JEE Main 2023 Chemistry Question Paper with Answer and Solution

(C) Boric acid $(H_3BO_3)$ molecules are linked together by strong intermolecular hydrogen bonding,which results in a layered solid structure at room temperature.
In contrast,$BF_3$ is a monomeric covalent molecule with weak van der Waals forces of attraction between its molecules,making it a gas at room temperature.

(D) For an adiabatic process,$q = 0$.
According to the first law of thermodynamics,$\Delta U = q + w$.
Since $q = 0$,$\Delta U = w$.
Given $w = -3000 \, \text{J} \, \text{mol}^{-1}$ (work done by the system).
We know $\Delta U = n C_{V} \Delta T$.
Substituting the values: $1 \times 20 \times (T_{2} - 300) = -3000$.
$T_{2} - 300 = -150$.
$T_{2} = 300 - 150 = 150 \, \text{K}$.

(D) The given reaction is $2 SO_{2(g)} + O_{2(g)} \rightleftharpoons 2 SO_{3(g)}$ with $\Delta H = -190 \ kJ$.
$1$. Increasing temperature: Since the reaction is exothermic $(\Delta H < 0)$,increasing the temperature shifts the equilibrium to the left,decreasing the yield of $SO_3$.
$2$. Increasing pressure: The reaction involves a decrease in the number of moles of gas $(3 \text{ moles of reactants} \rightarrow 2 \text{ moles of products})$. According to Le Chatelier's principle,increasing pressure shifts the equilibrium to the right,increasing the yield of $SO_3$.
$3$. Adding more $SO_2$ or $O_2$: Increasing the concentration of reactants shifts the equilibrium to the right,increasing the yield of $SO_3$.
$4$. Addition of catalyst: $A$ catalyst only increases the rate of reaction and does not change the position of equilibrium or the yield of products.
Therefore,the factors that increase the yield are $B, C,$ and $D$. The total number of such factors is $3$.

(D) The decomposition of $H_2O_2$ is given by: $2H_2O_2 \rightarrow 2H_2O + O_2$.
At $STP$,$2 \ mol$ of $H_2O_2$ $(2 \times 34 \ g = 68 \ g)$ produces $22.7 \ L$ of $O_2$.
Strength of $V$ volume $H_2O_2$ solution is given by the formula: $\text{Strength} (g/L) = \frac{V \times 34}{11.35}$.
Given $V = 50$,so $\text{Strength} = \frac{50 \times 34}{11.35}$.
$\text{Strength} = \frac{1700}{11.35} \approx 149.77 \ g/L$.
Rounding to the nearest integer,we get $150 \ g/L$.

(C) The reaction of chlorine monoxide $(ClO)$ with nitrogen dioxide $(NO_2)$ produces chlorine nitrate $(ClONO_2)$.
$ClO + NO_2 \rightarrow ClONO_2 \, (X)$
Chlorine nitrate $(ClONO_2)$ reacts with water $(H_2O)$ to form hypochlorous acid $(HOCl)$ and nitric acid $(HNO_3)$.
$ClONO_2 + H_2O \rightarrow HOCl \, (Y) + HNO_3 \, (Z)$
Therefore,the correct identification is $X = ClONO_2, Y = HOCl, Z = HNO_3$.

(A) The Rydberg formula for the wavelength $\lambda$ is given by $\frac{1}{\lambda} = R Z^2 \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right)$.
For $He^+$ $(Z = 2)$,the transition from $n_2 = 4$ to $n_1 = 2$ is:
$\frac{1}{\lambda} = R (2)^2 \left( \frac{1}{2^2} - \frac{1}{4^2} \right) = 4R \left( \frac{1}{4} - \frac{1}{16} \right) = 4R \left( \frac{3}{16} \right) = R \left( \frac{3}{4} \right)$.
For Hydrogen $(Z = 1)$,we want the transition from $n_2$ to $n_1$ such that:
$\frac{1}{\lambda} = R (1)^2 \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right) = R \left( \frac{3}{4} \right)$.
Comparing the two,we need $\frac{1}{n_1^2} - \frac{1}{n_2^2} = \frac{3}{4} = 1 - \frac{1}{4} = \frac{1}{1^2} - \frac{1}{2^2}$.
Thus,the transition is from $n = 2$ to $n = 1$.

(C) . $H_2O / CH_2Cl_2 \rightarrow II$. $CH_2Cl_2$ and $H_2O$ are immiscible liquids with different densities,so they can be separated by differential solvent extraction.
$B$. $\alpha$-tetralone / $p$-nitrophenol $\rightarrow III$. Due to the presence of an $-OH$ group in $p$-nitrophenol,it exhibits hydrogen bonding and has different polarity compared to $\alpha$-tetralone,allowing separation by column chromatography.
$C$. Kerosene / Naphthalene $\rightarrow IV$. Due to the difference in boiling points,they can be separated by fractional distillation.
$D$. $C_6H_{12}O_6 / NaCl \rightarrow I$. $NaCl$ (ionic compound) can be separated from glucose by crystallization based on solubility differences.

(D) For isoelectronic species (species with the same number of electrons),the ionic radius decreases as the nuclear charge $(Z)$ increases.
All the given ions,$S^{2-}$,$Cl^{-}$,$K^{+}$,and $Ca^{2+}$,have $18$ electrons.
Their respective nuclear charges $(Z)$ are $16, 17, 19,$ and $20$.
As the nuclear charge increases,the attraction between the nucleus and the electrons increases,causing the ionic radius to decrease.
Therefore,the correct increasing order of ionic radii is: $Ca^{2+} < K^{+} < Cl^{-} < S^{2-}$.

(A) reducing agent is a substance that undergoes oxidation (increase in oxidation state) and reduces another substance.
In the reaction $2 NaOCl + H_2O_2 \rightarrow 2 NaCl + H_2O + O_2$,the oxidation state of oxygen in $H_2O_2$ increases from $-1$ to $0$ (in $O_2$).
Since $H_2O_2$ is oxidized,it acts as a reducing agent.
In the other reactions (options $B$,$C$,and $D$),$H_2O_2$ acts as an oxidizing agent because the oxidation state of oxygen decreases from $-1$ to $-2$.

(D) To determine the geometry of the molecules,we use the $VSEPR$ theory:
$1$. $XeF_4$: The central atom $Xe$ has $8$ valence electrons. It forms $4$ bonds with $F$ atoms and has $2$ lone pairs. The steric number is $6$ ($sp^3d^2$ hybridization),resulting in a square planar geometry $(A-II)$.
$2$. $SF_4$: The central atom $S$ has $6$ valence electrons. It forms $4$ bonds with $F$ atoms and has $1$ lone pair. The steric number is $5$ ($sp^3d$ hybridization),resulting in a see-saw geometry $(B-I)$.
$3$. $NH_4^{+}$: The central atom $N$ has $5$ valence electrons. It forms $4$ bonds with $H$ atoms and has no lone pairs. The steric number is $4$ ($sp^3$ hybridization),resulting in a tetrahedral geometry $(C-IV)$.
$4$. $BrF_3$: The central atom $Br$ has $7$ valence electrons. It forms $3$ bonds with $F$ atoms and has $2$ lone pairs. The steric number is $5$ ($sp^3d$ hybridization),resulting in a bent $T$-shaped geometry $(D-III)$.
Thus,the correct matching is $A-II, B-I, C-IV, D-III$.

(B) $1$. First,the $trans$-alkene reacts with $Br_2$ to form a vicinal dibromide: $Ph-CH(Br)-CH(Br)-CH_3$.
$2$. Next,dehydrohalogenation occurs using $alc. KOH$ followed by $NaNH_2$ to remove two moles of $HBr$,resulting in the formation of an alkyne: $Ph-C \equiv C-CH_3$.
$3$. Finally,the partial hydrogenation of the alkyne to a $cis$-alkene is achieved using $H_2$ with $Lindlar \text{ } Catalyst$ (a poisoned $Pd$ catalyst).
$4$. Therefore,the correct sequence is $Br_2, \text{alc. } KOH, NaNH_2, H_2, \text{Lindlar Catalyst}$.

(D) The process involves three steps:
$1. \text{Dissociation of } \frac{1}{2} Cl_{2(g)}$ $\rightarrow Cl_{(g)}: \Delta H_1 = \frac{1}{2} \times 240 = 120 \, kJ \, mol^{-1}$
$2. \text{Electron gain of } Cl_{(g)} \rightarrow Cl^{-}_{(g)}: \Delta H_2 = -350 \, kJ \, mol^{-1}$
$3. \text{Hydration of } Cl^{-}_{(g)} \rightarrow Cl^{-}_{(aq)}: \Delta H_3 = -380 \, kJ \, mol^{-1}$
Total enthalpy change $\Delta H^{\circ} = \Delta H_1 + \Delta H_2 + \Delta H_3$
$\Delta H^{\circ} = 120 + (-350) + (-380) = -610 \, kJ \, mol^{-1}$
The magnitude is $610$.

(D) The mass of $C$ in $CO_2$ is calculated using the molar mass ratio:
Mass of $C = (12 / 44) \times 0.792 \ g = 0.216 \ g$.
The percentage of carbon in the organic compound is given by:
$\% \text{ of } C = (\text{Mass of } C / \text{Mass of compound}) \times 100$.
$\% \text{ of } C = (0.216 / 0.492) \times 100 = 43.90 \%$.
Rounding to the nearest integer,we get $44 \%$.
Thus,the correct option is $D$.

(D) The balanced chemical equation is: $Zn(s) + 2HCl(aq) \rightarrow ZnCl_2(aq) + H_2(g) \uparrow$
Moles of $Zn$ used $= \frac{\text{mass}}{\text{molar mass}} = \frac{11.5 \ g}{65.4 \ g \ mol^{-1}} \approx 0.1758 \ mol$.
According to the stoichiometry of the reaction,$1 \ mol$ of $Zn$ produces $1 \ mol$ of $H_2$.
Therefore,moles of $H_2$ produced $= 0.1758 \ mol$.
Volume of $H_2$ at $STP = \text{moles} \times \text{molar volume at } STP = 0.1758 \ mol \times 22.7 \ L \ mol^{-1} \approx 3.99 \ L$.
Rounding to the nearest integer,the volume is $4 \ L$.

(D) The relationship between $K_P$ and $K_C$ is given by the equation: $K_P = K_C(RT)^{\Delta n_g}$.
For the reaction $SO_{2(g)} + \frac{1}{2} O_{2(g)} \rightleftharpoons SO_{3(g)}$,the change in the number of moles of gaseous species is $\Delta n_g = 1 - (1 + 0.5) = -0.5$.
Given $T = 27^{\circ} C = 300 \ K$,$R = 0.082 \ L \ atm \ K^{-1} \ mol^{-1}$,and $K_P = 2 \times 10^{12}$.
Substituting the values: $2 \times 10^{12} = K_C \times (0.082 \times 300)^{-0.5}$.
$2 \times 10^{12} = K_C \times (24.6)^{-0.5} = K_C \times \frac{1}{\sqrt{24.6}}$.
$K_C = 2 \times 10^{12} \times \sqrt{24.6} \approx 2 \times 10^{12} \times 4.96 = 9.92 \times 10^{12}$.
Expressing in terms of $10^{13}$: $K_C = 0.992 \times 10^{13}$.
Rounding to the nearest integer,we get $1 \times 10^{13}$.

(D) The chemical formula for hypophosphoric acid is $H_4P_2O_6$.
To calculate the oxidation state of phosphorus $(P)$,let its oxidation state be $x$.
The oxidation state of hydrogen $(H)$ is $+1$ and oxygen $(O)$ is $-2$.
Applying the rule: $4(+1) + 2(x) + 6(-2) = 0$
$4 + 2x - 12 = 0$
$2x - 8 = 0$
$2x = 8$
$x = +4$
Therefore,the oxidation state of phosphorus in hypophosphoric acid is $+4$.

(D) Step $1$: Calculate the number of moles of each gas.
$n_X = \frac{0.6 \ g}{20 \ g \ mol^{-1}} = 0.03 \ mol$
$n_Y = \frac{0.45 \ g}{45 \ g \ mol^{-1}} = 0.01 \ mol$
Step $2$: Calculate the mole fraction of gas $X$ $(\chi_X)$.
$\chi_X = \frac{n_X}{n_X + n_Y} = \frac{0.03}{0.03 + 0.01} = \frac{0.03}{0.04} = 0.75$
Step $3$: Calculate the partial pressure of gas $X$ $(P_X)$.
$P_X = \chi_X \times P_{total} = 0.75 \times 740 \ mm \ Hg = 555 \ mm \ Hg$

(A)

Indicator	$pH$ range
Methyl orange	$3.2-4.5$
Phenolphthalein	$8.3-10.5$

Indicators are chosen based on the $pH$ range of the equivalence point of the titration.
$1$. Methyl orange is suitable for titrations where the equivalence point falls in the acidic range $(pH \approx 3-5)$,such as strong acid vs. weak base.
$2$. Phenolphthalein is suitable for titrations where the equivalence point falls in the basic range $(pH \approx 8-10)$,such as weak acid vs. strong base.
$3$. Both indicators can be used for strong acid vs. strong base titrations because the $pH$ change at the equivalence point is very sharp $(pH \approx 4-10)$.
$4$. For weak acid vs. weak base titrations,there is no sharp change in $pH$ at the equivalence point,so no simple indicator is suitable. Therefore,the statement that Methyl orange may be used for a weak acid vs. weak base titration is incorrect.

(D) $1$. The hydrocarbon '$X$' has the molecular formula $C_6H_8$.
$2$. It consumes two moles of $H_2$ upon catalytic hydrogenation,indicating the presence of two double bonds.
$3$. Ozonolysis of '$X$' yields two moles of methane dicarbaldehyde $(OHC-CH_2-CHO)$.
$4$. Methane dicarbaldehyde is also known as malonaldehyde.
$5$. Cyclohexa$-1, 4-$diene undergoes ozonolysis to break the two double bonds,resulting in two molecules of $OHC-CH_2-CHO$.
$6$. Therefore,the hydrocarbon '$X$' is cyclohexa$-1, 4-$diene.

(B) The energy of an orbital is determined by the $(n+l)$ rule.
$A$. $n = 3, l = 0, m = 0 \implies (n+l) = 3 + 0 = 3$ ($3s$ orbital)
$B$. $n = 4, l = 0, m = 0 \implies (n+l) = 4 + 0 = 4$ ($4s$ orbital)
$C$. $n = 3, l = 1, m = 0 \implies (n+l) = 3 + 1 = 4$ ($3p$ orbital)
$D$. $n = 3, l = 2, m = 1 \implies (n+l) = 3 + 2 = 5$ ($3d$ orbital)
Comparing $(n+l)$ values: $D (5) > B (4) = C (4) > A (3)$.
For $B$ and $C$,both have $(n+l) = 4$. According to the rule,the orbital with the higher $n$ value has higher energy. Thus,$B (n=4) > C (n=3)$.
The decreasing order of energy is $D > B > C > A$.

(D) The Lewis acidity of boron trihalides is determined by the extent of $p\pi-p\pi$ back bonding between the halogen atom and the boron atom.
In $BF_3$,the $2p-2p$ overlap is most effective,resulting in the strongest back bonding,which reduces the electron deficiency of the boron atom.
As we move from $BF_3$ to $BI_3$,the size of the halogen atom increases,making the $p\pi-p\pi$ overlap less effective $(2p-2p > 2p-3p > 2p-4p > 2p-5p)$.
Consequently,the extent of back bonding decreases,and the Lewis acidity increases in the order: $BF_3 < BCl_3 < BBr_3 < BI_3$.

(B) $Ca^{2+}$ ions play a significant role in neuromuscular function,interneuronal transmission,and cell membrane integrity.
They are essential for the release of neurotransmitters at the synaptic junction.

(A) $H_2O_2$ is widely used in industries for various purposes.
It is used in the synthesis of hydroquinone,tartaric acid,and certain food products and pharmaceuticals,including $Cephalosporin$.
Additionally,it is used for the restoration of aerobic conditions to sewage wastes.
Therefore,both Statement $I$ and Statement $II$ are correct.

(A) Normally,rain water has a $pH$ of $5.6$ due to the presence of $H^{+}$ ions formed by the reaction of rain water with atmospheric carbon dioxide.
$H_2O_{(l)} + CO_{2_{(g)}} \rightleftharpoons H_2CO_{3_{(aq)}}$
$H_2CO_{3_{(aq)}} \rightleftharpoons H^{+}_{(aq)} + HCO_{3^{-(aq)}}$

(A) The general combustion reaction for a hydrocarbon $C_xH_y$ is given by:
$C_xH_y + (x + \frac{y}{4}) O_2 \rightarrow xCO_2 + \frac{y}{2} H_2O$
Given that the reaction produces $4$ equivalents of water:
$\frac{y}{2} = 4 \implies y = 8$
Given that the reaction requires $11$ equivalents of oxygen:
$x + \frac{y}{4} = 11$
Substituting $y = 8$:
$x + \frac{8}{4} = 11$
$x + 2 = 11$
$x = 9$
Therefore,the molecular formula of the hydrocarbon $A$ is $C_9H_8$.

(B) In the Dumas method,the nitrogen-containing organic compound is heated with $CuO$ in an atmosphere of $CO_2$,which yields free $N_2$ gas along with $CO_2$ and $H_2O$.
The chemical reaction is: $C_xH_yN_z + (2x + \frac{y}{2}) CuO \rightarrow x CO_2 + \frac{y}{2} H_2O + \frac{z}{2} N_2 + (2x + \frac{y}{2}) Cu$.
Any traces of nitrogen oxides $(NO_x)$ that may be formed during the combustion are reduced back to $N_2$ gas by passing the gaseous mixture over a heated copper gauze.

(C) To determine the shape of the given species,we use the $VSEPR$ theory:
$1$. $XeF_2$: $sp^3d$ hybridization,$3$ lone pairs on $Xe$,shape is linear.
$2$. $I_3^{+}$: $sp^3$ hybridization,$2$ lone pairs on central $I$,shape is bent ($V$-shape).
$3$. $C_3O_2$: $O=C=C=C=O$,central carbon is $sp$ hybridized,shape is linear.
$4$. $I_3^{-}$: $sp^3d$ hybridization,$3$ lone pairs on central $I$,shape is linear.
$5$. $CO_2$: $O=C=O$,$sp$ hybridization,shape is linear.
$6$. $SO_2$: $sp^2$ hybridization,$1$ lone pair on $S$,shape is bent ($V$-shape).
$7$. $BeCl_2$: $sp$ hybridization,no lone pairs on $Be$,shape is linear.
$8$. $BCl_2^{\Theta}$: $sp^2$ hybridization,$1$ lone pair on $B$,shape is bent ($V$-shape).
The linear species are: $XeF_2, C_3O_2, I_3^{-}, CO_2, BeCl_2$.
Total number of linear species = $5$.

(C) The molar mass of $AgCl = 107.9 + 35.5 = 143.4 \ g \ mol^{-1}$.
Solubility in $mol \ L^{-1}$ $(S)$ = $\frac{1.434 \times 10^{-3} \ g \ L^{-1}}{143.4 \ g \ mol^{-1}} = 10^{-5} \ mol \ L^{-1}$.
For $AgCl_{(s)} \rightleftharpoons Ag^{+}_{(aq)} + Cl^{-}_{(aq)}$,the solubility product $K_{sp} = [Ag^{+}][Cl^{-}] = S^2$.
$K_{sp} = (10^{-5})^2 = 10^{-10}$.
Therefore,$-\log K_{sp} = -\log(10^{-10}) = 10$.

(C) Let the number of $M^{2+}$ ions be $x$. Then the number of $M^{3+}$ ions is $(0.83 - x)$.
Since the compound is electrically neutral,the total positive charge must equal the total negative charge:
$2x + 3(0.83 - x) = 2$
$2x + 2.49 - 3x = 2$
$-x = -0.49$
$x = 0.49$
Thus,the number of $M^{2+}$ ions is $0.49$.
The percentage of $M^{2+}$ ions is given by:
$\% M^{2+} = \frac{0.49}{0.83} \times 100 \approx 59.036\%$
Rounding to the nearest integer,we get $59\%$.

(C) The balanced chemical equation is: $2C_{(s)} + O_{2(g)} \rightarrow 2CO_{(g)}$
Moles of $C = \frac{12 \ g}{12 \ g \ mol^{-1}} = 1 \ mol$.
Moles of $O_2 = \frac{48 \ g}{32 \ g \ mol^{-1}} = 1.5 \ mol$.
According to the stoichiometry,$2 \ mol$ of $C$ requires $1 \ mol$ of $O_2$. Therefore,$1 \ mol$ of $C$ requires $0.5 \ mol$ of $O_2$.
Since we have $1.5 \ mol$ of $O_2$,$C$ is the limiting reagent.
From the equation,$2 \ mol$ of $C$ produces $2 \ mol$ of $CO$,so $1 \ mol$ of $C$ produces $1 \ mol$ of $CO$.
Volume of $1 \ mol$ of $CO$ at $STP$ = $22.7 \ L = 227 \times 10^{-1} \ L$.

(C) The ionization enthalpies of the given alkali metals are: $Li \approx 520 \ kJ \ mol^{-1}$,$K \approx 419 \ kJ \ mol^{-1}$,$Rb \approx 403 \ kJ \ mol^{-1}$,and $Cs \approx 376 \ kJ \ mol^{-1}$.
Among these,$Li$,$K$,and $Rb$ have ionization enthalpy greater than $400 \ kJ \ mol^{-1}$.
Regarding the formation of stable superoxides $(MO_2)$,only the larger alkali metals $K$,$Rb$,and $Cs$ form stable superoxides.
Comparing both conditions,only $K$ and $Rb$ satisfy both criteria (ionization enthalpy $> 400 \ kJ \ mol^{-1}$ and formation of stable superoxide).
Therefore,the number of such metals is $2$.

(C) The enthalpy of reaction is calculated using the formula: $\Delta_{r}H^{\circ} = \sum \Delta_{f}H^{\circ}(\text{products}) - \sum \Delta_{f}H^{\circ}(\text{reactants})$
For the reaction: $CCl_{4(g)} + 2H_2O_{(g)} \rightarrow CO_{2(g)} + 4HCl_{(g)}$
$\Delta_{r}H^{\circ} = [\Delta_{f}H^{\circ}(CO_{2(g)}) + 4 \times \Delta_{f}H^{\circ}(HCl_{(g)})] - [\Delta_{f}H^{\circ}(CCl_{4(g)}) + 2 \times \Delta_{f}H^{\circ}(H_2O_{(g)})]$
Substituting the given values:
$\Delta_{r}H^{\circ} = [-394 + 4 \times (-92)] - [-105 + 2 \times (-242)]$
$\Delta_{r}H^{\circ} = [-394 - 368] - [-105 - 484]$
$\Delta_{r}H^{\circ} = -762 - (-589)$
$\Delta_{r}H^{\circ} = -762 + 589 = -173 \ kJ \ mol^{-1}$
The magnitude of the enthalpy of the reaction is $173 \ kJ \ mol^{-1}$.

(C) . Tranquilizers are used as antidepressant drugs $(A-III)$.
$B$. Aspirin acts as an anti-blood clotting agent $(B-I)$.
$C$. Salvarsan is an early antibiotic $(C-II)$.
$D$. Soframicine is an antiseptic $(D-IV)$.
Therefore,the correct matching is $A-III, B-I, C-II, D-IV$.

(A) Statement $I$ is true: Chlorine oxides like $Cl_2O$,$ClO_2$,$Cl_2O_6$,and $Cl_2O_7$ are highly reactive oxidizing agents and are known to be unstable and prone to explosion.
Statement $II$ is true: The chemical reactivity of elements is often assessed by observing their ability to react with oxygen and halogens,which helps in determining their oxidation states and bonding characteristics.
Therefore,both statements are correct.

(D) Resonating structures are hypothetical and do not exist in reality.
Resonance hybrid is the real structure which is the weighted average of all the resonating structures.
Therefore,the $CO_3^{2-}$ ion exists as a single resonance hybrid structure.

(C) Photochemical smog is caused by the presence of nitrogen oxides $(NO_x)$ and hydrocarbons in the atmosphere.
To control it,we must reduce the emission of these precursors.
Using catalytic converters in automobiles and industries helps in controlling the release of nitrogen oxides and hydrocarbons,thereby reducing the formation of photochemical smog.
Therefore,option $C$ is the correct answer.

(C) The correct matches are as follows:
$A$. Slaked lime is $Ca(OH)_2$ $(II)$.
$B$. Dead burnt plaster is $CaSO_4$ $(IV)$.
$C$. Caustic soda is $NaOH$ $(I)$.
$D$. Washing soda is $Na_2CO_3 \cdot 10H_2O$ $(III)$.
Therefore,the correct matching is $A-II, B-IV, C-I, D-III$.

(B) $A.$ Beryllium oxide $(BeO)$ is amphoteric in nature, not purely acidic.
$B.$ Beryllium carbonate $(BeCO_3)$ is thermally unstable and decomposes easily, so it is kept in an atmosphere of $CO_2$ to prevent decomposition.
$C.$ Beryllium sulphate $(BeSO_4)$ is readily soluble in water due to the high hydration enthalpy of the small $Be^{2+}$ ion.
$D.$ Beryllium shows anomalous behavior compared to other alkaline earth metals due to its small atomic size, high ionization energy, and high polarising power $(\phi)$.
Therefore, statements $B, C,$ and $D$ are correct.

(C, D) The reaction of $but-2-yne$ with $Na/liq. NH_3$ (Birch reduction) yields $trans-but-2-ene$ $(B)$.
The reaction of $but-2-yne$ with $Pd/C$ (Lindlar's catalyst) yields $cis-but-2-ene$ $(A)$.
$A$ $(cis-but-2-ene)$ has a dipole moment $\mu \neq 0$,while $B$ $(trans-but-2-ene)$ has $\mu = 0$.
Statement $A$: $cis$ isomers are generally more polar and have higher boiling points,but solubility depends on crystal packing; $trans$ isomers $(B)$ pack better,leading to higher melting points.
Statement $B$: $A$ $(cis)$ has a higher boiling point than $B$ $(trans)$,but $B$ has a higher melting point due to better symmetry.
Statement $C$: $A$ is more polar than $B$ because $A$ has a non-zero dipole moment,while $B$ is zero. The statement claims $A$ is more polar because its dipole moment is zero,which is false.
Statement $D$: Both $A$ and $B$ are alkenes and react similarly with $Br_2$; there is no significant difference in ease of addition. Thus,$D$ is also incorrect.
Since $C$ and $D$ are incorrect,and no option matches,the question is flawed.

(B) Assertion $A$ is true because the combustion of hydrogen produces only water $(H_2O)$ as a byproduct,making it an environment-friendly fuel.
Reason $R$ is also true because the atomic number of hydrogen is $1$ and it is indeed a very light element.
However,the fact that hydrogen is a light element with atomic number $1$ is not the reason why it is environment-friendly; its environment-friendly nature is due to the clean combustion product $(H_2O)$.
Therefore,both $A$ and $R$ are true,but $R$ is not the correct explanation of $A$.

(A) Calculation of de Broglie wavelength: $\lambda = \frac{h}{mv} = \frac{6 \times 10^{-34} \, Js}{(9 \times 10^{-31} \, kg) \times (1000 \, m \, s^{-1})} = 6.6667 \times 10^{-7} \, m = 666.67 \, nm$. Thus,statement $(A)$ is true.
$(B)$ The characteristics of cathode rays (electrons) are independent of the material of the electrodes. Thus,statement $(B)$ is false.
$(C)$ Cathode rays originate from the cathode and travel towards the anode. Thus,statement $(C)$ is true.
$(D)$ The nature of cathode rays is independent of the nature of the gas present in the tube. Thus,statement $(D)$ is false.
Therefore,there are $2$ true statements ($A$ and $C$).

(A) In $HBrO_3$ (bromic acid),let the oxidation state of $Br$ be $x$.
$1 + x + 3(-2) = 0 \implies x - 5 = 0 \implies x = +5$.
In $HBrO_4$ (perbromic acid),let the oxidation state of $Br$ be $y$.
$1 + y + 4(-2) = 0 \implies y - 7 = 0 \implies y = +7$.
The sum of the oxidation states is $5 + 7 = 12$.

(A) For reaction $(i): X(g) \rightleftharpoons Y(g) + Z(g)$

$Initial \ moles$	$n$	$0$	$0$
$Equilibrium \ moles$	$n(1-\alpha)$	$n\alpha$	$n\alpha$

Total moles at equilibrium $= n(1-\alpha) + n\alpha + n\alpha = n(1+\alpha)$.
Partial pressures: $p_X = \frac{1-\alpha}{1+\alpha} p_1$,$p_Y = \frac{\alpha}{1+\alpha} p_1$,$p_Z = \frac{\alpha}{1+\alpha} p_1$.
$K_{p1} = \frac{p_Y p_Z}{p_X} = \frac{\alpha^2 p_1}{1-\alpha^2} = 3$
For reaction $(ii): A(g) \rightleftharpoons 2B(g)$

$Initial \ moles$	$n$	$0$
$Equilibrium \ moles$	$n(1-\alpha)$	$2n\alpha$

Total moles at equilibrium $= n(1-\alpha) + 2n\alpha = n(1+\alpha)$.
Partial pressures: $p_A = \frac{1-\alpha}{1+\alpha} p_2$,$p_B = \frac{2\alpha}{1+\alpha} p_2$.
$K_{p2} = \frac{p_B^2}{p_A} = \frac{4\alpha^2 p_2}{1-\alpha^2} = 1$
Dividing the two expressions: $\frac{K_{p1}}{K_{p2}} = \frac{3}{1} = \frac{\alpha^2 p_1 / (1-\alpha^2)}{4\alpha^2 p_2 / (1-\alpha^2)} = \frac{p_1}{4p_2}$
Therefore,$\frac{p_1}{p_2} = 3 \times 4 = 12$.
Thus,$x = 12$.

(A) $1$. The first compound is $4$-benzylidenecyclohexanecarboxylic acid. It lacks a plane of symmetry $(POS)$ and center of symmetry $(COS)$,making it chiral.
$2$. The second compound is a substituted ether derivative. It contains a chiral center and lacks $POS$ and $COS$,making it chiral.
$3$. The third compound is tartaric acid (meso form). It has a plane of symmetry $(POS)$,making it achiral.
$4$. The fourth compound is myo-inositol. It has a plane of symmetry $(POS)$,making it achiral.
$5$. The fifth compound is $2$-methylenecyclopropane-$1,3$-dicarboxylic acid. It has a plane of symmetry $(POS)$,making it achiral.
$6$. Thus,there are $2$ chiral compounds.

(A) The radioactive decay follows first-order kinetics: $[N]_t = [N]_0 e^{-kt}$,where $k = \frac{\ln 2}{t_{1/2}}$.
For substance $A$: $[A]_t = [A]_0 e^{-k_A t}$,where $k_A = \frac{\ln 2}{15}$.
For substance $B$: $[B]_t = [B]_0 e^{-k_B t}$,where $k_B = \frac{\ln 2}{5}$ and $[B]_0 = 4[A]_0$.
We want to find time $t$ such that $[A]_t = [B]_t$.
$[A]_0 e^{-(\frac{\ln 2}{15})t} = 4[A]_0 e^{-(\frac{\ln 2}{5})t}$.
Dividing both sides by $[A]_0 e^{-(\frac{\ln 2}{5})t}$:
$e^{(\frac{\ln 2}{5} - \frac{\ln 2}{15})t} = 4$.
Taking the natural logarithm on both sides:
$t(\frac{\ln 2}{5} - \frac{\ln 2}{15}) = \ln 4$.
$t(\ln 2)(\frac{1}{5} - \frac{1}{15}) = 2 \ln 2$.
$t(\frac{3-1}{15}) = 2$.
$t(\frac{2}{15}) = 2$.
$t = 15 \ min$.

(A) To find the enthalpy change for the reaction $H_2O_{(g)} \rightarrow H_{(g)} + OH_{(g)}$, we use Hess's Law by manipulating the given equations:
$1$. $2 H_2O_{(g)} \rightarrow 2 H_{2(g)} + O_{2(g)} \quad \Delta H^{\circ} = -2 \times (-242) = +484 \ kJ \ mol^{-1}$
$2$. $H_{2(g)} + O_{2(g)} \rightarrow 2 OH_{(g)} \quad \Delta H^{\circ} = +78 \ kJ \ mol^{-1}$
$3$. $H_{2(g)} \rightarrow 2 H_{(g)} \quad \Delta H^{\circ} = +436 \ kJ \ mol^{-1}$
Adding these equations:
$2 H_2O_{(g)} + H_{2(g)} + O_{2(g)} + H_{2(g)} \rightarrow 2 H_{2(g)} + O_{2(g)} + 2 OH_{(g)} + 2 H_{(g)}$
Simplifying, we get: $2 H_2O_{(g)} \rightarrow 2 H_{(g)} + 2 OH_{(g)}$
$\Delta H^{\circ} = 484 + 78 + 436 = +998 \ kJ \ mol^{-1}$
For the reaction $H_2O_{(g)} \rightarrow H_{(g)} + OH_{(g)}$, the enthalpy change is:
$X = \frac{998}{2} = 499 \ kJ \ mol^{-1}$

(D) Bond enthalpy is inversely proportional to the atomic size of the elements involved in the bond. \\ As we move down the group $14$ from $C$ to $Sn$,the atomic size increases,which leads to a decrease in the strength of the covalent bond formed between the atoms. \\ The atomic radii order is $C < Si < Ge < Sn$. \\ Therefore,the bond enthalpy order is $C - C > Si - Si > Ge - Ge > Sn - Sn$.

(A) The reaction involves the electrophilic addition of benzene to methylenecyclohexane in the presence of an acid catalyst $(HF)$.
$1$. First,the $HF$ protonates the double bond of methylenecyclohexane to form a stable tertiary carbocation,the $1$-methylcyclohexyl cation.
$2$. This carbocation then acts as an electrophile and undergoes an electrophilic aromatic substitution $(EAS)$ reaction with benzene.
$3$. The final product formed is $1$-methyl-$1$-phenylcyclohexane.

(D) Sulphanilic acid exists as a zwitterion $(H_3N^+-C_6H_4-SO_3^-)$. Due to the presence of the strong acidic $-SO_3H$ group and basic $-NH_2$ group,it does not behave as a typical carboxylic acid and does not undergo esterification.
Sulphanilic acid contains $C, H, N, O,$ and $S$. In Lassaigne's test,if both nitrogen and sulphur are present,they react to form sodium thiocyanate $(NaSCN)$. This compound reacts with $Fe^{3+}$ ions to form a blood-red colouration $([Fe(SCN)]^{2+})$,which confirms the presence of both $N$ and $S$.

(D) Assertion $(A)$ is true because gypsum $(CaSO_4 \cdot 2H_2O)$ is used in the construction of fireproof wall boards due to its low thermal conductivity and the presence of water of crystallization.
Reason $(R)$ is also true because gypsum loses its water of crystallization when heated to high temperatures (forming Plaster of Paris at $393 \ K$ and anhydrous calcium sulfate at higher temperatures),which makes it lose its structural integrity.
Since the fireproof nature of gypsum boards is specifically due to the endothermic release of water vapor when exposed to heat,which prevents the temperature of the wall from rising rapidly,the reason $R$ correctly explains why it is used for fireproofing. Thus,both $A$ and $R$ are true and $R$ is the correct explanation of $A$.

(A) van't Hoff factor,$i = \frac{\text{Normal molar mass}}{\text{Abnormal molar mass}}$,which corresponds to $III$.
$(B)$ $k_{f}$ is the Cryoscopic constant (molal depression constant),which corresponds to $I$.
$(C)$ Solutions with the same osmotic pressure are known as Isotonic solutions,which corresponds to $II$.
$(D)$ Azeotropes are binary mixtures having the same composition in liquid and vapour phases,which corresponds to $IV$.
Therefore,the correct matching is $A-III, B-I, C-II, D-IV$.

(C) The unit of the rate constant for a reaction of order $n$ is given by $(\text{mol} \ L^{-1})^{1-n} \ s^{-1}$.
Given the unit is $L \ mol^{-1} \ s^{-1}$,which is equivalent to $(\text{mol} \ L^{-1})^{-1} \ s^{-1}$.
Comparing the exponents: $1 - n = -1$.
Therefore,$n = 2$.
The reaction is of second order.

(C) The nature of the given oxides is as follows:
$1$. $N_2O_3$: Acidic
$2$. $NO_2$: Acidic
$3$. $N_2O$: Neutral
$4$. $Cl_2O_7$: Acidic
$5$. $SO_2$: Acidic
$6$. $CO$: Neutral
$7$. $CaO$: Basic
$8$. $Na_2O$: Basic
$9$. $NO$: Neutral
Therefore,the acidic oxides are $N_2O_3, NO_2, Cl_2O_7$ and $SO_2$.
The total number of acidic oxides is $4$.

(A) Fehling's reagent contains a copper$(II)$ complex with tartrate ions $(C_4H_4O_6^{2-})$.
In this complex,the tartrate ion acts as a bidentate ligand,meaning it coordinates to the central metal ion through two donor atoms (typically the carboxylate oxygen and the hydroxyl oxygen).
Therefore,the denticity of the tartrate ligand is $2$.

(C) In a hexagonal close-packed $(hcp)$ structure,the number of octahedral voids is equal to the number of atoms $(N)$,and the number of tetrahedral voids is equal to $2N$.
Total number of voids per atom $= N + 2N = 3N$.
Given amount of metal $= 0.02 \ mol$.
Number of atoms $= 0.02 \times 6.02 \times 10^{23} = 1.204 \times 10^{22}$.
Total number of voids $= 3 \times (1.204 \times 10^{22}) = 3.612 \times 10^{22} = 36.12 \times 10^{21}$.
The nearest integer is $36$.

(C) The cell reaction is $Zn_{(s)} + Sn^{2+}_{(aq)} \rightleftharpoons Zn^{2+}_{(aq)} + Sn_{(s)}$.
For this reaction,the number of electrons transferred is $n = 2$.
The relationship between standard cell potential and equilibrium constant is given by $E_{cell}^0 = \frac{0.059}{n} \log_{10} K_{eq}$.
Substituting the values: $E_{cell}^0 = \frac{0.059}{2} \log_{10} (1 \times 10^{20}) = \frac{0.059 \times 20}{2} = 0.59 \ V$.
We know that $E_{cell}^0 = E_{cathode}^0 - E_{anode}^0 = E_{Sn^{2+}/Sn}^0 - E_{Zn^{2+}/Zn}^0$.
$0.59 \ V = E_{Sn^{2+}/Sn}^0 - (-0.76 \ V)$.
$E_{Sn^{2+}/Sn}^0 = 0.59 - 0.76 = -0.17 \ V$.
The standard electrode potential for $Sn/Sn^{2+}$ is the negative of $E_{Sn^{2+}/Sn}^0$,so $E_{Sn/Sn^{2+}}^0 = -(-0.17 \ V) = 0.17 \ V$.
$0.17 \ V = 17 \times 10^{-2} \ V$.
Thus,the magnitude is $17$.

(C) Seliwanoff's test is a chemical test used to distinguish between aldose and ketose sugars.
Ketoses are more rapidly dehydrated in the presence of concentrated acid to form $5$-hydroxymethylfurfural compared to aldoses.
This intermediate then reacts with resorcinol to produce a cherry-red colored complex.
Therefore,Assertion $(A)$ is true.
However,the mechanism involves acid-catalyzed dehydration,not $\beta$-elimination to form furfural. Thus,Reason $(R)$ is false.

(D) The copper ore contains iron as an impurity.
It is mixed with silica $(SiO_2)$ before heating in a reverberatory furnace.
$FeO$ acts as a basic impurity and reacts with acidic silica to form iron silicate slag $(FeSiO_3)$,which can be easily removed.
$FeO + SiO_2 \longrightarrow FeSiO_3$

(A) $1.$ Ranitidine: Antacid
$2.$ Meprobamate: Tranquilizer
$3.$ Terfenadine: Antihistamine
$4.$ Brompheniramine: Antihistamine

(C) The synthesis of benzyl isocyanide $(C_6H_5CH_2NC)$ can be achieved through the following reactions:
$(A)$ Benzyl bromide $(C_6H_5CH_2Br)$ reacts with $AgCN$ to form benzyl isocyanide $(C_6H_5CH_2NC)$ as the major product because $AgCN$ is a covalent compound.
$(B)$ Benzylamine $(C_6H_5CH_2NH_2)$ is a primary amine. Primary amines undergo the carbylamine reaction when heated with chloroform $(CHCl_3)$ and alcoholic $KOH$ to form isocyanides $(R-NC)$. Thus,benzylamine gives benzyl isocyanide.
$(C)$ $N$-methylbenzylamine is a secondary amine and does not undergo the carbylamine reaction.
$(D)$ Benzyl tosylate $(C_6H_5CH_2OTs)$ reacts with $KCN$ (an ionic compound) to form benzyl cyanide $(C_6H_5CH_2CN)$ as the major product.
Therefore,benzyl isocyanide is obtained in reactions $(A)$ and $(B)$.

(C) Silica gel is a well-known adsorbent used to remove moisture from the surrounding environment.
In expensive scientific instruments,it is placed in watch-glasses or semipermeable bags to adsorb water vapor from the air.
By adsorbing moisture,it prevents the corrosion of metallic parts and protects sensitive electronic components from malfunctioning due to humidity.
Therefore,both the Assertion $(A)$ and the Reason $(R)$ are true,and $(R)$ is the correct explanation of $(A)$.

(A) The correct matches are:
$A$. The reaction of an aryl halide with an alkyl halide in the presence of sodium is known as the Wurtz-Fitting reaction. Thus,$A-I$.
$B$. The reaction of two molecules of aryl halide with sodium to form a diaryl is known as the Fitting reaction. Thus,$B-II$.
$C$. The conversion of benzene diazonium chloride to chlorobenzene using $Cu_2Cl_2$ is the Sandmeyer reaction. Thus,$C-III$.
$D$. The reaction of an alkyl halide with $NaI$ in acetone to form an alkyl iodide is the Finkelstein reaction. Thus,$D-IV$.
Therefore,the correct matching is $A-I, B-II, C-III, D-IV$.

(D) Caprolactam is heated with water at a high temperature $(533-543 \ K)$ to undergo ring-opening polymerization.
This process results in the formation of the polyamide known as Nylon $6$.
The reaction is represented as:
Caprolactam $\xrightarrow{H_2O, \Delta} [CO-(CH_2)_5-NH]_n$ (Nylon $6$).

(A) The spectrochemical series arranges ligands in the increasing order of their crystal field splitting energy (field strength).
The correct order is: $S^{2-} < C_2O_4^{2-} < NH_3 < en < CO$.
Thus,option $A$ (and $B$) represents the correct sequence.

(C) When $SO_{2}$ gas is passed through an acidified solution of $K_{2}Cr_{2}O_{7}$,the orange-colored dichromate ion $(Cr_{2}O_{7}^{2-})$ is reduced to the green-colored chromium$(III)$ ion $(Cr^{3+})$.
The balanced chemical equation is:
$Cr_{2}O_{7}^{2-} + 3SO_{2} + 2H^{+} \longrightarrow 2Cr^{3+} + 3SO_{4}^{2-} + H_{2}O$
Thus,the solution turns green.

(A) $Cis-Platin$ is a coordination compound of $Pt$ used in chemotherapy to inhibit the growth of tumors.
Its chemical formula is $cis-[Pt(NH_3)_2Cl_2]$.
Therefore,both $B$ and $D$ are correct as $Cis-Platin$ is a specific type of coordination compound of $Pt$.

(A) In qualitative inorganic analysis,cations are classified into groups based on their precipitation reactions with specific reagents.
$Zn^{2+}$,$Co^{2+}$,and $Ni^{2+}$ belong to group $IV$,which are precipitated as sulfides in the presence of $NH_4OH$ and $H_2S$.
$Fe^{3+}$ belongs to group $III$,which is precipitated as a hydroxide in the presence of $NH_4Cl$ and $NH_4OH$.
Therefore,$Fe^{3+}$ does not belong to group $IV$.

(B) The cell reaction is: $H_2(g) + 2Fe^{3+}(aq) \longrightarrow 2H^+(aq) + 2Fe^{2+}(aq)$.
The standard cell potential $E^{\circ}_{cell} = E^{\circ}_{cathode} - E^{\circ}_{anode} = 0.771\,V - 0\,V = 0.771\,V$.
Using the Nernst equation: $E_{cell} = E^{\circ}_{cell} - \frac{0.06}{n} \log Q$,where $n = 2$.
$0.712 = 0.771 - \frac{0.06}{2} \log \frac{[H^+]^2 [Fe^{2+}]^2}{[H_2] [Fe^{3+}]^2}$.
Given $[H^+] = 1\,M$,$P_{H_2} = 1\,atm$,so $0.712 = 0.771 - 0.03 \log \frac{[Fe^{2+}]^2}{[Fe^{3+}]^2}$.
$0.712 = 0.771 - 0.06 \log \frac{[Fe^{2+}]}{[Fe^{3+}]}$.
$0.06 \log \frac{[Fe^{2+}]}{[Fe^{3+}]} = 0.771 - 0.712 = 0.059$.
$\log \frac{[Fe^{2+}]}{[Fe^{3+}]} = \frac{0.059}{0.06} \approx 0.983 \approx 1$.
$\frac{[Fe^{2+}]}{[Fe^{3+}]} = 10^1 = 10$.

(B) The elevation in boiling point is given by $\Delta T_b = T_b - T_b^\circ = 373.52 \ K - 373 \ K = 0.52 \ K$.
Using the formula $\Delta T_b = K_b \times m$,where $m$ is the molality of the solution.
Molality $m = \frac{\text{mass of solute}}{\text{molar mass of solute}} \times \frac{1000}{\text{mass of solvent in } g}$.
Substituting the values: $0.52 = 0.52 \times \frac{2}{M} \times \frac{1000}{20}$.
$1 = \frac{2}{M} \times 50$.
$M = 100 \ g \ mol^{-1}$.

(B) For a first order reaction,the time $t$ is given by:
$t = \frac{2.303}{k} \log \frac{[A]_0}{[A]_t}$
Given:
$k = 2.011 \times 10^{-3} \ s^{-1}$
$[A]_0 = 7 \ g$
$[A]_t = 2 \ g$
$\log 7 = 0.845$
$\log 2 = 0.301$
Substituting the values:
$t = \frac{2.303}{2.011 \times 10^{-3}} \log \left( \frac{7}{2} \right)$
$t = \frac{2.303}{2.011 \times 10^{-3}} (\log 7 - \log 2)$
$t = \frac{2.303}{2.011 \times 10^{-3}} (0.845 - 0.301)$
$t = \frac{2.303 \times 0.544}{2.011} \times 10^3$
$t = \frac{1.252832}{2.011} \times 1000$
$t \approx 622.989 \ s$
The nearest integer is $623$.

(B) $1$. Compound '$A$' is trisubstituted and gives a positive neutral $FeCl_3$ test,indicating it is a phenol derivative.
$2$. The formula $C_{10}H_{12}O_2$ with a benzene ring ($6$ carbons) leaves $4$ carbons for substituents. The degree of unsaturation is $10 - 12/2 + 1 = 5$. Since the benzene ring accounts for $4$ degrees of unsaturation ($3$ $\pi$ bonds + $1$ ring),there is $1$ additional $\pi$ bond or ring present in the side chains.
$3$. The compound decolorises alkaline $KMnO_4$,confirming the presence of an alkene or an oxidizable side chain.
$4$. The reaction with $HI$ gives methyl iodide,indicating the presence of a methoxy group $(-OCH_3)$ or a similar ether linkage. However,the initial test with $NaOH$ and $CH_3Br$ suggests the presence of a phenolic $-OH$ group which gets methylated.
$5$. Given the structure is trisubstituted on a benzene ring,the total $\pi$ bonds are $3$ (from the benzene ring) + $1$ (from the alkene side chain) = $4$ $\pi$ bonds.

(B) $1$. Calculate the molar mass of $CH_2Cl_2$: $12 + 2(1) + 2(35.5) = 85 \ g \ mol^{-1}$.
$2$. Calculate the moles of $CH_2Cl_2$ in $0.671141 \ L$ of solution: $\text{moles} = \text{Molarity} \times \text{Volume} = 2.6 \times 10^{-3} \ mol \ L^{-1} \times 0.671141 \ L = 1.745 \times 10^{-3} \ mol$.
$3$. Calculate the mass of $CH_2Cl_2$ $(x)$: $x = \text{moles} \times \text{molar mass} = 1.745 \times 10^{-3} \ mol \times 85 \ g \ mol^{-1} = 0.1483 \ g$.
$4$. Calculate the mass of the solvent $(CHCl_3)$: $\text{mass} = \text{density} \times \text{volume} = 1.49 \ g \ cm^{-3} \times 671.141 \ cm^3 = 1000 \ g$.
$5$. Calculate the concentration in $ppm$: $\text{ppm} = \frac{\text{mass of solute}}{\text{mass of solvent}} \times 10^6 = \frac{0.1483 \ g}{1000 \ g} \times 10^6 = 148.3 \ ppm \approx 148 \ ppm$.

(B) The acidity of phenols is determined by the stability of the phenoxide ion formed after the loss of a proton.
Electron-withdrawing groups $(-EWG)$ increase acidity,while electron-donating groups $(-EDG)$ decrease acidity.
$1$. Compound $(c)$ has a $-NO_{2}$ group,which exerts a strong $-M$ and $-I$ effect,making it the most acidic.
$2$. Compound $(a)$ is phenol,which has no substituent.
$3$. Compound $(d)$ has an isopropyl group,which exerts a $+I$ effect and hyperconjugation,making it less acidic than phenol.
$4$. Compound $(b)$ has a $-N(CH_{3})_{2}$ group,which exerts a strong $+M$ effect,making it the least acidic.
Therefore,the order of acidic strength is $(c) > (a) > (d) > (b)$.
Since $pK_{a} \propto \frac{1}{\text{acidic strength}}$,the order of $pK_{a}$ values is $(b) > (d) > (a) > (c)$.

(C) The rate of $S_N 1$ reaction depends upon the stability of the carbocation intermediate formed after the departure of the leaving group $(Cl^-)$.
Electron-donating groups (like $-OCH_3$) increase the stability of the carbocation,while electron-withdrawing groups (like $-NO_2$ and $-Cl$) decrease its stability.
The stability order of the carbocations is:
$p-methoxybenzyl$ carbocation $(b)$ $>$ $benzyl$ carbocation $(d)$ $>$ $p-chlorobenzyl$ carbocation $(c)$ $>$ $p-nitrobenzyl$ carbocation $(a)$.
Therefore,the decreasing order of reactivity towards $S_N 1$ reaction is:
$(b) > (d) > (c) > (a)$.

(D) The conversion of $p$-nitrotoluene $(X)$ to $3,5$-dibromotoluene $(Y)$ involves the following steps:
$1$. Reduction of the nitro group to an amino group: $p$-nitrotoluene is reduced using $Fe/H^{+}$ to form $p$-toluidine.
$2$. Bromination: $p$-toluidine is treated with $Br_2 (aq)$ to form $2,6$-dibromo-$4$-methylaniline due to the strong activating effect of the $-NH_2$ group.
$3$. Diazotization: The amino group is converted to a diazonium salt using $HNO_2$ at $0-5 \ ^{\circ}C$.
$4$. Deamination: The diazonium group is removed using $H_3PO_2$ and water to yield $3,5$-dibromotoluene $(Y)$.

(D) $1$. $[Ni(CO)_4]$: $Ni$ is in $0$ oxidation state $(3d^8 4s^2)$. $CO$ is a strong field ligand,causing pairing. Hybridisation is $sp^3$ $(I)$.
$2$. $[Cu(NH_3)_4]^{2+}$: $Cu^{2+}$ is $3d^9$. The unpaired electron is in $4p_z$,and the hybridisation is $dsp^2$ $(II)$.
$3$. $[Fe(NH_3)_6]^{2+}$: $Fe^{2+}$ is $3d^6$. $NH_3$ is a strong field ligand,causing pairing. Hybridisation is $d^2sp^3$ $(IV)$.
$4$. $[Fe(H_2O)_6]^{2+}$: $Fe^{2+}$ is $3d^6$. $H_2O$ is a weak field ligand,no pairing. Hybridisation is $sp^3d^2$ $(III)$.
Thus,the correct match is $A-I, B-II, C-IV, D-III$.

(B) In a $fac-[Co(NH_3)_3Cl_3]$ complex,the three $Cl^-$ ligands occupy one face of the octahedral geometry.
This means the three $Cl^-$ ligands are at the corners of a triangular face.
In this arrangement,each $Cl^-$ ligand is adjacent to the other two $Cl^-$ ligands.
The bond angle between any two adjacent $Cl^-$ ligands in an octahedral geometry is $90^{\circ}$.
Therefore,the only $Cl-Co-Cl$ bond angle present is $90^{\circ}$.

(A) Assertion $A$ is false because the molecule contains an alcohol group which is acid-sensitive. In the presence of $HCl$ (a strong acid),the $-OH$ group will undergo dehydration or substitution,making Clemmensen reduction unsuitable for this specific conversion.
Reason $R$ is true because $Zn-Hg / HCl$ (Clemmensen reduction) is indeed a standard reagent used to reduce carbonyl groups to methylene $(-CH_2-)$ groups in acid-stable compounds.

(B) Assertion $A$ is true because antihistamines (like brompheniramine) are used to treat allergies by blocking histamine receptors,but they do not affect the secretion of acid in the stomach.
Reason $R$ is true because antacids (like cimetidine) work by preventing the interaction of histamine with receptors present in the stomach wall,which is a different mechanism compared to antiallergic drugs.
Since antihistamines used for allergies do not block the specific receptors in the stomach that trigger acid secretion,the reason correctly explains why the assertion is true.

(A) As we move down the group $16$ from $O$ to $Te$,the atomic size of the central element $E$ increases.
Due to the increase in atomic size,the bond length of the $E-H$ bond increases.
As the bond length increases,the bond strength decreases,which leads to a decrease in the bond dissociation energy.
Therefore,the order of bond dissociation energy is $H_2O > H_2S > H_2Se > H_2Te$.

(A) In electrolytic refining,the impure metal is made to act as the anode and the pure metal is made to act as the cathode. Thus,Statement $I$ is incorrect.
In the Hall-Heroult process,$Al_2O_3$ is mixed with cryolite $(Na_3AlF_6)$ and fluorspar $(CaF_2)$. This mixture lowers the melting point of the electrolyte and increases its electrical conductivity. Thus,Statement $II$ is correct.

(A) Nessler's reagent is an alkaline solution of potassium tetraiodomercurate$(II)$,which has the chemical formula $K_2[HgI_4]$.

(B) The initial moles of $[Co(NH_3)_5SO_4]Br$ are $1\,L \times 0.02\,M = 0.02\,mol$.
The initial moles of $[Co(NH_3)_5Br]SO_4$ are $1\,L \times 0.02\,M = 0.02\,mol$.
The total volume of the mixture is $2\,L$.
When the mixture is divided into two equal parts $(X)$,each part has a volume of $1\,L$.
In $1\,L$ of solution $(X)$,the moles of each complex are halved: $0.01\,mol$ of $[Co(NH_3)_5SO_4]Br$ and $0.01\,mol$ of $[Co(NH_3)_5Br]SO_4$.
The complex $[Co(NH_3)_5SO_4]Br$ dissociates to give $Br^-$ ions,and $[Co(NH_3)_5Br]SO_4$ dissociates to give $SO_4^{2-}$ ions.
Moles of $Br^-$ in $1\,L$ of $(X) = 0.01\,mol$.
Moles of $SO_4^{2-}$ in $1\,L$ of $(X) = 0.01\,mol$.
Reaction with $AgNO_3$: $Br^- + AgNO_3 \rightarrow AgBr(Y) + NO_3^-$. Moles of $Y (AgBr) = 0.01\,mol$.
Reaction with $BaCl_2$: $SO_4^{2-} + BaCl_2 \rightarrow BaSO_4(Z) + 2Cl^-$. Moles of $Z (BaSO_4) = 0.01\,mol$.

(D) The density formula for a unit cell is given by $d = \frac{Z \times M}{N_{A} \times a^3}$.
Here,$d = 4.0 \ g \ cm^{-3}$,$M = 56 + 16 = 72 \ g \ mol^{-1}$,$a = 5.0 \ \mathring{A} = 5.0 \times 10^{-8} \ cm$,and $N_{A} = 6.0 \times 10^{23} \ mol^{-1}$.
Substituting the values: $4.0 = \frac{Z \times 72}{6.0 \times 10^{23} \times (5.0 \times 10^{-8})^3}$.
$4.0 = \frac{Z \times 72}{6.0 \times 10^{23} \times 125 \times 10^{-24}}$.
$4.0 = \frac{Z \times 72}{0.075}$.
$Z = \frac{4.0 \times 0.075}{72} \approx 4.166$.
The nearest integer value for the number of $FeO$ units per unit cell is $4$.

(D) For a first order reaction,the rate constant $k$ is given by $k = \frac{1}{t} \ln \frac{a_0}{a_t}$.
For $60 \%$ decomposition,$a_t = a_0 - 0.6 a_0 = 0.4 a_0$ and $t_1 = 540 \ s$.
$k = \frac{1}{540} \ln \frac{a_0}{0.4 a_0} = \frac{1}{540} \ln \frac{10}{4} = \frac{1}{540} (\ln 10 - \ln 4) = \frac{1}{540} (2.3 - 2 \times 0.3 \times 2.3 / 2.3) = \frac{1}{540} (2.3 - 1.38) = \frac{0.92}{540}$.
For $90 \%$ decomposition,$a_t = a_0 - 0.9 a_0 = 0.1 a_0$.
$t_2 = \frac{1}{k} \ln \frac{a_0}{0.1 a_0} = \frac{1}{k} \ln 10 = \frac{1}{k} \times 2.3$.
Substituting $k$: $t_2 = \frac{540}{0.92} \times 2.3 = \frac{540 \times 2.3}{0.92} = \frac{1242}{0.92} = 1350 \ s$.

(D) The formula for freezing point depression is $\Delta T_f = i \cdot K_f \cdot m$.
Here,$i = 2.67$,$K_f = 1.8 \, K \, kg \, mol^{-1}$.
The molality $m$ is calculated as: $m = \frac{\text{mass of solute}}{\text{molar mass of solute}} \times \frac{1000}{\text{mass of solvent in grams}}$.
For $38\%$ by weight $H_2SO_4$,mass of $H_2SO_4 = 38 \, g$,mass of water = $100 - 38 = 62 \, g$.
$m = \frac{38}{98} \times \frac{1000}{62} \approx 6.254 \, mol \, kg^{-1}$.
$\Delta T_f = 2.67 \times 1.8 \times 6.254 \approx 30.05 \, K$.
Freezing point of pure water = $273.15 \, K$.
Freezing point of solution = $273.15 - 30.05 = 243.1 \, K$.
Rounding to the nearest integer,we get $243 \, K$.

(D) The Freundlich adsorption isotherm is given by the equation: $\log \frac{x}{m} = \log k + \frac{1}{n} \log P$.
Comparing this with the equation of a straight line $y = mx + c$,we have the slope $\frac{1}{n} = \tan 45^{\circ} = 1$ and the intercept $c = \log k = 0.6020$.
Since $\log 2 = 0.3010$,then $2 \times \log 2 = 0.6020$,which implies $\log 2^2 = \log 4 = 0.6020$. Thus,$k = 4$.
Substituting these values into the isotherm equation: $\frac{x}{m} = k \cdot P^{1/n} = 4 \times (0.4)^1 = 1.6$.
Expressing $1.6$ in the form $......... \times 10^{-1}$,we get $1.6 = 16 \times 10^{-1}$.
Therefore,the nearest integer is $16$.

(D) The compounds are analyzed for their solubility in cold $NaHCO_3$,cold $NaOH$,and hot $NaOH$:
$1$. $p$-Toluic acid: Dissolves in $NaHCO_3$ and $NaOH$.
$2$. Methyl benzoate: Does not dissolve in cold $NaHCO_3$ or $NaOH$,but undergoes hydrolysis in hot $NaOH$ to form benzoic acid and methanol,thus dissolving.
$3$. $p$-Cresyl acetate: Does not dissolve in cold $NaHCO_3$ or $NaOH$,but undergoes hydrolysis in hot $NaOH$ to form $p$-cresol and acetic acid,thus dissolving.
$4$. $o$-Hydroxyacetophenone: Dissolves in $NaOH$ (phenolic group) but not $NaHCO_3$.
$5$. $3$-Hydroxy$-5-$methylbenzaldehyde: Dissolves in $NaOH$ (phenolic group) but not $NaHCO_3$.
$6$. $1$-Isopropyl$-3-$methoxybenzene: Does not dissolve in any of these.
$7$. Phenyl acetate: Does not dissolve in cold $NaHCO_3$ or $NaOH$,but undergoes hydrolysis in hot $NaOH$ to form phenol and acetic acid,thus dissolving.
Therefore,the compounds that do not dissolve in cold $NaHCO_3$ and $NaOH$ but dissolve in hot $NaOH$ are $2$,$3$,and $7$. The total count is $3$.

(D) The total number of amino acid residues in the peptide is the sum of the moles of each amino acid produced upon hydrolysis.
Total amino acid residues $= 3 (G) + 2 (L) + 2 (V) = 7$.
For a peptide chain containing $n$ amino acid residues,the number of peptide linkages is given by $(n - 1)$.
Therefore,the number of peptide linkages $= 7 - 1 = 6$.

(D) The cell reaction is: $X(s) + Y^{2+}(aq) \rightarrow X^{2+}(aq) + Y(s)$.
The standard cell potential is $E^0_{cell} = E^0_{cathode} - E^0_{anode} = 0.36 \ V - (-2.36 \ V) = 2.72 \ V$.
Using the Nernst equation: $E_{cell} = E^0_{cell} - \frac{0.06}{n} \log \frac{[X^{2+}]}{[Y^{2+}]}$.
Here,$n = 2$,$[X^{2+}] = 0.001 \ M$,and $[Y^{2+}] = 0.01 \ M$.
$E_{cell} = 2.72 - \frac{0.06}{2} \log \frac{0.001}{0.01} = 2.72 - 0.03 \log(0.1) = 2.72 - 0.03(-1) = 2.72 + 0.03 = 2.75 \ V$.
Thus,$E_{cell} = 275 \times 10^{-2} \ V$.

(B) The atomic number of Neodymium $(Nd)$ is $60$.
The ground state electronic configuration of $Nd$ is $[Xe] 4f^4 6s^2$.
When $Nd$ forms the $Nd^{2+}$ ion,it loses two electrons from the outermost $6s$ orbital.
Therefore,the electronic configuration of $Nd^{2+}$ is $[Xe] 4f^4$.

(C) The concentration of ore (also known as beneficiation or dressing) involves removing gangue from the ore. The common methods are:
$(i)$ Hydraulic Washing (Gravity separation)
$(ii)$ Froth Flotation
$(iii)$ Magnetic Separation
$(iv)$ Leaching
Liquation $(A)$ is a method used for the refining of metals with low melting points.
Electrolysis $(C)$ is a method used for the refining of metals.
Therefore,the methods $NOT$ involved in the concentration of ore are $A$ (Liquation) and $C$ (Electrolysis).

(C) $1$. The reaction of $\text{Propanal} (CH_3CH_2CHO)$ and $\text{Methanal} (HCHO)$ in the presence of $\text{dil. NaOH}$ followed by heating $(\Delta)$ undergoes a cross-aldol condensation to form $2-\text{methylacrolein}$ $(CH_2=C(CH_3)CHO)$.
$2$. The subsequent reaction with $\text{NaCN}$ followed by acid hydrolysis $(\text{H}_3\text{O}^ )$ is a cyanohydrin formation reaction,which converts the aldehyde group $(-CHO)$ into a carboxylic acid group $(-COOH)$ with an adjacent hydroxyl group $(-OH)$.
$3$. The final product $B$ is $2-\text{hydroxy}-2-\text{methylbut}-3-\text{enoic acid}$ (or a similar structure depending on the specific pathway,but the key is the presence of the $-COOH$ group and the chiral center).
$4$. The product contains a chiral center (marked with $*$ in the mechanism),and since the reaction conditions do not involve chiral catalysts,it is formed as a racemic mixture.
$5$. The presence of the carboxylic acid group $(-COOH)$ ensures that it reacts with $\text{NaHCO}_3$ to release $\text{CO}_2$ gas.

(A) The basicity of oxides of a transition metal decreases as the oxidation state of the metal increases.
In $V_2O_3$,the oxidation state of $V$ is $+3$.
In $V_2O_4$,the oxidation state of $V$ is $+4$.
In $V_2O_5$,the oxidation state of $V$ is $+5$.
As the oxidation state increases,the acidic character increases and the basic character decreases.
Therefore,the order of basicity is $V_2O_3 > V_2O_4 > V_2O_5$.

(B) When $Cu^{2+}$ ions react with $KI$,$CuI_2$ is initially formed,which is unstable and decomposes to form a white precipitate of $Cu_2I_2$ and iodine $(I_2)$:
$2Cu^{2+} + 4I^-$ $\rightarrow 2CuI_2$ $\rightarrow Cu_2I_2 \downarrow + I_2$
The liberated $I_2$ dissolves in excess $KI$ to form $KI_3$ (triiodide ion),which gives a brown color to the solution:
$I_2 + I^- \rightarrow I_3^-$
This solution is then titrated against sodium thiosulphate $(Na_2S_2O_3)$,where $I_2$ is reduced to $I^-$ and thiosulphate is oxidized to tetrathionate $(Na_2S_4O_6)$:
$I_2 + 2Na_2S_2O_3 \rightarrow 2NaI + Na_2S_4O_6$
Thus,$X$ is $Cu_2I_2$ and $Y$ is $Na_2S_4O_6$.

(D) The reaction sequence is as follows:
$1$. Nitrobenzene $(C_6H_5NO_2)$ is reduced using $H_2/Pd$ in ethanol $(C_2H_5OH)$ to form aniline $(C_6H_5NH_2)$,which is compound $[A]$.
$2$. Aniline $(C_6H_5NH_2)$ reacts with acetic anhydride $((CH_3CO)_2O)$ in the presence of pyridine to undergo acetylation,forming $N$-phenylacetamide (acetanilide),$C_6H_5NHCOCH_3$,which is compound $[B]$.

(A) When $CoCl_2$ is dissolved in water,it forms the pink-colored hexaaquacobalt$(II)$ ion,$[Co(H_2O)_6]^{2+}$,which has an octahedral geometry $(X)$.
Upon adding concentrated $HCl$,the water ligands are replaced by chloride ions to form the deep blue tetrachlorocobaltate$(II)$ ion,$[CoCl_4]^{2-}$ $(Y)$,which has a tetrahedral geometry $(Z)$.

(B) The melting point of $1,4-$dichlorobenzene is higher than those of $1,2-$ and $1,3-$ isomers due to its symmetrical structure,which allows for better packing in the crystal lattice.
Comparing the melting points:
$1,4-$dichlorobenzene: $323 \ K$
$1,2-$dichlorobenzene: $256 \ K$
$1,3-$dichlorobenzene: $249 \ K$
Thus,the correct order is $1,4- > 1,2- > 1,3-$dichlorobenzene.

(B) Proteins are polymers of $\alpha$-amino acids. An $\alpha$-amino acid is one in which the amino group $(-NH_2)$ is attached to the carbon atom adjacent to the carboxylic acid group $(-COOH)$,i.e.,the $\alpha$-carbon.
In option $(B)$,the structure is $CH_3-CH(CH_3)-CH_2-CH(NH_2)-COOH$. Here,the $-NH_2$ group is attached to the carbon atom directly adjacent to the $-COOH$ group. This is the structure of the amino acid Leucine.
In the other options,the $-NH_2$ group is attached to a carbon atom further away from the $-COOH$ group,meaning they are not $\alpha$-amino acids.

(C) The sweetness values of artificial sweeteners relative to cane sugar are as follows:
$1$. $Alitame$: $2000$ times
$2$. $Sucralose$: $600$ times
$3$. $Saccharin$: $550$ times
$4$. $Aspartame$: $100$ times
Therefore,the order of sweetness is $Alitame > Sucralose > Saccharin > Aspartame$.
Thus,$Alitame$ has the highest sweetness value.