4 : 1.8 : 7.5 = \frac{8}{n_A} : \frac{18}{n_B} : \frac{50}{n_C}$
For $A$: $\frac{8}{n_A} = 2.4 \implies n_A = \frac{8}{2.4} = 3.33$ (This approach assumes proportionality constant $k=1$ for simplicity,let's use ratios).
$n_A : n_B : n_C = \frac{8}{2.4} : \frac{18}{1.8} : \frac{50}{7.5}$
$n_A : n_B : n_C = 3.33 : 10 : 6.66$
Dividing by $3.33$:
$n_A : n_B : n_C = 1 : 3 : 2$
Thus,the valencies are $1, 3$ and $2$.

(B) This question is based on Kohlrausch's law of independent migration of ions.
According to this law,at infinite dilution,the molar conductivity of an electrolyte is the sum of the molar conductivities of its constituent ions.
$\lambda^{\circ}_{MgSO_4} = \lambda^{\circ}_{Mg^{2+}} + \lambda^{\circ}_{SO_4^{2-}}$
Substituting the given values:
$\lambda^{\circ}_{MgSO_4} = 106 \ S \ cm^2 \ mole^{-1} + 160 \ S \ cm^2 \ mole^{-1} = 266 \ S \ cm^2 \ mole^{-1}$

(B) Step $I$: Calculate the total volume of $Cu$ to be deposited.
$V = \text{Area} \times \text{thickness} \times 2 \text{ (for both sides)}$
$V = (1 \times 2 \ cm^2) \times 0.01 \ cm \times 2 = 0.04 \ cm^3$
Step $II$: Calculate the mass of $Cu$ deposited.
$m = \text{Volume} \times \text{Density}$
$m = 0.04 \ cm^3 \times 9 \ g/cm^3 = 0.36 \ g$
Step $III$: Use Faraday's law of electrolysis to find time $t$.
$m = Z \times i \times t$
$0.36 \ g = 0.0003 \ g/C \times 1.5 \ A \times t$
$t = \frac{0.36}{0.0003 \times 1.5} = \frac{0.36}{0.00045} = 800 \ s$

(C) In the determination of Chemical Oxygen Demand $(COD)$,a known volume of the water sample is refluxed with a known excess of potassium dichromate $(K_2Cr_2O_7)$ in the presence of concentrated sulphuric acid $(H_2SO_4)$.
After the reaction,the unreacted excess dichromate is back-titrated against a standard solution of ferrous ammonium sulphate $(FAS)$.
During this titration,$1,10$-phenanthroline ferrous sulphate,commonly known as $Ferroin$,is used as the redox indicator.
The end point is marked by a sharp colour change from blue-green to reddish-brown.

(D) Zwitter ion is a dipolar ion that contains both a positively charged group and a negatively charged group within the same molecule,resulting in a net neutral charge.
In amino acids,the amino group $(-NH_2)$ acts as a base and accepts a proton to become $-NH_3^+$,while the carboxylic acid group $(-COOH)$ acts as an acid and donates a proton to become $-COO^-$.
Thus,the structure $R-CH(NH_3^+)-COO^-$ represents the Zwitter ion form of an amino acid.

(A) $I$. Piperidine is the most basic. The $N$ lone pair is in an $sp^3$ hybrid orbital,there is no resonance,and due to the $+I$ effect of the alkyl group attached,it becomes more basic than $sp^3$ hybridized ammonia.
$II$. In pyridine,the lone pair is in an $sp^2$ hybridized orbital. The $sp^2$ hybrid is smaller than $sp^3$,so there is a stronger attraction to the nucleus,making it less basic.
$III$. $NH_3$ is more basic than aniline because in ammonia,the lone pair is localized and easily available for donation.
$IV$. The basicity of aniline is lower than that of ammonia because in aniline,the lone pair of electrons on $N$ is involved in resonance with the benzene ring.
Therefore,the correct order of basic strength is $I > III > II > IV$.

(C) The reaction proceeds as follows:
$1$. The first step is the anti-Markovnikov addition of $HBr$ to methylenecyclohexane in the presence of a peroxide,$(C_6H_5CO)_2O_2$. This yields (cyclohexylmethyl) bromide,$C_6H_{11}CH_2Br$.
$2$. The second step is a nucleophilic substitution reaction with $KCN$,where the bromide ion is replaced by a cyanide group,yielding cyclohexylacetonitrile,$C_6H_{11}CH_2CN$.
$3$. The third step is the acid-catalyzed hydrolysis of the nitrile group followed by heating,which converts the $-CN$ group into a carboxylic acid group,$-COOH$. The final product is cyclohexylacetic acid,$C_6H_{11}CH_2COOH$.

(B) $PBr_3$ converts ethanol into ethyl bromide. The compound $X$ is $CH_3CH_2Br$.
Alcoholic $KOH$ causes dehydrohalogenation of ethyl bromide to form ethene. The compound $Y$ is $CH_2=CH_2$.
Ethene reacts with cold concentrated $H_2SO_4$ followed by hydrolysis to form ethanol. The compound $Z$ is $CH_3CH_2OH$.
Reaction sequence:
$CH_3CH_2OH$ $\xrightarrow{PBr_3} (X) CH_3CH_2Br$ $\xrightarrow{alc. KOH} (Y) CH_2=CH_2$ $\xrightarrow[{(ii) H_2O, \text{heat}}]{{(i) H_2SO_4, \text{RT}}} (Z) CH_3CH_2OH$
Therefore,the compound $Z$ is $CH_3CH_2OH$.

(A) The rate of nucleophilic substitution reaction depends on the nucleophilicity of the nucleophile $Nu^{\Theta}$.
Stronger nucleophiles react faster.
Comparing the basicity and nucleophilicity:
$CH_3O^{\Theta}$ is a stronger base and nucleophile than $OH^{\Theta}$ due to the $+I$ effect of the $CH_3$ group.
$OH^{\Theta}$ is a stronger nucleophile than $PhO^{\Theta}$ because the negative charge in $PhO^{\Theta}$ is delocalized over the benzene ring.
$PhO^{\Theta}$ is a stronger nucleophile than $CH_3COO^{\Theta}$ because the negative charge in $CH_3COO^{\Theta}$ is delocalized over two oxygen atoms.
Thus,the order of nucleophilicity is $(IV) CH_3O^{\Theta} > (III) OH^{\Theta} > (I) PhO^{\Theta} > (II) CH_3COO^{\Theta}$.

(C) Dehydrohalogenation involves the elimination of a hydrogen atom and a halogen atom from adjacent carbon atoms to form an alkene in the presence of a strong base like alcoholic $KOH$.
For a series of alkyl halides with the same alkyl group,the rate of dehydrohalogenation depends primarily on the strength of the $C-X$ bond and the leaving group ability of the halide ion.
The leaving group ability follows the order: $I^- > Br^- > Cl^- > F^-$.
Since the iodide ion $(I^-)$ is the best leaving group among the given halides,the alkyl iodide $(III)$ will undergo dehydrohalogenation the fastest.
The chloride ion $(Cl^-)$ is the poorest leaving group among the three,so the alkyl chloride $(II)$ will react the slowest.
Therefore,the correct order of the rate of dehydrohalogenation is $(III) > (I) > (II)$.

(A) The rate of $SN^1$ reaction depends on the stability of the carbocation intermediate formed after the departure of the leaving group $(Cl^-)$. The more stable the carbocation,the faster the $SN^1$ reaction.
$(I)$ forms a highly stable carbocation due to resonance stabilization by two $p$-methoxyphenyl groups. The $+R$ effect of the methoxy group significantly stabilizes the positive charge.
$(IV)$ is $tert$-butyl chloride,which forms a stable tertiary carbocation.
$(VI)$ is $2$-chloro-$2$-phenylpropane,which forms a tertiary carbocation stabilized by resonance with the phenyl ring and the $+I$ effect of methyl groups.
$(II)$ is a primary alkyl halide (unstable carbocation).
$(III)$ is a primary benzylic halide with a strong electron-withdrawing $-NO_2$ group,which destabilizes the carbocation.
$(V)$ is chlorobenzene,which does not undergo $SN^1$ reaction easily due to partial double bond character of the $C-Cl$ bond and the instability of the phenyl cation.
Therefore,$(I), (IV)$ and $(VI)$ are the most reactive towards $SN^1$ reaction.

(B) The Swarts reaction is used to prepare alkyl fluorides from alkyl chlorides or alkyl bromides.
This is achieved by heating the alkyl chloride or bromide in the presence of a metallic fluoride,such as $AgF$,$Hg_2F_2$,$CoF_2$,or $SbF_3$.
For example:
$CH_3-Br + AgF \rightarrow CH_3F + AgBr$
Thus,the Swarts reaction yields fluorohydrocarbons.

(B) The addition of $HI$ to propene follows Markownikoff's rule,proceeding via the formation of a $2^{\circ}$ carbocation,which is more stable than a $1^{\circ}$ carbocation. The major product is $2\text{-iodopropane}$ $(III)$.
$(B)$ The reaction of ethylbenzene $(C_6H_5CH_2CH_3)$ with $Br_2$ under heating $(\Delta)$ is a free radical substitution at the benzylic position. The $1\text{-phenylethyl}$ radical is stabilized by resonance,leading to $1\text{-bromo-}1\text{-phenylethane}$ $(IV)$.
$(C)$ The addition of $HBr$ to styrene $(C_6H_5CH=CH_2)$ follows Markownikoff's rule. The electrophilic addition of $H $ forms a stable benzylic carbocation $(C_6H_5CH^ CH_3)$,which then reacts with $Br^-$ to form $1\text{-bromo-}1\text{-phenylethane}$ $(IV)$.
Therefore,the correct match is $A-III, B-IV, C-IV$.

(A) The reaction proceeds as follows:
$(i)$ $CH_3OH$ reacts with $KI$ to form $CH_3I$.
(ii) $CH_3I$ reacts with $Mg$ in dry ether to form the Grignard reagent $CH_3MgI$.
(iii) $CH_3MgI$ undergoes nucleophilic addition to the carbonyl group of $1-$indanone ($2$,$3$-dihydro-1H-inden$-1-$one).
(iv) Subsequent hydrolysis with $H_2O$ yields $1-$methyl$-2,3-$dihydro-1H-inden$-1-$ol.
$(v)$ Dehydration of the alcohol using $20\% H_3PO_4$ at $358 \ K$ results in the formation of the more stable conjugated alkene,$1$-methyl-1H-indene,as the major product.

(C) The reaction sequence is as follows:
$1$. Toluene reacts with $Br_2$ in the presence of $UV$ light to undergo free-radical substitution at the benzylic position,forming benzyl bromide $(C_6H_5CH_2Br)$.
$2$. Benzyl bromide reacts with $Mg$ in dry ether to form the Grignard reagent,benzylmagnesium bromide $(C_6H_5CH_2MgBr)$.
$3$. The Grignard reagent reacts with formaldehyde $(CH_2O)$ followed by acidic hydrolysis $(H_3O^+)$ to form $2-$phenylethanol $(C_6H_5CH_2CH_2OH)$.
$4$. Finally,$2$-phenylethanol reacts with $PBr_3$ to replace the hydroxyl group with a bromine atom,yielding ($2$-bromoethyl)benzene $(C_6H_5CH_2CH_2Br)$.

(B) The reaction proceeds as follows:
$1$. Bromocyclohexane reacts with $Mg$ in dry ether to form cyclohexylmagnesium bromide (a Grignard reagent).
$2$. The Grignard reagent reacts with $CO_2$ followed by acidic workup $(H_3O^+)$ to form cyclohexanecarboxylic acid.
$3$. Cyclohexanecarboxylic acid reacts with $SOCl_2$ to form cyclohexanecarbonyl chloride.
$4$. Finally,the reaction of cyclohexanecarbonyl chloride with dimethylcadmium,$(CH_3)_2Cd$,yields cyclohexyl methyl ketone as the major product. This is a standard method for the preparation of ketones from acid chlorides.

(D) Step $1$: Conversion of toluene to $(P)$. Toluene reacts with $KMnO_4 / KOH$ followed by $H_3O^+$ to form $p$-methylbenzoic acid. This acid then reacts with $PCl_5$ to form $p$-methylbenzoyl chloride,which is $(P)$.
Step $2$: Conversion of bromobenzene to $(Q)$. Bromobenzene reacts with $Mg$ to form phenylmagnesium bromide,which then reacts with $CdCl_2$ to form diphenylcadmium. This organocadmium compound reacts with $(P)$ ($p$-methylbenzoyl chloride) to form the final product $(Q)$,which is $4$-methylbenzophenone.

(A) $1$. Toluene reacts with $Br_2$ in the presence of $Fe$ (dark) via electrophilic aromatic substitution to form a mixture of $o$-bromotoluene $(P)$ and $p$-bromotoluene $(Q)$.
$2$. These isomers react with $Mg$ in dry ether to form the corresponding Grignard reagents: $o$-tolylmagnesium bromide and $p$-tolylmagnesium bromide.
$3$. These Grignard reagents react with formaldehyde $(HCHO)$ followed by acidic hydrolysis $(H_2O)$ to form primary alcohols.
$4$. The final products $R$ and $S$ are $o$-methylbenzyl alcohol and $p$-methylbenzyl alcohol,respectively.

(B) Chlorobenzene undergoes electrophilic aromatic substitution (nitration) with conc. $HNO_3$ and conc. $H_2SO_4$ to form $p$-nitrochlorobenzene as the major product $(P)$ and $o$-nitrochlorobenzene as the minor product $(Q)$.
In the next step,$p$-nitrochlorobenzene $(P)$ reacts with $NaOH$ at $443 \ K$ followed by acidification $(H^+)$. The presence of the electron-withdrawing $-NO_2$ group at the para position activates the ring towards nucleophilic aromatic substitution,replacing the $-Cl$ atom with an $-OH$ group to form $p$-nitrophenol $(R)$.

(D) The reaction sequence is as follows:
$1$. Allylic bromination of cyclohexene with $Br_2$ in the presence of $h\nu$ gives $3$-bromocyclohexene.
$2$. Formation of Grignard reagent: $3$-bromocyclohexene reacts with $Mg$ in dry ether to form $3$-cyclohexenylmagnesium bromide.
$3$. Reaction with $CdCl_2$: The Grignard reagent reacts with $CdCl_2$ to form an organocadmium compound,$(C_6H_9)_2Cd$.
$4$. Reaction with $CH_3COCl$: The organocadmium compound reacts with acetyl chloride $(CH_3COCl)$ to form a ketone,$1$-(cyclohex$-2-$en$-1-$yl)ethanone.
Thus,the final product is $1$-(cyclohex$-2-$en$-1-$yl)ethanone.

(B) Mustard gas,also known as sulphur mustard,is a chemical compound belonging to the family of cytotoxic and blister agent chemical warfare agents.
The chemical formula of mustard gas is $C_4H_8Cl_2S$,and its structural formula is $ClCH_2CH_2SCH_2CH_2Cl$.
It can be synthesized by the reaction of ethylene with sulphur dichloride:
$2C_2H_4 + SCl_2 \rightarrow (ClCH_2CH_2)_2S$

(B) Step $(i)$: Benzene reacts with $Br_2$ in the presence of anhydrous $FeBr_3$ to form bromobenzene via electrophilic aromatic substitution.
Step (ii): Bromobenzene undergoes Friedel-Crafts alkylation with $CH_3Cl$ and anhydrous $AlCl_3$. Since the $-Br$ group is ortho/para directing,the major product is $p$-bromotoluene.
Step (iii): $p$-Bromotoluene reacts with $Mg$ in dry ether to form the Grignard reagent,$p$-tolylmagnesium bromide $(CH_3-C_6H_4-MgBr)$.
Step (iv) and $(v)$: The Grignard reagent reacts with acetaldehyde $(CH_3CHO)$ followed by hydrolysis $(H_2O)$ to form a secondary alcohol. The nucleophilic carbon of the Grignard reagent attacks the carbonyl carbon of acetaldehyde,resulting in $1-(4-methylphenyl)ethanol$.

(D) Step $1$: Phenol reacts with zinc dust $(Zn)$ under heating $(\Delta)$ to undergo reduction,resulting in the formation of benzene.
$C_6H_5OH Zn \xrightarrow{\Delta} C_6H_6 ZnO$
Step $2$: Benzene then undergoes electrophilic aromatic substitution (chlorination) in the presence of anhydrous ferric chloride $(FeCl_3)$ and chlorine $(Cl_2)$ to form chlorobenzene.
$C_6H_6 Cl_2 \xrightarrow{\text{anhyd. } FeCl_3} C_6H_5Cl HCl$
Therefore,the major product is chlorobenzene.

(A) The reaction proceeds as follows:
$1$. Hydroboration-oxidation of methylenecyclopentane with $(i) B_2H_6$ and $(ii) H_2O_2/NaOH$ yields cyclopentylmethanol.
$2$. Oxidation of the primary alcohol with $PCC$ (Pyridinium chlorochromate) yields cyclopentanecarbaldehyde.
$3$. Nucleophilic addition of the Grignard reagent $PhMgBr$ to the aldehyde,followed by hydrolysis with $(v) H_2O$,yields the secondary alcohol,cyclopentylphenylmethanol.

(A) The reaction proceeds as follows:
$(i)$ Phenol reacts with $NaOH$ and $CH_3I$ to form anisole (methoxybenzene) via Williamson ether synthesis.
(ii) Anisole undergoes Friedel-Crafts acylation with propionyl chloride $(CH_3CH_2COCl)$ in the presence of anhydrous $AlCl_3$. Since the $-OCH_3$ group is ortho/para-directing,and the para-position is sterically less hindered,the major product is $p$-methoxypropiophenone.
(iii) Reduction of the ketone group with $NaBH_4$ yields the corresponding secondary alcohol,$1$-($4$-methoxyphenyl)propan$-1-$ol.
Therefore,the correct option is $A$.

(C) The reaction proceeds as follows:
$(i)$ $Sn/HCl$ reduces the $-NO_2$ group to an $-NH_2$ group,forming $p$-toluidine.
(ii) $Br_2$ $(1 \ eq.)$ in the presence of the strongly activating $-NH_2$ group leads to ortho-bromination,yielding $2$-bromo-$4$-methylaniline.
(iii) $NaNO_2/HCl$ at $273-278 \ K$ converts the $-NH_2$ group into a diazonium salt,$-N_2^+Cl^-$.
(iv) $Cu_2Br_2/HBr$ (Sandmeyer reaction) replaces the diazonium group with a $-Br$ atom,resulting in $1,2$-dibromo-$4$-methylbenzene (also known as $3,4$-dibromotoluene).

(C) The basic strength of an anion is inversely proportional to the acidic strength of its conjugate acid.
First,identify the conjugate acids of the given bases:
$CH_{3}^{-}$'s conjugate acid is $CH_{4}$ $(pKa \approx 50)$.
$NH_{2}^{-}$'s conjugate acid is $NH_{3}$ $(pKa \approx 38)$.
$CN^{-}$'s conjugate acid is $HCN$ $(pKa \approx 9.2)$.
$HS^{-}$'s conjugate acid is $H_{2}S$ $(pKa \approx 7)$.
Since the acidic strength order is $CH_{4} < NH_{3} < HCN < H_{2}S$,the basic strength order of their conjugate bases is the reverse: $CH_{3}^{-} > NH_{2}^{-} > CN^{-} > HS^{-}$.

(A) The reducing character of the hydrides of group $15$ elements depends upon the thermal stability of the hydrides.
As we move down the group,the bond dissociation energy decreases due to an increase in the size of the central atom,which leads to a decrease in the stability of the hydrides.
Consequently,the ability to release hydrogen atoms increases,making the reducing character increase down the group.
Therefore,the correct order of reducing ability is: $BiH_3 > SbH_3 > AsH_3 > PH_3 > NH_3$.
Among the given options,the correct order is $BiH_3 > SbH_3 > PH_3 > NH_3$.

(B) Aluminium reacts with nitrogen at high temperatures $(1073 \ K - 1473 \ K)$ to form aluminium nitride $(AlN)$:
$2 \ Al + N_2 \rightarrow 2 \ AlN$
Aluminium nitride $(AlN)$ reacts with water to produce aluminium hydroxide $(Al(OH)_3)$ as a precipitate and ammonia gas $(NH_3)$:
$AlN + 3 \ H_2O \rightarrow Al(OH)_3 (ppt) + NH_3 (g)$
Therefore,the compound $C$ is ammonia $(NH_3)$.

(A) $(A)\ NO + NO_2 \xrightarrow{< 250 K} N_2O_3$. $N_2O_3$ is blue in liquid or solid state $(P = \text{Blue})$.
$(B)\ 2 NO_2 \longleftrightarrow N_2O_4$. $N_2O_4$ is a colourless gas/liquid $(Q = \text{Colourless})$.
$(C)\ 2 Pb(NO_3)_2 \xrightarrow{673 K} 4 NO_2 + 2 PbO + O_2$. $NO_2$ is a reddish-brown gas $(R = \text{Brown})$.
$(D)\ 4 HNO_3 + P_4O_{10} \longrightarrow 2 N_2O_5 + 4 HPO_3$. $N_2O_5$ is a colourless crystalline solid $(S = \text{Colourless})$.
Thus,the correct sequence is $P$: Blue,$Q$: Colourless,$R$: Brown,$S$: Colourless.