KCET 2024 Chemistry Question Paper with Answer and Solution

(B) The number of subsets of a set with $k$ elements is $2^k$.
Given that the first set has $m$ elements and the second set has $n$ elements,the number of subsets are $2^m$ and $2^n$ respectively.
According to the problem,$2^m - 2^n = 56$.
$2^n(2^{m-n} - 1) = 56$.
We can write $56$ as $8 \times 7 = 2^3 \times (8 - 1) = 2^3 \times (2^3 - 1)$.
Comparing $2^n(2^{m-n} - 1) = 2^3(2^3 - 1)$,we get $n = 3$ and $m - n = 3$.
Substituting $n = 3$ into $m - n = 3$,we get $m - 3 = 3$,so $m = 6$.
Thus,the values are $m = 6$ and $n = 3$.

(B) The number of subsets of a set with $k$ elements is $2^k$.
Given,$2^m - 2^n = 56$.
$2^n(2^{m-n} - 1) = 56$.
$2^n(2^{m-n} - 1) = 8 \times 7 = 2^3 \times (8 - 1) = 2^3 \times (2^3 - 1)$.
Comparing both sides,we get $n = 3$ and $m - n = 3$.
Substituting $n = 3$,we get $m - 3 = 3$,which implies $m = 6$.
Thus,the values are $m = 6$ and $n = 3$.

(A) The complete reaction sequence involved is as follows:
$CH_3COONa + NaOH \xrightarrow[\Delta]{CaO} CH_4 + Na_2CO_3$
(Sodium ethanoate) (Methane $X$)
$2CH_3COONa + 2H_2O \xrightarrow{\text{Electrolysis}} CH_3-CH_3 + 2CO_2 + H_2 + 2NaOH$
(Ethane $Y$)
Thus,the compounds $X$ and $Y$ are methane and ethane respectively.

(C) To determine the bond order,we calculate the number of electrons and use the molecular orbital theory formula: $B.O. = \frac{N_b - N_a}{2}$.
For $O_2^{2-}$,the total number of electrons is $8 + 8 + 2 = 18 \ e^{-}$. The bond order is $\frac{10 - 8}{2} = 1$.
For $F_2$,the total number of electrons is $9 + 9 = 18 \ e^{-}$. The bond order is $\frac{10 - 8}{2} = 1$.
Both species are isoelectronic $(18 \ e^{-})$ and have a bond order of $1$.

(A) The balanced chemical equation for the reaction between moist $SO_2$ and acidified permanganate solution is:
$2MnO_4^{-} + 5SO_2 + 2H_2O \rightarrow 2Mn^{2+} + 5SO_4^{2-} + 4H^{+}$
In this redox reaction,$SO_2$ (where $S$ is in $+4$ oxidation state) is oxidised to $SO_4^{2-}$ (where $S$ is in $+6$ oxidation state).
Simultaneously,$MnO_4^{-}$ (where $Mn$ is in $+7$ oxidation state) is reduced to $Mn^{2+}$ (where $Mn$ is in $+2$ oxidation state).
Therefore,option $A$ is correct.

(C) In the analysis of $III$ group basic radicals,$NH_4Cl$ is added to $NH_4OH$ to suppress the dissociation of $NH_4OH$ due to the common ion effect ($NH_4^+$ ions).
This ensures that the concentration of $OH^-$ ions is low enough to precipitate only the $III$ group hydroxides (like $Fe(OH)_3$,$Al(OH)_3$,$Cr(OH)_3$) while preventing the precipitation of hydroxides of subsequent groups.

(C) Given that $P = \text{adj}(A)$ and $|A| = 4$.
For a $3 \times 3$ matrix $A$,the property of the adjoint matrix is $|\text{adj}(A)| = |A|^{n-1}$,where $n$ is the order of the matrix.
Here,$n = 3$,so $|\text{adj}(A)| = |A|^{3-1} = |A|^2$.
Given $|A| = 4$,we have $|\text{adj}(A)| = 4^2 = 16$.
Since $P = \text{adj}(A)$,we have $|P| = 16$.
Calculating the determinant of $P$:
$|P| = \begin{vmatrix} 1 & \alpha & 3 \\ 1 & 3 & 3 \\ 2 & 4 & 4 \end{vmatrix} = 1(12 - 12) - \alpha(4 - 6) + 3(4 - 6) = 16$.
$0 - \alpha(-2) + 3(-2) = 16$.
$2\alpha - 6 = 16$.
$2\alpha = 22$.
$\alpha = 11$.

(D) Given,$f(x)=x e^{x(1-x)}$.
To find the intervals of increase and decrease,we calculate the derivative $f^{\prime}(x)$:
$f^{\prime}(x) = \frac{d}{dx} [x e^{x-x^2}] = e^{x-x^2} + x e^{x-x^2} (1-2x)$
$f^{\prime}(x) = e^{x-x^2} [1 + x(1-2x)] = e^{x-x^2} (1 + x - 2x^2)$
$f^{\prime}(x) = -e^{x-x^2} (2x^2 - x - 1) = -e^{x-x^2} (2x^2 - 2x + x - 1)$
$f^{\prime}(x) = -e^{x-x^2} [2x(x-1) + 1(x-1)] = -e^{x-x^2} (x-1)(2x+1)$.
Since $e^{x-x^2} > 0$ for all $x \in R$,the sign of $f^{\prime}(x)$ depends on $-(x-1)(2x+1)$.
Setting $f^{\prime}(x) = 0$,we get critical points $x = 1$ and $x = -\frac{1}{2}$.
Analyzing the sign of $f^{\prime}(x)$ on the number line:
For $x \in \left(-\infty, -\frac{1}{2}\right)$,$f^{\prime}(x) < 0$ (decreasing).
For $x \in \left[-\frac{1}{2}, 1\right]$,$f^{\prime}(x) \geq 0$ (increasing).
For $x \in (1, \infty)$,$f^{\prime}(x) < 0$ (decreasing).
Thus,$f(x)$ is increasing in the interval $\left[-\frac{1}{2}, 1\right]$.

(D) The reaction is a dehydration of a tertiary alcohol ($tert$-butyl alcohol) via $\beta$-elimination.
When $2$-methylpropan$-2-$ol $(A)$ is heated with $20\% \ H_3PO_4$ at $358 \ K$,it undergoes dehydration to form $2$-methylpropene $(B)$ as the major product.
The reaction is:
$(CH_3)_3C-OH \xrightarrow{20\% \ H_3PO_4, 358 \ K} CH_2=C(CH_3)_2 + H_2O$
Thus,the $IUPAC$ name of the product $B$ is $2$-methylpropene.

(B) $Propanone$ $(CH_3COCH_3)$ and $propanal$ $(CH_3CH_2CHO)$ have the same molecular formula,$C_3H_6O$.
$Propanone$ contains a ketone functional group $(-CO-)$,whereas $propanal$ contains an aldehyde functional group $(-CHO)$.
Since they possess the same molecular formula but different functional groups,they are classified as functional isomers.

(D) The complete reaction sequence is as follows:
$1$. $CH_2=CH_2 + Br_2 \xrightarrow{P=Br_2} CH_2(Br)-CH_2(Br)$ (Electrophilic addition)
$2$. $CH_2(Br)-CH_2(Br) \xrightarrow{Q=Alc. KOH} CH_2=CH-Br$ (Dehydrohalogenation)
$3$. $CH_2=CH-Br \xrightarrow{R=NaNH_2} CH \equiv CH$ (Dehydrohalogenation)
$4$. $3 CH \equiv CH \xrightarrow{S=Red \ hot \ iron \ tube} C_6H_6$ (Cyclic polymerization)
Thus,the reagents $P, Q, R$ and $S$ are $Br_2$,alc. $KOH, NaNH_2$,and Red hot iron tube respectively.

(A) The hydration of alkynes in the presence of $Hg^{2+}$ and $H_2SO_4$ follows Markovnikov's rule. The reaction proceeds as follows:
$CH_3-CH_2-C \equiv CH + H_2O \xrightarrow{Hg^{2+}, H_2SO_4, 333 \ K} [CH_3-CH_2-C(OH)=CH_2]$
This enol intermediate undergoes tautomerization to form a stable ketone:
$[CH_3-CH_2-C(OH)=CH_2] \rightleftharpoons CH_3-CH_2-CO-CH_3$ (Butan$-2-$one).

(A) Given,$K_{sp} = 4 \times 10^{-9} \ (mol \ L^{-1})^2$.
For the dissociation of $CaC_2O_4$:
$CaC_2O_4(s) \rightleftharpoons Ca^{2+}(aq) + C_2O_4^{2-}(aq)$.
Let the solubility be $S \ mol \ L^{-1}$.
Then,$[Ca^{2+}] = S$ and $[C_2O_4^{2-}] = S$.
$K_{sp} = [Ca^{2+}][C_2O_4^{2-}] = S \times S = S^2$.
$S^2 = 4 \times 10^{-9}$.
$S = \sqrt{4 \times 10^{-9}} = \sqrt{40 \times 10^{-10}} = 6.32 \times 10^{-5} \ mol \ L^{-1}$.
Thus,the solubility of $CaC_2O_4$ is $6.3 \times 10^{-5} \ mol \ L^{-1}$.

(D) Among the given nitrates,only $LiNO_3$ decomposes to give $NO_2$ on heating,whereas the nitrates of $Na, K,$ and $Rb$ decompose to form their respective nitrites and release oxygen gas.
The decomposition reaction for $LiNO_3$ is:
$4 LiNO_3 \xrightarrow{\Delta} 2 Li_2O + 4 NO_2 + O_2$
The decomposition reaction for other alkali metal nitrates $(M = Na, K, Rb)$ is:
$2 MNO_3 \xrightarrow{\Delta} 2 MNO_2 + O_2$

(D) Statement given in option $(D)$ is incorrect while all other statements are correct.
Ozone $(O_3)$ primarily absorbs high-energy $UV$ radiation from the Sun,which protects the Earth's surface.
It does not primarily absorb infrared radiation; rather,greenhouse gases like $CO_2$ and $CH_4$ are responsible for absorbing infrared radiation.

(A) Among the given halides,only $CCl_4$ cannot be hydrolysed.
This is because carbon lacks vacant $d$-orbitals in its valence shell,which are required to accept the lone pair of electrons from water molecules during the hydrolysis process.
In contrast,$SiCl_4$,$GeCl_4$,and $SnCl_4$ possess vacant $d$-orbitals,allowing them to undergo hydrolysis.

(D) Ionic hydrides are formed by the transfer of electrons from electropositive metals to hydrogen.
They are typically crystalline solids with high melting and boiling points,making them non-volatile.
They do not conduct electricity in the solid state but conduct in the molten state or in solution.
Therefore,being volatile is not a property of ionic hydrides.

(D) Homogeneous mixtures have uniform composition throughout. Thus,among the given options,a dilute aqueous solution of sugar is a homogeneous mixture and has a uniform composition.

(C) Given,
Mass of organic compound $= 0.48 \ g$
Mass of $CO_2$ produced $= 0.22 \ g$
$\% C = ?$
Using the formula:
$\% C = \frac{12}{44} \times \frac{\text{mass of } CO_2}{\text{mass of organic compound}} \times 100$
$= \frac{12}{44} \times \frac{0.22}{0.48} \times 100$
$= \frac{1}{44} \times \frac{0.22}{0.04} \times 100$
$= 12.5 \%$

(A) The wrong relation for real gases is given in option $A$.
The compressibility factor $Z$ is defined as the ratio of the molar volume of a real gas to the molar volume of an ideal gas at the same temperature and pressure.
Therefore,the correct form is $Z = \frac{V_{real}}{V_{ideal}}$.
Option $B$ represents the pressure correction term in the van der Waals equation.
Option $C$ represents the volume correction term where $V_{real} = V_{ideal} + nb$ is generally used for the excluded volume correction,but in the context of the van der Waals equation $(p + \frac{an^2}{V^2})(V - nb) = nRT$,the relation $V_{real} = V_{ideal} + nb$ is standard.
Option $D$ is the van der Waals equation for $1 \text{ mole}$ of gas.

(B) The energy of an electron in the $n^{th}$ orbit of a hydrogen-like species is given by the formula:
$E_n = -2.18 \times 10^{-18} \times \frac{Z^2}{n^2} \ J$
For $He^{+}$,the atomic number $Z = 2$ and for the first orbit $n = 1$.
Substituting these values into the formula:
$E_1 = -2.18 \times 10^{-18} \times \frac{2^2}{1^2} \ J$
$E_1 = -2.18 \times 10^{-18} \times 4 \ J$
$E_1 = -8.72 \times 10^{-18} \ J$

$\textbf{(C) According to Hess's Law, the enthalpy change for a cyclic process is zero.}$
$\text{For the cycle } A \rightarrow 2B \rightarrow C \rightarrow A, \text{ we have:}$
$\Delta H_{A \to 2B} + \Delta H_{2B \to C} + \Delta H_{C \to A} = 0$
$\text{Given } \Delta H_{A \to 2B} = +10 \text{ J and } \Delta H_{2B \to C} = +25 \text{ J}.$
$\text{Substituting these values:}$
$10 \text{ J} + 25 \text{ J} + \Delta H_{C \to A} = 0$
$35 \text{ J} + \Delta H_{C \to A} = 0$
$\Delta H_{C \to A} = -35 \text{ J}$

(D) $PCC$ stands for pyridinium chlorochromate.
It is a complex of chromium trioxide with pyridine and $HCl$,represented as $[C_5H_5NH]^{+}[CrO_3Cl]^{-}$.
It is a reagent in organic synthesis used primarily for the oxidation of primary alcohols to aldehydes and secondary alcohols to ketones.

(B) The reaction of monohydric alcohol $(R-OH)$ with ethyl magnesium iodide $(C_2H_5MgI)$ is: $R-OH + C_2H_5MgI \rightarrow R-OMgI + C_2H_6 \uparrow$.
At $STP$,$1 \ mol$ of ethane occupies $22400 \ cm^3$.
Given $2240 \ cm^3$ of ethane is liberated,which corresponds to $2240 / 22400 = 0.1 \ mol$ of ethane.
Since $1 \ mol$ of alcohol produces $1 \ mol$ of ethane,$0.1 \ mol$ of alcohol has a mass of $8.8 \ g$.
Thus,the molar mass of the alcohol is $8.8 / 0.1 = 88 \ g/mol$.
The general formula for a saturated monohydric alcohol is $C_nH_{2n+1}OH$,so $12n + (2n+1) + 16 + 1 = 88$,which gives $14n = 70$,so $n = 5$.
The alcohol is a pentanol isomer $(C_5H_{12}O)$.
Since the alcohol forms an aldehyde upon oxidation with $PCC$ (which gives a positive Tollen's test),it must be a primary $(1^{\circ})$ alcohol.
Among the options,$2, 2$-dimethylpropan-$1$-ol has the formula $(CH_3)_3CCH_2OH$,which is a primary alcohol with molar mass $88 \ g/mol$.

(B) Adrenaline and ephedrine are biologically active compounds used to increase blood pressure.
In the structure of adrenaline,the nitrogen atom is bonded to one methyl group and one alkyl chain,making it a secondary amine.
Similarly,ephedrine also contains a secondary amino group.
Therefore,both contain a secondary amino group.

(B) Nucleotides are joined together by a phosphodiester linkage between the $5^{\prime}$ and $3^{\prime}$ carbon atoms of the pentose sugar.

(B) Given,
The reaction is $PCl_5 \longrightarrow PCl_3 + Cl_2$.
Rate $= 1.02 \times 10^{-4} \ mol \ L^{-1} \ s^{-1}$.
Rate constant $(k) = 3.4 \times 10^{-5} \ s^{-1}$.
The rate law for this first-order reaction is:
$\text{Rate} = k [PCl_5]$
Substituting the given values:
$1.02 \times 10^{-4} = 3.4 \times 10^{-5} \times [PCl_5]$
$[PCl_5] = \frac{1.02 \times 10^{-4}}{3.4 \times 10^{-5}}$
$[PCl_5] = 3 \ mol \ L^{-1}$.

(A) The enthalpy change of a reaction is given by the difference between the activation energy of the forward reaction $(E_a)_f$ and the activation energy of the backward reaction $(E_a)_b$: $\Delta H = (E_a)_f - (E_a)_b$.
Given that the initial $(E_a)_f = 50 \ kJ \ mol^{-1}$ and the catalyst decreases it by $10 \ kJ \ mol^{-1}$,the new forward activation energy is $(E_a)_f = 50 - 10 = 40 \ kJ \ mol^{-1}$.
The enthalpy change $\Delta H$ remains constant at $-20 \ kJ \ mol^{-1}$ because a catalyst does not change the energy of the reactants or products.
Substituting the values into the formula: $-20 = 40 - (E_a)_b$.
Solving for $(E_a)_b$: $(E_a)_b = 40 + 20 = 60 \ kJ \ mol^{-1}$.

(D) The Arrhenius equation is given by $k = A e^{\frac{-E_a}{RT}}$.
Taking natural logarithm on both sides,we get $\ln k = \ln A - \frac{E_a}{RT}$.
Converting to base $10$ logarithm,we get $\log k = \log A - \frac{E_a}{2.303 RT}$.
Comparing these with the given options,option $(D)$ i.e.,$k = A e^{\frac{E_a}{RT}}$ is incorrect because the exponent should be negative.

(C) The nuclear reaction between ${}_{98}^{249}Cf$ and ${}_{20}^{48}Ca$ is given by:
${}_{98}^{249}Cf + {}_{20}^{48}Ca \longrightarrow {}_{118}^{294}Og + 3{}_0^1n$
Here,${}_{118}^{294}Og$ represents Oganesson,which is a synthetic radioactive noble gas with atomic number $Z = 118$.

(B) Estradiol is the primary female sex hormone. It is responsible for the development of secondary female characteristics and plays a crucial role in the regulation of the menstrual cycle.

(D) The first chlorinated organic insecticide prepared was $DDT$ (Dichlorodiphenyltrichloroethane).
Its structure is shown below:
$Cl-C_6H_4-CH(CCl_3)-C_6H_4-Cl$

(C) The structure provided in the image is $Phthalimide$.
None of the given options ($sucralose$,$aspartame$,$saccharin$,$alitame$) represent $Phthalimide$.
$Saccharin$ is a common artificial sweetener with the structure $1,2-benzisothiazol-3(2H)-one \ 1,1-dioxide$.
Since the question asks to identify the compound and the provided options do not contain the correct name,this question is technically flawed. However,if this were a multiple-choice question where one must be selected,it is often misidentified in some contexts due to structural similarities in bicyclic systems,but strictly speaking,none of the options are correct.

(B) metalloid is defined as a chemical element whose physical and chemical properties lie between those of metals and non-metals.
Among the given options,only antimony $(Sb)$ is a metalloid.

(D) The complex compounds $[Co(NH_3)_5 SO_4] Br$ and $[Co(NH_3)_5 Br] SO_4$ are ionisation isomers of each other.
- Ionisation isomerism occurs when the counter ion in a coordination compound is a potential ligand and can exchange places with a ligand present in the coordination sphere.
- In $[Co(NH_3)_5 SO_4] Br$,the $Br^-$ ion is the counter ion,while in $[Co(NH_3)_5 Br] SO_4$,the $SO_4^{2-}$ ion is the counter ion.

(B) For the complex $[CoF_6]^{3-}$:
$1$. The coordination number is $6$,so it has an octahedral geometry. Statement $I$ is true.
$2$. The oxidation state of $Co$ is $x + 6(-1) = -3$,which gives $x = +3$. The coordination number is $6$. Statement $II$ is false.
$3$. $F^-$ is a weak field ligand,so it forms an outer orbital complex with $sp^3d^2$ hybridization. Statement $III$ is true.
$4$. Since it is an outer orbital complex with weak field ligands,it remains a high spin complex. Statement $IV$ is true.
Therefore,statements $I$,$III$,and $IV$ are correct.

(B) The correct match is $I \to (ii), II \to (iii), III \to (i)$.
- $Zn^{2+} (Z=30)$: The electronic configuration is $[Ar] 3d^{10}$. Since all $d$-orbitals are fully filled,there are no unpaired electrons,making it colourless.
- $Cu^{2+} (Z=29)$: The electronic configuration is $[Ar] 3d^9$. It has one unpaired electron. The magnetic moment is $\mu = \sqrt{n(n+2)} = \sqrt{1(1+2)} = \sqrt{3} \approx 1.73 \ \text{BM}$.
- $Ni^{2+} (Z=28)$: The electronic configuration is $[Ar] 3d^8$. Thus,it has a $d^8$ configuration.

(C) Given,Molarity of the complex $= 0.1 \ M$.
Volume of the complex $= 100 \ mL$.
Mass of $AgCl$ obtained $= 2.86 \ g$.
Molar mass of $AgCl = 108 + 35.5 = 143.5 \ g/mol$.
Number of moles of complex $= \frac{0.1 \times 100}{1000} = 0.01 \ mol$.
Number of moles of $AgCl$ precipitated $= \frac{2.86 \ g}{143.5 \ g/mol} \approx 0.02 \ mol$.
Since $1 \ mol$ of complex gives $2 \ mol$ of $AgCl$,there must be $2$ ionizable $Cl^-$ ions outside the coordination sphere.
Therefore,the complex is $[Cr(H_2 O)_5 Cl] Cl_2 \cdot H_2 O$.

(B) $MnO$ exhibits antiferromagnetism.
They have a domain structure similar to ferromagnetic substances,but their domains are oppositely oriented and cancel out each other's magnetic moment.

(B) Misch metal is an alloy that consists of a lanthanoid metal (approximately $95 \%$) and iron (approximately $5 \%$).
It also contains traces of sulfur $(S)$,carbon $(C)$,calcium $(Ca)$,and aluminum $(Al)$.

(D) The incorrect statement regarding lanthanoids is given in option $(D)$.
Lanthanoids exhibit colour in solution due to $f-f$ transitions,not $d-d$ transitions,because the $4f$ orbitals are partially filled in their ions.

(B) The oxidation reaction of water is: $H_2O \longrightarrow \frac{1}{2}O_2 + 2H^+ + 2e^-$.
From the stoichiometry of the reaction,$1 \ mole$ of $H_2O$ requires $2 \ moles$ of electrons for oxidation.
Since $1 \ mole$ of electrons carries $96500 \ C$ (Faraday's constant),$2 \ moles$ of electrons carry $2 \times 96500 \ C = 193000 \ C$.
Therefore,for $0.1 \ mole$ of $H_2O$,the charge required is $0.1 \times 193000 \ C = 19300 \ C$.
This is equal to $1.93 \times 10^4 \ C$.

(D) Given:
$t = 1 \ hr \ 47 \ min \ 13 \ s = (1 \times 3600) + (47 \times 60) + 13 = 3600 + 2820 + 13 = 6433 \ s$
$Molar \ mass \ of \ Ca = 40 \ g \ mol^{-1}$
$Current \ (I) = 3 \ A$
$n-factor \ for \ Ca^{2+} + 2e^- \rightarrow Ca \text{ is } 2$.
Using Faraday's law of electrolysis:
$w = \frac{M \times I \times t}{n \times 96500}$
$w = \frac{40 \times 3 \times 6433}{2 \times 96500}$
$w = \frac{771960}{193000} \approx 4.0 \ g$

(C) The equation $\lambda_{m} = \lambda_{m}^{\circ} - A \sqrt{C}$ is the Debye-Huckel-Onsager equation.
In this equation,'$A$' is a constant that depends on the nature of the solvent and the temperature,but it also depends on the type of electrolyte (i.e.,the charges on the ions,such as $1:1$,$1:2$,$2:1$,or $2:2$ electrolytes).
$NaCl$ and $KBr$ are both $1:1$ electrolytes.
Since they are of the same type,they will have the same value of '$A$'.

(D) $\alpha-D-(+)$-glucose and $\beta-D-(+)$-glucose are called anomers.
These are the two cyclic hemiacetal forms of glucose that differ only in the configuration of the hydroxyl group at $C1$.

(D) The correct statement is given in option $(D)$.
- Roasting involves heating the ore in a regular supply of air.
- Calcination involves heating the ore below its melting point.
- Smelting is the process by which a metal is obtained at temperatures above its melting point.
- Calcination of calcium carbonate $(CaCO_3 \rightarrow CaO + CO_2)$ is an endothermic process as it requires heat to proceed.

(D) The incorrect statement about the Hall$-$Heroult process is given in option $(D)$.
In this process,the overall cell reaction is $2Al_2O_3 + 3C \longrightarrow 4Al + 3CO_2$.
In $Al_2O_3$,the oxidation state of oxygen is $-2$,and in $CO_2$,it is also $-2$.
Therefore,the oxidation state of oxygen does not change during the overall cell reaction.

(C) The given reaction scheme shows three different reactions of ethyl chloride $(C_2H_5Cl)$:
$1$. The conversion of $C_2H_5Cl$ to $C_2H_5F$ is a Swarts reaction,which uses a metallic fluoride like $Hg_2F_2$ or $AgF$. Thus,$X = Hg_2F_2$.
$2$. The conversion of $C_2H_5Cl$ to ethene $(CH_2=CH_2)$ is a dehydrohalogenation reaction,which occurs in the presence of alcoholic $KOH$. Thus,$Y = \text{alcoholic } KOH$.
$3$. The conversion of $C_2H_5Cl$ to butane $(C_4H_{10})$ is a Wurtz reaction,which uses $Na$ in dry ether. Thus,$Z = Na \text{ in dry ether}$.
Therefore,the correct sequence is $Hg_2F_2$,alcoholic $KOH$ and $Na$ in dry ether.

(A) - The mechanism of nucleophilic substitution in haloalkanes,whether $S_{N}2$ or $S_{N}1$,is significantly influenced by the nature of the solvent used in the reaction.
- $S_{N}2$ reactions are favored in polar aprotic solvents (e.g.,acetone,$DMSO$,$DMF$) because they do not solvate the nucleophile strongly.
- $S_{N}1$ reactions are favored in polar protic solvents (e.g.,$H_2O$,alcohols) because they stabilize the carbocation intermediate and the leaving group through solvation.

(A) To prepare $2-$methylpropane by the Wurtz reaction,we need to couple a methyl group with an isopropyl group. Therefore,the haloalkanes required are chloromethane $(CH_3Cl)$ and $2-$chloropropane $(CH_3CHClCH_3)$.
The reaction is as follows:
$CH_3Cl + CH_3CHClCH_3 + 2Na \xrightarrow{\text{dry ether}} CH_3CH(CH_3)CH_3 + 2NaCl$
Thus,the correct option is $A$.

(C) The reaction sequence is as follows:
$1$. Aniline reacts with $NaNO_2$ and $\text{Dil. } HCl$ at $0-5^{\circ}C$ to form benzenediazonium chloride $(P)$:
$C_6H_5NH_2 + NaNO_2 + 2HCl \rightarrow C_6H_5N_2^+Cl^- + NaCl + 2H_2O$
$2$. Benzenediazonium chloride $(P)$ then undergoes an electrophilic substitution (coupling reaction) with phenol in the presence of $NaOH$ ($pH$ $9-10$). The coupling occurs primarily at the $para$-position due to steric hindrance at the $ortho$-position,forming $p$-hydroxyazobenzene $(Q)$:
$C_6H_5N_2^+Cl^- + C_6H_5OH \xrightarrow{NaOH} C_6H_5-N=N-C_6H_4-OH(p) + HCl$

(C) $NO_2$ (Nitrogen dioxide) gas is brown in colour and acidic in nature.

(C) $HF$ exhibits strong intermolecular hydrogen bonding,which results in a significantly higher boiling point compared to other hydrogen halides.
For the remaining hydrogen halides $(HCl, HBr, HI)$,the boiling point is primarily determined by the magnitude of van der Waals' forces,which increase with an increase in molecular mass.
Therefore,the order of boiling points is $HI > HBr > HCl$.
Combining these,the overall decreasing order of boiling points is $HF > HI > HBr > HCl$.

(C) Fibres are thread-forming solids that possess high tensile strength and high modulus. These characteristics are due to strong intermolecular forces like hydrogen bonds or dipole-dipole interactions.
Among the given polymers:
$(i)$ Teflon is an elastomer/plastic.
$(ii)$ Starch is a natural polymer (carbohydrate).
$(iii)$ Terylene is a polyester fibre.
$(iv)$ Orlon (polyacrylonitrile) is a synthetic fibre.
Therefore,the correct set is $(iii)$ and $(iv)$.

(C) The biodegradable polymer nylon-$2$-nylon-$6$ is an alternating polyamide copolymer of glycine $[H_2N-CH_2-COOH]$ and aminocaproic acid $[H_2N-(CH_2)_5-COOH]$.

(B) The given unit cell parameters $a=b \neq c$ and $\alpha=\beta=90^{\circ}$,$\gamma=120^{\circ}$ correspond to the hexagonal crystal system.
Among the given options,graphite crystallizes in the hexagonal system.

(D) For a face-centred cubic $(FCC)$ crystal, the number of atoms per unit cell is $Z = 4$.
Edge length $a = 300 \ pm = 300 \times 10^{-10} \ cm = 3 \times 10^{-8} \ cm$.
Density $d = 10 \ g \ cm^{-3}$.
Using the density formula: $d = \frac{Z \times M}{N_A \times a^3}$, where $M$ is the molar mass.
$M = \frac{d \times N_A \times a^3}{Z} = \frac{10 \times 6.022 \times 10^{23} \times (3 \times 10^{-8})^3}{4}$.
$M = \frac{10 \times 6.022 \times 10^{23} \times 27 \times 10^{-24}}{4} = \frac{162.594}{4} = 40.6485 \ g \ mol^{-1}$.
Now, the number of atoms in $4.5 \ g$ is calculated as:
$\text{Number of atoms} = \frac{\text{mass}}{\text{molar mass}} \times N_A = \frac{4.5}{40.6485} \times 6.022 \times 10^{23} \approx 6.66 \times 10^{22}$ atoms.

(B) Relative lowering of vapour pressure is equal to the mole fraction of glucose.
$\frac{p_0 - p_s}{p_0} = \chi_{\text{glucose}}$ ...$(I)$
Number of moles of glucose $= \frac{18 \ g}{180 \ g/mol} = 0.1 \ mol$
Number of moles of water $= \frac{178.2 \ g}{18 \ g/mol} = 9.9 \ mol$
Mole fraction of glucose $(\chi_{\text{glucose}})$ $= \frac{0.1}{0.1 + 9.9} = \frac{0.1}{10} = 0.01$
Substituting the value of mole fraction in Eq. $(I)$:
$\frac{760 - p_s}{760} = 0.01$
$760 - p_s = 7.6$
$p_s = 760 - 7.6 = 752.4 \ torr$
Rounding to the nearest option,$p_s \approx 752.0 \ torr$.

(C) mixture of phenol $(C_6H_5OH)$ and aniline $(C_6H_5NH_2)$ shows negative deviation from Raoult's law.
This occurs because the intermolecular hydrogen bonding between the phenolic hydrogen and the lone pair of nitrogen in aniline is stronger than the individual intermolecular forces present in the pure components.
This stronger interaction leads to a decrease in the tendency of molecules to escape into the vapor phase,resulting in a lower vapor pressure than expected.

(B) Positive deviation from Raoult's law occurs when the solute-solvent intermolecular forces are weaker than the solute-solute and solvent-solvent forces.
In the mixture of benzene and methanol,the hydrogen bonding between methanol molecules is disrupted by the addition of non-polar benzene,leading to weaker interactions in the solution.
Therefore,the vapour pressure of the solution is higher than that predicted by Raoult's law.
Water $-$ $HCl$ and Water $-$ $HNO_3$ show negative deviation due to strong hydrogen bonding,and Acetone $-$ Chloroform shows negative deviation due to hydrogen bond formation between them.

(A) Gold sols are formed by the aggregation of a large number of gold atoms,making them $multimolecular$ colloids.
They are $lyophobic$ in nature and carry a $negative$ charge due to the adsorption of $OH^-$ ions.
They are not $macromolecular$ colloids,as macromolecular colloids consist of large molecules like proteins or starch.

(A) Cationic detergents are quaternary ammonium salts of amines with acetates,chlorides,or bromides as anions.
Among the given options,$Cetyltrimethylammonium \text{ } bromide$ is a popular cationic detergent.
It is commonly used in hair conditioners.

(D) The incorrect statement is given in option $(D)$.
Brownian movement is caused by the unbalanced bombardment of molecules of the dispersion medium on the colloidal particles,which results in a zig-zag motion.
Therefore,the statement claiming it is due to balanced bombardment is incorrect.

(D) According to the $Hardy-Schulze$ rule,the coagulating (or flocculating) power of an ion depends on its valency.
For a positively charged sol,the coagulating power of the anions increases with the increase in the magnitude of their negative charge.
The valencies of the given ions are: $[Fe(CN)_6]^{4-}$ $(4-)$,$PO_4^{3-}$ $(3-)$,$SO_4^{2-}$ $(2-)$,and $Cl^{-}$ $(1-)$.
Therefore,the order of flocculating power is $[Fe(CN)_6]^{4-} > PO_4^{3-} > SO_4^{2-} > Cl^{-}$.