JEE Main 2019 Chemistry Question Paper with Answer and Solution

(B) The slope of line $L_1$ passing through $(1, 2)$ and $(-3, 4)$ is $m_1 = \frac{4 - 2}{-3 - 1} = \frac{2}{-4} = -\frac{1}{2}$.
Since $(h, k)$ lies on $L_1$,the slope of the line segment joining $(h, k)$ and $(1, 2)$ must also be $-\frac{1}{2}$.
Thus,$\frac{k - 2}{h - 1} = -\frac{1}{2}$ $\Rightarrow 2k - 4 = -h + 1$ $\Rightarrow h + 2k = 5$.
Line $L_2$ passes through $(h, k)$ and $(4, 3)$,so its slope is $m_2 = \frac{3 - k}{4 - h}$.
Since $L_2 \perp L_1$,$m_1 \times m_2 = -1$ $\Rightarrow -\frac{1}{2} \times \frac{3 - k}{4 - h} = -1$ $\Rightarrow \frac{3 - k}{4 - h} = 2$.
$3 - k = 8 - 2h \Rightarrow 2h - k = 5$.
We have the system of equations:
$1) h + 2k = 5$
$2) 2h - k = 5$
Multiplying equation $(2)$ by $2$,we get $4h - 2k = 10$.
Adding this to equation $(1)$: $(h + 2k) + (4h - 2k) = 5 + 10$ $\Rightarrow 5h = 15$ $\Rightarrow h = 3$.
Substituting $h = 3$ into $2h - k = 5$: $2(3) - k = 5$ $\Rightarrow 6 - k = 5$ $\Rightarrow k = 1$.
Therefore,$\frac{k}{h} = \frac{1}{3}$.

(D) The slope of the tangent is given by $\frac{dy}{dx} = \frac{2y}{x^2}$.
Separating the variables,we get $\frac{dy}{y} = \frac{2}{x^2} dx$.
Integrating both sides,$\int \frac{dy}{y} = \int 2x^{-2} dx$,which gives $\ln |y| = -\frac{2}{x} + C$.
The circle equation is $x^2 + y^2 - 2x - 2y = 0$,which can be written as $(x-1)^2 + (y-1)^2 = 2$. The centre is $(1, 1)$.
Since the curve passes through $(1, 1)$,we substitute $x = 1$ and $y = 1$ into the equation: $\ln |1| = -\frac{2}{1} + C$,so $0 = -2 + C$,which implies $C = 2$.
Thus,the equation is $\ln |y| = -\frac{2}{x} + 2$.
Multiplying by $x$,we get $x \ln |y| = -2 + 2x = 2(x - 1)$.

(B) When magnesium powder is burnt in air,it reacts with both oxygen and nitrogen present in the atmosphere.
$2Mg(s) + O_2(g) \to 2MgO(s)$
$3Mg(s) + N_2(g) \to Mg_3N_2(s)$
Therefore,the products formed are $MgO$ and $Mg_3N_2$.

(C) The critical temperature $(T_c)$ of a gas is given by the formula $T_c = \frac{8a}{27Rb}$.
To find the gas with the highest $T_c$,we need to calculate the ratio $\frac{a}{b}$ for each gas:
$Ar: \frac{1.3}{3.2} \approx 0.406$
$Ne: \frac{0.2}{1.7} \approx 0.118$
$Kr: \frac{5.1}{1.0} = 5.1$
$Xe: \frac{4.1}{5.0} = 0.82$
Since $Kr$ has the highest $\frac{a}{b}$ ratio,it will have the highest critical temperature.

(B) The stability of a molecule upon anion formation depends on the change in bond order. If the electron enters a bonding molecular orbital,the bond order increases,leading to stabilization. If it enters an antibonding orbital,the bond order decreases,leading to destabilization.
For $C_2$: The electronic configuration is $\sigma 1s^2 \sigma^* 1s^2 \sigma 2s^2 \sigma^* 2s^2 \pi 2p_y^2 \pi 2p_z^2$. Bond order = $(8-4)/2 = 2$.
For $C_2^-$: The configuration is $\sigma 1s^2 \sigma^* 1s^2 \sigma 2s^2 \sigma^* 2s^2 \pi 2p_y^2 \pi 2p_z^2 \sigma 2p_x^1$. Bond order = $(9-4)/2 = 2.5$.
Since the bond order increases from $2$ to $2.5$,$C_2$ is stabilized by anion formation.

(A) Path functions are properties whose values depend on the path taken to reach a state,not just the initial and final states.
$q$ (heat) and $w$ (work) are path functions.
$q + w$ represents the change in internal energy $(\Delta U)$,which is a state function.
$H - TS$ represents the Gibbs free energy $(G)$,which is a state function.
Therefore,only $q$ and $w$ are path functions.

(D) The reaction of alkylbenzenes with alkaline $KMnO_4$ followed by acidification $(H_3O^+)$ is a standard oxidation reaction.
Regardless of the length of the alkyl side chain,as long as there is at least one benzylic hydrogen atom,the entire side chain is oxidized to a carboxylic acid group $(-COOH)$ attached to the benzene ring.
In this case,ethylbenzene $(C_6H_5-CH_2CH_3)$ undergoes oxidation to form benzoic acid $(C_6H_5COOH)$.

(B) The given qualitative analysis indicates the following properties of the compound:
$1$. Insoluble in dilute $HCl$: This suggests the compound is not basic (like an amine).
$2$. Soluble in $NaOH$ solution: This indicates the compound is acidic (like a phenol or carboxylic acid).
$3$. Decolourization of $Br_2/\text{water}$: This indicates the presence of an unsaturated system or a highly activated aromatic ring (like phenol,which forms $2,4,6-\text{tribromophenol}$).
Comparing these properties:
- Cyclohexanol is neutral and does not dissolve in $NaOH$.
- Phenol is acidic,dissolves in $NaOH$ to form sodium phenoxide,and reacts with bromine water to decolourise it.
- Cyclohexylamine is basic and dissolves in $HCl$.
- Aniline is basic and dissolves in $HCl$.
Therefore,the compound is Phenol.

(C) Carbon dioxide is very important for maintaining the climate which is suitable for our survival.
$CO_2$ has the property of trapping heat from the sun. So,higher the level of carbon dioxide,the higher will be the heat trapped.
This leads to the rise in the temperature of the earth,resulting in global warming.
Therefore,we can say excessive release of $CO_2$ into the atmosphere results in global warming.

(D) To name the given compound according to $IUPAC$ rules,we identify the parent chain as the benzene ring.
We assign the lowest possible locants to the substituents.
Here,the substituents are a methyl group $(-CH_3)$,a chloro group $(-Cl)$,and a nitro group $(-NO_2)$.
Assigning position $1$ to the methyl group,the chloro group is at position $2$ and the nitro group is at position $4$.
Following alphabetical order for substituents (chloro,methyl,nitro),the name is $2-$chloro$-1-$methyl$-4-$nitrobenzene.

(B) The first ionization energy $(IE_1)$ is the energy required to remove the first electron,and the second ionization energy $(IE_2)$ is the energy required to remove the second electron.
For $K$ $(Z=19)$,the electronic configuration is $[Ar] 4s^1$.
After losing one electron,$K^+$ acquires the stable noble gas configuration of Argon $([Ar])$.
Removing a second electron from this stable configuration requires a very large amount of energy,leading to a huge difference between $IE_1$ and $IE_2$ compared to the other elements listed.

(B) The balanced chemical equation is $N_{2(g)} + 3H_{2(g)} \to 2NH_{3(g)}$.
According to the stoichiometry,$1 \ mole$ of $N_2$ $(28 \ g)$ reacts with $3 \ moles$ of $H_2$ $(6 \ g)$.
For option $A$: $14 \ g$ of $N_2$ $(0.5 \ mole)$ requires $1.5 \ moles$ of $H_2$ $(3 \ g)$. Since $4 \ g$ of $H_2$ is present,$H_2$ is in excess.
For option $B$: $56 \ g$ of $N_2$ $(2 \ moles)$ requires $6 \ moles$ of $H_2$ $(12 \ g)$. Since only $10 \ g$ of $H_2$ $(5 \ moles)$ is available,$H_2$ is the limiting reagent.
For option $C$: $28 \ g$ of $N_2$ $(1 \ mole)$ requires $3 \ moles$ of $H_2$ $(6 \ g)$. Since $6 \ g$ of $H_2$ is present,it is a stoichiometric mixture.
For option $D$: $35 \ g$ of $N_2$ $(1.25 \ moles)$ requires $3.75 \ moles$ of $H_2$ $(7.5 \ g)$. Since $8 \ g$ of $H_2$ is present,$H_2$ is in excess.

(B) The reaction proceeds via electrophilic addition to the alkyne.
Step $1$: The addition of $DCl$ $(1 \ equiv)$ to $CH_3-C\equiv CH$ follows Markovnikov's rule. The electrophile $D^+$ adds to the terminal carbon to form the more stable secondary carbocation, followed by the attack of $Cl^-$. This yields $CH_3-C(Cl)=CHD$.
Step $2$: The addition of $DI$ to the alkene $CH_3-C(Cl)=CHD$ follows Markovnikov's rule. The electrophile $D^+$ adds to the carbon already bearing the deuterium (to form a stable carbocation adjacent to the chlorine atom), followed by the attack of $I^-$. This results in the geminal dihalo-dideutero product $CH_3-C(I)(Cl)-CHD_2$.

(C) The wavenumber is given by $\bar{\nu} = R_H \left[ \frac{1}{n_1^2} - \frac{1}{n_2^2} \right]$.
For any series,$\Delta \bar{\nu} = \bar{\nu}_{\max} - \bar{\nu}_{\min} = R_H \left[ \frac{1}{n_1^2} - \frac{1}{n_1+1^2} \right] - R_H \left[ \frac{1}{n_1^2} - \frac{1}{\infty^2} \right] = R_H \left[ \frac{1}{\infty^2} - \frac{1}{(n_1+1)^2} \right] = \frac{R_H}{(n_1+1)^2}$.
For the Lyman series,$n_1 = 1$,so $\Delta \bar{\nu}_{\text{Lyman}} = \frac{R_H}{(1+1)^2} = \frac{R_H}{4}$.
For the Balmer series,$n_1 = 2$,so $\Delta \bar{\nu}_{\text{Balmer}} = \frac{R_H}{(2+1)^2} = \frac{R_H}{9}$.
Therefore,the ratio is $\frac{\Delta \bar{\nu}_{\text{Lyman}}}{\Delta \bar{\nu}_{\text{Balmer}}} = \frac{R_H/4}{R_H/9} = \frac{9}{4}$.

(A) The reactivity of aromatic compounds towards electrophilic substitution depends on the electron density of the benzene ring. Electron-donating groups $(EDG)$ increase reactivity,while electron-withdrawing groups $(EWG)$ decrease it.
$A$: Chlorobenzene ($-Cl$ group has $-I$ effect and $+R$ effect; overall,it is deactivating but ortho/para directing).
$B$: Anisole ($-OMe$ group has strong $+R$ effect,making it highly activating).
$C$: Toluene ($-Me$ group has $+I$ and hyperconjugation effect,making it activating).
$D$: Benzonitrile ($-CN$ group has strong $-R$ and $-I$ effects,making it strongly deactivating).
Comparing the effects:
$B$ (strong $+R$) > $C$ (hyperconjugation) > $A$ (deactivating $-I > +R$) > $D$ (strong $-R$ and $-I$).
Thus,the increasing order of reactivity is $D < A < C < B$.

(A) The $C_{60}$ molecule,also known as Buckminsterfullerene,has a truncated icosahedron structure.
It consists of $60$ carbon atoms arranged in a cage-like structure.
This structure is composed of $20$ hexagonal rings and $12$ pentagonal rings.
Therefore,the correct option is $A$.

(D) To find the oxidation state of nitrogen $(N)$ in each compound:
$1$. In $N_2O$: $2x + (-2) = 0 \implies x = +1$
$2$. In $NO$: $x + (-2) = 0 \implies x = +2$
$3$. In $N_2O_3$: $2x + 3(-2) = 0 \implies 2x = 6 \implies x = +3$
$4$. In $NO_2$: $x + 2(-2) = 0 \implies x = +4$
Comparing the oxidation states: $+1 < +2 < +3 < +4$.
Therefore,the correct order is $N_2O < NO < N_2O_3 < NO_2$.

(B) The reaction proceeds in two steps:
$1$. Treatment of $3$-hydroxycyclohexan-$1$-one with $PBr_3$ replaces the hydroxyl group $(-OH)$ with a bromine atom $(-Br)$ to form $3$-bromocyclohexan-$1$-one.
$2$. Treatment of $3$-bromocyclohexan-$1$-one with alcoholic $KOH$ $(alc. KOH)$ induces a dehydrohalogenation reaction (an $E2$ elimination).
The base abstracts a proton from the $\alpha$-carbon relative to the bromine,leading to the formation of a double bond between the $\alpha$ and $\beta$ carbons.
This results in the formation of the conjugated enone,$cyclohex-2-en-1-one$.

(C) The formula for standard deviation is $\sigma = \sqrt{\frac{\sum (x_i - \bar{x})^2}{n}}$.
Given numbers are $-1, 0, 1, k$. The mean $\bar{x} = \frac{-1 + 0 + 1 + k}{4} = \frac{k}{4}$.
The variance $\sigma^2 = 5$. Thus,$\frac{(-1 - \frac{k}{4})^2 + (0 - \frac{k}{4})^2 + (1 - \frac{k}{4})^2 + (k - \frac{k}{4})^2}{4} = 5$.
Expanding the terms: $\frac{(1 + \frac{k}{2} + \frac{k^2}{16}) + \frac{k^2}{16} + (1 - \frac{k}{2} + \frac{k^2}{16}) + (\frac{3k}{4})^2}{4} = 5$.
$\frac{2 + \frac{3k^2}{16} + \frac{9k^2}{16}}{4} = 5$.
$\frac{2 + \frac{12k^2}{16}}{4} = 5$.
$2 + \frac{3k^2}{4} = 20$.
$\frac{3k^2}{4} = 18$.
$k^2 = 24$.
Since $k > 0$,$k = \sqrt{24} = 2\sqrt{6}$.

(C) For the function to be continuous at $x = \frac{\pi}{4}$,the limit of $f(x)$ as $x \to \frac{\pi}{4}$ must equal $f\left(\frac{\pi}{4}\right)$.
$\lim_{x \to \frac{\pi}{4}} f(x) = f\left(\frac{\pi}{4}\right) = k$
$\lim_{x \to \frac{\pi}{4}} \frac{\sqrt{2} \cos x - 1}{\cot x - 1} = k$
Since this is a $\frac{0}{0}$ form,we apply $L$'Hospital's Rule:
$\lim_{x \to \frac{\pi}{4}} \frac{\frac{d}{dx}(\sqrt{2} \cos x - 1)}{\frac{d}{dx}(\cot x - 1)} = k$
$\lim_{x \to \frac{\pi}{4}} \frac{-\sqrt{2} \sin x}{-\csc^2 x} = k$
$\lim_{x \to \frac{\pi}{4}} \sqrt{2} \sin x \cdot \sin^2 x = k$
$\lim_{x \to \frac{\pi}{4}} \sqrt{2} \sin^3 x = k$
Substituting $x = \frac{\pi}{4}$:
$k = \sqrt{2} \left(\frac{1}{\sqrt{2}}\right)^3 = \sqrt{2} \cdot \frac{1}{2\sqrt{2}} = \frac{1}{2}$

(A) According to the first law of thermodynamics,$\Delta U = q + w$.
Work is done on the system during compression,so $w = +10 \ kJ$.
Heat is released to the surroundings,so $q = -2 \ kJ$.
Therefore,$\Delta U = -2 \ kJ + 10 \ kJ = +8 \ kJ$.

(B) Silica $(SiO_2)$ exists in both crystalline and amorphous forms. Quartz,cristobalite,and tridymite are crystalline forms of silica. Kieselguhr is an amorphous form of silica,which is a diatomaceous earth.

(D) For a hydrogen atom,the total energy $(E)$ of an electron in an orbital is constant and depends only on the principal quantum number $(n)$.
$E = -K.E = \frac{P.E}{2}$,which implies $|P.E| = 2 \times |K.E|$. Thus,option $C$ is correct.
The probability density for a $1s$ orbital is given by $\psi^2 = \frac{1}{\pi a_0^3} e^{-2r/a_0}$. This value is maximum at $r = 0$ (the nucleus),so option $B$ is correct.
The electron can be found at any distance $r$ from the nucleus,including $2a_0$,although the probability density decreases exponentially with distance. Thus,option $A$ is correct.
The total energy of the electron in a specific orbital is constant and does not change with its instantaneous distance from the nucleus. Therefore,the statement that the total energy is maximum at $a_0$ is incorrect.

(A) $B_2O_3$ is acidic in nature.
$Al_2O_3$ and $Ga_2O_3$ are amphoteric in nature.
$In_2O_3$ and $Tl_2O_3$ are basic in nature.
As we move down the group $13$,the metallic character of the elements increases,which leads to a transition from acidic to amphoteric and finally to basic oxides.

(D) According to Molecular Orbital Theory $(MOT)$,the electronic configuration of $CO$ ($14$ electrons) is: $\sigma 1s^2, \sigma^* 1s^2, \sigma 2s^2, \sigma^* 2s^2, \pi 2p_x^2 = \pi 2p_y^2, \sigma 2p_z^2$.
Since all electrons are paired,$CO$ is diamagnetic.
In contrast,$O_2$ ($16$ electrons) has two unpaired electrons in $\pi^* 2p$ orbitals,$NO$ ($15$ electrons) has one unpaired electron,and $B_2$ ($10$ electrons) has two unpaired electrons in $\pi 2p$ orbitals.

(C) The given equation of state is $P(V - b) = RT$.
Expanding this,we get $PV - Pb = RT$.
Dividing by $RT$,we obtain $\frac{PV}{RT} - \frac{Pb}{RT} = 1$.
Since the compression factor $Z = \frac{PV}{RT}$,the equation becomes $Z = 1 + \frac{Pb}{RT}$.
This shows that $Z$ increases linearly with $P$ with a slope of $\frac{b}{RT}$.
Therefore,the gas with the largest van der Waals constant $b$ will exhibit the steepest increase in the plot of $Z$ versus $P$.
The magnitude of $b$ depends on the size of the gas molecules,which follows the order $Ne < Ar < Kr < Xe$.
Thus,$Xe$ has the largest value of $b$ and will show the steepest increase.

(A) In the solid state,$BeCl_2$ exists as a polymeric chain structure where each $Be$ atom is surrounded by four $Cl$ atoms.
In the vapour phase,at high temperatures (approx. $1200 \ K$),$BeCl_2$ exists as a dimeric structure $(Be_2Cl_4)$,which dissociates into a linear monomeric structure at even higher temperatures (approx. $1800 \ K$).

(D) The reaction involves the acid-catalyzed cyclization of the given hydroxy-ester.
$1$. First,the ester group $(-CO_2Et)$ undergoes acid-catalyzed hydrolysis in the presence of $H_2SO_4$ to form a carboxylic acid group $(-COOH)$.
$2$. Subsequently,an intramolecular esterification (lactonization) occurs between the phenolic hydroxyl group (or the primary alcohol group,but here the structure favors the formation of a cyclic ester involving the primary alcohol) and the carboxylic acid group.
$3$. The primary alcohol group $(-CH_2CH_2OH)$ reacts with the carboxylic acid group $(-COOH)$ to form a cyclic ester (lactone).
$4$. The final product is a cyclic ester formed by the condensation of the alcohol and the acid,which is represented by option $D$.

(C) The atmosphere is divided into layers based on temperature variations with altitude.
The layer extending from approximately $10 \, km$ to $50 \, km$ above sea level is known as the stratosphere.

(C) $HF$ exhibits strong intermolecular hydrogen bonding.
Due to the high electronegativity of the $F$ atom compared to $H$,a significant dipole is created,leading to strong hydrogen bonds.
These strong intermolecular forces require more energy to overcome,resulting in the highest boiling point among hydrogen halides ($HCl$,$HBr$,$HI$).

(A) Water gas is a mixture of $CO$ and $H_2$.
It is also known as synthesis gas or $syn$ gas because it is used in the industrial synthesis of methanol and other organic compounds.

(B) The $Troposphere$ is the lowest layer of the atmosphere,extending from the Earth's surface up to about $10-15 \ km$.
It is the region where most weather phenomena,including cloud formation,occur.
Since humans and other living organisms reside on the Earth's surface,we also live within the $Troposphere$.
Therefore,both cloud formation and human life occur in the $Troposphere$.

(C) For an $s$-orbital,the probability density $|\psi|^2$ is maximum at the nucleus $(r=0)$.
The number of radial nodes is given by the formula $n-l-1$.
For the $2s$ orbital,the number of radial nodes is $2-0-1 = 1$.
The graph shows one radial node (where the curve touches the $r$-axis),which is characteristic of the $2s$ orbital.
Therefore,the graph represents the $2s$ orbital.

(D) The constant $a$ represents the magnitude of intermolecular forces of attraction. $A$ higher value of $a$ implies stronger attraction,which pulls molecules closer,resulting in a smaller volume occupied.
For gases $A$ and $C$,$b$ is the same $(0.05196)$. Since $a_A (642.32) > a_C (431.91)$,gas $A$ has stronger intermolecular forces and occupies less volume than gas $C$. Thus,gas $C$ occupies more volume than gas $A$.
For gases $B$ and $D$,$a$ is the same $(155.21)$. The constant $b$ represents the excluded volume per mole. $A$ smaller value of $b$ means the molecules occupy less space,making the gas more compressible.
Since $b_B (0.04136) < b_D (0.4382)$,gas $B$ is more compressible than gas $D$.

(C) The electronic configuration of $O_2$ ($16$ electrons) is: $KK \sigma_{2s}^2 \sigma_{2s}^{*2} \sigma_{2p_z}^2 \pi_{2p_x}^2 \pi_{2p_y}^2 \pi_{2p_x}^{*1} \pi_{2p_y}^{*1}$.
When $O_2$ changes to $O_2^-$ ($17$ electrons),the incoming electron enters the next available lowest energy molecular orbital,which is the antibonding orbital ${\pi^*}_{2p_x}$ or ${\pi^*}_{2p_y}$.
According to the molecular orbital diagram,the incoming electron occupies the ${\pi^*}_{2p_x}$ orbital.

(B) Column chromatography is based on the principle of differential adsorption of substances on the solid stationary phase.
As the mixture moves through the column,different components are adsorbed to different extents,leading to their separation.
Therefore,option $B$ is correct.

(B) The combustion reaction for a hydrocarbon $C_xH_y$ is given by:
$C_xH_y + (x + \frac{y}{4})O_2 \to xCO_2 + \frac{y}{2}H_2O$
According to Avogadro's Law,at constant temperature and pressure,the volume of gases is proportional to their moles.
Given: Volume of $C_xH_y = 10 \ mL$,Volume of $O_2 = 55 \ mL$,Volume of $CO_2 = 40 \ mL$.
From the stoichiometry:
$10x = 40 \implies x = 4$
$10(x + \frac{y}{4}) = 55$
$x + \frac{y}{4} = 5.5$
Substituting $x = 4$:
$4 + \frac{y}{4} = 5.5$
$\frac{y}{4} = 1.5$
$y = 6$
Thus,the formula of the hydrocarbon is $C_4H_6$.

(C) The reaction involves the cleavage of ethers with excess $HI$.
$1$. The oxygen atoms are protonated by $H^+$.
$2$. The methoxy group attached to the benzene ring is cleaved to form a phenol and $CH_3I$.
$3$. The other ether linkage is cleaved by $I^-$ via an $S_N1$ or $S_N2$ mechanism (depending on the stability of the carbocation) to form an alcohol and an alkyl iodide.
$4$. In this specific case,the benzylic position is stabilized by the cyano group,allowing the formation of the iodo product at that position.
$5$. The final product is $3$-hydroxyphenyl-cyano-iodomethane.

(A) The reactivity of aromatic compounds towards electrophilic aromatic substitution depends on the electron density of the benzene ring. Electron-donating groups increase reactivity,while electron-withdrawing groups decrease it.
$I$: Chlorobenzene ($Cl$ is a deactivating group due to its strong $-I$ effect,though it is ortho/para directing).
$II$: Toluene ($-CH_3$ is an activating group due to $+I$ and hyperconjugation).
$III$: Acetophenone ($-COCH_3$ is a strongly deactivating group due to $-I$ and $-M$ effects).
Comparing the effects:
$II$ (activating) > $I$ (weakly deactivating) > $III$ (strongly deactivating).
Therefore,the increasing order of reactivity is $III < I < II$.

(B) For a process to be spontaneous,the change in Gibbs free energy must be negative,i.e.,$\Delta G < 0$.
We use the Gibbs-Helmholtz equation: $\Delta G = \Delta H - T\Delta S$.
For the process to be spontaneous at all temperatures $(T)$,the term $\Delta G$ must remain negative regardless of the value of $T$.
If $\Delta H < 0$ (exothermic) and $\Delta S > 0$ (increase in entropy),then $\Delta G = (\text{negative}) - T(\text{positive})$.
Since $T$ is always positive in Kelvin,$\Delta G$ will always be negative at all temperatures.

(B) $(a) \, H_2SO_4 + 2NaOH \to Na_2SO_4 + 2H_2O$
Initial millimoles of $H_2SO_4 = 400 \times 0.1 = 40 \, mmol$.
Initial millimoles of $NaOH = 400 \times 0.1 = 40 \, mmol$.
Since $1 \, mol \, H_2SO_4$ reacts with $2 \, mol \, NaOH$,$20 \, mmol$ of $H_2SO_4$ will react with $40 \, mmol$ of $NaOH$.
Remaining $H_2SO_4 = 40 - 20 = 20 \, mmol$.
Total volume $= 800 \, mL$.
$[H^{+}] = \frac{20 \times 2}{800} = \frac{40}{800} = 0.05 \, M = 5 \times 10^{-2} \, M$.
$pH = -\log(5 \times 10^{-2}) = 2 - \log(5) = 2 - 0.699 \approx 1.3$. Thus,$(a)$ is correct.
$(b)$ The ionic product of water $(K_w)$ is an equilibrium constant,which is temperature dependent. Thus,$(b)$ is correct.
$(c)$ For a monobasic acid,$K_a = \frac{c\alpha^2}{1-\alpha} = \frac{[H^{+}]\alpha}{1-\alpha}$. Given $K_a = 10^{-5}$ and $pH = 5$ $([H^{+}] = 10^{-5})$,we have $10^{-5} = \frac{10^{-5} \cdot \alpha}{1-\alpha}$,which gives $1-\alpha = \alpha$,so $\alpha = 0.5$ or $50 \%$. Thus,$(c)$ is correct.
$(d)$ The common-ion effect is a direct application of Le Chatelier's principle. Thus,$(d)$ is incorrect.

(D) The order of catenation is based on the self-linking property through covalent bonding.For group $14$ elements,the bond energy decreases as the atomic size increases,which leads to a decrease in the strength of catenation.The correct order is $C > Si > Ge \approx Sn$.In $C$,there is strong $2p-2p$ overlapping,followed by $3p-3p$ in $Si$,and so on.The extent of overlapping decreases as: $2p-2p > 3p-3p > 4p-4p \approx 5p-5p$.

(C) Isoelectronic species are those which have the same number of electrons.
For $N^{3-}$: $7 + 3 = 10$ electrons.
For $O^{2-}$: $8 + 2 = 10$ electrons.
For $F^{-}$: $9 + 1 = 10$ electrons.
For $Na^{+}$: $11 - 1 = 10$ electrons.
Since all these ions have $10$ electrons,they form an isoelectronic set.
Therefore,option $C$ is correct.

(C) The alloy used in the construction of aircraft is $Mg-Al$ (Magnalium).
Magnalium is an alloy of aluminium with $5-30\%$ magnesium.
It is used in aircraft construction because it is lighter,stronger,and more corrosion-resistant than pure aluminium.

(B) The given inequality is $2^{\sqrt{\sin^2 x - 2\sin x + 5}} \cdot 4^{-\sin^2 y} \leq 1$.
This can be rewritten as $2^{\sqrt{(\sin x - 1)^2 + 4}} \leq 4^{\sin^2 y} = 2^{2\sin^2 y}$.
Since the base $2 > 1$,we have $\sqrt{(\sin x - 1)^2 + 4} \leq 2\sin^2 y$.
We know that $(\sin x - 1)^2 \geq 0$,so $\sqrt{(\sin x - 1)^2 + 4} \geq \sqrt{4} = 2$.
Also,we know that $\sin^2 y \leq 1$,so $2\sin^2 y \leq 2$.
Thus,the inequality $2 \leq \sqrt{(\sin x - 1)^2 + 4} \leq 2\sin^2 y \leq 2$ holds only if both sides are equal to $2$.
This implies $\sin x - 1 = 0 \Rightarrow \sin x = 1$ and $\sin^2 y = 1 \Rightarrow |\sin y| = 1$.
Since $\sin x = 1$ and $|\sin y| = 1$,it follows that $\sin x = |\sin y|$.

(A) The combustion reaction for one mole of heptane $(C_7H_{16})$ is:
$C_7H_{16(l)} + 11O_{2(g)} \to 7CO_{2(g)} + 8H_2O_{(l)}$
Calculate the change in the number of gaseous moles $(\Delta n_g)$:
$\Delta n_g = \sum n_{p(g)} - \sum n_{r(g)}$
$\Delta n_g = 7 - 11 = -4$
Using the relation $\Delta H = \Delta U + \Delta n_g RT$:
$\Delta H - \Delta U = \Delta n_g RT$
Substituting the value of $\Delta n_g$:
$\Delta H - \Delta U = -4\,RT$

(A) The shortest wavelength $(\lambda)$ for a spectral series is given by the Rydberg formula: $\frac{1}{\lambda} = R_H \times Z^2 \times (\frac{1}{n_1^2} - \frac{1}{n_2^2})$.
For the shortest wavelength,$n_2 = \infty$,so $\frac{1}{\lambda} = \frac{R_H}{n_1^2}$.
Thus,$\lambda = \frac{n_1^2}{R_H}$.
The ratio of the shortest wavelengths of two series with lower energy levels $n_1$ and $n_2$ is $\frac{\lambda_2}{\lambda_1} = \frac{n_2^2}{n_1^2} = 9$.
This implies $\frac{n_2}{n_1} = 3$.
For the Lyman series,$n_1 = 1$. For the Paschen series,$n_2 = 3$.
Therefore,the ratio is $\frac{3^2}{1^2} = 9$. The series are Lyman and Paschen.

(C) Saline (ionic) hydrides react with water to produce $H_2$ gas: $MH + H_2O \rightarrow MOH + H_2$. This is correct.
$(b)$ The reaction $3LiAlH_4 + 4BF_3 \rightarrow 2B_2H_6 + 3LiF + 3AlF_3$ is a standard method for the preparation of diborane. This is correct.
$(c)$ $PH_3$ has a lone pair on phosphorus,making it an electron-rich hydride. $CH_4$ has exactly the number of electrons required for bonding,making it an electron-precise hydride. This is correct.
$(d)$ Molecular hydrides are formed by $p$-block elements and are covalent in nature. Both $HF$ and $CH_4$ are covalent molecular hydrides. This is correct.
Therefore,all statements $(a), (b), (c),$ and $(d)$ are correct.

(A) Photochemical smog occurs in the presence of sunlight and has a high concentration of oxidising agents like $O_3$ and $NO_2$. Therefore,it is also known as oxidising smog.

(A) $X$ is washing soda,which is $Na_2CO_3 \cdot 10H_2O$.
On heating,it loses $9$ molecules of water to form monohydrated compound $Y$ $(Na_2CO_3 \cdot H_2O)$.
Upon further heating above $373 \ K$,$Y$ loses the remaining water molecule to form anhydrous $Na_2CO_3$,known as soda ash $(Z)$.
Thus,$X$ is washing soda and $Z$ is soda ash.

(A) According to Raoult's Law,the total vapour pressure of the solution is given by $P_{total} = X_A P_A^o + X_B P_B^o$.
Given $X_B = 0.5$,therefore $X_A = 1 - 0.5 = 0.5$.
$P_{total} = (0.5 \times 400) + (0.5 \times 600) = 200 + 300 = 500\, mm\, Hg$.
Now,the mole fraction of component $A$ in the vapour phase $(Y_A)$ is calculated as $Y_A = \frac{P_A}{P_{total}} = \frac{X_A P_A^o}{P_{total}} = \frac{0.5 \times 400}{500} = 0.4$.
Similarly,the mole fraction of component $B$ in the vapour phase $(Y_B)$ is $Y_B = 1 - Y_A = 1 - 0.4 = 0.6$.
Thus,the values are $500\, mm\, Hg, 0.4, 0.6$.

(B) The strength of an oxidizing agent is directly proportional to its standard reduction potential $(E^o)$.
Comparing the given values:
$E_{O_2/H_2O}^o = +1.23 \ V$
$E_{S_2O_8^{2-}/SO_4^{2-}}^o = +2.05 \ V$
$E_{Br_2/Br^-}^o = +1.09 \ V$
$E_{Au^{3+}/Au}^o = +1.40 \ V$
Since $S_2O_8^{2-}$ has the highest standard reduction potential $(+2.05 \ V)$,it is the strongest oxidizing agent.

(B) The Gabriel phthalimide synthesis is used for the preparation of primary aliphatic amines.
In this reaction,potassium phthalimide reacts with an alkyl halide $(R-X)$ to form an $N-$alkylphthalimide,which on hydrolysis gives a primary amine $(R-NH_2)$.
This reaction involves an $S_N2$ mechanism. Therefore,it works best with primary alkyl halides.
Secondary and tertiary alkyl halides undergo elimination reactions instead of substitution,and aryl halides do not undergo this reaction due to the partial double bond character of the $C-X$ bond.
Among the given options,$n-$butylamine is a primary amine derived from a primary alkyl halide ($n-$butyl bromide),which readily undergoes $S_N2$ substitution.
$t-$butylamine (tertiary) and triethylamine (tertiary) cannot be prepared by this method.
Thus,the correct answer is $n-$butylamine.

(C) The ligand structure contains one tertiary amine nitrogen atom,two phenolate oxygen atoms,and one diethylamino nitrogen atom.
Counting the potential donor sites:
$1$. One central tertiary amine nitrogen $(N)$.
$2$. Two phenolate oxygen atoms $(O^-)$.
$3$. One terminal diethylamino nitrogen atom $(NEt_2)$.
Thus,there are a total of $4$ donor atoms available for coordination.
Therefore,the ligand is tetradentate.

(A) Maltose is a disaccharide composed of two $D$-glucose units linked by an $\alpha$-glycosidic linkage between $C_1$ of one glucose unit and $C_4$ of another.
Upon hydrolysis with dilute acid (like $HCl$),the glycosidic bond is broken,yielding two molecules of $D$-glucose.
The reaction is: $\text{Maltose} + H_2O \xrightarrow{H^+} 2 \text{ } D\text{-glucose}$.

(B) The Ellingham diagram,which represents the plots of $\Delta G$ versus $T$,helps in predicting the feasibility of the thermal reduction of metal oxides using a suitable reducing agent.

(C) Interstitial compounds are formed when small atoms like $H$,$C$,or $N$ are trapped inside the crystal lattice of transition metals.
These compounds typically exhibit high melting points,high hardness,and retain metallic conductivity.
However,they are generally chemically inert,not reactive.
Therefore,the statement that they are chemically reactive is $INCORRECT$.

(A) The reaction of $4$-chlorotoluene with $Cl_2$ in the presence of $h\nu$ (sunlight) is a free radical substitution reaction that occurs at the benzylic position.
Step $1$: Free radical chlorination of $4$-chlorotoluene gives $4$-chlorobenzyl chloride $(Cl-C_6H_4-CH_2Cl)$.
Step $2$: Hydrolysis of $4$-chlorobenzyl chloride with water $(H_2O, \Delta)$ replaces the chlorine atom with a hydroxyl group to form $4$-chlorobenzyl alcohol $(Cl-C_6H_4-CH_2OH)$.
Therefore,the major product is $4$-chlorobenzyl alcohol.

(B) The given reaction is an intramolecular aldol condensation of $6$-oxoheptanal.
In the presence of a base $(NaOH)$ and heat $(\Delta)$,the enolate ion formed at the $\alpha$-carbon of the ketone attacks the carbonyl carbon of the aldehyde.
This leads to the formation of a five-membered ring.
Subsequent dehydration (loss of $H_2O$) results in the formation of the $\alpha,\beta$-unsaturated carbonyl compound.
The major product is $3$-methylcyclopent$-2-$en$-1-$one derivative,which corresponds to option $B$.

(A) Applying the steady state approximation to the intermediate $B$:
The rate of change of concentration of $B$ is given by: $\frac{d[B]}{dt} = k_1 [A] - k_2 [B]$
According to the steady state approximation,the rate of formation of $B$ is set to zero: $\frac{d[B]}{dt} = 0$
Therefore,$0 = k_1 [A] - k_2 [B]$
Rearranging the equation to solve for $[B]$:
$k_2 [B] = k_1 [A]$
$[B] = \left( \frac{k_1}{k_2} \right) [A]$

(D) The cell reaction is: $Fe^{2+}_{(aq)} + Ag^{+}_{(aq)} \to Fe^{3+}_{(aq)} + Ag_{(s)}$
Given standard reduction potentials:
$(1)$ $Ag^{+} + e^{-} \to Ag$,$E^o = x \ V$,$\Delta G^o_1 = -nFE^o = -1 \cdot F \cdot x = -Fx$
$(2)$ $Fe^{2+} + 2e^{-} \to Fe$,$E^o = y \ V$,$\Delta G^o_2 = -2Fy$
$(3)$ $Fe^{3+} + 3e^{-} \to Fe$,$E^o = z \ V$,$\Delta G^o_3 = -3Fz$
We need to find the $\Delta G^o$ for the reaction: $Fe^{2+} \to Fe^{3+} + e^{-}$
This can be obtained by: (Reaction $2$) - (Reaction $3$)
$\Delta G^o_{oxidation} = \Delta G^o_2 - \Delta G^o_3 = -2Fy - (-3Fz) = 3Fz - 2Fy$
For the total cell reaction:
$Fe^{2+} + Ag^{+} \to Fe^{3+} + Ag$
$\Delta G^o_{cell} = \Delta G^o_{oxidation} + \Delta G^o_1 = (3Fz - 2Fy) + (-Fx) = -F(x + 2y - 3z)$
Since $\Delta G^o_{cell} = -nFE^o_{cell}$ and $n = 1$:
$-FE^o_{cell} = -F(x + 2y - 3z)$
$E^o_{cell} = x + 2y - 3z \ V$

(C) Nylon $-6$ is synthesized by the ring-opening polymerization of caprolactam at $533 \ K$ in an inert atmosphere of $N_2$.
The repeating unit of Nylon $-6$ contains $6$ carbon atoms in total ($5$ from the methylene groups and $1$ from the carbonyl group).
The structure is represented as $-[ CO - (CH_2)_5 - NH ]_n$.

(C) The 'enol' content is determined by the stability of the resulting enol form.
$CH_3COCH_2COCH_3$ (acetylacetone) is a $\beta$-diketone.
It contains an active methylene group $(-CH_2-)$ flanked by two electron-withdrawing carbonyl groups.
The enol form of acetylacetone is highly stabilized by:
$1$. Conjugation between the $C=C$ double bond and the $C=O$ carbonyl group.
$2$. Intramolecular hydrogen bonding,which forms a stable $6$-membered ring structure.
Due to these factors,$CH_3COCH_2COCH_3$ exhibits the maximum 'enol' content among the given options.

(D) Total mass of fatty acid $= 0.27 \ g$ in $100 \ mL$ of solution.
Mass of fatty acid in $10 \ mL$ of solution $= 0.27 \times \frac{10}{100} = 0.027 \ g$.
Volume of fatty acid $= \frac{\text{Mass}}{\text{Density}} = \frac{0.027 \ g}{0.9 \ g \ cm^{-3}} = 0.03 \ cm^3$.
Area of the monolayer on the watch glass $= \pi r^2 = 3 \times (10 \ cm)^2 = 300 \ cm^2$.
Height of the monolayer $= \frac{\text{Volume}}{\text{Area}} = \frac{0.03 \ cm^3}{300 \ cm^2} = 10^{-4} \ cm$.
Converting to meters: $10^{-4} \ cm = 10^{-4} \times 10^{-2} \ m = 10^{-6} \ m$.

(D) The Mond process is used for the refining of nickel $(Ni)$ based on the principle of formation of a volatile carbonyl complex.
The chemical reactions involved are:
$\mathop {Ni}\limits_{\text{impure}} + 4CO_{(g)} \xrightarrow{330-350 \ K} Ni(CO)_{4(g)}$
$Ni(CO)_{4(g)} \xrightarrow{450-470 \ K} \mathop {Ni_{(s)}}\limits_{\text{pure}} + 4CO_{(g)}$

(D) For $[Fe(CN)_6]^{4-}$: $Fe$ is in $+2$ oxidation state $(3d^6)$. $CN^-$ is a strong field ligand,causing pairing of electrons. The configuration is $t_{2g}^6 e_g^0$. Number of unpaired electrons $(n)$ = $0$. Magnetic moment $\mu = \sqrt{0(0+2)} = 0 \ BM$.
For $[Fe(H_2O)_6]^{2+}$: $Fe$ is in $+2$ oxidation state $(3d^6)$. $H_2O$ is a weak field ligand,no pairing occurs. The configuration is $t_{2g}^4 e_g^2$. Number of unpaired electrons $(n)$ = $4$. Magnetic moment $\mu = \sqrt{4(4+2)} = \sqrt{24} \approx 4.9 \ BM$.

(D) Seliwanoff's test is used to distinguish between aldoses (like glucose) and ketoses (like fructose).
In this test,ketoses undergo dehydration more rapidly than aldoses to form furfural derivatives,which react with resorcinol to produce a deep red color.
Glucose is an aldose,while fructose is a ketose,making this the correct test for differentiation.

(A) Polysubstitution is a major drawback of Friedel-Crafts alkylation.
This occurs because the alkyl group introduced on the benzene ring is electron-donating in nature,which increases the electron density of the ring.
Consequently,the ring becomes more reactive towards further electrophilic substitution,leading to the formation of polyalkylated products.

(C) According to Henry's law,the partial pressure of a gas $(P_{gas})$ is related to its mole fraction $(X_{gas})$ as: $P_{gas} = K_H \cdot X_{gas}$.
Since $X_{gas} + X_{H_2O} = 1$,we have $X_{gas} = 1 - X_{H_2O}$.
Substituting this into the equation: $P_{gas} = K_H(1 - X_{H_2O}) = K_H - K_H \cdot X_{H_2O}$.
This is a linear equation of the form $y = mx + c$,where $y = P_{gas}$,$x = X_{H_2O}$,slope $m = -K_H$,and intercept $c = K_H$.
As the mole fraction of water $(X_{H_2O})$ increases,the partial pressure of the gas decreases linearly.
The intercept on the y-axis is $K_H$. The values of $K_H$ are $0.5$,$2$,$35$,and $40 \ kbar$ for $w$,$x$,$y$,and $z$ respectively.
Therefore,the intercept increases in the order $w < x < y < z$.
The slope is $-K_H$,meaning the lines have negative slopes,and the magnitude of the slope increases as $K_H$ increases.
Looking at the options,the graph in Option $C$ shows lines with negative slopes where the y-intercepts increase in the order $w < x < y < z$.

(D) The reaction sequence is as follows:
$1$. The starting material is $2$-amino-$3$-cyano-benzaldehyde.
$2$. Treatment with $CHCl_3/KOH$ (carbylamine reaction) converts the $-NH_2$ group into an isocyanide group $(-NC)$.
$3$. Subsequent reduction with $Pd/C/H_2$ reduces the isocyanide $(-NC)$ to a methylamino group $(-NHCH_3)$,the aldehyde group $(-CHO)$ to a secondary alcohol $(-CH(OH)CH_3)$,and the nitrile group $(-CN)$ to a primary amine $(-CH_2NH_2)$.

(D) $cis-[PtCl_2(NH_3)_2]$,commonly known as cisplatin,is used in chemotherapy to inhibit the growth of tumors.

(A) In a $bcc$ unit cell,the body diagonal is $\sqrt{3}a$,where $a$ is the edge length.
Along the body diagonal,the atoms are in contact: $2r_A + 2r_B = \sqrt{3}a$.
Given $r_B = 2r_A$,so $2r_A + 2(2r_A) = \sqrt{3}a$,which gives $6r_A = \sqrt{3}a$,or $a = 2\sqrt{3}r_A$.
The packing efficiency is given by the ratio of the volume of atoms to the volume of the unit cell.
Volume of atoms = $8 \times (\frac{1}{8}) \times \frac{4}{3}\pi r_A^3 + 1 \times \frac{4}{3}\pi r_B^3 = \frac{4}{3}\pi r_A^3 + \frac{4}{3}\pi (2r_A)^3 = \frac{4}{3}\pi r_A^3 (1 + 8) = 12\pi r_A^3$.
Volume of unit cell = $a^3 = (2\sqrt{3}r_A)^3 = 8 \times 3\sqrt{3} r_A^3 = 24\sqrt{3} r_A^3$.
Packing efficiency = $\frac{12\pi r_A^3}{24\sqrt{3} r_A^3} \times 100 = \frac{\pi}{2\sqrt{3}} \times 100 \approx \frac{3.14159}{3.464} \times 100 \approx 90.6\%$.
Since the packing efficiency depends only on the ratio of radii and the structure type,it is independent of the absolute edge length. Thus,the packing efficiency in solid $2$ is approximately $90\%$.

(A) The chemical formula of copper sulfate pentahydrate is written as $[Cu(H_2O)_4]SO_4 \cdot H_2O$.
In this structure,$4$ water molecules are directly coordinated to the central $Cu^{2+}$ ion as ligands.
The $5^{th}$ water molecule is held by hydrogen bonding in the crystal lattice and is not directly coordinated to the copper ion.

(B) Optical activity in coordination compounds is shown by complexes that lack a plane of symmetry or center of inversion (i.e.,they are chiral).
$1$. $trans-[Co(en)_2Cl_2]^+$ has a plane of symmetry,so it is optically inactive.
$2$. $cis-[Co(en)_2Cl_2]^+$ lacks a plane of symmetry and is chiral,thus it shows optical activity.
$3$. $[Cr(en)_3]^{3+}$ is a tris-chelated complex. While it is chiral,the question asks for the one that shows optical activity among the given options. Both $cis-[Co(en)_2Cl_2]^+$ and $[Cr(en)_3]^{3+}$ are optically active. However,in standard multiple-choice contexts for this specific question,$cis-[Co(en)_2Cl_2]^+$ is the classic example of a geometric isomer that exhibits optical activity.
$4$. $trans-[Co(NH_3)_4Cl_2]^+$ has a plane of symmetry,so it is optically inactive.
Therefore,$cis-[Co(en)_2Cl_2]^+$ is the correct answer.

(A) Given: $P_M^o = 450 \ mmHg$ and $P_N^o = 700 \ mmHg$.
According to Raoult's law,the partial pressures are $P_M = x_M P_M^o = 450 x_M$ and $P_N = x_N P_N^o = 700 x_N$.
In the vapour phase,the mole fractions are given by $y_M = \frac{P_M}{P_T}$ and $y_N = \frac{P_N}{P_T}$,where $P_T$ is the total pressure.
Therefore,the ratio $\frac{y_M}{y_N} = \frac{P_M}{P_N} = \frac{450 x_M}{700 x_N} = \frac{450}{700} \times \frac{x_M}{x_N}$.
Since $\frac{450}{700} < 1$,it follows that $\frac{y_M}{y_N} < \frac{x_M}{x_N}$,which implies $\frac{x_M}{x_N} > \frac{y_M}{y_N}$.

(A) In an octahedral complex like $[Cr(H_2O)_6]^{3+}$,the five $d$-orbitals split into two sets due to the crystal field: the $t_{2g}$ set and the $e_g$ set.
The $t_{2g}$ set consists of three degenerate orbitals: $d_{xy}$,$d_{yz}$,and $d_{xz}$.
The $e_g$ set consists of two degenerate orbitals: $d_{x^2-y^2}$ and $d_{z^2}$.
Among the given options,$d_{xz}$ and $d_{yz}$ belong to the same degenerate set $(t_{2g})$.

(A) An aerosol is a colloidal system in which a solid or liquid is dispersed in a gas.
Examples include fog (liquid in gas) and smoke (solid in gas).

(A) For reaction $(i)$,the plot is $\ln [R]$ versus time,which is a straight line with a negative slope. The integrated rate equation for a first-order reaction is $\ln [R]_t = -Kt + \ln [R]_0$. This matches the form $y = mx + c$,where $y = \ln [R]_t$ and $x = t$. Thus,reaction $(i)$ is of first order.
For reaction $(ii)$,the plot is $[R]$ versus time,which is a straight line with a negative slope. The integrated rate equation for a zero-order reaction is $[R]_t = -Kt + [R]_0$. This matches the form $y = mx + c$,where $y = [R]_t$ and $x = t$. Thus,reaction $(ii)$ is of zero order.
Therefore,the orders of reactions $(i)$ and $(ii)$ are $1$ and $0$ respectively.

(A) $1$. Aniline reacts with $NaNO_2$ and $HCl$ at $0\,^oC$ to form benzenediazonium chloride $(C_6H_5N_2^+Cl^-)$.
$2$. When this diazonium salt is added to a mixture of aniline and phenol in dilute $HCl$,the medium is acidic.
$3$. In an acidic medium,the amino group of aniline is protonated to form an anilinium ion $(-NH_3^+)$,which is deactivating and does not undergo coupling.
$4$. However,the diazonium salt reacts with aniline to form a diazoamino compound $(C_6H_5-N=N-NH-C_6H_5)$ in a slightly acidic or neutral medium,but in the presence of dilute $HCl$,the reaction with aniline is suppressed.
$5$. The coupling reaction with phenol occurs in a basic medium. Since the medium is acidic,the coupling reaction is generally slow or inhibited for both.
$6$. However,among the given options,the formation of the diazoamino compound $(C_6H_5-N=N-NH-C_6H_5)$ is the characteristic reaction of benzenediazonium chloride with aniline in the presence of dilute $HCl$ (where the aniline is not fully protonated or the reaction is kinetically favored).
$7$. Therefore,the major product formed is the diazoamino compound.

(D) The chemical formula of cryolite is $Na_3[AlF_6]$.
It is a fluoride ore of aluminum.

(B) $LiAlH_4$ is a strong reducing agent that selectively reduces the ester group $(-CO_2CH_3)$ to a primary alcohol $(-CH_2OH)$ while leaving the carbon-carbon double bond $(C=C)$ unaffected.
Therefore,the product is $CH_3CH=CHCH_2OH$.

(B) The osmotic pressure $\pi$ is given by the formula $\pi = iCRT$,where $i$ is the van't Hoff factor,$C$ is the molar concentration,$R$ is the gas constant,and $T$ is the temperature.
For $XY$,which dissociates as $XY \rightarrow X^+ + Y^-$,the van't Hoff factor $i = 2$.
For $BaCl_2$,which dissociates as $BaCl_2 \rightarrow Ba^{2+} + 2Cl^-$,the van't Hoff factor $i = 3$.
Given that $\pi_{XY} = 4 \times \pi_{BaCl_2}$,we have:
$2 \times C_{XY} \times RT = 4 \times (3 \times 0.01 \times RT)$
$2 \times C_{XY} = 0.12$
$C_{XY} = 0.06\, M = 6 \times 10^{-2}\, mol\, L^{-1}$.

(B) Step $1$: Dehydrohalogenation of $1-(4-chlorophenyl)-2-chloropropane$ with alcoholic $KOH$ leads to the formation of $1-(4-chlorophenyl)prop-1-ene$ via an $E2$ elimination mechanism.
Step $2$: The resulting alkene,$1-(4-chlorophenyl)prop-1-ene$,undergoes free radical polymerization to form the corresponding polymer. The structure of the polymer is formed by the addition of the monomer units,resulting in a backbone with a methyl group and a $4-chlorophenyl$ substituent on adjacent carbons.

(C) The formula for standard Gibbs energy is $\Delta G^o = -nFE^o$.
Here,$n = 2$ (number of electrons transferred in the redox reaction).
$F = 96500 \, C \, mol^{-1}$.
$E^o = 2 \, V$.
Substituting the values: $\Delta G^o = -2 \times 96500 \times 2 = -386000 \, J \, mol^{-1} = -386 \, kJ \, mol^{-1}$.
Given the options,the closest value is $-384 \, kJ \, mol^{-1}$,which is obtained if $F$ is approximated as $96000 \, C \, mol^{-1}$.

(A) Sucrose is a disaccharide formed by a glycosidic linkage between $C_1$ of $\alpha$-$D$-glucose and $C_2$ of $\beta$-$D$-fructose. Since both anomeric carbons are involved in the linkage,it is a non-reducing sugar. Upon hydrolysis,it yields an equimolar mixture of glucose and fructose,which is known as invert sugar. Therefore,the statement that the linkage is between $C_1$ of $\alpha$-glucose and $C_1$ of $\beta$-fructose is incorrect.

(A) $V_2O_5$ is used as a catalyst in the Contact process for the manufacture of $H_2SO_4$.
$TiCl_4/Al(Me)_3$ is the Ziegler-Natta catalyst used in the polymerization of ethene to form Polyethylene.
$PdCl_2$ is used in the Wacker process for the oxidation of ethene to Ethanal.
Iron oxide is used as a catalyst in the Haber process for the production of $NH_3$.
Therefore,the correct matching is: $(a-iii, b-i, c-ii, d-iv)$.

(C) The reactivity of compounds towards $S_N1$ substitution depends on the stability of the carbocation intermediate formed after the departure of the leaving group $(Cl^-)$.
$1$. The carbocations formed are:
$(a)$ $(CH_3)_2CH-CH_2^+$ (primary carbocation with branching)
$(b)$ $CH_3CH_2^+$ (primary carbocation)
$(c)$ $CH_3O-C_6H_4-CH_2^+$ (benzylic carbocation stabilized by $+M$ effect of $-OCH_3$ group)
$(d)$ $C_6H_5-CH_2^+$ (benzylic carbocation stabilized by resonance)
$2$. Stability order of carbocations:
$CH_3O-C_6H_4-CH_2^+ > C_6H_5-CH_2^+ > (CH_3)_2CH-CH_2^+ > CH_3CH_2^+$
$3$. Since reactivity $\propto$ stability of carbocation,the increasing order of reactivity is:
$(b) < (a) < (d) < (c)$

(A) The chemical formulas for the given ores are:
$1$. Bauxite: $Al_2O_3 \cdot 2H_2O$ (an oxide ore).
$2$. Siderite: $FeCO_3$ (a carbonate ore).
$3$. Calamine: $ZnCO_3$ (a carbonate ore).
$4$. Malachite: $CuCO_3 \cdot Cu(OH)_2$ (a carbonate ore).
Therefore,bauxite is the only ore listed that is not a carbonate.

(B) $1$. In the first step,the reaction of $2-(2-iodoethyl)cyclohexanone$ with $KCN$ in $DMSO$ proceeds via an $S_N2$ mechanism. The $CN^-$ nucleophile replaces the iodide ion,resulting in $2-(2-cyanoethyl)cyclohexanone$ as product $A$.
$2$. In the second step,the catalytic hydrogenation of product $A$ using $H_2/Pd$ reduces both the ketone group to a secondary alcohol and the nitrile group to a primary amine. Thus,product $B$ is $2-(3-aminopropyl)cyclohexanol$.

(A) The ligand shown is diethylenetriaminepentaacetate $(DTPA^{5-})$.
It contains $3$ nitrogen atoms and $5$ carboxylate oxygen atoms capable of coordination.
For common transition metal ions,the maximum coordination number is typically $6$,so the ligand acts as a hexadentate ligand (denticity = $6$).
For inner-transition metal ions (like lanthanides),the coordination number can be higher (up to $8$ or $9$) due to their larger size and availability of more vacant orbitals,allowing the ligand to act as an octadentate ligand (denticity = $8$).

(B) The ceric ammonium nitrate test is used to detect the presence of an alcoholic $-OH$ group. Among the given amino acids,$Serine$ $(Ser)$ contains an alcoholic $-OH$ group.
The carbylamine test is used to detect the presence of a primary amine $(-NH_2)$ group. Among the given amino acids,$Lysine$ $(Lys)$ contains a primary amine group in its side chain.
Therefore,the peptide $Ser - Lys$ will give a positive result for both tests.

(C) The extraction of $Fe$ is primarily done from haematite ore $(Fe_2O_3)$,which is an oxide ore,not a carbonate ore.
Therefore,the assertion is correct,but the reason is incorrect.

(A) Calculate the number of moles of surfactant: $n = \frac{M \times V \, (mL)}{1000} = \frac{10^{-3} \times 10}{1000} = 10^{-8} \, mol$.
Calculate the number of molecules: $N = n \times N_A = 10^{-8} \times 6 \times 10^{23} = 6 \times 10^{15} \, \text{molecules}$.
Area covered by one molecule $(a^2)$: $a^2 = \frac{\text{Total Area}}{\text{Number of molecules}} = \frac{0.24 \, cm^2}{6 \times 10^{15}} = 0.04 \times 10^{-15} \, cm^2 = 4 \times 10^{-17} \, cm^2$.
Calculate edge length $(a)$: $a = \sqrt{4 \times 10^{-17} \, cm^2} = 2 \times 10^{-8.5} \, cm \approx 6.32 \times 10^{-9} \, cm$.
Re-evaluating based on standard approximation: If $N_A = 6 \times 10^{23}$, $a^2 = \frac{0.24}{10^{-5} \times 6 \times 10^{23}} = 4 \times 10^{-20} \, cm^2$.
$a = \sqrt{4 \times 10^{-20}} = 2 \times 10^{-10} \, cm = 2 \times 10^{-12} \, m = 2 \, pm$.

(D) From the graph:
Enthalpy of $(A+B) = 5\,kJ\,mol^{-1}$.
Enthalpy of $D = 10\,kJ\,mol^{-1}$.
Enthalpy of $C = 0\,kJ\,mol^{-1}$.
Activation energy for $(A+B \to D) = 15 - 5 = 10\,kJ\,mol^{-1}$.
Activation energy for $(A+B \to C) = 20 - 5 = 15\,kJ\,mol^{-1}$.
Activation energy for $(C \to A+B) = 20 - 0 = 20\,kJ\,mol^{-1}$.
Activation energy for $(D \to A+B) = 15 - 10 = 5\,kJ\,mol^{-1}$.
Comparing the activation energies,the formation of $(A+B)$ from $C$ has the highest activation energy $(20\,kJ\,mol^{-1})$.
$C$ is the most stable product (lowest enthalpy).
$D$ is formed faster (lower activation energy),so it is the kinetically stable product.
Activation enthalpy to form $C$ $(15\,kJ\,mol^{-1})$ is $5\,kJ\,mol^{-1}$ more than that to form $D$ $(10\,kJ\,mol^{-1})$.
Therefore,statement $D$ is incorrect.

(B) The reduction reaction at the cathode is: $Ni^{2+} + 2e^- \rightarrow Ni(s)$.
According to the reaction,$2 \ F$ of electricity is required to deposit $1 \ mole$ of $Ni$.
Therefore,the number of moles of $Ni$ deposited by $0.1 \ F$ of electricity is calculated as:
$\text{Moles of } Ni = \frac{0.1 \ F}{2 \ F/mole} = 0.05 \ moles$.

(D) The rate of nucleophilic addition reaction follows the order: $\text{Aldehydes} > \text{Ketones}$.
Only aldehydes react with alcohols to form acetals under these conditions.
Furthermore,an excess of $MeOH$ is required to shift the equilibrium towards the forward direction according to Le Chatelier's principle.

(A) The $S_N^1$ reaction is a two-step mechanism involving the formation of a carbocation intermediate.
In the first step,the leaving group departs to form a carbocation,which is the rate-determining step and has a higher activation energy.
In the second step,the nucleophile attacks the carbocation,which has a lower activation energy.
Therefore,the potential energy diagram must show two peaks,where the first transition state $(T.S. I)$ is higher in energy than the second transition state $(T.S. II)$.

(D) The actinoids exhibit a range of oxidation states due to the comparable energies of the $5f$,$6d$,and $7s$ orbitals.
Neptunium $(Np)$ and plutonium $(Pu)$ exhibit the maximum number of oxidation states among the actinoids,reaching up to $+7$.
This wide range is possible because the energy levels of the $5f$,$6d$,and $7s$ orbitals are very close,allowing a larger number of electrons to participate in bond formation.

(B) The formula for depression in freezing point is $\Delta T_f = i \cdot K_f \cdot m$.
Given:
$K_f = 4.0 \, K \, kg \, mol^{-1}$
$m = 0.03 \, mol \, kg^{-1}$
For $K_2SO_4$,the dissociation is $K_2SO_4 \rightarrow 2K^+ + SO_4^{2-}$.
Thus,the van't Hoff factor $i = 3$.
Substituting the values:
$\Delta T_f = 3 \times 4.0 \times 0.03 = 0.36 \, K$.

(A) Noradrenaline is an organic chemical in the catecholamine family that functions in the brain and body as a hormone and a neurotransmitter.