JEE Main 2024 Chemistry Question Paper with Answer and Solution

(C) The combustion reaction of methane is: $CH_{4(g)} + 2O_{2(g)} \rightarrow CO_{2(g)} + 2H_2O_{(l)}$
From the stoichiometry of the reaction,$1 \ mole$ of $CH_4$ produces $1 \ mole$ of $CO_2$.
The molar mass of $CO_2$ is $12 + 2 \times 16 = 44 \ g/mol$.
Number of moles of $CO_2$ produced $= \frac{22 \ g}{44 \ g/mol} = 0.5 \ mol$.
Since $1 \ mole$ of $CH_4$ produces $1 \ mole$ of $CO_2$,the moles of $CH_4$ required $= 0.5 \ mol$.
We are given that the moles of $CH_4$ required is $x \times 10^{-2}$.
$0.5 = x \times 10^{-2} \implies x = 0.5 \times 10^2 = 50$.
Thus,the value of $x$ is $50$.

(A) To determine the hybridization of the central atom,we calculate the steric number $(SN)$ = (Number of sigma bonds) + (Number of lone pairs).
$1. NH_3$: $SN = 3 + 1 = 4 \rightarrow sp^3$
$2. SO_2$: $SN = 2 + 1 = 3 \rightarrow sp^2$
$3. SiO_2$: $SN = 4 + 0 = 4 \rightarrow sp^3$
$4. BeCl_2$: $SN = 2 + 0 = 2 \rightarrow sp$
$5. CO_2$: $SN = 2 + 0 = 2 \rightarrow sp$
$6. H_2O$: $SN = 2 + 2 = 4 \rightarrow sp^3$
$7. CH_4$: $SN = 4 + 0 = 4 \rightarrow sp^3$
$8. BF_3$: $SN = 3 + 0 = 3 \rightarrow sp^2$
The species with $sp^3$ hybridization are $NH_3, SiO_2, H_2O,$ and $CH_4$.
Therefore,the total number of such species is $4$.

(C) The electrolysis of a mixture of sodium salts of carboxylic acids (Kolbe's electrolysis) produces free radicals at the anode.
The radicals formed are $CH_3^{\bullet}$ and $C_2H_5^{\bullet}$.
These radicals combine in all possible ways to form alkanes:
$1. CH_3^{\bullet} + CH_3^{\bullet} \rightarrow CH_3-CH_3$ (Ethane)
$2. C_2H_5^{\bullet} + C_2H_5^{\bullet} \rightarrow CH_3-CH_2-CH_2-CH_3$ (n-Butane)
$3. CH_3^{\bullet} + C_2H_5^{\bullet} \rightarrow CH_3-CH_2-CH_3$ (Propane)
Thus,a total of $3$ alkanes are obtained.

(D) The standard Gibbs free energy change is given by the formula: $\Delta_{r} G^{\ominus} = -RT \ln K_{P}$.
Given $R = 8.314 \ J \ K^{-1} \ mol^{-1} = 8.314 \times 10^{-3} \ kJ \ K^{-1} \ mol^{-1}$,$T = 298 \ K$,and $K_{P} = 2.47 \times 10^{-29}$.
Substituting the values:
$\Delta_{r} G^{\ominus} = -(8.314 \times 10^{-3} \ kJ \ K^{-1} \ mol^{-1}) \times (298 \ K) \times \ln(2.47 \times 10^{-29})$.
$\ln(2.47 \times 10^{-29}) = \ln(2.47) + \ln(10^{-29}) \approx 0.904 - 66.776 = -65.872$.
$\Delta_{r} G^{\ominus} = -8.314 \times 10^{-3} \times 298 \times (-65.872) \approx 163.29 \ kJ$.
Rounding to the nearest integer,we get $163 \ kJ$.

(A) The energy of a photon is given by $E = \frac{hc}{\lambda}$.
Using the formula $E = \frac{1240}{\lambda (nm)} \ eV$:
$E = \frac{1240}{242} \ eV \approx 5.124 \ eV$.
To convert this to $J \ mol^{-1}$,we multiply by the conversion factor $1.602 \times 10^{-19} \ J/eV$ and Avogadro's number $N_A = 6.022 \times 10^{23} \ mol^{-1}$.
$E = 5.124 \times 1.602 \times 10^{-19} \times 6.022 \times 10^{23} \ J \ mol^{-1}$.
$E \approx 494,300 \ J \ mol^{-1} = 494 \ kJ \ mol^{-1}$.

(A) $CaCO_{3(s)} \xrightarrow{\Delta} CaO_{(s)} + CO_{2(g)}$
$MgCO_{3(s)} \xrightarrow{\Delta} MgO_{(s)} + CO_{2(g)}$
Let the mass of $CaCO_3$ be $x \ g$.
Then the mass of $MgCO_3 = (2.21 - x) \ g$.
Mass of $CaO$ formed $= \frac{x}{100} \times 56 = 0.56x \ g$.
Mass of $MgO$ formed $= \frac{2.21 - x}{84} \times 40 = 0.4762(2.21 - x) \ g$.
Total mass of residue $= 0.56x + 0.4762(2.21 - x) = 1.152$.
$0.56x + 1.0524 - 0.4762x = 1.152$.
$0.0838x = 0.0996$.
$x \approx 1.188 \ g$ of $CaCO_3$.
Mass of $MgCO_3 = 2.21 - 1.188 = 1.022 \ g$.
Rounding to the nearest option,the composition is $1.187 \ g \ CaCO_3$ and $1.023 \ g \ MgCO_3$.

(A) Statement $I$: $S_8$ in alkaline medium undergoes disproportionation: $3S_8 + 24OH^{-} \rightarrow 16S^{2-} + 8S_2O_3^{2-} + 12H_2O$. This statement is correct.
Statement $II$: In $ClO_4^{-}$,the oxidation state of $Cl$ is $+7$,which is its maximum possible oxidation state. Therefore,it cannot be further oxidized and cannot undergo disproportionation. This statement is incorrect.

(D) The reaction between benzene and benzoyl chloride in the presence of anhydrous $AlCl_3$ is a Friedel-Crafts acylation reaction.
$1$. The anhydrous $AlCl_3$ acts as a Lewis acid and reacts with benzoyl chloride to generate the electrophile,the benzoyl cation $(C_6H_5CO^+)$.
$2$. This electrophile then attacks the benzene ring.
$3$. The final product formed is benzophenone $(C_6H_5COC_6H_5)$.

(D) The reaction is an electrophilic addition of $D-Cl$ to $1-methylcyclopentene$.
First,the electrophile $D^+$ attacks the double bond to form the more stable tertiary carbocation.
The $D^+$ adds to the $CH$ group (the carbon with more hydrogens) to form a tertiary carbocation at the carbon bearing the $CH_3$ group.
Then,the nucleophile $Cl^-$ attacks the planar carbocation from either the top or bottom face,resulting in a racemic mixture of $1-chloro-1-methyl-2-deuteriocyclopentane$ (where $D$ and $CH_3$ are trans or cis).
Based on the options provided,the structure representing the addition of $D$ and $Cl$ across the double bond is the correct product.

(C) To identify the structure of $2,3$-dibromo-$1$-phenylpentane,we break down the $IUPAC$ name:
$1$. The parent chain is a pentane,which is a $5$-carbon chain.
$2$. There is a phenyl group attached at the $1$-st position.
$3$. There are two bromine atoms attached at the $2$-nd and $3$-rd positions.
Following this,the structure consists of a phenyl group attached to a $CH_2$ group (position $1$),which is connected to a $CH(Br)$ group (position $2$),which is connected to another $CH(Br)$ group (position $3$),followed by a $CH_2$ group (position $4$) and a $CH_3$ group (position $5$).
This corresponds to the structure shown in option $C$.

(A) For the reaction: $A_{(g)} \rightleftharpoons B_{(g)} + \frac{1}{2} C_{(g)}$
Initial moles: $1, 0, 0$
Moles at equilibrium: $(1-\alpha), \alpha, \frac{\alpha}{2}$
Total moles at equilibrium: $n_{total} = 1 - \alpha + \alpha + \frac{\alpha}{2} = 1 + \frac{\alpha}{2} = \frac{2+\alpha}{2}$
Partial pressures are given by $P_i = x_i \cdot P$:
$P_A = \frac{1-\alpha}{(2+\alpha)/2} \cdot P = \frac{2(1-\alpha)P}{2+\alpha}$
$P_B = \frac{\alpha}{(2+\alpha)/2} \cdot P = \frac{2\alpha P}{2+\alpha}$
$P_C = \frac{\alpha/2}{(2+\alpha)/2} \cdot P = \frac{\alpha P}{2+\alpha}$
$K_P = \frac{P_B \cdot (P_C)^{1/2}}{P_A} = \frac{(\frac{2\alpha P}{2+\alpha}) \cdot (\frac{\alpha P}{2+\alpha})^{1/2}}{\frac{2(1-\alpha)P}{2+\alpha}}$
Simplifying the expression:
$K_P = \frac{2\alpha P}{2+\alpha} \cdot \frac{\alpha^{1/2} P^{1/2}}{(2+\alpha)^{1/2}} \cdot \frac{2+\alpha}{2(1-\alpha)P} = \frac{\alpha^{3/2} P^{1/2}}{(1-\alpha)(2+\alpha)^{1/2}}$

(B) According to Fajan's rule,the ionic character depends on the polarizability of the cation.
$Ag^+$ has a pseudo-inert gas configuration ($18$ electrons in the outermost shell),which makes it highly polarizing compared to $Ba^{2+}$,$K^+$,and $Co^{2+}$.
Greater polarization leads to more covalent character and less ionic character.
The order of ionic character is $KCl > BaCl_2 > CoCl_2 > AgCl$.
Therefore,$AgCl$ is the least ionic.

(D) Steam distillation is a technique used to separate substances that are steam volatile and immiscible with water. Since essential oils are steam volatile and insoluble in water,this method is the most suitable for their extraction.

(D) Statement $I$ is true: Group $13$ trivalent halides (except $AlF_3$) are covalent in nature and undergo hydrolysis in water.
Statement $II$ is true: In acidified aqueous solution,$AlCl_3$ exists as the octahedral $[Al(H_2O)_6]^{3+}$ ion,where $Al$ is $sp^3d^2$ hybridized.
Therefore,both statements are correct.

(B) The electronic configuration of potassium $(Z=19)$ is $1s^2, 2s^2, 2p^6, 3s^2, 3p^6, 4s^1$.
The outermost electron enters the $4s$ orbital.
For the $4s$ orbital,the principal quantum number $n=4$.
Since it is an $s$-orbital,the azimuthal quantum number $l=0$.
Consequently,the magnetic quantum number $m=0$.
The spin quantum number $s$ can be $+\frac{1}{2}$ or $-\frac{1}{2}$.
Comparing this with the given options,option $B$ is correct.

(D) The reactivity towards electrophilic substitution depends on the electron density of the benzene ring.
$A$ is benzene (reference).
$B$ is toluene ($-CH_3$ group): It shows $+M$ (hyperconjugation) and $+I$ effects,which increase electron density,making it more reactive than benzene.
$C$ is chlorobenzene ($-Cl$ group): It shows $+M$ and $-I$ effects. The $-I$ effect dominates,making it less reactive than benzene.
$D$ is nitrobenzene ($-NO_2$ group): It shows $-M$ and $-I$ effects,which strongly decrease electron density,making it the least reactive.
Therefore,the order of reactivity is $B > A > C > D$.

(B) Based on the provided image,the elements are arranged as follows:
Row $1$: $A^{\prime}, B^{\prime}$ (left to right)
Row $2$: $C^{\prime}, D^{\prime}$ (left to right)
$1$. Atomic Radii: In a period,size decreases from left to right. In a group,size increases from top to bottom.
Thus,$B^{\prime} < A^{\prime}$ and $D^{\prime} < C^{\prime}$. Also,elements in the second row are larger than those in the first row ($A^{\prime} < C^{\prime}$ and $B^{\prime} < D^{\prime}$).
The correct order is $B^{\prime} < A^{\prime} < D^{\prime} < C^{\prime}$. Statement $A$ is true.
$2$. Metallic Character: Metallic character decreases from left to right in a period and increases from top to bottom in a group.
Thus,$B^{\prime} < A^{\prime}$ and $D^{\prime} < C^{\prime}$. Also,$A^{\prime} < C^{\prime}$ and $B^{\prime} < D^{\prime}$.
The correct order is $B^{\prime} < A^{\prime} < D^{\prime} < C^{\prime}$. Statement $B$ is true.
$3$. Size of the element: As established,$B^{\prime} < A^{\prime} < D^{\prime} < C^{\prime}$. Statement $C$ is false.
$4$. Ionic Radii: For cations,the trend follows the atomic radii trend. Thus,$B^{\prime+} < A^{\prime+} < D^{\prime+} < C^{\prime+}$. Statement $D$ is true.
Therefore,statements $A, B,$ and $D$ are true.

(A) The fractional charge is calculated as the ratio of experimental dipole moment to the theoretical dipole moment (assuming $100\%$ ionic character).
$\text{Fractional charge} = \frac{\mu_{\text{exp.}}}{\mu_{\text{cal.}}}$
Given $\mu_{\text{exp.}} = 1.2 \ D = 1.2 \times 10^{-18} \ esu \ cm$.
The bond distance $d = 1 \ \mathring{A} = 10^{-8} \ cm$.
The theoretical dipole moment $\mu_{\text{cal.}}$ for a full electronic charge $(e = 4.8 \times 10^{-10} \ esu)$ is:
$\mu_{\text{cal.}} = q \times d = (4.8 \times 10^{-10} \ esu) \times (10^{-8} \ cm) = 4.8 \times 10^{-18} \ esu \ cm$.
$\text{Fractional charge} = \frac{1.2 \times 10^{-18}}{4.8 \times 10^{-18}} = 0.25$.
Expressing this as $\times 10^{-2}$:
$0.25 = 25 \times 10^{-2}$.

(D) The structure of $2$-methylbutane is $CH_3-CH(CH_3)-CH_2-CH_3$.
Monochlorination can occur at four different types of hydrogen atoms:
$1$. At $C_1$ (terminal methyl group): $CH_2Cl-CH(CH_3)-CH_2-CH_3$ ($1$-chloro$-2-$methylbutane,achiral).
$2$. At $C_2$ (tertiary carbon): $CH_3-CCl(CH_3)-CH_2-CH_3$ ($2$-chloro$-2-$methylbutane,achiral).
$3$. At $C_3$ (chiral center): $CH_3-CH(CH_3)-CHCl-CH_3$ ($2$-chloro$-3-$methylbutane). This product has a chiral center,so it exists as a pair of enantiomers ($d$ and $l$ forms).
$4$. At $C_4$ (terminal methyl group): $CH_3-CH(CH_3)-CH_2-CH_2Cl$ ($1$-chloro$-3-$methylbutane). This product has a chiral center at $C_2$,so it exists as a pair of enantiomers ($d$ and $l$ forms).
Total isomeric products = $1$ (from $C_1$) + $1$ (from $C_2$) + $2$ (from $C_3$) + $2$ (from $C_4$) = $6$.

(A) The balanced chemical equation for the reaction between permanganate ion and oxalate ion in acidic medium is:
$2 MnO_4^{-} + 5 C_2O_4^{2-} + 16 H^{+} \longrightarrow 2 Mn^{2+} + 10 CO_2 + 8 H_2O$
From the stoichiometry of the balanced equation,$2 \ moles$ of $MnO_4^{-}$ require $16 \ moles$ of $H^{+}$ ions.
Therefore,$1 \ mole$ of $MnO_4^{-}$ requires $\frac{16}{2} = 8 \ moles$ of $H^{+}$ ions.

(B) The reaction is: $K_2Cr_2O_7 + 4KCl + 6H_2SO_4 \rightarrow 2CrO_2Cl_2 + 6KHSO_4 + 3H_2O$.
This reaction is known as the chromyl chloride test.
In the product chromyl chloride $(CrO_2Cl_2)$,let the oxidation state of $Cr$ be $x$.
$x + 2(-2) + 2(-1) = 0$
$x - 4 - 2 = 0$
$x = +6$.
Thus,the oxidation state of chromium in the product is $+6$.

(C) Specific gravity (density) $= 1.54 \ g \ cm^{-3}$.
Volume of solution $= 1 \ L = 1000 \ mL$.
Mass of solution $= \text{Density} \times \text{Volume} = 1.54 \ g \ cm^{-3} \times 1000 \ cm^3 = 1540 \ g$.
Since the solution is $70 \%$ pure by weight,the mass of $H_3PO_4 = 0.70 \times 1540 \ g = 1078 \ g$.
Moles of $H_3PO_4 = \frac{\text{Mass}}{\text{Molar mass}} = \frac{1078 \ g}{98 \ g \ mol^{-1}} = 11 \ mol$.
Molarity $= \frac{\text{Moles of solute}}{\text{Volume of solution in } L} = \frac{11 \ mol}{1 \ L} = 11 \ M$.

(B) For an isothermal reversible expansion,the work done is given by the formula:
$w = -2.303 \ nRT \log \left( \frac{V_2}{V_1} \right)$
Substituting the given values:
$n = 5 \ mol$,$R = 8.314 \ J \ K^{-1} \ mol^{-1}$,$T = 300 \ K$,$V_1 = 10 \ L$,$V_2 = 100 \ L$
$w = -2.303 \times 5 \times 8.314 \times 300 \times \log \left( \frac{100}{10} \right)$
$w = -2.303 \times 5 \times 8.314 \times 300 \times \log(10)$
Since $\log(10) = 1$:
$w = -2.303 \times 5 \times 8.314 \times 300 = -28720.713 \ J$
Given $w = -x \ J$,we have $-x = -28720.713$,so $x \approx 28721$.

(D) The balanced half-reaction for the reduction of dichromate in acidic medium is:
$Cr_2O_7^{2-} + 14H^+ + 6e^- \rightarrow 2Cr^{3+} + 7H_2O$
Comparing this with the given equation $Cr_2O_7^{2-} + XH^+ + Ye^- \rightarrow 2A + ZH_2O$,we get:
$X = 14$
$Y = 6$
$Z = 7$
$A = Cr^{3+}$
Therefore,the correct values are $14, 6, 7$ and $Cr^{3+}$.

(A) disproportionation reaction is a redox reaction in which the same element is simultaneously oxidized and reduced.
In reaction $(A)$: $2 Cu^{+} \rightarrow Cu^{2+} + Cu$. The oxidation state of $Cu$ changes from $+1$ to $+2$ (oxidation) and from $+1$ to $0$ (reduction). Thus,it is a disproportionation reaction.
In reaction $(B)$: $3 MnO_4^{2-} + 4 H^{+} \rightarrow 2 MnO_4^{-} + MnO_2 + 2 H_2 O$. The oxidation state of $Mn$ changes from $+6$ to $+7$ (oxidation) and from $+6$ to $+4$ (reduction). Thus,it is a disproportionation reaction.
In reaction $(C)$: This is a thermal decomposition reaction,not a disproportionation reaction.
In reaction $(D)$: This is a comproportionation reaction (the reverse of disproportionation),where $Mn$ in $+7$ and $+2$ oxidation states reacts to form $Mn$ in $+4$ oxidation state.

(D) The species $F^{-}$,$Ne$,and $Na^{+}$ are isoelectronic,as each contains $10$ electrons with the electronic configuration $1s^2, 2s^2, 2p^6$.
Since they have the same number of electrons,the variation in their ionic/atomic radii is determined by the nuclear charge $(Z)$.
As the nuclear charge increases (from $F^{-}$ to $Ne$ to $Na^{+}$),the force of attraction between the nucleus and the electrons increases,leading to a decrease in size.

(A) According to de-Broglie's relation,the wavelength $(\lambda)$ is inversely proportional to the momentum $(p)$:
$\lambda = \frac{h}{p}$
where $h$ is Planck's constant.
This equation can be rewritten as $\lambda p = h$.
Since $h$ is a constant,the product of $\lambda$ and $p$ is constant.
This represents the equation of a rectangular hyperbola.
Therefore,the graph of $\lambda$ versus $p$ is a rectangular hyperbola.

(C) Ionic reactions involve the transfer of electrons between species,which occurs through heterolytic bond cleavage.
In heterolytic cleavage,the shared pair of electrons remains with one of the fragments,resulting in the formation of ions (carbocations or carbanions).
Therefore,ionic reactions proceed through $(B)$ only.

(C) The ionic character of a bond depends on the difference in electronegativity between the bonded atoms. Larger electronegativity difference leads to higher ionic character.
$1$. $N_2$ (non-polar covalent,$\Delta EN = 0$)
$2$. $SO_2$ (polar covalent,$\Delta EN \approx 0.9$)
$3$. $ClF_3$ (polar covalent,$\Delta EN \approx 1.0$)
$4$. $K_2O$ (ionic,$\Delta EN \approx 2.6$)
$5$. $LiF$ (ionic,$\Delta EN \approx 3.0$)
Therefore,the increasing order of ionic character is: $N_2 < SO_2 < ClF_3 < K_2O < LiF$.

(B) In the Kjeldahl's method,the organic compound is heated with concentrated $H_2SO_4$.
$CuSO_4$ is added to the reaction mixture to act as a catalyst,which accelerates the digestion process of the organic nitrogenous compound.

(D) For the free expansion of an ideal gas,the external pressure $P_{ext} = 0$,therefore the work done $w = -P_{ext} \Delta V = 0$.
Since the process is adiabatic,the heat exchange $q = 0$.
According to the first law of thermodynamics,$\Delta U = q + w$.
Since $q = 0$ and $w = 0$,the change in internal energy $\Delta U = 0$.
For an ideal gas,$\Delta U = nC_v \Delta T$. Since $\Delta U = 0$,it follows that $\Delta T = 0$.

(A) The structure of $2$-chlorobutane is $CH_3-CHCl-CH_2-CH_3$.
It contains one chiral carbon atom (indicated by an asterisk).
For a molecule with $n$ chiral centers and no internal plane of symmetry,the number of optical isomers is given by $2^n$.
Here,$n = 1$,so the number of optical isomers $= 2^1 = 2$.
These two isomers are a pair of enantiomers.

(A) The geometries of the given species are as follows:
$PF_5$: Trigonal bipyramidal ($sp^3d$ hybridization).
$BrF_5$: Square pyramidal ($sp^3d^2$ hybridization).
$PCl_5$: Trigonal bipyramidal ($sp^3d$ hybridization).
$[PtCl_4]^{2-}$: Square planar ($dsp^2$ hybridization).
$BF_3$: Trigonal planar ($sp^2$ hybridization).
$Fe(CO)_5$: Trigonal bipyramidal ($dsp^3$ hybridization).
Thus,the species with trigonal bipyramidal shape are $PF_5, PCl_5$,and $Fe(CO)_5$.
The total count is $3$.

(B) In aromatic electrophilic substitution reactions,groups that withdraw electron density from the benzene ring are called deactivating groups. These typically exhibit a $-M$ (negative mesomeric) or $-I$ (negative inductive) effect.
Let us analyze the given groups:
$1$. $-COOCH_3$: This group has a carbonyl group attached to the ring,which exerts a $-M$ effect. It is a deactivating group.
$2$. $-NHCOCH_3$: The nitrogen atom has a lone pair that can be donated to the ring via resonance ($+M$ effect). It is an activating group.
$3$. $-NHCH_3$: The nitrogen atom has a lone pair that can be donated to the ring via resonance ($+M$ effect). It is an activating group.
$4$. $-CN$: The cyano group has a triple bond between carbon and nitrogen,exerting a strong $-M$ effect. It is a deactivating group.
$5$. $-OCH_3$: The oxygen atom has lone pairs that can be donated to the ring via resonance ($+M$ effect). It is an activating group.
The deactivating groups are $-COOCH_3$ and $-CN$.
Therefore,the total number of deactivating groups is $2$.

(C) In the compound $A_2 B$,the atom $B$ exists as $B^{-2}$ ion.
For an atom to achieve the lowest oxidation state of $-2$,it must gain $2$ electrons to complete its octet.
This implies that the atom $B$ has $6$ electrons in its valence shell in its neutral atomic state ($6 + 2 = 8$ electrons).
Therefore,the number of valence electrons in atom $B$ is $6$.

(D) The nature of the given oxides is as follows:
Acidic oxides: $Cl_2O_7, SiO_2, N_2O_5$
Neutral oxides: $CO, NO, N_2O$
Amphoteric oxides: $Al_2O_3, SnO_2, PbO_2$
Therefore,the number of amphoteric oxides is $3$.

(A) The balanced chemical equation is: $3 PbCl_2 + 2 (NH_4)_3 PO_4 \rightarrow Pb_3(PO_4)_2 + 6 NH_4 Cl$
Calculate the moles required for each reactant:
For $PbCl_2$: $72 \ mmol / 3 = 24 \ mmol$
For $(NH_4)_3 PO_4$: $50 \ mmol / 2 = 25 \ mmol$
Since $24 < 25$,$PbCl_2$ is the limiting reagent.
The amount of $Pb_3(PO_4)_2$ formed is determined by the limiting reagent:
$mmol \text{ of } Pb_3(PO_4)_2 = \frac{1}{3} \times mmol \text{ of } PbCl_2 = \frac{72}{3} = 24 \ mmol$.

(B) Ammonium acetate $(CH_{3}COONH_{4})$ is a salt of a weak acid $(CH_{3}COOH)$ and a weak base $(NH_{4}OH)$.
The formula for the $pH$ of a salt of a weak acid and a weak base is given by: $pH = \frac{1}{2} (pK_{w} + pK_{a} - pK_{b})$.
Given that $K_{a} = 1.8 \times 10^{-5}$ and $K_{b} = 1.8 \times 10^{-5}$,we have $pK_{a} = pK_{b}$.
Substituting this into the formula: $pH = \frac{1}{2} (pK_{w} + pK_{a} - pK_{a}) = \frac{pK_{w}}{2}$.
Since $pK_{w} = 14$ at $25^{\circ}C$,$pH = \frac{14}{2} = 7$.

(C) Statement $I$ is false because a $\pi$ bonding molecular orbital $(MO)$ has higher electron density above and below the inter-nuclear axis.
Statement $II$ is true because the $\pi^*$ antibonding molecular orbital has a nodal plane perpendicular to the inter-nuclear axis between the two nuclei,where the electron density is zero.
Therefore,Statement $I$ is false but Statement $II$ is true.

(B) $I$. In $p-$block,both metals and non-metals are present,but in $d-$block,only metals are present. Therefore,Statement $I$ is false.
$II$. Non-metals generally have higher ionisation enthalpy $(IE)$ and higher electronegativity $(EN)$ compared to metals due to their smaller atomic size and higher effective nuclear charge. Therefore,Statement $II$ is true.
Conclusion: Statement $I$ is false and Statement $II$ is true.

(C) Meta-directing groups are those that withdraw electrons from the benzene ring through the inductive effect $(-I)$ and/or resonance effect $(-M)$.
In option $C$,all the groups $(-NO_2, -CHO, -SO_3H, -COR)$ are electron-withdrawing groups that exhibit a $-M$ effect,making them meta-directing.
In other options,groups like $-NH_2, -NHR, -OCH_3$ and $-NHCOCH_3$ are ortho/para-directing due to their $+M$ effect.

(D) $H_2O$,$NH_3$,and $C_2H_5OH$ exhibit intermolecular hydrogen bonding.
In $o$-nitrophenol,the hydrogen atom of the hydroxyl group is close to the oxygen atom of the nitro group,which leads to the formation of a stable six-membered ring through intramolecular hydrogen bonding.

(D) Lassaigne's test is a standard qualitative analysis method used to detect the presence of elements such as Nitrogen $(N)$,Sulphur $(S)$,Phosphorus $(P)$,and halogens ($X$,where $X = Cl, Br, I$) in organic compounds.

(A) $1$. The first reaction is the catalytic hydrogenation of an alkyne using $Pd/C$ (Lindlar's catalyst is typically poisoned,but $Pd/C$ generally reduces alkynes to alkenes). The product shown is a cis-alkene,specifically $cis$-$3$-methylpent-$2$-ene (or similar structure based on the image). Looking at the reactant $A$ in the image,it is $CH_3-C\equiv C-C_2H_5$,which is $2$-pentyne.
$2$. The second reaction is the Birch reduction of an internal alkyne ($CH_3-C\equiv C-CH_3$,which is but-$2$-yne) using $Na/\text{liquid } NH_3$. This reaction is stereospecific and yields the trans-alkene,which is trans-$2$-butene.
$3$. Therefore,$A$ is $2$-pentyne and $B$ is trans-$2$-butene.

(A) For $3p$ orbital,the principal quantum number $n = 3$ and the azimuthal quantum number $\ell = 1$.
The formula for calculating the number of radial nodes is given by $n - \ell - 1$.
Substituting the values: $3 - 1 - 1 = 1$.
Therefore,the number of radial nodes for $3p$ orbital is $1$.

(C) The functional group $-COOH$ contains a $\pi$-bond conjugated with an electronegative oxygen atom,which allows it to withdraw electron density from the system,thus showing a negative resonance $(-R)$ effect.
Conversely,the groups $-NH_2$,$-OH$,and $-OR$ possess lone pairs of electrons on the atom directly attached to the conjugated system,which they donate,thus showing a positive resonance $(+R)$ effect.

(C) $SiO_2$ and $GeO_2$ are acidic,whereas $SnO$ and $PbO$ are amphoteric in nature. This statement is true.
Carbon does not possess $d$-orbitals,so it cannot form $p \pi-d \pi$ bonds. Allotropic forms of carbon arise due to the property of catenation and $p \pi-p \pi$ bond formation. Thus,Statement $II$ is false.

(B) The solubility $S$ in $mol \ L^{-1}$ is calculated as: $S = \frac{W \times 1000}{M \times 100} = \frac{10W}{M} \ mol \ L^{-1}$.
The dissociation of calcium phosphate is: $Ca_3(PO_4)_{2(s)} \rightleftharpoons 3Ca^{2+}(aq) + 2PO_4^{3-}(aq)$.
The solubility product expression is: $K_{sp} = [Ca^{2+}]^3 [PO_4^{3-}]^2 = (3S)^3 (2S)^2$.
$K_{sp} = 27S^3 \times 4S^2 = 108S^5$.
Substituting $S = \frac{10W}{M}$: $K_{sp} = 108 \times \left(\frac{10W}{M}\right)^5 = 108 \times 10^5 \times \left(\frac{W}{M}\right)^5 = 1.08 \times 10^7 \left(\frac{W}{M}\right)^5$.
Rounding to the nearest order of magnitude,the value is approximately $10^7 \left(\frac{W}{M}\right)^5$.

(A) The structure of $2$-methylbutane is $CH_3-CH(CH_3)-CH_2-CH_3$.
Monochlorination can occur at four different types of hydrogen atoms:
$1$. At $C_1$ (terminal methyl group): $CH_2Cl-CH(CH_3)-CH_2-CH_3$. This molecule has a chiral center at $C_2$,so it exists as $2$ enantiomers.
$2$. At $C_2$ (tertiary carbon): $CH_3-CCl(CH_3)-CH_2-CH_3$. This molecule is achiral.
$3$. At $C_3$ (secondary carbon): $CH_3-CH(CH_3)-CHCl-CH_3$. This molecule has a chiral center at $C_3$,so it exists as $2$ enantiomers.
$4$. At $C_4$ (terminal methyl group): $CH_3-CH(CH_3)-CH_2-CH_2Cl$. This molecule has a chiral center at $C_2$,so it exists as $2$ enantiomers.
Total number of isomers = $2$ (from $C_1$) + $1$ (from $C_2$) + $2$ (from $C_3$) + $2$ (from $C_4$) = $7$.
However,considering the provided options,the intended answer is $6$.

(B) The reaction is: $H_2SO_4 + 2NH_3 \rightarrow (NH_4)_2SO_4$.
Millimoles of $H_2SO_4$ used = $10 \ mL \times 2 \ M = 20 \ mmol$.
Since $1 \ mol$ of $H_2SO_4$ reacts with $2 \ mol$ of $NH_3$,the millimoles of $NH_3$ produced = $20 \times 2 = 40 \ mmol$.
Since $1 \ mol$ of $NH_3$ contains $1 \ mol$ of $N$,the millimoles of $N$ = $40 \ mmol$.
Mass of $N$ = $\frac{40}{1000} \times 14 \ g = 0.56 \ g$.
Percentage of $N$ = $\frac{\text{Mass of } N}{\text{Mass of compound}} \times 100 = \frac{0.56}{1} \times 100 = 56 \%$.

(D) . The chromate ion $(CrO_4^{2-})$ is tetrahedral,not square planar.
$B$. Dichromates are prepared from chromates by adding acid $(2CrO_4^{2-} + 2H^{+} \rightarrow Cr_2O_7^{2-} + H_2O)$. This statement is correct.
$C$. The green manganate ion $(MnO_4^{2-})$ has $1$ unpaired electron ($d^1$ configuration),making it paramagnetic,not diamagnetic.
$D$. $K_2MnO_4$ disproportionates in neutral or acidic solution to give permanganate $(MnO_4^-)$ and manganese dioxide $(MnO_2)$. This statement is correct.
$E$. As the oxidation number of the transition metal increases,the ionic character of the oxides decreases (covalent character increases). This statement is correct.
Therefore,statements $B, D,$ and $E$ are correct.

(B) The fundamental principle used in $Chromatography$ is $Adsorption$. In this technique,different components of a mixture are adsorbed to different extents on a stationary phase.

(A) For the reaction $A(g) \rightarrow B(g) + C(g)$:

Time $t=0$	$P_i$	$0$	$0$
Time $t$	$P_i - x$	$x$	$x$

Total pressure at time $t$ is $P_t = (P_i - x) + x + x = P_i + x$.
Therefore,$x = P_t - P_i$.
The partial pressure of reactant $A$ at time $t$ is $P_A = P_i - x = P_i - (P_t - P_i) = 2 P_i - P_t$.
Substituting into the first-order rate equation $k = \frac{2.303}{t} \log \frac{P_i}{P_A}$:
$k = \frac{2.303}{t} \log \frac{P_i}{2 P_i - P_t}$.

(A) The $pK_a$ value is inversely proportional to the acid strength. $A$ lower $pK_a$ value indicates a stronger acid.
Phenol has a $pK_a$ of $10.0$,whereas ethanol has a $pK_a$ of $15.9$. Thus,phenol is a stronger acid than ethanol.
This is because the phenoxide ion (conjugate base of phenol) is stabilized by resonance,whereas the ethoxide ion (conjugate base of ethanol) is destabilized by the $+I$ effect of the ethyl group.
Therefore,$Assertion$ $A$ is true,but $Reason$ $R$ is false.

(D) . Incorrect. Crystal field theory explains the splitting of $d$-orbitals but does not explain the relative strength of ligands (spectrochemical series).
$B$. Correct. Valence bond theory $(VBT)$ provides a qualitative description of bonding but fails to provide a quantitative interpretation of kinetic stability.
$C$. Correct. In $\left[Ni(CN)_4\right]^{2-}$,$Ni$ is in $+2$ oxidation state $(3d^8)$. $CN^-$ is a strong field ligand,causing pairing of electrons,leading to $dsp^2$ hybridization.
$D$. Incorrect. The complex $cis-\left[PtCl_2(en)_2\right]^{2+}$ is a specific isomer. The question asks for the number of possible isomers of this specific configuration,which is just one (the $cis$ form itself).
Therefore,statements $B$ and $C$ are correct.

(B) $1$. Glucose does not react with $NaHCO_3$ because it is not acidic enough to evolve $CO_2$ gas. Thus,$A-II$.
$2$. Glucose on oxidation with concentrated $HNO_3$ gives saccharic acid (glucaric acid). Thus,$B-IV$.
$3$. Glucose on prolonged heating with $HI$ gives $n$-hexane,indicating the presence of a straight chain of six carbon atoms. Thus,$C-III$.
$4$. Glucose on oxidation with bromine water gives gluconic acid. Thus,$D-I$.
Therefore,the correct matching is $A-II, B-IV, C-III, D-I$.

(A) $CaF_2$ reacts with concentrated $H_2SO_4$ to form $HF$ gas,which is colourless.
$NaBr + H_2SO_4 \rightarrow NaHSO_4 + HBr$ (followed by oxidation to reddish-brown $Br_2$ vapours).
$NaNO_3 + H_2SO_4 \rightarrow NaHSO_4 + HNO_3$ (followed by decomposition to reddish-brown $NO_2$ vapours).
$KI + H_2SO_4 \rightarrow KHSO_4 + HI$ (followed by oxidation to violet $I_2$ vapours).
$CaF_2$ is the only salt that does not evolve coloured vapours.
Molar mass of $CaF_2 = 40 + 2 \times 19 = 40 + 38 = 78 \ g \ mol^{-1}$.

(B) $NH_3$ acts as a weak field ligand with $Ni^{2+}$.
The electronic configuration of $Ni$ $(Z=28)$ is $[Ar] 3d^8 4s^2$.
For $Ni^{2+}$,the configuration is $3d^8$.
The $3d$ orbitals are filled as follows: two orbitals are doubly occupied,and two orbitals are singly occupied (total $2$ unpaired electrons).
The spin-only magnetic moment is calculated as:
$\mu = \sqrt{n(n+2)} \ BM$
Where $n = 2$ (number of unpaired electrons).
$\mu = \sqrt{2(2+2)} = \sqrt{8} \approx 2.82 \ BM$.
To express this as $x \times 10^{-1} \ BM$,we have $2.82 = 28.2 \times 10^{-1}$.
Rounding to the nearest integer,$x = 28$.

(A) The reactant is $4-hydroxybenzaldehyde$,which contains both an aldehyde group $(-CHO)$ and a phenolic hydroxyl group $(-OH)$.
When $PhMgBr$ (a Grignard reagent) is added,it acts as a strong base and reacts with the acidic phenolic $-OH$ group first.
$4-hydroxybenzaldehyde + PhMgBr \rightarrow 4-formylphenoxide magnesium bromide + Benzene$
The $Ph^-$ from $PhMgBr$ abstracts the acidic proton from the phenolic $-OH$ group to form benzene $(C_6H_6)$ and the magnesium salt of the phenol.
Upon workup with $aq. NH_4Cl$,the magnesium salt is protonated back to the original $4-hydroxybenzaldehyde$.
Therefore,the final product $P$ is $4-hydroxybenzaldehyde$,which contains $1$ hydroxyl group.

(C) The reaction of $CH_3CH_2Br$ with $NaOH$ in the presence of $C_2H_5OH$ (alcoholic $NaOH$) undergoes dehydrohalogenation to form ethene $(CH_2=CH_2)$ as product $A$.
Product $A$ has $4$ hydrogen atoms.
The reaction of $CH_3CH_2Br$ with $NaOH$ in the presence of $H_2O$ (aqueous $NaOH$) undergoes nucleophilic substitution to form ethanol $(CH_3CH_2OH)$ as product $B$.
Product $B$ has $6$ hydrogen atoms.
The total number of hydrogen atoms in product $A$ and product $B$ is $4 + 6 = 10$.

(B) The electrode reaction for the reduction of copper ions is: $Cu^{2+} + 2e^{-} \rightarrow Cu$.
According to the stoichiometry of the reaction,$2 \ Faraday$ of electricity is required to deposit $1 \ mol$ (or $1 \ \text{gram atom}$) of copper.
Therefore,$1 \ Faraday$ of electricity will deposit $1/2 = 0.5 \ \text{mol}$ of copper.
Since $0.5 \ \text{mol} = 0.5 \ \text{gram atom}$,we can express this as $5 \times 10^{-1} \ \text{gram atom}$.
Comparing this with $x \times 10^{-1}$,we get $x = 5$.

(D) The electronic configurations of the metal ions in the given octahedral complexes are as follows:
$A. [Cr(H_2O)_6]^{3+}$: $Cr^{3+}$ is $3d^3$. In an octahedral field,this corresponds to $t_{2g}^3 e_g^0$. Thus,$A-II$.
$B. [Fe(H_2O)_6]^{3+}$: $Fe^{3+}$ is $3d^5$. With a weak field ligand like $H_2O$,it is high spin: $t_{2g}^3 e_g^2$. Thus,$B-III$.
$C. [Ni(H_2O)_6]^{2+}$: $Ni^{2+}$ is $3d^8$. In an octahedral field,this corresponds to $t_{2g}^6 e_g^2$. Thus,$C-IV$.
$D. [V(H_2O)_6]^{3+}$: $V^{3+}$ is $3d^2$. In an octahedral field,this corresponds to $t_{2g}^2 e_g^0$. Thus,$D-I$.
Therefore,the correct matching is $A-II, B-III, C-IV, D-I$.

(A) The reaction of bromobenzene with concentrated $HNO_3$ (nitration) leads to the formation of $2,4,6-$trinitrobromobenzene $(A)$ due to the strong activating effect of the three nitro groups.
Subsequent treatment with $NaOH$ followed by $HCl$ (nucleophilic aromatic substitution) replaces the bromine atom with a hydroxyl group to form $2,4,6-$trinitrophenol,commonly known as picric acid $(B)$.

(D) For $[Ni(CO)_4]$: The oxidation state of $Ni$ is $0$. The electronic configuration is $[Ar] 3d^8 4s^2$. Since $CO$ is a strong field ligand,it causes pairing of electrons,resulting in $sp^3$ hybridization with $0$ unpaired electrons,making it diamagnetic.
For $[NiCl_4]^{2-}$: The oxidation state of $Ni$ is $+2$. The electronic configuration is $[Ar] 3d^8$. Since $Cl^-$ is a weak field ligand,it does not cause pairing of electrons,resulting in $sp^3$ hybridization with $2$ unpaired electrons,making it paramagnetic.

(D) The reaction of sulphanilic acid with $NaNO_2$ and $CH_3COOH$ forms a diazonium salt intermediate (or a related diazo-acetate species $X$).
This intermediate $X$ then undergoes an electrophilic aromatic substitution (coupling reaction) with $1$-naphthylamine.
The coupling occurs at the position para to the $-NH_2$ group in the naphthylamine ring,resulting in the formation of a red azo-dye $(Y)$.
The correct structure for the product $(Y)$ is shown in option $D$.

(D) Statement $I$ is true: Aniline reacts with conc. $H_2SO_4$ to form anilinium hydrogen sulphate,which on heating at $453-473 \ K$ undergoes rearrangement to form $p$-aminobenzene sulphonic acid (sulphanilic acid). Since this molecule contains both $N$ and $S$,it gives a positive Lassaigne's test for both elements,resulting in the formation of $[Fe(SCN)]^{2+}$,which is blood red in colour.
Statement $II$ is true: Aniline is a Lewis base due to the lone pair on the nitrogen atom. It reacts with the Lewis acid $AlCl_3$ (used in Friedel-Crafts reactions) to form an acid-base adduct (salt). The nitrogen atom acquires a positive charge,which makes the $-NH_3^+$ group strongly electron-withdrawing (deactivating),thereby inhibiting the Friedel-Crafts reaction.

(D) Correct: Group $16$ elements form $EO_2$ and $EO_3$ type oxides which are acidic.
$(B)$ Correct: $SO_2$ acts as a reducing agent,whereas $TeO_2$ acts as an oxidising agent.
$(C)$ Incorrect: The reducing property increases from $H_2S$ to $H_2Te$ down the group due to the decrease in bond dissociation enthalpy.
$(D)$ Incorrect: The ozone molecule $(O_3)$ has a bent structure with resonance. The total number of lone pairs is $6$ (terminal oxygen atoms have $2$ and $3$ lone pairs respectively,and the central oxygen has $1$ lone pair).
Therefore,statements $A$ and $B$ are correct.

(C) The given reaction involves the treatment of benzene with carbon monoxide $(CO)$ and hydrogen chloride $(HCl)$ in the presence of anhydrous aluminum chloride $(AlCl_3)$ and cuprous chloride $(CuCl)$ to form benzaldehyde. This specific reaction is known as the Gatterman-Koch reaction.

(C) $Mn_2O_7$ is a green oil at room temperature. This statement is correct.
$(B)$ $V_2O_4$ dissolves in acids to give $VO^{2+}$ salts. This statement is correct.
$(C)$ $CrO$ is a basic oxide. This statement is correct.
$(D)$ $V_2O_5$ is amphoteric; it reacts with both acids and bases. Therefore,the statement that it does not react with acid is incorrect.
Thus,statements $(A), (B),$ and $(C)$ are correct.

(B) The rate law $r = k[A]$ indicates that the reaction is of the first order.
For a first-order reaction,the half-life is given by $t_{1/2} = \frac{0.693}{k} = 120 \ \text{min}$.
Thus,$k = \frac{0.693}{120} \ \text{min}^{-1}$.
For $90 \%$ decomposition,the amount remaining is $100 - 90 = 10 \%$.
The time $t$ is given by the formula $t = \frac{2.303}{k} \log \left( \frac{[A]_0}{[A]_t} \right)$.
Substituting the values: $t = \frac{2.303}{(0.693 / 120)} \log \left( \frac{100}{10} \right)$.
$t = \frac{2.303 \times 120}{0.693} \times \log(10)$.
Since $\frac{2.303}{0.693} \approx \frac{1}{0.301} \approx 3.32$,we get $t = 120 \times 3.32 \approx 399 \ \text{minutes}$.

(C) Acetylation involves the replacement of a hydrogen atom $(-H)$ by an acetyl group $(-COCH_3)$.
The net change in molar mass for each amino group $(-NH_2)$ reacting is the replacement of one $H$ atom (mass $1$) with one $COCH_3$ group (mass $43$).
Gain in molecular weight per amino group $= 43 - 1 = 42 \ g \ mol^{-1}$.
Total increase in molecular weight $= 192 - 108 = 84 \ g \ mol^{-1}$.
Therefore,the number of amino groups in $(x) = \frac{84}{42} = 2$.

(B) Materials with high conductivity (typically $> 10^2 \ S \ m^{-1}$) are classified as conductors.
Given values in $S \ m^{-1}$ are: $2.1 \times 10^3$,$1.0 \times 10^{-16}$,$1.2 \times 10$,$3.91$,$1.5 \times 10^{-2}$,$1 \times 10^{-1}$,$1.0 \times 10^3$.
$1$. $2.1 \times 10^3$ (Conductor)
$2$. $1.0 \times 10^{-16}$ (Insulator)
$3$. $1.2 \times 10$ (Semiconductor)
$4$. $3.91$ (Semiconductor)
$5$. $1.5 \times 10^{-2}$ (Semiconductor)
$6$. $1 \times 10^{-1}$ (Semiconductor)
$7$. $1.0 \times 10^3$ (Conductor)
There are $2$ materials with high conductivity values ($2.1 \times 10^3$ and $1.0 \times 10^3$).
Therefore,the number of conductors is $2$.

(A) Vitamins are classified into two categories: water-soluble and fat-soluble.
Fat-soluble vitamins $(A, D, E, K)$ are stored in the liver and adipose tissues.
Among water-soluble vitamins,Vitamin $B_{12}$ is also stored in the liver.
Therefore,the vitamins that can be stored in our body are $A, D, E, K$ and $B_{12}$.
The total count of these vitamins is $5$.

(B) According to the base-pairing rules in $DNA$:
$Adenine$ $(A)$ pairs with $Thymine$ $(T)$ via $2$ hydrogen bonds.
$Cytosine$ $(C)$ pairs with $Guanine$ $(G)$ via $3$ hydrogen bonds.
Given the sequence $ATGCTTCA$:
$A$ pairs with $T$
$T$ pairs with $A$
$G$ pairs with $C$
$C$ pairs with $G$
$T$ pairs with $A$
$T$ pairs with $A$
$C$ pairs with $G$
$A$ pairs with $T$
Therefore,the complementary sequence is $TACGAAGT$.

(A) $KCN$ is an ionic compound and provides cyanide ions $(CN^-)$ in solution. Since carbon is more nucleophilic,it attacks the alkyl halide to form alkyl cyanide $(R-CN)$ as the major product.
$AgCN$ is predominantly covalent in nature. In $AgCN$,the carbon atom is linked to silver,making the nitrogen atom the only site available for nucleophilic attack. Therefore,it forms alkyl isocyanide $(R-NC)$ as the major product.
Thus,Assertion $(A)$ is correct,but Reason $(R)$ is incorrect because $AgCN$ is covalent,not ionic.

(D) In $[Ni(H_2O)_6]^{2+}$,$Ni$ is in $+2$ oxidation state with $d^8$ configuration. $H_2O$ is a weak field ligand,so it forms an octahedral complex with two unpaired electrons,which undergoes $d-d$ transition,making the solution green.
In $[Ni(CN)_4]^{2-}$,$Ni$ is in $+2$ oxidation state with $d^8$ configuration. $CN^-$ is a strong field ligand,causing pairing of electrons. The complex is square planar and diamagnetic. However,it is yellow in colour due to charge transfer or specific electronic transitions,but it is often cited in textbooks as being colourless or having very pale colour compared to other complexes; however,strictly speaking,it is yellow. Given the standard context of such questions,Statement $I$ is correct and Statement $II$ is incorrect.

(D) Assertion $(A)$ is correct because $NH_3$ exhibits intermolecular hydrogen bonding due to the high electronegativity of nitrogen,whereas $PH_3$ does not exhibit hydrogen bonding.
Reason $(R)$ is incorrect because it states the opposite: $NH_3$ molecules are associated through hydrogen bonding,while $PH_3$ molecules are associated only through weak van der Waals forces.
Therefore,$(A)$ is correct but $(R)$ is not correct.

(B) The reaction of toluene with $Cl_2$ in the presence of light $(hv)$ is a free radical substitution reaction.
It proceeds as follows:
$C_6H_5CH_3 + 2Cl_2 \xrightarrow{hv} C_6H_5CHCl_2 + 2HCl$
Here,$A$ is benzal chloride $(C_6H_5CHCl_2)$.
Next,the hydrolysis of benzal chloride with water at $373 \ K$ yields benzaldehyde:
$C_6H_5CHCl_2 + H_2O \xrightarrow{373 \ K} C_6H_5CHO + 2HCl$
Thus,$A$ is benzal chloride and $B$ is benzaldehyde.

(B) Aniline is the common name for the chemical compound $C_6H_5NH_2$.
It is also systematically referred to as aminobenzene because it consists of an amino group $(-NH_2)$ attached to a benzene ring.
Therefore,Statement $I$ is correct and Statement $II$ is incorrect.

(A) homoleptic complex is a coordination compound in which all the ligands attached to the central metal atom or ion are of the same type.
In the complex $[Ni(CN)_4]^{2-}$,all four ligands are cyanide $(CN^-)$ ions.
Therefore,$[Ni(CN)_4]^{2-}$ is a homoleptic complex.

(D) The reactivity of a benzene ring towards an electrophilic substitution reaction depends on the electron density of the ring.
Groups that donate electrons to the ring (via resonance or inductive effect) increase the electron density,making it more susceptible to electrophilic attack.
- In $C_6H_6$ (Benzene),there is no substituent.
- In $C_6H_5CH_3$ (Toluene),the $-CH_3$ group is electron-donating via the inductive effect and hyperconjugation.
- In $C_6H_5Cl$ (Chlorobenzene),the $-Cl$ group is electron-withdrawing via the inductive effect,although it donates electrons via resonance.
- In $C_6H_5OH$ (Phenol),the $-OH$ group is strongly electron-donating via resonance (+$M$ effect),which significantly increases the electron density of the benzene ring.
Therefore,phenol is the most reactive towards electrophilic attack.

(A) The van 't Hoff factor $(i)$ for a strong electrolyte like $NaCl$ increases as the concentration decreases.
In more concentrated solutions (like $0.1 \ M$),interionic attractions are stronger,which reduces the effective dissociation.
As the solution becomes more dilute (from $A$ to $C$),the ions move further apart,and $i$ approaches its theoretical maximum value of $2$.
Therefore,the order of the van 't Hoff factor is $i_A < i_B < i_C$.

(A) Statement $I$ is correct: Potassium hydrogen phthalate $(KHP)$ is a widely used primary standard for the standardization of sodium hydroxide $(NaOH)$ solutions because it is a stable,non-hygroscopic solid with a high molar mass.
Statement $II$ is correct: The titration of a weak acid $(KHP)$ with a strong base $(NaOH)$ results in a basic equivalence point $(pH > 7)$. Phenolphthalein,which changes color in the $pH$ range of $8.3$ to $10.1$,is an ideal indicator for this titration.
Therefore,both statements are correct.

(B) The correct matches are as follows:
$(A)$ $CH_3(CH_2)_5COOC_2H_5 \xrightarrow{DIBAL-H, H_2O} CH_3(CH_2)_5CHO$. This is a partial reduction of an ester to an aldehyde using $DIBAL-H$. Thus,$A-(IV)$.
$(B)$ $C_6H_5COC_6H_5 \xrightarrow{Zn(Hg) \& conc. HCl} C_6H_5CH_2C_6H_5$. This is a Clemmensen reduction of a ketone to an alkane. Thus,$B-(II)$.
$(C)$ $C_6H_5CHO \xrightarrow[H_3O^+]{CH_3MgBr} C_6H_5CH(OH)CH_3$. This is a Grignard reaction where an aldehyde is converted to a secondary alcohol. Thus,$C-(I)$.
$(D)$ $CH_3COCH_2COOC_2H_5 \xrightarrow{NaBH_4, H^+} CH_3CH(OH)CH_2COOC_2H_5$. $NaBH_4$ selectively reduces the ketone group to an alcohol in the presence of an ester. Thus,$D-(III)$.
Therefore,the correct sequence is $A-(IV), B-(II), C-(I), D-(III)$.

(A) The $NH_2$ group in Aniline is ortho and para directing and a powerful activating group because the lone pair on the nitrogen atom exhibits a strong $+M$ effect.
Aniline does not undergo Friedel-Craft's reaction (alkylation and acylation) because the lone pair on the nitrogen atom of Aniline reacts with the Lewis acid $AlCl_3$ to form a salt (complex). This results in a positive charge on the nitrogen atom,which acts as a strong deactivating group for the benzene ring,thereby preventing the reaction.

(B) The Nernst equation for the given half-cell reaction is:
$E = E^0_{H^+/H_2} - \frac{0.06}{n} \log \frac{P_{H_2}}{[H^+]^2}$
Given that $E^0_{H^+/H_2} = 0.00 \ V$,$n = 2$,$P_{H_2} = 2 \ atm$,and $[H^+] = 1 \ M$:
$E = 0.00 - \frac{0.06}{2} \log \frac{2}{(1)^2}$
$E = -0.03 \times \log 2$
$E = -0.03 \times 0.3 = -0.009 \ V$
Converting to the form $x \times 10^{-2} \ V$:
$E = -0.9 \times 10^{-2} \ V$
Thus,the value is $0.9$,which is closest to $1$ among the given options.

(D) $SrSO_4$ is white.
$Mg(NH_4)PO_4$ is white.
$BaCrO_4$ is yellow.
$Mn(OH)_2$ is white.
$PbSO_4$ is white.
$PbCrO_4$ is yellow.
$AgBr$ is pale yellow.
$PbI_2$ is yellow.
$CaC_2O_4$ is white.
$[Fe(OH)_2(CH_3COO)]$ is brown-red.
The white coloured salts are $SrSO_4$,$Mg(NH_4)PO_4$,$Mn(OH)_2$,$PbSO_4$,and $CaC_2O_4$.
Total number of white salts = $5$.

(D) The radioactive decay follows first-order kinetics: $\lambda t = \ln \frac{N_0}{N_t}$.
Given that the ratio in the wood sample is $\frac{1}{8}$ of the atmospheric ratio,we have $\frac{N_t}{N_0} = \frac{1}{8}$,which implies $\frac{N_0}{N_t} = 8$.
Using the relation $t = \frac{2.303}{\lambda} \log \frac{N_0}{N_t}$ or $t = \frac{t_{1/2}}{0.693} \ln \frac{N_0}{N_t}$.
Since $\ln 8 = \ln 2^3 = 3 \ln 2$,we get $\lambda t = 3 \ln 2$.
Substituting $\lambda = \frac{\ln 2}{t_{1/2}}$,we have $\frac{\ln 2}{t_{1/2}} \times t = 3 \ln 2$.
Therefore,$t = 3 \times t_{1/2} = 3 \times 5730 \text{ years} = 17190 \text{ years}$.

(B) The $3^{rd}$ ionisation enthalpy $(IE_3)$ corresponds to the removal of an electron from the $M^{2+}$ ion.
Electronic configurations of $M^{2+}$ ions:
$V^{2+}: [Ar] 3d^3$
$Cr^{2+}: [Ar] 3d^4$
$Mn^{2+}: [Ar] 3d^5$
$Fe^{2+}: [Ar] 3d^6$
$Mn^{2+}$ has a stable half-filled $d$-orbital configuration $(3d^5)$. Removing an electron from this stable configuration requires a significantly higher amount of energy compared to the others.
Experimental values $(kJ/mol)$:
$V: 2833$
$Cr: 2990$
$Mn: 3260$
$Fe: 2962$
Thus,$Mn$ has the highest $3^{rd}$ ionisation enthalpy among the given elements.

(A) $Cr^{2+}$ $(d^4)$ is a strong reducing agent because it changes to $Cr^{3+}$ $(d^3)$,which has a stable half-filled $t_{2g}$ configuration.
$Mn^{3+}$ $(d^4)$ is an oxidizing agent because it changes to $Mn^{2+}$ $(d^5)$,which has a stable half-filled $d$-orbital configuration.
Both Assertion and Reason are true,and the Reason correctly explains the stability associated with half-filled configurations that drives these redox processes.

(C) The reactions are as follows:
$A$. Phenol reacts with $Zn$ dust upon heating to form Benzene $(III)$.
$B$. Phenol reacts with $CHCl_3$ and $NaOH$ followed by acidification $(HCl)$ to form Salicylaldehyde $(I)$ (Reimer-Tiemann reaction).
$C$. Phenol reacts with $CO_2$ and $NaOH$ followed by acidification $(HCl)$ to form Salicylic acid $(II)$ (Kolbe-Schmidt reaction).
$D$. Phenol reacts with concentrated $HNO_3$ to form $2,4,6$-trinitrophenol,commonly known as Picric acid $(IV)$.
Therefore,the correct matching is $A-III, B-I, C-II, D-IV$.

(C) The reducing character of hydrides of group $15$ elements increases down the group.
This is because the bond dissociation enthalpy of $E-H$ bonds decreases as the atomic size of the central atom increases.
$BiH_3$ has the longest and weakest $Bi-H$ bond,making it the strongest reducing agent among the given hydrides.

(A) The compound $CuSO_4 \cdot 5 H_2 O$ exhibits colour due to $d-d$ transition.
In $CuSO_4 \cdot 5 H_2 O$,the copper ion is in the $Cu^{2+}$ oxidation state.
The electronic configuration of $Cu^{2+}$ is $[Ar] 3 d^9$.
Since there is an unpaired electron in the $d$-orbital,the $d-d$ transition is possible,which results in the observed colour.
In contrast,$K_2 Cr_2 O_7$,$K_2 CrO_4$,and $KMnO_4$ exhibit colour primarily due to charge transfer transitions,as the metal ions ($Cr^{6+}$ and $Mn^{7+}$) have a $d^0$ configuration.

(B) The boiling point of organic compounds depends on intermolecular forces.
$CH_3CH_2CH_2CH_2OH$ is an alcohol,which exhibits strong intermolecular hydrogen bonding.
$CH_3CH_2CH_2CH_3$ (butane) is a non-polar alkane with weak van der Waals forces.
$CH_3CH_2CH_2CHO$ (butanal) and $C_2H_5OC_2H_5$ (diethyl ether) exhibit dipole-dipole interactions,which are weaker than hydrogen bonding.
Therefore,the alcohol has the highest boiling point.

(B) $Carbon \ tetrachloride \ (CCl_4)$ is used as a fire extinguisher $(A-III)$.
$Methylene \ chloride \ (CH_2Cl_2)$ is used as a paint remover $(B-I)$.
$DDT$ is a non-biodegradable insecticide $(C-IV)$.
$Freons$ are used in refrigerators and air conditioners $(D-II)$.
Therefore,the correct matching is $A-(III), B-(I), C-(IV), D-(II)$.

(B) For $[Co(NH_3)_6]^{3+}$: The central metal ion is $Co^{3+}$ ($3d^6$ configuration). $NH_3$ is a strong field ligand,which causes pairing of electrons in the $d$-orbitals. This results in a $d^2sp^3$ hybridization,forming an inner orbital complex (spin-paired complex).
For $[CoF_6]^{3-}$: The central metal ion is $Co^{3+}$ ($3d^6$ configuration). $F^-$ is a weak field ligand,which does not cause pairing of electrons. This results in an $sp^3d^2$ hybridization,forming an outer orbital complex (spin-free complex).
Therefore,the correct sequence is spin-paired complex and spin-free complex.

(B) The reaction sequence is as follows:
$1$. $C_2H_5Br$ reacts with alcoholic $KOH$ (dehydrohalogenation) to form ethene $(CH_2=CH_2)$ as product $A$.
$2$. Ethene $(A)$ reacts with $Br_2$ in $CCl_4$ (electrophilic addition) to form $1,2-$dibromoethane $(BrCH_2-CH_2Br)$ as product $B$.
$3$. $1,2-$dibromoethane $(B)$ reacts with excess $KCN$ (nucleophilic substitution) to form succinonitrile $(NC-CH_2-CH_2-CN)$ as product $C$.
$4$. Succinonitrile $(C)$ undergoes acid hydrolysis $(H_3O^+)$ to form succinic acid $(HOOC-CH_2-CH_2-COOH)$ as product $D$.

(A) Statement $I$ is false. When $Ni^{2+}$ reacts with dimethyl glyoxime $(dmg)$ in the presence of $NH_4OH$,it forms a complex $[Ni(dmg)_2]$. In this complex,the chelate rings formed are five-membered,not six-membered.
Statement $II$ is true. The chemical formula for Prussian blue is $Fe_4[Fe(CN)_6]_3$. In this compound,the iron outside the coordination sphere is in the $(+3)$ oxidation state $(Fe^{3+})$,and the iron inside the coordination sphere is in the $(+2)$ oxidation state $([Fe(CN)_6]^{4-})$.

(B) For the reaction $A_{(g)} \rightarrow 2 B_{(g)} + C_{(g)}$:
Let the initial pressure of $A$ be $P_0 = 0.1 \ atm$.
At time $t = 115 \ s$,let the decrease in pressure of $A$ be $x$.
Then,the pressures are: $P_A = P_0 - x$,$P_B = 2x$,and $P_C = x$.
The total pressure $P_t = (P_0 - x) + 2x + x = P_0 + 2x$.
Given $P_t = 0.28 \ atm$ at $t = 115 \ s$,we have $0.1 + 2x = 0.28$,which gives $2x = 0.18$,so $x = 0.09 \ atm$.
The pressure of $A$ at $t = 115 \ s$ is $P_A = 0.1 - 0.09 = 0.01 \ atm$.
For a first-order reaction,the rate constant $k = \frac{1}{t} \ln \frac{P_0}{P_A} = \frac{1}{115} \ln \frac{0.1}{0.01} = \frac{1}{115} \ln(10)$.
Using $\ln(10) \approx 2.303$,$k = \frac{2.303}{115} \approx 0.02002 \ s^{-1} = 2.002 \times 10^{-2} \ s^{-1}$.
The nearest integer is $2$.

(C) Let the $3$ different amino acids be $A, B,$ and $C$.
Since each amino acid is used exactly once,the number of possible tripeptides is equivalent to the number of permutations of $3$ distinct items taken $3$ at a time.
This is calculated as $3! = 3 \times 2 \times 1 = 6$.
The possible combinations are: $ABC, ACB, BAC, BCA, CAB, CBA$.