JEE Main 2023 Chemistry Question Paper with Answer and Solution

(D) In the industrial production of urea,ammonia $(NH_3)$ and carbon dioxide $(CO_2)$ are used as raw materials. The reaction is: $2NH_3 + CO_2 \rightarrow NH_2CONH_2 + H_2O$. Since $CO_2$ is consumed in this process,it is considered the least responsible for global warming among the given options.

(A) The laboratory preparation of deuterated hydrogen peroxide $(D_2O_2)$ is conveniently carried out by the reaction of potassium persulfate $(K_2S_2O_8)$ with heavy water $(D_2O)$.
The chemical equation is: $K_2S_2O_{8(s)} + 2D_2O_{(l)} \rightarrow 2KDSO_{4(aq)} + D_2O_2$.

(D) Isoelectronic species are those that have the same number of electrons.
Number of electrons in:
$K^{+} = 19 - 1 = 18$
$Cl^{-} = 17 + 1 = 18$
$Ca^{2+} = 20 - 2 = 18$
$Sc^{3+} = 21 - 3 = 18$
Since all these ions have $18$ electrons,they represent a collection of isoelectronic species.

(D) For the reaction: $PCl_{5(g)} \rightleftharpoons PCl_{3(g)} + Cl_{2(g)}$
$1$. At constant pressure: The addition of an inert gas like helium increases the volume of the system. According to Le Chatelier's principle,the equilibrium shifts towards the side with a greater number of moles of gas,which is the forward direction. Thus,more $PCl_3$ and $Cl_2$ are produced.
$2$. At constant volume: The addition of an inert gas does not change the partial pressures or concentrations of the reacting species. Therefore,the equilibrium remains unaffected.
Since the condition (constant pressure or constant volume) is not specified,in standard textbook problems,if not mentioned,it is often assumed to be at constant volume,or the question is considered ambiguous. However,based on the options provided,if the pressure is constant,$A$ is correct; if the volume is constant,$D$ is correct. In many competitive contexts,if the condition is not specified,the addition of an inert gas at constant volume is the standard assumption,leading to $D$.

(B) The electron gain enthalpy $(\Delta_{eg} H)$ becomes less negative as we move down a group due to an increase in atomic size.
$1$. For $Cl$ and $F$: $\Delta_{eg} H (Cl) = -349 \ kJ/mol$ and $\Delta_{eg} H (F) = -328 \ kJ/mol$. Since $-349 < -328$,the statement $\Delta_{eg} H (Cl) < \Delta_{eg} H (F)$ is correct.
$2$. For $Se$ and $S$: $\Delta_{eg} H (Se) = -195 \ kJ/mol$ and $\Delta_{eg} H (S) = -200 \ kJ/mol$. Since $-195 > -200$,the statement $\Delta_{eg} H (Se) < \Delta_{eg} H (S)$ is incorrect.
$3$. For $I$ and $At$: $\Delta_{eg} H (I) = -295 \ kJ/mol$ and $\Delta_{eg} H (At) = -270 \ kJ/mol$. Since $-295 < -270$,the statement $\Delta_{eg} H (I) < \Delta_{eg} H (At)$ is correct.
$4$. For $Te$ and $Po$: $\Delta_{eg} H (Te) = -190 \ kJ/mol$ and $\Delta_{eg} H (Po) = -183 \ kJ/mol$. Since $-190 < -183$,the statement $\Delta_{eg} H (Te) < \Delta_{eg} H (Po)$ is correct.

(D) The $O-O$ bond length in $H_2O_2$ is $1.48 \ \mathring{A}$,while in $F_2O_2$ it is $1.22 \ \mathring{A}$. Thus,the $O-O$ bond in $H_2O_2$ is longer $(X = \text{longer})$.
The $O-H$ bond length in $H_2O_2$ is approximately $0.95 \ \mathring{A}$,whereas the $O-F$ bond length in $F_2O_2$ is approximately $1.41 \ \mathring{A}$. Thus,the $O-H$ bond is shorter $(Y = \text{shorter})$.
Therefore,the correct option is $D$.

(B) $1$. Bomb calorimeter measures heat at constant volume,so the heat released is $\Delta U$.
$2$. Moles of ethane $(C_2H_6)$ = $\frac{0.3 \ g}{30 \ g \ mol^{-1}} = 0.01 \ mol$.
$3$. Heat released for $0.01 \ mol$ = $C \times \Delta T = 20 \ kJ \ K^{-1} \times 0.5 \ K = 10 \ kJ$.
$4$. Heat released for $1 \ mol$ $(\Delta U)$ = $\frac{10 \ kJ}{0.01 \ mol} = -1000 \ kJ \ mol^{-1}$.
$5$. Combustion reaction: $C_2H_6(g) + 3.5 O_2(g) \rightarrow 2CO_2(g) + 3H_2O(l)$.
$6$. Change in gaseous moles $(\Delta n_g)$ = $2 - (1 + 3.5) = -2.5$.
$7$. Relation between $\Delta H$ and $\Delta U$: $\Delta H = \Delta U + \Delta n_g RT$.
$8$. $\Delta H = -1000 \ kJ \ mol^{-1} + (-2.5 \times 8.3 \times 10^{-3} \ kJ \ K^{-1} \ mol^{-1} \times 300 \ K)$.
$9$. $\Delta H = -1000 - 6.225 = -1006.225 \ kJ \ mol^{-1}$.
$10$. The heat evolved is $1006 \ kJ \ mol^{-1}$ (nearest integer).

(C) The concentration of $Ag^{+}$ ions from $AgNO_3$ is $[Ag^{+}] = 10^{-5} \ M$.
The concentration of $NO_3^{-}$ ions is $[NO_3^{-}] = 10^{-5} \ M$.
Due to the common ion effect,the concentration of $Br^{-}$ ions is determined by the solubility product constant: $[Br^{-}] = \frac{K_{sp}}{[Ag^{+}]} = \frac{4.9 \times 10^{-13}}{10^{-5}} = 4.9 \times 10^{-8} \ M$.
The conductivity $\kappa$ is given by $\kappa = \sum C_i \lambda_i$.
For $Ag^{+}$: $\kappa_{Ag^{+}} = 10^{-5} \times 6 \times 10^{-3} = 6 \times 10^{-8} \ S \ m^{-1}$.
For $Br^{-}$: $\kappa_{Br^{-}} = 4.9 \times 10^{-8} \times 8 \times 10^{-3} = 39.2 \times 10^{-11} \ S \ m^{-1} = 0.392 \times 10^{-8} \ S \ m^{-1}$.
For $NO_3^{-}$: $\kappa_{NO_3^{-}} = 10^{-5} \times 7 \times 10^{-3} = 7 \times 10^{-8} \ S \ m^{-1}$.
Total conductivity $\kappa = (6 + 0.392 + 7) \times 10^{-8} \ S \ m^{-1} = 13.392 \times 10^{-8} \ S \ m^{-1}$.
Rounding to the nearest integer,we get $13 \times 10^{-8} \ S \ m^{-1}$.

(C) The general term in the expansion of $\left(2 x^3-\frac{1}{3 x^2}\right)^5$ is given by $T_{r+1} = { }^5 C_r \left(2 x^3\right)^{5-r} \left(-\frac{1}{3 x^2}\right)^r$.
Simplifying the expression,we get $T_{r+1} = { }^5 C_r \cdot 2^{5-r} \cdot x^{3(5-r)} \cdot (-1)^r \cdot 3^{-r} \cdot x^{-2r} = { }^5 C_r \cdot \frac{2^{5-r}}{(-3)^r} \cdot x^{15-5r}$.
To find the coefficient of $x^5$,we set the exponent of $x$ equal to $5$:
$15 - 5r = 5$
$5r = 10$
$r = 2$.
Substituting $r = 2$ into the expression for the coefficient:
Coefficient $= { }^5 C_2 \cdot \frac{2^{5-2}}{(-3)^2} = 10 \cdot \frac{2^3}{9} = 10 \cdot \frac{8}{9} = \frac{80}{9}$.

(D) The correct matches are:
$A$. Nitrogen: Detected by forming Prussian blue,$Fe_4[Fe(CN)_6]_3$ $(III)$.
$B$. Sulphur: Detected by Sodium nitroprusside,$Na_2[Fe(CN)_5NO]$ $(I)$.
$C$. Phosphorus: Detected by ammonium molybdate,$(NH_4)_2MoO_4$ $(IV)$.
$D$. Halogen: Detected by $AgNO_3$ $(II)$.
Therefore,the correct sequence is $A-III, B-I, C-IV, D-II$.

(B) The electron gain enthalpy $(\Delta H_{eg})$ is the energy change when an electron is added to a neutral gaseous atom.
$Cl$ has the most negative (exothermic) electron gain enthalpy among all elements due to its high effective nuclear charge and small size.
$Ne$ is a noble gas with a stable octet configuration,making the addition of an electron highly unfavorable,resulting in a large positive (endothermic) electron gain enthalpy.
Therefore,the difference between the most positive value $(Ne)$ and the most negative value $(Cl)$ results in the maximum difference.

(D) Photochemical smog occurs in warm,dry,and sunny climates.
It is formed by the action of sunlight on unsaturated hydrocarbons and nitrogen oxides $(NO_x)$ emitted by automobiles and factories.
Therefore,industrial areas with high vehicular traffic and industrial emissions are the most prone to photochemical smog formation.

(D) The setting time of cement is increased by adding $Gypsum$ $(CaSO_4 \cdot 2H_2O)$. It acts as a retarder,which slows down the hydration process of tricalcium aluminate $(C_3A)$,thereby preventing the cement from setting too quickly.

(D) The Assertion $(A)$ is correct because the hydrogen atom consists of one proton and one electron. When the electron is lost,only the nucleus (a proton) remains,which has a size of approximately $1.5 \times 10^{-3} \ pm$.
The Reason $(R)$ is also correct because a proton $(H^+)$ is highly reactive and cannot exist independently in the gas phase or aqueous solution; it always associates with other molecules or ions (e.g.,$H_3O^+$).
However,the Reason $(R)$ does not explain why the size of the proton is $1.5 \times 10^{-3} \ pm$. Therefore,both statements are correct,but $R$ is not the correct explanation of $A$.

(A) The given reaction is $2[Au(CN)_2]_{(aq)}^{-} + Zn_{(s)} \rightarrow 2Au_{(s)} + [Zn(CN)_4]_{(aq)}^{2-}$.
Assigning oxidation states: $2[\stackrel{+1}{Au}(CN)_2]_{(aq)}^{-} + \stackrel{0}{Zn}_{(s)}$ $\rightarrow 2\stackrel{0}{Au}_{(s)} + [\stackrel{+2}{Zn}(CN)_4]_{(aq)}^{2-}$.
Here,$Zn$ is oxidized from $0$ to $+2$ and $Au$ is reduced from $+1$ to $0$. Since both oxidation and reduction occur,it is a redox reaction.
$Zn$ displaces $Au$ from the complex,making it a displacement reaction.
Therefore,both $A$ and $B$ are correct.

(A) The dissociation reaction is: $A_2B_3 (aq.) \rightleftharpoons 2 A^{3+} (aq.) + 3 B^{2-} (aq.)$
At equilibrium,the concentrations are: $[A_2B_3] = c(1 - \alpha)$,$[A^{3+}] = 2c\alpha$,and $[B^{2-}] = 3c\alpha$.
The equilibrium constant expression is: $K_{eq} = \frac{[A^{3+}]^2 [B^{2-}]^3}{[A_2B_3]} = \frac{(2c\alpha)^2 (3c\alpha)^3}{c(1 - \alpha)}$.
For a weak electrolyte,$\alpha \ll 1$,so $(1 - \alpha) \approx 1$.
Thus,$K_{eq} = \frac{(4c^2\alpha^2)(27c^3\alpha^3)}{c} = 108c^4\alpha^5$.
Solving for $\alpha$: $\alpha^5 = \frac{K_{eq}}{108c^4} \implies \alpha = \left(\frac{K_{eq}}{108 c^4}\right)^{\frac{1}{5}}$.

(D) The de Broglie wavelength is given by the formula $\lambda = \frac{h}{p} = \frac{h}{\sqrt{2mKE}}$.
Substituting the given values:
$\lambda = \frac{6.6 \times 10^{-34}}{\sqrt{2 \times (9 \times 10^{-31} \ kg) \times (4.50 \times 10^{-29} \ J)}}$
$\lambda = \frac{6.6 \times 10^{-34}}{\sqrt{81 \times 10^{-60}}}$
$\lambda = \frac{6.6 \times 10^{-34}}{9 \times 10^{-30}}$
$\lambda = 0.733 \times 10^{-4} \ m = 7.33 \times 10^{-5} \ m$.
The nearest integer is $7$.

(D) When ethane $(CH_3-CH_3)$ reacts with excess $Br_2$ in the presence of diffused sunlight,free radical substitution occurs,replacing hydrogen atoms with bromine atoms.
$1$. Monobromoethane: $CH_3CH_2Br$ ($1$ isomer)
$2$. Dibromoethane: $CH_2BrCH_2Br$ and $CH_3CHBr_2$ ($2$ isomers)
$3$. Tribromoethane: $CH_2BrCHBr_2$ and $CH_3CBr_3$ ($2$ isomers)
$4$. Tetrabromoethane: $CHBr_2CHBr_2$ and $CH_2BrCBr_3$ ($2$ isomers)
$5$. Pentabromoethane: $CHBr_2CBr_3$ ($1$ isomer)
$6$. Hexabromoethane: $CBr_3CBr_3$ ($1$ isomer)
Total number of bromo derivatives = $1 + 2 + 2 + 2 + 1 + 1 = 9$.

(D) For a spontaneous process,the slope of the $G$ vs. extent of reaction graph is negative,i.e.,$dG/d\xi < 0$.
For a process at equilibrium,the slope is zero,i.e.,$dG/d\xi = 0$.
For a non-spontaneous process,the slope is positive,i.e.,$dG/d\xi > 0$.
At point $(a)$,the slope is negative,so the reaction is spontaneous.
At point $(b)$,the slope is zero,so the reaction is at equilibrium.
At point $(c)$,the slope is positive,so the reaction is non-spontaneous.
Evaluating the statements:
$A$. Reaction is spontaneous at $(a)$ and $(b)$: False (at $(b)$ it is at equilibrium).
$B$. Reaction is at equilibrium at point $(b)$ and non-spontaneous at point $(c)$: True.
$C$. Reaction is spontaneous at $(a)$ and non-spontaneous at $(c)$: True.
$D$. Reaction is non-spontaneous at $(a)$ and $(b)$: False.
Therefore,there are $2$ true statements ($B$ and $C$).

(D) The relationship between Gibbs free energy change and the equilibrium constant is given by $\Delta G^{\circ} = -RT \ln K = -2.303 \ RT \log K$.
First,calculate $\Delta G^{\circ}$ using $\Delta G^{\circ} = \Delta H^{\circ} - T \Delta S^{\circ}$.
Given $\Delta H^{\circ} = -54.07 \ kJ \ mol^{-1} = -54070 \ J \ mol^{-1}$,$T = 298 \ K$,and $\Delta S^{\circ} = 10 \ J \ K^{-1} \ mol^{-1}$.
$\Delta G^{\circ} = -54070 - (298 \times 10) = -54070 - 2980 = -57050 \ J \ mol^{-1}$.
Now,substitute into the equilibrium equation:
$-57050 = -2.303 \times 8.314 \times 298 \times \log K$.
Using the given value $2.303 \times 8.314 \times 298 = 5705$:
$-57050 = -5705 \times \log K$.
$\log K = \frac{57050}{5705} = 10$.

(D) To determine the number of species with a square pyramidal structure,we analyze the hybridization and geometry of each:
$1$. $PF_{5}$: $sp^{3}d$ hybridization,trigonal bipyramidal geometry.
$2$. $BrF_{4}^{-}$: $sp^{3}d^{2}$ hybridization with $2$ lone pairs,square planar geometry.
$3$. $IF_{5}$: $sp^{3}d^{2}$ hybridization with $1$ lone pair,square pyramidal geometry.
$4$. $BrF_{5}$: $sp^{3}d^{2}$ hybridization with $1$ lone pair,square pyramidal geometry.
$5$. $XeOF_{4}$: $sp^{3}d^{2}$ hybridization with $1$ lone pair,square pyramidal geometry.
$6$. $ICl_{4}^{-}$: $sp^{3}d^{2}$ hybridization with $2$ lone pairs,square planar geometry.
The species with a square pyramidal structure are $IF_{5}$,$BrF_{5}$,and $XeOF_{4}$.
Therefore,the total number of such species is $3$.

(D) The balanced chemical equation for the reaction is:
$3 BaCl_2 + 2 Na_3PO_4 \rightarrow Ba_3(PO_4)_2 + 6 NaCl$
From the stoichiometry of the reaction:
$3$ moles of $BaCl_2$ react with $2$ moles of $Na_3PO_4$ to produce $1$ mole of $Ba_3(PO_4)_2$.
Given:
$5$ moles of $BaCl_2$ and $2$ moles of $Na_3PO_4$.
To find the limiting reagent:
For $2$ moles of $Na_3PO_4$,we require $(3/2) \times 2 = 3$ moles of $BaCl_2$.
Since we have $5$ moles of $BaCl_2$ (which is more than $3$ moles),$BaCl_2$ is in excess and $Na_3PO_4$ is the limiting reagent.
Calculation of product:
$2$ moles of $Na_3PO_4$ produce $1$ mole of $Ba_3(PO_4)_2$.
Therefore,the maximum number of moles of $Ba_3(PO_4)_2$ formed is $1$.

(A) Hydration enthalpy is inversely proportional to the ionic size: $\text{Hydration Enthalpy} \propto \frac{1}{\text{size}}$.
As we move down the group,the ionic size increases,which causes the hydration enthalpy to decrease.
The order of hydration enthalpy for alkaline earth metal ions is: $Be^{2+} > Mg^{2+} > Ca^{2+} > Sr^{2+} > Ba^{2+}$.
Therefore,$Be^{2+}$ has the highest hydration enthalpy.

(B) In the solid state,$BeCl_2$ exists as a chain polymer.
In the vapour phase,it exists as a chloro-bridged dimer $(Be_2Cl_4)$.
At very high temperatures (above $1200 \ K$),it exists as a linear monomeric molecule $(BeCl_2)$.

(C) The basic character of oxides of Group-$13$ elements increases down the group as the metallic or electropositive character of the elements increases.
The trend in the nature of oxides is as follows:
$B_2O_3$ (acidic) < $Al_2O_3$ (amphoteric) < $Ga_2O_3$ (amphoteric) < $In_2O_3$ (basic) < $Tl_2O_3$ (most basic).
Therefore,$Tl_2O_3$ is the most basic oxide.

(B) In a neutral or weakly alkaline medium,the oxidation state of $Mn$ in $MnO_4^-$ (where $Mn$ is $+7$) changes to $MnO_2$ (where $Mn$ is $+4$).
This represents a change in oxidation state of $7 - 4 = 3$ units.
Therefore,the reaction is favoured in neutral or faintly alkaline media.

(A) In column chromatography,the stationary phase is typically polar (e.g.,silica gel).
The compound that is least polar moves faster and elutes first,while the most polar compound moves slower and elutes last.
Based on the figure,compound '$a$' is at the top,'$b$' is in the middle,and '$c$' is at the bottom (closest to the exit).
Therefore,the order of elution is '$c$' (first),then '$b$',then '$a$' (last).
This implies that '$c$' is the least polar and '$a$' is the most polar.
Let's evaluate the statements:
$A.$ '$c$' is more polar than '$a$' and '$b$': Incorrect,'$c$' is the least polar.
$B.$ '$a$' is least polar: Incorrect,'$a$' is the most polar.
$C.$ '$b$' comes out of the column before '$c$' and after '$a$': Incorrect,'$c$' comes out first,then '$b$',then '$a$.'
$D.$ '$a$' spends more time in the column: Correct,as it is the most polar and interacts most strongly with the stationary phase.
The incorrect statements are $A, B,$ and $C$.

(D) Statement-$I$: Morphine is a narcotic analgesic that relieves pain and produces sleep. The claim that it relieves pain 'without producing sleep' is incorrect.
Statement-$II$: Morphine and its derivatives (like codeine and heroin) are indeed obtained from the opium poppy plant $(Papaver \text{ } somniferum)$. This statement is correct.
Therefore,Statement-$I$ is false and Statement-$II$ is true.

(C) The neutralization reaction is: $Ba(OH)_2 + 2HBr \rightarrow BaBr_2 + 2H_2O$.
Using the law of equivalence,$n_{eq}(HBr) = n_{eq}(Ba(OH)_2)$.
Since $n_{eq} = M \times V \times n_{factor}$,where $n_{factor}$ for $HBr$ is $1$ and for $Ba(OH)_2$ is $2$:
$0.02 \times V_1 \times 1 = 0.01 \times 10 \times 2$.
$0.02 \times V_1 = 0.2$.
$V_1 = \frac{0.2}{0.02} = 10 \, mL$.

(C) Pesticides are chemical substances used to kill or control pests.
$DDT$ (Dichlorodiphenyltrichloroethane) and Aldrin are well-known organochlorine compounds used as pesticides.
Therefore,the group consisting of $DDT$ and Aldrin represents pesticides.

(B) $Li$ and $Cs$ are highly reactive metals that react vigorously with water.
$Br_2$ has a boiling point of $58^{\circ}C$,so it will change into a vapour state in boiling water $(100^{\circ}C)$.
$Ga$ has a melting point of $29^{\circ}C$ and a boiling point of $2400^{\circ}C$. Therefore,it remains in the liquid state within the temperature range of boiling water $(100^{\circ}C)$.

(B) The radius of the $n^{th}$ orbit of a hydrogen atom is given by $r_n = a_0 n^2$.
For the $3^{rd}$ orbit $(n = 3)$, the radius is $r_3 = a_0 \times 3^2 = 9 a_0$.
According to Bohr's postulate, the circumference of the orbit is an integral multiple of the de Broglie wavelength: $2 \pi r_n = n \lambda$.
Substituting the values for the $3^{rd}$ orbit: $2 \pi (9 a_0) = 3 \lambda$.
Solving for $\lambda$: $\lambda = \frac{18 \pi a_0}{3} = 6 \pi a_0$.

(C) In an ice crystal,the structure is highly ordered and three-dimensional.
Each water molecule is hydrogen bonded to $4$ neighbouring water molecules in a tetrahedral arrangement.

(A) The initial equilibrium constant is $K_c = \frac{[PCl_5]}{[PCl_3][Cl_2]} = \frac{0.40}{0.2 \times 0.1} = 20$.
When $0.2 \, mol$ of $Cl_2$ is added to $1 \, L$ volume,the new concentration of $Cl_2$ becomes $0.1 + 0.2 = 0.3 \, M$.
Let $x$ be the amount of $PCl_3$ and $Cl_2$ consumed to reach the new equilibrium.
The new equilibrium concentrations are: $[PCl_3] = 0.2 - x$,$[Cl_2] = 0.3 - x$,and $[PCl_5] = 0.4 + x$.
Substituting into the $K_c$ expression: $\frac{0.4 + x}{(0.2 - x)(0.3 - x)} = 20$.
$0.4 + x = 20(0.06 - 0.5x + x^2) = 1.2 - 10x + 20x^2$.
$20x^2 - 11x + 0.8 = 0$.
Using the quadratic formula $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$,we get $x \approx 0.086$.
Thus,$[PCl_5]_{eq} = 0.4 + 0.086 = 0.486 \, M = 48.6 \times 10^{-2} \, mol \, L^{-1}$.
Rounding to the nearest integer,the value is $49$.

(C) The formation reaction is: $2C_{(s)} + 3H_{2(g)} + \frac{1}{2}O_{2(g)} \rightarrow C_2H_5OH_{(l)}$
The heat of formation is calculated as: $(\Delta H_f)_{C_2H_5OH_{(l)}} = [2 \times \Delta H_c(C_{(s)}) + 3 \times \Delta H_c(H_{2(g)})] - \Delta H_c(C_2H_5OH_{(l)})$
Substituting the values: $= [2 \times (-393.5) + 3 \times (-241.8)] - (-1234.7)$
$= [-787.0 - 725.4] + 1234.7$
$= -1512.4 + 1234.7 = -277.7 \ kJ \ mol^{-1}$
The nearest integer value is $-278 \ kJ \ mol^{-1}$.

(D) To balance the chemical equation,the number of atoms of each element must be equal on both sides.
For Iodine $(I)$:
Reactant side: $2$ (from $2 IO_3^-$) $+ x$ (from $x I^-$) $= 2 + x$
Product side: $6 \times 2 = 12$ (from $6 I_2$)
Equating the two sides:
$2 + x = 12$
$x = 12 - 2$
$x = 10$
Thus,the balanced equation is $2 IO_3^- + 10 I^- + 12 H^+ \rightarrow 6 I_2 + 6 H_2 O$.

(D) Water gas is a mixture of $CO$ and $H_2$. When water gas reacts with $H_2$ in the presence of a cobalt catalyst,it produces methanol.
$CO + 2H_2 \xrightarrow{Co} CH_3OH$

(C) When water is added to cement,it sets very quickly,which makes it difficult to work with. Adding a small amount of gypsum $(CaSO_4 \cdot 2H_2O)$ to the cement clinker during the grinding process helps to retard or slow down the initial setting time of the cement,allowing sufficient time for mixing,placing,and finishing the concrete.

(D) According to the standards for drinking water:
$A$. $F^{-}$: The maximum concentration is $< 2 \, ppm$ $(III)$.
$B$. $SO_{4}^{2-}$: The maximum concentration is $< 500 \, ppm$ $(IV)$.
$C$. $NO_{3}^{-}$: The maximum concentration is $< 50 \, ppm$ $(I)$.
$D$. $Zn$: The maximum concentration is $< 5 \, ppm$ $(II)$.
Therefore,the correct matching is $A-III, B-IV, C-I, D-II$.

(D) The electronegativity $(E.N.)$ values on the Pauling scale for the given elements are as follows:

Element	$E.N.$ Value
$Br$	$3.0$
$C$	$2.5$
$At$	$2.2$
$P$	$2.1$

Based on these values,the correct order of electronegativity is $Br > C > At > P$. Therefore,the correct option is $D$.

(C) Statement $I$ is correct: Lithium $(Li)$ and Magnesium $(Mg)$ have high charge density and small ionic sizes,which prevents the formation of larger superoxide ions. They typically form normal oxides ($Li_2O$ and $MgO$).
Statement $II$ is correct: The ionic radius of $Li^{+}$ is $76 \ pm$ $(0.76 \ \mathring{A})$ and the ionic radius of $Mg^{2+}$ is $72 \ pm$ $(0.72 \ \mathring{A})$. Since $76 \ pm$ > $72 \ pm$,the ionic radius of $Li^{+}$ is indeed larger than that of $Mg^{2+}$.
Therefore,both statements are correct.

(B) The ozonolysis of an alkene involves the cleavage of the $C=C$ double bond to form carbonyl compounds.
Ethanal is $CH_3CHO$ and propanone is $(CH_3)_2CO$.
By joining these two fragments at the carbonyl carbons,we get the structure of the alkene $(X)$:
$CH_3-CH=C(CH_3)_2$.
The chemical formula of this hydrocarbon is $C_5H_{10}$.
The molar mass of $C_5H_{10} = (5 \times 12) + (10 \times 1) = 60 + 10 = 70 \ g\ mol^{-1}$.

(B) According to Fajan's rule,the covalent character in an ionic bond is determined by the following factors:
$1$. Polarising power of the cation $(a)$.
$2$. Polarisability of the anion $(c)$.
$3$. The extent of distortion of the anion $(b)$,which is a direct consequence of the first two factors.
Factor $(d)$,the polarising power of the anion,is generally negligible as anions are typically large and have low charge density compared to cations.
Therefore,there are $3$ factors ($a$,$b$,and $c$) that affect the covalent character.

(B) Power of the heater $P = 60 \, W = 60 \, J \, s^{-1}$.
Total energy supplied $Q = P \times t = 60 \, J \, s^{-1} \times 100 \, s = 6000 \, J$.
Since the container is adiabatic and of constant volume,the heat supplied is equal to the heat absorbed by the gas.
$Q = C \times \Delta T$,where $C$ is the heat capacity and $\Delta T$ is the change in temperature.
$6000 \, J = C \times 5 \, K$.
$C = \frac{6000}{5} = 1200 \, J \, K^{-1}$.

(B) The percentage of carbon in an organic compound is given by the formula:
$\% \text{Carbon} = \frac{12}{44} \times \frac{\text{mass of } CO_2 \text{ formed}}{\text{mass of compound taken}} \times 100$
Given,$\% \text{Carbon} = 60$,$\text{mass of compound} = 0.5 \ g$.
Substituting the values:
$60 = \frac{12}{44} \times \frac{\text{mass of } CO_2}{0.5} \times 100$
$\text{mass of } CO_2 = \frac{60 \times 44 \times 0.5}{12 \times 100} \ g$
$\text{mass of } CO_2 = \frac{1320}{1200} \ g = 1.1 \ g$
Expressing $1.1 \ g$ in the form $........ \times 10^{-1} \ g$:
$1.1 \ g = 11 \times 10^{-1} \ g$
Thus,the missing value is $11$.

(B) $pK_{In} = -\log(4 \times 10^{-10}) = -(\log 4 + \log 10^{-10}) = -(0.6 - 10) = 9.4$.
The indicator range is given by $pK_{In} \pm 1$,which is $8.4$ to $10.4$.
Statement $A$: Incorrect. Phenolphthalein is not suitable for weak acid-weak base titrations because the pH change is not sharp enough.
Statement $B$: Correct. The colour change range starts at $pH = pK_{In} - 1 = 9.4 - 1 = 8.4$.
Statement $C$: Incorrect. Phenolphthalein is a weak organic acid $(HIn)$.
Statement $D$: Correct. In an acidic medium,the equilibrium $HIn \rightleftharpoons H^+ + In^-$ shifts to the left,keeping it in the unionized $HIn$ form,which is colourless.
Therefore,statements $B$ and $D$ are correct. The total number of correct statements is $2$.

(B) Using the ideal gas equation $PV = nRT$,since $T$ is constant,$n = \frac{PV}{RT}$.
For $CH_4$: $n_1 = \frac{2 \times 2}{RT} = \frac{4}{RT}$.
For $CO_2$: $n_2 = \frac{4 \times 3}{RT} = \frac{12}{RT}$.
For $Ne$: $n_3 = \frac{3 \times 4}{RT} = \frac{12}{RT}$.
Total moles $n_T = n_1 + n_2 + n_3 = \frac{4 + 12 + 12}{RT} = \frac{28}{RT}$.
Total volume $V_T = 2 + 3 + 4 = 9 \ L$.
Final pressure $P_T = \frac{n_T RT}{V_T} = \frac{28}{RT} \times \frac{RT}{9} = \frac{28}{9} \approx 3.11 \ atm$.
The nearest integer is $3$.

(B) Statement $A$ is correct: Line emission spectra provide information about the electronic structure of atoms.
Statement $B$ is incorrect: Atoms in the gas phase emit line spectra (discrete wavelengths),not a continuous spectrum.
Statement $C$ is correct: An absorption spectrum consists of dark lines in a continuous spectrum,which corresponds to the bright lines in an emission spectrum,acting like a photographic negative.
Statement $D$ is correct: Helium was indeed discovered in the sun's spectrum during a solar eclipse.
Therefore,only $1$ statement (Statement $B$) is incorrect.

(A) To determine the number of significant figures,we follow these rules:
$1$. All non-zero digits are significant.
$2$. Zeros between non-zero digits are significant.
$3$. Leading zeros are not significant.
$4$. Trailing zeros in a number with a decimal point are significant.
Analysis of the given values:
- $0.00253$: The leading zeros are not significant. The significant figures are $2, 5, 3$. Total $= 3$.
- $1.0003$: Zeros between non-zero digits are significant. Total $= 5$.
- $15.0$: Trailing zeros in a decimal number are significant. Total $= 3$.
- $163$: All digits are non-zero. Total $= 3$.
Thus,$A, C,$ and $D$ have $3$ significant figures each.
Therefore,the correct option is $A$.

(C) The breakdown of ozone in the stratosphere by chlorofluorocarbons $(CFCs)$ involves the following steps:
$1$. Photodissociation of $CFCs$: $CF_2Cl_{2(g)} \stackrel{uv}{\longrightarrow} \dot{Cl}_{(g)} + \dot{C}F_2Cl_{(g)}$
$2$. Reaction of chlorine radical with ozone: $\dot{Cl}_{(g)} + O_{3(g)} \longrightarrow Cl\dot{O}_{(g)} + O_{2(g)}$
$3$. Reaction of chlorine monoxide radical with atomic oxygen: $Cl\dot{O}_{(g)} + O_{(g)} \longrightarrow \dot{Cl}_{(g)} + O_{2(g)}$
The reaction $2Cl\dot{O} \longrightarrow ClO_{2(g)} + Cl_{(g)}$ is not a standard step in the catalytic cycle of ozone depletion by chlorine radicals.

(A) In a non-polar solvent,the surfactant molecules form reverse micelles. The polar heads,which are hydrophilic,aggregate in the center to minimize contact with the non-polar solvent,while the non-polar tails,which are lipophilic,extend outwards into the non-polar solvent. This structure is represented by option $A$.

(A) The empirical formula $C_6H_6O$ corresponds to phenol $(C_6H_5OH)$.
Phenol is an aromatic compound,which burns with a sooty flame.
When phenol reacts with bromine $(Br_2)$ in a low polarity solvent like $CS_2$ or $CHCl_3$ at low temperature,the electrophilic substitution occurs primarily at the para-position due to steric hindrance at the ortho-position.
Thus,the major product $B$ is $4$-bromophenol.

(D) The electrolysis of brine ($NaCl$ solution) involves the following reactions:
$NaCl_{(aq)} \longrightarrow Na^{+}_{(aq)} + Cl^{-}_{(aq)}$
At the anode: $2Cl^{-}_{(aq)} \longrightarrow Cl_{2_{(g)}} + 2e^-$
At the cathode: $2H_2O_{(\ell)} + 2e^- \longrightarrow H_{2_{(g)}} + 2OH^{-}_{(aq)}$
As a result,$H_2$ gas is evolved at the cathode,and $OH^{-}$ ions are produced in the solution near the cathode,leading to the formation of $NaOH$.

(D) The cell reaction is the sum of the reduction of $Pd^{2+}$ and the oxidation of $Pd_{(s)}$ to $PdCl_4^{2-}$.
$Pd^{2+}{(aq)} + 2e^{-} \rightleftharpoons Pd_{(s)} \quad E^{\circ}_{red} = 0.83 \ V$
$Pd_{(s)} + 4Cl^{-}{(aq)} \rightleftharpoons PdCl_4^{2-}{(aq)} + 2e^{-} \quad E^{\circ}_{ox} = -0.65 \ V$
Adding these,the net cell reaction is $Pd^{2+}{(aq)} + 4Cl^{-}{(aq)} \rightleftharpoons PdCl_4^{2-}{(aq)}$ with $E^{\circ}_{cell} = 0.83 - 0.65 = 0.18 \ V$.
The relationship between equilibrium constant $K$ and standard cell potential is $\log K = \frac{n E^{\circ}_{cell}}{0.0591} \approx \frac{n E^{\circ}_{cell}}{0.06}$.
Here,$n = 2$ (number of electrons transferred).
$\log K = \frac{2 \times 0.18}{0.06} = \frac{0.36}{0.06} = 6$.
Thus,the logarithm of the equilibrium constant is $6$.

(D) Using the Arrhenius equation: $\log \frac{K_2}{K_1} = \frac{E_a}{2.303 R} \left( \frac{T_2 - T_1}{T_1 T_2} \right)$
Given: $K_1 = 0.03 \, min^{-1}$ at $T_1 = 200 \, K$,$K_2 = 0.05 \, min^{-1}$ at $T_2 = 300 \, K$,$R = 8.3 \, J \, K^{-1} \, mol^{-1}$,$\ln 10 = 2.3$
$\log \frac{0.05}{0.03} = \frac{E_a}{2.3 \times 8.3} \left( \frac{300 - 200}{200 \times 300} \right)$
$\log \left( \frac{5}{3} \right) = \frac{E_a}{19.09} \times \frac{100}{60000}$
$\log 5 - \log 3 = \frac{E_a}{19.09} \times \frac{1}{600}$
$0.70 - 0.48 = \frac{E_a}{11454}$
$0.22 = \frac{E_a}{11454}$
$E_a = 0.22 \times 11454 = 2519.88 \, J \approx 2520 \, J$

(D) The osmotic pressure formula is $\pi = CRT$,where $C = \frac{n}{V}$.
Given: $\pi = 400 \ Pa = 400 \times 10^{-5} \ bar = 0.004 \ bar$.
$V = 250 \ mL = 0.25 \ L$.
$T = 27 + 273 = 300 \ K$.
$R = 0.083 \ L \ bar \ K^{-1} \ mol^{-1}$.
$n = \frac{mass}{Molar \ mass} = \frac{2.5}{M_o}$.
Substituting the values: $0.004 = \frac{2.5 / M_o}{0.25} \times 0.083 \times 300$.
$0.004 = \frac{10}{M_o} \times 24.9$.
$M_o = \frac{249}{0.004} = 62250 \ g \ mol^{-1}$.

(C) Let us analyze each reaction:
$1$. Benzamide reacts with $Br_2$ and $NaOH$ (Hofmann bromamide degradation) to form aniline,which is an aromatic amine.
$2$. Potassium phthalimide does not react with chlorobenzene because aryl halides are poor substrates for $S_N2$ reactions due to the partial double bond character of the $C-Cl$ bond.
$3$. Nitrobenzene undergoes catalytic hydrogenation with $H_2/Pd-C$ to form aniline,which is an aromatic amine.
$4$. Acetanilide undergoes acid-catalyzed hydrolysis with $dil. H_2SO_4$ and heat to form aniline and acetic acid. Aniline is an aromatic amine.
Thus,reactions $1$,$3$,and $4$ result in the formation of aromatic amines. The total count is $3$.

(B) To determine the number of chlorine atoms in each compound:
$1$. Chloral $(CCl_3CHO)$: Contains $3$ chlorine atoms.
$2$. Gammaxene (Benzene hexachloride,$C_6H_6Cl_6$): Contains $6$ chlorine atoms.
$3$. Chloropicrin $(CCl_3NO_2)$: Contains $3$ chlorine atoms.
$4$. Freon $-12$ $(CCl_2F_2)$: Contains $2$ chlorine atoms.
Comparing these,Gammaxene has the maximum number of chlorine atoms $(6)$.
Therefore,the correct option is $B$.

(D) Chloroxylenol,Bithional,and Terpineol are used as disinfectants or antiseptics.
Veronal is a barbiturate derivative used as a tranquilizer (neurological medicine).
Prontosil is an antibacterial drug (antibiotic).
Therefore,$C$ and $D$ are not used as disinfectants.

(B) $1$. Cyclohexylamine $(C_6H_{11}NH_2)$ reacts with nitrous acid $(HNO_2)$ to form cyclohexanol $(P)$ $(C_6H_{11}OH)$.
$2$. Cyclohexanol $(P)$ on oxidation with $PCC$ (Pyridinium chlorochromate) gives cyclohexanone $(Q)$ $(C_6H_{10}O)$.
$3$. Cyclohexanone $(Q)$ undergoes an aldol condensation reaction when heated with dilute $NaOH$ to form the final product $(R)$. The reaction involves the formation of a $\beta$-hydroxy ketone which then undergoes dehydration to form an $\alpha,\beta$-unsaturated ketone. The product $(R)$ is $2$-cyclohexylidenecyclohexanone.

(C) In the Borax Bead Test,cupric sulphate $(CuSO_4)$ reacts with boric anhydride $(B_2O_3)$ to form cupric metaborate $(Cu(BO_2)_2)$,which is blue in an oxidizing flame.
In a luminous (reducing) flame,the cupric metaborate $(Cu(BO_2)_2)$ is reduced to cuprous metaborate $(CuBO_2)$,which is colourless,or to metallic copper $(Cu)$,which appears red and opaque.
Statement $I$ is false because the bead does not become green in a luminous flame; it becomes colourless or red.
Statement $II$ is false because the formation of copper$(I)$ metaborate $(CuBO_2)$ results in a colourless bead,not a green one.
Therefore,both statements are false.

(D) Incorrect. $\Delta T_b = i \times K_b \times m$. For $NaCl$,$i=2$,and for urea,$i=1$. Thus,elevation is not the same.
$(B)$ Correct. Azeotropic mixtures boil at a constant temperature and have the same composition in both liquid and vapour phases.
$(C)$ Incorrect. Osmosis always takes place from a hypotonic solution (lower concentration) to a hypertonic solution (higher concentration) through a semi-permeable membrane.
$(D)$ Correct. Molarity $M = \frac{\% \times 10 \times d}{M_{solute}} = \frac{32 \times 10 \times 1.26}{98} \approx 4.11 \ M$,which is approximately $4.09 \ M$.
$(E)$ Incorrect. When $KI$ is added to $AgNO_3$ (excess $AgNO_3$),$AgI$ adsorbs $Ag^+$ ions to form a positively charged sol $(AgI/Ag^+)$.
Therefore,only statements $(B)$ and $(D)$ are correct.

(A) $(i)$ Formation of a tetraacetate with $Ac_2O$ indicates that compound $A$ contains four $-OH$ groups.
$(ii)$ Oxidation of $A$ with $Br_2-H_2O$ yields an acid $C_5H_{10}O_6$,which confirms the presence of an aldehyde group $(-CHO)$ in $A$,as $Br_2-H_2O$ specifically oxidizes aldoses to aldonic acids.
$(iii)$ Reduction of $A$ with $HI$ gives isopentane $(2-methylbutane)$,which indicates the carbon skeleton is branched (isopentane structure).
$(iv)$ Combining these facts,the structure must be a branched pentose. The structure in option $A$ represents $2,3,4-trihydroxy-2-(hydroxymethyl)butanal$,which fits the molecular formula $C_5H_{10}O_5$ and the branched skeleton.

(A) . Physisorption involves weak van der Waals forces with low enthalpy of adsorption $(20-40 \, kJ \, mol^{-1})$.
$B$. Chemisorption involves strong chemical bonds and is typically unimolecular (single layer).
$C$. The reaction $N_{2(g)} + 3H_{2(g)} \xrightarrow{Fe(s)} 2NH_{3(g)}$ is an example of heterogeneous catalysis where the catalyst $(Fe)$ is in a solid phase and reactants are in a gaseous phase.
$D$. Chromatography is a common analytical application based on the principle of differential adsorption.
Therefore,the correct matching is: $A-II, B-I, C-IV, D-III$.

(C) The first ionization enthalpy of $3d$ series elements is generally comparable to or lower than that of group $2$ metals (alkaline earth metals) for the early members of the series. For example,the first ionization energy of $Mg$ is $737 \ kJ/mol$,while for $Sc$ it is $631 \ kJ/mol$ and for $Ti$ it is $656 \ kJ/mol$. Thus,Assertion $(A)$ is false.
Reason $(R)$ states that in the $3d$ series of elements,successive filling of $d$-orbitals takes place. This is a correct statement as the electronic configuration of $3d$ series elements involves the filling of the $3d$ subshell.
Therefore,$(A)$ is false but $(R)$ is true.

(D) $1$. The molecular formula $C_4H_{11}N$ corresponds to a saturated amine.
$2$. The reaction with $HNO_2$ to release $N_2$ gas indicates that the compound is a primary aliphatic amine $(R-NH_2)$.
$3$. The reaction with $PhSO_2Cl$ (Hinsberg reagent) produces a sulfonamide that is soluble in $KOH$. This confirms that the amine is a primary amine,as the resulting sulfonamide contains an acidic hydrogen on the nitrogen atom.
$4$. The compound must be optically active. Among the primary amines with the formula $C_4H_{11}N$,butan$-2-$amine $(CH_3CH_2CH(NH_2)CH_3)$ contains a chiral carbon atom and is therefore optically active.
$5$. Butan$-1-$amine is achiral.

(B) The Van-Arkel method is primarily used for the purification of metals like $Ti$,$Zr$,$Hf$,and $B$. It is not used for tungsten $(W)$. Therefore,the statement in option $B$ is incorrect.

(A) The electronic configurations of the given lanthanoids in their ground state are as follows:
$1.$ ${}_{62}Sm: [Xe] 4f^6 6s^2$
$2.$ ${}_{63}Eu: [Xe] 4f^7 6s^2$
$3.$ ${}_{64}Gd: [Xe] 4f^7 5d^1 6s^2$
$4.$ ${}_{65}Tb: [Xe] 4f^9 6s^2$
$5.$ ${}_{61}Pm: [Xe] 4f^5 6s^2$
$A$ half-filled $f$-orbital contains $7$ electrons $(f^7)$.
From the configurations above,$Eu$ $(4f^7 6s^2)$ and $Gd$ $(4f^7 5d^1 6s^2)$ both possess a half-filled $4f$ subshell.
Therefore,the correct elements are $B$ $(Eu)$ and $D$ $(Gd)$.

(C) For $[Ti(H_2O)_6]^{3+}$,the central metal ion is $Ti^{3+}$,which has a $3d^1$ electronic configuration.
The $CFSE$ for a $d^1$ octahedral complex is given by $CFSE = -0.4 \times \Delta_0$.
Given $CFSE = -96.0 \ kJ / mol = -96000 \ J / mol$.
Therefore,$\Delta_0 = \frac{96000}{0.4} = 240000 \ J / mol$.
To find the wavelength $(\lambda)$ in $nm$,we use the relation $\Delta_0 = \frac{N_A \times h \times c}{\lambda}$.
$\lambda = \frac{N_A \times h \times c}{\Delta_0} = \frac{6 \times 10^{23} \times 6.4 \times 10^{-34} \times 3.0 \times 10^8}{240000}$.
$\lambda = \frac{115.2 \times 10^{-3}}{240000} = 0.48 \times 10^{-6} \ m = 480 \ nm$.

(C) The formula for molar conductivity is $\Lambda_{m} = \frac{\kappa \times 1000}{M}$.
Given that conductivity $\kappa = \frac{1}{\rho}$,where $\rho = 5 \times 10^{-3} \ \Omega \ cm$.
Substituting the values: $\Lambda_{m} = \frac{1}{5 \times 10^{-3}} \times \frac{1000}{0.8}$.
$\Lambda_{m} = 200 \times 1250 = 250,000 \ \Omega^{-1} \ cm^2 \ mol^{-1}$.
Expressing in the required form: $250,000 = 25 \times 10^4 \ \Omega^{-1} \ cm^2 \ mol^{-1}$.
Thus,the nearest integer is $25$.

(D) The haloform test is given by compounds containing the $CH_3CO-$ group or the $CH_3CH(OH)-$ group.
Let us analyze the given molecules:
$1$. $2,4$-dimethylacetophenone: Contains $CH_3CO-$ group attached to an aromatic ring. It gives a positive haloform test.
$2$. Ethyl acetoacetate: Contains $CH_3CO-$ group. It gives a positive haloform test.
$3$. Acetophenone: Contains $CH_3CO-$ group attached to a phenyl ring. It gives a positive haloform test.
$4$. Pentan-$3$-ol: Does not contain $CH_3CH(OH)-$ group. It does not give the haloform test.
$5$. Butan-$2$-ol: Contains $CH_3CH(OH)-$ group. It gives a positive haloform test.
$6$. $3,3,5,5$-tetramethylcyclohexanone: Does not contain the required group. It does not give the haloform test.
Thus,there are $4$ molecules that give a positive haloform test.

(C) For a first order reaction,the rate constant $k$ is given by $k = \frac{2.303}{t} \log \frac{[A]_0}{[A]_t}$.
Given $[A]_t = \frac{[A]_0}{32}$ and $k = 20 \, min^{-1}$.
Substituting the values: $20 = \frac{2.303}{t} \log \frac{[A]_0}{[A]_0 / 32} = \frac{2.303}{t} \log 32$.
Since $32 = 2^5$,$\log 32 = 5 \log 2 = 5 \times 0.3010 = 1.505$.
$t = \frac{2.303 \times 1.505}{20} = \frac{3.466}{20} = 0.1733 \, min$.
Converting to $10^{-2} \, min$: $0.1733 \, min = 17.33 \times 10^{-2} \, min$.
The nearest integer is $17$.

(D) In the metal deficiency defect of $A_{0.95} O$,the total charge must be neutral. Let the number of $A^{3+}$ ions be $x$. Then the number of $A^{2+}$ ions is $(0.95 - x)$.
Equating the total positive charge to the total negative charge (which is $2$ for $O^{2-}$):
$3x + 2(0.95 - x) = 2$
$3x + 1.9 - 2x = 2$
$x = 0.1$
So,there are $0.1$ $A^{3+}$ ions and $0.85$ $A^{2+}$ ions for every $O^{2-}$ ion.
This defect involves the replacement of $A^{2+}$ ions by $A^{3+}$ ions,creating cation vacancies to maintain electrical neutrality. Structure $C$ correctly shows the presence of $A^{2+}$,$A^{3+}$,and a vacant cation site (represented by an empty circle with a dot).

(A) The given Fischer projection of sugar $[X]$ is $D-$mannose. In $D-$mannose,the $-OH$ groups at $C-2$,$C-3$,$C-4$ and $C-5$ are on the left,left,right and right respectively in the Fischer projection. When converting to the Haworth structure,groups on the left side of the Fischer projection are placed above the ring,and groups on the right side are placed below the ring. For $\alpha -D-$mannopyranose,the $-OH$ group at the anomeric carbon $(C-1)$ is below the ring. Thus,the correct Haworth structure corresponds to option $A$.

(A) The structure of $Mn_2O_7$ consists of two $MnO_4$ tetrahedra sharing a common oxygen atom at one corner.
Thus,each $Mn$ atom is tetrahedrally surrounded by four oxygen atoms (statement $A$ is correct).
The structure contains an $Mn-O-Mn$ bridge (statement $C$ is correct).
There is no direct $Mn-Mn$ bond in $Mn_2O_7$ (statement $D$ is incorrect).
Therefore,statements $A$ and $C$ are correct.

(B) The rate of dehydration of alcohols is directly proportional to the stability of the carbocation formed as an intermediate.
$(a)$ Phenol: The carbocation formed would be highly unstable due to the $sp^2$ hybridized carbon atom.
$(b)$ Cyclohexa$-2,4-$dien$-1-$ol: Forms a resonance-stabilized carbocation (aromatic character can be achieved).
$(c)$ Cyclohex$-2-$en$-1-$ol: Forms a resonance-stabilized allylic carbocation.
$(d)$ Cyclohexanol: Forms a secondary carbocation.
Comparing the stability: $(b) > (c) > (d) > (a)$.
Therefore,the decreasing order of dehydration is $b > c > d > a$.

(A) Assertion $A$ is true: The extent of adsorption of gases on activated charcoal increases with an increase in the magnitude of van der Waals forces of attraction. Since $Kr$ has the largest atomic size among the given noble gases,it has the strongest van der Waals forces,leading to maximum adsorption.
Reason $R$ is false: The critical temperature $T_c$ is the primary factor determining the ease of liquefaction and adsorption. While $V_c$ and $P_c$ values vary,the statement regarding $Z_c$ is incorrect because for all real gases following the van der Waals equation,the compressibility factor at the critical point $Z_c = \frac{P_c V_c}{R T_c}$ is a constant value of $\frac{3}{8} = 0.375$.

(B) The reaction involves the nucleophilic attack of the $-NH_2$ group on the electrophilic carbonyl carbon of diethyl carbonate,followed by the loss of an ethanol molecule to form a carbamate intermediate.
Subsequently,the lone pair of the oxygen atom from the $-CH_2OH$ group attacks the carbonyl carbon of the carbamate,leading to the elimination of another molecule of ethanol and the formation of a cyclic five-membered oxazolidin$-2-$one derivative.

(B) In option $B$,the reaction involves a primary alkyl halide $(CH_3)_3CCH_2Br$ with alcoholic $KOH$. Alcoholic $KOH$ acts as a strong base and promotes an elimination reaction ($E2$ mechanism) to form an alkene. However,the reaction shown is a substitution reaction,which is incorrect for alcoholic $KOH$. Aqueous $KOH$ would be required for substitution to form an alcohol.

(D) The reaction between $FeCl_3$ and $K_4[Fe(CN)_6]$ is a standard test for $Fe^{3+}$ ions.
When $Fe^{3+}$ ions react with potassium ferrocyanide,$K_4[Fe(CN)_6]$,they form a deep blue complex known as Prussian blue.
The chemical reaction is: $4Fe^{3+} + 3[Fe(CN)_6]^{4-} \rightarrow Fe_4[Fe(CN)_6]_3$.
The resulting compound is ferric ferrocyanide,$Fe_4[Fe(CN)_6]_3$,which is responsible for the Prussian blue color.

(A) Double salts are addition compounds that dissociate into their constituent ions completely when dissolved in water.
$FeSO_4 \cdot (NH_4)_2SO_4 \cdot 6H_2O$ (Mohr's salt) and $K_2SO_4 \cdot Al_2(SO_4)_3 \cdot 24H_2O$ (Potash alum) are examples of double salts.
$CuSO_4 \cdot 4NH_3 \cdot H_2O$ and $Fe(CN)_2 \cdot 4KCN$ (which is $K_4[Fe(CN)_6]$) are coordination compounds (complex salts) that do not dissociate completely into their constituent ions in solution.
Therefore,$A$ and $C$ are double salts.

(C) The magnitude of crystal field splitting $(\Delta_o)$ depends on the strength of the ligand.
According to the spectrochemical series,the strength of ligands is $F^- < C_2O_4^{2-} < NH_3 < CN^-$.
Since $CN^-$ is the strongest field ligand among the given options,it causes the maximum splitting of $d$-orbitals.

(D) The reaction for the oxidation of carbon is $2C(s) + O_2(g) \rightarrow 2CO(g)$.
The entropy change $\Delta_{r}S^{\circ}$ is positive because the number of moles of gaseous products is greater than the number of moles of gaseous reactants.
According to the Gibbs-Helmholtz equation,$\Delta_{r}G^{\circ} = \Delta_{r}H^{\circ} - T\Delta_{r}S^{\circ}$. Since $\Delta_{r}S^{\circ} > 0$,the slope of the line in the Ellingham diagram (which is $-\Delta_{r}S^{\circ}$) is negative.
Therefore,Assertion $A$ is correct.
Carbon monoxide $(CO)$ is a very stable compound at high temperatures,and its tendency to decompose decreases as temperature increases. Thus,Reason $R$ is incorrect.

(C) . Molisch's Test is a general test for carbohydrates $(II)$.
$B$. Biuret Test is used to detect the presence of peptide bonds in proteins $(I)$.
$C$. Carbylamine Test is a chemical test for the detection of primary amines $(III)$.
$D$. Schiff's Test is used for the detection of aldehydes $(IV)$.
Therefore,the correct matching is $A-II, B-I, C-III, D-IV$.

(A) The molarity $(M)$ is $3 \ M$ and density $(d)$ is $1.0 \ g \ mL^{-1}$.
The molar mass of $NaCl$ is $23 + 35.5 = 58.5 \ g \ mol^{-1}$.
The formula for molality $(m)$ is $m = \frac{1000 \times M}{1000 \times d - M \times M_{solute}}$.
Substituting the values: $m = \frac{1000 \times 3}{1000 \times 1 - 3 \times 58.5} = \frac{3000}{1000 - 175.5} = \frac{3000}{824.5} \approx 3.6385 \ m$.
Expressing in $10^{-2} \ m$: $3.6385 \times 10^2 \times 10^{-2} \approx 364 \times 10^{-2} \ m$.
The nearest integer is $364$.

(A) The reduction half-reaction is: $MnO_4^{-} + 8H^{+} + 5e^{-} \rightleftharpoons Mn^{2+} + 4H_2O$
Using the Nernst equation: $E = E^{\circ} - \frac{0.059}{5} \log \frac{[Mn^{2+}]}{[MnO_4^{-}] [H^{+}]^8}$
Substituting the given values: $1.282 = 1.54 - \frac{0.059}{5} \log \frac{10^{-3}}{10^{-1} \times [H^{+}]^8}$
$1.282 - 1.54 = -\frac{0.059}{5} \log \frac{10^{-2}}{[H^{+}]^8}$
$-0.258 = -\frac{0.059}{5} \log \frac{10^{-2}}{[H^{+}]^8}$
$\frac{0.258 \times 5}{0.059} = \log (10^{-2}) - \log ([H^{+}]^8)$
$21.86 = -2 + 8 \ pH$
$8 \ pH = 23.86$
$pH = 2.98 \approx 3$

(A) The molecular formula $C_9H_{10}O$ has a degree of unsaturation of $6$.
Given conditions:
$(i)$ Does not dissolve in $NaOH$ (not a phenol).
$(ii)$ Does not dissolve in $HCl$ (not an amine).
$(iii)$ Does not give orange precipitate with $2,4-DNP$ (no carbonyl group,i.e.,no aldehyde or ketone).
$(iv)$ On hydrogenation,it gives an identical compound $C_9H_{12}O$.
Since it is not a carbonyl compound and not a phenol,it is likely an ether.
The structure $Ph-CH=CH-O-CH_3$ (methyl styryl ether) satisfies these conditions.
It exists as $cis$ and $trans$ isomers.
Both isomers on hydrogenation yield $Ph-CH_2-CH_2-O-CH_3$ $(C_9H_{12}O)$.
Thus,there are $2$ such isomers.

(A) The reaction is: $KCl + AgNO_3 \rightarrow AgCl + KNO_3$.
At the equivalence point,the millimoles of $KCl$ equal the millimoles of $AgNO_3$:
$n(KCl) = 20 \ mL \times 1 \ M = 20 \ mmol = 0.02 \ mol$.
Mass of the solution $= 25 \ mL \times 1 \ g \ mL^{-1} = 25 \ g$.
Mass of $KCl$ solute $= 0.02 \ mol \times 74.5 \ g \ mol^{-1} = 1.49 \ g$.
Mass of solvent $= 25 \ g - 1.49 \ g = 23.51 \ g = 0.02351 \ kg$.
Molality $(m) = \frac{0.02 \ mol}{0.02351 \ kg} \approx 0.8507 \ mol \ kg^{-1}$.
Since $KCl$ undergoes $100 \%$ ionization,the van't Hoff factor $i = 2$.
Depression in freezing point $\Delta T_f = i \times K_f \times m = 2 \times 2.0 \times 0.8507 = 3.4028 \ K$.
The nearest integer is $3$.

(A) The starting material is salicylic acid ($2$-hydroxybenzoic acid).
$1$. Reaction with '$X$' leads to the acetylation of the phenolic $-OH$ group to form aspirin (acetylsalicylic acid). The reagent used for acetylation is acetic anhydride $((CH_3CO)_2O)$ in the presence of an acid catalyst $(H^+)$.
$2$. Reaction with '$Y$' leads to the esterification of the carboxylic acid $-COOH$ group to form methyl salicylate. The reagent used for esterification is methanol $(CH_3OH)$ in the presence of an acid catalyst $(H^+)$ and heat $(\Delta)$.
Therefore,'$X$' is $(CH_3CO)_2O / H^{+}$ and '$Y$' is $CH_3OH / H^{+}, \Delta$.

(A) Vitamin $C$ (ascorbic acid) is most stable in the form where the enediol group is present,which allows for resonance stabilization and intramolecular hydrogen bonding.
Structure $A$ represents the naturally occurring form of $L$-ascorbic acid,which is stabilized by intramolecular $H$-bonding between the carbonyl oxygen and the hydroxyl hydrogen.

(C) The given reaction is a nucleophilic substitution reaction of a tertiary alkyl halide,$(C_6H_5)_3C-Cl$,which proceeds via an $SN_1$ mechanism.
In an $SN_1$ reaction,the rate-determining step is the formation of the carbocation,which depends only on the concentration of the alkyl halide.
The rate law for this reaction is: $\text{Rate} = k[(C_6H_5)_3C-Cl]$.
Since the rate is directly proportional to the concentration of the alkyl halide,the graph of rate versus $[(C_6H_5)_3C-Cl]$ should be a straight line passing through the origin,as shown in option $C$.

(C) The complex cation $[Co(NH_3)_5NO_2]^{2+}$ exhibits linkage isomerism.
This occurs because the $NO_2^-$ ligand is an ambidentate ligand,which can coordinate to the central metal atom through either the nitrogen atom (nitro isomer) or the oxygen atom (nitrito isomer).

(D) The chemical formula for Nessler's reagent is $K_2[HgI_4]$.
It consists of potassium $(K)$,mercury $(Hg)$,and iodine $(I)$.
Therefore,oxygen $(O)$ is not present in Nessler's reagent.

(B) Assertion $(A)$ is correct. The reaction of $\alpha$-halocarboxylic acids with aqueous $NH_3$ is a standard method for preparing $\alpha$-amino acids because the reaction is controlled and the product is stable. In contrast,the reaction of alkyl halides with $NH_3$ (ammonolysis) often leads to a mixture of primary,secondary,and tertiary amines,resulting in a low yield of the primary amine.
Reason $(R)$ is correct. Amino acids do exist as zwitter ions in aqueous medium due to the internal acid-base reaction between the $-COOH$ and $-NH_2$ groups.
However,the zwitter ion nature of amino acids is not the reason why $\alpha$-halocarboxylic acids give a better yield of amino acids compared to alkyl halides. The yield difference is due to the nature of the substitution reaction and the stability of the products formed.
Therefore,both $(A)$ and $(R)$ are correct,but $(R)$ is not the correct explanation of $(A)$.

(D) $1$. The reaction of the aldehyde with $NaCN$ and $H_2O$ (cyanohydrin formation) yields the cyanohydrin product $[A]$. The structure of $[A]$ is a $2$-hydroxy-$3$-methylhexanenitrile derivative.
$2$. The reduction of the nitrile group $(-CN)$ in $[A]$ using $LiAlH_4$ produces a primary amine $(-CH_2NH_2)$. Thus, $[B]$ is the corresponding amino-alcohol.
$3$. The acid-catalyzed hydrolysis of the nitrile group $(-CN)$ in $[A]$ using $HCl/H_2O$ and heat $(\Delta)$ converts it into a carboxylic acid group $(-COOH)$. Thus, $[C]$ is the corresponding hydroxy-acid.
$4$. Comparing these structures with the given options, option $D$ correctly represents the structures of $[A]$, $[B]$, and $[C]$.

(B) The stability of $Cu^{2+}(aq)$ in aqueous solution is higher than that of $Cu^{+}(aq)$.
This is because the enthalpy of hydration $(\Delta_{hyd}H)$ for $Cu^{2+}$ is much more negative (higher magnitude) than that of $Cu^{+}$.
This large negative hydration energy of $Cu^{2+}$ more than compensates for the energy required for the second ionization enthalpy of $Cu$,making $Cu^{2+}$ more stable in water.
Therefore,Assertion $(A)$ is correct,but Reason $(R)$ is incorrect because the enthalpy of hydration for $Cu^{2+}$ is much more negative (greater in magnitude) than that of $Cu^{+}$,not less.

(B) The Freundlich adsorption isotherm is given by the equation: $\frac{x}{m} = K \cdot P^{1/n}$.
Taking logarithm on both sides,we get: $\log(\frac{x}{m}) = \log K + \frac{1}{n} \log P$.
Comparing this with the equation of a straight line $y = mx + c$,where $y = \log(\frac{x}{m})$,$x = \log P$,the slope $m = \frac{1}{n}$,and the intercept $c = \log K$.
Given the equation $y = 0.3x + 0.7033$,we have:
Slope $(m)$ = $\frac{1}{n} = 0.3$.
Intercept $(c)$ = $\log K = 0.7033$.
Thus,the values of $\frac{1}{n}$ and $\log K$ are $0.3$ and $0.7033$ respectively.

(C) $KOH$ is a strong base that reacts with atmospheric $CO_2$ to form potassium carbonate $(K_2CO_3)$.
$2KOH + CO_2 \rightarrow K_2CO_3 + H_2O$
Due to this reaction,the concentration of the $KOH$ solution decreases over time when exposed to air.
Therefore,it is necessary to standardize or check the concentration of $KOH$ solution before using it in volumetric analysis to ensure accuracy.
Thus,both Assertion $(A)$ and Reason $(R)$ are true,and $(R)$ is the correct explanation of $(A).$

(D) Copper matte is a mixture of $Cu_2S$ and $FeS$.
Among the given options,only $Cu_2S$ is present in copper matte.
Therefore,the number of compounds present in copper matte from the given list is $1$.

(B) Tranquilizers are a class of chemical compounds used for the treatment of stress,and mild or even severe mental diseases.
$1.$ Chlordiazepoxide is a well-known tranquilizer.
$2.$ Veronal is a derivative of barbituric acid and acts as a tranquilizer.
$3.$ Valium (Diazepam) is a widely used tranquilizer.
$4.$ Salvarsan is an antibiotic used for the treatment of syphilis.
Therefore,there are $3$ tranquilizers among the given options.