KCET 2019 Chemistry Question Paper with Answer and Solution

(C) The force acting on the particle in a uniform electric field is given by $F = qE$.
According to Newton's second law,the acceleration $a$ of the particle is $a = F/m = (qE)/m$.
Using the equation of motion $v^2 = u^2 + 2ay$,where initial velocity $u = 0$ and distance $s = y$,we get $v^2 = 0 + 2(qE/m)y = 2(qE/m)y$.
The kinetic energy $(KE)$ is given by $KE = (1/2)mv^2$.
Substituting the value of $v^2$,we get $KE = (1/2)m[2(qE/m)y] = qEy$.

(D) permanent magnet is a substance that retains its ferromagnetic properties for a long period at room temperature.
In ferromagnetic materials,individual atoms possess a magnetic dipole moment.
These atoms interact with each other such that they spontaneously align in a common direction over a macroscopic region known as a domain.
In a permanent magnet,these domains are all perfectly aligned in the direction of the external magnetizing field,resulting in a strong net magnetic moment.

(C) To determine the number of lone pairs on the central atom,we use the formula: $\text{Lone pairs} = \frac{1}{2} (V - B)$,where $V$ is the number of valence electrons of the central atom and $B$ is the number of bonding electrons (or atoms bonded to it).
$1$. $I_{3}^{+}$: The central $I$ atom has $7$ valence electrons. It forms $2$ bonds with other $I$ atoms and has a positive charge,so $V = 7 - 1 = 6$. Bonding electrons $B = 2$. Lone pairs = $\frac{1}{2} (6 - 2) = 2$.
$2$. $H_{2}O$: The central $O$ atom has $6$ valence electrons. It forms $2$ bonds with $H$ atoms. Lone pairs = $\frac{1}{2} (6 - 2) = 2$.
Both $I_{3}^{+}$ and $H_{2}O$ have $2$ lone pairs on their central atoms. Therefore,option $C$ is correct.

(D) The chemical formula of naphthalene is $C_{10}H_8$.
In the structure of naphthalene,there are two fused benzene rings.
The number of $\pi$-bonds is equal to the number of double bonds in the structure,which is $5$.
The number of $\sigma$-bonds can be calculated by counting all single bonds and one bond from each double bond.
Total $\sigma$-bonds = $19$.
Therefore,the number of $\pi$-bonds and $\sigma$-bonds are $5$ and $19$ respectively.
The correct option is $(D)$.

(C) $BeCl_2$ and $CO_2$ have a linear geometry,and $BF_3$ has a trigonal planar geometry; therefore,they all have a net dipole moment of zero.
$SO_2$ has a bent (angular) geometry due to the presence of a lone pair on the sulfur atom,which results in a non-zero net dipole moment.
Thus,the correct option is $C$.

(A) The first ionization enthalpy $(\Delta H_{IE})$ generally increases across a period from left to right and decreases down a group.
Comparing the elements $C$ (Group $14$,Period $2$),$N$ (Group $15$,Period $2$),$Si$ (Group $14$,Period $3$),and $P$ (Group $15$,Period $3$):
$1$. Elements in Period $2$ $(C, N)$ have higher ionization enthalpy than those in Period $3$ $(Si, P)$ due to smaller atomic size.
$2$. Within the same period,Group $15$ elements have higher ionization enthalpy than Group $14$ elements due to the stable half-filled $p$-orbital configuration $(ns^2 np^3)$.
$3$. Combining these trends,the order is $Si < P < C < N$.

(A) Given $ f(x) = x^2 $ and $ g(x) = \sqrt{x} $.
Check option $ (A) $: $ f \circ g(-4) = f(g(-4)) $. Since the domain of $ g $ is $ [0, \infty) $,$ g(-4) $ is not defined. Thus,$ f \circ g(-4) $ is not defined. This statement is not true.
Check option $ (B) $: $ g \circ f(-2) = g(f(-2)) = g((-2)^2) = g(4) = \sqrt{4} = 2 $. This is true.
Check option $ (C) $: $ g \circ f(4) = g(f(4)) = g(4^2) = g(16) = \sqrt{16} = 4 $. This is true.
Check option $ (D) $: $ f \circ g(2) = f(g(2)) = f(\sqrt{2}) = (\sqrt{2})^2 = 2 $. This is true.
Therefore,the statement that is not true is $ (A) $.

(A) The force acting on the particle is $F = qE$.
According to Newton's second law,the acceleration of the particle is $a = \frac{F}{m} = \frac{qE}{m}$.
Using the kinematic equation $v^2 = u^2 + 2ay$,where initial velocity $u = 0$:
$v^2 = 0 + 2 \left(\frac{qE}{m}\right)y = \frac{2qEy}{m}$.
The kinetic energy $K$ is given by $K = \frac{1}{2}mv^2$.
Substituting the value of $v^2$:
$K = \frac{1}{2} \times m \times \left(\frac{2qEy}{m}\right) = qEy$.
Therefore,the correct option is $A$.

(A)
Eutrophication refers to the excessive enrichment of water bodies with minerals and nutrients,which promotes the growth of algae.
This excessive algal growth leads to the consumption of dissolved oxygen by microorganisms during decomposition,resulting in a significant reduction in dissolved oxygen levels.

(B)
In $CH_2=CH-CH_2-NH_2$,the lone pair on the nitrogen atom is separated from the $\pi$ bond by an $sp^3$ hybridized carbon atom ($-CH_2-$ group).
Since there is no continuous conjugation,the resonance effect is not observed in this molecule.

(D)
The reduction of an alkyne to a $trans-alkene$ is achieved using dissolving metal reduction,such as $Na$ in liquid $NH_3$ (Birch reduction).
The reaction is:
$CH_3-C \equiv C-CH_3 \xrightarrow{Na \mid liq. NH_3} trans-CH_3-CH=CH-CH_3$

(B)
Hydrogen is highly inflammable and hence storage is difficult.
Hydrogen has a low critical temperature and is therefore not easily liquefiable.

(B) . Boric acid $(B(OH)_{3})$ is a weak monobasic acid. It does not dissociate to give $H^{+}$ ions directly; instead,it acts as a Lewis acid by accepting a lone pair of electrons from the oxygen atom of a water molecule to form the $[B(OH)_{4}]^{-}$ complex.

(A) In a permanent magnet,the magnetic domains are not perfectly aligned due to thermal agitations at room temperature. While the material exhibits a net magnetic moment,the alignment is only partial because thermal energy prevents complete orientation of all domains in a single direction.

(A) The reaction of gold with aqua regia is given by the equation:
$Au + 4H^{+} + NO_{3}^{-} (+5) + 4Cl^{-} \rightarrow AuCl_{4}^{-} + NO (+2) + 2H_{2}O$
In this reaction,the nitrogen atom in the nitrate ion $(NO_{3}^{-})$ has an oxidation state of $+5$.
It is reduced to nitric oxide $(NO)$,where the oxidation state of nitrogen is $+2$.
Therefore,the oxidation state of nitrogen changes from $+5$ to $+2$.

(D) The thermal decomposition of alkali metal nitrates follows different patterns based on the size of the cation.
$LiNO_{3}$ decomposes to form its oxide,nitrogen dioxide,and oxygen: $4LiNO_{3} \rightarrow 2Li_{2}O + 4NO_{2} + O_{2}$.
In contrast,nitrates of other alkali metals like $KNO_{3}$,$RbNO_{3}$,and $NaNO_{3}$ decompose to form nitrites and oxygen gas $(2MNO_{3} \rightarrow 2MNO_{2} + O_{2})$ and do not release $NO_{2}$ gas.
Therefore,the correct option is $D$.

(C) The reduction half-reaction is: $Cr_2O_7^{2-} + 14H^{+} + 6e^{-} \rightarrow 2Cr^{3+} + 7H_2O$.
In $Cr_2O_7^{2-}$,the oxidation state of $Cr$ is $+6$.
The change in oxidation state for $2$ atoms of $Cr$ is $2 \times (6 - 3) = 6$.
Thus,$1 \ mol$ of $Cr_2O_7^{2-}$ requires $6 \ mol$ of electrons.
For $0.2 \ mol$ of $Cr_2O_7^{2-}$,the number of moles of electrons required is $0.2 \times 6 = 1.2 \ mol$.

(B) The solubility of a sparingly soluble salt like $AgCl$ is governed by the common ion effect.
According to the common ion effect,the presence of a common ion $(Cl^-)$ decreases the solubility of the salt.
The concentration of $Cl^-$ ions provided by the electrolytes are:
$0.1 \ M \ BaCl_2$ provides $0.2 \ M \ Cl^-$ ions.
$0.1 \ M \ AlCl_3$ provides $0.3 \ M \ Cl^-$ ions.
$0.1 \ M \ NaCl$ provides $0.1 \ M \ Cl^-$ ions.
Since $AlCl_3$ provides the highest concentration of common $Cl^-$ ions $(0.3 \ M)$,the equilibrium $AgCl(s) \rightleftharpoons Ag^+(aq) + Cl^-(aq)$ shifts furthest to the left.
Therefore,the solubility of $AgCl$ is least in $0.1 \ M \ AlCl_3$.

(A) The balanced chemical equation is:
$NaCl + AgNO_3 \rightarrow AgCl + NaNO_3$
Molar mass of $AgNO_3 = 108 + 14 + (3 \times 16) = 170 \ g/mol$
Molar mass of $NaCl = 23 + 35.5 = 58.5 \ g/mol$
Molar mass of $AgCl = 108 + 35.5 = 143.5 \ g/mol$
Given amounts:
$NaCl = 11.70 \ g$
$AgNO_3 = 3.4 \ g$
From the stoichiometry,$1 \ mol$ of $AgNO_3$ reacts with $1 \ mol$ of $NaCl$.
$170 \ g$ of $AgNO_3$ requires $58.5 \ g$ of $NaCl$.
$3.4 \ g$ of $AgNO_3$ requires $\frac{58.5}{170} \times 3.4 = 1.17 \ g$ of $NaCl$.
Since we have $11.70 \ g$ of $NaCl$ (which is more than $1.17 \ g$),$AgNO_3$ is the limiting reagent.
The amount of $AgCl$ produced depends on the limiting reagent $(AgNO_3)$:
$170 \ g$ of $AgNO_3$ produces $143.5 \ g$ of $AgCl$.
$3.4 \ g$ of $AgNO_3$ produces $\frac{143.5}{170} \times 3.4 = 2.87 \ g$ of $AgCl$.
Thus,the mass of $AgCl$ precipitated is $2.87 \ g$.

(A) Charles's law states that for a given mass of gas at constant pressure,the volume $V$ is directly proportional to the absolute temperature $T$,expressed as $V/T = K$ (where $K$ is a constant).
Taking the natural logarithm on both sides: $\ln V = \ln K + \ln T$.
Taking the common logarithm on both sides: $\log V = \log K + \log T$.
Rearranging the logarithmic equation: $\log V - \log T = \log K$.
Option $A$ represents $\log K = \log V + \log T$,which implies $\log K = \log(VT)$,or $K = VT$. This contradicts Charles's law $(V/T = K)$.
Therefore,option $A$ does not represent Charles's law.

(A) According to the de Broglie equation,$\lambda = h / p$,where $\lambda$ is the wavelength,$h$ is Planck's constant,and $p$ is the momentum.
Given $\lambda_{A} = 5 \times 10^{-8} \ m$ and $p_{B} = 0.5 \ p_{A}$.
Since $\lambda \propto 1 / p$,we have $\lambda_{B} / \lambda_{A} = p_{A} / p_{B}$.
Substituting $p_{B} = 0.5 \ p_{A}$,we get $\lambda_{B} / \lambda_{A} = p_{A} / (0.5 \ p_{A}) = 2$.
Therefore,$\lambda_{B} = 2 \times \lambda_{A} = 2 \times 5 \times 10^{-8} \ m = 10^{-7} \ m$.
Converting to $\mathring{A}$: $1 \ m = 10^{10} \ \mathring{A}$,so $10^{-7} \ m = 10^{-7} \times 10^{10} \ \mathring{A} = 1000 \ \mathring{A}$.

(D) The relationship between enthalpy change and internal energy change is given by $\Delta H = \Delta U + \Delta n_g RT$.
For $\Delta H > \Delta U$,the value of $\Delta n_g$ (change in the number of moles of gaseous products and reactants) must be positive,i.e.,$\Delta n_g > 0$.
In option $(d)$: $CaCO_{3(s)} \rightarrow CaO_{(s)} + CO_{2(g)}$.
Here,$\Delta n_g = (\text{moles of gaseous products}) - (\text{moles of gaseous reactants}) = 1 - 0 = 1$.
Since $\Delta n_g$ is positive,$\Delta H > \Delta U$.

(B)
$1$. Ethanol is a very weak acid,much weaker than water. It does not react with bases like $NaHCO_{3}$ or $Na_{2}CO_{3}$ and does not change the color of litmus paper.
$2$. However,it reacts with active metals like sodium $(Na)$ to evolve hydrogen gas $(H_{2})$,which confirms its acidic hydrogen atom.
$3$. The reaction is: $2C_{2}H_{5}OH + 2Na \rightarrow 2C_{2}H_{5}ONa + H_{2} \uparrow$

(A) The starting material is $HOCH_{2}-CH=CH-CHO$.
Step $A$: The product is $HOCH_{2}-CH=CH-CH_{2}OH$. This indicates the reduction of the aldehyde group to a primary alcohol while leaving the $C=C$ double bond intact. $NaBH_{4}$ is a selective reducing agent that reduces aldehydes/ketones but not $C=C$ double bonds.
Step $B$: The product is $HOCH_{2}-CH=CH-CHO$. This indicates that the aldehyde group remains unchanged. $PCC$ (Pyridinium chlorochromate) is an oxidizing agent used to oxidize alcohols to aldehydes,but it does not affect the aldehyde group itself.
Step $C$: The product is $HOCH_{2}-CH_{2}-CH_{2}-CH_{2}OH$. This indicates the reduction of both the aldehyde group and the $C=C$ double bond. $H_{2} / Pd$ is a catalytic hydrogenation reagent that reduces both aldehydes and $C=C$ double bonds.
Therefore,the reagents are $A = NaBH_{4}$,$B = PCC$,$C = H_{2} / Pd$.

(D) The solubility of amines in water depends on their ability to form hydrogen bonds with water molecules.
$CH_3NH_2$ and $(CH_3)_2NH$ are primary and secondary amines,respectively,which can form hydrogen bonds with water.
$C_6H_5NH_2$ (aniline) is less soluble due to the large hydrophobic phenyl group,but it still has an $-NH_2$ group capable of hydrogen bonding.
$(CH_3)_3N$ is a tertiary amine and lacks a hydrogen atom attached to the nitrogen atom,meaning it cannot act as a hydrogen bond donor to water molecules.
Therefore,$(CH_3)_3N$ is the least soluble in water among the given options.

(D)
In nucleic acids,the individual nucleotides are linked together through the $3'-5'$ phosphodiester bonds,which connect the $5'$ carbon atom of one sugar molecule to the $3'$ carbon atom of the next sugar molecule via a phosphate group.

(B) . Vitamin $K$ is essential for the synthesis of proteins required for blood coagulation (clotting). It acts as a cofactor for the enzyme that carboxylates glutamate residues in clotting factors.

(C)
Fibrous proteins are generally insoluble in water due to the presence of strong intermolecular hydrogen bonds and disulfide linkages that hold the polypeptide chains together.

(D) The given reaction is a haloform reaction (specifically iodoform test).
$CH_3-CH=CH-CH_2-CO-CH_3$ contains a methyl ketone group $(-COCH_3)$.
Methyl ketones react with $I_2$ in the presence of $NaOH$ to form a carboxylate salt and iodoform $(CHI_3)$.
This reaction is specific to methyl ketones and does not affect the double bond present in the chain.
Therefore,the correct reagent is $I_2$ and $NaOH$ solution.

(C) The $HVZ$ (Hell-Volhard-Zelinsky) reaction of propanoic acid $(CH_3CH_2COOH)$ with $Cl_2/P$ yields $2-$chloropropanoic acid $(CH_3CHClCOOH)$.
In $2-$chloropropanoic acid,the chlorine atom is an electron-withdrawing group ($-I$ effect).
Electron-withdrawing groups stabilize the carboxylate anion $(RCOO^-)$ by dispersing the negative charge,thereby increasing the acidity of the carboxylic acid.
Therefore,$2-$chloropropanoic acid is a stronger acid than propanoic acid.

(C) Acidity is inversely proportional to the $pKa$ value.
$1$. Picric acid ($2,4,6$-trinitrophenol) is the strongest acid among the given options due to the presence of three electron-withdrawing $-NO_2$ groups,so it has the lowest $pKa$ value $(0.78)$.
$2$. $p$-Nitrophenol is more acidic than phenol due to the $-I$ and $-M$ effect of the $-NO_2$ group $(pKa = 7.1)$.
$3$. Phenol is more acidic than ethanol because the phenoxide ion is resonance stabilized $(pKa = 10)$.
$4$. Ethanol is the least acidic $(pKa = 16)$.
Therefore,the correct match is: $a-iii, b-iv, c-i, d-ii$.

(C) Esterification is a second-order reaction.
$CH_3COOH + C_2H_5OH \rightarrow CH_3COOC_2H_5 + H_2O$
The rate law for this reaction is $r = k[CH_3COOH][C_2H_5OH]$.
When each solution is diluted with an equal volume of water,the total volume doubles,so the concentration of each reactant is reduced to half of its initial value.
Let the initial concentrations be $[A]_0$ and $[B]_0$. The initial rate is $r = k[A]_0[B]_0$.
After dilution,the new concentrations are $[A]' = \frac{[A]_0}{2}$ and $[B]' = \frac{[B]_0}{2}$.
The new rate is $r' = k \times (\frac{[A]_0}{2}) \times (\frac{[B]_0}{2}) = \frac{1}{4} \times k[A]_0[B]_0 = \frac{1}{4}r$.
Thus,the rate of reaction becomes $0.25$ times the initial rate.

(B) For a reaction,the half-life $t_{1/2}$ is related to the initial concentration $[R]_0$ by the expression $t_{1/2} \propto [R]_0^{1-n}$,where $n$ is the order of the reaction.
If the plot of $t_{1/2}$ versus $[R]_0$ is a straight line parallel to the $x$-axis,it implies that $t_{1/2}$ is independent of the initial concentration $[R]_0$.
This occurs when $1-n = 0$,which means $n = 1$.
Therefore,the reaction is a first-order reaction.
The unit of the rate constant for a first-order reaction is $s^{-1}$.

(B)
According to the Arrhenius equation,a catalyst provides an alternative reaction pathway with a lower activation energy $(E_a)$.
Therefore,the presence of a catalyst changes the value of $E_a$.

(C)
Bromopheniramine is a well-known anti-histamine drug used to treat allergic symptoms.

(D) is the correct answer.
Sodium stearate $(C_{17}H_{35}COONa)$ is a soap. In hard water,it reacts with calcium or magnesium ions ($Ca^{2+}$ or $Mg^{2+}$) to form insoluble precipitates (scum),which makes it ineffective as a cleansing agent.
The reaction with calcium ions is:
$2 C_{17}H_{35}COONa + CaCl_{2} \rightarrow (C_{17}H_{35}COO)_{2}Ca + 2 NaCl$

(A)
$CO$ is the strongest ligand among the given options due to the synergic bonding effect.
It acts as a $\sigma$-donor and a $\pi$-acceptor,which strengthens the metal-ligand bond significantly.

(A) The name pentaaquanitratochromium$(III)$ nitrate indicates the following components:
$1$. The central metal ion is Chromium $(Cr)$ with an oxidation state of $+3$.
$2$. The ligands are five water molecules ($H_2O$,aqua) and one nitrate ion ($NO_3^-$,nitrato).
$3$. The coordination sphere is $[Cr(H_2O)_5(NO_3)]$.
$4$. To balance the charge: $Cr$ is $+3$,$H_2O$ is $0$,and $NO_3$ is $-1$. The net charge on the coordination sphere is $(+3) + 0 + (-1) = +2$.
$5$. To neutralize the $+2$ charge of the complex,two nitrate ions $(NO_3^-)$ are required outside the coordination sphere.
$6$. Thus,the formula is $[Cr(H_2O)_5(NO_3)](NO_3)_2$.

(B) The correct answer is $(B)$.
$Zn$ (Zinc) has a completely filled $d$-orbital $(3d^{10}4s^2)$ and does not exhibit variable oxidation states; it only shows a $+2$ oxidation state.

(A) The reaction is the Chromyl Chloride test: $4Cl^{-} + Cr_{2}O_{7}^{2-} + 6H^{+} \rightarrow 2CrO_{2}Cl_{2} + 3H_{2}O$.
The deep red vapours consist of $CrO_{2}Cl_{2}$ (Chromyl chloride).
Statement $A$ is false because the vapours do not contain $Cl_{2}$ gas.
When these vapours are passed into $NaOH$,a yellow solution of sodium chromate is formed: $CrO_{2}Cl_{2} + 4OH^{-} \rightarrow CrO_{4}^{2-} + 2Cl^{-} + 2H_{2}O$.
This solution,when treated with lead acetate,gives a yellow precipitate of lead chromate: $CrO_{4}^{2-} + Pb^{2+} \rightarrow PbCrO_{4} \downarrow$ (yellow).

(A) The elements in which the $4f$ orbitals are progressively filled are known as Lanthanoids.
These elements range from atomic number $58$ $(Ce)$ to $71$ $(Lu)$.
They are also known as $4f$-block elements or rare earth elements.
Therefore,the correct option is $A$.

(C) $Ce^{4+}$ acts as a strong oxidizing agent because it has a strong tendency to gain an electron to attain the stable $+3$ oxidation state $(Ce^{4+} + e^- \rightarrow Ce^{3+})$. Therefore,the statement that $Ce^{4+}$ is a reducing agent is incorrect.

(C) The electrolysis reaction for $MgCl_2$ is: $Mg^{2+} + 2e^- \rightarrow Mg(s)$.
The total charge passed is $Q = I \times t = 1 \ A \times (16 \times 60 + 5) \ s = 965 \ C$.
According to Faraday's law,the number of moles of $Mg$ deposited is $n = \frac{Q}{n_f \times F} = \frac{965}{2 \times 96500} = 0.005 \ \text{mol}$.
Since the volume of the solution is $1 \ L$,the molarity $M = \frac{n}{V} = \frac{0.005 \ \text{mol}}{1 \ L} = 5 \times 10^{-3} \ M$.

(A) The electrolysis of an aqueous solution of $CuSO_{4}$ with inert electrodes involves the following reactions:
At the cathode: $Cu^{2+}_{(aq)} + 2e^{-} \rightarrow Cu_{(s)}$
At the anode: $2H_{2}O_{(l)} \rightarrow O_{2(g)} + 4H^{+}_{(aq)} + 4e^{-}$
Since $H^{+}$ ions are produced at the anode,the concentration of $H^{+}$ ions in the solution increases.
As $pH = -\log[H^{+}]$,an increase in $[H^{+}]$ leads to a decrease in the $pH$ of the solution.

(A) For the reaction $Mn^{+7} + 5e^{-} \rightarrow Mn^{2+}$,$\Delta G^{0}_{1} = -nFE^{0} = -5 \times F \times 1.5 = -7.5F$.
For the reaction $Mn^{+4} + 2e^{-} \rightarrow Mn^{2+}$,$\Delta G^{0}_{2} = -nFE^{0} = -2 \times F \times 1.2 = -2.4F$.
To obtain the reaction $Mn^{+7} + 3e^{-} \rightarrow Mn^{4+}$,we subtract the second reaction from the first:
$(Mn^{+7} + 5e^{-}$ $\rightarrow Mn^{2+}) - (Mn^{+4} + 2e^{-}$ $\rightarrow Mn^{2+})$ $\Rightarrow Mn^{+7} + 3e^{-}$ $\rightarrow Mn^{4+}$.
Thus,$\Delta G^{0}_{3} = \Delta G^{0}_{1} - \Delta G^{0}_{2} = -7.5F - (-2.4F) = -5.1F$.
Using $\Delta G^{0}_{3} = -nFE^{0}_{3}$ where $n=3$:
$-3FE^{0}_{3} = -5.1F
\therefore E^{0}_{3} = \frac{5.1}{3} = 1.7 \ V$.

(A) The reaction of $1$ mole of $PdCl_2 \cdot 4 NH_3$ with excess $AgNO_3$ produces $2$ moles of $AgCl$ precipitate,which indicates that $2$ chloride ions $(Cl^-)$ are present outside the coordination sphere.
Thus,the complex can be formulated as $[Pd(NH_3)_4]Cl_2$.
In aqueous solution,it dissociates as: $[Pd(NH_3)_4]Cl_2 \rightarrow [Pd(NH_3)_4]^{2+} + 2Cl^-$.
This produces $1$ cation and $2$ anions,making it a $1:2$ electrolyte.

(D) The oxidation state of copper in the given ores is as follows:
$1$. Azurite: $Cu_3(CO_3)_2(OH)_2$ (Copper is in $ +2 $ state)
$2$. Chalcopyrite: $CuFeS_2$ (Copper is in $ +2 $ state)
$3$. Malachite: $CuCO_3 \cdot Cu(OH)_2$ (Copper is in $ +2 $ state)
$4$. Cuprite: $Cu_2O$ (Copper is in $ +1 $ state)
Therefore,the correct option is $D$.

(B) In the blast furnace,the extraction of iron from haematite $(Fe_2O_3)$ involves the following key processes:
$1$. Reduction of iron oxide: $Fe_2O_3$ is reduced by $CO$ to form metallic iron $(Fe)$ and $CO_2$. This is represented by reaction $i$: $Fe_2O_3 + 3CO \rightarrow 2Fe + 3CO_2$.
$2$. Slag formation: The impurity $SiO_2$ (silica) is removed by adding limestone $(CaCO_3)$,which decomposes to $CaO$. $CaO$ reacts with $SiO_2$ to form calcium silicate slag $(CaSiO_3)$. This is represented by reaction $iv$: $CaO + SiO_2 \rightarrow CaSiO_3$.
Therefore,reactions $i$ and $iv$ are the main reactions occurring in the blast furnace.

(C) The Wurtz reaction involves the coupling of two alkyl halide molecules in the presence of sodium metal in dry ether to form a symmetrical alkane.
To prepare $2,3-$dimethylbutane,which has a symmetric structure,we need two molecules of $2-$bromopropane:
$2 CH_3-CH(Br)-CH_3 + 2 Na \xrightarrow{\text{dry ether}} CH_3-CH(CH_3)-CH(CH_3)-CH_3 + 2 NaBr$
Thus,the required alkyl halide is $2-$bromopropane.

(B) The correct answer is $(B)$.
Nucleophilic aromatic substitution in haloarenes is facilitated by the presence of electron-withdrawing groups (like $-NO_2$) at the ortho and para positions.
These groups stabilize the carbanion intermediate formed during the reaction.
$2,4,6$-trinitrochlorobenzene has three electron-withdrawing $-NO_2$ groups at the ortho and para positions,which makes it highly reactive towards nucleophilic substitution.
Therefore,it can undergo hydrolysis even under mild conditions like warming with water or dilute aqueous $NaOH$.

(D) The bond length of $C-Cl$ depends on the hybridization of the carbon atom attached to the chlorine atom.
$1$. In $p-Nitrochlorobenzene$,$Chlorobenzene$,and $CH_2=CH-Cl$,the carbon atom attached to $Cl$ is $sp^2$ hybridized.
$2$. In $3-Chlorocyclohexene$,the carbon atom attached to $Cl$ is $sp^3$ hybridized.
$3$. The bond length increases as the $s$-character of the hybrid orbital decreases. $sp^3$ hybridized carbon has $25\%$ $s$-character,while $sp^2$ hybridized carbon has $33.3\%$ $s$-character.
$4$. Therefore,the $C-Cl$ bond in $3-Chlorocyclohexene$ is the longest due to the $sp^3$ hybridization of the carbon atom.

(D)
In the presence of strong acids like conc. $H_{2}SO_{4}$,the $-NH_{2}$ group of aniline gets protonated to form the anilinium ion $(-NH_{3}^{+})$.
The $-NH_{3}^{+}$ group is electron-withdrawing and meta-directing.
However,because the reaction is an equilibrium and some unprotonated aniline (which is ortho/para-directing) remains,a mixture of products is obtained,including a significant amount of $m$-nitroaniline due to the presence of the anilinium ion.

(A) $1$. The reaction $P \stackrel{H_{2} / Pd-BaSO_{4}}{\longrightarrow} Q$ is the Rosenmund reduction,which converts an acid chloride $(RCOCl)$ to an aldehyde $(RCHO)$. Thus,$P$ is $C_{6}H_{5}COCl$ (benzoyl chloride) and $Q$ is $C_{6}H_{5}CHO$ (benzaldehyde).
$2$. The reaction of $Q$ (benzaldehyde) with conc. $NaOH$ is the Cannizzaro reaction,which produces a carboxylic acid salt and an alcohol. Upon acidification,we get $R$ ($C_{6}H_{5}COOH$,benzoic acid) and $S$ ($C_{6}H_{5}CH_{2}OH$,benzyl alcohol).
$3$. Benzoic acid $(R)$ and benzyl alcohol $(S)$ undergo esterification to form benzyl benzoate $(C_{6}H_{5}COOCH_{2}C_{6}H_{5})$.
$4$. Therefore,$P$ is $C_{6}H_{5}COCl$.

(A) The molar mass of $H_2O$ is $18 \ g/mol$.
Given mass of $H_2O = 1.8 \ g$.
Number of moles of $H_2O = \frac{1.8 \ g}{18 \ g/mol} = 0.1 \ mol$.
The balanced chemical equation for the reaction of $XeF_6$ with $H_2O$ in a $1:1$ molar ratio is:
$XeF_6 + H_2O \rightarrow XeOF_4 + 2HF$.
Since $0.1 \ mol$ of $XeF_6$ reacts with $0.1 \ mol$ of $H_2O$,the product formed is $XeOF_4$.

(D) The correct answer is $(D)$.
Nylon-$6$ is prepared by the polymerization of caprolactam.
The monomer,caprolactam,contains a seven-membered ring with one nitrogen atom and six carbon atoms.
In the repeating unit of Nylon-$6$,which is $-[NH-(CH_2)_5-CO]-$,there are $5$ methylene $(-CH_2-)$ groups.

(B) $Cis-1,4-polyisoprene$ is the chemical name for natural rubber. It is a polymer of $2-methyl-1,3-butadiene$ (isoprene). The structure is as follows:
$[-CH_2-C(CH_3)=CH-CH_2-]_n$
Due to the $cis$ configuration of the double bonds,the polymer chains are coiled and exhibit elastic properties,making it natural rubber.

(A) $F_2$ is a very strong oxidizing agent $(E^\circ = +2.87 \ V)$ and oxidizes water to oxygen: $2F_2 + 2H_2O \rightarrow 4HF + O_2$.
In contrast,$Cl_2$ reacts with water to form a mixture of $HCl$ and $HOCl$.
Fluoride $(F^-)$ is a weak reducing agent,not an oxidizing agent.
$F_2$ is the strongest oxidizing agent among the halogens.

(D) network crystalline solid,also known as a covalent network solid,consists of atoms linked by a continuous network of covalent bonds throughout the crystal.
$NaCl$ is an ionic solid.
$Ice$ is a molecular solid held by hydrogen bonds.
$I_2$ is a molecular solid held by London dispersion forces.
$AlN$ (Aluminum Nitride) is a covalent network solid with a wurtzite-like structure.
Therefore,the correct option is $(D)$.

(C) For a $BCC$ crystal,the number of atoms per unit cell,$Z = 2$.
The density formula is $d = \frac{Z \times M}{a^3 \times N_A}$.
Given: $d = 10 \ g \ cm^{-3}$,$a = 200 \ pm = 200 \times 10^{-10} \ cm = 2 \times 10^{-8} \ cm$,$N_A = 6 \times 10^{23} \ mol^{-1}$,$Z = 2$.
Calculating molar mass $(M)$:
$M = \frac{d \times a^3 \times N_A}{Z} = \frac{10 \times (2 \times 10^{-8})^3 \times 6 \times 10^{23}}{2} = \frac{10 \times 8 \times 10^{-24} \times 6 \times 10^{23}}{2} = 24 \ g \ mol^{-1}$.
Number of moles in $2.4 \ g$ = $\frac{2.4 \ g}{24 \ g \ mol^{-1}} = 0.1 \ mol$.
Number of atoms = $\text{moles} \times N_A = 0.1 \times 6 \times 10^{23} = 6 \times 10^{22} \ atoms$.

(C) When $SrCl_{2}$ is added to $NaCl$,each $Sr^{2+}$ ion replaces two $Na^{+}$ ions to maintain electrical neutrality.
One $Sr^{2+}$ ion creates one cationic vacancy.
Therefore,$1 \ \text{mole}$ of $SrCl_{2}$ creates $1 \ \text{mole}$ of cationic vacancies.
$10^{-5} \ \text{mole}$ of $SrCl_{2}$ will create $10^{-5} \ \text{mole}$ of cationic vacancies.
Number of cationic vacancies $= 10^{-5} \times 6.022 \times 10^{23} = 6.022 \times 10^{18}$.

(B) Solution $A$ consists of acetone and chloroform. The interaction between acetone and chloroform is stronger than the individual interactions due to hydrogen bonding,leading to a negative deviation from Raoult's law.
Solution $B$ consists of acetone and carbon disulphide. The interaction between acetone and carbon disulphide is weaker than the individual interactions,leading to a positive deviation from Raoult's law.
Therefore,the correct sequence is negative and positive.

(A) The relative lowering of vapour pressure is given by the formula: $\frac{P^{\circ}-P}{P^{\circ}} = \frac{n_2}{n_1 + n_2} \approx \frac{n_2}{n_1}$ for a dilute solution.
Here,$n_2$ is the number of moles of solute (glucose) and $n_1$ is the number of moles of solvent (water).
Given: $\frac{P^{\circ}-P}{P^{\circ}} = 0.002$ and mass of water $(W_1)$ = $1000 \ g$.
Since $n_1 = \frac{1000 \ g}{18 \ g/mol} = 55.55 \ mol$,we have $0.002 = \frac{n_2}{55.55}$.
Therefore,$n_2 = 0.002 \times 55.55 = 0.111 \ mol$.
Molality $(m)$ is defined as the number of moles of solute per kilogram of solvent: $m = \frac{n_2}{W_1 (\text{in } kg)} = \frac{0.111 \ mol}{1 \ kg} = 0.111 \ m$.

(C) Given: Degree of association $\alpha = 80\% = 0.8$ and $n = 4$ (for tetramerization).
Van't Hoff factor $i = 1 + (\frac{1}{n} - 1)\alpha = 1 + (\frac{1}{4} - 1) \times 0.8 = 1 - 0.6 = 0.4$.
Depression in freezing point $\Delta T_f = i \times K_f \times m$.
$0.3 = 0.4 \times 1.86 \times \frac{2.5 \times 1000}{M_A \times 100}$.
$0.3 = 0.4 \times 1.86 \times \frac{25}{M_A}$.
$M_A = \frac{0.4 \times 1.86 \times 25}{0.3} = \frac{18.6}{0.3} = 62 \text{ g mol}^{-1}$.

(A)
$2 SO_{2(g)} + O_{2(g)} \xrightarrow{NO_{(g)}} 2 SO_{3(g)}$
In this reaction,the reactants ($SO_2$ and $O_2$) and the catalyst $(NO)$ are all in the gaseous phase,which defines homogeneous catalysis.

(C) The correct option is $C$.
Micelles are formed only when the concentration of the surfactant (soap) in the solution is greater than the $CMC$ value.
Given $CMC = 1.5 \times 10^{-4} \ mol \ L^{-1}$.
Comparing the given options,only $2.0 \times 10^{-3} \ mol \ L^{-1}$ is greater than $1.5 \times 10^{-4} \ mol \ L^{-1}$.