JEE Main 2016 Chemistry Question Paper with Answer and Solution

(A) Given that the total length of the wire is $2$ units.
Perimeter of the square $= 4x$ and circumference of the circle $= 2\pi r$.
Thus,$4x + 2\pi r = 2$,which simplifies to $2x + \pi r = 1$.
From this,we get $r = \frac{1 - 2x}{\pi}$.
The sum of the areas $A$ is given by $A = x^2 + \pi r^2$.
Substituting the value of $r$,we get $A = x^2 + \pi \left( \frac{1 - 2x}{\pi} \right)^2 = x^2 + \frac{(1 - 2x)^2}{\pi}$.
To find the minimum area,we differentiate $A$ with respect to $x$ and set it to zero:
$\frac{dA}{dx} = 2x + \frac{2(1 - 2x)(-2)}{\pi} = 0$.
$2x - \frac{4(1 - 2x)}{\pi} = 0$.
Multiplying by $\pi$,we get $2\pi x - 4 + 8x = 0$.
$x(2\pi + 8) = 4$,so $x = \frac{4}{2\pi + 8} = \frac{2}{\pi + 4}$.
Now,substitute $x$ back into the perimeter equation: $2(\frac{2}{\pi + 4}) + \pi r = 1$.
$\pi r = 1 - \frac{4}{\pi + 4} = \frac{\pi + 4 - 4}{\pi + 4} = \frac{\pi}{\pi + 4}$.
Therefore,$r = \frac{1}{\pi + 4}$.
Comparing $x$ and $r$,we see that $x = \frac{2}{\pi + 4}$ and $r = \frac{1}{\pi + 4}$,which implies $x = 2r$.

(C) The equation for the line passing through points $(V_0, 2P_0)$ and $(2V_0, P_0)$ is:
Slope $m = \frac{P_0 - 2P_0}{2V_0 - V_0} = \frac{-P_0}{V_0}$.
Using the point-slope form $P - P_0 = m(V - V_0)$:
$P - 2P_0 = \frac{-P_0}{V_0}(V - V_0)$
$P = \frac{-P_0}{V_0}V + P_0 + 2P_0 = \frac{-P_0}{V_0}V + 3P_0$.
From the ideal gas law,$PV = nRT$,so $T = \frac{PV}{nR}$.
Substituting $P$:
$T = \frac{1}{nR} \left( \frac{-P_0}{V_0}V + 3P_0 \right)V = \frac{1}{nR} \left( \frac{-P_0}{V_0}V^2 + 3P_0V \right)$.
For maximum temperature,$\frac{dT}{dV} = 0$:
$\frac{dT}{dV} = \frac{1}{nR} \left( \frac{-2P_0}{V_0}V + 3P_0 \right) = 0$.
$\frac{2P_0}{V_0}V = 3P_0 \implies V = \frac{3}{2}V_0$.
Substituting $V = \frac{3}{2}V_0$ back into the equation for $T$:
$T_{\text{max}} = \frac{1}{nR} \left( \frac{-P_0}{V_0} \left( \frac{3}{2}V_0 \right)^2 + 3P_0 \left( \frac{3}{2}V_0 \right) \right)$
$T_{\text{max}} = \frac{1}{nR} \left( \frac{-9}{4}P_0V_0 + \frac{9}{2}P_0V_0 \right) = \frac{1}{nR} \left( \frac{9}{4}P_0V_0 \right) = \frac{9P_0V_0}{4nR}$.

(B) The kinetic energy acquired by an electron accelerated through a potential difference $V$ is given by $K.E. = eV = \frac{1}{2}mu^2$, where $u$ is the velocity of the electron.
Rearranging for velocity: $u = \sqrt{\frac{2eV}{m}}$.
According to the de Broglie relation, the wavelength $\lambda$ is given by $\lambda = \frac{h}{mu}$, which implies $\frac{h}{\lambda} = mu$.
Substituting the expression for $u$ into the equation for $\frac{h}{\lambda}$:
$\frac{h}{\lambda} = m \times \sqrt{\frac{2eV}{m}} = \sqrt{m^2 \times \frac{2eV}{m}} = \sqrt{2meV}$.

(C) To determine the hybridization of the $N$ atom,we use the formula: $\text{Steric Number} = \frac{1}{2} [V + M - C + A]$,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.
For $NO_{2}^{+}$: $\text{Steric Number} = \frac{1}{2} [5 + 0 - 1 + 0] = \frac{4}{2} = 2$. $A$ steric number of $2$ corresponds to $sp$ hybridization.
For $NO_{3}^{-}$: $\text{Steric Number} = \frac{1}{2} [5 + 0 - 0 + 1] = 3$ ($sp^2$ hybridization).
For $NO_{2}$: The $N$ atom has one unpaired electron and is $sp^2$ hybridized.
For $NO_{2}^{-}$: $\text{Steric Number} = \frac{1}{2} [5 + 0 - 0 + 1] = 3$ ($sp^2$ hybridization).
Thus,the $N$ atom is $sp$ hybridized in $NO_{2}^{+}$.
Therefore,the correct option is $C$.

(A) Initially,the number of moles of gas in each bulb is $n_1 = \frac{P_i V}{R T_1}$ and $n_2 = \frac{P_i V}{R T_1}$.
After the temperature of the second bulb is raised to $T_2$,the number of moles of gas in both bulbs are $n_1' = \frac{P_f V}{R T_1}$ and $n_2' = \frac{P_f V}{R T_2}$.
Since the total number of moles of gas in both bulbs remains constant:
$n_1 + n_2 = n_1' + n_2'$
$\frac{P_i V}{R T_1} + \frac{P_i V}{R T_1} = \frac{P_f V}{R T_1} + \frac{P_f V}{R T_2}$
$\frac{2 P_i V}{R T_1} = \frac{P_f V}{R} \left( \frac{1}{T_1} + \frac{1}{T_2} \right)$
$\frac{2 P_i}{T_1} = P_f \left( \frac{T_2 + T_1}{T_1 T_2} \right)$
$P_f = \frac{2 P_i T_2}{T_1 + T_2}$

(B) The heat of formation of $CO$ corresponds to the reaction: $C_{(s)} + \frac{1}{2} O_{2(g)} \rightarrow CO_{(g)}$.
Given reactions:
$1) C_{(s)} + O_{2(g)} \rightarrow CO_{2(g)}; \Delta H_1 = -393.5 \ kJ \ mol^{-1}$
$2) CO_{(g)} + \frac{1}{2} O_{2(g)} \rightarrow CO_{2(g)}; \Delta H_2 = -283.5 \ kJ \ mol^{-1}$
To obtain the target reaction,subtract equation $(2)$ from equation $(1)$:
$(C_{(s)} + O_{2(g)}) - (CO_{(g)} + \frac{1}{2} O_{2(g)}) \rightarrow CO_{2(g)} - CO_{2(g)}$
$C_{(s)} + \frac{1}{2} O_{2(g)} \rightarrow CO_{(g)}$
Therefore,$\Delta H_f = \Delta H_1 - \Delta H_2 = -393.5 - (-283.5) = -110 \ kJ \ mol^{-1}$.

(A) For the reaction $A + B \rightleftharpoons C + D$,the equilibrium constant $K_c = 100$.
Initial concentrations: $[A]_0 = 1 \ M, [B]_0 = 1 \ M, [C]_0 = 1 \ M, [D]_0 = 1 \ M$.
Let $x$ be the change in concentration at equilibrium.
At equilibrium: $[A] = 1-x, [B] = 1-x, [C] = 1+x, [D] = 1+x$.
$K_c = \frac{[C][D]}{[A][B]} = \frac{(1+x)(1+x)}{(1-x)(1-x)} = \left(\frac{1+x}{1-x}\right)^2 = 100$.
Taking the square root of both sides: $\frac{1+x}{1-x} = 10$.
$1+x = 10 - 10x \implies 11x = 9 \implies x = \frac{9}{11} \approx 0.818$.
Equilibrium concentration of $D$ is $[D] = 1 + x = 1 + 0.818 = 1.818 \ M$.

(A) The statement regarding intramolecular hydrogen bonding is $FALSE$. In the condensed phase (liquid or solid),water molecules exhibit extensive $intermolecular$ hydrogen bonding,not $intramolecular$ hydrogen bonding.

(B) The first ionization energy is the energy required to remove the most loosely bound electron from an isolated gaseous atom.
$K$ $(Z=19)$ and $Na$ $(Z=11)$ are alkali metals (Group $1$),which have the lowest ionization energies in their respective periods.
$Rb$ $(Z=37)$ is also an alkali metal with an even lower ionization energy than $K$ due to its larger atomic size.
$Sc$ $(Z=21)$ is a $d$-block transition metal.
Transition metals generally have higher first ionization energies compared to alkali metals in the same or adjacent periods because of their smaller atomic radii and higher effective nuclear charge,which holds the valence electrons more tightly.

(B) When alkali metals are heated in excess of air,they form different types of oxides based on their size and polarizability:
$1$. Lithium forms the monoxide: $4Li + O_2 \longrightarrow 2Li_2O$
$2$. Sodium forms the peroxide: $2Na + O_2 \longrightarrow Na_2O_2$
$3$. Potassium forms the superoxide: $K + O_2 \longrightarrow KO_2$
Thus,the correct sequence is $Li_2O, Na_2O_2$ and $KO_2$.

(D) To determine the absolute configuration,we assign priorities to the groups attached to each chiral carbon using the Cahn-Ingold-Prelog $(CIP)$ rules.
For the $C-2$ carbon:
The groups are $-OH$ (priority $1$),$-CH(Cl)CH_3$ (priority $2$),$-CO_2H$ (priority $3$),and $-H$ (priority $4$).
Since the lowest priority group $(-H)$ is on a horizontal bond,the configuration is reversed. The sequence $1$ $\rightarrow 2$ $\rightarrow 3$ is clockwise,which would be $R$,but due to the horizontal $-H$,it becomes $S$. Thus,$C-2$ is $2S$.
For the $C-3$ carbon:
The groups are $-Cl$ (priority $1$),$-CH(OH)CO_2H$ (priority $2$),$-CH_3$ (priority $3$),and $-H$ (priority $4$).
Since the lowest priority group $(-H)$ is on a horizontal bond,the configuration is reversed. The sequence $1$ $\rightarrow 2$ $\rightarrow 3$ is counter-clockwise,which would be $S$,but due to the horizontal $-H$,it becomes $R$. Thus,$C-3$ is $3R$.
Therefore,the absolute configuration is $(2S, 3R)$.

(B) Spent-lye and glycerol are separated by distillation under reduced pressure.
Under reduced pressure,the liquid boils at a lower temperature,preventing thermal decomposition.
For example,glycerol boils at $290^{\circ}C$ with decomposition at atmospheric pressure,but at reduced pressure,it boils at $180^{\circ}C$ without decomposition.

(A) Bunsen flame consists of several zones based on the extent of combustion.
Region $1$ is the innermost dark zone containing unburnt fuel.
Region $2$ is the luminous zone where incomplete combustion occurs.
Region $3$ is the non-luminous zone where complete combustion occurs,making it the hottest part of the flame.
Region $4$ is the tip of the flame.
Therefore,the hottest region is region $3$.

(D) The combustion reaction is: $C_xH_y + (x + \frac{y}{4}) O_2 \rightarrow x CO_2 + \frac{y}{2} H_2O_{(l)}$
Volume of $O_2$ used $= 375 \times 0.20 = 75 \ mL$.
From the stoichiometry: $15(x + \frac{y}{4}) = 75$,which simplifies to $x + \frac{y}{4} = 5$,or $4x + y = 20$.
The final volume $330 \ mL$ consists of $CO_2$ produced and the remaining $N_2$ from the air.
Volume of $N_2 = 375 - 75 = 300 \ mL$.
Volume of $CO_2 = 330 - 300 = 30 \ mL$.
Since $15 \ mL$ of hydrocarbon produces $30 \ mL$ of $CO_2$,$x = 2$. Wait,checking stoichiometry: $15x = 30 \implies x = 2$.
Substituting $x=2$ into $4x + y = 20$: $4(2) + y = 20 \implies y = 12$. This suggests $C_2H_{12}$ which is impossible.
Re-evaluating: The total volume $330 \ mL$ includes $CO_2$ and $N_2$. $15x = 30 \implies x = 2$. If $x=3$,$15x = 45 \ mL$ of $CO_2$. Then $N_2 = 300 \ mL$,total $= 345 \ mL$. If $x=3$,$4(3) + y = 20 \implies y = 8$. This gives $C_3H_8$. $15 \ mL$ of $C_3H_8$ produces $45 \ mL$ of $CO_2$. Total volume $= 45 + 300 = 345 \ mL$. Given $330 \ mL$,let's check $C_3H_6$: $x=3, y=6$. $4(3)+6 = 18 \neq 20$. The correct hydrocarbon is $C_3H_8$.

(D) The reaction of propene with $HOCl$ involves the electrophilic attack of $Cl^+$ on the double bond.
The $Cl^+$ ion attacks the double bond to form a cyclic chloronium ion intermediate,which is then opened by the nucleophilic attack of $H_2O$ or $OH^-$.
However,in terms of carbocation character during the ring opening,the positive charge is stabilized on the more substituted carbon atom.
The intermediate formed is $CH_3-CH^+-CH_2-Cl$.

(A) First,convert all concentrations to $ppm$ $(1 \ ppm = 1000 \ ppb)$:
$1.$ Fluoride: $1000 \ ppb = 1 \ ppm$ (Permissible limit is $1 \ ppm$)
$2.$ Lead: $40 \ ppb = 0.04 \ ppm$ (Permissible limit is $0.05 \ ppm$)
$3.$ Nitrate: $100 \ ppm$ (Permissible limit is $45 \ ppm$)
$4.$ Iron: $0.2 \ ppm$ (Permissible limit is $0.2 \ ppm$)
The concentration of nitrate $(100 \ ppm)$ is significantly higher than the permissible limit of $45 \ ppm$. Excess nitrate in drinking water causes methemoglobinemia,also known as blue baby syndrome.

(C) Let the equation of the incident ray be $y - 1 = m(x - 0)$,which is $mx - y + 1 = 0$.
The angle between the incident ray and the mirror line $7x - y + 1 = 0$ is equal to the angle between the reflected ray $2x + y - 1 = 0$ and the mirror line.
The slope of the incident ray is $m_1 = m$,the slope of the mirror is $m_2 = 7$,and the slope of the reflected ray is $m_3 = -2$.
Using the formula $\tan \theta = |\frac{m_1 - m_2}{1 + m_1 m_2}| = |\frac{m_2 - m_3}{1 + m_2 m_3}|$:
$|\frac{m - 7}{1 + 7m}| = |\frac{7 - (-2)}{1 + 7(-2)}| = |\frac{9}{1 - 14}| = |\frac{9}{-13}| = \frac{9}{13}$.
Case $1$: $\frac{m - 7}{1 + 7m} = \frac{9}{13}$ $\Rightarrow 13m - 91 = 9 + 63m$ $\Rightarrow -50m = 100$ $\Rightarrow m = -2$ (This is the reflected ray).
Case $2$: $\frac{m - 7}{1 + 7m} = -\frac{9}{13}$ $\Rightarrow 13m - 91 = -9 - 63m$ $\Rightarrow 76m = 82$ $\Rightarrow m = \frac{41}{38}$.
The equation of the incident ray is $y - 1 = \frac{41}{38}(x - 0)$ $\Rightarrow 38y - 38 = 41x$ $\Rightarrow 41x - 38y + 38 = 0$.

(C) For a polytropic process $PV^n = \text{constant}$, the molar heat capacity $C$ is given by the formula:
$C = C_V + \frac{R}{1 - n}$
Rearranging the formula to solve for $n$:
$C - C_V = \frac{R}{1 - n}$
Since $R = C_P - C_V$, we substitute this into the equation:
$C - C_V = \frac{C_P - C_V}{1 - n}$
$1 - n = \frac{C_P - C_V}{C - C_V}$
$n = 1 - \frac{C_P - C_V}{C - C_V}$
$n = \frac{C - C_V - (C_P - C_V)}{C - C_V}$
$n = \frac{C - C_V - C_P + C_V}{C - C_V}$
$n = \frac{C - C_P}{C - C_V}$

(D) Let $v_1$ be the velocity of water at the hole and $v_2$ be the velocity of water when it hits the ground.
Using Torricelli's Law,the velocity of water at the hole is $v_1 = \sqrt{2gH}$.
Using the equation of motion for a freely falling body,the velocity at the ground is $v_2 = \sqrt{v_1^2 + 2gh} = \sqrt{2gH + 2gh} = \sqrt{2g(H+h)}$.
According to the equation of continuity,$A_1 v_1 = A_2 v_2$,where $A_1 = \pi r^2$ is the area of the hole and $A_2 = \pi x^2$ is the area of the stream at the ground.
So,$\pi r^2 \sqrt{2gH} = \pi x^2 \sqrt{2g(H+h)}$.
Dividing both sides by $\pi \sqrt{2g}$,we get $r^2 \sqrt{H} = x^2 \sqrt{H+h}$.
Squaring both sides,$r^4 H = x^4 (H+h)$.
Therefore,$x^4 = r^4 \left( \frac{H}{H+h} \right)$,which gives $x = r \left( \frac{H}{H+h} \right)^{\frac{1}{4}}$.

(A) $CH_3MgBr$ is a Grignard reagent,which acts as a strong base.
Methanol $(CH_3OH)$ contains an acidic hydrogen atom attached to the oxygen atom.
When $CH_3MgBr$ reacts with $CH_3OH$,the methyl group $(CH_3^-)$ abstracts the acidic proton from the methanol to form methane $(CH_4)$ gas.
The reaction is: $CH_3MgBr + CH_3OH \rightarrow CH_4 \uparrow + Mg(OCH_3)Br$.

(B) The total number of orbitals in a shell is given by the formula $n^2$,where $n$ is the principal quantum number.
For $n = 5$,the number of orbitals $= (5)^2 = 25$.

(C) $BOD$ stands for Biochemical Oxygen Demand. It is a measure of the amount of dissolved oxygen required by aerobic microorganisms to decompose the organic matter present in a given water sample at a specific temperature over a specific time period.

(B) Heavy water $(D_2O)$ is used as a moderator in nuclear reactors to slow down fast-moving neutrons,not as a coolant.
Chemical reactions of $D_2O$ are analogous to $H_2O$:
$1$. $SO_3 + D_2O \rightarrow D_2SO_4$
$2$. $CaC_2 + 2D_2O \rightarrow C_2D_2 + Ca(OD)_2$
$3$. $Al_4C_3 + 12D_2O \rightarrow 3CD_4 + 4Al(OD)_3$
Therefore,the statement that it is used as a coolant is incorrect.

(A) For one mole of a real gas,the van der Waals equation is given by:
$(P + \frac{a}{V^2})(V - b) = RT$
At very high pressures,the volume $V$ is small,so the term $\frac{a}{V^2}$ becomes negligible compared to $P$.
Thus,the equation simplifies to:
$P(V - b) = RT$
$PV - Pb = RT$
$PV = RT + Pb$
Dividing both sides by $RT$:
$\frac{PV}{RT} = 1 + \frac{Pb}{RT}$
Since the compressibility factor $Z = \frac{PV}{RT}$,we get:
$Z = 1 + \frac{Pb}{RT}$

(C) Fluorine is the most electronegative element in the periodic table.
Due to its highest electronegativity,it always attracts electrons and never loses them to form positive oxidation states.
Therefore,it exhibits an oxidation state of $-1$ only.

(D) The general combustion reaction for an alkane $C_nH_{2n+2}$ is:
$C_nH_{2n+2} + (\frac{3n+1}{2})O_2 \to nCO_2 + (n+1)H_2O$
According to Avogadro's law,at constant temperature and pressure,the volume of a gas is directly proportional to the number of moles.
Therefore,the ratio of volumes is equal to the ratio of stoichiometric coefficients:
$\frac{V_{alkane}}{V_{O_2}} = \frac{1}{\frac{3n+1}{2}} = \frac{2}{3n+1}$
Given $V_{alkane} = 5 \, L$ and $V_{O_2} = 25 \, L$:
$\frac{5}{25} = \frac{2}{3n+1}$
$\frac{1}{5} = \frac{2}{3n+1}$
$3n+1 = 10$
$3n = 9$
$n = 3$
Since $n=3$,the alkane is propane $(C_3H_8)$.

(A) The solubility of alkaline earth metal sulphates decreases down the group from $Be$ to $Ba$.
This is because the hydration energy decreases more rapidly than the lattice energy as the size of the cation increases down the group.
Since the sulphate ion is very large,the lattice energy remains relatively constant,while the hydration energy decreases significantly.
Therefore,the correct order of solubility is $MgSO_4 > CaSO_4 > SrSO_4 > BaSO_4$.

(D) To determine the shape,we calculate the hybridization and number of lone pairs for each molecule in option $D$:
$ClF_3$: Hybridization $= 3 + \frac{1}{2}[7 - 3] = 5$ $(sp^3d)$,with $2$ lone pairs,resulting in a $T$-shaped geometry.
$XeOF_2$: Hybridization $= 3 + \frac{1}{2}[8 - 4] = 5$ $(sp^3d)$,with $2$ lone pairs,resulting in a $T$-shaped geometry.
$XeF_3^+$: Hybridization $= 3 + \frac{1}{2}[8 - 3 - 1] = 5$ $(sp^3d)$,with $2$ lone pairs,resulting in a $T$-shaped geometry.
Since all three molecules have $sp^3d$ hybridization and $2$ lone pairs,they all possess an identical $T$-shaped geometry.

(A) The noble gas $Xe$ is non-polar and consists of individual atoms.
For non-polar species,the only intermolecular forces present are London dispersion forces,also known as instantaneous dipole-induced dipole forces.
These weak forces are responsible for the liquefaction of $Xe$ gas at low temperatures.

(A) . Silica gel is used as a drying agent to absorb moisture.
$B$. Silicon is a semiconductor used in the manufacturing of transistors.
$C$. Silicones are synthetic polymers used as sealants,greases,and electrical insulators.
$D$. Silicates are used as ion-exchangers in water softening processes.
Therefore,the correct matching is $A-iii, B-i, C-iv, D-ii$.

(B) neopentyl group is $(CH_3)_3C-CH_2-$ and a vinyl group is $CH_2=CH-$.
Combining these two groups,we get the structure: $CH_2=CH-CH_2-C(CH_3)_3$.
According to $IUPAC$ rules,the numbering of the parent chain starts from the end closer to the double bond.
The longest chain contains $5$ carbon atoms.
Thus,the name is $4, 4-$dimethyl$-1-$pentene.

(C) The percentage of an element in a compound is given by the formula:
$\% \text{ of Sulphur} = \frac{\text{Atomic weight of } S \times \text{number of atoms}}{\text{Molecular weight of compound}} \times 100$
For the minimum molecular weight,we assume there is at least $1$ atom of sulphur per molecule.
$8 = \frac{32 \times 1}{\text{Molecular weight}} \times 100$
$\text{Molecular weight} = \frac{32 \times 100}{8} = 400\,g\,mol^{-1}$

(D) The balanced chemical equation is:
$2H_3AsO_4 + 5H_2S \xrightarrow{\text{conc. } HCl} As_2S_5 + 8H_2O$
First,calculate the molar mass of arsenic acid $(H_3AsO_4)$:
$M = (3 \times 1) + 74.92 + (4 \times 16) \approx 142 \ g/mol$.
Number of moles of $H_3AsO_4 = \frac{\text{mass}}{\text{molar mass}} = \frac{35.5 \ g}{142 \ g/mol} = 0.25 \ mol$.
According to the stoichiometry of the reaction,$2 \ mol$ of $H_3AsO_4$ produce $1 \ mol$ of $As_2S_5$.
Therefore,$0.25 \ mol$ of $H_3AsO_4$ will produce:
$\frac{0.25}{2} = 0.125 \ mol$ of $As_2S_5$.

(C) Given: $\Delta H^o = -29.8 \, kJ \, mol^{-1}$,$\Delta S^o = -0.100 \, kJ \, K^{-1} \, mol^{-1}$,and $T = 298 \, K$.
Using the Gibbs free energy equation: $\Delta G^o = \Delta H^o - T\Delta S^o$.
Substituting the values: $\Delta G^o = -29.8 \, kJ \, mol^{-1} - (298 \, K \times -0.100 \, kJ \, K^{-1} \, mol^{-1})$.
$\Delta G^o = -29.8 + 29.8 = 0 \, kJ \, mol^{-1}$.
Since $\Delta G^o = -RT \ln K_{eq}$,and $\Delta G^o = 0$,we have $0 = -RT \ln K_{eq}$.
This implies $\ln K_{eq} = 0$,which means $K_{eq} = e^0 = 1$.

(A) The spontaneity of a reaction is determined by the Gibbs free energy equation: $\Delta G = \Delta H - T\Delta S$.
For a reaction to be spontaneous,$\Delta G$ must be negative $(\Delta G < 0)$.
If $\Delta H > 0$ (endothermic) and $\Delta S > 0$ (increase in entropy),then at high temperatures,the term $T\Delta S$ becomes larger than $\Delta H$.
Consequently,$\Delta G = \Delta H - T\Delta S$ becomes negative,making the reaction spontaneous at high temperatures.
At low temperatures,$T\Delta S < \Delta H$,so $\Delta G$ is positive,making the reaction non-spontaneous.

(A) Let a point on the curve be $P(\alpha, \alpha^2 - 4)$. The distance $D$ of this point from the origin $(0,0)$ is given by $D = \sqrt{\alpha^2 + (\alpha^2 - 4)^2}$.
To minimize $D$,we minimize $f(\alpha) = D^2 = \alpha^2 + (\alpha^2 - 4)^2$.
$f(\alpha) = \alpha^2 + \alpha^4 - 8\alpha^2 + 16 = \alpha^4 - 7\alpha^2 + 16$.
Taking the derivative with respect to $\alpha$ and setting it to zero:
$f'(\alpha) = 4\alpha^3 - 14\alpha = 0$.
$2\alpha(2\alpha^2 - 7) = 0$.
This gives $\alpha = 0$ or $\alpha^2 = \frac{7}{2}$.
If $\alpha^2 = 0$,$f(0) = 16$.
If $\alpha^2 = \frac{7}{2}$,$f(\alpha) = (\frac{7}{2})^2 - 7(\frac{7}{2}) + 16 = \frac{49}{4} - \frac{49}{2} + 16 = -\frac{49}{4} + 16 = \frac{15}{4}$.
Since $\frac{15}{4} < 16$,the minimum value of $D^2$ is $\frac{15}{4}$.
Thus,the minimum distance $D = \sqrt{\frac{15}{4}} = \frac{\sqrt{15}}{2}$.

(A) We have the integral $I = \int \frac{dx}{\cos^3 x \sqrt{2 \sin 2x}}$.
Using $\sin 2x = 2 \sin x \cos x$,the integral becomes:
$I = \int \frac{dx}{\cos^3 x \sqrt{4 \sin x \cos x}} = \int \frac{dx}{2 \cos^3 x \sqrt{\sin x \cos x}} = \int \frac{dx}{2 \cos^4 x \sqrt{\tan x}}$.
Since $\frac{1}{\cos^4 x} = \sec^4 x = (1 + \tan^2 x) \sec^2 x$,we have:
$I = \frac{1}{2} \int \frac{(1 + \tan^2 x) \sec^2 x}{\sqrt{\tan x}} dx$.
Let $u = \tan x$,then $du = \sec^2 x dx$:
$I = \frac{1}{2} \int \frac{1 + u^2}{\sqrt{u}} du = \frac{1}{2} \int (u^{-1/2} + u^{3/2}) du$.
Integrating term by term:
$I = \frac{1}{2} \left( \frac{u^{1/2}}{1/2} + \frac{u^{5/2}}{5/2} \right) + k = u^{1/2} + \frac{1}{5} u^{5/2} + k$.
Substituting $u = \tan x$ back:
$I = (\tan x)^{1/2} + \frac{1}{5} (\tan x)^{5/2} + k$.
Comparing this with the given form $(\tan x)^A + C(\tan x)^B + k$,we get $A = 1/2$,$B = 5/2$,and $C = 1/5$.
Therefore,$A + B + C = \frac{1}{2} + \frac{5}{2} + \frac{1}{5} = 3 + \frac{1}{5} = \frac{16}{5}$.

(A) According to the law of equivalence,the number of equivalents of acid must equal the number of equivalents of base.
Equivalents of base = (moles of $OH^-$) $\times$ (acidity of base) = $0.04 \times 1 = 0.04 \, eq$.
Equivalents of acid = Normality $\times$ Volume (in $L$).
$0.1 \times V = 0.04$.
$V = \frac{0.04}{0.1} = 0.4 \, L$.
Converting to $mL$: $0.4 \times 1000 = 400 \, mL$.

(A) The electronic configuration $1s^2 \, 2s^2 \, 2p^6 \, 3s^2 \, 3p^6$ corresponds to the $Ar$ (Argon) noble gas core,which has $18$ electrons.
In $NaF$,the ions are $Na^+$ ($10$ electrons: $1s^2 \, 2s^2 \, 2p^6$) and $F^-$ ($10$ electrons: $1s^2 \, 2s^2 \, 2p^6$). Neither ion has the $18$-electron configuration.
In $KBr$,$K^+$ ($18$ electrons) and $Br^-$ ($36$ electrons) are present. $K^+$ has the $18$-electron configuration.
In $NaCl$,$Cl^-$ ($18$ electrons) is present.
In $CaI_2$,$Ca^{2+}$ ($18$ electrons) is present.
Therefore,the aqueous solution of $NaF$ does not contain any ions with the $1s^2 \, 2s^2 \, 2p^6 \, 3s^2 \, 3p^6$ configuration.

(C) Sodium extract is heated with conc. $HNO_3$ to convert any $NaCN$ formed during fusion into $HCN$ and sulphide into $H_2S$.
This is done when the compound contains $N$ or $S$ along with halogens.
$HCN$ and $H_2S$ formed escape as they are volatile.
Otherwise,they interfere with the test for halogens as they also form a precipitate with $AgNO_3$.
$NaCN + HNO_3 \rightarrow NaNO_3 + HCN \uparrow$
$Na_2S + 2HNO_3 \rightarrow 2NaNO_3 + H_2S \uparrow$

(A) The chemical formula for calcium oxide is $CaO$.
It is commercially known as Quick lime.

(B) Liquefied $CO_2$ is considered a better strategy for dry cleaning because it is environmentally friendly and replaces the use of hazardous solvents like $Cl_2C=CCl_2$ (Tetrachloroethylene),which is a suspected carcinogen and groundwater pollutant. Therefore,$CO_2$ is the preferred choice for sustainable dry cleaning.

(B) The bond angle depends on the hybridization of the central atom and the repulsion between lone pairs and bond pairs.
In $CH_4$,the central atom $C$ is $sp^3$ hybridized with no lone pair,resulting in a tetrahedral geometry with a bond angle of $109.5^{\circ}$.
In $NH_3$,$H_2O$,and $PH_3$,the central atoms have lone pairs which cause bond pair-lone pair repulsion,leading to a decrease in bond angle from the ideal tetrahedral angle.
Specifically,$PH_3$ has a bond angle of approximately $93.5^{\circ}$ due to the lack of hybridization and the large size of the $P$ atom.
Therefore,$CH_4$ has the greatest bond angle among the given options.

(A) In the reaction of lithium hydride $(LiH)$ with diborane $(B_2H_6)$,lithium borohydride is formed without the evolution of hydrogen gas:
$2LiH + B_2H_6 \longrightarrow 2Li[BH_4]$
In contrast,the other reactions listed liberate hydrogen gas:
$1$. Electrolysis of acidified water: $2H_2O \longrightarrow 2H_2 + O_2$
$2$. Zinc with aqueous alkali: $Zn + 2NaOH + 2H_2O \longrightarrow Na_2[Zn(OH)_4] + H_2$
$3$. Sodium in liquid ammonia: $2Na + 2NH_3 \longrightarrow 2NaNH_2 + H_2$

(A) The decomposition reaction is: $XY_{(s)} \leftrightarrow X_{(g)} + Y_{(g)}$
Let the partial pressure of each gas $X$ and $Y$ at equilibrium be $P$.
Total pressure at equilibrium is given by: $P_{total} = P_X + P_Y = P + P = 2P$
Given that the total equilibrium pressure is $10 \, bar$,we have: $2P = 10 \, bar$
Therefore,$P = 5 \, bar$.
The equilibrium constant $K_p$ is defined as: $K_p = P_X \times P_Y = P \times P = P^2$
Substituting the value of $P$: $K_p = (5)^2 = 25$

(A) The decomposition reaction is: $2H_2O_{2(l)} \rightarrow 2H_2O_{(l)} + O_{2(g)}$
From the stoichiometry,$2$ moles of $H_2O_2$ produce $1$ mole of $O_{2(g)}$.
Therefore,$100$ moles of $H_2O_2$ produce $50$ moles of $O_{2(g)}$.
The work done during expansion against constant external pressure is given by $W = -P_{ext} \Delta V$.
Since $P_{ext} \Delta V = \Delta n_g RT$,we have $W = -(\Delta n_g)RT$.
Here,$\Delta n_g = 50$ moles of $O_{2(g)}$.
$W = -(50 \ mol) \times (8.3 \ J \ K^{-1} \ mol^{-1}) \times (300 \ K) = -124500 \ J$.
Converting to $kJ$,$W = -124.5 \ kJ$.
The magnitude of work done is $124.5 \ kJ$.

(A) In the presence of sunlight $(hv)$,alkenes undergo free radical substitution at the allylic position.
$1$. The $Br_2$ molecule undergoes homolytic cleavage to form bromine radicals $(Br\cdot)$.
$2$. The bromine radical abstracts an allylic hydrogen atom from cyclohexene to form an allylic radical.
$3$. The allylic radical then reacts with another $Br_2$ molecule to form $3-$bromocyclohexene.
Therefore,the product is $3-$bromocyclohexene.

(B) In diamond,each $C$ atom is covalently bonded to four other $C$ atoms to give a tetrahedral unit. Each $C$ atom is $sp^3$ hybridized,and all four valence electrons are involved in sigma bond formation,leaving no free electrons. Thus,diamond is an insulator.
In graphite,each $C$ atom is covalently bonded to three other $C$ atoms in a trigonal planar geometry. Each $C$ atom is $sp^2$ hybridized. Three valence electrons are used in sigma bond formation,while the fourth electron remains in an unhybridized $p$-orbital,which is delocalized over the layers. This free electron makes graphite a good conductor of electricity.
Since the difference in electrical conductivity is directly due to the difference in hybridization and the resulting availability of free electrons,the reason is the correct explanation for the assertion.

(C) $1$. The reaction of anisole with succinic anhydride in the presence of $AlCl_3$ (Friedel-Crafts acylation) occurs primarily at the para-position due to the activating and ortho/para-directing nature of the $-OCH_3$ group. This yields $4-(4-methoxyphenyl)-4-oxobutanoic$ acid as intermediate $A$.
$2$. Clemmensen's reduction $(Zn-Hg/conc. HCl)$ reduces the ketone group to a methylene group,yielding $4-(4-methoxyphenyl)butanoic$ acid.
$3$. Subsequent cyclization (often using $PCl_5$ followed by $AlCl_3$ or polyphosphoric acid) leads to the formation of $6-methoxytetralone$. Further reduction of the ketone gives $6-methoxytetralin$.
$4$. Based on the provided options and the final structure in the solution image,the product $X$ is $6-methoxytetralin$.

(A) The formula for $rms$ velocity is $v_{rms} = \sqrt{\frac{3RT}{M}}$.
Initially,for $N_2$ molecules at temperature $T$,$u = \sqrt{\frac{3RT}{28 \times 10^{-3} \ kg/mol}}$.
After the temperature is doubled $(T' = 2T)$ and $N_2$ dissociates into $N$ atoms,the molar mass becomes $M' = 14 \times 10^{-3} \ kg/mol$.
The new $rms$ velocity $v'$ is given by $v' = \sqrt{\frac{3R(2T)}{14 \times 10^{-3}}} = \sqrt{4 \times \frac{3RT}{28 \times 10^{-3}}} = 2 \times \sqrt{\frac{3RT}{28 \times 10^{-3}}}$.
Therefore,$v' = 2u$.

(A) The correct answer is $(A)$.
Froth-flotation is a process specifically used for the concentration of sulphide ores.
Galena $(PbS)$ is a sulphide ore.
Cassiterite $(SnO_2)$ is an oxide ore.
Magnetite $(Fe_3O_4)$ is an oxide ore.
Malachite $(Cu(OH)_2 \cdot CuCO_3)$ is a carbonate ore.
Therefore,only Galena is concentrated by the froth-flotation method,which relies on the preferential wetting of sulphide particles by oil (frothing agent) and gangue particles by water.

(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: $\log\left(\frac{x}{m}\right) = \log k + \frac{1}{n} \log p$.
Comparing this with the linear equation $y = mx + c$,where $y = \log(x/m)$,$x = \log p$,the slope is $1/n$ and the intercept is $\log k$.
Therefore,only $1/n$ appears as the slope.

(A) According to Raoult's Law for non-volatile solutes:
$\frac{P^{\circ} - P_s}{P^{\circ}} = \frac{n_B}{n_A + n_B}$
Given:
Mass of glucose $(W_B) = 18 \ g$,Molar mass $(M_B) = 180 \ g/mol$
Moles of glucose $(n_B) = \frac{18}{180} = 0.1 \ mol$
Mass of water $(W_A) = 178.2 \ g$,Molar mass $(M_A) = 18 \ g/mol$
Moles of water $(n_A) = \frac{178.2}{18} = 9.9 \ mol$
Vapour pressure of pure water at $100 \ ^{\circ}C$ $(P^{\circ}) = 760 \ torr$
Substituting the values:
$\frac{760 - P_s}{760} = \frac{0.1}{9.9 + 0.1} = \frac{0.1}{10} = 0.01$
$760 - P_s = 760 \times 0.01 = 7.6$
$P_s = 760 - 7.6 = 752.4 \ torr$

(B) Galvanization is the process of applying a protective $Zn$ coating to steel or iron to prevent rusting.
The most common method is hot-dip galvanizing,in which the parts are submerged in a bath of molten $Zn$.

(D) The decomposition reaction is: $H_2O_{2(aq)} \rightarrow H_2O_{(l)} + \frac{1}{2} O_{2(g)}$
For a first order reaction,the rate constant $k$ is given by: $k = \frac{2.303}{t} \log \frac{[A]_0}{[A]_t}$
Given $[A]_0 = 0.5 \ M$,$[A]_t = 0.125 \ M$,and $t = 50 \ min$:
$k = \frac{2.303}{50} \log \frac{0.5}{0.125} = \frac{2.303}{50} \log(4) = \frac{2.303 \times 0.602}{50} \approx 0.0277 \ min^{-1}$
When $[H_2O_2] = 0.05 \ M$,the rate of disappearance of $H_2O_2$ is:
$Rate_{H_2O_2} = k[H_2O_2] = 0.0277 \times 0.05 = 1.385 \times 10^{-3} \ mol \ L^{-1} \ min^{-1}$
From the stoichiometry,$-\frac{d[H_2O_2]}{dt} = 2 \frac{d[O_2]}{dt}$,so the rate of formation of $O_2$ is:
$\frac{d[O_2]}{dt} = \frac{1}{2} \times Rate_{H_2O_2} = \frac{1.385 \times 10^{-3}}{2} = 6.925 \times 10^{-4} \ mol \ L^{-1} \ min^{-1} \approx 6.93 \times 10^{-4} \ mol \ min^{-1}$

(A) The froth floatation method is primarily used for the concentration of sulphide ores because sulphide ores are preferentially wetted by oil,while gangue particles are wetted by water.
$1$. Malachite: $Cu(OH)_2 \cdot CuCO_3$ (Carbonate ore)
$2$. Magnetite: $Fe_3O_4$ (Oxide ore)
$3$. Siderite: $FeCO_3$ (Carbonate ore)
$4$. Galena: $PbS$ (Sulphide ore)
Since $PbS$ is a sulphide ore,it is best concentrated by the froth floatation method.

(C) $1.$ Orthophosphorous acid $(H_3PO_3)$: $3(+1) + x + 3(-2) = 0 \Rightarrow x = +3$
$2.$ Pyrophosphorous acid $(H_4P_2O_5)$: $4(+1) + 2x + 5(-2) = 0$ $\Rightarrow 2x = 6$ $\Rightarrow x = +3$
$3.$ Hypophosphoric acid $(H_4P_2O_6)$: $4(+1) + 2x + 6(-2) = 0$ $\Rightarrow 2x = 8$ $\Rightarrow x = +4$
$4.$ Pyrophosphoric acid $(H_4P_2O_7)$: $4(+1) + 2x + 7(-2) = 0$ $\Rightarrow 2x = 10$ $\Rightarrow x = +5$
Therefore,the pair with $+3$ oxidation state is Orthophosphorous and pyrophosphorous acids.

(B) The reaction of $Zn$ with dilute $HNO_3$ is given by:
$4 Zn + 10 HNO_3 (dil.) \longrightarrow 4 Zn(NO_3)_2 + 5 H_2O + N_2O$
Thus,dilute $HNO_3$ produces $N_2O$.
The reaction of $Zn$ with concentrated $HNO_3$ is given by:
$Zn + 4 HNO_3 (conc.) \longrightarrow Zn(NO_3)_2 + 2 H_2O + 2 NO_2$
Thus,concentrated $HNO_3$ produces $NO_2$.
Therefore,the products are $N_2O$ and $NO_2$ respectively.

(D) Among the given metallic oxides,$CrO_2$ is strongly attracted by a magnetic field.
This magnetic property persists even after the removal of the external magnetic field.
Therefore,$CrO_2$ is classified as a metallic and ferromagnetic compound.

(D) Optical isomerism is shown by complexes that are chiral,meaning they do not have a plane of symmetry or a center of inversion.
In the case of octahedral complexes of the type $[M(AA)_2X_2]$,the $cis$-isomer is chiral and exists as a pair of enantiomers (non-superimposable mirror images).
The $trans$-isomer,however,has a plane of symmetry and is achiral.
Therefore,$cis-[Co(en)_2Cl_2]Cl$ shows optical isomerism.

(D) The magnetic moment is calculated using the formula $\mu = \sqrt{n(n+2)} \text{ B.M.}$,where $n$ is the number of unpaired electrons.
$1$. $[Cr(H_2O)_6]^{2+}$: $Cr$ is in $+2$ oxidation state $(3d^4)$. It has $4$ unpaired electrons $(n=4)$.
$2$. $[Fe(H_2O)_6]^{2+}$: $Fe$ is in $+2$ oxidation state $(3d^6)$. In a weak field ligand like $H_2O$,it has $4$ unpaired electrons $(n=4)$.
$3$. $[Mn(H_2O)_6]^{2+}$: $Mn$ is in $+2$ oxidation state $(3d^5)$,having $5$ unpaired electrons $(n=5)$.
$4$. $[CoCl_4]^{2-}$: $Co$ is in $+2$ oxidation state $(3d^7)$,having $3$ unpaired electrons $(n=3)$.
Since $[Cr(H_2O)_6]^{2+}$ and $[Fe(H_2O)_6]^{2+}$ both have $4$ unpaired electrons,they have the same magnetic moment.

(A) The thiol group $(-SH)$ is a functional group consisting of a sulfur atom bonded to a hydrogen atom.
Among the given options,$Cysteine$ is an amino acid that contains a thiol group in its side chain.
The structure of $Cysteine$ is $HS-CH_2-CH(NH_2)-COOH$.

(D) The reaction proceeds in two steps:
$1$. Allylic bromination using $NBS$ ($N$-Bromosuccinimide) under $hv$ (light) conditions,which introduces a bromine atom at the allylic position.
$2$. Nucleophilic substitution of the bromine atom with a hydroxyl group using $H_2O/K_2CO_3$,resulting in the formation of an allylic alcohol.
The final product is $4$-tert-butylcyclohex-$2$-en-$1$-ol.

(C) $2-$chloro$-2-$methylpentane is a tertiary $(3^{\circ})$ alkyl halide.
In the presence of a strong base/nucleophile like sodium methoxide $(CH_3ONa)$ in methanol $(CH_3OH)$,it can undergo both substitution and elimination reactions.
$1.$ Substitution $(S_N1)$: The $Cl^-$ is replaced by $-OCH_3$ to form $2-$methoxy$-2-$methylpentane (Product $(1)$).
$2.$ Elimination $(E2)$: Removal of $HCl$ from $\beta-$carbons.
$-$ From the terminal methyl group,it forms $2-$methylpent$-1-$ene (Product $(2)$).
$-$ From the internal methylene group,it forms $2-$methylpent$-2-$ene (Product $(3)$).
Thus,all three products are obtained.

(B) The $Hofmann$ bromamide degradation reaction is represented by the following chemical equation:
$RCONH_2 + 4NaOH + Br_2 \rightarrow RNH_2 + Na_2CO_3 + 2NaBr + 2H_2O$
From the stoichiometry of the balanced equation,it is clear that for every $1$ mole of amine $(RNH_2)$ produced,$4$ moles of $NaOH$ and $1$ mole of $Br_2$ are consumed.

(D) Anionic detergents are sodium salts of sulphonated long chain alcohols or hydrocarbons.
In anionic detergents,the anionic part of the molecule is involved in the cleansing action.
$Sodium \ lauryl \ sulphate$ $(CH_3(CH_2)_{11}SO_4^-Na^+)$ is an anionic detergent.
$Cetyltrimethylammonium \ bromide$ is a popular cationic detergent used in hair conditioners.
$Glyceryl \ oleate$ is an ester of glycerin and oleic acid.
$Sodium \ stearate$ is a soap,which is the sodium salt of stearic acid.

(B) Low density polythene $(LDPE)$ is a poor conductor of electricity.
Its synthesis involves the polymerization of ethene at high pressure $(1000-2000 \ atm)$ and temperatures $(350-570 \ K)$ in the presence of traces of dioxygen or a peroxide initiator as a catalyst.
$LDPE$ is chemically inert and tough but flexible,making it suitable for squeeze bottles,toys,and flexible pipes.
High density polythene $(HDPE)$ is used in the manufacture of buckets,dust-bins,and bottles.
Therefore,the statement that $LDPE$ is used in the manufacture of buckets and dust-bins is $FALSE$.

(D) The consumption of $AgNO_3$ depends on the number of ionizable $Cl^-$ ions present outside the coordination sphere.
$Na_2[CrCl_5(H_2O)]$ provides $0$ ionizable $Cl^-$ ions.
$Na_3[CrCl_6]$ provides $0$ ionizable $Cl^-$ ions.
$[Cr(H_2O)_5Cl]Cl_2$ provides $2$ ionizable $Cl^-$ ions.
$[Cr(H_2O)_6]Cl_3$ provides $3$ ionizable $Cl^-$ ions.
Since $[Cr(H_2O)_6]Cl_3$ furnishes the highest number of $Cl^-$ ions ($3$ moles per mole of complex),it will consume the most equivalents of $AgNO_3$.

(D) The characteristics given suggest that this is physical adsorption (physisorption).
Physical adsorption is characterized by weak van der Waals forces,reversibility,and the formation of multimolecular layers.
It is an exothermic process that typically decreases with an increase in temperature.
The enthalpy of adsorption is low (usually $20-40 \ kJ \ mol^{-1}$).
Since it involves weak forces and no chemical bond formation,the energy of activation required is very low.

(C) The Bouveault-Blanc reduction involves the reduction of esters to primary alcohols using sodium metal in the presence of an alcohol (usually ethanol).
The general reaction is: $R-COOR' + 4[H] \xrightarrow{Na/C_2H_5OH} R-CH_2OH + R'OH$.

(B) For an acidic amino acid like aspartic acid,the isoelectric point $(pI)$ is calculated as the average of the two $pK_a$ values that flank the zwitterionic form.
The zwitterionic form of aspartic acid exists between the first deprotonation of the $\alpha$-carboxyl group $(pK_1 = 1.88)$ and the deprotonation of the side-chain carboxyl group $(pK_R = 3.65)$.
$pI = \frac{pK_1 + pK_R}{2}$
$pI = \frac{1.88 + 3.65}{2} = \frac{5.53}{2} = 2.765 \approx 2.77$

(C) No reaction will occur because $Zn$ is placed above $Cu$ in the electrochemical series.
Since $Cu$ is a weaker reducing agent than $Zn$,it cannot displace $Zn^{2+}$ ions from the $ZnSO_4$ solution.
Therefore,no displacement reaction takes place.

(B) Both the assertion and the reason are correct statements,and the reason is the correct explanation for the assertion.
Assertion:
Rayon (specifically cellulose acetate) is a semi-synthetic polymer that exhibits superior properties compared to natural cotton fiber.
Reason:
The mechanical and aesthetic properties of cellulose are significantly enhanced through the process of acetylation. Since rayon is produced via the acetylation of cellulose,it possesses improved characteristics over natural cotton.
Conclusion:
Therefore,the reason correctly explains why rayon has better properties than natural cotton.

(D) The relationship between Gibbs free energy change and equilibrium constant is $\Delta G^\circ = -RT \ln K_p$,which can be rewritten as $-\ln K_p = \frac{\Delta G^\circ}{RT}$.
For a reaction to be spontaneous,$\Delta G^\circ < 0$,which implies $-\ln K_p < 0$.
From the Ellingham diagram,at $T > 1200 \ K$,the value of $-\ln K_p$ for the reaction $C_{(s)} + \frac{1}{2} O_{2(g)} \to CO_{(g)}$ is lower than that for $M_{(s)} + \frac{1}{2} O_{2(g)} \to MO_{(s)}$.
This indicates that the formation of $CO$ is more thermodynamically stable than $MO$ at $T > 1200 \ K$.
Therefore,carbon can reduce $MO_{(s)}$ to $M_{(s)}$ at temperatures above $1200 \ K$ according to the reaction $MO_{(s)} + C_{(s)} \to M_{(s)} + CO_{(g)}$.

(A) Preparation of egg albumin sol is done in the following two steps:
$(i)$ Prepare $100 \ mL$ of $5 \ \% \ (w/V)$ solution of $NaCl$ in water in a $250 \ mL$ beaker.
$(ii)$ Break one egg in a porcelain dish,pipette out the albumin (the transparent white part),and pour it into the sodium chloride solution. Stir well to ensure the preparation of the sol.

(A) The overall reaction is obtained by adding the two elementary steps:
Step $(i): O_{3(g)} + Cl_{(g)}^{\bullet} \to O_{2(g)} + ClO_{(g)}^{\bullet}$
Step $(ii): ClO_{(g)}^{\bullet} + O_{(g)}^{\bullet} \to O_{2(g)} + Cl_{(g)}^{\bullet}$
Adding these gives: $O_{3(g)} + O_{(g)}^{\bullet} \to 2O_{2(g)}$
The rate of the overall reaction is determined by the product of the rate constants of the elementary steps in a mechanism where the intermediate $(ClO^{\bullet})$ cancels out.
$K_{overall} = K_i \times K_{ii}$
$K_{overall} = (5.2 \times 10^9) \times (2.6 \times 10^{10})$
$K_{overall} \approx 1.352 \times 10^{20} \approx 1.4 \times 10^{20} \ L \ mol^{-1} \ s^{-1}$

(D) Alitame is a high-potency artificial sweetener that is approximately $2,000$ times as sweet as cane sugar.
It is much sweeter than other common artificial sweeteners like Aspartame ($100$ times),Saccharin ($550$ times),and Sucralose ($600$ times).

(B) The stability of species in aqueous solution depends on their tendency to undergo disproportionation or oxidation-reduction reactions.
$Cr^{2+}$ is a strong reducing agent and is easily oxidized to $Cr^{3+}$ in the presence of air or water.
$Cu^{+}$ undergoes disproportionation in aqueous solution: $2Cu^{+} \to Cu^{2+} + Cu$.
$MnO_4^{3-}$ is highly unstable and disproportionates rapidly.
$MnO_4^{2-}$ (manganate ion) is stable in strongly alkaline solutions,whereas it disproportionates in neutral or acidic media. Among the given options,$MnO_4^{2-}$ is the most stable species in specific aqueous conditions (alkaline medium).

(D) Primary,secondary,and tertiary amines are distinguished by the Hinsberg reagent test.
The Hinsberg reagent is benzene sulphonyl chloride,represented as $C_6H_5SO_2Cl$.

(D) According to Henry's law,the solubility $(S)$ of a gas is directly proportional to its partial pressure $(P)$: $S \propto P$ or $S = kP$.
Given: $P_1 = 500 \ torr$,$S_1 = 0.01 \ g \ L^{-1}$,$P_2 = 750 \ torr$.
Using the relation $\frac{S_1}{S_2} = \frac{P_1}{P_2}$:
$\frac{0.01}{S_2} = \frac{500}{750}$
$S_2 = \frac{0.01 \times 750}{500} = 0.015 \ g \ L^{-1}$.

(C) The crystal field splitting energy $(\Delta _0)$ is directly proportional to the oxidation state of the central metal ion. $A$ higher oxidation state results in a greater electrostatic attraction between the ligands and the metal,leading to larger splitting.
Therefore,$\Delta _0$ of $[Ti(H_2O)_6]^{3+} > \Delta _0$ of $[Ti(H_2O)_6]^{2+}$.
Additionally,$\Delta _0$ increases down a group because the $4d$ orbitals are more extended than $3d$ orbitals,leading to stronger interaction with ligands. Thus,$\Delta _0$ of $[Mo(H_2O)_6]^{2+} > \Delta _0$ of $[Cr(H_2O)_6]^{2+}$.
Combining these,the correct trend is $\Delta _0$ of $[Cr(H_2O)_6]^{2+} < [Mo(H_2O)_6]^{2+}$ and $\Delta _0$ of $[Ti(H_2O)_6]^{3+} > [Ti(H_2O)_6]^{2+}$.

(D) Teflon (polytetrafluoroethene) is an addition polymer formed by the free radical polymerization of its monomer,tetrafluoroethene $(CF_2 = CF_2)$.
In contrast,Terylene,Melamine polymer,and Nylon-$6, 6$ are condensation polymers formed by the elimination of small molecules like water or alcohol.

(C) The reaction described is the Balz-Schiemann reaction.
In this reaction,an aromatic diazonium salt is treated with fluoroboric acid $(HBF_4)$ to form an insoluble diazonium fluoroborate salt $(ArN_2^ BF_4^-)$.
Upon heating this salt,it undergoes thermal decomposition to yield an aryl fluoride,nitrogen gas $(N_2)$,and boron trifluoride $(BF_3)$.
Therefore,the condition required for the conversion of the diazonium fluoroborate to the aryl fluoride is simply heating $(\Delta)$.

(A) The protective power of a colloid is inversely proportional to its gold number.
$\text{Gold number} \propto \frac{1}{\text{Protective power}}$
This means that the smaller the value of the gold number of a lyophilic sol,the greater is its protective action.
Comparing the given values: Gelatin $(0.005-0.01)$ has the lowest gold number among the options.
Therefore,Gelatin is the best protective colloid.

(A) The given compound is purine.
In purine,the lone pair of electrons on the $N-9$ atom is involved in the aromatic sextet of the imidazole ring.
Therefore,this lone pair is not available for protonation,and $N-9$ does not contribute to the basicity of the compound.
Other nitrogen atoms $(N-1, N-3, N-7)$ have lone pairs in $sp^2$ hybridized orbitals that are not involved in aromaticity,making them available for protonation.

(A) The reaction shown is the dehydrohalogenation of an alkyl halide to form an alkene,which is an elimination reaction.
$NaOEt/EtOH$ (sodium ethoxide in ethanol) is a strong base and is highly effective for $E2$ elimination reactions.
$NaOH/EtOH$ (sodium hydroxide in ethanol) is also a strong base and is commonly used for elimination reactions.
$NaOH/H_2O$ (aqueous sodium hydroxide) primarily favors substitution ($S_N2$ or $S_N1$) over elimination because water is a polar protic solvent that solvates the hydroxide ion,reducing its basicity.
$NaI$ (sodium iodide) is a nucleophile and is typically used in substitution reactions,such as the Finkelstein reaction,to convert alkyl chlorides or bromides into alkyl iodides. It is not a base and therefore cannot promote an elimination reaction.
Thus,both $NaI$ and $NaOH/H_2O$ are not suitable for elimination,but $NaI$ is the most distinct reagent that acts as a nucleophile rather than a base.

(C) ruby is a crystal of alumina,aluminum oxide $(Al_2O_3)$,containing a trace of chromium $(III)$ ions replacing some of the aluminum ions.
In ruby,each $Al^{3+}$ ion and $Cr^{3+}$ ion is surrounded by six oxide ions in an octahedral arrangement.
The origin of the color of emeralds is similar to that of the color of rubies.
However,the bulk of an emerald crystal is composed of beryl,beryllium aluminum silicate $(Be_3Al_2(SiO_3)_6)$,instead of the alumina which forms rubies.
The color is produced by chromium $(III)$ ions,which replace some of the aluminum ions in the crystal.
In emeralds,the $Cr^{3+}$ is surrounded by six silicate ions,rather than the six oxide ions in ruby.
Therefore,both ruby and emerald derive their color from the presence of $Cr^{3+}$ ions.

(A) The dissociation of the salt $MX_2$ is represented as: $MX_2 \rightleftharpoons M^{2+} + 2X^-$.
Let the initial concentration be $1$ mole.
At equilibrium,the moles are: $M^{2+} = \alpha$,$X^- = 2\alpha$,and $MX_2 = 1 - \alpha$,where $\alpha$ is the degree of dissociation.
The total number of particles at equilibrium is $(1 - \alpha) + \alpha + 2\alpha = 1 + 2\alpha$.
The van't Hoff factor $i$ is given by the ratio of the total number of particles at equilibrium to the initial number of particles: $i = \frac{1 + 2\alpha}{1}$.
Given $i = 2$,we have $1 + 2\alpha = 2$.
Solving for $\alpha$: $2\alpha = 1$,which gives $\alpha = 0.50$.

(A) At Anode: $^-OOC-CH_2-CH_2-COO^- \rightarrow CH_2=CH_2 + 2CO_2 + 2e^-$. For $2 \ F$ electricity,$3 \ moles$ of gas ($1 \ mole \ C_2H_4$ and $2 \ moles \ CO_2$) are produced.
At Cathode: $2H_2O + 2e^- \rightarrow H_2 + 2OH^-$. For $2 \ F$ electricity,$1 \ mole \ H_2$ gas is produced.
Total moles of gas for $2 \ F = 3 + 1 = 4 \ moles$.
Total moles of gas for $0.2 \ F = (4/2) \times 0.2 = 0.4 \ moles$.
Total volume at $STP = 0.4 \times 22.4 = 8.96 \ L$.

(C) Ninhydrin is often used to detect $\alpha$-amino acids and also free amino and carboxylic acid groups on proteins and peptides.
When about $0.5 \ mL$ of a $0.1\%$ solution of ninhydrin is boiled for one or two minutes with a few $mL$ of dilute amino acid or protein solution,a blue color develops.
Ninhydrin degrades amino acids into aldehydes,ammonia,and $CO_2$ through a series of reactions; the net result is ninhydrin in a partially reduced form,hydrindantin.
Ninhydrin then condenses with ammonia and hydrindantin to produce an intensely blue or purple pigment,sometimes called Ruhemann's purple.

(A) Pentaerythritol is produced by a base-catalyzed reaction of acetaldehyde with excess formaldehyde.
First,three moles of formaldehyde undergo aldol condensation with one mole of acetaldehyde to form pentaerythrose: $CH_3CHO + 3HCHO \to C(CH_2OH)_3CHO$.
This is followed by a crossed Cannizzaro reaction between the intermediate and another mole of formaldehyde to yield pentaerythritol and sodium formate: $C(CH_2OH)_3CHO + HCHO + NaOH \to C(CH_2OH)_4 + HCOONa$.
Thus,the process involves three aldol condensations and one Cannizzaro reaction.

(A) Antibiotics are classified as either bactericidal (kill bacteria) or bacteriostatic (inhibit the growth of bacteria).
$Ofloxacin$ is a bactericidal antibiotic.
$Tetracycline$,$Chloramphenicol$,and $Erythromycin$ are examples of bacteriostatic antibiotics.

(A) The $S_6$ molecule has a chair-form hexagon ring with the same bond length as that in $S_8$,but with somewhat smaller bond angles. Therefore,the bond angles in $S_8$ (crown shape,$\approx 107^{\circ}$) and $S_6$ (chair shape,$\approx 102^{\circ}$) are different. Thus,statement $A$ is incorrect.

(C) In the extraction of copper,$FeO$ is present as a basic gangue impurity.
$SiO_2$ is added as an acidic flux to remove this impurity.
They react to form a fusible slag,$FeSiO_3$.
The reaction is: $\mathop {FeO}\limits_{\text{Basic impurity}} + \mathop {SiO_2}\limits_{\text{Acidic flux}}$ $\longrightarrow \mathop {FeSiO_3}\limits_{\text{Slag}}$

(D) The rate constant $(k)$ is a characteristic property of a reaction at a given temperature.
It does not depend on the concentration of the reactants.
Therefore,changing the concentration of $B$ will change the reaction rate,but the rate constant $(k)$ remains unchanged.

(B) Group $15$ elements have a stable half-filled $ns^2 np^3$ electronic configuration.
Due to this extra stability,they have higher ionization enthalpy values compared to Group $16$ elements in the corresponding periods.
Therefore,Group $16$ elements have lower ionization enthalpy values compared to Group $15$ elements.

(A) homoleptic complex is a coordination compound in which the metal atom or ion is bonded to only one type of donor atom or ligand.
In the complex $[Co(NH_3)_6]Cl_3$,the central metal ion $Co^{3+}$ is bonded to six $NH_3$ ligands,which are all of the same type.
Therefore,$[Co(NH_3)_6]Cl_3$ is a homoleptic complex.

(C) Corrosion of iron is an electrochemical process that requires both oxygen and moisture.
It can be prevented or minimized by creating an impermeable barrier (like paint or galvanization) on the surface of the iron to prevent contact with air and moisture.
Option $A$ is incorrect because iron should be in contact with a metal having a lower reduction potential (like $Zn$) for cathodic protection.
Option $B$ is incorrect as iron does not corrode significantly in oxygen-free water.
Option $D$ is incorrect because salt water acts as an electrolyte,increasing the conductivity and rate of the electrochemical reaction,not because the potential of iron changes.