IIT JEE 2019 Chemistry Question Paper with Answer and Solution

(A) $U_{rms} = \sqrt{\frac{3RT}{M}}$. Since $U_{rms} \propto \sqrt{T}$,if $T$ becomes $4T$,$U_{rms}$ becomes $\sqrt{4} = 2$ times. Thus,statement $(1)$ is correct.
$\varepsilon_{av} = \frac{3}{2}kT$. This expression depends only on temperature $T$ and not on molecular mass $M$. Thus,statement $(2)$ is correct.
From $U_{rms} = \sqrt{\frac{3RT}{M}}$,$U_{rms} \propto \frac{1}{\sqrt{M}}$. Thus,statement $(3)$ is correct.
Since $\varepsilon_{av} = \frac{3}{2}kT$,$\varepsilon_{av} \propto T$. If $T$ is increased four times,$\varepsilon_{av}$ increases four times,not doubled. Thus,statement $(4)$ is incorrect.
Therefore,statements $(1), (2),$ and $(3)$ are correct.

(C) $Q$ is $SnCl_2$.
$1. SnCl_2 + Cl^{-} \rightarrow [SnCl_3]^{-} (X)$. The $Sn$ atom in $[SnCl_3]^{-}$ has $3$ bond pairs and $1$ lone pair,so it is $sp^3$ hybridized with pyramidal geometry. Statement $(1)$ is correct.
$2. SnCl_2 + Me_3N \rightarrow SnCl_2 \cdot NMe_3 (Y)$. This is an adduct where $N$ donates a lone pair to $Sn$,forming a coordinate bond. Statement $(4)$ is correct.
$3. SnCl_2 + 2CuCl_2 \rightarrow SnCl_4 (Z) + 2CuCl$. The oxidation state of $Sn$ in $SnCl_4$ is $+4$. Statement $(2)$ is incorrect.
$4. SnCl_4$ has $4$ bond pairs and $0$ lone pairs. Statement $(3)$ is incorrect.
Therefore,statements $(1)$ and $(4)$ are correct.

(D) molecule possesses a permanent dipole moment if it is polar,meaning its net dipole moment $\mu \neq 0$.
In set $(1)$,$BeCl_2$ (linear),$CO_2$ (linear),and $BCl_3$ (trigonal planar) are symmetrical and have $\mu = 0$.
In set $(2)$,$SO_2$ (bent),$C_6H_5Cl$ (polar),$H_2Se$ (bent),and $BrF_5$ (square pyramidal) are all polar molecules with $\mu \neq 0$.
In set $(3)$,$BF_3$ (trigonal planar) and $SF_6$ (octahedral) are symmetrical and have $\mu = 0$.
In set $(4)$,$NO_2$ (bent),$NH_3$ (pyramidal),$POCl_3$ (distorted tetrahedral),and $CH_3Cl$ (distorted tetrahedral) are all polar molecules with $\mu \neq 0$.
Thus,sets $(2)$ and $(4)$ contain molecules that all possess a permanent dipole moment.

(A) The standard enthalpy of formation $(\Delta_f H^\circ)$ is defined as the enthalpy change when $1 \text{ mole}$ of a compound is formed from its constituent elements in their most stable standard states.
$(1)$ $\frac{3}{2} O_{2(g)} \rightarrow O_{3(g)}$: $O_2$ is the most stable form of oxygen. $1 \text{ mole}$ of $O_3$ is formed. This represents $\Delta_f H^\circ$.
$(2)$ $\frac{1}{8} S_{8(s)} + O_{2(g)} \rightarrow SO_{2(g)}$: $S_8$ is the most stable form of sulfur and $O_2$ is the most stable form of oxygen. $1 \text{ mole}$ of $SO_2$ is formed. This represents $\Delta_f H^\circ$.
$(3)$ $2 H_{2(g)} + O_{2(g)} \rightarrow 2 H_2O_{(l)}$: Here,$2 \text{ moles}$ of $H_2O$ are formed,so this is $2 \times \Delta_f H^\circ$,not $\Delta_f H^\circ$.
$(4)$ $2 C_{(g)} + 3 H_{2(g)} \rightarrow C_2H_{6(g)}$: Carbon is not in its most stable state (graphite is the standard state). Thus,this is not $\Delta_f H^\circ$.
Therefore,reactions $(1)$ and $(2)$ satisfy the condition.

(C) The reaction is $Fe^{2+}_{(aq)} + S^{2-}_{(aq)} \rightleftharpoons FeS_{(s)}$.
Since equal volumes are mixed,the new concentrations are $0.03 \ M \ Fe^{2+}$ and $0.1 \ M \ S^{2-}$.
Given $K_{c} = 1.6 \times 10^{17}$,which is very large,the reaction proceeds almost to completion.
Let $y$ be the equilibrium concentration of $Fe^{2+}_{(aq)}$.
Initial concentrations: $[Fe^{2+}] = 0.03 \ M$,$[S^{2-}] = 0.1 \ M$.
At equilibrium: $[Fe^{2+}] = y$,$[S^{2-}] = 0.1 - 0.03 = 0.07 \ M$.
$K_{c} = \frac{1}{[Fe^{2+}][S^{2-}]} = \frac{1}{y \times 0.07} = 1.6 \times 10^{17}$.
$y = \frac{1}{1.6 \times 0.07} \times 10^{-17} = \frac{1}{0.112} \times 10^{-17} \approx 8.928 \times 10^{-17} \ M$.
Thus,$Y \approx 8.93$.

(B) Let us analyze the structures of the given molecules to identify those with a covalent bond between two atoms of the same kind:
$1$. $B_2H_6$ (Diborane): Contains $B-H-B$ bridge bonds. There is no $B-B$ bond.
$2$. $B_3N_3H_6$ (Borazine): Contains $B-N$ bonds. There is no $B-B$ or $N-N$ bond.
$3$. $N_2O$: The structure is $N \equiv N \rightarrow O$. It contains an $N-N$ bond.
$4$. $N_2O_4$: The structure is $O_2N-NO_2$. It contains an $N-N$ bond.
$5$. $H_2S_2O_3$ (Thiosulfuric acid): The structure contains an $S-S$ bond.
$6$. $H_2S_2O_8$ (Peroxodisulfuric acid): The structure contains an $O-O$ (peroxide) bond.
Thus,the molecules containing a covalent bond between two atoms of the same kind are $N_2O, N_2O_4, H_2S_2O_3$,and $H_2S_2O_8$.
The total number of such molecules is $4$.

(C) Let us analyze each reaction:
$(1)$ Kolbe's electrolysis of sodium butanoate $(CH_3CH_2CH_2COONa)$ produces $n$-hexane $(CH_3CH_2CH_2CH_2CH_2CH_3)$.
$(2)$ Decarboxylation of sodium butanoate $(CH_3CH_2CH_2COONa)$ with soda lime $(NaOH + CaO)$ produces propane $(CH_3CH_2CH_3)$.
$(3)$ Reduction of $1$-chloropropane $(CH_3CH_2CH_2Cl)$ with $Zn$ and dilute $HCl$ produces propane $(CH_3CH_2CH_3)$.
$(4)$ Dehalogenation of $1,2$-dibromopropane with $Zn$ produces propene $(CH_3CH=CH_2)$.
Thus,reactions $(2)$ and $(3)$ produce propane as the major product.

(A) For a hydrogen-like species,the energy is given by $E_n = -13.6 \times \frac{Z^2}{n^2} \ eV$.
For $He^+$,$Z = 2$. Given $E = -3.4 \ eV$,we have $-3.4 = -13.6 \times \frac{2^2}{n^2}$,which simplifies to $n^2 = 16$,so $n = 4$.
Given azimuthal quantum number $\ell = 2$,the state is $4d$.
Number of angular nodes $= \ell = 2$.
Number of radial nodes $= n - \ell - 1 = 4 - 2 - 1 = 1$.
Since $He^+$ is a single-electron species,the nuclear charge experienced is the full nuclear charge $Z = 2e$,so statement $(4)$ is incorrect.
Thus,statements $(1)$ and $(3)$ are true.

(C) Analysis of the given reactions:
$(1)$ Benzene reacts with excess $Cl_2$ in the presence of $UV$ light at $500 \ K$ to form benzene hexachloride,which is non-aromatic.
$(2)$ $1,2$-dibromopropane reacts with $alc. KOH$ followed by $NaNH_2$ to form propyne $(CH_3-C \equiv CH)$. Cyclotrimerization of propyne using a red hot iron tube at $873 \ K$ yields $1,3,5$-trimethylbenzene (mesitylene),which is aromatic.
$(3)$ $3$-bromocyclobutene reacts with $NaOEt$ to give a substitution product and an elimination product (cyclobutadiene),which dimerizes to a non-aromatic compound.
$(4)$ Cyclopentadiene reacts with $NaOMe$ to form the cyclopentadienyl anion,which is aromatic ($6 \pi$ electrons,Huckel's rule).
Therefore,reactions $(2)$ and $(4)$ yield aromatic products.

(D) The balanced reactions are:
$Zn + 2H_2SO_4 \text{ (hot conc.)} \rightarrow ZnSO_4 (G) + SO_2 (R) + 2H_2O (X)$
$Zn + 2NaOH \text{ (conc.)} \rightarrow Na_2ZnO_2 (T) + H_2 (Q)$
$ZnSO_4 (G) + H_2S + 2NH_4OH \rightarrow ZnS (Z) + 2H_2O (X) + (NH_4)_2SO_4 (Y)$
Analysis of statements:
$(1)$ In $Na_2ZnO_2$ $(T)$,the oxidation state of $Zn$ is $+2$,not $+1$. So,statement $(1)$ is incorrect.
$(2)$ $Q$ is $H_2$. The bond order of $H_2$ is $1$. So,statement $(2)$ is correct.
$(3)$ $Z$ is $ZnS$,which is white in colour. So,statement $(3)$ is correct.
$(4)$ $R$ is $SO_2$. $SO_2$ has a bent or $V$-shaped geometry. So,statement $(4)$ is correct.
Thus,statements $(2, 3, 4)$ are correct.

(C) The molecular formula $C_4H_8O$ corresponds to a degree of unsaturation of $1$. For cyclic ethers,we consider the following structures:
$1$. Tetrahydrofuran $(THF)$: $1$ isomer.
$2$. $2$-methyl-oxetane: Has a chiral center,so it exists as a pair of enantiomers ($R$ and $S$). Total = $2$ isomers.
$3$. $3$-methyl-oxetane: Achiral. Total = $1$ isomer.
$4$. $2$-ethyl-oxirane: Has a chiral center,so it exists as a pair of enantiomers ($R$ and $S$). Total = $2$ isomers.
$5$. $2,3$-dimethyl-oxirane: Has two chiral centers. It exists as $(R,R)$,$(S,S)$ (enantiomeric pair) and $(R,S)$ (meso compound). Total = $3$ isomers.
$6$. $1,1$-dimethyl-oxirane: Achiral. Total = $1$ isomer.
Summing these up: $1 + 2 + 1 + 2 + 3 + 1 = 10$ isomers.

(C) The oxidation of rhombic sulphur $(S_8)$ by concentrated $HNO_3$ produces sulphuric acid $(H_2SO_4)$ as the product with the highest oxidation state of sulphur $(+6)$.
The balanced chemical equation is:
$S_8 + 48 \ HNO_3 \longrightarrow 8 \ H_2SO_4 + 48 \ NO_2 + 16 \ H_2O$
From the stoichiometry of the reaction,$1$ mole of $S_8$ produces $16$ moles of $H_2O$.
Since $1$ mole of rhombic sulphur is $S_8$,we must consider the reaction for $1$ mole of $S_8$ atoms (as implied by the question context of $1$ mole of sulphur atoms,or if $1$ mole of $S_8$ is meant,the stoichiometry adjusts accordingly. Given the standard interpretation of $1$ mole of $S$ atoms):
For $1$ mole of $S$ atoms,the equation is: $S + 6 \ HNO_3 \longrightarrow H_2SO_4 + 6 \ NO_2 + 2 \ H_2O$.
Thus,$1$ mole of $S$ produces $2$ moles of $H_2O$.
Mass of $H_2O = 2 \ mol \times 18 \ g \ mol^{-1} = 36 \ g$.
However,if the question refers to $1$ mole of $S_8$ molecules,then $16$ moles of $H_2O$ are produced.
Mass of $H_2O = 16 \ mol \times 18 \ g \ mol^{-1} = 288 \ g$.

(A) $1$. The starting material contains one alkyne group and two alkene groups (one in a cyclopentene ring and one in a cyclohexene ring).
$2$. The first step uses $H_2, Pd-BaSO_4$,and quinoline (Lindlar's catalyst),which selectively reduces the alkyne to a $cis$-alkene without affecting the other alkene groups.
$3$. The second step uses dilute $KMnO_4$ at $273 \ K$ (Baeyer's reagent),which performs syn-dihydroxylation of all double bonds present in the molecule.
$4$. The molecule has three double bonds: one newly formed $cis$-alkene from the alkyne and the two original cyclic alkenes.
$5$. Each double bond undergoes dihydroxylation,adding two $-OH$ groups per double bond.
$6$. Total $-OH$ groups $= 3 \times 2 = 6$.

(A) According to Bohr's model for a one-electron atom:
$1.$ Radius of $n^{\text{th}}$ orbit,$r_n = \frac{n^2 h^2 \epsilon_0}{\pi m Z e^2} \propto n^2$. Thus,$(I)$ matches with $(T)$.
$2.$ Angular momentum,$L = \frac{nh}{2\pi} \propto n^1$. Thus,$(II)$ matches with $(S)$.
$3.$ Kinetic energy,$KE = \frac{m Z^2 e^4}{8 \epsilon_0^2 n^2 h^2} \propto n^{-2}$. Thus,$(III)$ matches with $(P)$.
$4.$ Potential energy,$PE = - \frac{m Z^2 e^4}{4 \epsilon_0^2 n^2 h^2} \propto n^{-2}$. Thus,$(IV)$ matches with $(P)$.
For question $(1)$,the correct combination is $(I), (T)$,which is option $(3)$.
For question $(2)$,the correct combination is $(III), (P)$,which is option $(4)$.
Therefore,the answer is $3, 4$.

(A) Sodium stearate is a surfactant that behaves as a strong electrolyte at low concentrations.
As the concentration increases,the molar conductivity $\left(\Lambda_m\right)$ decreases due to the interionic attractions,similar to other strong electrolytes.
However,once the concentration reaches the Critical Micelle Concentration $(CMC)$,the individual ions aggregate to form large,bulky micelles.
These micelles have a much lower mobility compared to individual ions,leading to a sharp decrease in the rate of change of molar conductivity with respect to $\sqrt{c}$ after the $CMC$ point.
This results in a plot where the slope changes significantly at the $CMC$,which is best represented by the graph in option $A$.

(A) Acid strength is inversely proportional to the $pK_a$ value.
Lower $pK_a$ indicates higher acid strength.
The given compounds are:
$(I)$ Propiolic acid $(pK_a = 1.86)$
$(II)$ Acrylic acid $(pK_a = 4.3)$
$(III)$ $p$-Methoxybenzoic acid $(pK_a = 4.5)$
$(IV)$ Propanoic acid $(pK_a = 4.88)$
Comparing the $pK_a$ values: $1.86 < 4.3 < 4.5 < 4.88$.
Therefore,the order of acid strength is $I > II > III > IV$.

(B) The chemical formulas for the given ores are as follows:

Ore	Formula
Calamine	$ZnCO_3$
Malachite	$CuCO_3 \cdot Cu(OH)_2$
Magnetite	$Fe_3O_4$
Cryolite	$Na_3AlF_6$

Therefore,the correct option is $B$.

(A) In the borax bead test,the chromium $(III)$ salt decomposes upon heating to form chromium $(III)$ oxide $(Cr_2O_3)$.
Borax $(Na_2B_4O_7 \cdot 10H_2O)$ on heating loses water and forms sodium metaborate $(NaBO_2)$ and boric anhydride $(B_2O_3)$.
The reaction is: $Cr_2O_3 + 3B_2O_3 \longrightarrow 2Cr(BO_2)_3$.
The green colour is due to the formation of chromium metaborate,$Cr(BO_2)_3$.

(B) $MnO_2 + 2 KOH + \frac{1}{2} O_2 \xrightarrow{\Delta} K_2MnO_4 + H_2O$ $(W)$
$W = K_2MnO_4 \rightleftharpoons 2K^{\oplus} + MnO_4^{2-} (Y)$
$K_2MnO_4 + H_2O \xrightarrow{\text{Electrolytic}} H_2 + KOH + KMnO_4 (X)$
Anion of $X = MnO_4^{-} (Z)$
$1$. $Y$ $(MnO_4^{2-})$ is $d^1$ (paramagnetic) and $Z$ $(MnO_4^{-})$ is $d^0$ (diamagnetic). Statement $(1)$ is incorrect.
$2$. Both $MnO_4^{2-}$ and $MnO_4^{-}$ are coloured due to charge transfer and have tetrahedral geometry. Statement $(2)$ is correct.
$3$. In both,$\pi$-bonding occurs between $p$-orbitals of oxygen and $d$-orbitals of manganese ($p\pi-d\pi$ bonding). Statement $(3)$ is correct.
$4$. In acidic solution,$MnO_4^{2-}$ undergoes disproportionation: $3MnO_4^{2-} + 4H^{\oplus} \longrightarrow 2MnO_4^{-} + MnO_2 + 2H_2O$. Statement $(4)$ is correct.
Thus,statements $(2, 3, 4)$ are correct.

(A) Analyze the decay sequence:
$1$. $^{238}_{92}U \xrightarrow{- x_1} ^{234}_{90}Th$: Mass number decreases by $4$ and atomic number by $2$,so $x_1$ is an $\alpha$-particle $(^{4}_{2}He^{2+})$. $\alpha$-particles are positively charged and deflect towards the negatively charged plate. Statement $(3)$ is correct.
$2$. $^{234}_{90}Th \xrightarrow{- x_2} ^{234}_{91}Pa$: Atomic number increases by $1$,so $x_2$ is a $\beta^{-}$-particle $(^{0}_{-1}e)$. Statement $(2)$ is correct.
$3$. $^{234}_{91}Pa \xrightarrow{- x_3} ^{234}_{92}Z$: Atomic number increases by $1$,so $x_3$ is a $\beta^{-}$-particle. Statement $(4)$ is incorrect.
$4$. $^{234}_{92}Z$ has atomic number $92$,same as $U$,so $Z$ is an isotope of uranium. Statement $(1)$ is correct.
Thus,statements $(1)$,$(2)$,and $(3)$ are correct.

(B) $(1)$ $FALSE$: Oxidation of glucose with bromine water ($Br_2$ water) gives gluconic acid,not glutamic acid.
$(2)$ $TRUE$: The two six-membered cyclic hemiacetal forms of $D-( )-$glucose ($\alpha-D-$glucose and $\beta-D-$glucose) differ in configuration at the $C_1$ carbon and are called anomers.
$(3)$ $TRUE$: Hydrolysis of sucrose yields $D-( )-$glucose (dextrorotatory) and $D-(-)-$fructose (laevorotatory).
$(4)$ $TRUE$: Monosaccharides are the simplest carbohydrates and cannot be further hydrolysed into smaller polyhydroxy aldehyde or ketone units.

(A) $1$. The starting material $C_6H_{10}O$ is cyclohexanone. Reaction with $CH_3MgBr$ followed by $H_2O$ gives $1$-methylcyclohexanol $(Q)$.
$2$. $Q$ reacts with $conc. HCl$ to form $1$-chloro-$1$-methylcyclohexane $(S)$. This matches structure $(2)$.
$3$. $Q$ undergoes acid-catalyzed dehydration with $20\% H_3PO_4$ at $360 \ K$ to form $1$-methylcyclohexene $(R)$.
$4$. $R$ reacts with $HBr$ in the presence of benzoyl peroxide (anti-Markovnikov addition) to form $1$-bromo-$2$-methylcyclohexane $(U)$. This matches structure $(4)$.
$5$. $R$ undergoes hydrogenation $(H_2, Ni)$ to form methylcyclohexane,followed by radical bromination $(Br_2, h\nu)$ to form $1$-bromo-$1$-methylcyclohexane $(T)$.
$6$. Comparing the structures: $S$ is $(2)$,$U$ is $(4)$,and $T$ is $(1)$. The correct set is $(2, 4)$.

(A) The rate law is given by $r = K [A]^{n_1} [B]^{n_2} [C]^{n_3}$.
Comparing experiments $1$ and $2$,$[A]$ and $[C]$ are constant,but $[B]$ changes and the rate remains constant,so $n_2 = 0$.
Comparing experiments $1$ and $3$,$[A]$ and $[B]$ are constant,$[C]$ doubles,and the rate doubles,so $n_3 = 1$.
Comparing experiments $1$ and $4$,$[B]$ and $[C]$ are constant,$[A]$ increases by $1.5$ times,and the rate increases by $1.5$ times,so $n_1 = 1$.
Thus,the rate law is $r = K [A] [C]$.
Using experiment $1$: $6.0 \times 10^{-5} = K \times 0.2 \times 0.1$,which gives $K = 3.0 \times 10^{-3} \ L \ mol^{-1} \ s^{-1}$.
Now,for $[A] = 0.15 \ mol \ dm^{-3}, [B] = 0.25 \ mol \ dm^{-3}, [C] = 0.15 \ mol \ dm^{-3}$:
$r = (3.0 \times 10^{-3}) \times 0.15 \times 0.15 = 6.75 \times 10^{-5} \ mol \ dm^{-3} \ s^{-1}$.
Comparing this with $Y \times 10^{-5}$,we get $Y = 6.75$.

(A) According to Raoult's law for relative lowering of vapor pressure:
$\frac{P^{\circ} - P_s}{P^{\circ}} = \frac{n_2}{n_1 + n_2} \approx \frac{n_2}{n_1}$
Given: $P^{\circ} = 650 \ mm \ Hg$,$P_s = 640 \ mm \ Hg$,$W_2 = 0.5 \ g$,$W_1 = 39 \ g$,$M_1 = 78 \ g \ mol^{-1}$,$K_f = 5.12 \ K \ kg \ mol^{-1}$.
$\frac{650 - 640}{650} = \frac{0.5 / M_2}{39 / 78} = \frac{0.5 / M_2}{0.5} = \frac{1}{M_2}$
$\frac{10}{650} = \frac{1}{M_2} \implies M_2 = 65 \ g \ mol^{-1}$.
Now,calculate molality $(m)$:
$m = \frac{W_2 \times 1000}{M_2 \times W_1} = \frac{0.5 \times 1000}{65 \times 39} \approx 0.197 \ mol \ kg^{-1}$.
Depression in freezing point $\Delta T_f = K_f \times m = 5.12 \times 0.197 \approx 1.01 \ K$.

(C) Step $1$: Conversion of $P$ (aniline) to $Q$ ($2$,$4$,$6$-tribromobenzoyl chloride).
Aniline reacts with excess $Br_2/H_2O$ to form $2,4,6$-tribromoaniline. Diazotization followed by reaction with $CuCN/KCN$ gives $2,4,6$-tribromobenzonitrile. Hydrolysis gives $2,4,6$-tribromobenzoic acid,which reacts with $SOCl_2$ to form $Q$ ($2,4,6$-tribromobenzoyl chloride).
Step $2$: Conversion of $R$ (benzene) to $S$ ($4$-bromophenol).
Benzene reacts with oleum to form benzenesulfonic acid,which on fusion with $NaOH$ followed by acidification gives phenol. Bromination of phenol with $Br_2/CS_2$ at $273 \ K$ gives $4$-bromophenol $(S)$ as the major product.
Step $3$: Synthesis of $T$ from $Q$ and $S$.
$S$ ($4$-bromophenol) reacts with $NaOH$ to form sodium $4$-bromophenoxide,which acts as a nucleophile and attacks the carbonyl carbon of $Q$ ($2,4,6$-tribromobenzoyl chloride) to form an ester $T$ ($4$-bromophenyl $2,4,6$-tribromobenzoate).
Structure of $T$ contains $3$ $Br$ atoms from the benzoyl part and $1$ $Br$ atom from the phenoxy part.
Total number of $Br$ atoms in $T = 3 + 1 = 4$.

(A) The chemical reaction is: $XeF_4 + O_2F_2 \rightarrow XeF_6 + O_2$.
The compound $Y$ is $XeF_6$.
In $XeF_6$,there are $6$ fluorine atoms. Each fluorine atom has $3$ lone pairs of electrons,contributing $6 \times 3 = 18$ lone pairs.
The xenon atom in $XeF_6$ has $1$ lone pair of electrons.
Therefore,the total number of lone pairs of electrons in the $XeF_6$ molecule is $18 + 1 = 19$.

(D) The extraction of gold involves the following reactions:
$4Au_{(s)} + 8CN^{-}_{(aq)} + 2H_2O_{(l)} + O_{2(g)} \rightarrow 4[Au(CN)_2]^{-}_{(aq)} + 4OH^{-}_{(aq)}$
Here,$Q = O_2$ and $R = [Au(CN)_2]^{-}$.
The complex $R$ is then treated with zinc $(T = Zn)$ to displace gold:
$2[Au(CN)_2]^{-}_{(aq)} + Zn_{(s)} \rightarrow [Zn(CN)_4]^{2-}_{(aq)} + 2Au_{(s)}$
Here,$Z = [Zn(CN)_4]^{2-}$.
All statements $1, 2, 3,$ and $4$ are correct.

(D) $(1)$ The reaction of gold with aqua regia produces $NO$ (nitric oxide),not $NO_2$. Thus,statement $(1)$ is incorrect.
$(2)$ Aqua regia is indeed prepared by mixing concentrated $HCl$ and concentrated $HNO_3$ in a $3:1$ volume ratio. Thus,statement $(2)$ is correct.
$(3)$ The reaction is $Au + 4H^+ + NO_3^- + 4Cl^- \rightarrow AuCl_4^- + NO + 2H_2O$. The complex ion $AuCl_4^-$ contains gold in the $+3$ oxidation state. Thus,statement $(3)$ is correct.
$(4)$ The yellow-orange colour of aqua regia is due to the formation of nitrosyl chloride $(NOCl)$ and chlorine $(Cl_2)$. Thus,statement $(4)$ is correct.

(C) $(1)$ Natural rubber is cis-polyisoprene. Thus,statement $(1)$ is incorrect.
$(2)$ Nylon-$6$ is formed by the polymerization of caprolactam and contains amide linkages $(-NH-CO-)$. Thus,statement $(2)$ is correct.
$(3)$ Cellulose is a linear polymer of $\beta-D$-glucose units joined by $\beta$-glycosidic linkages. Thus,statement $(3)$ is incorrect.
$(4)$ Teflon is prepared by heating tetrafluoroethene $(CF_2=CF_2)$ in the presence of a persulphate catalyst at high pressure. Thus,statement $(4)$ is correct.
Therefore,statements $(2)$ and $(4)$ are correct.

(A) $1$. The starting material is $4$-methoxyphenylbut-$3$-ynal. Treatment with $Hg^{2+}, dil. H_2SO_4$ hydrates the alkyne to a ketone,yielding $4$-methoxyphenyl-$4$-oxobutanal.
$2$. Oxidation with $AgNO_3, NH_4OH$ (Tollens' reagent) converts the aldehyde to a carboxylic acid,followed by Clemmensen reduction $(Zn-Hg, conc. HCl)$ which reduces the ketone to a methylene group,resulting in $4-(4-methoxyphenyl)butanoic$ acid $(Q)$.
$3$. Treatment of $Q$ with $SOCl_2$ followed by $AlCl_3$ (Friedel-Crafts acylation) leads to intramolecular cyclization to form $7-methoxy-1-tetralone$ $(R)$.
$4$. Finally,Clemmensen reduction $(Zn-Hg, conc. HCl)$ of the ketone $R$ gives $6-methoxytetralin$ $(S)$.
$5$. Comparing with the given structures,$Q$ matches structure $(2)$,$R$ matches structure $(4)$,and $S$ matches structure $(4)$. Thus,the correct options are $2$ and $4$.

(B) For the reaction $2N_2O_{5(g)} \rightarrow 2N_2O_{4(g)} + O_{2(g)}$ at constant $V$ and $T$:
At $t = 0$,$P_{N_2O_5} = 1 \ atm$,$P_{N_2O_4} = 0$,$P_{O_2} = 0$.
At $t = Y \times 10^3 \ s$,let the pressure of $O_2$ formed be $P$. Then $P_{N_2O_5} = 1 - 2P$,$P_{N_2O_4} = 2P$,and $P_{O_2} = P$.
The total pressure $P_T = (1 - 2P) + 2P + P = 1 + P = 1.45 \ atm$.
Thus,$P = 0.45 \ atm$.
The initial pressure of $N_2O_5$ is $P_0 = 1 \ atm$ and the pressure at time $t$ is $P_t = 1 - 2P = 1 - 2(0.45) = 0.1 \ atm$.
For a first-order reaction,$k = \frac{2.303}{t} \log \left(\frac{P_0}{P_t}\right)$.
Substituting the values: $5 \times 10^{-4} = \frac{2.303}{Y \times 10^3} \log \left(\frac{1}{0.1}\right)$.
$5 \times 10^{-4} = \frac{2.303}{Y \times 10^3} \times 1$.
$0.5 = \frac{2.303}{Y} \implies Y = \frac{2.303}{0.5} = 4.606$.
Given the options provided,there is a discrepancy in the calculation or the provided options. Based on standard calculation,$Y = 4.606$.

(A) In the $cis-[Mn(en)_2Cl_2]$ complex,the central metal ion $Mn$ is octahedrally coordinated.
There are two $Cl$ ligands and two $en$ (ethylenediamine) ligands.
Each $en$ ligand provides two $N$ donor atoms.
Let the $Cl$ atoms be $Cl_{(a)}$ and $Cl_{(b)}$,and the $N$ atoms be $N_{(1)}, N_{(2)}, N_{(3)}, N_{(4)}$.
By analyzing the octahedral geometry,the cis $N-Mn-Cl$ bond angles are:
$Cl_{(a)}-Mn-N_{(1)}$,$Cl_{(a)}-Mn-N_{(2)}$,$Cl_{(a)}-Mn-N_{(4)}$,$Cl_{(b)}-Mn-N_{(1)}$,$Cl_{(b)}-Mn-N_{(3)}$,and $Cl_{(b)}-Mn-N_{(4)}$.
Counting these,we find a total of $6$ such cis bond angles.

(D) Moles of water $(n_{H_2O})$ = $\frac{900}{18} = 50 \text{ mol}$.
Mole fraction of urea $(x_{urea})$ = $0.05$.
Using the formula $x_{urea} = \frac{n_{urea}}{n_{urea} + n_{H_2O}}$:
$0.05 = \frac{n_{urea}}{n_{urea} + 50}$.
$0.05n_{urea} + 2.5 = n_{urea} \implies 0.95n_{urea} = 2.5 \implies n_{urea} = \frac{2.5}{0.95} \approx 2.6316 \text{ mol}$.
Mass of urea = $2.6316 \text{ mol} \times 60 \text{ g mol}^{-1} = 157.896 \text{ g}$.
Total mass of solution = $157.896 \text{ g} + 900 \text{ g} = 1057.896 \text{ g}$.
Volume of solution $(V)$ = $\frac{\text{Mass}}{\text{Density}} = \frac{1057.896 \text{ g}}{1.2 \text{ g cm}^{-3}} = 881.58 \text{ mL} = 0.88158 \text{ L}$.
Molarity $(M)$ = $\frac{n_{urea}}{V(L)} = \frac{2.6316}{0.88158} \approx 2.98 \text{ M}$.

(A) For reaction $(III)$: The starting material is $o$-chloromethyl methyl benzoate. Treatment with $KCN$ gives $o$-cyanomethyl methyl benzoate. Hydrolysis $(H_3O^+, \Delta)$ yields $o$-carboxymethyl benzoic acid $(T)$. Reduction with $LiAlH_4$ gives the diol,and cyclization with $conc. H_2SO_4$ yields the ether $(R)$. Thus,$(III)$ matches $(T)$ and $(R)$.
For reaction $(IV)$: The starting material is dimethyl $o$-phenylenediacetate. Reduction with $LiAlH_4$ gives the diol $(Q)$. Cyclization with $conc. H_2SO_4$ gives the ether $(R)$. Thus,$(IV)$ matches $(Q)$ and $(R)$.
For reaction $(I)$: The starting material is $2$-(cyanomethyl)benzaldehyde ethylene acetal. $DiBAL-H$ followed by $dil. HCl$ gives the dialdehyde. $NaBH_4$ reduces it to the diol $(Q)$. Cyclization with $conc. H_2SO_4$ gives the ether $(R)$. Thus,$(I)$ matches $(Q)$ and $(R)$.
For reaction $(II)$: The starting material is $o$-allylbenzoic acid. Ozonolysis $(O_3/Zn, H_2O)$ gives the aldehyde-acid. $NaBH_4$ reduces the aldehyde to alcohol $(S)$. Cyclization with $conc. H_2SO_4$ gives the lactone $(U)$. Thus,$(II)$ matches $(S)$ and $(U)$.
Question $(1)$: $(III)$ matches $(T, R)$ and $(IV)$ matches $(Q, R)$. Option $(1)$ is $(III, T, R)$ and Option $(2)$ is $(IV, Q, R)$. Both are correct combinations.
Question $(2)$: $(I)$ matches $(Q, R)$ and $(II)$ matches $(S, U)$. Option $(1)$ is $(I, Q, R)$ and Option $(2)$ is $(II, S, U)$. Both are correct combinations.
Therefore,the correct answer is $(1, 2)$.