JEE Main 2024 Chemistry Question Paper with Answer and Solution

(A) In a carbocation,the positively charged carbon atom is $sp^2$ hybridized.
It is bonded to three other atoms and has no lone pair of electrons.
According to the Valence Shell Electron Pair Repulsion $(VSEPR)$ theory,the arrangement of three bonding pairs around the central atom results in a trigonal planar geometry with bond angles of approximately $120^{\circ}$.

(C) $1$. Identify the longest carbon chain: The longest chain contains $8$ carbon atoms,so the parent alkane is octane.
$2$. Number the chain: Number the chain from the end that gives the lowest locants to the substituents. Numbering from left to right gives substituents at positions $2, 5,$ and $6$.
$3$. Identify substituents: There are three methyl groups at positions $2, 5,$ and $6$.
$4$. Combine: The name is $2,5,6-$trimethyloctane.

(B) The given equilibrium is $Cr_2O_7^{2-} + H_2O \rightleftharpoons 2CrO_4^{2-} + 2H^{+}$.
According to Le Chatelier's principle,if we add a base ($OH^-$ ions) to the system,the $OH^-$ ions will react with the $H^+$ ions produced in the reaction to form water $(H^+ + OH^- \rightarrow H_2O)$.
This removal of $H^+$ ions from the product side decreases the concentration of $H^+$,which causes the equilibrium to shift to the right to produce more $H^+$ ions.
Therefore,the equilibrium shifts to the right in a basic medium.

(A) buffer solution is a mixture of a weak acid and its conjugate base or a weak base and its conjugate acid in specific proportions,not just any quantity.
Statement $I$ is false because it implies any quantity of salt and acid/base works,whereas buffer action depends on specific ratios.
Blood is a naturally occurring buffer system maintained by the $H_2CO_3 / HCO_3^{-}$ pair,which keeps the $pH$ constant.
Statement $II$ is true.
Therefore,Statement $I$ is false but Statement $II$ is true.

(B) Statement $(A)$ is correct: Non-metals have higher electronegativity compared to metals due to their smaller size and higher effective nuclear charge.
Statement $(B)$ is incorrect: Non-metals have higher ionisation enthalpy than metals because they hold their valence electrons more tightly.
Statement $(C)$ is correct: Highly reactive metals (low electronegativity) and highly reactive non-metals (high electronegativity) form ionic compounds due to a large electronegativity difference.
Statement $(D)$ is incorrect: Non-metal oxides are generally acidic in nature.
Statement $(E)$ is incorrect: Metal oxides are generally basic in nature.
Therefore,statements $(A)$ and $(C)$ are correct.

(A) The Kjeldahl method is not applicable to compounds containing nitrogen in a ring,such as pyridine,because the nitrogen atom in the ring is not easily converted to ammonium sulphate $(NH_4)_2SO_4$ under the conditions of the Kjeldahl method.
Therefore,both Statement $I$ and Statement $II$ are false.

(A) The vaporization reaction is: $H_2O(\ell) \rightarrow H_2O(g)$.
For this reaction,the change in the number of moles of gas is $\Delta n_g = 1 - 0 = 1$.
The temperature is $T = 100^{\circ} C = 373.15 \ K$.
The relationship between enthalpy change and internal energy change is: $\Delta H = \Delta U + \Delta n_g RT$.
Given $\Delta H = 40.49 \ kJ \ mol^{-1} = 40490 \ J \ mol^{-1}$.
Substituting the values: $40490 = \Delta U + (1 \times 8.3 \times 373.15)$.
$\Delta U = 40490 - 3097.145 = 37392.855 \ J \ mol^{-1} \approx 37.39 \ kJ \ mol^{-1}$.
Rounding to the nearest integer,we get $37 \ kJ \ mol^{-1}$. However,based on the provided options and standard calculation,the closest integer is $38$.

(B) The bond order is calculated using the formula: $\text{B.O.} = \frac{N_b - N_a}{2}$.
$C_2$ $(12 \ e^-)$: $\sigma 1s^2, \sigma^* 1s^2, \sigma 2s^2, \sigma^* 2s^2, \pi 2p_x^2 = \pi 2p_y^2$. $\text{B.O.} = \frac{8-4}{2} = 2$.
$O_2$ $(16 \ e^-)$: $\sigma 1s^2, \sigma^* 1s^2, \sigma 2s^2, \sigma^* 2s^2, \sigma 2p_z^2, \pi 2p_x^2 = \pi 2p_y^2, \pi^* 2p_x^1 = \pi^* 2p_y^1$. $\text{B.O.} = \frac{10-6}{2} = 2$.
$Be_2$ $(8 \ e^-)$: $\text{B.O.} = 0$.
$Li_2$ $(6 \ e^-)$: $\text{B.O.} = 1$.
$Ne_2$ $(20 \ e^-)$: $\text{B.O.} = 0$.
$N_2$ $(14 \ e^-)$: $\text{B.O.} = 3$.
$He_2$ $(4 \ e^-)$: $\text{B.O.} = 0$.
Thus,only $C_2$ and $O_2$ have a bond order of $2$. The total count is $2$.

(C) To determine the number of optically active compounds,we check for the presence of chiral centers and the absence of a plane of symmetry $(POS)$ or center of symmetry $(COS)$.
$1$. $2,3$-butanediol (first structure): The structure shown is the meso form (due to a plane of symmetry),so it is optically inactive.
$2$. $2,3,4$-hexanetriol (second structure): This molecule has chiral centers at $C2, C3,$ and $C4$. It does not possess a plane of symmetry or center of symmetry,making it optically active.
$3$. $1$-butanol $(CH_3-CH_2-CH_2-CH_2-OH)$: No chiral center,optically inactive.
$4$. $2$-chlorobutane $(CH_3-CH_2-CHCl-CH_3)$: This molecule has a chiral center at $C2$. It is optically active.
$5$. $1$-chlorobutane $(CH_3-CH_2-CH_2-CH_2-Cl)$: No chiral center,optically inactive.
$6$. $1$-chloro-$3$-methylbutane $((CH_3)_2CH-CH_2-CH_2-Cl)$: No chiral center,optically inactive.
Thus,there are $2$ optically active compounds in the given list.

(A) To determine the aromaticity of the given compounds,we check for $H$ückel's rule ($4n+2$ $\pi$ electrons),planarity,and cyclic conjugation:
$1$. $1,4$-dihydronaphthalene: Non-aromatic (not fully conjugated).
$2$. Fulvalene: Non-aromatic (not fully conjugated).
$3$. Cyclopentadienyl cation: Aromatic ($2$ $\pi$ electrons,$n=0$,planar,cyclic,conjugated).
$4$. Cyclooctatetraene: Non-aromatic (tub-shaped,non-planar,$8$ $\pi$ electrons).
$5$. Pyridine: Aromatic ($6$ $\pi$ electrons,$n=1$,planar,cyclic,conjugated).
$6$. Cycloheptatriene: Non-aromatic ($sp^3$ carbon present).
Thus,there are $2$ aromatic compounds (Cyclopentadienyl cation and Pyridine).

(A) The wavenumber $(\bar{\nu})$ is defined as the reciprocal of wavelength $(\lambda)$:
$\bar{\nu} = \frac{1}{\lambda}$
Given $\lambda = 5800 \ \mathring{A} = 5800 \times 10^{-8} \ cm = 5.8 \times 10^{-5} \ cm$.
$\bar{\nu} = \frac{1}{5.8 \times 10^{-5} \ cm} = 17241.37 \ cm^{-1}$.
We are given $\bar{\nu} = x \times 10 \ cm^{-1}$.
So,$x \times 10 = 17241.37$.
$x = 1724.137 \approx 1724$.

(C) The choice of a purification method for organic compounds depends on both the physical and chemical properties of the compound itself and the nature of the impurities present in the mixture.

(A) To determine the hybridization,we calculate the steric number using the formula: $\text{Steric Number} = \frac{1}{2} [V + M - C + A]$,where $V$ is the number of valence electrons,$M$ is the number of monovalent atoms,$C$ is the cationic charge,and $A$ is the anionic charge.
$1. BF_3$: Steric number = $\frac{1}{2} [3 + 3] = 3$ ($sp^2$ hybridization).
$2. NO_2^{-}$: Steric number = $\frac{1}{2} [5 + 0 + 1] = 3$ ($sp^2$ hybridization).
$3. NH_2^{-}$: Steric number = $\frac{1}{2} [5 + 2 + 1] = 4$ ($sp^3$ hybridization).
$4. H_2O$: Steric number = $\frac{1}{2} [6 + 2] = 4$ ($sp^3$ hybridization).
$5. NO_2$: The nitrogen atom has one unpaired electron,but for hybridization purposes,it is considered $sp^2$ due to the presence of one double bond and one lone electron.
Thus,both $BF_3$ and $NO_2^{-}$ exhibit $sp^2$ hybridization.

(A) The enamel on the surface of teeth is made of hydroxyapatite,which is $[3(Ca_3(PO_4)_2) \cdot Ca(OH)_2]$.
When $F^{-}$ ions are present in drinking water,they react with hydroxyapatite to form fluoroapatite,which is much harder and more resistant to decay.
The chemical formula for fluoroapatite is $[3(Ca_3(PO_4)_2) \cdot CaF_2]$.

(B) $(1)$ Neutral structures are more stable than charged ones. Therefore,structure $I$ is more stable than structures $II$ and $III$.
$(2)$ In charged structures,a positive charge on a less electronegative atom is more stable. In structure $II$,the positive charge is on the carbon atom $(C^{\oplus})$,whereas in structure $III$,the positive charge is on the oxygen atom $(O^{\oplus})$. Since carbon is less electronegative than oxygen,structure $II$ is more stable than structure $III$.
$\therefore$ The order of stability is $I > II > III$.

(D) Statement $(I)$ is incorrect because the oxidation state is defined based on electronegativity,not electron gain enthalpy.
Statement $(II)$ is correct because elements of the second period (like $C, N, O$) have small atomic sizes,which allow for effective side-on overlap of $2p$ orbitals,leading to strong $p\pi-p\pi$ bonding.
Therefore,Statement $(I)$ is incorrect but Statement $(II)$ is correct.

(A) The energy of an orbital in a multielectron system is determined by the $(n+\ell)$ rule.
$1$. Calculate $(n+\ell)$ for each set:
$A: 4+1 = 5$
$B: 4+2 = 6$
$C: 3+1 = 4$
$D: 3+2 = 5$
$E: 4+0 = 4$
$2$. Compare the $(n+\ell)$ values: $6 (B) > 5 (A, D) > 4 (C, E)$.
$3$. For orbitals with the same $(n+\ell)$ value,the orbital with the higher $n$ value has higher energy.
Comparing $A (n=4)$ and $D (n=3)$: $A > D$.
Comparing $E (n=4)$ and $C (n=3)$: $E > C$.
$4$. Combining these,the order of energy is: $(B) > (A) > (D) > (E) > (C)$.

(C) $(1)$ Acid-catalyzed hydration of propene follows Markovnikov's rule,where the electrophile $H^+$ adds to the terminal carbon to form the more stable secondary carbocation $(CH_3-CH^+-CH_3)$,which is then attacked by $H_2O$ to form propan-$2$-ol as the major product $A$.
$(2)$ Hydroboration-oxidation of propene follows anti-Markovnikov's rule,where the boron atom adds to the terminal carbon,and subsequent oxidation with $H_2O_2/OH^-$ replaces boron with an $-OH$ group,resulting in propan-$1$-ol as the major product $B$.

(A) The given reactions are:
$(1)$ $CuSO_4(s) + 5H_2O(l) \rightarrow CuSO_4 \cdot 5H_2O(s) \quad \Delta H = -x \ kJ \ mol^{-1}$
$(2)$ $CuSO_4 \cdot 5H_2O(s) + aq \rightarrow CuSO_4(aq) \quad \Delta H = +12 \ kJ \ mol^{-1}$
$(3)$ $CuSO_4(s) + aq \rightarrow CuSO_4(aq) \quad \Delta H = -70 \ kJ \ mol^{-1}$
According to Hess's Law,the heat of solution of anhydrous $CuSO_4$ is the sum of the heat of hydration and the heat of solution of the hydrated salt:
$\Delta H_3 = \Delta H_1 + \Delta H_2$
$-70 = -x + 12$
$x = 12 + 70$
$x = 82$
Therefore,the value of $x$ is $82$.

(C) To show all three effects (inductive,mesomeric,and hyperconjugation),a compound must have:
$1$. An electronegative atom or group (for inductive effect).
$2$. $A$ conjugated system or lone pair (for mesomeric effect).
$3$. An $\alpha$-hydrogen atom attached to an $sp^2$ hybridized carbon (for hyperconjugation).
Let's analyze the given compounds:
$1$. Anisole $(C_6H_5OCH_3)$: Shows inductive and mesomeric effects,but lacks $\alpha$-hydrogens on an $sp^2$ carbon for hyperconjugation.
$2$. $5$-methylhex-$3$-en-$2$-one: Shows inductive,mesomeric (conjugation with carbonyl),and hyperconjugation (due to $\alpha$-hydrogens on the isopropyl group).
$3$. Benzene: Shows only mesomeric effect (resonance).
$4$. Chlorocyclohexane: Shows only inductive effect.
$5$. $1$-isopropyl-$2$-nitrobenzene: Shows inductive,mesomeric (nitro group),and hyperconjugation (isopropyl group).
$6$. $1$-(o-nitrophenyl)prop-$2$-ene: Shows inductive,mesomeric,and hyperconjugation.
$7$. $m$-xylene: Shows inductive,hyperconjugation,but no mesomeric effect.
$8$. $1$-acetyl-$2$-methylcyclohexene: Shows inductive,mesomeric,and hyperconjugation.
The compounds that show all three effects are: $5$-methylhex-$3$-en-$2$-one,$1$-isopropyl-$2$-nitrobenzene,$1$-(o-nitrophenyl)prop-$2$-ene,and $1$-acetyl-$2$-methylcyclohexene.
Total count = $4$.

(B) The heat of neutralization is the heat liberated when $1 \ \text{mole}$ of $H^+$ ions reacts with $1 \ \text{mole}$ of $OH^-$ ions to form $1 \ \text{mole}$ of $H_2O$. This value is constant at $-57.1 \ \text{kJ/mol}$.
For $1 \ \text{M} \ HCl$ (a strong monoprotic acid),$1 \ \text{L}$ contains $1 \ \text{mole}$ of $H^+$. Neutralization produces $1 \ \text{mole}$ of $H_2O$,releasing $x \ \text{J}$ of heat.
For $1 \ \text{M} \ H_2SO_4$ (a strong diprotic acid),$1 \ \text{L}$ contains $2 \ \text{moles}$ of $H^+$. Neutralization produces $2 \ \text{moles}$ of $H_2O$,releasing $y \ \text{J}$ of heat.
Since $y$ corresponds to the heat released by $2 \ \text{moles}$ of $H_2O$ and $x$ corresponds to $1 \ \text{mole}$,we have $y = 2x$.
Therefore,$y / x = 2$.

(C) To determine the number of unpaired electrons,we use Molecular Orbital Theory $(MOT)$:
$N_2$: $(14 \ e^-) \rightarrow \sigma 1s^2, \sigma^* 1s^2, \sigma 2s^2, \sigma^* 2s^2, \pi 2p_x^2 = \pi 2p_y^2, \sigma 2p_z^2$ ($0$ unpaired $e^-$)
$O_2$: $(16 \ e^-) \rightarrow \dots, \sigma 2p_z^2, \pi 2p_x^2 = \pi 2p_y^2, \pi^* 2p_x^1 = \pi^* 2p_y^1$ ($2$ unpaired $e^-$)
$C_2^{-}$: $(13 \ e^-) \rightarrow \dots, \sigma 2p_z^2, \pi 2p_x^2 = \pi 2p_y^1$ ($1$ unpaired $e^-$)
$O_2^{-}$: $(17 \ e^-) \rightarrow \dots, \pi^* 2p_x^2 = \pi^* 2p_y^1$ ($1$ unpaired $e^-$)
$O_2^{2-}$: $(18 \ e^-) \rightarrow \dots, \pi^* 2p_x^2 = \pi^* 2p_y^2$ ($0$ unpaired $e^-$)
$H_2^{+}$: $(1 \ e^-) \rightarrow \sigma 1s^1$ ($1$ unpaired $e^-$)
$CN^{-}$: $(14 \ e^-) \rightarrow \dots, \sigma 2p_z^2$ ($0$ unpaired $e^-$)
$He_2^{+}$: $(3 \ e^-) \rightarrow \sigma 1s^2, \sigma^* 1s^1$ ($1$ unpaired $e^-$)
Species with one unpaired electron are $C_2^{-}, O_2^{-}, H_2^{+}, He_2^{+}$.
Total count = $4$.

(B) The definition of the candela,as established by the $SI$ system,is the luminous intensity in a given direction of a source that emits monochromatic radiation of frequency $540 \times 10^{12} \ Hz$ and that has a radiant intensity in that direction of $\frac{1}{683} \ W \ sr^{-1}$.
Comparing this with the given expression '$A$' $\times 10^{12} \ Hz$ and $\frac{1}{B} \ W \ sr^{-1}$,we get $A = 540$ and $B = 683$.

(B) The stability of resonance structures is determined by the following rules:
$1$. Structures with more covalent bonds are more stable.
$2$. Structures with complete octets for all atoms are more stable.
$3$. Structures with negative charge on a more electronegative atom and positive charge on a less electronegative atom are more stable.
Analyzing the given structures:
- Structure $III$ $(CH_3-CH=CH-CHO)$ is a non-polar structure with the maximum number of covalent bonds and complete octets for all atoms,making it the most stable.
- Structure $II$ $(CH_3-CH^+-CH=CH-O^-)$ has a negative charge on the oxygen atom,which is more electronegative than carbon,making it more stable than structure $I$.
- Structure $I$ $(CH_3-CH^--CH=CH-O^+)$ has a negative charge on a carbon atom and a positive charge on an oxygen atom,which is unfavorable due to the electronegativity difference,making it the least stable.
Therefore,the correct stability order is $III > II > I$.

(A) The given structure contains three stereogenic elements:
$1$. One chiral center (the carbon atom attached to the $Br$ atom).
$2$. Two double bonds capable of exhibiting geometrical isomerism ($E/Z$ isomerism).
Since all three stereogenic centers are independent and the molecule is unsymmetrical,the total number of stereoisomers is given by $2^n$,where $n$ is the number of stereocenters.
Here,$n = 3$.
Therefore,total stereoisomers $= 2^3 = 8$.

(C) To determine the bond angles,we analyze the hybridisation and molecular geometry of each molecule:
$1$. $BF_3$: Boron is $sp^2$ hybridized with a trigonal planar geometry. The bond angle is $120^{\circ}$.
$2$. $PF_3$: Phosphorus is $sp^3$ hybridized with one lone pair,resulting in a trigonal pyramidal geometry. Due to the lone pair-bond pair repulsion,the bond angle is approximately $97^{\circ}$.
$3$. $ClF_3$: Chlorine is $sp^3d$ hybridized with two lone pairs,resulting in a $T$-shaped geometry. Due to the repulsion from two lone pairs,the bond angle is distorted to approximately $87.5^{\circ}$.
Comparing these values: $87.5^{\circ} (ClF_3) < 97^{\circ} (PF_3) < 120^{\circ} (BF_3)$.
Therefore,the correct increasing order is $ClF_3 < PF_3 < BF_3$.

(B) The dissociation of the salt is given by: $AB_{2(s)} \rightleftharpoons A^{2+}_{(aq)} + 2B^{-}_{(aq)}$
The expression for the solubility product constant $(K_{sp})$ is: $K_{sp} = [A^{2+}] [B^{-}]^2$
Given concentrations: $[A^{2+}] = 1.2 \times 10^{-4} \ M$ and $[B^{-}] = 0.24 \times 10^{-3} \ M = 2.4 \times 10^{-4} \ M$
Substituting the values: $K_{sp} = (1.2 \times 10^{-4}) \times (2.4 \times 10^{-4})^2$
$K_{sp} = (1.2 \times 10^{-4}) \times (5.76 \times 10^{-8})$
$K_{sp} = 6.912 \times 10^{-12} \ M^3$
Thus,the correct option is $B$.

(D) In ethyne $(HC \equiv CH)$,the carbon-carbon bond is a triple bond,which is shorter and stronger than the double bond in ethene $(CH_2=CH_2)$.
Ethyne is a linear molecule where both carbon atoms are $sp$ hybridised.
Therefore,the statement that the carbon-carbon bond in ethyne is weaker than that in ethene is incorrect.

(B) The electronic configurations are as follows:
$(A)$ Nitrogen ($N$,$Z=7$): $[He] 2s^2 2p^3$ (Matches $III$)
$(B)$ Sulfur ($S$,$Z=16$): $[Ne] 3s^2 3p^4$ (Matches $II$)
$(C)$ Bromine ($Br$,$Z=35$): $[Ar] 3d^{10} 4s^2 4p^5$ (Matches $I$)
$(D)$ Krypton ($Kr$,$Z=36$): $[Ar] 3d^{10} 4s^2 4p^6$ (Matches $IV$)
Therefore,the correct matching is $A-III, B-II, C-I, D-IV$.

(C) The correct matches are as follows:
$A$. Melting point $[K]$: The order is $B > Al > Tl > In > Ga$,which corresponds to $IV$.
$B$. Ionic Radius $[M^{+3} / pm]$: The order is $Tl > In > Ga > Al > B$,which corresponds to $I$.
$C$. $\Delta_{i} H_1 [kJ \ mol^{-1}]$: The order is $B > Tl > Al \approx Ga > In$,which corresponds to $II$.
$D$. Atomic Radius $[pm]$: The order is $Tl > In > Al > Ga > B$,which corresponds to $III$.
Therefore,the correct sequence is $A-IV, B-I, C-II, D-III$.

(A) Fuming sulphuric acid,also known as oleum,is a solution of sulphur trioxide $(SO_3)$ in concentrated sulphuric acid $(H_2SO_4)$.
Its chemical formula is $H_2S_2O_7$ (pyrosulphuric acid).
In the formula $H_2S_2O_7$,the number of oxygen atoms is $7$.

(B) $R_f = \frac{\text{Distance moved by substance from base line}}{\text{Distance moved by solvent from base line}}$
For spot $A$:
$(R_f)_A = \frac{4 \ cm}{8 \ cm} = 0.5$
For spot $B$:
$(R_f)_B = \frac{6 \ cm}{8 \ cm} = 0.75$
Given that $(R_f)_B = (x \times 10^{-1}) \times (R_f)_A$:
$0.75 = (x \times 10^{-1}) \times 0.5$
$0.75 = x \times 0.05$
$x = \frac{0.75}{0.05} = 15$

(C) According to Heisenberg's uncertainty principle:
$m \Delta V \cdot \Delta x \geq \frac{h}{4 \pi}$
Given:
$\Delta x = 10^{-15} \ m$
$m = 9.1 \times 10^{-31} \ kg$
$h = 6.626 \times 10^{-34} \ Js$
$\pi = 3.14$
Substituting the values:
$\Delta V = \frac{h}{4 \pi m \Delta x}$
$\Delta V = \frac{6.626 \times 10^{-34}}{4 \times 3.14 \times 9.1 \times 10^{-31} \times 10^{-15}}$
$\Delta V = \frac{6.626 \times 10^{-34}}{114.296 \times 10^{-46}}$
$\Delta V = 0.05797 \times 10^{12} \ ms^{-1}$
$\Delta V = 57.97 \times 10^9 \ ms^{-1}$
Rounding to the nearest integer,we get $58 \times 10^9 \ ms^{-1}$.

(A) The molecular orbital configuration for $O_2$ $(16 \ e^-)$ is $(\sigma_{1s})^2(\sigma_{1s}^*)^2(\sigma_{2s})^2(\sigma_{2s}^*)^2(\sigma_{2p_z})^2(\pi_{2p_x})^2(\pi_{2p_y})^2(\pi_{2p_x}^*)^1(\pi_{2p_y}^*)^1$.
Number of electrons in $(\pi^*)$ of $O_2 = 2$.
For $O_2^{+}$ $(15 \ e^-)$,one electron is removed from the $(\pi^*)$ orbital,so number of electrons in $(\pi^*)$ of $O_2^{+} = 1$.
For $O_2^{-}$ $(17 \ e^-)$,one electron is added to the $(\pi^*)$ orbital,so number of electrons in $(\pi^*)$ of $O_2^{-} = 3$.
Total number of electrons in $(\pi^*)$ orbitals $= 2 + 1 + 3 = 6$.

(B) At equilibrium,the change in Gibbs free energy is $\Delta G = 0$.
Using the relation $\Delta G = \Delta H - T \Delta S$,at equilibrium we have $\Delta H_{vap} = T \Delta S_{vap}$.
Given $\Delta H_{vap} = 30 \ kJ \ mol^{-1} = 30000 \ J \ mol^{-1}$ and $\Delta S_{vap} = 75 \ J \ mol^{-1} \ K^{-1}$.
Substituting the values: $30000 \ J \ mol^{-1} = T \times 75 \ J \ mol^{-1} \ K^{-1}$.
$T = \frac{30000}{75} \ K = 400 \ K$.

(D) For a first-order reaction,the half-life is given by $t_{1/2} = \frac{\ln 2}{k}$.
Given $\frac{(t_{1/2})_1}{(t_{1/2})_2} = \frac{k_2}{k_1} = \frac{5}{2}$.
For a first-order reaction,the time $t$ to complete a fraction $x$ is $t = \frac{1}{k} \ln \frac{1}{1-x}$.
For Reaction $1$,$t_1 = \frac{1}{k_1} \ln \frac{1}{1 - 2/3} = \frac{1}{k_1} \ln 3$.
For Reaction $2$,$t_2 = \frac{1}{k_2} \ln \frac{1}{1 - 4/5} = \frac{1}{k_2} \ln 5$.
Taking the ratio: $\frac{t_1}{t_2} = \frac{k_2}{k_1} \times \frac{\ln 3}{\ln 5} = \frac{5}{2} \times \frac{\log_{10} 3}{\log_{10} 5}$.
Substituting the given values: $\frac{t_1}{t_2} = \frac{5}{2} \times \frac{0.477}{0.699} = 2.5 \times 0.6824 = 1.706$.
Rounding to the nearest integer,$1.706 \approx 1.7 = 17 \times 10^{-1}$.

(B) The oxidizing ability of these oxoanions depends on the oxidation state and the stability of the lower oxidation states of the central metal atom. The order of oxidizing power is $VO_2^{+} < Cr_2O_7^{2-} < MnO_4^{-}$.
Thus,$VO_2^{+}$ has the least oxidizing ability.
In $VO_2^{+}$,the oxidation state of $V$ is $+5$.
The electronic configuration of $V$ $(Z=23)$ is $[Ar] 3d^3 4s^2$.
For $V^{+5}$,the configuration is $[Ar] 3d^0 4s^0$.
Since there are no unpaired electrons $(n=0)$,the spin-only magnetic moment $\mu = \sqrt{n(n+2)} = \sqrt{0(0+2)} = 0 \ BM$.

(C) The rate of $S_{N}2$ reaction depends on the steric hindrance around the electrophilic carbon atom. The order of reactivity is: $Methyl \ halide > 1^{\circ} \ halide > 2^{\circ} \ halide > 3^{\circ} \ halide$.
In the given options:
$(A)$ is a $3^{\circ}$ alkyl halide.
$(B)$ is a $3^{\circ}$ alkyl halide.
$(C)$ is a $1^{\circ}$ alkyl halide (cyclobutyl-methyl bromide).
$(D)$ is a $2^{\circ}$ alkyl halide.
Since $1^{\circ}$ alkyl halides have the least steric hindrance,they undergo $S_{N}2$ reactions the fastest. Therefore,cyclobutyl-methyl bromide will undergo the fastest $S_{N}2$ reaction.

(A) The reaction of $1\text{-methylcyclohexanol}$ with $\text{conc. } H_2SO_4$ at high temperature involves acid-catalyzed dehydration. This proceeds via an $E1$ mechanism,where the hydroxyl group is protonated and leaves as water to form a tertiary carbocation. Subsequent loss of a proton from the adjacent carbon leads to the formation of the more stable,more substituted alkene,which is $1\text{-methylcyclohexene}$ $(A)$.
The reaction of $1\text{-methylcyclohexanol}$ with $CH_3COCl$ in the presence of pyridine is an acetylation reaction. The alcohol acts as a nucleophile,attacking the acetyl chloride to form an ester,$1\text{-methylcyclohexyl acetate}$ $(B)$.

(A) The correct matches are as follows:
$A$. Borax bead test: $MSO_4$ $\xrightarrow[\Delta]{Na_2B_4O_7} M(BO_2)_2$ $\rightarrow MBO_2$ $\rightarrow M$ (Matches $III$)
$B$. Charcoal cavity test: $MSO_4$ $\xrightarrow[\Delta]{Na_2CO_3} MCO_3$ $\rightarrow MO$ $\rightarrow M$ (Matches $IV$)
$C$. Cobalt nitrate test: $MCO_3$ $\rightarrow MO$ $\xrightarrow[+\Delta]{Co(NO_3)_2} CoO \cdot MO$ (Matches $I$)
$D$. Flame test: $MCO_3$ $\rightarrow MCl_2$ $\rightarrow M^{2+}$ (Matches $II$)
Therefore,the correct sequence is $A-III, B-IV, C-I, D-II$.

(C) The reaction of the complex with excess $AgNO_3$ yields $2$ moles of $AgCl$,which indicates that there are $2$ chloride ions outside the coordination sphere.
The formula of the complex is $[Co(NH_3)_nCl]Cl_2$.
Since the complex is octahedral,the coordination number of $Co$ is $6$. Thus,$n + 1 = 6$,which gives $n = 5$.
The complex is $[Co(NH_3)_5Cl]Cl_2$.
Let the oxidation state of $Co$ be $x$. The sum of oxidation states in the complex is $x + 5(0) + 1(-1) + 2(-1) = 0$,which gives $x - 3 = 0$,so $x = +3$.
Therefore,$x + n = 3 + 5 = 8$.

(A) The structure provided is $D$-glucose.
$1$. $Br_2$ water is a mild oxidizing agent that oxidizes the $-CHO$ group to a $-COOH$ group,forming gluconic acid (a monocarboxylic acid),not a dicarboxylic acid. Thus,this statement is incorrect.
$2$. Glucose exists primarily in cyclic hemiacetal forms in solution,where the concentration of the free aldehyde group is extremely low. Therefore,it does not give Schiff's test.
$3$. $D$-glucose has $4$ chiral (asymmetric) carbon atoms (at $C_2, C_3, C_4, C_5$). This statement is correct.
$4$. In aqueous solution,$D$-glucose exists in equilibrium with its $\alpha$ and $\beta$ pyranose cyclic forms. This statement is correct.

(B) disproportionation reaction is a type of redox reaction where the same element is simultaneously oxidized and reduced.
$F_2$ is the most electronegative element and cannot exhibit a positive oxidation state,so it cannot undergo disproportionation.
$Cl_2$,$Br_2$,and $I_2$ can exist in various positive oxidation states (such as $+1, +3, +5, +7$) and thus undergo disproportionation in alkaline media.
Therefore,$Cl_2$,$Br_2$,and $I_2$ can undergo disproportionation reactions.

(C) Statement $I$ is correct because both $N(CH_3)_3$ and $P(CH_3)_3$ possess a lone pair of electrons on the central atom ($N$ or $P$),allowing them to act as $\sigma$-donor ligands.
Statement $II$ is incorrect because,although $N$ and $P$ belong to the same group,their bonding behavior differs. $P(CH_3)_3$ has vacant $d$-orbitals that can accept electron density from the metal $d$-orbitals (back-bonding),exhibiting $\pi$-acceptor character,whereas $N(CH_3)_3$ lacks such orbitals and acts only as a $\sigma$-donor.

(D) The catalytic reaction proceeds in two steps:
$1$. $2 Fe^{3+} + 2 I^{-} \longrightarrow 2 Fe^{2+} + I_2$
$2$. $2 Fe^{2+} + S_2 O_8^{2-} \longrightarrow 2 Fe^{3+} + 2 SO_4^{2-}$
In the first step,$Fe^{3+}$ oxidises the iodide ion $(I^-)$ to iodine $(I_2)$ and is reduced to $Fe^{2+}$.
In the second step,$Fe^{2+}$ reduces the persulphate ion $(S_2 O_8^{2-})$ to sulphate $(SO_4^{2-})$ and is oxidised back to $Fe^{3+}$.
Thus,statements $A$ and $D$ are correct.

(A) $Fe_4[Fe(CN)_6]_3 \cdot xH_2O$ is Prussian Blue.
$[Fe(CN)_5NOS]^{4-}$ is Violet.
$[Fe(SCN)]^{2+}$ is Blood Red.
$(NH_4)_3PO_4 \cdot 12MoO_3$ is Yellow.
Therefore,the correct matching is $A-III, B-I, C-II, D-IV$.

(B) We analyze the $d$-electron configuration of each complex in an octahedral field (weak field ligand $H_2O$):
$1$. $[Fe(H_2O)_6]^{2+}$: $Fe^{2+}$ is $d^6$. Configuration is $t_{2g}^4 e_g^2$. Number of electrons in $t_{2g} = 4$ (even).
$2$. $[Co(H_2O)_6]^{2+}$: $Co^{2+}$ is $d^7$. Configuration is $t_{2g}^5 e_g^2$. Number of electrons in $t_{2g} = 5$ (odd).
$3$. $[Co(H_2O)_6]^{3+}$: $Co^{3+}$ is $d^6$. Configuration is $t_{2g}^6 e_g^0$. Number of electrons in $t_{2g} = 6$ (even).
$4$. $[Cu(H_2O)_6]^{2+}$: $Cu^{2+}$ is $d^9$. Configuration is $t_{2g}^6 e_g^3$. Number of electrons in $t_{2g} = 6$ (even).
$5$. $[Cr(H_2O)_6]^{2+}$: $Cr^{2+}$ is $d^4$. Configuration is $t_{2g}^3 e_g^1$. Number of electrons in $t_{2g} = 3$ (odd).
The complexes with an even number of electrons in $t_{2g}$ orbitals are $[Fe(H_2O)_6]^{2+}$,$[Co(H_2O)_6]^{3+}$,and $[Cu(H_2O)_6]^{2+}$.
Total count = $3$.

(A) The reaction shown is the Hell-Volhard-Zelinsky $(HVZ)$ reaction.
In this reaction,carboxylic acids having an $\alpha$-hydrogen atom are reacted with $Br_2$ or $Cl_2$ in the presence of a small amount of red phosphorus to form $\alpha$-halo carboxylic acids.
The $\alpha$-carbon is the carbon atom directly attached to the carboxyl group.
In cyclopentanecarboxylic acid,the carbon atom of the cyclopentane ring attached to the $-COOH$ group is the $\alpha$-carbon.
Therefore,the bromine atom replaces the hydrogen atom at the $\alpha$-position,resulting in $1$-bromocyclopentanecarboxylic acid.

(C) Hinsberg's reagent $(C_6H_5SO_2Cl)$ reacts with primary amines $(R-NH_2)$ to form $N$-alkylbenzenesulfonamide,which contains an acidic hydrogen atom attached to the nitrogen. This acidic hydrogen makes the product soluble in aqueous $NaOH$.
Secondary amines $(R_2NH)$ form $N,N$-dialkylbenzenesulfonamide,which lacks an acidic hydrogen and is insoluble in $NaOH$.
Tertiary amines $(R_3N)$ do not react with Hinsberg's reagent.
Let us analyze the given compounds:
$1$. Aniline $(C_6H_5NH_2)$: Primary amine,forms soluble product.
$2$. $o$-methoxyaniline $(o-CH_3OC_6H_4NH_2)$: Primary amine,forms soluble product.
$3$. Ethanamine $(CH_3CH_2NH_2)$: Primary amine,forms soluble product.
$4$. $N$-phenyl-$p$-phenylenediamine $(C_6H_5NH-C_6H_4-NH_2)$: Contains a primary amine group,forms soluble product.
$5$. Cyclohexanamine $(C_6H_{11}NH_2)$: Primary amine,forms soluble product.
Other compounds like amides,ureas,and secondary/tertiary amines do not meet the criteria.
There are $5$ such primary amine compounds.

(A) The metallic radii of the given elements generally decrease across the period due to increasing effective nuclear charge,but $Zn$ has a larger radius due to metallic bonding characteristics. Among the transition metals $Sc$ to $Mn$,the atomic radius decreases. However,the question asks for the least metallic radii among the listed elements. $Cr$ has a smaller radius than $Sc, Ti, V$ and $Mn$.
In $CrO_4^{2-}$,the oxidation state of $Cr$ is $x + 4(-2) = -2$,so $x = +6$.
The electronic configuration of $Cr$ is $[Ar] 3d^5 4s^1$.
The electronic configuration of $Cr^{6+}$ is $[Ar] 3d^0$.
Since there are no unpaired electrons $(n=0)$,the spin-only magnetic moment $\mu = \sqrt{n(n+2)} = \sqrt{0(0+2)} = 0 \ BM$.

(C) The reaction between aniline $(C_6H_5NH_2)$ and benzenediazonium chloride $(C_6H_5N_2^+Cl^-)$ is an electrophilic substitution reaction (coupling reaction) that produces $p$-aminoazobenzene (aniline yellow).
The molar mass of aniline $(C_6H_5NH_2)$ is $93 \ g/mol$.
The number of moles of aniline given is $n = \frac{279 \ g}{93 \ g/mol} = 3 \ mol$.
Since the reaction is with one equivalent of benzenediazonium chloride,and assuming complete conversion,$3 \ moles$ of aniline will produce $3 \ moles$ of aniline yellow $(C_{12}H_{11}N_3)$.
The molar mass of aniline yellow $(C_{12}H_{11}N_3)$ is $(12 \times 12) + (11 \times 1) + (3 \times 14) = 144 + 11 + 42 = 197 \ g/mol$.
The mass of aniline yellow formed is $3 \ mol \times 197 \ g/mol = 591 \ g$.

(D) For a reaction,if the time taken for the concentration to become $1/4$ of its initial value $(t_{75\%})$ is twice the time taken to become $1/2$ of its initial value $(t_{50\%})$,it indicates a first-order reaction with respect to $A$.
Mathematically,for first-order: $t_{75\%} = 2 \times t_{50\%}$.
Regarding $B$,the graph of concentration $[B]$ versus time $t$ is a straight line with a negative slope and a positive intercept. This represents a zero-order reaction with respect to $B$,as $[B]_t = [B]_0 - kt$.
Therefore,the overall order of the reaction $= 1 + 0 = 1$.

(B) The dissociation reaction is $AB_2 \rightarrow A^{2+} + 2B^-$. The van't Hoff factor $i$ is given by $i = 1 + (n-1)\alpha$,where $n=3$. So,$i = 1 + 2\alpha$.
The boiling point elevation is $\Delta T_b = T_b - T_b^{\circ} = 100.52^{\circ} C - 100^{\circ} C = 0.52 \ K$.
The molality $m$ is calculated as $m = \frac{\text{moles of solute}}{\text{mass of solvent in kg}} = \frac{10/200}{100/1000} = \frac{0.05}{0.1} = 0.5 \ mol \ kg^{-1}$.
Using the formula $\Delta T_b = i \cdot K_b \cdot m$:
$0.52 = (1 + 2\alpha) \times 0.52 \times 0.5$
$1 = (1 + 2\alpha) \times 0.5$
$2 = 1 + 2\alpha$
$2\alpha = 1$
$\alpha = 0.5 = 5 \times 10^{-1}$.
Thus,the value is $5$.

(D) The rate of formation of $B$ is given by the expression:
$\frac{d[B]}{dt} = K_1[A] - K_2[B]$
According to the steady-state approximation,the rate of formation of $B$ is set to zero:
$0 = K_1[A] - K_2[B]$
Rearranging the equation to solve for $[B]$:
$K_2[B] = K_1[A]$
$[B] = \frac{K_1}{K_2}[A]$

(D) Phenol $(C_6H_5OH)$ is more acidic than water and aliphatic alcohols. Therefore,it reacts with a strong base like $NaOH$ to form sodium phenoxide and water.
$C_6H_5OH + NaOH \rightarrow C_6H_5O^-Na^+ + H_2O$
Aliphatic alcohols ($C_6H_5CH_2OH$,$C_2H_5OH$,$(CH_3)_3COH$) are less acidic than water and do not react with dilute $NaOH$.

(C) $S_N2$ reactions proceed via a backside attack,leading to the inversion of configuration (Walden inversion). Since only one specific stereoisomer is formed,they are stereospecific.
$S_N1$ reactions proceed via the formation of a planar carbocation intermediate. The nucleophile can attack from either side with equal probability,leading to the formation of a racemic mixture (a mixture of both enantiomers).
Therefore,both Statement $I$ and Statement $II$ are true.

(D) The reactions are matched as follows:
$(A)$ Aniline reacts with $NaNO_2 + HCl$ followed by warm $H_2O$ to form phenol. This corresponds to product $(II)$.
$(B)$ Phenol reacts with $Na_2Cr_2O_7 / H_2SO_4$ to form $p$-benzoquinone. This corresponds to product $(IV)$.
$(C)$ Phenol reacts with $CHCl_3 + aq. NaOH$ followed by $H^+$ (Reimer-Tiemann reaction) to form salicylaldehyde. This corresponds to product $(I)$.
$(D)$ Phenol reacts with $NaOH$,$CO_2$,and $H^+$ (Kolbe's reaction) to form salicylic acid. This corresponds to product $(III)$.
Therefore,the correct matching is $A-II, B-IV, C-I, D-III$.

(D) . Bayer's test $\rightarrow$ Unsaturation $(IV)$
$B$. Ceric ammonium nitrate test $\rightarrow$ Alcoholic-$OH$ group $(III)$
$C$. Phthalein dye test $\rightarrow$ Phenol $(I)$
$D$. Schiff's test $\rightarrow$ Aldehyde $(II)$
Therefore,the correct matching is $A-IV, B-III, C-I, D-II$.

(C) . Incorrect: Dinitrogen is a diatomic gas,but it is not inert; it is relatively unreactive at room temperature due to the high bond enthalpy of the $N \equiv N$ bond.
$B$. Correct: The common oxidation states are $-3, +3, +5$.
$C$. Correct: Nitrogen has a small size and high electronegativity,allowing it to form $p\pi-p\pi$ multiple bonds.
$D$. Incorrect: The stability of the $+5$ oxidation state decreases down the group due to the inert pair effect.
$E$. Incorrect: Nitrogen has a maximum covalency of $4$ because it lacks $d$-orbitals in its valence shell.

(A) The acidic strength of carboxylic acids depends on the electron-donating effect of the alkyl group attached to the $-COOH$ group.
Alkyl groups exhibit a $+I$ (positive inductive) effect,which destabilizes the carboxylate anion and decreases the acidic strength.
As the size of the alkyl group increases,the $+I$ effect increases,thereby decreasing the acidic strength.
$HCOOH$ has no alkyl group,so it is the strongest acid.
Comparing the others: $CH_3COOH$ (methyl group) > $CH_3CH_2COOH$ (ethyl group) > $CH_3CH_2CH_2COOH$ (propyl group).
Therefore,the correct decreasing order is: $HCOOH > CH_3COOH > CH_3CH_2COOH > CH_3CH_2CH_2COOH$.

(A) $p$-Toluenesulfonyl chloride (Hinsberg reagent) reacts only with primary $(1^{\circ})$ and secondary $(2^{\circ})$ amines.
$C_6H_5NH_2$ ($1^{\circ}$ amine) reacts to form $N$-phenyl-$p$-toluenesulfonamide,which is soluble in aqueous $NaOH$ due to the acidic hydrogen on the nitrogen atom.
$(C_6H_5)_2NH$ ($2^{\circ}$ amine) reacts to form $N,N$-diphenyl-$p$-toluenesulfonamide,which is insoluble in aqueous $NaOH$ because it lacks an acidic hydrogen.
$(C_6H_5)_3N$ ($3^{\circ}$ amine) does not react with $p$-toluenesulfonyl chloride.
Therefore,Statement $I$ is false because not all compounds react,and Statement $II$ is false because the products are not all soluble in $NaOH$.

(D) The cell reaction is: $2Tl_{(s)} + Cu^{2+}_{(aq)} \rightarrow 2Tl^+_{(aq)} + Cu_{(s)}$
According to the Nernst equation:
$E_{cell} = E^o_{cell} - \frac{0.0591}{2} \log \frac{[Tl^+]^2}{[Cu^{2+}]}$
To increase $E_{cell}$,the value of the logarithmic term $\frac{[Tl^+]^2}{[Cu^{2+}]}$ must decrease.
This can be achieved by increasing the concentration of the reactant $[Cu^{2+}]$ or decreasing the concentration of the product $[Tl^+]$.
Therefore,increasing the concentration of $Cu^{2+}$ ions will increase the $E_{cell}$.

(A) The given redox reaction is: $\frac{1}{2} H_{2(g)} + AgCl_{(s)} \rightarrow H_{(aq)}^{+} + Cl_{(aq)}^{-} + Ag_{(s)}$
Anodic half-cell reaction (Oxidation):
$\frac{1}{2} H_{2(g)} \rightarrow H_{(aq)}^{+} + e^{-}$
This corresponds to the hydrogen electrode: $Pt|H_{2(g)}|H_{(aq)}^{+}$
Cathodic half-cell reaction (Reduction):
$AgCl_{(s)} + e^{-} \rightarrow Ag_{(s)} + Cl_{(aq)}^{-}$
This corresponds to the silver-silver chloride electrode: $Cl_{(aq)}^{-}|AgCl_{(s)}|Ag$
Combining these,the cell representation is:
$Pt|H_{2(g)}|H_{(aq)}^{+}, Cl_{(aq)}^{-}|AgCl_{(s)}|Ag$
Since $HCl$ or $KCl$ provides the $H^{+}$ and $Cl^{-}$ ions,the representation $Pt|H_{2(g)}|HCl_{(soln.)}|AgCl_{(s)}|Ag$ is correct. Option $A$ represents this cell.

(A) $MnO_2 + 2 KOH + \frac{1}{2} O_2 \rightarrow K_2MnO_4 + H_2O$ (Dark green color).
Statement $I$ is true because the fusion of $MnO_2$ with $KOH$ in the presence of an oxidizing agent like $KNO_3$ or atmospheric oxygen yields potassium manganate $(K_2MnO_4)$,which is dark green in color.
Statement $II$ is true because the manganate ion $(MnO_4^{2-})$ undergoes electrolytic oxidation in an alkaline medium to form the purple-colored permanganate ion $(MnO_4^-)$.
The reaction at the anode is: $MnO_4^{2-} \rightarrow MnO_4^- + e^-$.

(A) The spin-only magnetic moment is calculated using the formula $\mu = \sqrt{n(n+2)} \ B.M.$,where $n$ is the number of unpaired electrons.
$(A) [Cr(NH_3)_6]^{3+}$: $Cr^{3+}$ is $3d^3$. Number of unpaired electrons $n=3$. $\mu = \sqrt{3(3+2)} = \sqrt{15} \approx 3.87 \ B.M. \ (II)$.
$(B) [NiCl_4]^{2-}$: $Ni^{2+}$ is $3d^8$. In a tetrahedral field,$n=2$. $\mu = \sqrt{2(2+2)} = \sqrt{8} \approx 2.83 \ B.M. \ (IV)$.
$(C) [CoF_6]^{3-}$: $Co^{3+}$ is $3d^6$. $F^-$ is a weak field ligand,so $n=4$. $\mu = \sqrt{4(4+2)} = \sqrt{24} \approx 4.90 \ B.M. \ (I)$.
$(D) [Ni(CN)_4]^{2-}$: $Ni^{2+}$ is $3d^8$. $CN^-$ is a strong field ligand,causing pairing,so $n=0$. $\mu = 0 \ B.M. \ (III)$.
Thus,the correct match is $A-II, B-IV, C-I, D-III$.

(D) The chemical structure of tyrosine is $HO-C_6H_4-CH_2-CH(NH_2)-COOH$.
Counting the carbon atoms:
$1$ carbon atom in the carboxylic acid group $(-COOH)$,
$1$ carbon atom in the alpha-carbon position $(-CH(NH_2)-)$,
$1$ carbon atom in the methylene bridge $(-CH_2-)$,
$6$ carbon atoms in the benzene ring.
Total number of carbon atoms $= 1 + 1 + 1 + 6 = 9$.

(A) The reaction between two moles of benzaldehyde $(C_6H_5CHO)$ and one mole of acetone $(CH_3COCH_3)$ in the presence of aqueous $NaOH$ and heat is a Claisen-Schmidt condensation reaction.
The product $x$ formed is dibenzylideneacetone,which has the structure: $C_6H_5-CH=CH-CO-CH=CH-C_6H_5$.
To calculate the number of $\pi$ bonds in dibenzylideneacetone:
$1$. Each phenyl ring $(C_6H_5)$ contains $3$ $\pi$ bonds.
$2$. There are two phenyl rings,so $2 \times 3 = 6$ $\pi$ bonds.
$3$. There are two $C=C$ double bonds,contributing $2$ $\pi$ bonds.
$4$. There is one $C=O$ double bond,contributing $1$ $\pi$ bond.
Total $\pi$ bonds = $6 + 2 + 1 = 9$.

(C) Molality $(m)$ is defined as the number of moles of solute per kilogram of solvent.
Given $m = 4.44 \ m$,this means $4.44 \ mol$ of urea is dissolved in $1000 \ g$ $(1 \ kg)$ of water.
Moles of water $(n_{water})$ = $\frac{1000 \ g}{18 \ g/mol} \approx 55.56 \ mol$.
Mole fraction of urea $(X_{urea})$ = $\frac{n_{urea}}{n_{urea} + n_{water}} = \frac{4.44}{4.44 + 55.56} = \frac{4.44}{60} = 0.074$.
We are given $X_{urea} = x \times 10^{-5}$.
$0.074 = 7400 \times 10^{-5}$.
Therefore,$x = 7400$.

(D) For $[Co(NH_3)_6]^{3+}$: $Co$ is in $+3$ oxidation state. The electronic configuration of $Co^{3+}$ is $[Ar] 3d^6$. Since $NH_3$ is a strong field ligand,it causes pairing of electrons in $d$-orbitals,resulting in $t_{2g}^6 e_g^0$. Thus,the number of unpaired electrons is $0$.
For $[NiCl_4]^{2-}$: $Ni$ is in $+2$ oxidation state. The electronic configuration of $Ni^{2+}$ is $[Ar] 3d^8$. Since $Cl^-$ is a weak field ligand,the electrons remain unpaired in the $d$-orbitals. The configuration is $e^4 t_2^4$,which results in $2$ unpaired electrons.
Total number of unpaired electrons = $0 + 2 = 2$.

(B) Mass of solute (ethyl alcohol) $= 1 \ mole \times 46 \ g \ mol^{-1} = 46 \ g$.
Mass of solvent (water) $= 9 \ mole \times 18 \ g \ mol^{-1} = 162 \ g$.
Total mass of solution $= 46 \ g + 162 \ g = 208 \ g$.
Mass percent of solute $= \frac{\text{Mass of solute}}{\text{Total mass of solution}} \times 100$.
Mass percent $= \frac{46}{208} \times 100 = \frac{4600}{208} \approx 22.11 \%$.
The integer value is $22$.

(C) For strong electrolytes,molar conductivity $(\Lambda_m)$ varies linearly with $C^{1/2}$ according to the Kohlrausch equation: $\Lambda_m = \Lambda_m^0 - A \sqrt{C}$. This is represented by the straight line $B$.
For weak electrolytes,molar conductivity increases sharply with dilution (decrease in $C^{1/2}$) due to an increase in the degree of dissociation,as shown by the curve $A$.

(A) $1$. The reaction of $1$-bromo-$3$-chlorocyclobutane with $Na/Et_2O$ is a Wurtz reaction. Since the $Br$ atom is a better leaving group than $Cl$,the coupling occurs at the $Br$ position,forming $1,1'$-dichloro-$3,3'$-bicyclobutane (compound $A$).
$2$. Compound $A$ reacts with $(i) Mg/Et_2O$ followed by $(ii) D_2O$. The $Mg$ forms a Grignard reagent with the $Cl$ atoms,which is then quenched by $D_2O$ to replace $Cl$ with $D$,yielding $B$ as $1,1'$-dideuterio-$3,3'$-bicyclobutane.
$3$. Compound $A$ reacts with $CoF_2$ (a fluorinating agent used in the Swarts reaction) to replace the $Cl$ atoms with $F$ atoms,yielding $C$ as $1,1'$-difluoro-$3,3'$-bicyclobutane.
$4$. Therefore,the products $B$ and $C$ are $1,1'$-dideuterio-$3,3'$-bicyclobutane and $1,1'$-difluoro-$3,3'$-bicyclobutane respectively.

(D) The basic strength depends on the availability of the lone pair on the nitrogen atom.
$1$. In Piperidine,the nitrogen is $sp^3$ hybridized and the lone pair is localized,making it the most basic.
$2$. In Pyridine,the nitrogen is $sp^2$ hybridized and the lone pair is localized in an $sp^2$ orbital,making it less basic than Piperidine.
$3$. In Pyrrole,the lone pair on the nitrogen is involved in the aromatic sextet (delocalized),making it the least basic.
Therefore,the correct order is: $\text{Piperidine} > \text{Pyridine} > \text{Pyrrole}$.

(C) In an $S_N2$ reaction,the nucleophile attacks from the backside of the leaving group,leading to a trigonal bipyramidal transition state.
In the case of benzyl bromide $(C_6H_5CH_2Br)$,the transition state involves a partially bonded unhybridized $p$-orbital at the benzylic carbon.
This $p$-orbital is stabilized by conjugation with the $\pi$-electron system of the phenyl ring,which lowers the activation energy of the reaction.
Therefore,the $S_N2$ reaction of $C_6H_5CH_2Br$ occurs more readily than that of $CH_3CH_2Br$.
Both Assertion $(A)$ and Reason $(R)$ are correct,and $(R)$ is the correct explanation of $(A)$.

(A) The acidic strength of phenols is increased by electron-withdrawing groups $(-NO_2)$ and decreased by electron-donating groups $(-OCH_3)$.
$pK_{a}$ is inversely proportional to acidic strength.
$(A)$ Phenol
$(B)$ $p$-Nitrophenol (Strongest acid due to $-M$ and $-I$ effect of $-NO_2$)
$(C)$ $m$-Methoxyphenol ($-I$ effect of $-OCH_3$ increases acidity compared to phenol)
$(D)$ $o$-Nitrophenol (Strong acid due to $-I$ and $-M$ effect,but slightly weaker than $p$-nitrophenol due to intramolecular $H$-bonding)
$(E)$ $p$-Methoxyphenol (Strongest electron-donating effect $+M$ of $-OCH_3$ decreases acidity the most)
Acidic strength order: $(B) > (D) > (A) > (C) > (E)$
Therefore,the increasing order of $pK_{a}$ is: $(E) < (C) < (A) < (D) < (B)$.

(A) Oxygen has a small atomic size and high electronegativity,which allows it to form stable $p \pi-p \pi$ multiple bonds with itself $(O_2)$ and other small atoms like $C$ and $N$.
Sulphur has a larger atomic size,making $p \pi-p \pi$ overlap ineffective; therefore,it prefers to form single bonds and exists as $S_8$ puckered rings.
Thus,Assertion $(A)$ is correct because $S_8$ is the stable form of sulphur.
Reason $(R)$ is also correct as it explains why oxygen exists as a diatomic molecule $(O_2)$ while sulphur exists as a polyatomic molecule $(S_8)$.

(C) The complex ion $[Co(en)_2 Cl_2]^{+}$ has an octahedral geometry.
It exhibits geometrical isomerism,specifically $cis$ and $trans$ isomers.
Therefore,the total number of geometrical isomers is $2$,not $3$.
Thus,Assertion $A$ is incorrect,while Reason $R$ is correct.

(C) In aqueous solution,$Cu(II)$ is more stable than $Cu(I)$.
This is because the high hydration energy of the $Cu^{2+}$ ion compensates for the second ionization energy $(IE_2)$ required to convert $Cu^+$ to $Cu^{2+}$.

(D) The reaction sequence is as follows:
$1$. Friedel-Crafts acylation of benzene with succinic anhydride in the presence of $AlCl_3$ gives $4$-oxo-$4$-phenylbutanoic acid $(A)$.
$2$. Clemmensen reduction of $A$ using $Zn-Hg/HCl$ reduces the keto group to a methylene group,yielding $4$-phenylbutanoic acid $(B)$.
$3$. Intramolecular Friedel-Crafts acylation of $B$ in the presence of $conc. H_2SO_4$ leads to the formation of $\alpha$-tetralone $(C)$,which is $1,2,3,4$-tetrahydronaphthalen-$1$-one.

(C) The given reaction is the Gattermann-Koch reaction. In this reaction,benzene reacts with carbon monoxide $(CO)$ and hydrogen chloride $(HCl)$ in the presence of anhydrous aluminum chloride $(AlCl_3)$ and cuprous chloride $(CuCl)$ to form benzaldehyde as the major product. The reaction is: $C_6H_6 + CO + HCl \xrightarrow{Anhydrous AlCl_3/CuCl} C_6H_5CHO$.

(D) The balanced chemical equation for the reaction of $Lead$ $Sulphide$ $(PbS)$ with dilute nitric acid $(HNO_3)$ is:
$3PbS + 8HNO_3 \rightarrow 3Pb(NO_3)_2 + 2NO + 3S + 4H_2O$
From the reaction,it is clear that $Lead$ $nitrate$ $(Pb(NO_3)_2)$,$Nitric$ $oxide$ $(NO)$,and $Sulphur$ $(S)$ are formed.
$Nitrous$ $oxide$ $(N_2O)$ is not formed during this reaction.

(D) primary standard must be stable,non-hygroscopic,and available in a high state of purity.
Statement $(A)$ is correct: It should be available in a pure,dry form.
Statement $(B)$ is correct: It should not react with atmospheric components like moisture or $CO_2$.
Statement $(C)$ is incorrect: $A$ primary standard must $NOT$ be hygroscopic.
Statement $(D)$ is correct: It should be readily soluble in water.
Statement $(E)$ is incorrect: $KMnO_4$ and $NaOH$ are secondary standards because they are not stable or pure enough to be primary standards.
Therefore,the incorrect statements are $(C)$ and $(E)$.

(D) Osmosis is the movement of solvent molecules from a region of lower solute concentration to a region of higher solute concentration through a Semi Permeable Membrane $(SPM)$.
Here,$0.01 \ M \ K_2Cr_2O_7$ (Side $X$) has a lower concentration than $0.05 \ M \ CuSO_4$ (Side $Y$).
Therefore,solvent molecules move from Side $X$ to Side $Y$.
As solvent moves to Side $Y$,the $CuSO_4$ solution becomes more dilute,and the $K_2Cr_2O_7$ solution becomes more concentrated.
Since the solvent moves to Side $Y$,the $K_2Cr_2O_7$ molecules from Side $X$ cannot pass through the $SPM$ to react with $CuSO_4$ on Side $Y$. However,if we consider the net movement of solvent,the concentration of $CuSO_4$ on Side $Y$ decreases due to dilution.
Thus,the molarity of the $CuSO_4$ solution is lowered.

(B) For Statement $I$:
The rate law is $r = k[A]^2[B]$.
When concentrations of $A$ and $B$ are doubled,the new rate $r'$ is:
$r' = k[2A]^2[2B] = k(4[A]^2)(2[B]) = 8k[A]^2[B] = 8r$.
Thus,$x = 8$.
For Statement $II$:
The plot of concentration $[R]$ versus time $t$ is a straight line with a negative slope $-K$. This is characteristic of a zero-order reaction.
Thus,$y = 0$.
Therefore,$x + y = 8 + 0 = 8$.

(C) metal will be oxidized by $Cr_2O_7^{2-}$ if the standard reduction potential $(E^{\circ})$ of the metal is less than the $E^{\circ}$ of the $Cr_2O_7^{2-} / Cr^{3+}$ half-cell $(1.33 \ V)$.
Comparing the given values:
$1. E^{\circ}(Fe^{3+}/Fe) = -0.04 \ V < 1.33 \ V$ (Oxidized)
$2. E^{\circ}(Ni^{2+}/Ni) = -0.25 \ V < 1.33 \ V$ (Oxidized)
$3. E^{\circ}(Ag^+/Ag) = 0.80 \ V < 1.33 \ V$ (Oxidized)
$4. E^{\circ}(Au^{3+}/Au) = 1.40 \ V > 1.33 \ V$ (Not oxidized)
Thus,$Fe$,$Ni$,and $Ag$ will be oxidized.
The total number of such metals is $3$.