JEE Main 2024 Chemistry Question Paper with Answer and Solution

(B) The acceleration $a$ of an Atwood machine with masses $m_1$ and $m_2$ is given by the formula:
$a = \left( \frac{m_1 - m_2}{m_1 + m_2} \right) g$
Given that the acceleration $a = \frac{g}{8}$,we substitute this into the equation:
$\frac{g}{8} = \left( \frac{m_1 - m_2}{m_1 + m_2} \right) g$
Dividing both sides by $g$:
$\frac{1}{8} = \frac{m_1 - m_2}{m_1 + m_2}$
Cross-multiplying gives:
$m_1 + m_2 = 8(m_1 - m_2)$
$m_1 + m_2 = 8m_1 - 8m_2$
Rearranging the terms to group the masses:
$9m_2 = 7m_1$
Therefore,the ratio of the masses is:
$\frac{m_1}{m_2} = \frac{9}{7}$

(D) mixture shows positive deviation from Raoult's law when the solute-solvent intermolecular interactions are weaker than the solute-solute and solvent-solvent interactions.
In the mixture of $(CH_3)_2CO$ (acetone) and $CS_2$ (carbon disulfide),the dipole-dipole interactions between acetone molecules and the dispersion forces between $CS_2$ molecules are stronger than the interactions between acetone and $CS_2$ molecules.
Consequently,the vapor pressure of the solution is higher than that predicted by Raoult's law.
Therefore,$(CH_3)_2CO + CS_2$ exhibits positive deviation.

(B) The acceleration $a$ of a system of two masses $m_1$ and $m_2$ connected by a string over a pulley is given by the formula: $a = \left( \frac{m_2 - m_1}{m_1 + m_2} \right) g$,assuming $m_2 > m_1$.
Given that the acceleration $a = \frac{g}{8}$,we substitute this into the equation:
$\frac{g}{8} = \left( \frac{m_2 - m_1}{m_1 + m_2} \right) g$
Canceling $g$ from both sides:
$\frac{1}{8} = \frac{m_2 - m_1}{m_1 + m_2}$
Cross-multiplying gives:
$m_1 + m_2 = 8(m_2 - m_1)$
$m_1 + m_2 = 8m_2 - 8m_1$
Rearranging the terms to group $m_1$ and $m_2$:
$m_1 + 8m_1 = 8m_2 - m_2$
$9m_1 = 7m_2$
Therefore,the ratio of the masses is:
$\frac{m_2}{m_1} = \frac{9}{7}$

(A) -glucose is the naturally occurring isomer where the $-OH$ group on the chiral carbon furthest from the aldehyde group (i.e.,$C-5$) is on the right side in the Fischer projection.
$L$-glucose is the enantiomer of $D$-glucose.
To obtain the structure of $L$-glucose,we must invert the configuration at every chiral center $(C-2, C-3, C-4, C-5)$ of $D$-glucose.
In $D$-glucose,the $-OH$ groups are at: $C-2$ (right),$C-3$ (left),$C-4$ (right),$C-5$ (right).
Therefore,in $L$-glucose,the $-OH$ groups must be at: $C-2$ (left),$C-3$ (right),$C-4$ (left),$C-5$ (left).
This corresponds to the structure where the $-OH$ group on $C-5$ is on the left,which is the definition of the $L$-configuration.

(B) For an Atwood machine with masses $m_1$ and $m_2$ (where $m_2 > m_1$),the acceleration $a$ is given by the formula:
$a = \left(\frac{m_2 - m_1}{m_1 + m_2}\right) g$
Given that the acceleration $a = \frac{g}{8}$,we substitute this into the equation:
$\frac{g}{8} = \left(\frac{m_2 - m_1}{m_1 + m_2}\right) g$
$\frac{1}{8} = \frac{m_2 - m_1}{m_1 + m_2}$
Cross-multiplying gives:
$m_1 + m_2 = 8(m_2 - m_1)$
$m_1 + m_2 = 8m_2 - 8m_1$
Rearranging the terms to group $m_1$ and $m_2$:
$m_1 + 8m_1 = 8m_2 - m_2$
$9m_1 = 7m_2$
Therefore,the ratio of the masses is:
$\frac{m_2}{m_1} = \frac{9}{7}$

(C) The magnetic moment $(mu)$ is calculated using the formula $\mu = \sqrt{n(n+2)} \, BM$,where $n$ is the number of unpaired electrons.
For $[Ar] \, 3d^8$,$n = 2$ (unpaired electrons).
For $[Ar] \, 3d^3$,$n = 3$ (unpaired electrons).
For $[Ar] \, 3d^6$,$n = 4$ (unpaired electrons).
For $[Ar] \, 3d^7$,$n = 3$ (unpaired electrons).
Since $[Ar] \, 3d^6$ has the maximum number of unpaired electrons $(n=4)$,it will have the highest magnetic moment.

(B) The enol content is determined by the stability of the resulting enol form.
In the case of $Cyclohexane-1,3,5-trione$ (option $B$),the enolization leads to the formation of $benzene-1,3,5-triol$ (phloroglucinol).
This product is highly stable due to aromaticity.
Since aromatic compounds are significantly more stable than non-aromatic compounds,the equilibrium strongly favors the enol form in this case,resulting in the highest enol content among the given options.

(D) molecule is polar if it has a net dipole moment,i.e.,$\mu \neq 0$.
$1$. $CCl_4$ has a tetrahedral geometry where the bond dipoles of four $C-Cl$ bonds cancel each other out,making it non-polar $(\mu = 0)$.
$2$. $CO_2$ is a linear molecule where the two $C=O$ bond dipoles are equal and opposite,canceling each other out,making it non-polar $(\mu = 0)$.
$3$. $CH_2=CH_2$ (ethene) is a planar molecule with a symmetric distribution of charge,making it non-polar $(\mu = 0)$.
$4$. $CHCl_3$ (chloroform) has a tetrahedral geometry,but due to the presence of three $Cl$ atoms and one $H$ atom,the bond dipoles do not cancel each other out. Thus,it has a net dipole moment $(\mu \neq 0)$ and is a polar molecule.

(C) Statement $I$ is true because placing the $4f$ (lanthanoids) and $5f$ (actinoids) series separately prevents the periodic table from becoming unnecessarily wide and maintains the classification based on electronic configuration.
Statement $II$ is false because $S$-block elements (alkali and alkaline earth metals) are highly reactive due to their low ionization enthalpies and are therefore found in nature only in combined states (as ores,salts,etc.),not in their pure elemental form.

(B) Boron exists as a giant covalent polymer in its solid state,which results in a very strong crystalline lattice structure.
Due to this strong lattice,a large amount of energy is required to break the bonds,leading to an unusually high melting point $(2453 \ K)$ compared to other group $13$ elements.
Therefore,both Assertion $(A)$ and Reason $(R)$ are true,and $(R)$ is the correct explanation of $(A)$.

(D) Cyclohexene is a cyclic compound that does not contain a benzene ring and does not exhibit aromaticity. \\ Such cyclic compounds that resemble aliphatic compounds in their properties are classified as alicyclic compounds. \\ Therefore,cyclohexene is an alicyclic compound.

(A) Ammonium carbonate $(NH_4)_2CO_3$ is a salt of a weak base $(NH_4OH)$ and a weak acid $(H_2CO_3)$.
The $pH$ of a salt solution formed from a weak acid and a weak base is given by the formula: $pH = 7 + \frac{1}{2}(pK_a - pK_b)$.
For $(NH_4)_2CO_3$,the $pK_a$ of $H_2CO_3$ is approximately $6.35$ and the $pK_b$ of $NH_4OH$ is approximately $4.75$. Since $pK_b < pK_a$,the solution is slightly basic.
Thus,both Statement $I$ and Statement $II$ are correct.

(B) $1$. Identify the parent chain: The parent chain is a cyclohexane ring.
$2$. Number the ring: Start numbering to give the lowest possible locants to the substituents. If we start at the carbon with the two methyl groups as $1$,the ethyl group is at position $3$.
$3$. Alphabetical order: Ethyl comes before methyl. Thus,the name is $3-$ethyl$-1,1-$dimethylcyclohexane.

(C) The balanced chemical equation for the combustion of methane is:
$CH_4 + 2O_2 \rightarrow CO_2 + 2H_2O$
Calculate the molar mass of $CO_2$:
$M(CO_2) = 12.0 + 2 \times 16.0 = 44.0 \ g \ mol^{-1}$
Moles of $CO_2$ produced = $\frac{\text{Given mass}}{\text{Molar mass}} = \frac{22 \ g}{44 \ g \ mol^{-1}} = 0.5 \ mol$
From the stoichiometry of the reaction,$1 \ mol$ of $CH_4$ produces $1 \ mol$ of $CO_2$.
Therefore,$0.5 \ mol$ of $CO_2$ requires $0.5 \ mol$ of $CH_4$.
Calculate the molar mass of $CH_4$:
$M(CH_4) = 12.0 + 4 \times 1.0 = 16.0 \ g \ mol^{-1}$
Mass of $CH_4$ required = $\text{Moles} \times \text{Molar mass} = 0.5 \ mol \times 16.0 \ g \ mol^{-1} = 8 \ g$.

(D) For an isothermal process of an ideal gas,the change in internal energy $\Delta U = 0$.
According to the first law of thermodynamics,$\Delta U = Q + W$,which implies $Q = -W$.
The work done against a constant external pressure is given by $W = -P_{ext} \Delta V$.
Given $P_{ext} = 80 \ kPa = 80 \times 10^3 \ Pa$ and $\Delta V = (45 - 30) \ dm^3 = 15 \ dm^3 = 15 \times 10^{-3} \ m^3$.
$W = -80 \times 10^3 \ Pa \times 15 \times 10^{-3} \ m^3 = -1200 \ J$.
Since $Q = -W$,$Q = -(-1200 \ J) = 1200 \ J$.

(A) The reaction of $3-$methylhex$-2-$ene with $HBr$ in the presence of peroxide follows the anti-Markovnikov addition mechanism.
The product $(A)$ formed is $2-$bromo$-3-$methylhexane.
The structure of $2-$bromo$-3-$methylhexane is $CH_3-CH(Br)-CH(CH_3)-CH_2-CH_2-CH_3$.
This molecule contains $2$ chiral centers at $C-2$ and $C-3$.
Since the two chiral centers are not equivalent (the groups attached to them are different),the number of stereoisomers is given by $2^n$,where $n$ is the number of chiral centers.
Number of stereoisomers $= 2^2 = 4$.

(B) To determine if a compound is aromatic,it must be cyclic,planar,fully conjugated,and follow $H$ückel's rule ($4n+2$ $\pi$ electrons).
$(A)$ Decalin derivative: Not fully conjugated,non-aromatic.
$(B)$ Benzylcyclohexene: The benzene ring is aromatic,but the whole molecule is not considered a single aromatic system due to the $sp^3$ carbon bridge. However,in the context of identifying aromatic components,the benzene ring is aromatic.
$(C)$ Indene anion (indenyl anion): The cyclopentadienyl ring is conjugated with the benzene ring,and the lone pair on the $sp^3$ carbon becomes part of the $10$ $\pi$ electron system $(n=2)$,making it aromatic.
$(D)$ Fluorenyl anion: It has $14$ $\pi$ electrons $(n=3)$,is planar,and fully conjugated,making it aromatic.
Thus,compounds $(C)$ and $(D)$ are aromatic. The total number is $2$.

(B) Meta-directing groups are those that withdraw electron density from the benzene ring through the inductive effect or resonance,making the ring less reactive towards electrophilic substitution and directing the incoming electrophile to the meta position.
From the given list:
$1$. $-NO_2$ (Nitro group): Meta-directing
$2$. $-CN$ (Cyano group): Meta-directing
$3$. $-COR$ (Acyl group): Meta-directing
$4$. $-COOH$ (Carboxyl group): Meta-directing
The groups $-OCH_3$,$-CH_3$,$-NHCOCH_3$,$-OH$,and $-Cl$ are ortho/para-directing.
Thus,the total number of meta-directing groups is $4$.

(D) For $n=4$,the subshells are $4s, 4p, 4d,$ and $4f$.
The maximum number of electrons in these subshells are $2, 6, 10,$ and $14$ respectively.
Since all subshells are completely filled,the total number of electrons is $2 + 6 + 10 + 14 = 32$.
In any completely filled subshell,exactly half of the electrons have $s=+\frac{1}{2}$ and the other half have $s=-\frac{1}{2}$.
Therefore,the number of electrons with $s=+\frac{1}{2}$ is $\frac{32}{2} = 16$.

Subshell	$4s, 4p, 4d, 4f$
Total $e^-$	$2+6+10+14=32$
Total $e^-$ with $s=+\frac{1}{2}$	$1+3+5+7=16$

Thus,the correct answer is $16$.

(A) The bond order of $CO$ is $3$ because it has a triple bond $(C \equiv O)$.
The bond order of $NO^{+}$ is $3$ because it is isoelectronic with $CO$ and has a triple bond $(N \equiv O^{+})$.
The sum of the bond orders is $3 + 3 = 6$.

(B) To find the oxidation state of Sulphur $(S)$ in each compound:
$1$. $SO_3$: $x + 3(-2) = 0 \implies x = +6$
$2$. $H_2SO_3$: $2(+1) + x + 3(-2) = 0 \implies x = +4$
$3$. $SOCl_2$: $x + (-2) + 2(-1) = 0 \implies x = +4$
$4$. $SF_4$: $x + 4(-1) = 0 \implies x = +4$
$5$. $BaSO_4$: $(+2) + x + 4(-2) = 0 \implies x = +6$
$6$. $H_2S_2O_7$: $2(+1) + 2x + 7(-2) = 0 \implies 2x = +12 \implies x = +6$
The compounds with $+4$ oxidation state are $H_2SO_3, SOCl_2$,and $SF_4$.
Therefore,the total number of such compounds is $3$.

(A) The stability of resonating structures is determined by the following rules:
$1$. Neutral structures are more stable than charged structures. Structure $I$ is neutral,while $II$ and $III$ are charged. Thus,$I$ is the most stable.
$2$. Between charged structures,the structure with the negative charge on the more electronegative atom (Oxygen) and the positive charge on the less electronegative atom (Carbon) is more stable. In structure $II$,the negative charge is on the oxygen atom,whereas in structure $III$,the positive charge is on the oxygen atom.
$3$. Therefore,the stability order is $I > II > III$.

(D) Steam distillation is the technique used for the purification of substances that are steam volatile and immiscible with water.
In this process,the substance is heated by passing steam through it,which carries the volatile component along with it,leaving behind non-volatile impurities.

(D) The chemical formula is $HO-CH(CN)_2$.
In this molecule,the central carbon atom is bonded to one hydrogen atom $(H)$,one hydroxyl group $(-OH)$,and two cyano groups $(-CN)$.
The bond line formula represents the central carbon as a vertex where the bonds to $-OH$ and two $-CN$ groups originate.
Option $D$ correctly represents this connectivity in a simplified bond line notation.

(A) An oxidizing agent is a substance that gains electrons and undergoes reduction,meaning its oxidation state must decrease.
In the $N^{3-}$ ion,nitrogen is in its lowest possible oxidation state $(-3)$.
Since it cannot lose more electrons to be reduced further,it cannot function as an oxidizing agent.
In contrast,$S$ in $SO_4^{2-}$,$Br$ in $BrO_3^{-}$,and $Mn$ in $MnO_4^{-}$ are in high oxidation states and can be reduced.

(C) The eclipsed conformation of ethane has maximum torsional strain due to the proximity of $C-H$ bonds on adjacent carbon atoms,making it the least stable conformation. The staggered conformation is the most stable.

(A) noble gas configuration corresponds to a filled valence shell,typically represented as $ns^2 np^6$ (or the configuration of $He, Ne, Ar, Kr, Xe, Rn$).
$1. Sr^{2+} (Z=38)$: Electronic configuration is $[Kr] 5s^0$,which is the noble gas configuration of $Kr$.
$2. Cs^{+} (Z=55)$: Electronic configuration is $[Xe] 6s^0$,which is the noble gas configuration of $Xe$.
$3. La^{2+} (Z=57)$: Electronic configuration is $[Xe] 5d^1$,which is not a noble gas configuration.
$4. Pb^{2+} (Z=82)$: Electronic configuration is $[Xe] 4f^{14} 5d^{10} 6s^2$,which is not a noble gas configuration.
$5. Yb^{2+} (Z=70)$: Electronic configuration is $[Xe] 4f^{14}$,which is not a noble gas configuration.
$6. Fe^{2+} (Z=26)$: Electronic configuration is $[Ar] 3d^6$,which is not a noble gas configuration.
Thus,only $Sr^{2+}$ and $Cs^{+}$ have noble gas configurations. The total number of such ions is $2$.

(B) molecule is non-polar if its net dipole moment is zero.
$1$. $H_2$: Homonuclear diatomic molecule,dipole moment = $0$.
$2$. $CO_2$: Linear geometry,dipole moments of $C=O$ bonds cancel each other,net dipole moment = $0$.
$3$. $CH_4$: Tetrahedral geometry,bond dipoles cancel each other,net dipole moment = $0$.
$4$. $BF_3$: Trigonal planar geometry,bond dipoles cancel each other,net dipole moment = $0$.
Other molecules like $HF, H_2O, SO_2, NH_3, HCl,$ and $CHCl_3$ have a net dipole moment greater than zero.
Therefore,the non-polar molecules are $H_2, CO_2, CH_4,$ and $BF_3$.
The total count is $4$.

(A) Given: $\Delta H^{\ominus} = 77.2 \ kJ \ mol^{-1} = 77200 \ J \ mol^{-1}$,$T = 400 \ K$,$\Delta S = 122 \ J \ K^{-1} \ mol^{-1}$,$R = 8.314 \ J \ K^{-1} \ mol^{-1}$.
Using the relation $\Delta G^{\ominus} = \Delta H^{\ominus} - T \Delta S$:
$\Delta G^{\ominus} = 77200 - (400 \times 122) = 77200 - 48800 = 28400 \ J \ mol^{-1}$.
Using the relation $\Delta G^{\ominus} = -2.303 \ RT \log K$:
$28400 = -2.303 \times 8.314 \times 400 \times \log K$.
$28400 = -7657.6 \times \log K$.
$\log K = -28400 / 7657.6 \approx -3.708$.
Since $\log K = -3.708 = -37.08 \times 10^{-1}$,the value is approximately $-37$.

(B) The molarity $M$ is defined as the number of moles of solute per liter of solution.
$M = \frac{n_{NaOH}}{V_{sol} \text{ (in } L)}$
First,calculate the number of moles of $NaOH$:
$n_{NaOH} = \frac{\text{mass}}{\text{molar mass}} = \frac{84 \ g}{40 \ g \ mol^{-1}} = 2.1 \ mol$
Now,use the molarity formula to find the volume $V$:
$3 \ M = \frac{2.1 \ mol}{V \text{ (in } L)}$
$V = \frac{2.1}{3} \ L = 0.7 \ L$
Since $1 \ L = 1 \ dm^3$,$V = 0.7 \ dm^3 = 7 \times 10^{-1} \ dm^3$.
Thus,the value is $7$.

(C) The balanced chemical equation for the oxidation of $PbS$ by $O_3$ is:
$PbS + 4 O_3 \rightarrow PbSO_4 + 4 O_2$
From the balanced equation,we can see that $1$ mole of $PbS$ reacts with $4$ moles of $O_3$ to produce $4$ moles of $O_2$.
Therefore,$X = 4$ and $Y = 4$.
The value of $X + Y = 4 + 4 = 8$.

(A) The chemical reaction is: $C_6H_5NH_2 + (CH_3CO)_2O \rightarrow C_6H_5NHCOCH_3 + CH_3COOH$
Molar mass of aniline $(C_6H_5NH_2)$ = $(6 \times 12) + (7 \times 1) + 14 = 93 \ g \ mol^{-1}$.
Molar mass of acetanilide $(C_6H_5NHCOCH_3)$ = $(8 \times 12) + (9 \times 1) + 14 + 16 = 135 \ g \ mol^{-1}$.
Number of moles of aniline = $\frac{9.3 \ g}{93 \ g \ mol^{-1}} = 0.1 \ mol$.
Since the reaction is $100 \%$ complete,$0.1 \ mol$ of aniline will produce $0.1 \ mol$ of acetanilide.
Mass of acetanilide = $0.1 \ mol \times 135 \ g \ mol^{-1} = 13.5 \ g$.
Given that the mass is $x \times 10^{-1} \ g$,we have $13.5 = x \times 10^{-1}$,which implies $x = 135$.

(B) chiral carbon atom is a carbon atom bonded to four different groups.
$1$. $2-$methylcyclohexanone: The carbon at position $2$ is bonded to $-H$,$-CH_3$,$-C=O$ (part of ring),and $-CH_2$ (part of ring). It is chiral.
$2$. Cyclopentane$-1,3-$dione: All carbons are achiral due to symmetry.
$3$. $CH_3-CH_2-CH(NO_2)-COOH$: The central carbon is bonded to $-H$,$-NO_2$,$-COOH$,and $-CH_2CH_3$. It is chiral.
$4$. $CH_3-CH(I)-CH_2-NO_2$: The carbon bonded to $I$ is attached to $-H$,$-CH_3$,$-I$,and $-CH_2NO_2$. It is chiral.
$5$. $CH_3-CH_2-CH(OH)-CH_2OH$: The carbon bonded to $OH$ is attached to $-H$,$-OH$,$-CH_2CH_3$,and $-CH_2OH$. It is chiral.
$6$. $CH_3-CH_2-CH(I)-C_2H_5$: The carbon bonded to $I$ is attached to $-H$,$-I$,$-CH_2CH_3$,and $-CH_2CH_3$. Since two groups are identical $(-C_2H_5)$,it is achiral.
Thus,compounds $1, 3, 4,$ and $5$ contain at least one chiral carbon atom. The total number is $4$.

(C) The first ionisation enthalpy generally increases across a period from left to right because the atomic radius decreases and the effective nuclear charge increases.
Therefore,Assertion $A$ is false.
Reason $R$ states that the increasing nuclear charge outweighs the shielding effect across the period,which is a correct statement explaining why ionisation energy increases.
Thus,$A$ is false but $R$ is true.

(A) In the chromyl chloride test,$Cl^{-}$ reacts to form chromyl chloride $(CrO_2Cl_2)$,which dissolves in a basic medium to form a yellow solution containing chromate ions $(CrO_4^{2-})$.
When this yellow solution is acidified and treated with $10\% H_2O_2$ in the presence of amyl alcohol,it forms a blue-colored peroxo compound,chromium pentoxide $(CrO_5)$.
The structure of $CrO_5$ (butterfly structure) contains one double-bonded oxygen atom (oxidation state $-2$) and four peroxo oxygen atoms (each with oxidation state $-1$).
Let the oxidation state of $Cr$ be $x$.
$x + 1(-2) + 4(-1) = 0$
$x - 2 - 4 = 0$
$x = +6$
Thus,the oxidation state of chromium in $CrO_5$ is $+6$.

(B) The difference in energy between the actual structure and the lowest energy resonance structure for the given compound is known as resonance energy.

(A) The electronegativity values for group $14$ elements are as follows:
$C = 2.5$,$Si = 1.8$,$Ge = 1.8$,$Sn = 1.8$,$Pb = 1.9$.
From $Si$ to $Pb$,the electronegativity values do not decrease gradually; they remain almost constant ($1.8$ to $1.9$). Thus,Statement $I$ is false.
Group $14$ includes Carbon (non-metal),Silicon and Germanium (metalloids),and Tin and Lead (metals). Thus,Statement $II$ is true.

(A) The atomic number of rubidium $(Rb)$ is $37$.
The electronic configuration of $Rb$ is $[Kr] 5s^1$.
The valence electron is in the $5s$ orbital.
For the $5s$ orbital:
Principal quantum number $(n) = 5$.
Azimuthal quantum number $(l) = 0$ (for $s$-orbital).
Magnetic quantum number $(m) = 0$.
Spin quantum number $(s) = +\frac{1}{2}$ or $-\frac{1}{2}$.
Thus,the correct set is $(n=5, l=0, m=0, s=+\frac{1}{2})$.

(C) The $-NH_2$ group is an ortho/para directing group due to its strong $+M$ effect. During the electrophilic aromatic substitution (bromination) of aniline,the electrophile $(Br^+)$ attacks the ortho or para positions of the benzene ring to form stable arenium ions (sigma complexes).
The meta position is not activated by the $-NH_2$ group,and the arenium ion formed by the attack of an electrophile at the meta position is significantly less stable because the positive charge cannot be delocalized onto the nitrogen atom.
Among the given options,the structure representing the meta-substituted arenium ion is not involved in the major pathway of the reaction. Based on the provided options,the structure that does not represent a valid resonance-stabilized arenium ion intermediate for ortho/para substitution is the one corresponding to the meta attack.

(C) The sodium fusion extract contains $NaCN$ and $Na_2S$ if both nitrogen and sulphur are present in the organic compound.
These react to form sodium thiocyanate: $Na^{+} + CN^{-} + S^{2-} \rightarrow NaSCN$.
In the presence of $Fe^{3+}$ ions (formed by the oxidation of $Fe^{2+}$ by concentrated $H_2SO_4$),the thiocyanate ion forms a blood-red complex: $Fe^{3+} + SCN^{-} \rightarrow [Fe(SCN)]^{2+}$.
Thus,the appearance of a blood-red colour confirms the presence of both $N$ and $S$.

(D) $1$. In the presence of $h\nu$ (ultraviolet light),$Cl_2$ undergoes homolytic cleavage to form chlorine free radicals. These radicals perform allylic substitution on cyclohexene to form $1,4-\text{dichlorocyclohex-2-ene}$ as product $A$.
$2$. In the presence of $CCl_4$ (an inert solvent),$Cl_2$ undergoes electrophilic addition across the double bond of cyclohexene to form $1,2-\text{dichlorocyclohexane}$ as product $B$.

(A) The chemical formula for Fluorspar is $CaF_2$,not $BF_3$.
Therefore,the pair Fluorspar $- BF_3$ is incorrect.

(D) The interaction between a $\pi$ bond and a lone pair of electrons on an adjacent atom is a classic example of conjugation,which leads to the delocalization of electrons.
This phenomenon is known as the resonance effect (or mesomeric effect).

(B) For a spontaneous reaction,$\Delta G < 0$ (negative).
For a non-spontaneous reaction,$\Delta G > 0$ (positive).
For a reversible reaction at equilibrium,$\Delta G = 0$.
Therefore,the statement '$\Delta G$ is positive for a spontaneous reaction' is incorrect.

(A) The given reaction is a disproportionation reaction of chlorine in an alkaline medium:
$Cl_2 + 2OH^- \rightarrow Cl^- + ClO^- + H_2O$
Step $1$: Assign oxidation states.
$Cl_2$ $(0)$ $\rightarrow Cl^-$ $(-1)$ and $ClO^-$ $(+1)$.
Step $2$: Write half-reactions.
Reduction: $Cl_2 + 2e^- \rightarrow 2Cl^-$
Oxidation: $Cl_2 + 4OH^- \rightarrow 2ClO^- + 2H_2O + 2e^-$
Step $3$: Add the half-reactions.
$2Cl_2 + 4OH^- \rightarrow 2Cl^- + 2ClO^- + 2H_2O$
Step $4$: Simplify by dividing by $2$.
$Cl_2 + 2OH^- \rightarrow Cl^- + ClO^- + H_2O$
Comparing with $a Cl_2 + b OH^- \rightarrow c ClO^- + d Cl^- + e H_2O$,we get $a=1, b=2, c=1, d=1$.

(D) In an alkaline medium,the permanganate ion $(MnO_4^{-})$ acts as an oxidizing agent and oxidizes the iodide ion $(I^{-})$ to the iodate ion $(IO_3^{-})$.
The balanced chemical equation for this reaction is:
$2 MnO_4^{-} + H_2O + I^{-} \xrightarrow{\text{alkaline medium}} 2 MnO_2 + 2 OH^{-} + IO_3^{-}$

(A) To determine the number of lone pairs on the central atom,we calculate the number of lone pairs using the formula: $\text{Lone pair} = \frac{V - B}{2}$,where $V$ is the number of valence electrons of the central atom and $B$ is the number of bonding electrons (shared with surrounding atoms).
$1$. $O_3$: Central $O$ has $1$ lone pair.
$2$. $H_2O$: Central $O$ has $2$ lone pairs.
$3$. $SF_4$: Central $S$ has $1$ lone pair.
$4$. $ClF_3$: Central $Cl$ has $2$ lone pairs.
$5$. $NH_3$: Central $N$ has $1$ lone pair.
$6$. $BrF_5$: Central $Br$ has $1$ lone pair.
$7$. $XeF_4$: Central $Xe$ has $2$ lone pairs.
The compounds with exactly one lone pair on the central atom are $O_3$,$SF_4$,$NH_3$,and $BrF_5$.
Thus,the total count is $4$.

(D) The relationship between $K_p$ and $K_c$ is given by the formula: $K_p = K_c(RT)^{\Delta n_g}$.
For the reaction $N_2O_{4(g)} \rightleftharpoons 2NO_{2(g)}$,the change in the number of moles of gaseous species is $\Delta n_g = 2 - 1 = 1$.
Given $K_p = 0.492 \ atm$,$R = 0.082 \ L \ atm \ mol^{-1} \ K^{-1}$,and $T = 300 \ K$.
Substituting the values: $0.492 = K_c \times (0.082 \times 300)^1$.
$K_c = \frac{0.492}{24.6} = 0.02 = 2 \times 10^{-2}$.

(A) The mass percentage of $H_2SO_4$ is $31.4 \%$,which means $31.4 \ g$ of $H_2SO_4$ is present in $100 \ g$ of the solution.
The volume of $100 \ g$ of the solution is calculated using density $(d = 1.25 \ g \ mL^{-1})$:
$V = \frac{\text{mass}}{\text{density}} = \frac{100 \ g}{1.25 \ g \ mL^{-1}} = 80 \ mL$.
The number of moles of $H_2SO_4$ is:
$n = \frac{\text{mass}}{\text{molar mass}} = \frac{31.4 \ g}{98 \ g \ mol^{-1}} \approx 0.3204 \ mol$.
Molarity $(M)$ is defined as moles of solute per liter of solution:
$M = \frac{n}{V(L)} = \frac{0.3204 \ mol}{0.080 \ L} = 4.005 \ M$.
Rounding to the nearest integer,the molarity is $4 \ M$.

(B) heteroatom is any atom other than carbon or hydrogen in an organic molecule.
$1$. Furan: Contains oxygen as a heteroatom.
$2$. Thiophene: Contains sulphur as a heteroatom.
$3$. Pyridine: Contains nitrogen as a heteroatom.
$4$. Pyrrole: Contains nitrogen as a heteroatom.
$5$. Cysteine: Contains sulphur as a heteroatom.
$6$. Tyrosine: Does not contain sulphur.
The compounds containing sulphur as a heteroatom are Thiophene and Cysteine.
Therefore,the total number of such compounds is $2$.

(A) Nucleotides are the building blocks of nucleic acids like $DNA$ and $RNA$.
Two nucleotides are joined together by a $3'-5'$ phosphodiester linkage.
This linkage is formed between the $5'$ carbon atom of one sugar molecule and the $3'$ carbon atom of the adjacent sugar molecule through a phosphate group.

(D) Fluorine is the most electronegative element in the periodic table and has no $d$-orbitals in its valence shell. Due to its high electronegativity and lack of $d$-orbitals,it only exhibits an oxidation state of $-1$ in its compounds. Other halogens like $Cl$,$Br$,and $I$ exhibit variable oxidation states (e.g.,$-1, +1, +3, +5, +7$) due to the presence of vacant $d$-orbitals.

(D) Bronsted base is a substance that can accept a proton $(H^+)$. The strength of a base depends on the availability of the lone pair of electrons on the nitrogen atom.
$1$. In $A$ (Aniline),the lone pair on nitrogen is delocalized into the benzene ring due to resonance,making it less available for protonation.
$2$. In $B$ (Diphenylamine),the lone pair is delocalized into two benzene rings,making it even less basic than aniline.
$3$. In $C$ (Pyrrole),the lone pair on nitrogen is involved in the aromatic sextet (it is part of the $6\pi$ electron system),so it is not available for protonation.
$4$. In $D$ (Pyrrolidine),the nitrogen atom is $sp^3$ hybridized and the lone pair is localized (not involved in resonance). This makes the lone pair readily available for donation to a proton,making it the strongest base among the given options.

(D) The magnetic moment $(\mu)$ is calculated using the formula $\mu = \sqrt{n(n+2)} \ BM$,where $n$ is the number of unpaired electrons.

$3d$ configuration	Unpaired electrons $(n)$
$3d^7$	$3$
$3d^8$	$2$
$3d^3$	$3$
$3d^6$	$4$

Comparing the values:
For $3d^7$: $\mu = \sqrt{3(3+2)} = \sqrt{15} \ BM \approx 3.87 \ BM$
For $3d^8$: $\mu = \sqrt{2(2+2)} = \sqrt{8} \ BM \approx 2.83 \ BM$
For $3d^3$: $\mu = \sqrt{3(3+2)} = \sqrt{15} \ BM \approx 3.87 \ BM$
For $3d^6$: $\mu = \sqrt{4(4+2)} = \sqrt{24} \ BM \approx 4.90 \ BM$
Thus,the configuration $3d^6$ has the highest number of unpaired electrons $(n=4)$ and consequently the highest magnetic moment.

(D) The acidity of hydrogen atoms in carbonyl compounds depends on the stability of the resulting conjugate base (enolate ion).
In $Heptane-2,4-dione$ $(CH_3-CO-CH_2-CO-CH_2-CH_3)$,the central methylene $(-CH_2-)$ group is flanked by two carbonyl groups.
The conjugate base formed by removing a proton from this central methylene group is highly stabilized by resonance with both adjacent carbonyl groups.
This makes the hydrogen atoms of this active methylene group significantly more acidic than those in the other given options,where the resonance stabilization is less effective or absent.

(D) For a solution showing negative deviation from Raoult's law,the intermolecular forces between the components are stronger than those in the pure components.
This leads to a decrease in the tendency of molecules to escape into the vapour phase,resulting in a lower total vapour pressure: $P_{T} < P_{A}^{0} X_{A} + P_{B}^{0} X_{B}$.
Since the vapour pressure is decreased,the solution requires more heat to reach the atmospheric pressure,which leads to an increase in the boiling point.
Therefore,the correct characteristics are decreased vapour pressure and increased boiling point.

(C) $[FeF_6]^{3-}$: $Fe^{3+}$ is $[Ar] 3d^5$. $F^-$ is a weak field ligand,so the number of unpaired electrons is $5$. $\mu = \sqrt{5(5+2)} = \sqrt{35} \ B.M.$
$[V(H_2O)_6]^{2+}$: $V^{2+}$ is $3d^3$. The number of unpaired electrons is $3$. $\mu = \sqrt{3(3+2)} = \sqrt{15} \ B.M.$
$[Fe(H_2O)_6]^{2+}$: $Fe^{2+}$ is $3d^6$. $H_2O$ is a weak field ligand,so the number of unpaired electrons is $4$. $\mu = \sqrt{4(4+2)} = \sqrt{24} \ B.M.$
Comparing the values: $\sqrt{15} < \sqrt{24} < \sqrt{35}$.
Therefore,the correct order is $Q < R < P$.

(A) Statement $I$: The acidic strength order of nitrophenols is $p$-nitrophenol $>$ $o$-nitrophenol $>$ $m$-nitrophenol. $p$-nitrophenol is the most acidic due to the strong $-I$ and $-M$ effects of the nitro group at the para position,which stabilizes the phenoxide ion significantly. $o$-nitrophenol is less acidic than $p$-nitrophenol due to intramolecular hydrogen bonding,and $m$-nitrophenol is the least acidic as it only exhibits the $-I$ effect.
Statement $II$: Lucas reagent $(conc. HCl + ZnCl_2)$ is used to distinguish between primary,secondary,and tertiary alcohols. Tertiary alcohols give immediate turbidity,secondary alcohols give turbidity in $5-10$ minutes,and primary alcohols (like ethanol) do not give turbidity at room temperature; they only react upon heating.
Therefore,Statement $I$ is true and Statement $II$ is false.

(A) The acidity of phenols depends on the stability of the phenoxide ion formed after the loss of a proton. Electron-withdrawing groups $(-EWG)$ increase acidity by stabilizing the negative charge,while electron-donating groups $(-EDG)$ decrease acidity by destabilizing it.
$1$. $E$ ($2,4$-dinitrophenol): Has two $-NO_2$ groups (strong $-M$ and $-I$ effect),making it the most acidic.
$2$. $B$ ($p$-nitrophenol): Has one $-NO_2$ group ($-M$ and $-I$ effect),making it more acidic than phenol.
$3$. $D$ (Phenol): The reference compound.
$4$. $C$ ($p$-methoxyphenol): The $-OCH_3$ group has a $+M$ effect,which destabilizes the phenoxide ion,making it less acidic than phenol.
$5$. $A$ $(Bu-OH)$: Aliphatic alcohols are significantly less acidic than phenols because the alkoxide ion is not resonance-stabilized.
Thus,the order of acidity is: $A < C < D < B < E$.

(D) The reaction of lead$(II)$ chromate with hot excess sodium hydroxide is given by:
$PbCrO_4(s) + 4NaOH(aq) \rightarrow Na_2[Pb(OH)_4](aq) + Na_2CrO_4(aq)$
In the product $[Pb(OH)_4]^{2-}$,the lead ion is coordinated to $four$ hydroxide ligands.
This is a dianionic complex (charge $-2$) with a coordination number of $four$.

(A) $4NaCl + K_2Cr_2O_7 + 6H_2SO_4 \rightarrow 2CrO_2Cl_2 \ [(B) \text{Reddish brown fumes}] + 2KHSO_4 + 4NaHSO_4 + 3H_2O$
$CrO_2Cl_2 + 4NaOH \rightarrow Na_2CrO_4 \ [(C) \text{Yellow solution}] + 2NaCl + 2H_2O$
Therefore,$(B)$ is $CrO_2Cl_2$ and $(C)$ is $Na_2Cr_4$.

(D) In an $S_{N}1$ reaction,the carbocation intermediate formed is planar,allowing the nucleophile to attack from either side,which leads to the formation of a racemic mixture (racemisation).
In an $S_{N}2$ reaction,the nucleophile attacks from the side opposite to the leaving group,resulting in the inversion of configuration (Walden inversion).

(A) The atomic number of Neodymium $(Nd)$ is $60$.
The nearest noble gas is Xenon $(Xe)$ with atomic number $54$.
The remaining $6$ electrons are filled in the $4f$ and $6s$ orbitals.
According to the Aufbau principle and experimental observations for lanthanoids,the configuration is $[Xe] 4f^4 6s^2$.

(A) According to Faraday's law of electrolysis,the number of equivalents of substances deposited or liberated by the same quantity of electricity are equal.
$Eq. \text{ of } Ag = Eq. \text{ of } O_2$
For $O_2$ gas,the reaction is: $2H_2O \rightarrow O_2 + 4H^+ + 4e^-$.
The $n$-factor for $O_2$ is $4$.
Number of moles of $O_2 = \frac{5600 \ mL}{22400 \ mL \ mol^{-1}} = 0.25 \ mol$ (using $S.T.P.$ molar volume as $22.4 \ L \ mol^{-1}$).
$Eq. \text{ of } O_2 = \text{moles} \times n\text{-factor} = 0.25 \times 4 = 1$.
For $Ag^+ + e^- \rightarrow Ag$,the $n$-factor is $1$.
$Eq. \text{ of } Ag = \frac{\text{mass}}{\text{molar mass}} \times n\text{-factor} = \frac{x}{108} \times 1$.
Equating the two: $\frac{x}{108} = 1$.
$x = 108 \ g$.

(B) The rate law for the reaction is given by $R = k[HI]^n$.
Using data from Experiment $1$ and Experiment $2$:
$\frac{R_2}{R_1} = \frac{k[HI]_2^n}{k[HI]_1^n}$
$\frac{3.0 \times 10^{-3}}{7.5 \times 10^{-4}} = \left(\frac{0.01}{0.005}\right)^n$
$4 = (2)^n$
Since $2^2 = 4$,we get $n = 2$.
Thus,the order of the reaction is $2$.

(D) $S_{N}1$ reaction proceeds via the formation of a carbocation intermediate. The stability of the carbocation determines the reactivity.
$(A)$ $H_2C=CH-CH_2Cl$ forms an allyl carbocation $(H_2C=CH-CH_2^+)$,which is resonance stabilized,so it shows $S_{N}1$ reaction.
$(B)$ $CH_3-CH=CHCl$ is a vinylic halide. The carbocation formed $(CH_3-CH=CH^+)$ is highly unstable due to the positive charge on an $sp$-hybridized carbon atom,so it does not show $S_{N}1$ reaction.
$(C)$ $C_6H_5CH_2Cl$ (benzyl chloride) forms a benzyl carbocation $(C_6H_5CH_2^+)$,which is highly resonance stabilized,so it shows $S_{N}1$ reaction.
$(D)$ $(CH_3)_2CHCl$ (isopropyl chloride) forms a secondary carbocation $((CH_3)_2CH^+)$,which is relatively stable due to inductive effect and hyperconjugation,so it shows $S_{N}1$ reaction.
Therefore,only $(B)$ will not show $S_{N}1$ reaction.

(A) The correct answer is $A$. Rusting of iron is an electrochemical process. Iron is more reactive than tin. If the tin coating is peeled off,the iron is exposed to the atmosphere and acts as an anode,while the remaining tin acts as a cathode. This leads to the formation of a galvanic cell,which accelerates the corrosion of iron rather than preventing it. Therefore,the statement in option $A$ is incorrect.

(B) Statement $I$ is true: The electronic configuration of $Ce$ $(Z=58)$ is $[Xe] 4f^1 5d^1 6s^2$. Upon losing four electrons,$Ce^{+4}$ attains the stable noble gas configuration of $[Xe]$.
Statement $II$ is true: $Ce^{+4}$ has a high reduction potential $(E^{\circ} = +1.74 \ V)$ for the reaction $Ce^{+4} + e^- \rightarrow Ce^{+3}$. Due to this,it readily gains an electron to revert to the more stable $+3$ oxidation state,making it a strong oxidizing agent.

(B) The electronic configurations of the given elements are:
$Cu (Z=29): [Ar] 3d^{10} 4s^1$
$Zn (Z=30): [Ar] 3d^{10} 4s^2$
$Cd (Z=48): [Kr] 4d^{10} 5s^2$
$Ag (Z=47): [Kr] 4d^{10} 5s^1$
All these elements have a completely filled $d$-orbital ($d^{10}$ configuration) in their ground state.
Therefore,option $B$ is correct.

(A) The Phthalein dye test is used for the identification of phenols. When a phenol is heated with phthalic anhydride in the presence of concentrated $H_2SO_4$,a phthalein dye is formed.
$1$. Carbylamine test: Used for the identification of primary amines.
$2$. Lucas test: Used for the differentiation between $1^{\circ}$,$2^{\circ}$,and $3^{\circ}$ alcohols.
$3$. Tollen's test: Used for the identification of aldehydes.

(B) The general formula for mono carboxylic acids is $C_nH_{2n}O_2$ (where $n \geq 1$).
The first member $(n=1)$ is methanoic acid $(HCOOH)$,which is $CH_2O_2$.
The second member $(n=2)$ is ethanoic acid $(CH_3COOH)$,which has the molecular formula $C_2H_4O_2$.
Therefore,the second homologue is $C_2H_4O_2$.

(D) Step $1$: Hydroboration-oxidation of styrene $(Ph-CH=CH_2)$ gives anti-Markovnikov addition of water,resulting in $Ph-CH_2-CH_2-OH$.
Step $2$: Reaction with $HBr$ converts the alcohol to an alkyl bromide: $Ph-CH_2-CH_2-OH + HBr \rightarrow Ph-CH_2-CH_2-Br + H_2O$.
Step $3$: Formation of Grignard reagent and reaction with formaldehyde $(HCHO)$ followed by hydrolysis:
$Ph-CH_2-CH_2-Br + Mg \xrightarrow{\text{ether}} Ph-CH_2-CH_2-MgBr$
$Ph-CH_2-CH_2-MgBr + HCHO \rightarrow Ph-CH_2-CH_2-CH_2-OMgBr$
$Ph-CH_2-CH_2-CH_2-OMgBr \xrightarrow{H_3O^{+}} Ph-CH_2-CH_2-CH_2-OH$.
Thus,the final product $A$ is $Ph-CH_2-CH_2-CH_2-OH$.

(A) $A \rightarrow$ Kolbe-Schmidt reaction: Phenol reacts with $NaOH$ followed by $CO_2$ and $HCl$ to form salicylic acid. Matches with $IV$.
$B \rightarrow$ Reimer-Tiemann reaction: Phenol reacts with $CHCl_3$ and $NaOH$ followed by $HCl$ to form salicylaldehyde. Matches with $III$.
$C \rightarrow$ Oxidation of phenol: Phenol is oxidized by $Na_2Cr_2O_7$ and $H_2SO_4$ to $p$-benzoquinone. Matches with $I$.
$D \rightarrow$ Williamson ether synthesis: Phenol reacts with $NaOH$ to form sodium phenoxide,which then reacts with $CH_3Cl$ to form anisole $(PhOCH_3)$. Matches with $II$.
Therefore,the correct matching is $A-IV, B-III, C-I, D-II$.

(D) The reaction of an ether with $HI$ proceeds via the protonation of the ether oxygen atom,followed by the cleavage of the $C-O$ bond.
In the given ether,cyclohexyl tert-butyl ether,the cleavage occurs such that the more stable carbocation is formed.
The tert-butyl group forms a stable tertiary carbocation $(CH_3)_3C^+$,which is then attacked by the iodide ion $(I^-)$ to form tert-butyl iodide,$(CH_3)_3CI$.
The cyclohexyl group remains as the alcohol,cyclohexanol $(C_6H_{11}OH)$.
Thus,the major products are cyclohexanol and $2-$iodo$-2-$methylpropane (tert-butyl iodide).

(C) Statement-$I$: Oxygen exhibits oxidation states ranging from $-2$ to $+2$ (e.g.,in $OF_2$,it is $+2$). Thus,Statement-$I$ is incorrect.
Statement-$II$: Due to the inert pair effect,as we move down group $16$,the stability of the $+6$ oxidation state decreases,while the stability of the $+4$ oxidation state increases. Thus,Statement-$II$ is incorrect.
Therefore,both statements are incorrect.

(A) $[Co(NH_3)_6]^{3+}$: The central metal ion $Co^{3+}$ has a $d^6$ configuration. In the presence of strong field ligand $NH_3$,electrons pair up,providing two vacant $3d$ orbitals for $d^2 sp^3$ hybridization.
$BrF_5$: The central atom $Br$ undergoes $sp^3 d^2$ hybridization.
$[PtCl_4]^{2-}$: The central metal ion $Pt^{2+}$ undergoes $dsp^2$ hybridization.
$SF_6$: The central atom $S$ undergoes $sp^3 d^2$ hybridization.

(B) When $1,4$-dichlorobutane reacts with excess ammonia,it undergoes nucleophilic substitution to form the ammonium salt $A$ $([H_3N^{+}-(CH_2)_4-NH_3^{+}] 2Cl^{-})$.
Treatment of this salt with a strong base like $NaOH$ releases the free diamine $B$ $(H_2N-(CH_2)_4-NH_2)$,along with water and sodium chloride.
The reaction is:
$Cl-(CH_2)_4-Cl + 2NH_3 \rightarrow [H_3N^{+}-(CH_2)_4-NH_3^{+}] 2Cl^{-} (A)$
$[H_3N^{+}-(CH_2)_4-NH_3^{+}] 2Cl^{-} + 2NaOH \rightarrow H_2N-(CH_2)_4-NH_2 (B) + 2H_2O + 2NaCl$

(A) $Molarity = \frac{\text{Moles of solute}}{\text{Volume of solution (L)}}$
Since the volume of a solution is temperature-dependent,the molarity of the solution changes with a change in temperature.
Other concentration terms like molality,mass percentage,and mole fraction involve mass,which is independent of temperature.

(A) Boiling an egg causes denaturation of its protein.
During denaturation,the $Secondary$,$Tertiary$,and $Quaternary$ structures of the protein are disrupted due to the breaking of hydrogen bonds and other non-covalent interactions.
However,the $Primary$ structure,which consists of the specific sequence of amino acids held together by peptide bonds,remains intact.

(D) In the Wacker process,the catalyst used is $PdCl_2$ (palladium$(II)$ chloride) along with a copper$(II)$ chloride co-catalyst.
Therefore,the pair $Wacker \ process - PtCl_2$ is incorrect.

(C) For a first order reaction,the time $t$ is given by $t = \frac{2.303}{k} \log \left( \frac{a}{a-x} \right)$.
For $99.9 \%$ completion,$x = 0.999a$,so $a-x = 0.001a = \frac{a}{1000}$.
Thus,$t_{99.9 \%} = \frac{2.303}{k} \log \left( \frac{a}{a/1000} \right) = \frac{2.303}{k} \log(10^3) = \frac{2.303 \times 3}{k}$.
The half-life is $t_{1/2} = \frac{0.693}{k} = \frac{2.303 \times 0.301}{k}$.
Taking the ratio: $\frac{t_{99.9 \%}}{t_{1/2}} = \frac{2.303 \times 3 / k}{2.303 \times 0.301 / k} \approx \frac{3}{0.3} = 10$.

(D) The complex is $[Pt(NH_3)_2 Cl(NH_2 CH_3)] Cl$.
In this complex,the oxidation state of $Pt$ is $+2$.
$Pt$ $(Z=78)$ has an electronic configuration of $[Xe] 4f^{14} 5d^9 6s^1$. Thus,$Pt^{2+}$ has a $5d^8$ configuration.
Since the complex is square planar,it undergoes $dsp^2$ hybridization.
In a $5d^8$ square planar complex,the electrons pair up in the $5d$ orbitals,leaving no unpaired electrons.
Therefore,the number of unpaired electrons $(n)$ is $0$.
The spin-only magnetic moment is given by $\mu = \sqrt{n(n+2)} \ B.M.$
Substituting $n=0$,we get $\mu = \sqrt{0(0+2)} = 0 \ B.M.$

(D) The half-cell reaction for the hydrogen electrode is: $2 H^{+}_{(aq)} + 2 e^{-} \rightarrow H_{2(g)}$.
Using the Nernst equation: $E = E^{\circ} - \frac{0.059}{n} \log \frac{1}{[H^{+}]}$.
For the standard hydrogen electrode,$E^{\circ} = 0 \ V$ and $n = 1$ (for the reduction of $H^{+}$ to $\frac{1}{2} H_2$).
$E = 0 - 0.059 \times \log \frac{1}{[H^{+}]}$.
Since $pH = -\log[H^{+}]$,we have $\log \frac{1}{[H^{+}]} = pH$.
Therefore,$E = -0.059 \times pH$.
Given $pH = 3$,$E = -0.059 \times 3 = -0.177 \ V$.
Converting to the required format: $-0.177 \ V = -17.7 \times 10^{-2} \ V$.

(D) . Ziegler catalyst is $TiCl_4 + (C_2H_5)_3Al$,which contains Titanium $(IV)$.
$B$. Blood pigment (Hemoglobin) contains Iron $(III)$.
$C$. Wilkinson catalyst is $[RhCl(PPh_3)_3]$,which contains Rhodium $(I)$.
$D$. Vitamin $B_{12}$ (Cyanocobalamin) contains Cobalt $(II)$.
Therefore,the correct matching is $A-IV, B-III, C-I, D-II$.

(D) The reaction involves two steps with excess $HBr$:
$1$. Electrophilic addition of $HBr$ to the vinyl group $(-CH=CH_2)$ follows Markovnikov's rule,forming a stable benzylic carbocation intermediate,which then reacts with $Br^-$ to give a $1$-bromoethyl group.
$2$. Cleavage of the ether linkage $(-OCH_2CH_3)$ occurs via protonation of the oxygen followed by nucleophilic attack of $Br^-$ on the less hindered ethyl group,resulting in the formation of phenol and ethyl bromide.
Thus,the major product is $3-(1\text{-bromoethyl})phenol$.

(C) Assertion $(A)$ is true. The $C-O$ bond in phenol has partial double bond character due to resonance,making it very strong and difficult to break by nucleophilic substitution with halogen acids.
Reason $(R)$ is false. Phenols do not react with halogen acids under normal conditions because the $C-O$ bond is strong and the formation of a phenyl carbocation is highly unfavorable.
Therefore,$A$ is true but $R$ is false.

(D) The thermal decomposition of potassium permanganate $(KMnO_4)$ at $513 \ K$ is given by the following reaction:
$2KMnO_4 \xrightarrow{\Delta} K_2MnO_4 + MnO_2 + O_2$
Thus,$KMnO_4$ decomposes to form potassium manganate $(K_2MnO_4)$,manganese dioxide $(MnO_2)$,and oxygen gas $(O_2)$.

(A) In the structure of $[Co_2(CO)_8]$,there are two bridging $CO$ ligands between the two cobalt atoms,along with a $Co-Co$ bond.
In $[Mn_2(CO)_{10}]$,$[Os_3(CO)_{12}]$,and $[Ru_3(CO)_{12}]$,all $CO$ ligands are terminal,and there are no bridging $CO$ groups.
Therefore,the correct option is $A$.

(B) Proteins are natural polymers composed of $\alpha$-amino acids which are connected by peptide linkages.
Hence,proteins upon acidic hydrolysis produce $\alpha$-amino acids.

(A) Step $1$: Hydration of styrene $(C_6H_5CH=CH_2)$ with $H_2O, H^+$ follows Markovnikov's rule to yield $1$-phenylethanol $(C_6H_5CH(OH)CH_3)$.
Step $2$: Oxidation of $1$-phenylethanol with $CrO_3$ yields acetophenone $(C_6H_5COCH_3)$.
Step $3$: Wolff-Kishner reduction of acetophenone using $H_2N-NH_2, KOH$ and heating reduces the carbonyl group to a methylene group,yielding ethylbenzene $(C_6H_5CH_2CH_3)$ as the final product $A$.

(B) The reduction reaction at the cathode is: $Zn^{2+} + 2e^{-} \longrightarrow Zn(s)$.
According to Faraday's law of electrolysis,the mass $W$ deposited is given by $W = \frac{M \times I \times t}{n \times F}$,where $M = 65.4 \ g/mol$,$I = 0.015 \ A$,$t = 15 \times 60 \ s = 900 \ s$,$n = 2$,and $F = 96500 \ C/mol$.
$W = \frac{65.4 \times 0.015 \times 900}{2 \times 96500} \ g$.
$W = \frac{882.9}{193000} \ g \approx 0.0045746 \ g$.
$W \approx 45.75 \times 10^{-4} \ g$.
Rounding to the nearest integer,the value is $46$.

(C) The overall rate constant is given by $K = \frac{K_1 K_2}{K_3}$.
Substituting the Arrhenius equation $K = A \cdot e^{-E_a / RT}$ for each step:
$A \cdot e^{-E_a / RT} = \frac{(A_1 e^{-E_{a1} / RT}) \cdot (A_2 e^{-E_{a2} / RT})}{(A_3 e^{-E_{a3} / RT})}$
$A \cdot e^{-E_a / RT} = \frac{A_1 A_2}{A_3} \cdot e^{-(E_{a1} + E_{a2} - E_{a3}) / RT}$
Comparing the exponents,we get $E_a = E_{a1} + E_{a2} - E_{a3}$.
Substituting the given values: $E_a = 40 + 50 - 60 = 30 \ kJ/mol$.

(B) The osmotic pressure $\pi$ is given by the formula $\pi = CRT$.
Since the concentration $C$ and gas constant $R$ are constant for the same solution,we have $\frac{\pi_1}{T_1} = \frac{\pi_2}{T_2}$.
Given $\pi_1 = 7 \times 10^5 \ Pa$,$T_1 = 273 \ K$,and $T_2 = 283 \ K$.
Substituting the values: $\pi_2 = \frac{\pi_1 \times T_2}{T_1} = \frac{7 \times 10^5 \times 283}{273} \ Pa$.
$\pi_2 = 7.256 \times 10^5 \ Pa = 72.56 \times 10^4 \ Nm^{-2}$.
Rounding to the nearest integer,we get $73 \times 10^4 \ Nm^{-2}$.

(A) Fehling's test is given by aliphatic aldehydes only.
$1$. Benzaldehyde: Aromatic aldehyde (Negative)
$2$. Acetaldehyde $(CH_3CHO)$: Aliphatic aldehyde (Positive)
$3$. Acetone: Ketone (Negative)
$4$. Acetophenone: Aromatic ketone (Negative)
$5$. Methanal $(HCHO)$: Aliphatic aldehyde (Positive)
$6$. $4$-nitrobenzaldehyde: Aromatic aldehyde (Negative)
$7$. Cyclohexanecarbaldehyde: Aliphatic aldehyde (Positive)
Thus,the compounds that give a positive Fehling's test are Acetaldehyde,Methanal,and Cyclohexanecarbaldehyde.
The total number of such compounds is $3$.

(D) The correct matching is as follows:
$A$. Starch is a polymer of $\alpha$-glucose $(II)$.
$B$. Cellulose is a polymer of $\beta$-glucose $(III)$.
$C$. Nucleic acid is a polymer of nucleotides $(I)$.
$D$. Protein is a polymer of $\alpha$-amino acids $(IV)$.
Therefore,the correct sequence is $A-II, B-III, C-I, D-IV$.

(A) The $pK_a$ values for the given compounds are as follows:
$A$. Ethanol: $15.9$
$B$. Phenol: $10.0$
$C$. $m$-Nitrophenol: $8.3$
$D$. $p$-Nitrophenol: $7.1$
Therefore,the correct matching is $A-II, B-I, C-IV, D-III$.

(B) In $K_2MnO_4$,let the oxidation state of $Mn$ be $x$.
$2(+1) + x + 4(-2) = 0$
$2 + x - 8 = 0$
$x = +6$
The oxidation state of $Mn$ is $+6$.
According to $IUPAC$ nomenclature for coordination compounds and oxoanions,the name is Potassium tetraoxidomanganate $(VI)$.

(D) The reagent used for the detection of $Ni^{2+}$ ions is dimethyl glyoxime $(dmg)$.
In the presence of an ammonium hydroxide $(NH_4OH)$ basic medium,$Ni^{2+}$ reacts with $dmg$ to form a complex,$Ni(dmg)_2$,which appears as a brilliant red precipitate.
The chemical reaction is: $Ni^{2+} + 2C_4H_8N_2O_2 \rightarrow [Ni(C_4H_7N_2O_2)_2] + 2H^+$.
Thus,the correct option is $D$.

(D) The reaction described is the Reimer-Tiemann reaction.
In this reaction,phenol reacts with chloroform $(CHCl_3)$ in the presence of an aqueous base like sodium hydroxide $(NaOH)$ to form an intermediate,which upon acidic hydrolysis yields $2$-hydroxybenzaldehyde (also known as salicylaldehyde) as the major product.
The chemical equation is:
$C_6H_5OH + CHCl_3 + 3NaOH \rightarrow C_6H_4(OH)(CHO) + 3NaCl + 2H_2O$

(C) Nessler's reagent is an alkaline solution of potassium tetraiodomercurate$(II)$,$K_2[HgI_4]$.
When ammonia gas $(NH_3)$ is passed through Nessler's reagent,it forms a brown precipitate of iodide of Millon's base,$HgO \cdot Hg(NH_2)I$.
The chemical equation is:
$2 K_2[HgI_4] + NH_3 + 3 KOH \rightarrow HgO \cdot Hg(NH_2)I + 7 KI + 2 H_2O$
Thus,the gas $X$ is $NH_3$.