AIIMS 2017 Chemistry Question Paper with Answer and Solution

(C) Using the ideal gas equation: $PV = nRT = \frac{m}{M}RT$
Here,$P = 2 \ atm$,$V = 100 \ L$,$T = 20 + 273 = 293 \ K$,$R = 0.0821 \ L \ atm \ K^{-1} \ mol^{-1}$,and $M$ (molar mass of $H_2S$) $= 34 \ g \ mol^{-1}$.
Rearranging for mass $m$: $m = \frac{MPV}{RT}$
$m = \frac{34 \times 2 \times 100}{0.0821 \times 293} \approx 282.4 \ g$.

(C) The molar mass of water $(H_2O)$ is $18 \ g \ mol^{-1}$.
Given,molar heat capacity $(C_p)$ = $75 \ J \ K^{-1} \ mol^{-1}$.
Specific heat capacity $(c)$ = $\frac{C_p}{\text{Molar mass}} = \frac{75}{18} \approx 4.17 \ J \ g^{-1} \ K^{-1}$.
Heat supplied $(Q)$ = $1.0 \ kJ = 1000 \ J$.
Mass of water $(m)$ = $100 \ g$.
Using the formula $Q = m \cdot c \cdot \Delta T$:
$1000 = 100 \times 4.17 \times \Delta T$.
$\Delta T = \frac{1000}{417} \approx 2.4 \ K$.

(C) Zeolite is represented as $Na_2Al_2Si_2O_8 \cdot xH_2O$.
When hard water containing $Ca^{2+}$ or $Mg^{2+}$ ions is passed through zeolite,the $Na^{+}$ ions are exchanged with $Ca^{2+}$ or $Mg^{2+}$ ions.
The reaction is: $Na_2Al_2Si_2O_8 \cdot xH_2O + Ca^{2+} \to CaAl_2Si_2O_8 \cdot xH_2O + 2Na^{+}$.

(B) Only alkyl aryl ethers,such as $C_6H_5OCH_3$ (anisole),undergo electrophilic substitution reactions.
In these compounds,the $-OCH_3$ group is an ortho/para-directing group and activates the benzene ring towards electrophilic substitution.

(B) The formula for heat supplied at constant pressure is $Q = n \times C_p \times \Delta T$.
Here,$n$ is the number of moles of water,$C_p$ is the molar heat capacity at constant pressure,and $\Delta T$ is the change in temperature.
First,calculate the number of moles of water: $n = \frac{\text{mass}}{\text{molar mass}} = \frac{100 \, g}{18 \, g \, mol^{-1}} = \frac{100}{18} \, mol$.
Given $Q = 1.0 \, kJ = 1000 \, J$ and $C_p = 75 \, J \, K^{-1} \, mol^{-1}$.
Substituting the values into the equation: $1000 = (\frac{100}{18}) \times 75 \times \Delta T$.
Solving for $\Delta T$: $\Delta T = \frac{1000 \times 18}{100 \times 75} = \frac{10 \times 18}{75} = \frac{180}{75} = 2.4 \, K$.

(C) The molar mass of water $(H_2O)$ is $18 \ g \ mol^{-1}$.
Number of moles of water $n = \frac{100 \ g}{18 \ g \ mol^{-1}} = 5.55 \ mol$.
Given heat $Q = 1.0 \ kJ = 1000 \ J$.
Using the formula $Q = n \times C_p \times \Delta T$:
$1000 = 5.55 \times 75 \times \Delta T$.
$1000 = 416.25 \times \Delta T$.
$\Delta T = \frac{1000}{416.25} \approx 2.4 \ K$.

(D) The red end of the hydrogen spectrum corresponds to the visible region,which is the Balmer series.
For the Balmer series,the lower energy level is $n_1 = 2$.
The lines in the Balmer series are transitions from $n_2 = 3, 4, 5, 6, \dots$ to $n_1 = 2$.
The first line (closest to the red end) is $3 \rightarrow 2$.
The second line is $4 \rightarrow 2$.
The third line from the red end is $5 \rightarrow 2$.
Therefore,the transition is $5 \rightarrow 2$.

(D) According to Faraday's law of electromagnetic induction,the induced electromotive force $(EMF)$ is given by $e = -\frac{d\phi}{dt}$.
Given the flux $\phi = 6t^2 - 5t + 1$,we differentiate with respect to time $t$:
$\frac{d\phi}{dt} = \frac{d}{dt}(6t^2 - 5t + 1) = 12t - 5$.
The induced current $i$ is given by $i = \frac{|e|}{R} = \frac{1}{R} \left| \frac{d\phi}{dt} \right|$.
Given resistance $R = 10\,\Omega$,at $t = 0.25\,s$:
$|e| = |12(0.25) - 5| = |3 - 5| = |-2| = 2\,V$.
Therefore,the magnitude of the induced current is $i = \frac{2\,V}{10\,\Omega} = 0.2\,A$.

(A) Nitrogen belongs to the second period and lacks $d$-orbitals in its valence shell. Due to the absence of $d$-orbitals,it cannot expand its octet to form pentahalides like $NF_5$. Therefore,$NF_5$ does not exist,whereas $PF_5$,$AsF_5$,and $SbF_5$ are stable.

(A) For the water not to spill at the highest point of a vertical circle,the gravitational force must provide the necessary centripetal force.
At the highest point,the condition for the minimum speed $V$ is given by:
$\frac{mV^2}{r} = mg$
Where $m$ is the mass of the water,$V$ is the velocity,$r$ is the radius of the circle $(1.6\, m)$,and $g$ is the acceleration due to gravity $(10\, m/sec^2)$.
Canceling $m$ from both sides,we get:
$V^2 = rg$
$V = \sqrt{rg}$
Substituting the given values:
$V = \sqrt{1.6 \times 10} = \sqrt{16} = 4\, m/sec$.
Thus,the minimum speed required is $4\, m/sec$.

(A) Nitrogen belongs to the second period and has a valence shell configuration of $2s^2 2p^3$.
It lacks $d$-orbitals in its valence shell,which prevents it from expanding its octet to form $5$ bonds.
Therefore,$NF_5$ cannot exist,whereas $P$,$As$,and $Sb$ possess vacant $d$-orbitals and can form pentahalides.

(B) Using the dilution formula $N_1 V_1 = N_2 V_2$:
Given $N_1 = 10 \ N$,$V_1 = 10 \ mL$,$N_2 = 0.1 \ N$.
Substituting the values: $10 \times 10 = 0.1 \times V_2$.
$V_2 = \frac{100}{0.1} = 1000 \ mL$.
This $V_2$ is the final total volume of the solution.
The volume of water to be added is $V_{added} = V_2 - V_1$.
$V_{added} = 1000 \ mL - 10 \ mL = 990 \ mL$.

(C) One mole of any substance contains $6.022 \times 10^{23}$ particles (Avogadro's number). Therefore,equal moles of different substances contain the same number of constituent particles. The Assertion is correct.
The number of particles in a given weight depends on the molar mass of the substance $(n = \frac{mass}{molar \ mass})$. Since different substances have different molar masses,equal weights of different substances will contain different numbers of moles and thus different numbers of constituent particles. The Reason is incorrect.

(A) The lines falling in the visible region of the hydrogen spectrum constitute the Balmer series,where the electron jumps to the $n_1 = 2$ orbit.
The lines are ordered by wavelength,with the red end representing the longest wavelength (lowest energy transition).
The first line is $3 \to 2$,the second line is $4 \to 2$,and the third line is $5 \to 2$.
Therefore,the third line from the red end corresponds to the transition $5 \to 2$.

(A) The radius of the $n^{th}$ orbit of a hydrogen-like species is given by the formula $r_n = 0.529 \ \mathring{A} \times \frac{n^2}{Z}$.
For the hydrogen atom,$Z = 1$.
For the first orbit,$n = 1$.
Substituting these values,$r_1 = 0.529 \ \mathring{A} \times \frac{1^2}{1} = 0.529 \ \mathring{A}$.
Thus,both the assertion and the reason are correct,and the reason provides the correct formula used to derive the assertion.

(A) According to the law of triads,the atomic weight of the middle element is approximately the arithmetic mean of the atomic weights of the first and third elements in the triad.
For the triad $Cl, Br, I$:
Atomic weight of $Br = \frac{\text{Atomic weight of } Cl + \text{Atomic weight of } I}{2} = \frac{35.5 + 127}{2} = 81.25 \approx 80$.

(D) According to Fajan's rule,the covalent character of an ionic bond is directly proportional to the polarizing power of the cation and the polarizability of the anion.The polarizing power of a cation increases as its size decreases and its charge increases.Comparing the cations in the given compounds: $N^{3+}$ (in $NCl_3$),$O^{2+}$ (in $Cl_2O$),$Pb^{2+}$ (in $PbCl_2$),and $Ba^{2+}$ (in $BaCl_2$).Among these,$Ba^{2+}$ has the largest ionic radius and the lowest charge density.Since $Ba^{2+}$ has the lowest polarizing power,$BaCl_2$ exhibits the least covalent character and the highest ionic character.

(C) Using the ideal gas equation,$PV = nRT$,where $n = \frac{m}{M}$.
Given:
$P = 2 \, atm$
$V = 100 \, L$
$T = 20 + 273 = 293 \, K$
$R = 0.0821 \, L \cdot atm \cdot K^{-1} \cdot mol^{-1}$
$M$ (molar mass of $H_2S$) $= 2(1) + 32 = 34 \, g/mol$.
Substituting the values:
$m = \frac{MPV}{RT} = \frac{34 \times 2 \times 100}{0.0821 \times 293} \approx 282.68 \, g$.

(C) Given: Molar heat capacity at constant pressure $C_p = 75 \ J \ K^{-1} \ mol^{-1}$.
Heat supplied $Q = 1 \ kJ = 1000 \ J$.
Mass of water $m = 100 \ g$.
Molar mass of water $M = 18 \ g \ mol^{-1}$.
Number of moles $n = \frac{m}{M} = \frac{100}{18} \ mol$.
The formula for heat supplied at constant pressure is $Q = n \times C_p \times \Delta T$.
Substituting the values: $1000 = (\frac{100}{18}) \times 75 \times \Delta T$.
$\Delta T = \frac{1000 \times 18}{100 \times 75} = \frac{180}{75} = 2.4 \ K$.

(B) The standard enthalpy change of a reaction is calculated using the formula: $\Delta H^o = \sum \Delta H_f^o(\text{products}) - \sum \Delta H_f^o(\text{reactants})$.
For the reaction $CO_{2(g)} + H_{2(g)} \to CO_{(g)} + H_2O_{(g)}$,the expression is:
$\Delta H^o = [\Delta H_f^o(CO_{(g)}) + \Delta H_f^o(H_2O_{(g)})] - [\Delta H_f^o(CO_{2(g)}) + \Delta H_f^o(H_{2(g)})]$.
Given that $\Delta H_f^o(H_{2(g)}) = 0 \ kJ/mol$ (standard state of an element):
$\Delta H^o = [-110.5 + (-241.8)] - [-393.5 + 0]$.
$\Delta H^o = -352.3 - (-393.5) = 41.2 \ kJ$.

(B) The relationship between $K_p$ and $K_c$ is given by $K_p = K_c(RT)^{\Delta n_g}$.
For the reaction $CO_{(g)} + \frac{1}{2} O_{2(g)} \to CO_{2(g)}$,the change in the number of moles of gaseous species is $\Delta n_g = n_{p(g)} - n_{r(g)} = 1 - (1 + 0.5) = -0.5$ or $-\frac{1}{2}$.
Substituting this into the equation,we get $K_p = K_c(RT)^{-\frac{1}{2}}$.
Therefore,the ratio $\frac{K_p}{K_c} = (RT)^{-\frac{1}{2}}$.

(B) The dissociation of $Mg(OH)_2$ is given by: $Mg(OH)_2(s) \leftrightarrow Mg^{2+}(aq) + 2OH^{-}(aq)$.
The solubility product expression is $K_{sp} = [Mg^{2+}][OH^{-}]^2$.
Given $K_{sp} = 1.0 \times 10^{-11}$ and $[Mg^{2+}] = 0.001\, M = 10^{-3}\, M$.
Substituting the values: $1.0 \times 10^{-11} = (10^{-3})[OH^{-}]^2$.
$[OH^{-}]^2 = \frac{1.0 \times 10^{-11}}{10^{-3}} = 10^{-8}$.
$[OH^{-}] = \sqrt{10^{-8}} = 10^{-4}\, M$.
Now,$pOH = -\log[OH^{-}] = -\log(10^{-4}) = 4$.
Since $pH + pOH = 14$ at $25\,^{\circ}C$,we have $pH = 14 - 4 = 10$.

(D) The given skeletal equation is $2MnO_4^- (aq) + Br^- (aq) \to 2MnO_2 (s) + BrO_3^- (aq)$.
Step $1$: Balance the charge. The total charge on the left is $-3$ and on the right is $-1$. To balance the charge in a basic medium,we add $2 \ OH^-$ ions to the right side.
$2MnO_4^- (aq) + Br^- (aq) \to 2MnO_2 (s) + BrO_3^- (aq) + 2 \ OH^- (aq)$
Step $2$: Balance the hydrogen atoms. There are $2 \ H$ atoms on the right,so we add one $H_2O$ molecule to the left side.
$2MnO_4^- (aq) + Br^- (aq) + H_2O (l) \to 2MnO_2 (s) + BrO_3^- (aq) + 2 \ OH^- (aq)$
Thus,both $(a)$ and $(b)$ are correct steps to balance the reaction.

(B) Zeolite,represented as $Na_2Ze$,acts as an ion exchanger. When hard water containing $Ca^{2+}$ or $Mg^{2+}$ ions is passed through it,the sodium ions $(Na^+)$ in the zeolite are replaced by these hardness-causing ions.
The reaction is: $Na_2Ze(s) + Ca^{2+}(aq) \to CaZe(s) + 2Na^+(aq)$.

(A) Lithium carbonate $(Li_2CO_3)$ is thermally unstable compared to other alkali metal carbonates.
Due to the very small size of the $Li^+$ ion,it has a high polarizing power.
It polarizes the large $CO_3^{2-}$ ion,which weakens the $C-O$ bond and facilitates the decomposition into the more stable oxide $Li_2O$ and carbon dioxide gas $(CO_2)$: $Li_2CO_3 \xrightarrow{\Delta} Li_2O + CO_2$.

$(D)$ The atomic radius of $Al$ is $143 \ pm$ and that of $Ga$ is $135 \ pm$.
Thus, the atomic radius of $Ga$ is actually smaller than that of $Al$.
This occurs because, in $Ga$, the $3d^{10}$ electrons provide poor shielding of the nuclear charge, leading to a higher effective nuclear charge which pulls the valence shell closer to the nucleus.
Since the Assertion is false and the Reason is true, the correct option is $D$.

(A) In $CH_3^+$,the carbon atom is bonded to $3$ hydrogen atoms and has a positive charge,resulting in $3$ sigma bonds and $0$ lone pairs. The steric number is $3$,which corresponds to $sp^2$ hybridisation.
In $CH_4$,the carbon atom is bonded to $4$ hydrogen atoms,resulting in $4$ sigma bonds and $0$ lone pairs. The steric number is $4$,which corresponds to $sp^3$ hybridisation.

(B) The stability of free radicals is governed by resonance and inductive effects.
$1$. Resonance stabilization: The triphenylmethyl radical $(C_6H_5)_3\dot{C}$ is more stable than the diphenylmethyl radical $(C_6H_5)_2\dot{C}H$ because it has more phenyl groups to delocalize the unpaired electron.
$2$. Inductive effect: Alkyl radicals are stabilized by the $+I$ effect of alkyl groups. The tertiary butyl radical $(CH_3)_3\dot{C}$ is more stable than the isopropyl radical $(CH_3)_2\dot{C}H$.
$3$. Overall order: Resonance-stabilized radicals are significantly more stable than alkyl radicals.
Therefore,the increasing order of stability is: $(CH_3)_2\dot{C}H < (CH_3)_3\dot{C} < (C_6H_5)_2\dot{C}H < (C_6H_5)_3\dot{C}$.

(D) The reaction represents the oxidative cleavage of an alkene using a strong oxidizing agent.
$CH_3CH_2CH=CHCH_3$ undergoes oxidative cleavage with alkaline $KMnO_4$ followed by acidic workup.
The double bond breaks to form carboxylic acids from the corresponding fragments.
$CH_3CH_2CH=CHCH_3 \xrightarrow[(i) \ KMnO_4, OH^{-} \text{ (heat)}]{(ii) \ H^{+}} CH_3CH_2COOH + CH_3COOH$.
Therefore,$X$ is $KMnO_4/OH^{-}$.

(D) The Assertion is incorrect because benzene does not exhibit two different bond lengths; all $C-C$ bonds in benzene are equivalent due to resonance.
The Reason is correct because the actual structure of benzene is a resonance hybrid of the two Kekulé structures shown.
Benzene has a uniform $C-C$ bond distance of $139 \ pm$, which is an intermediate value between the $C-C$ single bond $(154 \ pm)$ and $C=C$ double bond $(134 \ pm)$ lengths.

(B) The Assertion is correct: $HClO_4$ is indeed a stronger acid than $HClO_3$.
The Reason is also correct: The oxidation state of $Cl$ in $HClO_4$ is $+VII$ and in $HClO_3$ is $+V$.
However,the Reason is not the correct explanation for the Assertion. The acidity of oxyacids is determined by the number of non-hydroxylated oxygen atoms $(n)$ in the general formula $(HO)_m ZO_n$.
For $HClO_4$ $(HOClO_3)$,$n = 3$. For $HClO_3$ $(HOClO_2)$,$n = 2$.
As the number of non-hydroxylated oxygen atoms $(n)$ increases,the electron-withdrawing effect on the $Cl$ atom increases,which in turn weakens the $O-H$ bond,facilitating the release of the $H^+$ ion. Thus,the acidity depends on the number of non-hydroxylated oxygen atoms,not directly on the oxidation state of the central atom.

(A) In eukaryotes,$DNA$ replication and transcription occur within the nucleus because the genetic material is enclosed by a nuclear envelope.
After transcription,the $mRNA$ molecule is transported out of the nucleus into the cytoplasm through nuclear pores.
Translation,the process of protein synthesis,occurs in the cytoplasm because the necessary machinery,including ribosomes,$tRNA$,and amino acids,is located there.
Therefore,the Assertion is correct,and the Reason provides the correct explanation for why translation is spatially separated from transcription in eukaryotes.

(A) Most fungi are filamentous, consisting of long, thread-like structures called hyphae. However, $Yeast$ (Saccharomyces) is a notable exception because it is a unicellular fungus. It is a microscopic organism consisting of a single oval cell that typically reproduces through the process of budding.

(C) Viruses are sub-microscopic infectious agents that replicate only inside the living cells of a host organism.
Because of their extremely small size,viruses can easily pass through bacterial-proof filters.
Therefore,the statement that viruses cannot pass through bacterial filters is incorrect.
Viruses are obligate parasites,they multiply only within living cells,and they consist of genetic material ($DNA$ or $RNA$) enclosed in a protein coat (nucleoprotein particles).

(B) Cyanobacteria are photosynthetic autotrophs,not chemosynthetic. They contain chlorophyll $a$ similar to green plants and perform oxygenic photosynthesis. They are prokaryotic organisms that can be unicellular,colonial,or filamentous,and are found in marine or terrestrial habitats. They often form blooms in polluted water bodies. Some species,like $Nostoc$ and $Anabaena$,possess specialized cells called Heterocysts to fix atmospheric nitrogen.

(C) To determine the correct statement, let's analyze the characteristics of each algal group:
\tau\tau\tau\tau\tau\tau\tau\tau\tau\tau\tau\tau\tau\tau\tau\tau\tau\tau\tau\tau\tau\tau

Group	Major Pigments	Stored Food
Chlorophyceae	Chlorophyll-$a$, $b$	Starch
Phaeophyceae	Chlorophyll-$a$, $c$, Fucoxanthin	Mannitol, Laminarin
Rhodophyceae	Chlorophyll-$a$, $d$, Phycoerythrin	Floridean starch

Analysis of statements:
$I.$ Chlorophyceae contains chlorophyll-$a$ and $b$, and stores food as starch. This statement is correct.
$II.$ Phaeophyceae contains chlorophyll-$a$ and $c$ (not $b$), and stores food as laminarin/mannitol. The statement mentions chlorophyll-$b$, so it is incorrect.
$III.$ Rhodophyceae contains chlorophyll-$a$ and $d$, and phycoerythrin, and stores food as floridean starch. This statement is correct.
Therefore, statements $I$ and $III$ are correct, while $II$ is incorrect.

(D) Both sexes of the cockroach possess $anal \ cerci$,which are paired,jointed,filamentous structures located at the $10^{th}$ abdominal segment.
However,the male cockroach possesses an additional pair of small,thread-like,unjointed structures called $anal \ styles$ on the $9^{th}$ abdominal sternum.
These $anal \ styles$ are absent in females.
Therefore,the presence of $anal \ styles$ is a key morphological feature used to distinguish male cockroaches from females.

(A) The directing nature of a group depends on its electronic effect on the benzene ring.
$-NH_2$ is a strong electron-donating group due to the $+M$ effect,making it the strongest ortho-para directing group.
$-NO_2$ is a strong electron-withdrawing group due to the $-M$ and $-I$ effects,making it the strongest meta-directing group.
Therefore,the correct pair is $-NH_2$ and $-NO_2$.

(C) $NF_3$ is a weaker ligand than $N(CH_3)_3$ because fluorine is highly electronegative,which withdraws electron density from the nitrogen atom,making the lone pair less available for donation.
In contrast,$N(CH_3)_3$ is a stronger ligand because the methyl groups are electron-releasing,increasing the electron density on the nitrogen atom.
$NF_3$ does not ionize to give $F^{-}$ ions in aqueous solution; it is a covalent molecule.
Therefore,the Assertion is correct,but the Reason is incorrect.

(C) Step $1$: Hydration of ethyne $(HC \equiv CH)$ in the presence of $1\% \ HgSO_4$ and $20\% \ H_2SO_4$ (Kucherov reaction) yields acetaldehyde $(CH_3CHO)$ as product $A$.
Step $2$: Reaction of acetaldehyde $(CH_3CHO)$ with methylmagnesium halide $(CH_3MgX)$ followed by hydrolysis $(H_2O)$ gives propan$-2-$ol $(CH_3CHOHCH_3)$ as product $B$.
Step $3$: Oxidation $([O])$ of propan$-2-$ol $(CH_3CHOHCH_3)$ yields acetone $(CH_3COCH_3)$ as the final product $C$.

(C) In a $bcc$ lattice,the atoms touch along the body diagonal.
The relationship between the edge length $(a)$ and the atomic radius $(r)$ is given by $4r = \sqrt{3}a$.
Given $a = 4.29 \ \mathring{A}$.
$r = \frac{\sqrt{3} \times 4.29}{4}$.
$r = \frac{1.732 \times 4.29}{4} \approx 1.857 \ \mathring{A}$.
Rounding to two decimal places,we get $1.86 \ \mathring{A}$.

(D) Colligative properties are properties of solutions that depend on the number of solute particles in a given amount of solvent,not on the nature of the solute.
$(i)$ $NaBr$ dissociates into $2$ ions $(Na^+, Br^-)$ and $BaCl_2$ dissociates into $3$ ions $(Ba^{2+}, 2Cl^-)$. Since $BaCl_2$ produces more particles,it lowers the vapour pressure more than $NaBr$. Thus,$NaBr$ has a higher vapour pressure. This is a colligative property.
$(ii)$ The freezing point of a pure substance is a characteristic property of that substance,not a colligative property. Comparing two different pure solvents is not a colligative phenomenon.
$(iii)$ The addition of a non-volatile solute $(NaOH)$ to a solvent (water) causes freezing point depression. This is a classic colligative property.
Therefore,$(i)$ and $(iii)$ are correct.

(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 $(P_i^o)$.

(C) The given reactions are oxidation half-reactions. To determine the oxidizing agent,we look at the reduction potentials $(E^o_{red})$.
For the first reaction: $[Fe(CN)_6]^{3-} + e^- \to [Fe(CN)_6]^{4-}$,$E^o_{red} = +0.35 \ V$.
For the second reaction: $Fe^{3+} + e^- \to Fe^{2+}$,$E^o_{red} = +0.77 \ V$.
$A$ higher reduction potential indicates a stronger tendency to get reduced,making the species a stronger oxidizing agent.
Comparing the two,$Fe^{3+}$ has a higher reduction potential $(+0.77 \ V > +0.35 \ V)$.
Therefore,$Fe^{3+}$ is the strongest oxidizing agent. The correct option is $(C)$.

(D) The given cell reaction is $2Fe_{(s)} + O_{2(g)} + 4H^{+}_{(aq)} \to 2Fe^{2+}_{(aq)} + 2H_2O_{(l)}$.
Here,$n = 4$ electrons are transferred.
Given: $[Fe^{2+}] = 10^{-3} \ M$,$p(O_2) = 0.1 \ atm$,$pH = 3$,so $[H^{+}] = 10^{-3} \ M$.
Using the Nernst equation:
$E_{cell} = E^o_{cell} - \frac{0.0591}{n} \log \frac{[Fe^{2+}]^2}{[H^{+}]^4 \cdot p(O_2)}$
$E_{cell} = 1.67 - \frac{0.0591}{4} \log \frac{(10^{-3})^2}{(10^{-3})^4 \cdot 0.1}$
$E_{cell} = 1.67 - 0.014775 \cdot \log \frac{10^{-6}}{10^{-12} \cdot 10^{-1}}$
$E_{cell} = 1.67 - 0.014775 \cdot \log 10^7$
$E_{cell} = 1.67 - 0.014775 \cdot 7 = 1.67 - 0.1034 = 1.5666 \ V \approx 1.57 \ V$.

(A) The recovery of $Ag$ from the complex $[Ag(CN)_2]^-$ involves the displacement of $Ag^+$ by a more reactive metal.
The standard reduction potentials are $E^o_{Zn^{2+}/Zn} = -0.76 \ V$ and $E^o_{Cu^{2+}/Cu} = +0.34 \ V$.
Since $Zn$ has a more negative reduction potential than $Cu$,it is a stronger reducing agent.
$Zn$ can displace $Ag$ from $[Ag(CN)_2]^-$ because it can reduce $Ag^+$ to $Ag(s)$,whereas $Cu$ cannot effectively reduce $Ag^+$ in this complex due to its lower reducing power.
Thus,both the Assertion and the Reason are correct,and the Reason correctly explains the Assertion.

(C) The relationship between standard Gibbs free energy change and equilibrium constant is $\Delta G^o = -RT \ln K = -2.303 RT \log K$.
Also,$\Delta G^o = -nF E^o$.
Equating the two: $-nF E^o = -2.303 RT \log K$.
Rearranging gives $\log K = \frac{nF E^o}{2.303 RT}$,which is statement $I$.
Since $\frac{1}{2.303} \approx 0.4342$,we can write $\log K = 0.4342 \frac{nF E^o}{RT}$,which is statement $IV$.
From $\ln K = \frac{nF E^o}{RT}$,we get $K = e^{\frac{nF E^o}{RT}}$,which is statement $II$.
Therefore,statements $I$,$II$,and $IV$ are correct.

(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.

(C) Electrodialysis involves the movement of ions towards oppositely charged electrodes under the influence of an electric field.
Urea $(NH_2CONH_2)$ is a covalent compound and does not dissociate into ions in an aqueous solution.
Since it does not form ions,it cannot be moved by an electric field and thus cannot be removed by electrodialysis.
In contrast,$NaCl$,$K_2SO_4$,and $CaCl_2$ are ionic compounds that dissociate into ions ($Na^+$,$Cl^-$,$K^+$,$SO_4^{2-}$,$Ca^{2+}$) and can be easily removed by this process.

(D) The enthalpy of chemisorption is high,typically in the range of $40-400 \ kJ \ mol^{-1}$,because it involves the formation of chemical bonds.
In contrast,physisorption involves weak van der Waal's forces,resulting in a low enthalpy of adsorption,typically in the range of $20-40 \ kJ \ mol^{-1}$.
Therefore,the enthalpy of physisorption is lower than that of chemisorption.
Thus,the Assertion is false,and the Reason is true.

(B) Roasting is the process in which the ore is strongly heated in the presence of excess air.
In the extraction of copper,after crushing and concentration of the ore,it is heated in a reverberatory furnace.
Copper pyrites $(CuFeS_2)$ is partially oxidized,and a part of the sulfur in the ore is removed as sulfur dioxide $(SO_2)$.
The given equation $2CuFeS_2 + O_2 \rightarrow Cu_2S + 2FeS + SO_2$ represents the partial roasting of copper pyrites.

(B) Both Assertion and Reason are true. Smelting is a process of reduction of roasted ore with carbon (coke) in the presence of a flux. The flux is added to remove the gangue as slag. While the definition provided in the Reason is accurate,the use of coke and flux in smelting is a specific requirement for the reduction and slag formation process,making the Reason a descriptive definition rather than the direct cause for why they are used.

(A) The standard reduction potential of $Zn^{2+}/Zn$ is $-0.76 \ V$,while that of $Fe^{2+}/Fe$ is $-0.44 \ V$.
Since zinc has a more negative electrode potential than iron,it acts as a sacrificial anode.
Zinc gets oxidized in preference to iron,thereby protecting the iron from rusting.
Thus,both the Assertion and the Reason are correct,and the Reason is the correct explanation of the Assertion.

(A) $NF_5$ does not exist because nitrogen $(N)$ cannot form pentahalides due to the absence of $d$-orbitals in its valence shell.
$P$,$As$,and $Sb$ can form pentahalides of the general formula $MX_5$ (where $M = P, As, Sb$) because they possess vacant $d$-orbitals in their respective valence shells,allowing for the expansion of their octet.

(A) Peroxoacids of sulphur are those that contain at least one peroxy linkage $(-O-O-)$.
$H_2SO_5$ (peroxomonosulphuric acid or Caro's acid) contains one peroxy linkage.
$H_2S_2O_8$ (peroxodisulphuric acid or Marshall's acid) also contains one peroxy linkage.
Therefore,both $H_2SO_5$ and $H_2S_2O_8$ are peroxoacids of sulphur.

(A) Oxygen has a small atomic size and small bond length,which allows it to form stable $p\pi-p\pi$ multiple bonds,resulting in the formation of diatomic $O_2$ molecules.
In contrast,sulphur has a larger atomic size and longer bond length,making effective $p\pi-p\pi$ overlap difficult. Consequently,sulphur prefers to form single bonds with other sulphur atoms,leading to the formation of puckered $S_8$ ring structures.
Thus,both the Assertion and the Reason are correct,and the Reason correctly explains why oxygen exists as $O_2$ while sulphur exists as $S_8$.

(A) The first ionization energy $(IE_1)$ generally increases with an increase in atomic number across a period due to the increase in effective nuclear charge.
However,the trend is irregular among $d-$block elements due to shielding effects and stable electronic configurations.
Comparing the given elements:
$Zn (3d^{10} 4s^2)$ has a completely filled $d-$orbital and $s-$orbital,making it very stable.
$Fe (3d^6 4s^2)$ has a relatively high $IE_1$.
$Cu (3d^{10} 4s^1)$ has a filled $d-$subshell but a single $s-$electron.
$Cr (3d^5 4s^1)$ has a half-filled $d-$subshell.
Based on experimental data for the first ionization potential,the order is $Zn > Fe > Cu > Cr$.

(C) The assertion is correct because transition metals have partially filled $d$-orbitals,allowing them to lose electrons from both the $ns$ and $(n-1)d$ subshells.
The reason is incorrect because the energy difference between the $ns$ and $(n-1)d$ electrons is actually very small,which is precisely why both sets of electrons can participate in bond formation,leading to variable valency.

(D) Optical isomerism is exhibited by coordination compounds that lack a plane of symmetry and a center of symmetry.
$A$. $[Co(NH_3)_5(NO_2)]I_2$ has a plane of symmetry and is achiral.
$B$. $[Pt(NH_3)_2Cl_2]$ is a square planar complex,which is inherently achiral.
$C$. The $trans$ isomer of $[Cr(en)_2(CN)_2]^+$ has a plane of symmetry and is achiral.
$D$. $[Co(en)_3]Br_3$ contains three bidentate ethylenediamine $(en)$ ligands. This complex does not have a plane of symmetry or a center of symmetry,making it chiral and optically active. It exists as a pair of non-superimposable mirror images (enantiomers).

(B) In $[Fe(CN)_6]^{3-}$,$Fe$ is in $+3$ oxidation state ($3d^5$ configuration). Due to the strong field ligand $CN^-$,the electrons pair up,leaving $1$ unpaired electron,making it paramagnetic.
In $[Fe(CN)_6]^{4-}$,$Fe$ is in $+2$ oxidation state ($3d^6$ configuration). Due to the strong field ligand $CN^-$,all $6$ electrons pair up in the $t_{2g}$ orbitals,leaving $0$ unpaired electrons,making it diamagnetic.
Both statements are true,but the magnetic property depends on the number of unpaired electrons resulting from the crystal field splitting and the nature of the ligand,not just the oxidation state. Thus,the Reason is not the correct explanation of the Assertion.

(B) Alkyl fluorides are most conveniently prepared by heating suitable chloro- or bromo-alkanes with inorganic fluorides such as $AsF_3$,$SbF_3$,$CoF_2$,$AgF$,$Hg_2F_2$,etc. This reaction is known as the $Swarts$ reaction.
$CH_3Br + AgF \rightarrow CH_3F + AgBr$
$2CH_3CH_2Cl + Hg_2F_2 \rightarrow 2CH_3CH_2F + Hg_2Cl_2$

(D) The Assertion is false because aryl halides do not undergo nucleophilic substitution reactions under ordinary conditions. This is due to the resonance effect,which gives the $C-Cl$ bond partial double bond character,making it shorter and stronger,and thus difficult to replace by nucleophiles.
The Reason is also false because $S_{N}2$ reactions proceed with inversion of configuration,not retention.

(A) Electrophilic substitution reactions (such as halogenation,nitration,or Friedel-Crafts reactions) occur on the benzene ring.
Only aromatic ethers (alkyl aryl ethers) contain a benzene ring that can be activated by the alkoxy group $(-OR)$ towards electrophilic substitution.
Among the given options,$C_6H_5OCH_3$ (Anisole) is an aromatic ether,while the others are aliphatic ethers.
Therefore,$C_6H_5OCH_3$ undergoes electrophilic substitution reactions.

(C) triglyceride is an ester derived from glycerol and three fatty acids.
These three fatty acids can be the same or different.
Therefore,a triglyceride can have up to $3$ different acyl groups attached to the glycerol backbone.

(D) The reaction sequence is the Kolbe-Schmitt reaction.
$1$. Phenol reacts with $NaOH$ to form sodium phenoxide $(A)$.
$2$. Sodium phenoxide reacts with $CO_2$ at $140^{\circ}C$ under pressure,followed by acidification with $HCl$,to yield salicylic acid $(B)$ as the major product.
Therefore,the correct option is $D$.

(A) In the Victor-Meyer's test,the alcohols are converted into nitroalkanes,which are then treated with nitrous acid $(HNO_2)$ and finally made alkaline with $NaOH$.
For $1^o$ alcohols,the product is a nitrolic acid,which gives a blood-red colour in alkaline solution.
For $2^o$ alcohols,the product is a pseudonitrol,which gives a blue colour in alkaline solution.
For $3^o$ alcohols,the product does not react with nitrous acid and remains colourless in alkaline solution.
Therefore,the colours are red,blue,and colourless respectively.

(A) Aldol condensation is observed only in carbonyl compounds (aldehydes and ketones) that possess at least one $\alpha$-hydrogen atom.
$(a)$ Chloral $(CCl_3CHO)$ has no $\alpha$-hydrogen atom because the $\alpha$-carbon is bonded to three chlorine atoms.
$(b)$ Phenylacetaldehyde $(C_6H_5CH_2CHO)$ has two $\alpha$-hydrogen atoms.
$(c)$ Hexanal $(CH_3(CH_2)_4CHO)$ has two $\alpha$-hydrogen atoms.
$(d)$ Nitromethane $(CH_3NO_2)$ is not an aldehyde or ketone,but it does possess $\alpha$-hydrogen atoms and can undergo a condensation reaction similar to aldol,but strictly speaking,the question refers to carbonyl compounds. Among the options,chloral is the classic example that does not undergo aldol condensation.

(B) The reaction of carboxylic acids having an $\alpha$-hydrogen with halogen ($Cl_2$ or $Br_2$) in the presence of a small amount of red phosphorus to give $\alpha$-halo carboxylic acids is known as the Hell-Volhard-Zelinsky $(HVZ)$ reaction.

(A) Aniline is a Lewis base due to the lone pair of electrons on the nitrogen atom of the $-NH_2$ group.
Friedel-Crafts reaction requires $AlCl_3$ as a Lewis acid catalyst.
When aniline reacts with $AlCl_3$,it forms a salt (an acid-base adduct) where the nitrogen atom coordinates with the aluminum atom.
This results in the nitrogen atom acquiring a positive charge,which acts as a strong deactivating group for the benzene ring,thereby inhibiting the electrophilic substitution reaction required for Friedel-Crafts alkylation or acylation.
Therefore,both the Assertion and the Reason are correct,and the Reason is the correct explanation for the Assertion.

(A) nucleophile is a species that donates a lone pair of electrons.
In aniline $(C_6H_5NH_2)$,the nitrogen atom has a lone pair of electrons available for donation.
In the anilinium ion $(C_6H_5NH_3^+)$,the lone pair of the nitrogen atom is involved in bond formation with a proton $(H^+)$,resulting in a positive charge on the nitrogen.
Because the anilinium ion carries a positive charge and lacks an available lone pair,it cannot act as a nucleophile.
Therefore,aniline is a much better nucleophile than the anilinium ion.
Both the assertion and the reason are correct,and the reason correctly explains why the anilinium ion is not a nucleophile.

(C) Anomers are a special type of diastereomers that differ in configuration only at the $C-1$ carbon atom (the anomeric carbon).
Since $\alpha - D - (+)$ glucose and $\beta - D - (+)$ glucose differ in configuration specifically at the $C-1$ atom,they are classified as anomers.

(D) Neoprene is an addition polymer formed by the polymerization of chloroprene $(CH_2=C(Cl)-CH=CH_2)$.
Melamine,Glyptal,and Dacron are condensation polymers.

(C) Vulcanisation is a process of treating natural rubber with sulphur or some compounds of sulphur under heat to modify its properties.
In this process,sulphur forms cross-links at the reactive sites of double bonds in the rubber chain,which provides mechanical strength and elasticity to the rubber.
Therefore,the Assertion is correct.
However,vulcanisation is not a free radical initiated chain reaction; it is a chemical cross-linking process.
Thus,the Reason is incorrect.

(A) $Phenelzine$ is an antidepressant,while $Ranitidine$,$Aluminium \ hydroxide$,and $Cimetidine$ are antacids.

(A) Sedatives are a class of drugs that act on the central nervous system $(CNS)$ to produce a calming effect,reduce anxiety,and induce sleep.
Patients who are mentally agitated or violent require drugs that can suppress the overactive functions of the $CNS$ to bring about relaxation and calmness.
Therefore,both the Assertion and the Reason are correct,and the Reason is the correct explanation for the Assertion.

(B) The reagent is $BaCl_2$,which imparts a green colour to the flame.
$BaCl_2$ reacts with $K_2Cr_2O_7$ and conc. $H_2SO_4$ to form chromyl chloride $(CrO_2Cl_2)$,which is a red gas.
The chemical reaction is:
$2BaCl_2 + K_2Cr_2O_7 + 3H_2SO_4 \to K_2SO_4 + 2BaSO_4 + 2CrO_2Cl_2 \text{ (red gas)} + 3H_2O$