AIIMS 2011 Chemistry Question Paper with Answer and Solution

(C) The $Spleen$ is known as the 'blood bank' of the human body.
It acts as a reservoir for blood,storing red blood cells and platelets.
In times of emergency,such as severe hemorrhage,the spleen can release these stored blood cells into the circulation to help maintain blood volume and oxygen-carrying capacity.

(C) In the given complex $[Fe(CN)_6]^{3-}$,the oxidation state of iron is $+3$.
The electronic configuration of $Fe^{3+}$ is $[Ar] 3d^5$.
Since $CN^-$ is a strong field ligand,it causes pairing of electrons in the $3d$ orbitals.
According to Crystal Field Theory,the $5$ electrons in $3d$ orbitals are arranged as $t_{2g}^5 e_g^0$,resulting in $1$ unpaired electron.
This complex has an octahedral geometry and is a low-spin complex.
Due to the strong field ligand $CN^-$,the complex is very stable.

(A) When a charged particle is released from rest in a region where electric field $\vec{E}$ and magnetic field $\vec{B}$ are parallel to each other,the magnetic force $\vec{F}_m = q(\vec{v} \times \vec{B})$ is zero because the initial velocity $\vec{v} = 0$.
As the particle starts moving due to the electric force $\vec{F}_e = q\vec{E}$,its velocity $\vec{v}$ becomes parallel to the electric field $\vec{E}$.
Since $\vec{E}$ is parallel to $\vec{B}$,the velocity $\vec{v}$ remains parallel to the magnetic field $\vec{B}$ at all times.
Because $\vec{v} \parallel \vec{B}$,the magnetic force $\vec{F}_m = q(\vec{v} \times \vec{B})$ remains zero throughout the motion.
The particle is only acted upon by the electric force,which causes it to accelerate in a straight line along the direction of the electric field.

(A) During the collision of two elastic bodies,the bodies undergo deformation.
At the point of maximum deformation,the relative velocity of the bodies becomes zero.
At this instant,a part of the kinetic energy of the system is converted into elastic potential energy due to the deformation of the bodies.
Therefore,the kinetic energy of the system decreases during the time of collision.
This decrease in kinetic energy is stored as elastic potential energy because the intermolecular distances change (decrease) during deformation.
Thus,both the $Assertion$ and $Reason$ are correct,and the $Reason$ is the correct explanation for the $Assertion$.

(A) In the absence of an electric current,the free electrons in a conductor are in a state of random motion,similar to molecules in a gas. Their average velocity is zero,meaning they do not have any net velocity in a specific direction. As a result,there is no net magnetic force on the free electrons in a magnetic field. When a current is passed,the free electrons acquire a drift velocity in a definite direction,and consequently,a magnetic force acts on them (provided the magnetic field has a component perpendicular to the direction of motion).

(A) The magnetic susceptibility of a ferromagnetic substance does not follow Curie's law $( \chi \propto 1/T )$ below the Curie temperature $( T_C )$.
Instead, it follows the Curie-Weiss law: $ \chi = \frac{C}{T - T_C} $ for $ T > T_C $.
At the Curie point $( T_C )$, the ferromagnetic substance undergoes a phase transition and begins to behave as a paramagnetic substance.
Since the assertion states that ferromagnetic substances do not obey Curie's law (which is true for the ferromagnetic state itself), and the reason correctly explains the transition at the Curie point, both are correct and the reason explains the assertion.

(C) The normality of a solution is given by the formula: $\text{Normality} = \text{Molarity} \times \text{n-factor}$.
For $H_3PO_3$ (phosphorous acid),the structure contains two $P-OH$ bonds,making it a dibasic acid. Thus,its n-factor is $2$.
Normality $= 0.3 \ M \times 2 = 0.6 \ N$. Therefore,the Assertion is correct.
The equivalent weight of an acid is defined as $\frac{\text{Molecular weight}}{\text{Basicity}}$.
Since the basicity of $H_3PO_3$ is $2$,its equivalent weight is $\frac{\text{Molecular weight}}{2}$.
The Reason states the denominator is $3$,which is incorrect. Therefore,the Reason is incorrect.

(D) For an orbital with $n = 3$ and $l = 1$,the orbital is $3p$.
The $p-$orbitals have a dumb-bell shape,not a double dumb-bell shape.
Double dumb-bell shape is characteristic of $d-$orbitals $(l = 2)$.
Therefore,the Assertion is incorrect.
The Reason states it belongs to the $p-$subshell,which is correct for $l = 1$.
Thus,the Assertion is incorrect but the Reason is correct.

(C) The Assertion is correct because elements tend to lose electrons to achieve a stable noble gas configuration.
The Reason is incorrect because ionization enthalpy is defined as the energy $required$ (absorbed) to remove an electron from an isolated gaseous atom,not the energy $released$.

(C) The bond angle of $H_2S$ $(92^o)$ is smaller than that of $H_2O$ $(104.5^o)$.
As the electronegativity of the central atom decreases,the bond pairs move further away from the central atom,reducing the repulsion between them and leading to a smaller bond angle.
Since oxygen is more electronegative than sulfur,the bond pairs in $H_2O$ are closer to the central atom,resulting in greater repulsion and a larger bond angle.
Therefore,the Assertion is correct,but the Reason is incorrect because the bond angle actually increases as the electronegativity of the central atom increases.

(B) Critical temperature $(T_C)$ is the maximum temperature at which a gas can be liquefied by the application of pressure alone.
Below the critical temperature,the thermal motion of the molecules is slow enough for the intermolecular forces to come into play,leading to the condensation of the gas.
At the critical temperature,the liquid and gaseous phases become indistinguishable.
Ideal gases do not have a characteristic critical temperature because they lack intermolecular forces of attraction.
Therefore,the correct statement is that below the critical temperature,thermal motion of the molecules is slow enough for the intermolecular forces to come into play leading to condensation of the gas.

(C) Entropy is a measure of the randomness or disorder of a system. $\Delta S < 0$ implies a decrease in entropy,which occurs when a system becomes more ordered.
In option $A$,solid to gas transition increases disorder.
In option $B$,dissolution increases entropy due to mixing.
In option $D$,mixing of gases increases entropy.
In option $C$,two moles of gaseous reactants form one mole of solid product. Since the solid state is much more ordered than the gaseous state,the entropy of the system decreases.

(C) The difference in enthalpy of combustion for each successive $CH_2$ group is calculated as follows:
$\Delta H_c(C_2H_6) - \Delta H_c(CH_4) = -368.4 - (-210.8) = -157.6 \ k \ cal \ mol^{-1}$
$\Delta H_c(C_3H_8) - \Delta H_c(C_2H_6) = -526.2 - (-368.4) = -157.8 \ k \ cal \ mol^{-1}$
Average value for $\Delta H_c(-CH_2-) = \frac{-157.6 + (-157.8)}{2} = -157.7 \ k \ cal \ mol^{-1}$
For hexane $(C_6H_{14})$,we add three $-CH_2-$ groups to propane $(C_3H_8)$:
$\Delta H_c(C_6H_{14}) = \Delta H_c(C_3H_8) + 3 \times \Delta H_c(-CH_2-)$
$\Delta H_c(C_6H_{14}) = -526.2 + 3(-157.7) = -526.2 - 473.1 = -999.3 \ k \ cal \ mol^{-1}$
Rounding to the nearest integer,the value is $-1000 \ k \ cal \ mol^{-1}$.

(B) For an isothermal process,the temperature remains constant. Since internal energy $(\Delta U)$ of an ideal gas depends only on temperature,$\Delta U = 0$.
According to the first law of thermodynamics,$\Delta U = Q + W$.
Substituting $\Delta U = 0$,we get $0 = Q + W$,which implies $Q = -W$.
This means the heat absorbed by the system is equal to the work done by the system.
However,for an isothermal process,enthalpy change $\Delta H = \Delta U + \Delta(PV) = 0 + \Delta(nRT)$. Since $n, R, T$ are constant,$\Delta H = 0$ for an ideal gas.
While $\Delta H = 0$ is true for an ideal gas in an isothermal process,the reason for $Q = -W$ is the first law of thermodynamics $(\Delta U = 0)$,not the fact that $\Delta H = 0$.

(C) For an isothermal process,the temperature remains constant. Since internal energy $(\Delta U)$ of an ideal gas is a function of temperature only,$\Delta U = 0$.
According to the first law of thermodynamics,$\Delta U = Q + W$.
Substituting $\Delta U = 0$,we get $0 = Q + W$,which implies $Q = -W$.
This means the heat absorbed by the system is equal to the work done by the system.
However,for an isothermal process,enthalpy change $\Delta H = \Delta U + \Delta(PV)$. For an ideal gas,$\Delta H = \Delta U + \Delta(nRT) = 0 + nR\Delta T = 0$. While $\Delta H$ is zero for an ideal gas,the reason provided is not the fundamental explanation for $Q = -W$,which is based on the first law and $\Delta U = 0$. In general,$\Delta H$ is not necessarily zero for all isothermal processes (e.g.,real gases). Thus,the Assertion is correct,but the Reason is incorrect.

(C) The reaction $N_{2(g)} + 3H_{2(g)} \longleftrightarrow 2NH_{3(g)}$ is exothermic $(\Delta H < 0)$ and involves a decrease in the number of moles of gaseous species (from $4$ moles to $2$ moles).
According to Le Chatelier's principle,high pressure favors the forward reaction.
Although low temperature favors the forward reaction for an exothermic process,the reaction has a high activation energy $(E_a)$.
Therefore,in a continuous flow process,an elevated optimum temperature is required to ensure a sufficient rate of reaction to produce $NH_3$ efficiently.

(C) The mixture contains a strong acid $(HCl)$ and a neutral salt $(NaCl)$,resulting in an acidic solution with $pH < 7$.
$(B)$ The reaction is $H_2SO_4 + 2NaOH \rightarrow Na_2SO_4 + 2H_2O$. Here,$50 \ mL \times 0.1 \ M \times 2 \ (\text{for } H^+) = 10 \ mmol$ of $H^+$ and $50 \ mL \times 0.2 \ M = 10 \ mmol$ of $OH^-$. This is complete neutralization,resulting in $pH = 7$.
$(C)$ The reaction is $CH_3COOH + KOH \rightarrow CH_3COOK + H_2O$. This results in the formation of $CH_3COOK$,which is a salt of a weak acid and a strong base. It undergoes anionic hydrolysis,resulting in a basic solution with $pH > 7$.
$(D)$ The reaction is $HNO_3 + NH_3 \rightarrow NH_4NO_3$. This results in the formation of $NH_4NO_3$,which is a salt of a strong acid and a weak base. It undergoes cationic hydrolysis,resulting in an acidic solution with $pH < 7$.

(C) The reaction quotient $(Q_c)$ is defined for a general reaction $aA + bB \rightleftharpoons cC + dD$ as $Q_c = \frac{[C]^c [D]^d}{[A]^a [B]^b}$ at any stage of the reaction,which is the same expression as the equilibrium constant $(K_c)$. Thus,the Assertion is correct.
If $Q_c < K_c$,the ratio of products to reactants is less than the equilibrium ratio,so the reaction proceeds in the forward direction (towards products) to reach equilibrium. The Reason states the reaction moves towards reactants,which is incorrect. Therefore,the Assertion is correct but the Reason is incorrect.

(C) The Assertion is correct because $KOH$ has a higher solubility in water compared to $NaOH$ due to the smaller hydration energy of $Na^+$ compared to $K^+$ and the lattice energy differences.
However,the Reason is incorrect because $KOH$ is a stronger base than $NaOH$. This is because the $K-OH$ bond is weaker than the $Na-OH$ bond,allowing $KOH$ to dissociate more readily into $K^+$ and $OH^-$ ions in aqueous solution.

(B) The melting point of alkaline earth metal chlorides depends on their ionic character.
As the size of the cation increases from $Be^{2+}$ to $Ba^{2+}$,the polarizing power of the cation decreases,leading to an increase in the ionic character of the $M-Cl$ bond.
Greater ionic character results in stronger electrostatic forces of attraction,which increases the melting point.
Therefore,the order of increasing melting points is $BeCl_2 < MgCl_2 < CaCl_2 < SrCl_2 < BaCl_2$.

(D) $Be$ (Beryllium) is the first member of the alkaline earth metals and exhibits anomalous properties compared to the rest of the group. Due to its small size and high charge density,it shows a diagonal relationship with $Al$ (Aluminium). Both $Be$ and $Al$ have an electronegativity value of approximately $1.5$ on the Pauling scale.

(C) The property of sodium atom to emit photons in the yellow region of the visible spectrum,due to electrically stimulated electron transitions,is used in street lights. This is why sodium vapor lamps emit a characteristic yellow light.

(C) $1$. Identify the longest carbon chain. The longest chain in the given structure contains $7$ carbon atoms (heptane).
$2$. Number the chain from the end that gives the lowest locants to the substituents. Numbering from right to left gives the substituents at positions $2, 3, 4,$ and $5$.
$3$. The substituents are three methyl groups at positions $2, 3,$ and $5$,and a propyl group at position $4$.
$4$. Combining these,the correct $IUPAC$ name is $2,3,5-$trimethyl$-4-$propylheptane.

(B) In the structure $R'-CH=O^{+}R$,all atoms (carbon and oxygen) have a complete octet.
In the structure $R'-C^{+}H-OR$,the carbon atom has an incomplete octet ($6$ electrons).
Resonance structures where all atoms have complete octets are significantly more stable than those with incomplete octets.

(C) $1-$Butyne is a terminal alkyne $(CH_3CH_2C\equiv CH)$,which contains an acidic hydrogen atom attached to the $sp$ hybridized carbon.
Tollen's reagent (ammoniacal silver nitrate) reacts with terminal alkynes to form a white precipitate of silver alkynide,whereas $2-$butyne $(CH_3C\equiv CCH_3)$ is an internal alkyne and does not react with Tollen's reagent.
The reaction is:
$CH_3CH_2C\equiv CH + [Ag(NH_3)_2]^+ + OH^- \to CH_3CH_2C\equiv CAg \downarrow (\text{white ppt.}) + 2NH_3 + H_2O$

(D) The molecular formula $C_4H_6$ corresponds to the degree of unsaturation $2$.
Ozonolysis of compound $Y$ yields only ethanoic acid $(CH_3COOH)$,which implies that $Y$ is $CH_3CH=CHCH_3$ (but$-2-$ene).
All three given compounds,when treated with one equivalent of $H_2$ in the presence of $Pt$,can form $CH_3CH=CHCH_3$:
$1$. $CH_2=CH-CH=CH_2 + H_2 \xrightarrow{Pt} CH_3CH=CHCH_3$
$2$. $CH_3C \equiv CCH_3 + H_2 \xrightarrow{Pt} CH_3CH=CHCH_3$
$3$. $CH_2=C=CHCH_3 + H_2 \xrightarrow{Pt} CH_3CH=CHCH_3$
Finally,$CH_3CH=CHCH_3 + O_3 \rightarrow 2CH_3COOH$.
Therefore,all three compounds satisfy the given conditions.

(C) Dicumarol is a natural anticoagulant found in spoilt sweet clover hay. It acts as a vitamin $K$ antagonist. Vitamin $K$ is essential for the synthesis of clotting factors in the liver. When cattle consume this hay,the action of vitamin $K$ is inhibited,leading to a deficiency in clotting factors,which results in prolonged bleeding and hemorrhagic syndrome.

(D) The reaction sequence is as follows:
$1$. Aniline $(C_6H_5NH_2)$ reacts with $Br_2$ in $CH_3COOH$ to form $2,4,6$-tribromoaniline $(Y)$.
$2$. $2,4,6$-tribromoaniline $(Y)$ reacts with $NaNO_2/HCl$ at $0-5 \ ^\circ C$ to form $2,4,6$-tribromobenzenediazonium chloride $(Z)$.
$3$. $2,4,6$-tribromobenzenediazonium chloride $(Z)$ reacts with ethanol $(C_2H_5OH)$ upon heating to form $1,3,5$-tribromobenzene.
Thus,the starting compound $[X]$ is aniline.

(D) Two solutions are isotonic if their osmotic pressures are equal at the same temperature. Osmotic pressure $\pi = iCRT$,where $i$ is the van't Hoff factor,$C$ is the molarity,$R$ is the gas constant,and $T$ is the temperature. For isotonic solutions,$i_1C_1 = i_2C_2$.
For option $D$:
For $0.1 \ M$ $BaCl_2$,$i = 3$ (as $BaCl_2 \rightarrow Ba^{2+} + 2Cl^-$),so $iC = 3 \times 0.1 = 0.3 \ M$.
For $0.075 \ M$ $FeCl_3$,$i = 4$ (as $FeCl_3 \rightarrow Fe^{3+} + 3Cl^-$),so $iC = 4 \times 0.075 = 0.3 \ M$.
Since $i_1C_1 = i_2C_2$,these solutions are isotonic.

(B) $1$. The metal salt solution reacts with potassium chromate $(K_2CrO_4)$ in acetic acid to form a yellow precipitate. This indicates the presence of $Ba^{2+}$ or $Pb^{2+}$ ions,as both form yellow chromates ($BaCrO_4$ and $PbCrO_4$).
$2$. It forms a white precipitate with dilute $H_2SO_4$,which indicates the presence of $Ba^{2+}$,$Pb^{2+}$,or $Ca^{2+}$ (as $BaSO_4$,$PbSO_4$,and $CaSO_4$ are white).
$3$. It gives no precipitate with $NaCl$. $Pb^{2+}$ forms a white precipitate of $PbCl_2$ with $NaCl$,whereas $Ba^{2+}$ does not form a precipitate with $NaCl$ because $BaCl_2$ is soluble.
$4$. Therefore,the metal salt solution contains $Ba^{2+}$ ions,such as $Ba(NO_3)_2$.

(B) For an $fcc$ lattice, the atoms touch along the face diagonal.
The relation between the edge length $(a)$ and the radius $(r)$ is given by $4r = a \sqrt{2}$.
The diameter $(d)$ of the atom is $2r$.
Therefore, $d = \frac{a \sqrt{2}}{2} = \frac{a}{\sqrt{2}}$.
Given $a = 407 \ pm$, we have $d = \frac{407}{1.414} \approx 287.8 \ pm$.

(B) In a Schottky defect,an equal number of cations and anions are missing from their lattice sites to maintain electrical neutrality and stoichiometry.
This creates vacancies,which causes the density of the crystal to decrease.
Since the formation of these defects is an endothermic process,the lattice energy of the crystal actually increases,not decreases.
Therefore,the statement that lattice energy decreases is incorrect.
The defect does increase electrical conductivity due to the migration of ions into the vacant holes.

(A) In a Frenkel defect,an ion (usually a smaller cation) leaves its lattice site and occupies an interstitial site within the same crystal.
Since no ions leave the crystal lattice entirely,the total mass and volume of the crystal remain unchanged.
Therefore,the density of the crystalline solid remains constant.
Both the Assertion and the Reason are correct,and the Reason correctly explains why the density remains unchanged.

(D) For $NaCl$,the van't Hoff factor $i_1 = 2$. The depression in freezing point is given by $\Delta T_f = i_1 K_f m = 0.372 \, K$.
Thus,$m = \frac{0.372}{2 \times 1.86} = 0.1 \, mol \, kg^{-1}$.
For $BaCl_2$,the van't Hoff factor $i_2 = 3$.
The elevation in boiling point is $\Delta T_b = i_2 K_b m = 3 \times 0.52 \times 0.1 = 0.156 \, K$.
The boiling point of the solution is $T_b = 100 + \Delta T_b = 100 + 0.156 = 100.156 \, ^oC$.

(D) According to the thermodynamic relationship between Raoult's law and Henry's law,if one component of a binary solution obeys Raoult's law $(P_i = x_i P_i^o)$ over the entire range of composition,the other component must also obey Raoult's law.
However,in dilute solutions,the solvent obeys Raoult's law $(P_1 = x_1 P_1^o)$ while the solute obeys Henry's law $(P_2 = K_H x_2)$.
The Assertion is incorrect because if one component obeys Raoult's law,the other component often obeys Henry's law in the dilute range.
The Reason is correct because Raoult's law is indeed a special case of Henry's law where the Henry's constant $(K_H)$ becomes equal to the pure component vapor pressure $(P_i^o)$.

(A) According to Kohlrausch's law,the molar conductance at infinite dilution is the sum of the ionic conductances of the constituent ions.
$\lambda_m^\infty (BaSO_4) = \lambda_{Ba^{2+}}^\infty + \lambda_{SO_4^{2-}}^\infty$
We can express this using the given values:
$\lambda_m^\infty (BaSO_4) = \lambda_m^\infty (BaCl_2) + \lambda_m^\infty (H_2SO_4) - 2\lambda_m^\infty (HCl)$
$\lambda_m^\infty (BaSO_4) = x_1 + x_2 - 2x_3$
Equivalent conductance $(\lambda_e^\infty)$ is related to molar conductance $(\lambda_m^\infty)$ by the formula $\lambda_e^\infty = \frac{\lambda_m^\infty}{n}$,where $n$ is the valency factor. For $BaSO_4$,$n = 2$.
Therefore,$\lambda_e^\infty (BaSO_4) = \frac{x_1 + x_2 - 2x_3}{2}$.

(B) reducing agent is a substance that undergoes oxidation (loses electrons). The strength of a reducing agent is determined by its standard oxidation potential,which is the negative of its standard reduction potential $(E^o_{ox} = -E^o_{red})$.
Comparing the standard reduction potentials $(E^o_{red})$:
$E^o_{K^{+}/K} = -2.93 \ V$
$E^o_{Zn^{2+}/Zn} = -0.76 \ V$
$E^o_{Fe^{2+}/Fe} = -0.44 \ V$
$E^o_{Cu^{2+}/Cu} = 0.34 \ V$
The species with the most negative standard reduction potential is the strongest reducing agent because it has the highest tendency to lose electrons.
Since $E^o_{K^{+}/K} = -2.93 \ V$ is the most negative value,$K_{(s)}$ is the strongest reducing agent.

(A) According to Kohlrausch law of independent migration of ions,the limiting molar conductivity of an electrolyte is the sum of the individual contributions of the anion and cation of the electrolyte.
Thus,for $NaCl$,$\Lambda^o_{NaCl} = \lambda^o_{Na^{+}} + \lambda^o_{Cl^{-}}$.
Both the Assertion and the Reason are correct,and the Reason is the correct explanation of the Assertion.

(B) For a $10 \ K$ rise in temperature,the collision frequency increases only by about $1 \%$ to $2 \%$.
However,the fraction of molecules possessing energy equal to or greater than the threshold energy increases significantly,typically by a factor of $2$ to $3$.
This leads to a doubling or tripling of the rate constant.

(D) The stoichiometry of the reaction is $2N_2O_5 \to 4NO_2 + O_2$.
From the stoichiometry,$1 \ mol$ of $O_2$ is formed from $2 \ mol$ of $N_2O_5$.
Therefore,the concentration of $N_2O_5$ reacted when $[O_2] = 0.1 \ mol \ L^{-1}$ is $2 \times 0.1 = 0.2 \ mol \ L^{-1}$.
The concentration of $N_2O_5$ remaining is $[N_2O_5] = 1.0 - 0.2 = 0.8 \ mol \ L^{-1}$.
The rate of reaction is given by $Rate = k[N_2O_5] = 3.0 \times 10^{-4} \ s^{-1} \times 0.8 \ mol \ L^{-1} = 2.4 \times 10^{-4} \ mol \ L^{-1} \ s^{-1}$.
From the rate expression,$Rate = -\frac{1}{2} \frac{d[N_2O_5]}{dt} = \frac{1}{4} \frac{d[NO_2]}{dt} = \frac{d[O_2]}{dt}$.
Thus,the rate of formation of $NO_2$ is $\frac{d[NO_2]}{dt} = 4 \times Rate = 4 \times 2.4 \times 10^{-4} \ mol \ L^{-1} \ s^{-1} = 9.6 \times 10^{-4} \ mol \ L^{-1} \ s^{-1}$.

(A) The rate expression $\frac{dX}{dt} = k[A]^m[B]^n$ indicates that the reaction rate depends only on the concentrations of reactants $A$ and $B$.
The order of the reaction with respect to $C$ is $0$,which means the rate of the reaction is independent of the concentration of $C$.
Since the rate law is experimentally determined and explicitly excludes $C$,the Reason correctly explains why $C$ does not appear in the rate expression.

(D) The reaction between $FeCl_3$ and excess $NaOH$ produces a negatively charged ferric hydroxide sol: $FeCl_3 + 3NaOH \to Fe(OH)_3(s) + 3NaCl$.
Due to the adsorption of $OH^-$ ions from the excess $NaOH$,the sol is negatively charged: $[Fe(OH)_3]OH^-$.
According to the Hardy-Schulze rule,the coagulating power of an ion is directly proportional to its valency. For a negatively charged sol,the coagulating power increases with the valency of the cation.
The valencies of the given cations are: $Ba^{2+}$ $(+2)$ and $Al^{3+}$ $(+3)$.
Since $Al^{3+}$ has the highest valency among the cations,it has the maximum coagulating power and,therefore,the minimum coagulating value.

(D) In the Hall-Heroult process,pure alumina $(Al_2O_3)$ has a very high melting point (approx. $2323 \ K$),which makes it difficult to melt and consume a lot of energy.
Cryolite $(Na_3AlF_6)$ is added to the molten alumina to lower its melting point to about $1140 \ K$ and to increase its electrical conductivity.
Therefore,the correct option is $D$.

(B) The central atom in $XeOF_4$ is Xenon $(Xe)$.
$Xe$ has $8$ valence electrons.
In $XeOF_4$,$Xe$ forms $4$ single bonds with $4$ Fluorine atoms and $1$ double bond with $1$ Oxygen atom.
Total electrons used in bonding = $4 + 2 = 6$ electrons.
Remaining electrons on $Xe$ = $8 - 6 = 2$ electrons.
These $2$ electrons form $1$ lone pair.
Therefore,the total number of lone pair of electrons on the central atom $Xe$ is $1$.

(C) $XeF_6$ reacts with silica $(SiO_2)$ present in glass or quartz vessels to form explosive xenon oxyfluorides and eventually $XeO_3$.
The reaction is as follows:
$2XeF_6 + SiO_2 \rightarrow 2XeOF_4 + SiF_4$
$2XeOF_4 + SiO_2 \rightarrow 2XeO_2F_2 + SiF_4$
$2XeO_2F_2 + SiO_2 \rightarrow 2XeO_3 + SiF_4$
$XeO_3$ is a highly explosive solid. Therefore,$XeF_6$ cannot be stored in glass or quartz.

(C) When ozone $(O_3)$ is passed over a silver surface,it oxidizes silver to silver oxide $(Ag_2O)$,which is black in color.
The reaction is as follows:
$2Ag + O_3 \to Ag_2O + O_2$
Thus,the blackening of the silver surface is due to the formation of $Ag_2O$.

(B) The stability of oxidation states in $d$-block elements is often determined by the electronic configuration.
$Mn^{2+}$ has a $3d^5$ configuration,which is a half-filled stable configuration.
$Mn^{3+}$ has a $3d^4$ configuration,which is less stable than the half-filled $3d^5$ state.
Therefore,$Mn^{2+} > Mn^{3+}$ is the correct representation of stability.

(B) The structure of $Fe_2(CO)_9$ consists of two $Fe(CO)_3$ units joined by three bridging $CO$ ligands.
Additionally,there is a direct $Fe-Fe$ bond between the two iron atoms.
Therefore,the two iron atoms are linked directly along with $3\,CO$ molecules as bridging ligands.

(D) These two compounds are coordination isomers.
In the first complex,$[Co(NH_3)_6][Cr(NO_2)_6]$,the cation is $[Co(NH_3)_6]^{3+}$ and the anion is $[Cr(NO_2)_6]^{3-}$.
In the second complex,$[Cr(NH_3)_6][Co(NO_2)_6]$,the cation is $[Cr(NH_3)_6]^{3+}$ and the anion is $[Co(NO_2)_6]^{3-}$.
Upon electrolysis of their aqueous solutions,the metal ions present in the cationic complex will migrate to the cathode and be deposited.
For $[Co(NH_3)_6][Cr(NO_2)_6]$,$Co$ metal is deposited at the cathode.
For $[Cr(NH_3)_6][Co(NO_2)_6]$,$Cr$ metal is deposited at the cathode.
Thus,they can be distinguished by the electrolysis of their aqueous solutions.

(C) In the $[Fe(CN)_6]^{3-}$ complex,the central metal ion is $Fe^{3+}$,which has a $d^5$ electronic configuration.
$CN^-$ is a strong field ligand,causing pairing of electrons,resulting in one unpaired electron $(t_{2g}^5 e_g^0)$.
Since $CN^-$ is a strong field ligand and $Fe^{3+}$ has a high charge density,the crystal field splitting energy is high,making the complex very stable.

(C) Low spin complexes are formed when strong field ligands cause pairing of electrons,resulting in a smaller number of unpaired electrons.
In $[FeF_6]^{3-}$,the oxidation state of $Fe$ is $+3$ ($3d^5$ configuration).
$F^{-}$ is a weak field ligand,which does not cause pairing of electrons.
Therefore,$[FeF_6]^{3-}$ is a high spin complex with $5$ unpaired electrons.
Thus,the Assertion is correct,but the Reason is incorrect.

(C) The cyanide ion $(CN^{-})$ is indeed a strong nucleophile because the negative charge is present on the carbon atom,which is less electronegative than nitrogen,making it a good electron donor. Thus,the Assertion is correct.
However,chlorobenzene does not undergo nucleophilic substitution with $KCN$ under normal conditions because the $C-Cl$ bond in chlorobenzene acquires partial double bond character due to resonance,making it resistant to nucleophilic attack. Therefore,benzonitrile cannot be prepared by this method. Thus,the Reason is incorrect.

(C) The reaction of anisole $(C_6H_5OCH_3)$ with $HI$ involves the protonation of the oxygen atom,followed by the nucleophilic attack of the iodide ion $(I^-)$.
The $I^-$ ion attacks the less sterically hindered methyl group $(CH_3)$ to form methyl iodide $(CH_3I)$ and phenol $(C_6H_5OH)$.
Therefore,the Assertion is correct as the iodide ion attacks the smaller group.
However,the Reason states that it gives iodobenzene and methyl alcohol,which is incorrect; it actually gives phenol and methyl iodide.
Thus,the Assertion is correct but the Reason is incorrect.

(D) The Assertion is incorrect because alkylbenzene $CAN$ be prepared by Friedel-Crafts alkylation,although it has limitations such as polyalkylation and rearrangement.
The Reason is also incorrect because alkyl halides are generally more reactive or comparable in reactivity to acyl halides in the context of Friedel-Crafts reactions,but the primary issue with alkylation is the formation of polyalkylated products due to the activating nature of the alkyl group added to the benzene ring.
Therefore,both the Assertion and the Reason are incorrect.

(B) Reaction $(i)$ involves the hydrolysis of ethyl bromide with water,which is a slow reversible reaction and produces $HBr$,which can push the equilibrium backward.
Reaction $(ii)$ uses moist silver oxide $(Ag_2O + H_2O \longrightarrow 2AgOH)$. The $AgOH$ reacts with $CH_3CH_2Br$ to form $CH_3CH_2OH$ and $AgBr$. The precipitation of $AgBr$ drives the reaction to completion,making it a much more efficient and easier method for the preparation of ethanol.

(D) The haloform reaction requires the presence of a methyl ketone group $(-COCH_3)$ or a group that can be converted into one,such as a secondary alcohol $(-CH(OH)CH_3)$.
In the presence of a base and halogen,the $\alpha$-hydrogens are replaced by halogen atoms to form a trihalo derivative $(-COCX_3)$.
This intermediate then undergoes nucleophilic attack by $OH^-$,followed by the cleavage of the $C-C$ bond to release the haloform $(CHX_3)$.
Compound $ (i) \ CH_3CH_2COCH_2Cl $ has $\alpha$-hydrogens on the $CH_2Cl$ group,which can be further halogenated to $COCCl_3$.
Compound $ (ii) \ C_6H_5COCH_3 $ is a methyl ketone.
Compound $ (iii) \ C_6H_5COCHCl_2 $ already has a dihalo group and can be further halogenated to $COCCl_3$.
Compound $ (iv) \ CH_3CH_2COCCl_3 $ already contains the $COCCl_3$ group,which is the necessary intermediate for the haloform cleavage.
Therefore,all four compounds can undergo the haloform reaction.

(B) Tertiary alkyl halides like $CH_3-C(CH_3)_2-Cl$ undergo elimination $(E2)$ when treated with $NaCN$ (which acts as a base) to form isobutylene $(CH_2=C(CH_3)_2)$.
Subsequent hydration with $dil. H_2SO_4$ follows Markovnikov's rule to yield tert-butyl alcohol $(CH_3-C(CH_3)_2-OH)$.
Reaction:
$CH_3-C(CH_3)_2-Cl$ $\xrightarrow{NaCN} CH_2=C(CH_3)_2 (A)$ $\xrightarrow{dil. H_2SO_4} CH_3-C(CH_3)_2-OH (B)$

(D) Ethane-$1, 2$-dioic acid (oxalic acid) undergoes dehydration in the presence of concentrated $H_2SO_4$ upon heating.
The reaction is as follows:
$(COOH)_2 \xrightarrow[\text{conc. } H_2SO_4]{\Delta} CO + CO_2 + H_2O$
Thus,the products formed are carbon monoxide,carbon dioxide,and water.

(D) . Formyl chloride $(HCOCl)$ is unstable at room temperature and decomposes into $CO$ and $HCl$.
$B$. Acetamide $(CH_3CONH_2)$ is amphoteric; it can be protonated by strong acids to form $CH_3CONH_3^+Cl^-$ (base) and deprotonated by strong bases (acid).
$C$. Reduction of acetamide with $LiAlH_4$ yields ethylamine $(CH_3CH_2NH_2)$.
Since all statements $A$,$B$,and $C$ are correct,the correct answer is $D$.

(C) The Cannizzaro reaction of formaldehyde $(HCHO)$ involves the nucleophilic attack of the hydroxide ion $(OH^-)$ on the carbonyl carbon to form the hydroxyalkoxide monoanion $(HO-CH_2-O^-)$.
In the presence of a high concentration of base,a second $OH^-$ can deprotonate the hydroxyl group to form the dianion $(O^--CH_2-O^-)$.
Both of these species can act as intermediates that transfer a hydride ion to another formaldehyde molecule to complete the disproportionation process.

(C) The Assertion is correct because the negative charge on the acetate ion $(CH_3COO^-)$ is delocalized over two oxygen atoms through resonance,which stabilizes the ion.
The Reason is incorrect because the acetate ion is a much weaker base than the methoxide ion $(CH_3O^-)$. This is because $CH_3COOH$ is a much stronger acid $(pK_a \approx 4.75)$ than $CH_3OH$ $(pK_a \approx 15.5)$. Since the conjugate base of a stronger acid is always a weaker base,the acetate ion is less basic than the methoxide ion.

(C) The reaction $CH_3COCl + 2NH_3 \rightarrow CH_3CONH_2 + NH_4Cl$ is an example of nucleophilic acyl substitution. The Assertion is correct because $NH_3$ acts as a nucleophile to replace the $Cl$ atom.
However,the Reason is incorrect. While $Cl^{-}$ is indeed a good leaving group,it is a very weak nucleophile compared to $NH_3$. The strength of a nucleophile is generally inversely related to the strength of its conjugate acid. Since $HCl$ is a strong acid,$Cl^{-}$ is a weak base and a weak nucleophile.

(B) The Assertion is correct because aldol condensation involves the formation of a nucleophile (either an enolate ion in base or an enol in acid) which attacks the carbonyl carbon.
Both acids and bases can catalyze this process.
The Reason is also correct because $\beta$-hydroxyaldehydes or ketones (aldols) are unstable and undergo dehydration to form $\alpha,\beta$-unsaturated carbonyl compounds in the presence of an acid.
However,the Reason does not explain why the condensation itself is catalyzed by both acids and bases; it describes a subsequent step (dehydration).
Therefore,both are correct,but the Reason is not the correct explanation of the Assertion.

(B) Benzamide $(C_6H_5CONH_2)$ is an amide and is neutral in nature,so it does not react with cold dilute $NaOH$ or $HCl$.
Benzylamine $(C_6H_5CH_2NH_2)$ is a primary aliphatic amine and is basic in nature.
It reacts with cold dilute $HCl$ to form a salt,$C_6H_5CH_2NH_3^+Cl^-$,whereas benzamide does not react.
Therefore,cold dilute $HCl$ can be used to distinguish between them.

(A) In a non-polar solvent like chlorobenzene,the solvation effect (hydrogen bonding) is absent.
Therefore,the basicity is determined solely by the inductive effect ($+I$ effect) of the alkyl groups.
As the number of ethyl groups increases,the electron density on the nitrogen atom increases,making the amine more basic.
Thus,the order of basicity is: $C_2H_5NH_2 < (C_2H_5)_2NH < (C_2H_5)_3N$.

(C) Reduction of the ketonic group in fructose $(C-2)$ creates a new chiral center at $C-2$.
This results in the formation of two isomeric alcohols,specifically sorbitol and mannitol.
These two alcohols differ in configuration only at the $C-2$ position,making them $C-2$ epimers.
Since they are epimers and have the same molecular formula but different spatial arrangements,they are also classified as diastereomers.
Therefore,both $(a)$ and $(b)$ are correct.

(B) The Assertion is correct because proteins are polymers of $\alpha-$ amino acids linked by peptide bonds.
The Reason is also correct because denaturation involves the unfolding of the protein's secondary and tertiary structures due to physical or chemical changes,while the primary structure remains intact.
However,the Reason does not explain why proteins are made of $\alpha-$ amino acids.
Therefore,both statements are correct,but the Reason is not the correct explanation of the Assertion.

(B) Phosphorus oxoacids containing at least one $P-H$ bond act as strong reducing agents and can reduce $AgNO_3$ solution to metallic silver $(Ag)$,forming a silver mirror.
Among the given options,$H_4P_2O_5$ (pyrophosphorous acid) contains two $P-H$ bonds.
Therefore,it can reduce $AgNO_3$ to $Ag$.