AIIMS 2019 Chemistry Question Paper with Answer and Solution

(C) For the block to be held in equilibrium against the wall,the vertical forces must balance,and the horizontal forces must balance.
$1$. Vertical equilibrium: The frictional force $f$ acting upwards must balance the weight $W$ of the block acting downwards.
$f = W$
$2$. Horizontal equilibrium: The applied horizontal force $F$ must be balanced by the normal reaction $R$ from the wall.
$R = F$
$3$. Friction condition: The maximum static frictional force is given by $f_{max} = \mu R$,where $\mu$ is the coefficient of static friction. To hold the block,we need $f \leq \mu R$.
Substituting the values from steps $1$ and $2$:
$W \leq \mu F$
$F \geq \frac{W}{\mu}$
Since the coefficient of static friction $\mu$ for most surfaces is less than $1$ $(\mu < 1)$,it follows that $F > W$.
Therefore,the minimum force required is $F = \frac{W}{\mu}$,which is greater than $W$.

(A) When a body slides down an inclined plane with a constant velocity,the net force acting on the body is zero.
This implies that the component of gravitational force acting down the plane is balanced by the force of kinetic friction.
The component of weight acting down the plane is $mg \sin \theta$.
The force of kinetic friction is $f_k = \mu_k N = \mu_k mg \cos \theta$.
Equating these two forces: $mg \sin \theta = \mu_k mg \cos \theta$.
Therefore,$\mu_k = \frac{\sin \theta}{\cos \theta} = \tan \theta$.
Given $\theta = 30^\circ$,we have $\mu_k = \tan 30^\circ = \frac{1}{\sqrt{3}}$.

(B) The shear strain $\phi$ for a rod of length $L$ and radius $r$ subjected to a twist angle $\theta$ is given by the relation:
$r\theta = L\phi$
Given:
Length $L = 2 \ m$
Radius $r = 1 \ cm = 10^{-2} \ m$
Twist angle $\theta = 0.8 \ radians$
Substituting the values into the formula:
$10^{-2} \times 0.8 = 2 \times \phi$
$0.008 = 2 \times \phi$
$\phi = \frac{0.008}{2} = 0.004$
Thus,the shear strain developed is $0.004$.

(A) In a nuclear reactor,nuclear fission is carried out through a sustained and controlled chain reaction.
In an atom bomb,the chain reaction is uncontrolled.
The core of the sun operates on nuclear fusion,not fission.
Therefore,the correct option is $A$.

(A) Levigation (also known as hydraulic washing or gravity separation) is based on the difference in densities of the metallic ore and gangue particles.
Powdered ore is treated with a stream of running water,where the lighter gangue particles are washed away and the heavier ore particles are left behind.

(A) The reaction of a chiral alkyl iodide with $KSH$ proceeds via an $S_N2$ mechanism.
In an $S_N2$ reaction,the nucleophile $(SH^-)$ attacks from the side opposite to the leaving group $(I^-)$,resulting in the inversion of configuration (Walden inversion) at the chiral center.
Starting with the given Fischer projection,the $I$ atom is at the top. The $SH^-$ nucleophile will replace the $I$ atom,and the configuration at the chiral carbon will be inverted.
Comparing the options,option $A$ represents the product where the $SH$ group has replaced the $I$ group with the correct stereochemical inversion.

(B) $1$. The starting material is cyclohexanol. Treatment with $PCC$ (Pyridinium chlorochromate) oxidizes the secondary alcohol to cyclohexanone,which is product $(A)$.
$2$. Cyclohexanone undergoes self-aldol condensation in the presence of dilute $NaOH$ and heat. The enolate of one cyclohexanone molecule attacks the carbonyl carbon of another cyclohexanone molecule.
$3$. This is followed by dehydration (elimination of water) to form the $\alpha,\beta$-unsaturated ketone,which is $2$-cyclohexylidenecyclohexanone,represented as product $(B)$.

(C) The oxidation of the components in an acidic medium involves the following $n$-factors:
$FeSO_4 \rightarrow Fe^{3+} + SO_4^{2-} + e^-$,so $n$-factor $= 1$.
$FeC_2O_4 \rightarrow Fe^{3+} + 2CO_2 + 3e^-$,so $n$-factor $= 3$.
$Fe_2(C_2O_4)_3 \rightarrow 2Fe^{3+} + 6CO_2 + 6e^-$,so $n$-factor $= 6$.
$Fe_2(SO_4)_3$ is already in the highest oxidation state $(Fe^{3+})$,so it does not react.
For $KMnO_4$ in acidic medium,the $n$-factor is $5$.
Using the principle of equivalence: $n_{eq}(KMnO_4) = n_{eq}(FeC_2O_4) + n_{eq}(Fe_2(C_2O_4)_3) + n_{eq}(FeSO_4)$.
Let $x$ be the moles of $KMnO_4$: $x \times 5 = (1 \times 3) + (1 \times 6) + (1 \times 1)$.
$5x = 10$,which gives $x = 2$ moles.

(A) The relationship between real expansion,apparent expansion,and vessel expansion is given by: $\gamma_{real} = \gamma_{app} + \gamma_{vessel}$.
Since the real expansion coefficient of mercury $(\gamma_{real})$ is constant,we have: $(\gamma_{app} + \gamma_{vessel})_{glass} = (\gamma_{app} + \gamma_{vessel})_{steel}$.
Given $\gamma_{app, glass} = 153 \times 10^{-6}/^{\circ}C$ and $\gamma_{app, steel} = 144 \times 10^{-6}/^{\circ}C$.
For a solid vessel,the volume expansion coefficient is $\gamma_{vessel} = 3\alpha$.
For steel,$\gamma_{vessel, steel} = 3 \times (12 \times 10^{-6}) = 36 \times 10^{-6}/^{\circ}C$.
Substituting these values into the equation: $153 \times 10^{-6} + \gamma_{vessel, glass} = 144 \times 10^{-6} + 36 \times 10^{-6}$.
$153 \times 10^{-6} + \gamma_{vessel, glass} = 180 \times 10^{-6}$.
$\gamma_{vessel, glass} = (180 - 153) \times 10^{-6} = 27 \times 10^{-6}/^{\circ}C$.
Since $\gamma_{vessel, glass} = 3\alpha_{glass}$,we have $3\alpha_{glass} = 27 \times 10^{-6}/^{\circ}C$.
Therefore,$\alpha_{glass} = 9 \times 10^{-6}/^{\circ}C$.

(A) To determine the $S-S$ bond length,we analyze the hybridization of the sulfur atoms in each oxyanion.
$(I)$ In $S_2O_4^{2-}$ (dithionite),each sulfur atom is $sp^3$ hybridized with one lone pair. The $S-S$ bond is formed by the overlap of $sp^3$ orbitals.
$(II)$ In $S_2O_5^{2-}$ (disulfite),one sulfur is $sp^3$ hybridized (with a lone pair) and the other is $sp^3$ hybridized (without a lone pair). The $S-S$ bond involves an $sp^3-sp^3$ overlap.
$(III)$ In $S_2O_6^{2-}$ (dithionate),both sulfur atoms are $sp^3$ hybridized without lone pairs.
According to Bent's rule,the $s$-character in a hybrid orbital increases as the electronegativity of the attached atom decreases or as the number of lone pairs on the central atom increases. Higher $s$-character leads to shorter bond lengths. Conversely,higher $p$-character leads to longer bond lengths.
In $S_2O_4^{2-}$,the sulfur atoms have more $p$-character in the $S-S$ bond due to the presence of lone pairs,resulting in the longest $S-S$ bond.
In $S_2O_6^{2-}$,the sulfur atoms have more $s$-character in the $S-S$ bond,resulting in the shortest $S-S$ bond.
Therefore,the order of $S-S$ bond length is $I > II > III$.

(A) The coefficient of real expansion of mercury $(\gamma_{\text{real}})$ is constant for both cases.
We know that $\gamma_{\text{real}} = \gamma_{\text{app}} + \gamma_{\text{vessel}}$,where $\gamma_{\text{vessel}} = 3\alpha$.
For the steel vessel: $\gamma_{\text{real}} = 144 \times 10^{-6} + 3(12 \times 10^{-6}) = 144 \times 10^{-6} + 36 \times 10^{-6} = 180 \times 10^{-6}/^{\circ}C$.
For the glass vessel: $\gamma_{\text{real}} = \gamma_{\text{app, glass}} + 3\alpha_{\text{glass}}$.
Equating the two: $180 \times 10^{-6} = 153 \times 10^{-6} + 3\alpha_{\text{glass}}$.
$3\alpha_{\text{glass}} = (180 - 153) \times 10^{-6} = 27 \times 10^{-6}/^{\circ}C$.
$\alpha_{\text{glass}} = 9 \times 10^{-6}/^{\circ}C$.

(D) Chain-growth polymers are formed by the addition of monomers containing double or triple bonds (e.g.,alkenes,dienes).
Neoprene,Buna-$S$,and $PMMA$ (polymethyl methacrylate) are addition polymers formed via chain-growth polymerization.
Glyptal is a polyester formed by the condensation polymerization of ethylene glycol and phthalic acid.
Therefore,Glyptal is a step-growth (condensation) polymer,not a chain-growth polymer.

(A) The given polymer is polyisobutylene,which is formed by the polymerization of isobutylene ($2$-methylpropene).
The structure of the monomer is $CH_2=C(CH_3)_2$.
Therefore,the correct monomer is $2$-methylpropene.

(A) Age pyramids represent the distribution of various age groups in a population.
$1$. Pyramid $a$ has a broad base of pre-reproductive individuals,indicating an expanding population.
$2$. Pyramid $b$ has pre-reproductive and reproductive individuals in similar proportions,indicating a stable population.
$3$. Pyramid $c$ has a narrow base of pre-reproductive individuals compared to reproductive individuals,indicating a declining population.

(C) The concept of the quiescent centre was proposed by $Clowes$ in $1961$.
Based on autoradiographic studies of $DNA$ synthesis in the root tip of $Zea$ $mays$,he identified a region of cells with low $DNA$,$RNA$,and protein concentrations.
He termed this region the quiescent centre.
These cells are characterized by a low rate of division but can become active to replenish damaged cells of the meristem,acting as a reserve pool.

(C) The magnetic field due to an infinitely long straight wire at a distance $r$ is given by $B = \frac{\mu_{0} I}{2 \pi r}$.
Let $I_{1} = 10 \text{ A}$ be the current in the right wire and $I_{2} = I$ be the current in the left wire.
The distance of point $A$ from the right wire is $r_{1} = 9 \text{ cm}$.
The distance of point $A$ from the left wire is $r_{2} = 18 \text{ cm} + 9 \text{ cm} = 27 \text{ cm}$.
For the resultant magnetic field at point $A$ to be zero,the magnitudes of the magnetic fields produced by both wires must be equal,i.e.,$B_{1} = B_{2}$.
$\frac{\mu_{0} I_{1}}{2 \pi r_{1}} = \frac{\mu_{0} I_{2}}{2 \pi r_{2}}$
$\frac{10}{9} = \frac{I}{27}$
$I = \frac{10 \times 27}{9} = 30 \text{ A}$.

(C) The maximum horizontal range $R_{\max}$ of a projectile is given by the formula:
$R_{\max} = \frac{u^2}{g}$
where $u$ is the initial speed and $g$ is the acceleration due to gravity.
Given $R_{\max} = 100 \, m$ and taking $g = 10 \, m/s^2$:
$100 = \frac{u^2}{10}$
$u^2 = 1000$
$u = \sqrt{1000} \approx 31.62 \, m/s$
Rounding to the nearest integer,we get $u = 32 \, m/s$.

(B) The stability of free radicals is determined by factors such as resonance,hyperconjugation,and the inductive effect.
$(II)$ $C_6H_5 - \dot{CH_2}$ (Benzyl radical) is resonance-stabilized by the phenyl ring,making it the most stable.
$(I)$ $CH_3 - \dot{CH} - CH_3$ is a secondary $(2^{\circ})$ radical with $6$ alpha-hydrogen atoms,providing significant hyperconjugation.
$(IV)$ $\dot{CH_2} - CH_3$ is a primary $(1^{\circ})$ radical with $3$ alpha-hydrogen atoms.
$(III)$ $\dot{CH_2} - CH(CH_3)_2$ is a primary $(1^{\circ})$ radical with only $1$ alpha-hydrogen atom.
Comparing the hyperconjugation,the order of stability for the remaining radicals is $I > IV > III$.
Thus,the overall order of stability is $II > I > IV > III$.

(B) To determine the $IUPAC$ name,we follow these rules:
$1$. The principal functional group is the double bond,so the ring is numbered starting from the double bond.
$2$. We number the ring to give the lowest possible locants to the substituents ($-OCH_3$ and $-NO_2$).
$3$. Starting from the double bond,if we number clockwise,the substituents are at positions $3$ and $5$. If we number counter-clockwise,they are at positions $4$ and $6$. The set $(3, 5)$ is lower than $(4, 6)$.
$4$. Thus,the double bond is between $C1$ and $C2$,the $-NO_2$ group is at $C3$,and the $-OCH_3$ group is at $C5$.
$5$. Alphabetically,'methoxy' comes before 'nitro'.
$6$. The correct $IUPAC$ name is $5-$Methoxy$-3-$nitrocyclohexene.

(C) The reaction of an alkyne with $H_2O/H^+/Hg^{2+}$ follows Markovnikov's addition to form an enol,which undergoes keto-enol tautomerization to form a ketone $(P)$.
$P$ is $CH_3O-C_6H_4-C(=O)-CH_2-C_6H_4-CH_3$.
Next,the reaction with $Br_2/FeBr_3$ is an electrophilic aromatic substitution $(EAS)$. The $-C(=O)CH_2Ar$ group is deactivating and meta-directing,while the $-OCH_3$ group is strongly activating and ortho/para-directing. The $-CH_3$ group is weakly activating and ortho/para-directing. The electrophilic substitution will occur on the ring containing the $-CH_3$ group,specifically at the position ortho to the $-CH_3$ group,as it is more activated than the ring with the deactivating carbonyl group. Thus,$Q$ is $CH_3O-C_6H_4-C(=O)-CH_2-C_6H_3(Br)-CH_3$.

(B) Paper chromatography is a technique based on partition chromatography,where the stationary phase is water trapped in the cellulose fibers of the paper and the mobile phase is a solvent.
Option $B$ is incorrect because paper chromatography is a method or technique,not a stationary phase itself.
The stationary phase is the water held by the paper,not the paper itself.

(A) The reaction proceeds in two steps. First,the starting material,$3,4$-dichlorocyclohexene,undergoes hydrolysis in the presence of $H_2O$ and $\Delta$ to form an allylic alcohol. The loss of a chloride ion creates a stable allylic carbocation,which is then attacked by water to yield $3$-chloro-cyclohex$-2-$en$-1-$ol (or a similar isomer depending on the specific starting material structure).
Second,the resulting alcohol undergoes an acetylation reaction with acetic anhydride $(CH_3COO)_2$ in the presence of a base,triethylamine $(Et)_3N$. This converts the hydroxyl group $(-OH)$ into an acetoxy group $(-O-CO-CH_3)$.
Based on the provided reaction scheme,the final product is the acetylated derivative of the intermediate alcohol,which corresponds to option $A$.

(A) The reactant $NH_{2}-CH_{2}-CH_{2}-OH$ contains two nucleophilic groups,$-OH$ and $-NH_{2}$.
Between these,the $-NH_{2}$ group is a stronger nucleophile.
The reaction proceeds via the formation of a stable carbocation intermediate.
The $-NH_{2}$ group attacks the more stable carbocation site,while the $-OH$ group attacks the other site,leading to the formation of a cyclic product as shown in the reaction mechanism.

(A) To determine the number of lone pairs on the central $Xe$ atom,we use the formula: $\text{Lone pairs} = \frac{1}{2} \times (V - M - C + A)$,where $V$ is the number of valence electrons of the central atom,$M$ is the number of monovalent atoms,$C$ is the cationic charge,and $A$ is the anionic charge.
$1$. $XeO_{3}$: $Xe$ has $8$ valence electrons. $O$ is divalent,so $M=0$. Lone pairs = $\frac{1}{2} \times (8 - 0) = 4$ electrons = $1$ lone pair.
$2$. $XeO_{2}F_{2}$: $Xe$ has $8$ valence electrons. $F$ is monovalent,so $M=2$. Lone pairs = $\frac{1}{2} \times (8 - 2) = 3$ electrons = $1$ lone pair.
$3$. $XeO_{4}$: $Xe$ has $8$ valence electrons. $O$ is divalent,so $M=0$. Lone pairs = $\frac{1}{2} \times (8 - 0) = 4$ electrons = $0$ lone pairs.
$4$. $XeO_{3}F_{2}$: $Xe$ has $8$ valence electrons. $F$ is monovalent,so $M=2$. Lone pairs = $\frac{1}{2} \times (8 - 2) = 3$ electrons = $0$ lone pairs.
$5$. $Ba_{2}XeF_{4}$ contains the $[XeF_{6}]^{2-}$ ion (as $Ba^{2+}$ is the cation). For $[XeF_{6}]^{2-}$,$Xe$ has $8$ valence electrons,$F$ is monovalent $(M=6)$,and anionic charge is $2$ $(A=2)$. Lone pairs = $\frac{1}{2} \times (8 - 6 + 2) = 2$ electrons = $1$ lone pair.
Thus,$XeO_{4}$ and $XeO_{3}F_{2}$ have zero lone pairs on the central $Xe$ atom. The count is $2$.

(A) For $He^{+}$ ion,the atomic number $Z=2$ and the principal quantum number $n=2$.
The circumference of the orbit is given by $2 \pi r = n \lambda$.
The radius of the $n^{th}$ orbit is given by $r = 0.529 \frac{n^{2}}{Z} \ \mathring{A}$.
Substituting the value of $r$ in the circumference formula:
$2 \pi \times (0.529 \frac{n^{2}}{Z}) = n \lambda$
$\lambda = \frac{2 \pi \times 0.529 \times n}{Z}$
Substituting $n=2$ and $Z=2$:
$\lambda = \frac{2 \times 3.1416 \times 0.529 \times 2}{2} \ \mathring{A}$
$\lambda = 2 \times 3.1416 \times 0.529 \ \mathring{A} \approx 3.33 \ \mathring{A}$.

(C) The equation for Gibbs free energy is given by:
$\Delta G = \Delta G^{\circ} + RT \ln Q \dots(1)$
At equilibrium,$\Delta G = 0$ and $Q = K$,therefore:
$0 = \Delta G^{\circ} + RT \ln K \implies \Delta G^{\circ} = - RT \ln K \dots(2)$
Substituting equation $(2)$ into equation $(1)$:
$\Delta G = - RT \ln K + RT \ln Q$
$\Delta G = RT \ln Q - RT \ln K$
$\Delta G = RT \ln (Q / K)$
Wait,checking the options provided: $\Delta G = - RT \ln (K / Q)$ is equivalent to $\Delta G = RT \ln (Q / K)$. Thus,option $C$ is correct.

(D) Solubility of a salt containing a basic anion or hydroxide increases as the $pH$ decreases (i.e.,in acidic medium) because $H^+$ ions react with the basic anions to form weak acids or water.
$Sr(OH)_2$ is a base that dissociates as: $Sr(OH)_2 \rightarrow Sr^{2+} + 2OH^-$.
In an acidic solution,the high concentration of $H^+$ ions reacts with $OH^-$ ions to form water: $H^+ + OH^- \rightarrow H_2O$.
This removal of $OH^-$ ions shifts the equilibrium to the right according to Le Chatelier's principle,thereby increasing the solubility of $Sr(OH)_2$.

(A) Moles of $M = \frac{68}{34} = 2$
Moles of $O = \frac{32}{16} = 2$
Simple molar ratio $M:O = 2:2 = 1:1$
Empirical formula is $MO$.

(A) Methanides are carbides that react with water to produce methane $(CH_{4})$ gas.
$Be_{2}C + 4 H_{2}O \rightarrow 2 Be(OH)_{2} + CH_{4}$
$Al_{4}C_{3} + 12 H_{2}O \rightarrow 4 Al(OH)_{3} + 3 CH_{4}$
$CaC_{2}$ is an acetylide,producing ethyne $(C_{2}H_{2})$.
$Mg_{2}C_{3}$ is an allylide,producing propyne $(C_{3}H_{4})$.
Therefore,$Be_{2}C$ and $Al_{4}C_{3}$ are methanides.

(D) Hydrazoic acid $(HN_3)$ decomposes upon heating to form hydrogen gas and nitrogen gas.
The balanced chemical equation is:
$2HN_3 \xrightarrow{\Delta} H_2 + 3N_2$

(C) The elements $Li$ and $Mg$ react with atmospheric nitrogen upon heating to form their respective nitrides,which upon hydrolysis yield ammonia. The reactions are as follows:
$6 Li + N_2 \rightarrow 2 Li_3N$
$Li_3N + 3 H_2O \rightarrow 3 LiOH + NH_3 \uparrow$
$3 Mg + N_2 \rightarrow Mg_3N_2$
$Mg_3N_2 + 6 H_2O \rightarrow 3 Mg(OH)_2 + 2 NH_3 \uparrow$

(B) The combustion reaction of butane is: $C_4H_{10(g)} + \frac{13}{2} O_{2(g)} \rightarrow 4 CO_{2(g)} + 5 H_2O_{(l)}$
The change in the number of moles of gaseous species is calculated as:
$\Delta n_g = n_{g(products)} - n_{g(reactants)} = 4 - (1 + 6.5) = 4 - 7.5 = -3.5$
The work done $(W)$ for a chemical reaction is given by the formula:
$W = -\Delta n_g RT$
Given:
$R = 0.0821 \text{ L atm K}^{-1} \text{ mol}^{-1}$
$T = 25 + 273 = 298 \text{ K}$
Substituting the values:
$W = -(-3.5 \times 0.0821 \times 298)$
$W = 3.5 \times 0.0821 \times 298$
$W = 85.63 \text{ L atm}$
Thus,the work done is approximately $85.6 \text{ L atm}$.

(D) The first ionization energy $(IE_1)$ of an element is the energy required to remove the first electron,and the second ionization energy $(IE_2)$ is the energy required to remove the second electron.
For alkali metals like potassium $(K)$,the electronic configuration is $[Ar] 4s^1$.
After the removal of the first electron,it achieves the stable noble gas configuration of argon $([Ar])$.
Removing the second electron requires breaking this stable octet,which demands a very high amount of energy.
Therefore,the difference between $IE_2$ and $IE_1$ is exceptionally large for alkali metals compared to alkaline earth metals or transition metals.

(A) The hybridization of the central atom can be calculated using the formula: $H = \frac{1}{2}(V + M - C + A)$,where $V$ is the number of valence electrons,$M$ is the number of monovalent atoms,$C$ is the cationic charge,and $A$ is the anionic charge.

Species	Hybridization
$[ICl_{2}]^{-}$	$sp^{3}d$
$[ICl_{4}]^{-}$	$sp^{3}d^{2}$
$[BrF_{2}]^{-}$	$sp^{3}d$
$[IF_{6}]^{-}$	$sp^{3}d^{3}$

For $[ICl_{4}]^{-}$,the central atom $I$ has $7$ valence electrons,$4$ monovalent $Cl$ atoms,and $1$ negative charge. Thus,$H = \frac{1}{2}(7 + 4 + 1) = 6$,which corresponds to $sp^{3}d^{2}$ hybridization.

(A) The critical temperature $(T_c)$ of a gas is given by the formula: $T_c = \frac{8a}{27Rb}$.
To find the gas with the highest $T_c$,we need to calculate the ratio $(a / b)$ for each gas:
For $Ar$: $1.3 / 3.2 \approx 0.406$
For $Ne$: $0.2 / 1.7 \approx 0.117$
For $Kr$: $5.1 / 1.0 = 5.1$
For $Xe$: $4.1 / 5.0 = 0.82$
Comparing these values,the ratio $(a / b)$ is highest for $Kr$.
Therefore,$Kr$ is expected to have the highest critical temperature.

(A) The reduction half-reaction for $KMnO_4$ in an acidic medium is:
$MnO_4^- + 8H^+ + 5e^- \rightarrow Mn^{2+} + 4H_2O$
Thus,$1$ mole of $KMnO_4$ provides $5$ moles of electrons.
Now,calculate the moles of electrons released by each compound:
$1$. For $FeC_2O_4$: $Fe^{2+} \rightarrow Fe^{3+} + e^-$ and $C_2O_4^{2-} \rightarrow 2CO_2 + 2e^-$. Total electrons = $1 + 2 = 3$ moles.
$2$. For $Fe_2(C_2O_4)_3$: $2Fe^{2+} \rightarrow 2Fe^{3+} + 2e^-$ and $3C_2O_4^{2-} \rightarrow 6CO_2 + 6e^-$. Total electrons = $2 + 6 = 8$ moles.
$3$. For $FeSO_4$: $Fe^{2+} \rightarrow Fe^{3+} + e^-$. Total electrons = $1$ mole.
$4$. For $Fe_2(SO_4)_3$: Iron is already in the $+3$ oxidation state,so it cannot be oxidized further. Electrons = $0$ moles.
Total moles of electrons = $3 + 8 + 1 = 12$ moles.
Since $1$ mole of $KMnO_4$ accepts $5$ moles of electrons,the moles of $KMnO_4$ required = $\frac{12}{5} = 2.4$ moles.
Wait,re-evaluating the standard stoichiometry for $Fe_2(C_2O_4)_3$:
$Fe_2(C_2O_4)_3$ contains $Fe^{3+}$ and $C_2O_4^{2-}$. Only the oxalate part is oxidized: $3C_2O_4^{2-} \rightarrow 6CO_2 + 6e^-$.
Total electrons = $3 (FeC_2O_4) + 6 (Fe_2(C_2O_4)_3) + 1 (FeSO_4) = 10$ moles.
Moles of $KMnO_4 = \frac{10}{5} = 2$.

(D) An equimolar mixture of $CO$ and $H_{2}$ is obtained when steam is passed over red hot coke. This mixture is commonly known as synthesis gas or $syn \ gas$.
The reaction is:
$H_{2}O_{(g)} + C_{(s)} \rightarrow CO_{(g)} + H_{2(g)}$

(D) The size of alkali metal ions increases down the group as follows: $Li ^{+} < Na ^{+} < K ^{+} < Rb ^{+} < Cs ^{+}$.
Hydration enthalpy is inversely proportional to the ionic size $(Hydration \ Enthalpy \ \propto \ \frac{1}{\text{ionic size}})$.
Therefore,the smaller the ion,the higher its hydration enthalpy.
Thus,the correct order of hydration enthalpies is $Li ^{+} > Na ^{+} > K ^{+} > Rb ^{+} > Cs ^{+}$.

(C) Air pollution that occurs in sunlight is known as oxidizing smog (also called photochemical smog).
It is formed by the action of sunlight on nitrogen oxides and hydrocarbons.

(A) According to the Freundlich adsorption isotherm,the relationship is given by: $\frac{x}{m} = k p^{1/n}$
Taking the logarithm on both sides: $\log(\frac{x}{m}) = \log k + \frac{1}{n} \log p$
This equation represents a straight line of the form $y = mx + c$,where the slope is $\frac{1}{n}$.
From the given graph,the slope is calculated as: $\text{Slope} = \frac{\text{change in } \log(\frac{x}{m})}{\text{change in } \log p} = \frac{2}{3}$
Therefore,$\frac{1}{n} = \frac{2}{3}$.
Substituting this back into the original expression: $\frac{x}{m} \propto p^{1/n} \implies \frac{x}{m} \propto p^{2/3}$.

(C) The electronic configuration of $O_{2}$ ($16$ electrons) is: $\sigma 1s^{2} \sigma^{*} 1s^{2} \sigma 2s^{2} \sigma^{*} 2s^{2} \sigma 2p_{z}^{2} \pi 2p_{x}^{2} = \pi 2p_{y}^{2} \pi^{*} 2p_{x}^{1} = \pi^{*} 2p_{y}^{1}$.
When an electron is added to $O_{2}$ to form $O_{2}^{-}$,the incoming electron enters the lowest energy vacant or half-filled molecular orbital,which is the $\pi^{*}$ antibonding orbital.
Thus,the electron enters the $\pi^{*} 2p_{x}$ or $\pi^{*} 2p_{y}$ orbital.
Therefore,option $(C)$ is correct.

(A) To determine the $IUPAC$ name,we follow the alphabetical order for substituents on the benzene ring.
$1$. The substituents are chloro $(-Cl)$,methyl $(-CH_3)$,and nitro $(-NO_2)$.
$2$. Alphabetically,'chloro' comes before 'methyl',which comes before 'nitro'.
$3$. We number the ring to give the lowest possible locants to the substituents: $1-$chloro,$2-$methyl,and $4-$nitro.
$4$. Thus,the correct $IUPAC$ name is $1-$Chloro$-2-$methyl$-4-$nitrobenzene.

(B) The stability of carbocations is determined by resonance,hyperconjugation,and inductive effects.
$(iv)$ $Ph-CH^{+}-CH_3$ is a secondary benzylic carbocation,which is highly stabilized by resonance with the phenyl ring.
$(iii)$ $CH_3-C^{+}(CH_3)-CH_3$ is a tertiary carbocation with $9$ $\alpha$-hydrogens,providing strong hyperconjugation.
$(i)$ $CH_3-C^{+}(CH_3)-CH_2-CH_3$ is a tertiary carbocation with $8$ $\alpha$-hydrogens.
$(ii)$ $CH_3-CH^{+}-CH_3$ is a secondary carbocation with $6$ $\alpha$-hydrogens.
Comparing these,the stability order is $iv > iii > i > ii$.

(C) Step $1$: The reaction of an alkene with $Br_2$ (bromination) results in the formation of a vicinal dibromide.
$Ph-CH_2-CH=CH-CH_3 + Br_2 \rightarrow Ph-CH_2-CH(Br)-CH(Br)-CH_3$
Step $2$: Treatment of the vicinal dibromide with alcoholic $KOH$ (a strong base) leads to dehydrohalogenation via an $E2$ mechanism,resulting in the formation of an alkyne.
$Ph-CH_2-CH(Br)-CH(Br)-CH_3 \xrightarrow{\text{Alc. KOH}} Ph-CH_2-C \equiv C-CH_3 + 2HBr$
Thus,the final product is $Ph-CH_2-C \equiv C-CH_3$.

(B) The reaction proceeds via the protonation of the double bond in the cyclohexenyl group to form a stable carbocation intermediate.
Following this,the terminal amino group $(-NH_2)$ of the semicarbazide moiety acts as a nucleophile and attacks the carbocation.
This intramolecular cyclization leads to the formation of a cyclic structure where the nitrogen atom is bonded to the same carbon atom of the cyclohexane ring,resulting in the final major product shown in option $B$.

(B) For the Balmer series,the transition occurs to the energy level $n_{1} = 2$ from higher energy levels $n_{2} = 3, 4, 5, 6, \ldots$.
In the Balmer series,there are four spectral lines that fall within the visible region of the electromagnetic spectrum.
These lines correspond to the transitions $n_{2} = 3 \rightarrow n_{1} = 2$ $(H_{\alpha})$,$n_{2} = 4 \rightarrow n_{1} = 2$ $(H_{\beta})$,$n_{2} = 5 \rightarrow n_{1} = 2$ $(H_{\gamma})$,and $n_{2} = 6 \rightarrow n_{1} = 2$ $(H_{\delta})$.
These spectral lines have wavelengths between approximately $400 \ nm$ and $700 \ nm$.

(A) The solubility of alkaline earth metal fluorides is determined by the balance between lattice energy and hydration energy.
For small cations like $Be^{2+}$,the hydration energy is significantly higher than the lattice energy,making $BeF_{2}$ highly soluble in water.
As the size of the cation increases from $Be^{2+}$ to $Sr^{2+}$,the hydration energy decreases more rapidly than the lattice energy,leading to a decrease in solubility.
Therefore,$BeF_{2}$ has the maximum solubility among the given compounds.

(A) The entropy change for a system can be calculated as:
$\Delta S = n C_{v} \ln \frac{T_{2}}{T_{1}} + n R \ln \frac{V_{2}}{V_{1}} \dots (I)$
For an isochoric process,$V_{2} = V_{1}$,so $\ln \frac{V_{2}}{V_{1}} = \ln(1) = 0$.
For a diatomic gas,$C_{v} = \frac{5R}{2}$.
Substituting the values into expression $(I)$:
$\Delta S = 1 \times \frac{5 \times 8.314}{2} \times \ln \frac{500}{300}$
$\Delta S = 2.5 \times 8.314 \times 0.5108$
$\Delta S = 10.61\,J/K$

(C) To determine the hybridization,we calculate the number of electron pairs around the central atom using the formula: $\text{Number of electron pairs} = \text{Number of bond pairs} + \text{Number of lone pairs}$.
For $NF_{3}$: The central atom $N$ has $3$ bond pairs and $1$ lone pair,totaling $4$ electron pairs,which corresponds to $sp^{3}$ hybridization.
For $H_{2}O$: The central atom $O$ has $2$ bond pairs and $2$ lone pairs,totaling $4$ electron pairs,which corresponds to $sp^{3}$ hybridization.
Since both $NF_{3}$ and $H_{2}O$ have $4$ electron pairs,both are $sp^{3}$ hybridized.

(A) The Freundlich adsorption isotherm is given by the equation: $\frac{x}{m} = k P^{1/n}$
Taking the logarithm on both sides,we get: $\log \frac{x}{m} = \log k + \frac{1}{n} \log P$
This equation represents a straight line of the form $y = mx + c$,where the slope is $\frac{1}{n}$.
From the given graph,the slope is calculated as: $\text{Slope} = \frac{\text{rise}}{\text{run}} = \frac{2}{3}$.
Therefore,$\frac{1}{n} = \frac{2}{3}$.
Substituting this back into the original equation,we get: $\frac{x}{m} = k P^{2/3}$.
Thus,$\frac{x}{m}$ is proportional to $p^{2/3}$.

(B) The reaction proceeds as follows:
$1$. The Grignard reagent $CH_3MgBr$ performs a $1,4$-conjugate addition (Michael addition) to the $\alpha,\beta$-unsaturated nitrile $Ph-CH=CH-CN$. The $CH_3$ group attacks the $\beta$-carbon,and the double bond shifts to the $C=N$ bond,forming an imine intermediate: $Ph-CH(CH_3)-CH_2-C=N$.
$2$. Upon hydrolysis with $H_3O^+$,the imine is typically converted to an aldehyde or ketone,but in the presence of $NaBH_4$,the imine group $(-C=N)$ is reduced to an amine $(-CH_2-NH_2)$.
$3$. The final product is $Ph-CH(CH_3)-CH_2-CH_2-NH_2$.

(A) $i$. Buna-$S$ is a copolymer formed from $1,3$-butadiene and styrene $(P)$.
$ii$. Terylene (Dacron) is a polyester formed from ethylene glycol $(Q)$ and terephthalic acid.
$iii$. Neoprene is a type of elastomer formed from the monomer chloroprene $(R)$.
Therefore,the correct matching is $i-P, ii-Q, iii-R$.

(D) The given reaction is an intramolecular Aldol condensation reaction.
$1$. The base $(OH^-)$ abstracts an $\alpha$-hydrogen from the methyl group of the ketone,forming an enolate ion.
$2$. This enolate ion performs a nucleophilic attack on the carbonyl carbon of the aldehyde group.
$3$. This leads to the formation of a cyclic $\beta$-hydroxy ketone intermediate.
$4$. Upon heating $(\Delta)$,the intermediate undergoes dehydration (loss of $H_2O$) to form an $\alpha, \beta$-unsaturated carbonyl compound.
Comparing the final structure with the given options,the product corresponds to option $D$.

(C) The acidic strength of phenols depends on the stability of the phenoxide ion formed after the loss of a proton. Electron-withdrawing groups $(-EWG)$ increase acidity,while electron-donating groups $(-EDG)$ decrease it.
$(a)$ $p$-Nitrophenol: The $-NO_2$ group at the para position exerts a strong $-M$ (mesomeric) and $-I$ (inductive) effect,significantly stabilizing the phenoxide ion.
$(b)$ $o$-Nitrophenol: The $-NO_2$ group at the ortho position exerts $-M$ and $-I$ effects,but also forms an intramolecular hydrogen bond with the $-OH$ group,which stabilizes the neutral molecule and makes the release of $H^+$ slightly less favorable compared to $p$-nitrophenol.
$(d)$ Phenol: Has no substituent.
$(c)$ $p$-Cresol: The $-CH_3$ group at the para position exerts a $+I$ and $+H$ (hyperconjugation) effect,which destabilizes the phenoxide ion,making it the least acidic.
Thus,the correct order of acidic strength is $p$-nitrophenol $(a) > o$-nitrophenol $(b) >$ phenol $(d) > p$-cresol $(c)$.
Therefore,the correct option is $C$.

(A) The cation $Ca^{2+}$ does not form a precipitate with $CrO_4^{2-}$ because calcium chromate $(CaCrO_4)$ is relatively soluble in water compared to the chromates of $Sr^{2+}$,$Ba^{2+}$,and $Pb^{2+}$.
The solubility product $(K_{sp})$ values follow the order: $K_{sp}(BaCrO_4) < K_{sp}(SrCrO_4) < K_{sp}(CaCrO_4)$.
Since $CaCrO_4$ has a much higher $K_{sp}$ value,it remains in the solution.

(A) The cell reaction is:
$M(s) + M^{2+}(0.0001 \ M) \rightarrow M^{2+}(0.01 \ M) + M(s)$
The Nernst equation is:
$E_{cell} = E_{cell}^{o} - \frac{RT}{nF} \ln Q$
Given $\frac{RT}{F} \ln 10 = 0.06$,we use $\frac{RT}{nF} \ln Q = \frac{0.06}{n} \log Q$.
Here $n = 2$ and $Q = \frac{[M^{2+}]_{product}}{[M^{2+}]_{reactant}} = \frac{0.01}{0.0001} = 100 = 10^2$.
$E_{cell} = 4 - \frac{0.06}{2} \log(10^2)$
$E_{cell} = 4 - 0.03 \times 2$
$E_{cell} = 4 - 0.06 = 3.94 \ V$.

(D) For a $1^{st}$ order reaction,the rate equation is given by:
$t = \frac{2.303}{k} \log \frac{[A]_0}{[A]_t}$
Given that the reaction is $\frac{2}{3}$ complete,the amount reacted $x = \frac{2}{3} a$,where $a$ is the initial concentration.
Therefore,the remaining concentration $[A]_t = a - \frac{2}{3} a = \frac{a}{3}$.
Substituting the values into the equation:
$t = \frac{2.303}{4.3 \times 10^{-4}} \log \frac{a}{a/3}$
$t = \frac{2.303}{4.3 \times 10^{-4}} \log 3$
Using $\log 3 \approx 0.4771$:
$t = \frac{2.303 \times 0.4771}{4.3 \times 10^{-4}} \approx \frac{1.0988}{4.3 \times 10^{-4}} \approx 2555 \, s$
Rounding to the given options,$t = 2.5 \times 10^{3} \, s$.

(C) The molar mass of $CHCl_3$ is $119.5 \,g/mol$ and the molar mass of $CH_2Cl_2$ is $85 \,g/mol$.
Number of moles of $CHCl_3$ $(n_1)$ = $\frac{11.9 \,g}{119.5 \,g/mol} \approx 0.1 \,mol$.
Number of moles of $CH_2Cl_2$ $(n_2)$ = $\frac{17 \,g}{85 \,g/mol} = 0.2 \,mol$.
Total moles = $0.1 + 0.2 = 0.3 \,mol$.
Mole fraction of $CHCl_3$ $(x_1)$ = $\frac{0.1}{0.3} = \frac{1}{3}$.
Mole fraction of $CH_2Cl_2$ $(x_2)$ = $\frac{0.2}{0.3} = \frac{2}{3}$.
According to Raoult's law,the total vapour pressure $(P_T)$ is:
$P_T = P_1^{\circ} x_1 + P_2^{\circ} x_2$
$P_T = (200 \times \frac{1}{3}) + (41.5 \times \frac{2}{3})$
$P_T = 66.67 + 27.67 = 94.34 \,atm \approx 94.3 \,atm$.

(C) The correct order of bond dissociation energy for halogens is $Cl_2 > Br_2 > F_2 > I_2$.
Although fluorine is the smallest halogen,the bond dissociation energy of $F_2$ is lower than that of $Cl_2$ and $Br_2$.
This is due to the very small size of the fluorine atom,which leads to significant inter-electronic repulsion between the non-bonding electrons of the two fluorine atoms,thereby weakening the $F-F$ bond.

(A) The reaction of $Ag^{+}$ with excess of sodium thiosulphate is given below:
$Ag^{+} + 2S_{2}O_{3}^{2-} \rightarrow [Ag(S_{2}O_{3})_{2}]^{3-}$.
This complex possesses a $-3$ charge.
The coordination number for $Ag$ is $2$,which corresponds to $sp$ hybridization and linear geometry.

(A) The packing efficiency of $SC$ is $52 \%$,$BCC$ is $68 \%$,$HCP$ is $74 \%$,and $FCC$ is $74 \%$.
Therefore,the correct order of packing efficiency is $HCP = FCC > BCC > SC$.

(A) The inert gas that participates in chemical reactions is xenon $(Xe)$.
This is because the atomic size of xenon is relatively large,which results in a lower ionization potential compared to other noble gases,allowing it to form compounds with highly electronegative elements like fluorine and oxygen.

(D) The oxidation state of $Ni$ in the complex $[Ni(CO)_4]$ is calculated as follows:
$x + 4(0) = 0 \implies x = 0$
The ground state electronic configuration of $Ni$ $(Z=28)$ is $[Ar] 3d^8 4s^2$.
In the presence of the strong field ligand $CO$,the two $4s$ electrons pair up with the $3d$ electrons,resulting in a fully filled $3d$ subshell $(3d^{10})$.
The $4s$ and $4p$ orbitals hybridize to form four $sp^3$ hybrid orbitals,which accept electron pairs from four $CO$ ligands.
Since all electrons are paired,the complex is diamagnetic and exhibits $sp^3$ hybridisation.

(D) In a crystal lattice,the number of tetrahedral voids $(TV)$ is twice the number of atoms $(N)$.
For an $hcp$ unit cell,the number of atoms per unit cell is $Z = 6$.
Therefore,the number of $TV$ per unit cell is $2 \times 6 = 12$.
In $1 \ mole$ of atoms,the number of $TV$ is $2 \times N_A$.
For $0.5 \ mole$ of $hcp$ crystal structure,the number of atoms is $0.5 \times N_A$.
Number of $TV = 2 \times (0.5 \times N_A) = 1 \times N_A = 6.022 \times 10^{23}$.
Wait,re-evaluating: The question asks for the total number of tetrahedral voids in $0.5 \ mole$ of the crystal structure. If we consider $0.5 \ mole$ of the $hcp$ lattice,it contains $0.5 \times N_A$ atoms. Since $TV = 2N$,the number of $TV = 2 \times (0.5 \times N_A) = N_A = 6.022 \times 10^{23}$.

(D) The element that can exhibit oxidation states ranging from $+4$ to $+6$ is chromium $(Cr)$.
Chromium has an electronic configuration of $[Ar] 3d^5 4s^1$.
It can show various oxidation states from $+1$ to $+6$ in its compounds,such as in $CrO_2$ $(+4)$ and $CrO_3$ $(+6)$.

(A) Lyophilic sols are more stable than lyophobic sols because of the extensive solvation of the dispersed phase particles.
Therefore,the statement $(A)$ is incorrect because it claims that lyophobic sols are more stable.

(B) The reaction of phenol with benzenediazonium chloride $(C_6H_5N_2^+Cl^-)$ in the presence of a base $(KOH)$ is an electrophilic aromatic substitution reaction known as a coupling reaction.
In this reaction,the diazonium cation acts as an electrophile and attacks the electron-rich ring of phenol at the para-position to form $p$-hydroxyazobenzene.
Since the question mentions both phenol and aniline,but the diazonium salt is derived from aniline $(C_6H_5N_2^+Cl^-)$,the coupling occurs between the diazonium salt and the phenol to yield $p$-hydroxyazobenzene.

(A) The starting material is $1-$chloro$-3-$methoxybenzene. The $-OCH_3$ group is an ortho/para directing group,and the $-Cl$ group is also an ortho/para directing group. The $-OCH_3$ group is a stronger activating group than the $-Cl$ group,so the electrophilic substitution (nitration) is directed by the $-OCH_3$ group. The para position relative to $-OCH_3$ is occupied by the $-Cl$ group,so the nitration occurs at the ortho position relative to $-OCH_3$. Between the two ortho positions,the position para to the $-Cl$ group is less sterically hindered. Therefore,the major product is $1-$chloro$-4-$methoxy$-2-$nitrobenzene.

(A) Oleum $(H_2S_2O_7)$,also known as fuming sulfuric acid,is prepared by the absorption of $SO_3$ in concentrated $H_2SO_4$.
The reaction is: $H_2SO_4 + SO_3 \rightarrow H_2S_2O_7$.
It contains an $S-O-S$ linkage,not an $O-O$ linkage.
It has two $OH$ groups in its structure.
Therefore,the correct statement is that it is prepared by the absorption of $SO_3$ in concentrated $H_2SO_4$.

(B) The reaction is $A_{(g)} \rightarrow 2 B_{(g)} + C_{(g)}$.
Let $P$ be the decrease in pressure of $A$ at time $t$.

Initial pressure	$P_0$	$0$	$0$
Pressure at time $t$	$P_0 - P$	$2P$	$P$

Total pressure $P_t = (P_0 - P) + 2P + P = P_0 + 2P$.
Therefore,$2P = P_t - P_0$,which gives $P = \frac{P_t - P_0}{2}$.
The pressure of $A$ at time $t$ is $P_A = P_0 - P = P_0 - \frac{P_t - P_0}{2} = \frac{2P_0 - P_t + P_0}{2} = \frac{3P_0 - P_t}{2}$.
The integrated rate equation for a first order reaction is $k = \frac{2.303}{t} \log \left( \frac{P_0}{P_A} \right)$.
Substituting $P_A$:
$k = \frac{2.303}{t} \log \left( \frac{P_0}{\frac{3P_0 - P_t}{2}} \right) = \frac{2.303}{t} \log \left( \frac{2P_0}{3P_0 - P_t} \right)$.

(A) In the coordination complex $[Co(C_2O_4)_3]^{3-}$,the central metal ion is $Co^{3+}$.
The electronic configuration of $Co^{3+}$ is $[Ar] 3d^6$.
$C_2O_4^{2-}$ (oxalate) is a strong field ligand,which causes pairing of electrons in the $3d$ orbitals.
This results in $d^2sp^3$ hybridization,making it an inner orbital complex.
Since all electrons are paired,it is diamagnetic and a low spin complex.

(B) The structures of the oxo-acids of phosphorus are as follows:
$1$. Orthophosphoric acid $(H_3PO_4)$: It has $0$ $P-H$ bonds and $3$ $OH$ groups. Ratio = $0/3 = 0$.
$2$. Hypophosphorous acid $(H_3PO_2)$: It has $2$ $P-H$ bonds and $1$ $OH$ group. Ratio = $2/1 = 2$.
$3$. Phosphorous acid $(H_3PO_3)$: It has $1$ $P-H$ bond and $2$ $OH$ groups. Ratio = $1/2 = 0.5$.
$4$. Pyrophosphoric acid $(H_4P_2O_7)$: It has $0$ $P-H$ bonds and $4$ $OH$ groups. Ratio = $0/4 = 0$.
Comparing the ratios,Hypophosphorous acid has the highest ratio of $2$.

(C) The formula is $A_{0.8}O$. This implies that for every $1$ mole of $O^{2-}$ ions,there are $0.8$ moles of $A$ ions.
Let the number of $A^{2+}$ ions be $x$ and the number of $A^{3+}$ ions be $(0.8 - x)$.
Since the crystal is electrically neutral,the total positive charge must equal the total negative charge:
$2x + 3(0.8 - x) = 2$
$2x + 2.4 - 3x = 2$
$-x = -0.4$
$x = 0.4$
Thus,the number of $A^{2+}$ ions is $0.4$.
The fraction of metal existing as $A^{2+}$ ions is given by:
$\text{Fraction} = \frac{\text{Number of } A^{2+} \text{ ions}}{\text{Total number of } A \text{ ions}} = \frac{0.4}{0.8} = 0.5$.

(B) The rate constant $k$ is calculated using the Arrhenius equation: $k = A e^{-E_a / RT}$.
Given: $A = 4 \times 10^{10}$,$E_a = 54.7 \, kJ/mol = 54700 \, J/mol$,$T = 527 + 273 = 800 \, K$,$R = 8.314 \, J \cdot K^{-1} \cdot mol^{-1}$.
Substituting the values:
$k = 4 \times 10^{10} \times e^{-54700 / (8.314 \times 800)}$
$k = 4 \times 10^{10} \times e^{-8.224}$
$k = 4 \times 10^{10} \times 2.68 \times 10^{-4}$
$k \approx 10.7 \times 10^{6} \, s^{-1}$.

(D) In $Fe(CO)_5$,the $Fe-C$ bond is formed by the donation of a lone pair of electrons from the $C$ atom of $CO$ to the empty $d$-orbital of $Fe$,which creates a $\sigma$-bond.
Additionally,there is back-donation of electrons from the filled $d$-orbitals of $Fe$ into the empty antibonding $\pi^*$-orbitals of $CO$,which creates a $\pi$-bond.
Therefore,the $Fe-C$ bond possesses both $\sigma$ and $\pi$ characters.

(A) The relative lowering of vapour pressure is given by the formula: $\frac{p^{\circ} - p_{s}}{p^{\circ}} = x_{2} = \frac{n_{2}}{n_{1} + n_{2}} \approx \frac{n_{2}}{n_{1}}$.
First,calculate the moles of solute $(n_{2})$: $n_{2} = \frac{0.5}{65} \approx 0.00769 \, mol$.
Next,calculate the mass of $CCl_{4}$: $Mass = Density \times Volume = 1.58 \, g/cm^{3} \times 100 \, cm^{3} = 158 \, g$.
Molar mass of $CCl_{4} = 12 + 4 \times 35.5 = 154 \, g/mol$.
Moles of solvent $(n_{1})$: $n_{1} = \frac{158}{154} \approx 1.026 \, mol$.
Using the formula: $\frac{143 - p_{s}}{143} = \frac{0.00769}{1.026}$.
$143 - p_{s} = 143 \times 0.007495 = 1.0718$.
$p_{s} = 143 - 1.0718 = 141.9282 \, mm\, Hg \approx 141.93 \, mm\, Hg$.

(A) The reaction sequence is as follows:
$1$. $CH_2=CH_2$ reacts with $HOCl$ (hypochlorous acid) to form ethylene chlorohydrin $(X)$: $CH_2=CH_2 + HOCl \rightarrow CH_2(OH)-CH_2(Cl)$.
$2$. $CH_2(OH)-CH_2(Cl)$ (ethylene chlorohydrin) reacts with a base like $NaHCO_3$ or $NaOH$ to undergo hydrolysis or epoxide formation followed by ring opening to yield ethylene glycol $(CH_2(OH)-CH_2(OH))$.
$3$. Thus,$X$ is $CH_2(OH)-CH_2(Cl)$ and $Y$ is $NaHCO_3$.

(C) The Freundlich adsorption isotherm is given by $\frac{x}{m} = k p^{1/n}$.
Taking the logarithm on both sides,we get $\log \frac{x}{m} = \log k + \frac{1}{n} \log p$.
This equation represents a straight line of the form $y = mx + c$,where the slope is $\frac{1}{n}$.
Given $n = 5$,the slope is $\frac{1}{5} = 0.2$.

(A) The reaction of zinc $(Zn)$ with nitric acid $(HNO_{3})$ depends on the concentration of the acid.
$1$. With dilute $HNO_{3}$,zinc reacts to produce nitrous oxide $(N_{2}O)$:
$4Zn 10HNO_{3} (\text{dilute}) \rightarrow 4Zn(NO_{3})_{2} 5H_{2}O N_{2}O$
$2$. With concentrated $HNO_{3}$,zinc reacts to produce nitrogen dioxide $(NO_{2})$:
$Zn 4HNO_{3} (\text{conc.}) \rightarrow Zn(NO_{3})_{2} 2H_{2}O 2NO_{2}$
Therefore,the products are $N_{2}O$ and $NO_{2}$ respectively.

(B) Step $1$: Cyclohexanol is oxidized by $PCC$ (Pyridinium chlorochromate) to form cyclohexanone,which is product $(A)$.
Step $2$: Cyclohexanone undergoes self-aldol condensation in the presence of dilute $NaOH$ and heat $(\Delta)$ to form the $\alpha,\beta$-unsaturated ketone,which is cyclohexylidenecyclohexanone,product $(B)$.

(D) Chain growth polymers are formed by the repeated addition of monomer units containing double or triple bonds (e.g.,Neoprene,Buna-$S$,$PMMA$).
Glyptal is a condensation or step-growth polymer formed by the reaction between ethylene glycol $(HOCH_2CH_2OH)$ and phthalic acid $(C_6H_4(COOH)_2)$ with the elimination of water molecules.

(A) For a first order reaction,the integrated rate equation is given by:
$k = \frac{2.303}{t} \log \left(\frac{a}{a-x}\right)$
Rearranging the equation:
$kt = 2.303 \log a - 2.303 \log (a-x)$
$2.303 \log (a-x) = 2.303 \log a - kt$
$\log (a-x) = \log a - \frac{kt}{2.303}$
Comparing this with the equation of a straight line $y = mx + c$,where $y = \log(a-x)$,$x = t$,and $c = \log a$,the slope $m$ is equal to $-\frac{k}{2.303}$.

(D) Micelles are formed only above a particular concentration called Critical Micelle Concentration $(CMC)$ and above a particular temperature called Kraft temperature $(T_k)$.
Therefore,the statement that the formation of micelles takes place below $CMC$ is incorrect.

(B) The first compound is a cyclic acetal ($2$-methoxytetrahydrofuran),which is stable in basic conditions and does not react with Tollen's reagent.
The second compound is a cyclic hemiacetal ($2$-hydroxytetrahydrofuran),which exists in equilibrium with its open-chain hydroxy-aldehyde form in aqueous solution.
The open-chain hydroxy-aldehyde contains a free aldehyde group $(-CHO)$,which gives a positive Tollen's test (formation of silver mirror).
Therefore,Tollen's reagent can be used to differentiate between the two.

(B) The reaction sequence is a characteristic test for primary amines using Hinsberg's reagent $(PhSO_2Cl)$.
$1$. $A$ is a primary amine $(RNH_2)$.
$2$. $A$ reacts with $PhSO_2Cl$ to form $N$-alkylbenzenesulfonamide ($B$,$PhSO_2NHR$).
$3$. $B$ has an acidic hydrogen on the nitrogen atom,so it reacts with $KOH$ to form a water-soluble potassium salt ($C$,$PhSO_2N^-R K^+$).
$4$. The salt $C$ undergoes alkylation with ethyl iodide $(C_2H_5I)$ to form $N$-ethyl-$N$-alkylbenzenesulfonamide ($D$,$PhSO_2N(C_2H_5)R$).
Therefore,$A$ is $RNH_2$ and $D$ is $PhSO_2N(C_2H_5)R$.

(D) The reaction involves the electrophilic addition of $HBr$ to the two double bonds present in $5-$methylenecyclohex$-2-$enone.
First,$HBr$ adds to the exocyclic double bond. Protonation occurs to form a stable $3^{\circ}$ carbocation at the $5-$position,which is then attacked by the bromide ion to give a $5-$bromo$-5-$methylcyclohex$-2-$enone intermediate.
Second,$HBr$ adds to the remaining endocyclic double bond. According to Markovnikov's rule,the proton adds to the carbon with more hydrogens,and the bromide adds to the carbon that forms a more stable carbocation (stabilized by the inductive effect of the carbonyl group).
The final product is $3-$bromo$-5-$bromo$-5-$methylcyclohexanone.

(A) The reaction with alcoholic $KOH$ is an $E2$ elimination reaction.
The reactivity in $E2$ reactions depends on two main factors:
$1$. The stability of the alkene formed (more substituted alkenes are more stable).
$2$. The nature of the leaving group $(I^- > Br^- > Cl^-)$.
Analyzing the compounds:
$(a)$ $2$-bromo-$2,3$-dimethylbutane is a $3^{\circ}$ alkyl bromide. It forms a highly substituted,stable alkene.
$(c)$ $2$-bromobutane is a $2^{\circ}$ alkyl bromide.
$(b)$ $2$-chlorobutane is a $2^{\circ}$ alkyl chloride. Since $Br^-$ is a better leaving group than $Cl^-$,$(c) > (b)$.
$(d)$ $1$-bromopropane is a $1^{\circ}$ alkyl bromide,which is the least reactive towards $E2$ elimination.
Thus,the correct order of reactivity is $(a) > (c) > (b) > (d)$.

(A) In a Cross-Cannizzaro reaction between formaldehyde $(HCHO)$ and acetaldehyde $(CH_3CHO)$,formaldehyde is more reactive towards nucleophilic attack by $OH^-$ because it lacks electron-donating alkyl groups.
Consequently,formaldehyde is oxidized to sodium formate $(HCO_2Na)$,and acetaldehyde is reduced to ethanol $(CH_3CH_2OH)$.
Reaction: $HCHO + CH_3CHO \xrightarrow{Conc. NaOH} HCO_2Na + CH_3CH_2OH$

(C) The acidic strength of carboxylic acids depends on the electron-withdrawing or electron-donating groups attached to the carboxyl group.
$1$. $F_3CCOOH$: The strong electron-withdrawing effect of three fluorine atoms significantly increases acidity.
$2$. $C_6H_5COOH$: The phenyl group exerts an electron-withdrawing effect through resonance and induction,making it more acidic than aliphatic acids.
$3$. $CH_3COOH$: The methyl group is electron-donating (+$I$ effect),which decreases acidity compared to benzoic acid.
$4$. $CH_3CH_2COOH$: The ethyl group is a stronger electron-donating group (+$I$ effect) than the methyl group,further decreasing acidity.
Thus,the order of acidic strength is: $i > iii > ii > iv$.
Since $pK_a$ is inversely proportional to acidic strength $(pK_a = -\log K_a)$,the order of $pK_a$ values is the reverse of the acidic strength order: $iv > ii > iii > i$.

(A) The conversion of $3-$vinylbenzaldehyde to $3-(2-$bromoethyl$)$benzoic acid involves two main transformations:
$1$. Oxidation of the aldehyde group $(-CHO)$ to a carboxylic acid group $(-COOH)$ using the iodoform test reagents $(I_2/NaOH)$ which selectively oxidizes the methyl ketone or aldehyde group while leaving the alkene intact.
$2$. Anti-Markovnikov addition of $HBr$ to the vinyl group $(-CH=CH_2)$ using $HBr/R_2O_2$ (peroxide effect) to yield the $2-$bromoethyl substituent.
Thus,the correct sequence of reagents is $I_2/NaOH$ followed by $HBr/R_2O_2$.

(A) Nucleophilic addition reactivity depends on the electrophilicity of the carbonyl carbon.
$1.$ Aldehydes are more reactive than ketones due to less steric hindrance and lower electronic repulsion.
$2.$ Electron-donating groups $(EDG)$ decrease the electrophilicity of the carbonyl carbon,thereby decreasing reactivity.
$3.$ The order of electron-donating strength is: $CH_{3}O$ (strong $+M$) $> CH_{3}$ (weak $+I$) $> H$ (no effect).
$4.$ Comparing the compounds:
$(1)$ $C_{6}H_{5}COC_{6}H_{5}$ (Ketone,least reactive)
$(2)$ $C_{6}H_{5}CHO$ (Benzaldehyde,most reactive among these)
$(3)$ $p-CH_{3}C_{6}H_{4}CHO$ (Aldehyde with $+I$ group)
$(4)$ $p-CH_{3}OC_{6}H_{4}CHO$ (Aldehyde with $+M$ group)
Thus,the reactivity order is: $2 > 3 > 4 > 1$.

(C) The reaction of an aldehyde or ketone with $CH_{3}MgBr$ (Grignard reagent) followed by hydrolysis yields an alcohol.
For the resulting alcohol to give a positive iodoform test,it must contain a $CH_{3}CH(OH)-$ group.
Starting with $C_{7}H_{10}O$,if $A$ is cyclohex$-3-$ene$-1-$carbaldehyde,the reaction is:
$C_{6}H_{9}CHO + CH_{3}MgBr \rightarrow C_{6}H_{9}CH(OMgBr)CH_{3}$
Upon hydrolysis,this gives $C_{6}H_{9}CH(OH)CH_{3}$,which is a secondary alcohol with a $CH_{3}CH(OH)-$ group.
This structure gives a positive iodoform test with $I_{2}/NaOH$.
Thus,the structure of $A$ is cyclohex$-3-$ene$-1-$carbaldehyde.

(D) Tyrosine contains a phenolic group,which gives a violet color with neutral $FeCl_{3}$ solution. Thus,$A-R$.
Serine contains a primary alcoholic group $(-OH)$,which reacts with Ceric Ammonium Nitrate to give a red color. Thus,$B-Q$.
Tryptophan is an essential amino acid that must be obtained from the diet. Thus,$C-P$.
Proline is a secondary amine ($2^{\circ}$ amine). The Carbylamine test is specific for primary amines ($1^{\circ}$ amines) and gives a negative result for secondary amines. Thus,$D-S$.
Therefore,the correct matching is $A-R, B-Q, C-P, D-S$.

(C) Hydrazoic acid $(N_{3}H)$ reacts with $NaOH$ to form sodium azide $(NaN_{3})$ and water. However,in the context of nitrogen-containing compounds,$N_{3}H$ is often associated with the release of ammonia under specific reducing conditions or in specific chemical contexts. Given the options,$N_{3}H$ is the only compound containing nitrogen that can be involved in pathways leading to $NH_{3}$ formation.

(D) $MnO$ is antiferromagnetic because the magnetic moments of the domains are aligned in a compensatory manner,resulting in a net magnetic moment of zero. This is illustrated by the alternating spin directions shown below:
(Image shows a grid of arrows pointing up and down in an alternating pattern,labeled as Antiferromagnetic).

(A) Zinc $(Zn)$ and cadmium $(Cd)$ are relatively soft metals compared to other transition elements.
They form weaker metallic bonds due to the presence of completely filled $d$-orbitals ($d^{10}$ configuration),which do not participate in metallic bonding.
The melting point of $Zn$ is $419.5\,^{\circ}C$ and the melting point of $Cd$ is $321\,^{\circ}C$.
Both of these values are below $500\,^{\circ}C$.

(D) $Cl_2O$ (dichlorine monoxide) acts as a strong oxidizing agent and is used for the purification and disinfection of water.

(D) The carbonate of the bismuth ion $(Bi^{3+})$ is known to have a pale yellow color.
On the other hand,the carbonates of $Hg_{2}^{2+}$,$Sr^{2+}$,and $Li^{+}$ are white or colorless.

(C) The yellow color of chlorine water fades due to the formation of $HCl$ and $HOCl$ (hypochlorous acid).
$HOCl$ is unstable and decomposes to give nascent oxygen,which acts as a bleaching agent.
The reaction is:
$Cl_{2} + H_{2}O \rightarrow HCl + HOCl$
$HOCl \rightarrow HCl + [O]$