JEE Main 2015 Chemistry Question Paper with Answer and Solution

(A) The vertex $O$ of the parabola $x^2 = 8y$ is $(0, 0)$.
Let the coordinates of point $Q$ on the parabola be $(4t, 2t^2)$,as it satisfies $x^2 = 8y$ (i.e.,$(4t)^2 = 16t^2 = 8(2t^2)$).
Let $P(h, k)$ be the point that divides the line segment $OQ$ internally in the ratio $1:3$.
Using the section formula,the coordinates of $P$ are:
$h = \frac{1 \times 4t + 3 \times 0}{1 + 3} = \frac{4t}{4} = t$
$k = \frac{1 \times 2t^2 + 3 \times 0}{1 + 3} = \frac{2t^2}{4} = \frac{t^2}{2}$
From $h = t$,we have $t = h$.
Substituting this into the expression for $k$:
$k = \frac{h^2}{2} \implies h^2 = 2k$.
Replacing $(h, k)$ with $(x, y)$,the locus of $P$ is $x^2 = 2y$.

(A) The given region is defined by the parabola $y^2 = 2x$ and the line $y = 4x - 1$.
To find the intersection points,substitute $x = \frac{y+1}{4}$ into $y^2 = 2x$:
$y^2 = 2\left(\frac{y+1}{4}\right) \Rightarrow y^2 = \frac{y+1}{2} \Rightarrow 2y^2 - y - 1 = 0$.
Solving the quadratic equation: $(2y+1)(y-1) = 0$,so $y = 1$ and $y = -\frac{1}{2}$.
For $y = 1$,$x = \frac{1+1}{4} = \frac{1}{2}$. For $y = -\frac{1}{2}$,$x = \frac{-1/2+1}{4} = \frac{1}{8}$.
The points of intersection are $(\frac{1}{2}, 1)$ and $(\frac{1}{8}, -\frac{1}{2})$.
The area is given by the integral of (right curve - left curve) with respect to $y$:
Area $= \int_{-1/2}^{1} \left( \frac{y+1}{4} - \frac{y^2}{2} \right) dy$
$= \left[ \frac{y^2}{8} + \frac{y}{4} - \frac{y^3}{6} \right]_{-1/2}^{1}$
$= \left( \frac{1}{8} + \frac{1}{4} - \frac{1}{6} \right) - \left( \frac{1/4}{8} - \frac{1/2}{4} - \frac{-1/8}{6} \right)$
$= \left( \frac{3+6-4}{24} \right) - \left( \frac{1}{32} - \frac{1}{8} + \frac{1}{48} \right)$
$= \frac{5}{24} - \left( \frac{3-12+2}{96} \right) = \frac{5}{24} - \left( -\frac{7}{96} \right) = \frac{20+7}{96} = \frac{27}{96} = \frac{9}{32}$ sq. units.

(A) The energy of an electron in the $n^{th}$ orbit of a hydrogen atom is given by the formula $E_n = -\frac{13.6}{n^2} \ eV$.
For the first excited state,the principal quantum number is $n = 2$.
Substituting $n = 2$ into the formula,we get $E_2 = -\frac{13.6}{2^2} = -\frac{13.6}{4} = -3.4 \ eV$.
Therefore,the energy of the first excited state of hydrogen is $-3.4 \ eV$.

(A) $N^{3-}$,$O^{2-}$,and $F^{-}$ are isoelectronic species,as each contains $10$ electrons.
For isoelectronic species,the ionic radius decreases as the nuclear charge (atomic number) increases.
The atomic numbers are $N (7)$,$O (8)$,and $F (9)$.
Therefore,the ionic radii decrease in the order $N^{3-} > O^{2-} > F^{-}$.
The values are $1.71 \ \mathring{A}$,$1.40 \ \mathring{A}$,and $1.36 \ \mathring{A}$ respectively.

(B) The interaction energy between two dipoles (dipole-dipole interaction) is proportional to $1/r^3$ for stationary dipoles,where $r$ is the distance between the molecules.
Hydrogen bonding is a specific,strong type of dipole-dipole interaction.
London dispersion forces depend on $1/r^6$.
Ion-ion interaction depends on $1/r$.
Ion-dipole interaction depends on $1/r^2$.

(D) Given: $\Delta G^{\circ} = 2494.2 \, J$,$T = 300 \, K$,$R = 8.314 \, J/K \cdot mol$.
First,calculate the equilibrium constant $K_c$ using $\Delta G^{\circ} = -RT \ln K_c$.
$2494.2 = -(8.314) \times 300 \times \ln K_c$.
$\ln K_c = -2494.2 / 2494.2 = -1$.
$K_c = e^{-1} \approx 0.367$.
Next,calculate the reaction quotient $Q$ for the given concentrations $[A] = 0.5, [B] = 2, [C] = 0.5$.
$Q = ([B][C]) / [A]^2 = (2 \times 0.5) / (0.5)^2 = 1 / 0.25 = 4$.
Since $Q = 4$ and $K_c \approx 0.367$,we have $Q > K_c$.
When $Q > K_c$,the reaction proceeds in the reverse direction to reach equilibrium.

(B) The standard free energy change for the reaction is given by $\Delta G^{\circ} = -RT \ln K_p$.
For the reaction $2 NO_{(g)} + O_{2(g)} \rightleftharpoons 2 NO_{2(g)}$,the standard free energy change is $\Delta G^{\circ} = 2 \Delta G^{\circ}_{f}(NO_2) - [2 \Delta G^{\circ}_{f}(NO) + \Delta G^{\circ}_{f}(O_2)]$.
Since $\Delta G^{\circ}_{f}(O_2) = 0$,we have $\Delta G^{\circ} = 2 \Delta G^{\circ}_{f}(NO_2) - 2 \Delta G^{\circ}_{f}(NO)$.
Given $\Delta G^{\circ}_{f}(NO) = 86.6 \, kJ/mol = 86600 \, J/mol$ and $T = 298 \, K$,we substitute these into the equation:
$-R(298) \ln (1.6 \times 10^{12}) = 2 \Delta G^{\circ}_{f}(NO_2) - 2(86600)$.
Rearranging for $\Delta G^{\circ}_{f}(NO_2)$:
$2 \Delta G^{\circ}_{f}(NO_2) = 2(86600) - R(298) \ln (1.6 \times 10^{12})$.
$\Delta G^{\circ}_{f}(NO_2) = 0.5 [2 \times 86600 - R(298) \ln (1.6 \times 10^{12})]$.

(D) The solubility of alkaline earth metal sulphates depends on the balance between lattice enthalpy and hydration enthalpy.
For a compound to be highly soluble,its hydration enthalpy must be greater than its lattice enthalpy.
As we move down the group,the size of the metal cation increases,which causes the hydration enthalpy to decrease significantly more than the lattice enthalpy.
$Be^{2+}$ is the smallest cation among the alkaline earth metals,resulting in a very high hydration enthalpy that outweighs its lattice enthalpy.
Therefore,$BeSO_4$ is highly soluble in water,while the solubility decreases down the group from $BeSO_4$ to $BaSO_4$.

(B) The ion exchange reaction is: $2(C_8H_7SO_3^- Na^+) + Ca^{2+} \rightarrow (C_8H_7SO_3^-)_2 Ca^{2+} + 2Na^+$.
From the stoichiometry,$2 \ mol$ of the resin are required to exchange $1 \ mol$ of $Ca^{2+}$ ions.
Therefore,$1 \ mol$ of the resin can take up $\frac{1}{2} \ mol$ of $Ca^{2+}$ ions.
Given the molar mass of the resin is $206 \ g/mol$,the uptake capacity per gram of resin is $\frac{1 \ mol \ Ca^{2+}}{2 \times 206 \ g \ resin} = \frac{1}{412} \ mol/g$.

(C) . It has to be stored in plastic or wax-lined glass bottles in the dark to prevent its decomposition. Thus,the statement is correct.
$B$. It has to be kept away from dust as it decomposes in the presence of dust. Thus,the statement is correct.
$C$. Hydrogen peroxide can act as both an oxidizing and a reducing agent. For example,in an acidic medium,it oxidizes ferrous ions to ferric ions: $2 Fe^{2+} + H_2O_2 + 2 H^{+} \rightarrow 2 Fe^{3+} + 2 H_2O$. In an acidic medium,it also reduces $HOCl$ to chloride ions: $HOCl + H_2O_2 \rightarrow H^{+} + Cl^{-} + H_2O + O_2$. Thus,the statement that it acts only as an oxidizing agent is incorrect.
$D$. It decomposes on exposure to light to form water and oxygen: $2 H_2O_2 \rightarrow 2 H_2O + O_2$. Thus,the statement is correct.

(C) For a compound to exhibit geometrical isomerism,each carbon atom of the $C=C$ double bond must be attached to two different groups.
In $1-$Phenyl$-2-$butene $(C_6H_5-CH_2-CH=CH-CH_3)$,the carbon atoms of the double bond are attached to different groups ($H$ and $CH_3$ on one side; $H$ and $CH_2C_6H_5$ on the other).
Therefore,it can exist as $cis-$ and $trans-$ isomers.

(C) Mass of organic compound $= 250 \ mg = 0.250 \ g$
Mass of $AgBr$ formed $= 141 \ mg = 0.141 \ g$
Molar mass of $AgBr = 108 + 80 = 188 \ g/mol$
$188 \ g$ of $AgBr$ contains $80 \ g$ of $Br$.
Mass of $Br$ in $0.141 \ g$ of $AgBr = \frac{80}{188} \times 0.141 \ g = 0.060 \ g$
Percentage of $Br = \frac{\text{Mass of } Br}{\text{Mass of organic compound}} \times 100$
Percentage of $Br = \frac{0.060}{0.250} \times 100 = 24 \%$

(A) Ozonolysis of a cyclic alkene involves the cleavage of the double bond to form a dicarbonyl compound.
For $1,3-$dimethylcyclopentene,the double bond is between $C_1$ and $C_2$.
Upon ozonolysis,the ring opens,and the carbons at the double bond become carbonyl groups.
Specifically,$1,3-$dimethylcyclopentene gives $5-$keto$-2-$methylhexanal as the product.
Therefore,the correct compound is $1,3-$dimethylcyclopentene.

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

(B) Alkyl fluorides are best prepared by the action of metallic fluorides like $Hg_2F_2$,$AgF$,or $SbF_3$ on alkyl halides. This specific method is known as the $Swarts$ reaction.
Example: $CH_3Br + AgF \rightarrow CH_3F + AgBr$.
In contrast,the $Finkelstein$ reaction is used to prepare alkyl iodides by treating alkyl chlorides or bromides with $NaI$ in acetone.
Free radical fluorination is highly exothermic and explosive,making it unsuitable for laboratory synthesis.

(A) Aluminium hydroxide $(Al(OH)_3)$,cimetidine,and ranitidine are antacids,while phenelzine is not.
Phenelzine is a tranquilizer,not an antacid. Phenelzine is used as an antidepressant drug.

(C) For two coaxial solenoids carrying current $I$ in the same direction,the magnetic field produced by the outer solenoid is uniform inside it and zero outside.
Since the inner solenoid is placed in a uniform magnetic field,the net magnetic force on it is zero because the forces on opposite sides of the current loops cancel each other out.
Similarly,the magnetic field produced by the inner solenoid is confined within its volume and is zero outside.
Therefore,the outer solenoid experiences no magnetic force from the inner one.
According to Newton's third law of motion,the action and reaction forces must be equal and opposite.
Since both forces are zero,they satisfy the condition $\vec{F}_1 = \vec{F}_2 = 0$.

(C) When $K_1$ is closed for a long time,the inductor acts as a short circuit. The initial current $I_0$ through the inductor is:
$I_0 = \frac{V}{R} = \frac{15 \ V}{0.15 \times 10^3 \ \Omega} = \frac{15}{150} = 0.1 \ A = 100 \ mA$.
When $K_1$ is opened and $K_2$ is closed,the battery is removed from the circuit,and the inductor discharges through the resistor. This is a decaying $LR$ circuit.
The current at time $t$ is given by $I(t) = I_0 e^{-Rt/L}$.
Given $R = 150 \ \Omega$,$L = 0.03 \ H$,and $t = 1 \ ms = 10^{-3} \ s$.
The exponent is $-\frac{Rt}{L} = -\frac{150 \times 10^{-3}}{0.03} = -\frac{0.15}{0.03} = -5$.
Thus,$I(t) = 100 \ mA \times e^{-5}$.
Using $e^5 \cong 150$,we have $e^{-5} = \frac{1}{e^5} \cong \frac{1}{150}$.
$I(1 \ ms) = 100 \times \frac{1}{150} = \frac{10}{15} = \frac{2}{3} \approx 0.67 \ mA$.

(D) The average force $F$ exerted by a molecule on the wall of a container is given by the rate of change of momentum,$F = \frac{\Delta p}{\Delta t}$.
For a molecule of mass $m$ moving with velocity $v$,the force is proportional to $\frac{mv^2}{L}$,where $L$ is the dimension of the container.
From the kinetic theory of gases,the average kinetic energy of a molecule is $\frac{1}{2}mv^2 = \frac{3}{2}kT$,which implies $v^2 \propto T$.
Substituting this into the force expression,we get $F \propto \frac{m(T)}{L}$,so $F \propto T^1$.
Comparing this with $F \propto T^q$,we find $q = 1$.

(B) For Rolle's theorem to hold on $[-1, 1]$,we must have $f(-1) = f(1)$.
$f(-1) = 2(-1)^3 + b(-1)^2 + c(-1) = -2 + b - c$
$f(1) = 2(1)^3 + b(1)^2 + c(1) = 2 + b + c$
Equating $f(-1) = f(1)$:
$-2 + b - c = 2 + b + c$
$-2c = 4 \implies c = -2$
Since Rolle's theorem holds at $x = \frac{1}{2}$,we have $f'(\frac{1}{2}) = 0$.
$f'(x) = 6x^2 + 2bx + c$
$f'(\frac{1}{2}) = 6(\frac{1}{4}) + 2b(\frac{1}{2}) + c = 0$
$\frac{3}{2} + b + c = 0$
Substituting $c = -2$:
$\frac{3}{2} + b - 2 = 0 \implies b = 2 - \frac{3}{2} = \frac{1}{2}$
Now,calculate $(2b + c)$:
$2b + c = 2(\frac{1}{2}) + (-2) = 1 - 2 = -1$.

(B) $Cimetidine$ and $Ranitidine$ are well-known antacids that reduce stomach acidity.
$Aluminium \ hydroxide$ is also used as an antacid.
$Phenelzine$ is an anti-depressant drug,not an antacid.

(B) According to the Bohr quantization condition,the angular momentum is given by $mvr = \frac{nh}{2\pi}$.
For a particle of mass $m$ and charge $q$ moving in a circular path in a magnetic field $B$,the magnetic force provides the centripetal force: $qvB = \frac{mv^2}{r}$,which implies $r = \frac{mv}{qB}$.
Substituting the expression for $r$ into the angular momentum equation: $mv \left( \frac{mv}{qB} \right) = \frac{nh}{2\pi}$.
This simplifies to $\frac{m^2v^2}{qB} = \frac{nh}{2\pi}$,or $mv^2 = \frac{nhqB}{2\pi m}$.
The kinetic energy $E$ of the particle is given by $E = \frac{1}{2}mv^2$.
Substituting the value of $mv^2$,we get $E = \frac{1}{2} \left( \frac{nhqB}{2\pi m} \right) = \frac{nhqB}{4\pi m}$.

(D) The boiling point of noble gases increases down the group as the atomic size increases.
As the atomic size increases,the magnitude of van der Waals forces of attraction increases.
Among the given noble gases ($He$,$Ne$,$Kr$,$Xe$),$Xe$ has the largest atomic size and therefore the strongest van der Waals forces.
Thus,$Xe$ has the highest boiling point.

(B) According to the $(n+l)$ rule,electrons fill orbitals in order of increasing $(n+l)$ values.
For $n=6$:
$6s: n+l = 6+0 = 6$
$4f: n+l = 4+3 = 7$
$5d: n+l = 5+2 = 7$
$6p: n+l = 6+1 = 7$
Comparing the orbitals with the same $(n+l)$ value,the one with the lower $n$ value fills first.
Thus,the order is $6s \to 4f \to 5d \to 6p$,which corresponds to $ns \to (n-2)f \to (n-1)d \to np$.

(D) Mass of hydrated salt = $61 \ g$
Mass of anhydrous $BaCl_2$ = $52 \ g$
Mass of water lost = $61 \ g - 52 \ g = 9 \ g$
Molar mass of $BaCl_2 = 137 + 2 \times 35.5 = 208 \ g/mol$
Molar mass of $H_2O = 18 \ g/mol$
Moles of $BaCl_2 = \frac{52}{208} = 0.25 \ mol$
Moles of $H_2O = \frac{9}{18} = 0.5 \ mol$
Ratio of moles of $H_2O$ to $BaCl_2 = \frac{0.5}{0.25} = 2$
Therefore,the formula is $BaCl_2 \cdot 2H_2O$.

(B) The reaction is an example of Kolbe's electrolysis of a dicarboxylic acid salt.
$1$. Upon electrolysis,the carboxylate ions $(R-COO^-)$ lose electrons at the anode to form carboxyl radicals $(R-COO^{\bullet})$.
$2$. These radicals undergo decarboxylation to release $CO_2$ and form alkyl radicals $(R^{\bullet})$.
$3$. In this specific case,the two radical centers formed on the cyclohexane ring combine to form a double bond,resulting in the formation of naphthalene,which is stable due to its aromatic character.
Therefore,the product $A$ is naphthalene.

(A) Milli-equivalents of $H_2SO_4$ taken $= 60 \ mL \times \frac{1}{10} \times 2 = 12 \ mEq$.
Milli-equivalents of $NaOH$ used for back titration $= 20 \ mL \times \frac{1}{10} = 2 \ mEq$.
Milli-equivalents of $NH_3$ evolved $= 12 - 2 = 10 \ mEq$.
Percentage of Nitrogen $= \frac{1.4 \times \text{milli-equivalents of } NH_3}{\text{mass of organic compound in } g} = \frac{1.4 \times 10}{1.4} = 10 \%$.

(C) The kinetic theory of gases assumes that gas particles occupy negligible volume and that collisions are perfectly elastic. However,it does not assume that gas particles are difficult to compress at high pressure. In reality,gases are compressible,and at high pressure,the volume of gas particles becomes significant,leading to deviations from ideal behavior.

(C) Photochemical smog is formed by the action of sunlight on nitrogen oxides and hydrocarbons.
It consists of primary pollutants like nitrogen oxides ($NO$ and $NO_2$) and secondary pollutants such as ozone $(O_3)$,formaldehyde,acrolein,and peroxyacetyl nitrate $(PAN)$.
Among the given options,$O_3$ is a major component of photochemical smog.

(C) Permanent hardness is caused by the presence of dissolved chlorides and sulfates of $Ca^{2+}$ and $Mg^{2+}$ ions.
Boiling is only effective for removing temporary hardness,which is caused by the presence of $HCO_3^-$ (bicarbonate) ions.
Methods like treatment with washing soda $(Na_2CO_3)$,Calgon's method,and the ion exchange method are used to remove permanent hardness.

(B) The assertion is correct because in a bonding molecular orbital $(MO)$, the electron density is concentrated in the region between the two nuclei, which leads to attraction and stability.
The reason is incorrect because the bonding $MO$ is formed by the constructive interference of the atomic orbitals $(\psi_A + \psi_B)$, not destructive interference. Destructive interference results in an antibonding molecular orbital $(\psi_A - \psi_B)$.

(C) The valence shell electronic configuration is given as $5s^2\,5p^4$.
The principal quantum number $n = 5$ indicates that the element belongs to period $5$.
The total number of valence electrons is $2 + 4 = 6$.
For $p$-block elements,the group number is calculated as $10 + \text{valence electrons} = 10 + 6 = 16$.
Therefore,the element belongs to group $16$ and period $5$.

(C) The thermal stability of alkaline earth metal hydroxides increases down the group as the size of the metal cation increases.
As the size of the cation increases,the lattice energy changes,and the polarizing power of the cation decreases.
Therefore,the thermal stability increases in the order: $Mg(OH)_2 < Ca(OH)_2 < Sr(OH)_2 < Ba(OH)_2$.

(B) In $CH_4,$ the heat of atomization corresponds to $4 \times BE_{(C-H)} = 360 \ kJ/mol.$
Therefore,$BE_{(C-H)} = 90 \ kJ/mol.$
In $C_2H_6,$ the heat of atomization corresponds to $BE_{(C-C)} + 6 \times BE_{(C-H)} = 620 \ kJ/mol.$
Substituting $BE_{(C-H)} = 90 \ kJ/mol,$ we get $BE_{(C-C)} + 6(90) = 620 \ kJ/mol.$
Therefore,$BE_{(C-C)} = 620 - 540 = 80 \ kJ/mol.$
The energy required per molecule is $E = \frac{80 \times 10^3 \ J/mol}{6.02 \times 10^{23} \ molecules/mol} \approx 1.329 \times 10^{-19} \ J/molecule.$
Using the relation $E = \frac{hc}{\lambda},$ the wavelength is $\lambda = \frac{hc}{E}.$
$\lambda = \frac{6.62 \times 10^{-34} \ J \cdot s \times 3 \times 10^8 \ m/s}{1.329 \times 10^{-19} \ J} \approx 1.494 \times 10^{-6} \ m.$
Converting to nanometers,$\lambda \approx 1.494 \times 10^3 \ nm.$
Thus,the correct option is $B$.

(C) Reaction: $N_2O_{4(g)} \rightleftharpoons 2NO_{2(g)}$
Initial moles $(t=0)$: $1 \ mol$ of $N_2O_4$ and $0 \ mol$ of $NO_2$.
At equilibrium $(t=eq)$: $(1-\alpha) \ mol$ of $N_2O_4$ and $2\alpha \ mol$ of $NO_2$,where $\alpha = 0.2$.
Total moles at equilibrium: $n_{total} = (1-\alpha) + 2\alpha = 1 + \alpha = 1 + 0.2 = 1.2 \ mol$.
Molar mass of $N_2O_4 = 92 \ g/mol$ and $NO_2 = 46 \ g/mol$.
Average molar mass of the mixture $(M_{mix})$:
$M_{mix} = \frac{(1 - \alpha) \times 92 + 2\alpha \times 46}{1 + \alpha} = \frac{0.8 \times 92 + 0.4 \times 46}{1.2} = \frac{73.6 + 18.4}{1.2} = \frac{92}{1.2} = 76.66 \ g/mol$.
Using the ideal gas law $PV = nRT$ or $PM = dRT$:
$d = \frac{P \times M_{mix}}{R \times T} = \frac{1 \times 76.66}{0.0821 \times 300} = \frac{76.66}{24.63} \approx 3.11 \ g/L$.

(C) In $XeOF_4$,the central atom $Xe$ has $8$ valence electrons.
It forms $4$ single bonds with $F$ atoms and $1$ double bond with the $O$ atom,utilizing $6$ electrons for bonding.
This leaves $2$ electrons,forming $1$ lone pair.
Total electron pairs = $5$ bond pairs + $1$ lone pair = $6$ electron pairs.
With $6$ electron pairs,the hybridization is $sp^3d^2$,which corresponds to an octahedral electron geometry.
Due to the presence of $1$ lone pair,the molecular geometry is square pyramidal.

(A) Given the differential equation: $(x + 2)\frac{dy}{dx} = x^2 + 4x - 9$.
Rearranging for $dy$:
$dy = \frac{x^2 + 4x - 9}{x + 2} dx$.
Perform polynomial division on the right side:
$\frac{x^2 + 4x - 9}{x + 2} = \frac{x(x + 2) + 2x - 9}{x + 2} = x + \frac{2x - 9}{x + 2} = x + \frac{2(x + 2) - 13}{x + 2} = x + 2 - \frac{13}{x + 2}$.
Integrating both sides:
$y = \int (x + 2 - \frac{13}{x + 2}) dx = \frac{x^2}{2} + 2x - 13 \ln|x + 2| + C$.
Using the initial condition $y(0) = 0$:
$0 = \frac{0^2}{2} + 2(0) - 13 \ln|0 + 2| + C \implies C = 13 \ln 2$.
So,the solution is $y(x) = \frac{x^2}{2} + 2x - 13 \ln|x + 2| + 13 \ln 2$.
Now,find $y(-4)$:
$y(-4) = \frac{(-4)^2}{2} + 2(-4) - 13 \ln|-4 + 2| + 13 \ln 2$
$y(-4) = \frac{16}{2} - 8 - 13 \ln 2 + 13 \ln 2$
$y(-4) = 8 - 8 = 0$.

(D) The reaction of $1$-methylcyclopentene with $D-Cl$ proceeds via an electrophilic addition mechanism.
$1$. The electrophile $D^+$ attacks the double bond to form the more stable tertiary carbocation at the $C-1$ position.
$2$. The resulting carbocation is $sp^2$ hybridized and planar.
$3$. The nucleophile $Cl^-$ can attack the planar carbocation from either the top or the bottom face with equal probability.
$4$. This leads to the formation of both possible stereoisomers (diastereomers) where the $Cl$ atom and the $CH_3$ group can be either cis or trans to each other relative to the $D$ atom.
$5$. Therefore,both products shown in $(b)$ and $(c)$ are formed.

(D) $H_2$ is a highly inflammable gas. Therefore,the statement that it is a non-inflammable gas is incorrect.

(D) The molecular formula $C_6H_{14}$ corresponds to an alkane. The structural isomers are:
$1$. $n$-hexane: $CH_3-CH_2-CH_2-CH_2-CH_2-CH_3$
$2$. $2$-methylpentane: $CH_3-CH(CH_3)-CH_2-CH_2-CH_3$
$3$. $3$-methylpentane: $CH_3-CH_2-CH(CH_3)-CH_2-CH_3$
$4$. $2,3$-dimethylbutane: $CH_3-CH(CH_3)-CH(CH_3)-CH_3$
$5$. $2,2$-dimethylbutane: $CH_3-C(CH_3)_2-CH_2-CH_3$
Thus,there are a total of $5$ structural isomers.

(A) In Lassaigne's test,fusion with sodium takes place and the following species are formed:
$A$. Aniline $(C_6H_5NH_2)$ contains $N$,which forms $CN^-$ ions. $CN^-$ reacts with hot and acidic solution of $FeSO_4$ to form $Fe_4[Fe(CN)_6]_3$ (Prussian blue),which is blue in colour. Thus,$A-iii$.
$B$. Benzene sulfonic acid $(C_6H_5SO_3H)$ contains $S$,which forms $S^{2-}$ ions. However,the question implies a phenolic derivative or specific test context. Given the options,if we consider the formation of a phenoxide-like species or specific test,$FeCl_3$ is used for phenol tests (red/violet colour). Thus,$B-i$.
$C$. Thiourea $(NH_2CSNH_2)$ contains both $N$ and $S$. The $S^{2-}$ ions formed react with sodium nitroprusside $(Na_2[Fe(CN)_5NO])$ to give a violet colour. Thus,$C-ii$.
Therefore,the correct match is $A-iii, B-i, C-ii$.

(D) The density of liquid water is higher than that of ice,which means the volume of ice is greater than the volume of an equivalent mass of liquid water $(V_{\text{ice}} > V_{\text{water}})$.
According to Le Chatelier's principle,an increase in pressure favors the direction that results in a decrease in volume.
Since the forward reaction (ice $\rightarrow$ water) involves a decrease in volume,the equilibrium will shift in the forward direction.

(D) During vigorous exercise,the demand for energy in the muscles increases significantly.
Since the supply of oxygen is insufficient to meet this demand through aerobic respiration,the muscles switch to anaerobic respiration.
In this process,$Glucose$ is converted into $Pyruvic \ acid$ via glycolysis.
Under anaerobic conditions,$Pyruvic \ acid$ is further reduced to $L^{-}$-lactic acid.
The accumulation of $L^{-}$-lactic acid in the muscle tissues leads to fatigue and muscle cramps.

(D) According to Fajan's rule,covalent character increases as the size of the cation decreases and its charge density increases.
Among the alkaline earth metal ions ($Mg^{2+}$,$Ca^{2+}$,$Sr^{2+}$,$Ba^{2+}$),the $Mg^{2+}$ ion has the smallest size.
Therefore,$MgCl_2$ exhibits the most significant covalent character among the given halides.

(C) gas deviates from ideal behaviour when the assumptions of the kinetic molecular theory are no longer valid.
At $high$ pressure,the volume occupied by the gas molecules becomes significant compared to the total volume of the container.
At $low$ temperature,the kinetic energy of the molecules decreases,making intermolecular forces of attraction significant.
Therefore,at $high$ pressure and $low$ temperature,the gas deviates the most from its ideal behaviour.

(C) The dipole moment $(\mu)$ is defined as the product of the magnitude of the charge $(q)$ and the distance $(d)$ between the centers of positive and negative charges,i.e.,$\mu = q \times d$.
In $p$-nitroaniline,the electron-donating $-NH_2$ group and the electron-withdrawing $-NO_2$ group are at the para positions.
Due to resonance,there is a significant charge separation where the lone pair on the nitrogen of the $-NH_2$ group is delocalized towards the $-NO_2$ group,creating a strong polar structure with a large distance between the partial positive and negative charges.
This resonance effect is most pronounced in the para isomer due to its linear geometry,leading to the maximum dipole moment compared to the ortho and meta isomers.

(D) The addition of phosphate-containing fertilisers to water bodies leads to a process known as $Eutrophication$.
Phosphates act as limiting nutrients for aquatic plants.
When these fertilisers enter water bodies,they promote the excessive growth of algae,which is known as an algal bloom.
This excessive growth eventually leads to the depletion of dissolved oxygen in the water,which can harm aquatic life.

(D) The kinetic energy of a particle is given by $K.E. = \frac{3}{2} KT = \frac{1}{2} mv^2$.
Thus,$v = \sqrt{\frac{3KT}{m}}$.
The de Broglie wavelength is $\lambda = \frac{h}{mv} = \frac{h}{m \sqrt{\frac{3KT}{m}}} = \frac{h}{\sqrt{3KTm}}$.
This implies $\lambda \propto \frac{1}{\sqrt{m}}$.
Since the mass of a photon is effectively zero (or it travels at $c$),it has the largest wavelength.
Comparing particles,$m_{\text{electron}} < m_{\text{neutron}} < m_{\text{proton}}$.
Therefore,$\lambda_{\text{electron}} > \lambda_{\text{neutron}} > \lambda_{\text{proton}}$.
Considering the photon,the order is $\lambda_{\text{photon}} > \lambda_{\text{electron}} > \lambda_{\text{neutron}}$.

(A) The balanced chemical equation is: $A + 2B + 3C \rightleftharpoons AB_2C_3$
Calculate the moles of reactants:
$n_A = \frac{6.0 \ g}{60 \ g/mol} = 0.1 \ mol$
$n_B = \frac{6.0 \times 10^{23}}{6 \times 10^{23}} = 1.0 \ mol$
$n_C = 0.036 \ mol$
Determine the limiting reagent:
For $A$: $0.1 / 1 = 0.1$
For $B$: $1.0 / 2 = 0.5$
For $C$: $0.036 / 3 = 0.012$
Since $C$ has the lowest ratio,it is the limiting reagent.
Calculate moles of product $AB_2C_3$ formed:
$n_{AB_2C_3} = \frac{n_C}{3} = \frac{0.036}{3} = 0.012 \ mol$
Calculate the molar mass of $AB_2C_3$:
$M.M. = \frac{\text{mass}}{\text{moles}} = \frac{4.8 \ g}{0.012 \ mol} = 400 \ g/mol$
Calculate atomic mass of $B$ $(x)$:
$400 = (1 \times 60) + (2 \times x) + (3 \times 80)$
$400 = 60 + 2x + 240$
$400 = 300 + 2x$
$2x = 100$
$x = 50 \ amu$

(A) Positional isomers are compounds that have the same molecular formula and the same functional group,but the functional group is attached to different positions on the carbon chain.
In option $A$,$CH_3-CH_2-CH_2-CO-CH_3$ is $pentan-2-one$ and $CH_3-CH_2-CO-CH_2-CH_3$ is $pentan-3-one$.
Both have the same molecular formula $C_5H_{10}O$ and the same ketone functional group,but the carbonyl group is at the $C-2$ position in the first compound and the $C-3$ position in the second compound.
Therefore,they are positional isomers.

(C) The initial amount of acetic acid in $50 \, mL$ of $0.06 \, N$ solution is calculated as:
$w_1 = \frac{N \times M_{wt} \times V}{1000} = \frac{0.06 \times 60 \times 50}{1000} = 0.18 \, g = 180 \, mg$.
The final amount of acetic acid in the filtrate after adsorption is:
$w_2 = \frac{0.042 \times 60 \times 50}{1000} = 0.126 \, g = 126 \, mg$.
The amount of acetic acid adsorbed by $3 \, g$ of charcoal is:
$w_{ads} = 180 \, mg - 126 \, mg = 54 \, mg$.
The amount of acetic acid adsorbed per gram of charcoal is:
$\frac{54 \, mg}{3 \, g} = 18 \, mg/g$.

(D) The relative lowering of vapour pressure is given by the formula: $\frac{P^o - P_s}{P_s} = \frac{w_2 \times M_1}{w_1 \times M_2}$
Here,$P^o = 185\,torr$ (vapour pressure of pure solvent),
$P_s = 183\,torr$ (vapour pressure of solution),
$w_1 = 100\,g$ (mass of solvent,acetone),
$M_1 = 58\,g\,mol^{-1}$ (molar mass of acetone,$CH_3COCH_3$),
$w_2 = 1.2\,g$ (mass of solute),
$M_2 = ?$ (molar mass of solute).
Substituting the values:
$\frac{185 - 183}{183} = \frac{1.2 \times 58}{100 \times M_2}$
$\frac{2}{183} = \frac{69.6}{100 \times M_2}$
$M_2 = \frac{69.6 \times 183}{2 \times 100} = \frac{12736.8}{200} = 63.684 \approx 64\,g\,mol^{-1}$.

(D) The reduction reaction at the cathode is: $Cu^{2+} + 2e^{-} \longrightarrow Cu(s)$.
According to the stoichiometry of the reaction,$2 \ mol$ of electrons ($2 \ F$ of electricity) are required to deposit $1 \ mol$ of $Cu$.
The molar mass of $Cu$ is $63.5 \ g/mol$.
Therefore,passing $2 \ F$ of electricity deposits $63.5 \ g$ of $Cu$ at the cathode.

(C) Higher order $(> 3)$ reactions are rare because the probability of simultaneous collision of all the reacting species is extremely low.
For a reaction to occur,molecules must collide with sufficient energy and proper orientation.
As the number of molecules involved in a single step increases,the likelihood of them colliding at the same time and place decreases significantly.

(B) In the Hall-Heroult process,$Al_2O_3$ is dissolved in molten cryolite $(Na_3AlF_6)$ and fluorspar $(CaF_2)$,which lowers the melting point of the mixture and increases electrical conductivity.
$Al^{3+}$ is reduced to $Al$ at the cathode.
Carbon anodes are oxidized to $CO$ and $CO_2$.
$Na_3AlF_6$ (cryolite) acts as a solvent for $Al_2O_3$,not as the electrolyte itself; the mixture of $Al_2O_3$ and cryolite acts as the electrolyte. Therefore,the statement that $Na_3AlF_6$ serves as the electrolyte is technically false as it is a component of the electrolytic mixture.

(C) Nitrogen $(N_2)$ and oxygen $(O_2)$ are the primary components of the atmosphere.
They do not react under normal atmospheric conditions to form nitrogen oxides $(NO_x)$ because the reaction is highly endothermic and requires a very high temperature (e.g.,lightning or internal combustion engines) to overcome the high bond dissociation energy of the $N \equiv N$ triple bond.
Therefore,both the assertion and the reason are correct,and the reason provides the correct explanation for the assertion.

(B) As we move down the group in the noble gases,the atomic size increases.
This leads to an increase in the magnitude of the van der Waals forces of attraction between the atoms.
Consequently,the boiling point increases as we move down the group.
Among the given options ($He$,$Ne$,$Kr$,$Xe$),$Xe$ is at the bottom of the group,thus it has the highest boiling point.

(C) The compound $Zn_2[Fe(CN)_6]$ is white in color.
$(NH_4)_3[As(Mo_3O_{10})_4]$ (Ammonium phosphomolybdate/arsenomolybdate derivative) is yellow.
$BaCrO_4$ (Barium chromate) is yellow due to ligand-to-metal charge transfer $(LMCT)$.
$K_3[Co(NO_2)_6]$ (Potassium hexanitrocobaltate$(III)$) is yellow.
$Zn_2[Fe(CN)_6]$ is a white precipitate formed during the test for $Zn^{2+}$ ions.

(D) The correct matches are as follows:
$A. TiCl_4$ is used in $ii. \text{Ziegler-Natta polymerization}$.
$B. PdCl_2$ is used in $i. \text{Wacker process}$.
$C. CuCl_2$ is used in $iv. \text{Deacon's process}$.
$D. V_2O_5$ is used in $iii. \text{Contact process}$.
Therefore,the correct sequence is $A-ii, B-i, C-iv, D-iii$.

(D) The complex $[Pt(Cl)(py)(NH_3)(NH_2OH)]^+$ is of the type $[M(abcd)]$,where $M=Pt$,$a=Cl^-$,$b=py$,$c=NH_3$,and $d=NH_2OH$.
For a square planar complex of the type $[M(abcd)]$,the number of geometric isomers is given by $3$.
These isomers are formed by fixing one ligand (e.g.,$Cl^-$) at one position and arranging the other three ligands ($py$,$NH_3$,$NH_2OH$) in the remaining three positions relative to the fixed ligand.
Thus,there are $3$ possible geometric isomers.

(A) $KMnO_4 \to K^{+} + MnO_4^-$.
In $MnO_4^-$,$Mn$ is in the $+7$ oxidation state,which means it has a $d^0$ electronic configuration (no electrons in $d$-orbitals).
The intense purple color of $KMnO_4$ is not due to $d-d$ transitions because there are no $d$-electrons.
Instead,it arises from $L \to M$ (Ligand to Metal) charge transfer,where an electron from the oxygen $2p$ orbital is excited to the vacant $Mn$ $3d$ orbital.
Since this charge transfer is both Laporte-allowed and spin-allowed,it results in an intense color.

(B) Step $1$: Toluene $(C_6H_5CH_3)$ undergoes oxidation with $KMnO_4$ to form benzoic acid $(C_6H_5COOH)$,which is product $A$.
Step $2$: Benzoic acid reacts with thionyl chloride $(SOCl_2)$ to form benzoyl chloride $(C_6H_5COCl)$,which is product $B$.
Step $3$: Benzoyl chloride undergoes Rosenmund reduction with $H_2/Pd$ and $BaSO_4$ to form benzaldehyde $(C_6H_5CHO)$,which is product $C$.

(A) The reaction proceeds in two steps:
$1$. The reaction of $p$-toluidine ($4$-methylaniline) with $NaNO_2/HCl$ at $0-5^{\circ}C$ is a diazotization reaction,which forms the diazonium salt,$4$-methylbenzenediazonium chloride $(D)$.
$2$. The reaction of the diazonium salt $(D)$ with $CuCN/KCN$ is a Sandmeyer reaction,where the diazonium group $(-N_2^+Cl^-)$ is replaced by a cyano group $(-CN)$ to form $4$-methylbenzonitrile $(E)$.

(C) Vitamins $B$ and $C$ are water soluble.
Vitamins $A, D, E$ and $K$ are fat soluble or water insoluble.

(D) Bakelite is used for making gears,protective coatings,and electrical fittings.
$(b)$ Glyptal is used in the manufacture of paints and lacquers.
$(c)$ $PP$ (Polypropene) is used in the manufacture of textiles,packaging materials,etc.
$(d)$ Polyvinyl chloride $(PVC)$ is used in the manufacture of raincoats,handbags,and leather clothes.

(B) Let $A$ be benzene and $B$ be toluene.
Given: $P^o_A = 74.7\,torr$,$P^o_B = 22.3\,torr$,$n_A = 1.5\,mol$,$n_B = 3.5\,mol$.
Total moles $n_{total} = 1.5 + 3.5 = 5.0\,mol$.
Mole fraction of benzene in liquid phase,$x_A = \frac{1.5}{5.0} = 0.3$.
Mole fraction of toluene in liquid phase,$x_B = \frac{3.5}{5.0} = 0.7$.
Total vapour pressure of the solution,$P_{total} = P^o_A x_A + P^o_B x_B = (74.7 \times 0.3) + (22.3 \times 0.7) = 22.41 + 15.61 = 38.02\,torr \approx 38.0\,torr$.
Mole fraction of benzene in vapour phase $(y_A)$ is given by $y_A = \frac{P_A}{P_{total}} = \frac{P^o_A x_A}{P_{total}} = \frac{22.41}{38.02} \approx 0.589$.

(B) The presence of $Fe^{3+}$ ions is indicated by the blood-red color formed with $CNS^{-}$ ions: $Fe^{3+} + 3CNS^{-} \to Fe(CNS)_3$ (Ferric thiocyanate,blood-red coloration).
$Fe^{3+}$ ions react with $K_4[Fe(CN)_6]$ to form a blue precipitate (Prussian blue): $4Fe^{3+} + 3[Fe(CN)_6]^{4-} \to Fe_4[Fe(CN)_6]_3$.
The chromyl chloride test is specific for the presence of chloride ions $(Cl^{-})$,where $X$ reacts with $K_2Cr_2O_7$ and concentrated $H_2SO_4$ to evolve reddish-brown vapors of chromyl chloride $(CrO_2Cl_2)$.
Since the salt $X$ contains both $Fe^{3+}$ and $Cl^{-}$ ions,the salt is $FeCl_3$.

(C) Compound $A$ $(C_{10}H_{13}Cl)$ gives a white precipitate with silver nitrate,indicating it is a tertiary halide or a reactive alkyl halide that forms a stable carbocation.
$C_6H_5-CH_2-C(Cl)(CH_3)_2 (A) \xrightarrow{alc. KOH} C_6H_5-CH=C(CH_3)_2 (B)$
$B$ on ozonolysis:
$C_6H_5-CH=C(CH_3)_2 \xrightarrow{O_3/H_2O} C_6H_5CHO (C) + CH_3COCH_3 (D)$
$C$ $(C_6H_5CHO)$ has no $\alpha$-hydrogen,so it gives Cannizzaro reaction but not aldol condensation.
$D$ $(CH_3COCH_3)$ has $\alpha$-hydrogens,so it gives aldol condensation but not Cannizzaro reaction.
Thus,$A$ is $C_6H_5-CH_2-C(Cl)(CH_3)_2$.

(D) The reaction described is the Hell-Volhard-Zelinsky $(HVZ)$ reaction.
In this reaction,aliphatic carboxylic acids containing at least one $\alpha$-hydrogen atom are reacted with chlorine or bromine in the presence of a small amount of red phosphorus.
The $\alpha$-hydrogen is replaced by a halogen atom to form $\alpha$-halo carboxylic acids.
The general chemical equation is: $R-CH_2-COOH + X_2 \xrightarrow{\text{red } P} R-CH(X)-COOH + HX$,where $X = Cl, Br$.

(C) Adsorption is a surface phenomenon where molecules of a substance accumulate on the surface of a solid or liquid.
During adsorption,the residual forces on the surface are satisfied by the incoming molecules,leading to a decrease in surface energy.
Therefore,the statement that residual forces increase is incorrect.
For adsorption,$\Delta H < 0$ (exothermic),$\Delta S < 0$ (decrease in randomness),and $\Delta G < 0$ (spontaneous process).

(A) ligand is an atom,molecule,or ion that donates a pair of electrons to a central metal atom or ion to form a coordinate bond.
For a species to act as a ligand,it must possess at least one lone pair of electrons.
In $CH_4$ (methane),the carbon atom is $sp^3$ hybridized and forms four covalent bonds with hydrogen atoms,completing its octet.
Since $CH_4$ has no lone pair of electrons available for donation,it cannot act as a ligand.

(A) For the reaction $2N_2O_5 \,(g) \to 4NO_2 \,(g) + O_2 \,(g)$,let the initial pressure of $N_2O_5$ be $P_0 = 50 \, mmHg$.
At $t = 30 \, min$,let the pressure of $N_2O_5$ decrease by $2p$. The total pressure is $P_t = (50 - 2p) + 4p + p = 50 + 3p = 87.5 \, mmHg$.
Thus,$3p = 37.5 \, mmHg$,so $p = 12.5 \, mmHg$.
The pressure of $N_2O_5$ remaining at $t = 30 \, min$ is $50 - 2(12.5) = 25 \, mmHg$.
Since the pressure of $N_2O_5$ halves in $30 \, min$,the half-life $t_{1/2} = 30 \, min$.
At $t = 60 \, min$ (which is $2 \times t_{1/2}$),the pressure of $N_2O_5$ remaining is $50 / 4 = 12.5 \, mmHg$.
Let $50 - 2p' = 12.5$,so $2p' = 37.5$,which means $p' = 18.75 \, mmHg$.
The total pressure at $t = 60 \, min$ is $50 + 3p' = 50 + 3(18.75) = 50 + 56.25 = 106.25 \, mmHg$.

(C) Calcination is the process of heating an ore strongly below its melting point in the absence or limited supply of air.
It is primarily used to convert metal carbonates and hydroxides into their respective oxides.
For example:
$CaCO_3 \xrightarrow{\Delta} CaO + CO_2 \uparrow$
$CuCO_3 \cdot Cu(OH)_2 \xrightarrow{\Delta} 2CuO + H_2O \uparrow + CO_2 \uparrow$
Thus,the product obtained is a metal oxide.

(D) An optically inactive compound is one that does not possess a chiral center (a carbon atom bonded to four different groups).
In $2-$chloropropanal $(CH_3-CHCl-CHO)$,the $C2$ carbon is chiral.
In $2-$chlorobutane $(CH_3-CH_2-CHCl-CH_3)$,the $C2$ carbon is chiral.
In $2-$chloropentane $(CH_3-CH_2-CH_2-CHCl-CH_3)$,the $C2$ carbon is chiral.
In $2-$chloro$-2-$methylbutane $(CH_3-CH_2-CCl(CH_3)_2)$,the $C2$ carbon is bonded to two identical methyl groups,making it achiral.
Therefore,$2-$chloro$-2-$methylbutane is optically inactive.

(D) The correct matching is as follows:

Polymer	Use
$(A)$ Polystyrene	$(iii)$ Manufacture of toys
$(B)$ Glyptal	$(i)$ Paints and lacquers
$(C)$ Polyvinyl chloride	$(ii)$ Rain coats
$(D)$ Bakelite	$(iv)$ Computer discs

Thus,the correct sequence is $(A)-(iii), (B)-(i), (C)-(ii), (D)-(iv)$.

(C) For complex $A$,$[Ni(H_2O)_5NH_3]^{2+}$,it is of the type $[MA_5B]$,which does not show geometrical or optical isomerism.
For complex $B$,$[Ni(H_2O)_4(NH_3)_2]^{2+}$,it is of the type $[MA_4B_2]$. It shows geometrical isomerism ($cis$ and $trans$ forms). The $cis$ form is optically active.
For complex $C$,$[Ni(H_2O)_3(NH_3)_3]^{2+}$,it is of the type $[MA_3B_3]$. It shows geometrical isomerism ($facial$ and $meridional$ forms). The $facial$ form is optically active.
Thus,both $B$ and $C$ exhibit geometrical and optical isomerism.

(A) The basicity of amines depends on the availability of the lone pair of electrons on the nitrogen atom.
Aliphatic amines (like $CH_3NH_2$) are more basic than aromatic amines because the lone pair on the nitrogen atom in aromatic amines is involved in resonance with the benzene ring,making it less available for protonation.
Among aromatic amines,electron-donating groups (like $-OCH_3$) increase the electron density on the nitrogen atom,thereby increasing basicity.
Electron-withdrawing groups (like $-NO_2$) decrease the electron density on the nitrogen atom,thereby decreasing basicity.
Therefore,the order of increasing basicity is: $p$-nitroaniline < aniline < $p$-methoxyaniline < $CH_3NH_2$.

(C) The $EMF$ of the given galvanic cell is $1.1 \ V$.
When $E_{ext} < 1.1 \ V$,the cell functions as a galvanic cell,and electrons flow from the anode $(Zn)$ to the cathode $(Cu)$.
When $E_{ext} > 1.1 \ V$,the external potential overcomes the cell potential,forcing the reaction to reverse; thus,the cell acts as an electrolytic cell and electrons flow from the cathode $(Cu)$ to the anode $(Zn)$.

(C) In qualitative inorganic analysis,cations are divided into groups based on their precipitation reactions.
$Pb^{2+}$ (Group $II$),$Cu^{2+}$ (Group $II$),and $As^{3+}$ (Group $II$) are precipitated as sulfides by $H_2S$ in the presence of $dil. \ HCl$ due to the low solubility product of their sulfides.
$Co^{2+}$ belongs to Group $IV$ and is not precipitated by $H_2S$ in the presence of $dil. \ HCl$ because the concentration of $S^{2-}$ ions is too low due to the common ion effect of $H^+$.
$Co^{2+}$ is precipitated as $CoS$ only in the presence of $NH_4OH$ and $H_2S$ (Group $IV$ reagent).

(A) The structure shown in the image is $2$-acetoxybenzoic acid,commonly known as Aspirin.
Drugs which relieve pain are called analgesics.
Analgesics are of two types: $(i)$ Narcotics and $(ii)$ Non-narcotics.
Aspirin (acetylsalicylic acid) is a well-known non-narcotic analgesic.

(A) Starch is a polysaccharide composed of $amylose$ and $amylopectin$. Both of these are polymers of $D-glucose$. Therefore,the complete hydrolysis of starch yields only $glucose$ units.

(C) Fluorine is the most electronegative element and has the least tendency to form $p\pi-d\pi$ double bonds. Due to its small size and high electronegativity,it forms only one oxyacid,which is $HOF$ (hypofluorous acid). In contrast,other halogens like $Cl$,$Br$,and $I$ form multiple oxyacids due to the availability of vacant $d$-orbitals.

(C) The initial rate is given by $R = K[A][B]$.
When the concentration of $A$ is kept constant and the concentration of $B$ is doubled,the new concentration of $B$ becomes $[B]' = 2[B]$.
The new rate $R'$ is given by $R' = K[A][B]' = K[A](2[B])$.
Substituting the initial rate expression into the new rate expression,we get $R' = 2 \times (K[A][B]) = 2R$.
Therefore,the rate of the reaction is doubled.

(A) The aqueous solution of $CoCl_2$ contains the octahedral complex $[Co(H_2O)_6]^{2+}$,which is reddish-pink in colour.
When concentrated $HCl$ is added,the chloride ions $(Cl^-)$ act as ligands and replace the water molecules to form the tetrahedral complex $[CoCl_4]^{2-}$.
The formation of the tetrahedral $[CoCl_4]^{2-}$ complex is responsible for the deep blue colour observed in the solution.
Therefore,the correct complex ion is $[CoCl_4]^{2-}$.

(D) The Critical Micelle Concentration $(CMC)$ of a surfactant decreases as the length of the hydrophobic hydrocarbon chain increases.
This is because a longer chain increases the hydrophobic effect and the van der Waals forces of attraction between the hydrocarbon tails,which stabilizes the micelle structure.
Comparing the given options:
$A$: $C_9$ chain
$B$: $C_{12}$ chain
$C$: $C_{14}$ chain
$D$: $C_{16}$ chain
Since $CH_3(CH_2)_{15}N^{+}(CH_3)_3Br^{-}$ has the longest hydrocarbon chain $(C_{16})$,it will have the strongest hydrophobic interactions and will form micelles at the lowest molar concentration.

(A) complex ion is considered to have symmetrically filled orbitals if the electrons are distributed equally among the degenerate orbitals of the $t_{2g}$ and $e_g$ sets.
$1$. For $[FeF_6]^{3-}$: $Fe^{3+}$ is $d^5$. $F^-$ is a weak field ligand,so the configuration is $t_{2g}^3 e_g^2$. Here,$t_{2g}$ has $3$ electrons (one in each orbital) and $e_g$ has $2$ electrons (one in each orbital). Both sets are symmetrically filled.
$2$. For $[Mn(CN)_6]^{4-}$: $Mn^{2+}$ is $d^5$. $CN^-$ is a strong field ligand,so the configuration is $t_{2g}^5 e_g^0$. This is unsymmetrical.
$3$. For $[CoF_6]^{3-}$: $Co^{3+}$ is $d^6$. $F^-$ is a weak field ligand,so the configuration is $t_{2g}^4 e_g^2$. This is unsymmetrical.
$4$. For $[Co(NH_3)_6]^{2+}$: $Co^{2+}$ is $d^7$. $NH_3$ is a strong field ligand,so the configuration is $t_{2g}^6 e_g^1$. This is unsymmetrical.
Therefore,$[FeF_6]^{3-}$ has symmetrically filled orbitals.

(B) The reduction half-reaction for permanganate is: $MnO_4^- + 8H^+ + 5e^- \to Mn^{2+} + 4H_2O$.
Using the Nernst equation: $E = E^\circ - \frac{0.059}{5} \log \frac{[Mn^{2+}]}{[MnO_4^-][H^+]^8}$.
Assuming standard concentrations for $Mn^{2+}$ and $MnO_4^-$ $(1 \ M)$,the potential at $pH = 3$ is: $E = 1.51 - \frac{0.059}{5} \times \log \frac{1}{(10^{-3})^8} = 1.51 - \frac{0.059 \times 24}{5} = 1.51 - 0.2832 = 1.2268 \ V$.
Permanganate can oxidize species whose standard reduction potential is lower than its own potential $(1.2268 \ V)$.
Comparing this with the given potentials: $Cl_2/Cl^- = 1.36 \ V$,$Br_2/Br^- = 1.07 \ V$,and $I_2/I^- = 0.54 \ V$.
Since $1.2268 \ V > 1.07 \ V$ and $1.2268 \ V > 0.54 \ V$,but $1.2268 \ V < 1.36 \ V$,permanganate will oxidize $Br^-$ and $I^-$.

(A) Calamine is a mineral consisting of zinc carbonate. Its chemical formula is $ZnCO_3$.

(C) Neoprene is a synthetic rubber formed by the free radical polymerization of chloroprene ($2$-chloro-$1,3$-butadiene).
The repeating unit of the polymer is $[CH_2-C(Cl)=CH-CH_2]_n$.

(A) Sucralose is a trichloro derivative of sucrose. Its structure contains three chlorine atoms,as shown in the provided chemical structure. Therefore,it is the artificial sweetener that contains chlorine.

(A) Given,$\frac{E_f}{E_b} = \frac{2}{3}$.
We know that for a reaction,$\Delta H = E_f - E_b$.
Given $\Delta H = -40 \ kJ/mol$,so $-40 = E_f - E_b$,which implies $E_b = E_f + 40$.
Substituting this into the ratio: $\frac{E_f}{E_f + 40} = \frac{2}{3}$.
Cross-multiplying gives $3E_f = 2(E_f + 40)$.
$3E_f = 2E_f + 80$.
$E_f = 80 \ kJ/mol$.
Then,$E_b = 80 + 40 = 120 \ kJ/mol$.

(C) Chlorine water is yellow in colour due to the presence of dissolved $Cl_2$.
On standing,$Cl_2$ reacts with water to form hydrochloric acid and hypochlorous acid.
The reaction is: $Cl_2 + H_2O \to HCl + HOCl$.
Since $HOCl$ is unstable,it further decomposes to give $HCl$ and nascent oxygen,which causes the bleaching action and loss of colour.

(D) Acetic acid $(CH_3COOH)$ contains a carboxylic group $(-COOH)$ which can form intermolecular hydrogen bonds in non-polar solvents like benzene.
Due to this,acetic acid molecules undergo association to form dimers.
This association reduces the number of effective particles in the solution,which affects the colligative property (freezing point depression) and leads to an abnormal molar mass.

(C) $H_4P_2O_6$ (Hypophosphoric acid) contains a $P-P$ bond.
Its structural formula is $(HO)_2P(O)-P(O)(OH)_2$.

(A, B) $Na_2Cr_2O_7$ is deliquescent and is $NOT$ used as a primary standard in volumetry,whereas $K_2Cr_2O_7$ is used as a primary standard.
$Na_2Cr_2O_7$ is more soluble than $K_2Cr_2O_7$ due to the higher hydration energy of $Na^+$ ions compared to $K^+$ ions.
$CrO_4^{2-}$ has a tetrahedral structure.
$Cr_2O_7^{2-}$ contains a $Cr-O-Cr$ bridge bond.

(B) The reaction is an Aldol condensation.
Step $1$: $2CH_3CHO \xrightarrow{OH^{-}} CH_3-CH(OH)-CH_2-CHO$ (Aldol,$A$)
Step $2$: $CH_3-CH(OH)-CH_2-CHO \xrightarrow{\Delta} CH_3-CH=CH-CHO$ (Crotonaldehyde,$B$)
Therefore,the product $B$ is $CH_3-CH=CH-CHO$.

(A) Hydrated cobalt$(II)$ salts,such as $CoCl_2 \cdot 6H_2O$,are pink in colour due to the presence of the $[Co(H_2O)_6]^{2+}$ complex.
Upon heating,the water of crystallization is lost,and the salt converts to anhydrous cobalt$(II)$ chloride $(CoCl_2)$,which is blue in colour.
Therefore,the presence of the $Co^{2+}$ cation is most likely.

(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 $1.86 \ \mathring{A}$.

(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.
$CH_3Br + AgF \rightarrow CH_3F + AgBr$
$2CH_3CH_2Cl + Hg_2F_2 \rightarrow 2CH_3CH_2F + Hg_2Cl_2$

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