JEE Main 2026 Chemistry Question Paper with Answer and Solution

(A) $1$. Calculate initial moles:
$n_A = \frac{36.0 \ g}{60 \ g \ mol^{-1}} = 0.6 \ mol$
$n_B = \frac{56.0 \ g}{80 \ g \ mol^{-1}} = 0.7 \ mol$
$2$. Determine limiting reagent:
According to the reaction $A + 2B \rightarrow AB_{2}$,$1 \ mol$ of $A$ requires $2 \ mol$ of $B$.
For $0.6 \ mol$ of $A$,required $B = 0.6 \times 2 = 1.2 \ mol$.
Since we have only $0.7 \ mol$ of $B$,'$B$' is the limiting reagent.
Statement $(A)$ is incorrect.
$3$. Calculate product formed:
$0.7 \ mol$ of $B$ will produce $\frac{0.7}{2} = 0.35 \ mol$ of $AB_{2}$.
Molar mass of $AB_{2} = 60 + (2 \times 80) = 220 \ g \ mol^{-1}$.
Mass of $AB_{2} = 0.35 \ mol \times 220 \ g \ mol^{-1} = 77.0 \ g$.
Statement $(B)$ and $(C)$ are correct.
$4$. Calculate unreacted '$A$':
$B$ is limiting,so $0.7 \ mol$ of $B$ reacts with $\frac{0.7}{2} = 0.35 \ mol$ of $A$.
Remaining moles of $A = 0.6 - 0.35 = 0.25 \ mol$.
Mass of remaining $A = 0.25 \ mol \times 60 \ g \ mol^{-1} = 15.0 \ g$.
Statement $(D)$ is correct.
Thus,statements $(B)$,$(C)$,and $(D)$ are correct. Based on the options provided,$(C)$ and $(D)$ are correct.

(B) Step $1$: Dehydrohalogenation of $1,2-$dibromopentane with $2NaNH_2$ gives pent$-1-$yne $(X = CH_3CH_2CH_2C \equiv CH)$.
Step $2$: Treatment of pent$-1-$yne with $NaNH_2$ forms the sodium acetylide salt $(CH_3CH_2CH_2C \equiv C^-Na^+)$.
Step $3$: The acetylide ion acts as a nucleophile and attacks isopropyl bromide $((CH_3)_2CHBr)$ via an $S_N2$ mechanism.
$CH_3CH_2CH_2C \equiv C^- + (CH_3)_2CHBr \rightarrow CH_3CH_2CH_2C \equiv C-CH(CH_3)_2$.
The resulting product is $5-$methylhex$-2-$yne.

(B) The reaction is $A_{2(g)} \rightleftharpoons 2A_{(g)}$.
First,calculate the standard Gibbs free energy change for the reaction: $\Delta_r G^{\circ} = 2 \times \Delta G_f^{\circ}(A) - \Delta G_f^{\circ}(A_2) = 2 \times (-50.832) - (-100.00) = -101.664 + 100.00 = -1.664 \ kJ \ mol^{-1} = -1664 \ J \ mol^{-1}$.
Using the relation $\Delta_r G^{\circ} = -RT \ln K_p$:
$-1664 = -8.3 \times 300 \times \ln K_p$
$\ln K_p = \frac{1664}{2490} \approx 0.668 \approx 0.693$ (using the provided $\ln 2 = 0.693$ approximation).
Thus,$K_p = 2$.
For the dissociation $A_2 \rightleftharpoons 2A$,if $\alpha$ is the degree of dissociation,$K_p = \frac{4\alpha^2 P}{(1-\alpha^2)}$.
Given $P = 1 \ bar$,$2 = \frac{4\alpha^2}{1-\alpha^2} \implies 2 - 2\alpha^2 = 4\alpha^2 \implies 6\alpha^2 = 2 \implies \alpha^2 = \frac{1}{3} = 0.3333$.
$\alpha = (33.33 \times 10^{-2})^{1/2}$.
Comparing with $(x \times 10^{-2})^{1/2}$,we get $x = 33$.

(B) The formation of a violet color in the $CHCl_3$ layer upon adding chlorine water indicates the presence of iodine $(I_2)$.
In the Carius method,the precipitate formed is silver iodide $(AgI)$.
The molar mass of $AgI = 107.87 + 126.9 = 234.77 \approx 235 \ g \ mol^{-1}$.
The percentage of iodine is calculated as:
$\% \text{ of } I = \frac{\text{Atomic weight of } I}{\text{Molecular weight of } AgI} \times \frac{\text{mass of precipitate}}{\text{mass of compound}} \times 100$
$\% \text{ of } I = \frac{127}{235} \times \frac{0.12}{0.15} \times 100$
$\% \text{ of } I = 0.5404 \times 0.8 \times 100 = 43.23 \%$.
The nearest integer is $43 \%$.

(B) The bromination of a cycloalkane $(X)$ consuming one mole of $Br_2$ implies an addition reaction across a double bond or ring opening,resulting in a dibromo derivative $(Y)$.
Let the formula of $(Y)$ be $C_n H_m Br_2$.
Given the $C:Br$ ratio is $3:1$,we have $\frac{n}{2} = \frac{3}{1}$,which gives $n = 6$.
The product $(Y)$ is $C_6 H_{10} Br_2$.
The molar mass of $C_6 H_{10} Br_2 = (6 \times 12) + (10 \times 1) + (2 \times 80) = 72 + 10 + 160 = 242 \ g \ mol^{-1}$.
The percentage of bromine in $(Y) = \frac{\text{Mass of } Br}{\text{Molar mass of } Y} \times 100 = \frac{160}{242} \times 100 \approx 66.11 \%$.
Rounding to the nearest integer,we get $66 \%$.

(B) Lewis acid is an electron-deficient species. $BF_3$ is a $p$-block fluoride where Boron has an incomplete octet ($6$ electrons),making it a Lewis acid. Its hybridization is $sp^2$ due to $3$ bond pairs and $0$ lone pairs.
$YF_3$ acts as a Lewis base because the central atom has a lone pair of electrons available for donation. $NF_3$ is a $p$-block fluoride where Nitrogen has $3$ bond pairs and $1$ lone pair,making it a Lewis base. Its hybridization is $sp^3$ due to $4$ electron domains ($3$ bond pairs + $1$ lone pair).
Therefore,the hybridization of $XF_3$ $(BF_3)$ is $sp^2$ and $YF_3$ $(NF_3)$ is $sp^3$.

(A) . Work done in reversible isothermal expansion: $W = -nRT \ln(\frac{V_2}{V_1}) = -2 \times 8.314 \times 300 \times \ln(10) \approx -11488 \ J = -11.5 \ kJ$. Magnitude is $11.5 \ kJ$ $(II)$.
$B$. Work done in irreversible isothermal expansion: $W = -P_{ext}(V_2 - V_1) = -3 \times 10^3 \ Pa \times (3 - 1) \ m^3 = -6000 \ J = -6 \ kJ$. Magnitude is $6 \ kJ$ $(III)$.
$C$. Change in internal energy: $\Delta U = nC_V \Delta T = 1 \times \frac{3}{2} \times 8.314 \times 320 \approx 3990 \ J \approx 4 \ kJ$. Magnitude is $4 \ kJ$ $(I)$.
$D$. Change in enthalpy: $\Delta H = nC_P \Delta T = 1 \times \frac{5}{2} \times 8.314 \times 337 \approx 7004 \ J \approx 7 \ kJ$. Magnitude is $7 \ kJ$ $(IV)$.
Thus,the correct match is $A-II, B-III, C-I, D-IV$.

(B) The balanced chemical equation is: $2Al_{(s)} + 6HCl_{(aq)} \rightarrow 2Al^{3+}_{(aq)} + 6Cl^{-}_{(aq)} + 3H_{2(g)}$.
From the stoichiometry,$2 \ \text{moles of } Al$ produce $3 \ \text{moles of } H_{2(g)}$.
Therefore,$1 \ \text{mole of } Al$ produces $\frac{3}{2} = 1.5 \ \text{moles of } H_{2(g)}$.
At $STP$,$1 \ \text{mole of any gas occupies } 22.4 \ L$.
So,$1.5 \ \text{moles of } H_{2(g)} = 1.5 \times 22.4 \ L = 33.6 \ L \ H_{2(g)}$.
Thus,$33.6 \ L \ H_{2(g)}$ at $STP$ is produced for every mole of $Al$ that reacts.

(B) The energy of an electron in the $n^{th}$ orbit is given by $E_{n} = -R_{H} \times \frac{Z^{2}}{n^{2}}$.
The energy difference between the first $(n=1)$ and second $(n=2)$ orbit for a hydrogen atom $(Z=1)$ is $\Delta E = E_{2} - E_{1} = R_{H} \times \left[ \frac{1}{1^{2}} - \frac{1}{2^{2}} \right]$.
Given $R_{H} = 2.18 \times 10^{-11} \ ergs = 2.18 \times 10^{-18} \ J$.
$\Delta E = 2.18 \times 10^{-18} \times \left[ 1 - \frac{1}{4} \right] = 2.18 \times 10^{-18} \times 0.75 = 1.635 \times 10^{-18} \ J \ atom^{-1}$.
To convert to $J \ mol^{-1}$,multiply by Avogadro's number $(N_{A} = 6.022 \times 10^{23} \ mol^{-1})$:
$\Delta E = 1.635 \times 10^{-18} \times 6.022 \times 10^{23} \approx 9.846 \times 10^{5} \ J \ mol^{-1}$.
Rounding to the provided option,the correct answer is $9.835 \times 10^{5} \ J \ mol^{-1}$.

(C) The formula for formal charge is: $\text{Formal charge} = \text{Valence electrons} - \text{Non-bonding electrons} - \frac{1}{2}(\text{Bonding electrons})$.
For atom $(1)$ (Oxygen): Valence $e^- = 6$,Non-bonding $e^- = 4$,Bonding $e^- = 4$. $\text{FC} = 6 - 4 - \frac{4}{2} = 0$.
For atom $(2)$ (Nitrogen): Valence $e^- = 5$,Non-bonding $e^- = 0$,Bonding $e^- = 8$. $\text{FC} = 5 - 0 - \frac{8}{2} = +1$.
For atom $(3)$ (Oxygen): Valence $e^- = 6$,Non-bonding $e^- = 4$,Bonding $e^- = 4$. $\text{FC} = 6 - 4 - \frac{4}{2} = 0$.
For atom $(4)$ (Oxygen): Valence $e^- = 6$,Non-bonding $e^- = 6$,Bonding $e^- = 2$. $\text{FC} = 6 - 6 - \frac{2}{2} = -1$.
Thus,the formal charges are $0, +1, 0, -1$.

(D) The reaction sequence illustrates the synthesis of phenolphthalein.
Phenolphthalein exists in different structural forms depending on the pH of the solution.
In a basic medium,the molecule exists in a quinonoid form,which is pink in colour.
When treated with excess $NaOH$,the molecule undergoes further deprotonation to form the trianion structure (compound '$X$').
This trianion form lacks the extended conjugated system required for visible light absorption in the visible region,making it colourless.

(D) The conversion of alcohol $(A)$ to alkene $(B)$ is an acid-catalyzed dehydration reaction,which requires a strong acid like $Conc. H_{2}SO_{4}$ or $H_{3}PO_{4}$ as reagent $D$.
Compound $C$ is $HO - CH_{2} - CH_{2} - CH(OH) - CH_{3}$,which is a diol.
When $C$ undergoes dehydration with excess reagent $D$,both hydroxyl $(-OH)$ groups are eliminated to form a conjugated diene.
The product $E$ formed is $CH_{2} = CH - CH = CH_{2}$ (buta$-1,3-$diene).

(A) The reaction sequence typically involves the oxidation of an alkyl side chain on a benzene ring. When an alkyl group (like an ethyl or propyl group) is attached to a benzene ring,treatment with strong oxidizing agents such as alkaline $KMnO_4$ followed by acidification results in the oxidation of the benzylic carbon to a carboxylic acid group $(-COOH)$. The remaining part of the side chain is cleaved. Therefore,the major product $P$ is a benzoic acid derivative.

(C) $n$-Butane is $CH_{3}CH_{2}CH_{2}CH_{3}$.
Monochlorination of $n$-butane yields $2$-chlorobutane $(CH_{3}CHClCH_{2}CH_{3})$,which contains a chiral center and is thus optically active ('$P$').
Further chlorination of $2$-chlorobutane can occur at different carbon positions to form dichloro derivatives.
The possible structural isomers for the dichloro products are:
$1,1$-dichlorobutane
$1,2$-dichlorobutane
$1,3$-dichlorobutane
$2,2$-dichlorobutane
$2,3$-dichlorobutane
Ignoring stereoisomers,there are $5$ distinct structural isomers.

(D) The reaction sequence illustrates the synthesis and structural changes of phenolphthalein in different pH conditions.
Phenolphthalein acts as an acid-base indicator.
In a basic medium,the molecule exists in a quinonoid form,which imparts a pink/red colour.
When excess $NaOH$ is added,the structure undergoes further deprotonation to form the trianion species labeled as '$X$'.
This trianion form of phenolphthalein is colourless because the extended conjugation of the quinonoid system is disrupted.
Therefore,the compound '$X$' is colourless.

(D) The conversion of alcohol $A$ $(pentan-1-ol)$ to alkene $B$ $(pent-1-ene)$ is a dehydration reaction.
Dehydration of alcohols is typically carried out using a strong acid catalyst like concentrated $H_2SO_4$ or $H_3PO_4$ at high temperatures.
Compound $C$ is $butane-1,3-diol$ $(HO-CH_2-CH_2-CH(OH)-CH_3)$.
When $C$ reacts with an excess of a dehydrating agent like concentrated $H_2SO_4$,both hydroxyl $(-OH)$ groups are removed to form two double bonds,resulting in the formation of a diene.
The product $E$ is $buta-1,3-diene$ $(CH_2=CH-CH=CH_2)$.

(C) $n$-Butane $(CH_3-CH_2-CH_2-CH_3)$ on monochlorination gives $2$-chlorobutane $(CH_3-CHCl-CH_2-CH_3)$,which is the optically active compound "$P$".
Further chlorination of $2$-chlorobutane involves the substitution of a hydrogen atom by a chlorine atom at different positions.
The possible dichloro products are:
$1,1$-dichlorobutane $(CH_3-CH_2-CH_2-CHCl_2)$
$1,2$-dichlorobutane $(CH_3-CH_2-CHCl-CH_2Cl)$
$1,3$-dichlorobutane $(CH_3-CHCl-CH_2-CH_2Cl)$
$2,2$-dichlorobutane $(CH_3-CCl_2-CH_2-CH_3)$
$2,3$-dichlorobutane $(CH_3-CHCl-CHCl-CH_3)$
Ignoring stereoisomers,there are $5$ distinct constitutional isomers.

(A) $K_2Cr_2O_7$ acts as a strong oxidizing agent in an acidic medium.
It reacts with reducing agents by oxidizing them,while the dichromate ion $(Cr_2O_7^{2-})$ is reduced to the green-colored chromium$(III)$ ion $(Cr^{3+})$.
Among the given species,$Fe^{2+}$,$Sn^{2+}$,$I^-$,and $S^{2-}$ are all reducing agents.
Therefore,they will reduce $K_2Cr_2O_7$ and change its orange color to green.
Thus,the correct set is $Fe^{2+}, Sn^{2+}, I^-, S^{2-}$.

(B) In an acidic medium,$MnO_4^-$ acts as an oxidizing agent and is reduced to $Mn^{2+}$,gaining $5$ electrons per mole $(MnO_4^- + 8H^+ + 5e^- \rightarrow Mn^{2+} + 4H_2O)$.
We calculate the moles of electrons lost by each component:
$1$) $FeC_2O_4 \rightarrow Fe^{3+} + 2CO_2 + 3e^-$ ($1 \text{ mole}$ loses $3 \text{ moles of } e^-$).
$2$) $Fe_2(C_2O_4)_3 \rightarrow 2Fe^{3+} + 6CO_2 + 6e^-$ ($1 \text{ mole}$ loses $6 \text{ moles of } e^-$).
$3$) $FeSO_4 \rightarrow Fe^{3+} + SO_4^{2-} + 1e^-$ ($1 \text{ mole}$ loses $1 \text{ mole of } e^-$).
$4$) $Fe_2(SO_4)_3$: Iron is already in the $+3$ oxidation state and sulfate is in its highest oxidation state,so no further oxidation occurs $(0 \text{ moles of } e^-)$.
Total moles of electrons lost = $3 + 6 + 1 + 0 = 10 \text{ moles}$.
Since $1 \text{ mole}$ of $KMnO_4$ accepts $5 \text{ moles}$ of electrons,the moles of $KMnO_4$ required = $\frac{10}{5} = 2 \text{ moles}$.

(C) Basic oxides are typically oxides of metals,such as $Na_2O$ and $K_2O$.
Acidic oxides are typically oxides of non-metals,such as $Cl_2O_7$.
Amphoteric oxides exhibit both acidic and basic properties,such as $Al_2O_3$ and $As_2O_3$.
Neutral oxides do not react with either acids or bases,such as $NO, N_2O$,and $CO$.
Option $C$ contains: $K_2O$ (Basic),$Cl_2O_7$ (Acidic),$As_2O_3$ (Amphoteric),and $NO$ (Neutral). Thus,it is the correct set.

(A) Statement $I$ is true: Aluminium reacts with $NaOH$ to form sodium aluminate,which is represented as $[Al(OH)_4]^-$ or in its hydrated form as $[Al(OH)_6]^{3-}$.
Statement $II$ is true:
$ICl_4^-$: The central iodine atom has $sp^3d^2$ hybridization with $2$ lone pairs,resulting in a square planar geometry.
$ClO_3^-$: The central chlorine atom has $sp^3$ hybridization with $1$ lone pair,resulting in a trigonal pyramidal geometry.
$IBr_2^-$: The central iodine atom has $sp^3d$ hybridization with $3$ lone pairs,resulting in a linear geometry.
Therefore,both statements are correct.

(C) The molar mass of $72 \text{ g mol}^{-1}$ corresponds to the molecular formula $C_5H_{12}$ (pentane isomers).
Let us analyze the number of primary carbon atoms in the isomers of $C_5H_{12}$:
$1$. $n$-pentane $(CH_3-CH_2-CH_2-CH_2-CH_3)$: It has two primary carbon atoms.
$2$. $2$-Methylbutane $(CH_3-CH(CH_3)-CH_2-CH_3)$: It has three primary carbon atoms.
$3$. $2,2$-Dimethylpropane $(C(CH_3)_4)$: It has four primary carbon atoms.
Therefore,the hydrocarbon $(x)$ with three primary carbon atoms is $2$-Methylbutane.

(A) The structures shown are mirror images of each other and are non-superimposable.
They represent the $(R)$ and $(S)$ configurations of the chiral center.
Since they are non-superimposable mirror images,they are enantiomers.

(C) The bond length is inversely proportional to the bond order.
In acetone $(y)$,the $C=O$ bond has a bond order of $2$.
In tropone $(x)$,the $C=O$ bond is conjugated within a $7$-membered aromatic ring,which gives it a partial single bond character,resulting in a bond order between $1$ and $2$ (lower than $2$,thus higher length than $y$).
In the acetate ion $(z)$,resonance leads to two equivalent $C-O$ bonds,each with a bond order of $1.5$.
Comparing the bond orders: $y$ $(2)$ > $x$ $(1 < \text{bond order} < 2)$ > $z$ $(1.5)$.
Since bond length is inversely proportional to bond order,the order of bond length is $z > x > y$.

(C) The electron-withdrawing power of functional groups is determined by their inductive effect ($-I$ effect) and electronegativity.
Comparing the given groups:
$1$. The $-I$ group has the weakest electron-withdrawing effect among the options provided.
$2$. The $-COOH$ group is more electron-withdrawing than $-I$.
$3$. The $-CN$ group is more electron-withdrawing than $-COOH$ due to the higher electronegativity of the nitrogen atom and the $sp$ hybridized carbon.
$4$. The $-NO_2$ group is the strongest electron-withdrawing group among these due to the presence of two highly electronegative oxygen atoms attached to a positively charged nitrogen atom.
Therefore,the increasing order is: $d < b < a < c$.

(A) Statement $I$ is true: The carbocation is electron-deficient. The $-OCH_3$ group exerts a $+R$ (resonance) effect,which donates electron density into the ring and stabilizes the adjacent positive charge.
Statement $II$ is true: The carbanion is electron-rich. The $-NO_2$ group exerts a $-R$ (resonance) effect,which withdraws electron density from the ring and stabilizes the adjacent negative charge.

(A) Mass of $CO_2 = 0.25$ g. Moles of $CO_2 = \frac{0.25}{44} \approx 0.00568$ mol.
Mass of $C = \text{Moles of } CO_2 \times 12 = 0.00568 \times 12 = 0.06818$ g.
Given percentage of $C = 25\%$.
So,$0.06818 = 0.25 \times X \implies X = \frac{0.06818}{0.25} = 0.2727$ g.
Similarly for $H$,Mass of $H_2O = 0.12$ g.
Moles of $H_2O = \frac{0.12}{18} \approx 0.00666$ mol.
Mass of $H = 0.00666 \times 2 = 0.01333$ g.
Given percentage of $H = 4.89\%$.
So,$0.01333 = 0.0489 \times X \implies X = \frac{0.01333}{0.0489} = 0.2726$ g.
Thus,$X \approx 0.273$ g.
Expressing as $273 \times 10^{-3}$ g.

(D) Statement $I$ is false because while decarboxylation of sodium ethanoate $(CH_3COONa + NaOH xrightarrow{CaO} CH_4 + Na_2CO_3)$ and the reaction of $CH_3MgBr$ with water $(CH_3MgBr + H_2O ightarrow CH_4 + Mg(OH)Br)$ produce methane,Kolbe's electrolysis of sodium acetate produces ethane $(2CH_3COONa + 2H_2O ightarrow CH_3-CH_3 + 2CO_2 + H_2 + 2NaOH)$.
Statement $II$ is true because methane $(CH_4)$ contains only one carbon atom,whereas the Wurtz reaction $(2R-X + 2Na ightarrow R-R + 2NaX)$ is used to prepare alkanes with an even number of carbon atoms (at least two carbons) by coupling two alkyl groups.

(D) The reaction of $2$-methylpropene with dilute $H_2SO_4$ is an acid-catalyzed hydration reaction which follows Markovnikov's rule to yield $2$-methylpropan-$2$-ol (tert-butyl alcohol) as the major product $(X)$.
In thin-layer chromatography $(TLC)$,the $R_f$ value is inversely related to the polarity of the compound when using a polar stationary phase like silica gel.
$2$-methylpropene is a non-polar hydrocarbon,whereas $2$-methylpropan-$2$-ol is a polar alcohol capable of forming hydrogen bonds with the stationary phase.
Due to stronger interactions with the polar stationary phase,the polar product $(X)$ will travel a shorter distance compared to the non-polar reactant.
Therefore,the $R_f$ value of $(X)$ must be significantly lower than $0.42$.
Among the given options,$0.12$ is the only value lower than $0.42$.

(A) $1$. The formation of a yellow precipitate with $I_2/NaOH$ (iodoform test) indicates the presence of a methyl ketone group $(CH_3CO-)$ or a secondary alcohol that can be oxidized to a methyl ketone.
$2$. "$P$" $(C_8H_{14})$ undergoes ozonolysis to form "$Q$".
$3$. "$Q$" undergoes an intramolecular aldol condensation (alkali under reflux) to form "$R$",which contains a methyl ketone moiety.
$4$. The structure "$S$" is a cyclic carboxylic acid derivative formed after the haloform reaction and acidification.
$5$. Analyzing the degree of unsaturation ($C_8H_{14}$ has $8 - 14/2 + 1 = 2$ degrees of unsaturation),and the reaction sequence,$1$,$2$-dimethylcyclohexene (option $A$) is the most suitable precursor that fits the molecular formula and the described chemical transformations.

(A) Ozonolysis cleaves $C=C$ bonds to form carbonyl compounds.
Given products are $2$ moles of $HCHO$ (formaldehyde),$HCOCHO$ (glyoxal),and $CH_3COCOCH_3$ (butane$-2,3-$dione).
By analyzing the fragments,we can reconstruct the original cyclic alkene structure.
$HCHO$ comes from terminal $=CH_2$ groups.
$HCOCHO$ and $CH_3COCOCH_3$ represent the internal segments of the ring.
By connecting these fragments,we find that the structure corresponds to $1,2,4,5$-tetramethylenecyclohexane or a similar derivative based on the provided options. Based on the standard chemical analysis of such ozonolysis products,the structure is identified as option $A$.

(C) Heptane $(C_7H_{16})$ undergoes isomerisation in the presence of $AlCl_3/HCl$ to give branched isomers like $2,4$-dimethylpentane.
Aromatisation (dehydrocyclization) of $C_7$ alkanes leads to the formation of toluene $(C_6H_5CH_3)$.
Cyclisation of heptane specifically yields methylcyclohexane.
Since all these chemical properties are characteristic of the straight-chain alkane heptane,the compound $(X)$ is heptane.

(C) The molecule $x$ is $N$-phenylbenzamide. It contains two phenyl rings attached to an amide group $(-CONH-)$.
In the ring attached to the nitrogen atom,the lone pair on the nitrogen atom is in conjugation with the benzene ring,making it an activating group. This group is ortho/para directing.
In the ring attached to the carbonyl carbon,the carbonyl group $(-C=O)$ is a deactivating group and is meta-directing.
Therefore,the electrophilic aromatic substitution will occur on the ring attached to the nitrogen atom. The ortho and para positions relative to the nitrogen atom are the most electron-rich sites.
In the given structure,$u$ represents the para position and $s$ represents the ortho position relative to the nitrogen atom. Thus,the electrophile will preferentially attack at these positions.

(D) To find the number of atoms,we calculate the moles of molecules and multiply by the atomicity and Avogadro's number $(N_A)$:
$A$. $2 \text{ g}$ of $O_2 = 2/32 = 0.0625 \text{ mol}$. Atoms = $0.0625 \times 2 \times N_A = 0.125 N_A$.
$B$. $4 \text{ g}$ of $SO_2 = 4/64 = 0.0625 \text{ mol}$. Atoms = $0.0625 \times 3 \times N_A = 0.1875 N_A$.
$C$. $1400 \text{ mL}$ of $O_2 = 1.4 \text{ L} / 22.4 \text{ L/mol} = 0.0625 \text{ mol}$. Atoms = $0.0625 \times 2 \times N_A = 0.125 N_A$.
$D$. $0.05 \text{ L}$ of $He = 0.05 / 22.4 \approx 0.00223 \text{ mol}$. Atoms = $0.00223 N_A$.
$E$. $0.0625 \text{ mol}$ of $H_2$. Atoms = $0.0625 \times 2 \times N_A = 0.125 N_A$.
Comparing the results,options $A, C$,and $E$ have the same number of atoms $(0.125 N_A)$.

(D) For $Ag_2CrO_4$ (sparingly soluble salt of type $A_2B$): $K_{sp} = (2s_1)^2(s_1) = 4s_1^3 = 32x$.
Thus,$s_1^3 = 8x$,which gives $s_1 = 2\sqrt[3]{x}$.
For $AgBr$ (sparingly soluble salt of type $AB$): $K_{sp} = s_2^2 = 4y$.
Thus,$s_2 = \sqrt{4y} = 2\sqrt{y}$.
The ratio of molarities is $\frac{s_1}{s_2} = \frac{2\sqrt[3]{x}}{2\sqrt{y}} = \frac{\sqrt[3]{x}}{\sqrt{y}}$.

(B) At the start,the solution contains only the weak acid $HX$. The concentration $C = 0.2 \ M$. The $pH$ of a weak acid is given by $pH = 0.5(pK_a - \log C)$. Substituting the values: $pH = 0.5(3.3 - \log 0.2) = 0.5(3.3 - (-0.699)) \approx 0.5(4.0) = 2.0$.
$(b)$ When $10 \ mL$ of $0.2 \ M$ $NaOH$ is added to $20 \ mL$ of $0.2 \ M$ $HX$,we reach the half-equivalence point because the moles of $NaOH$ added $(10 \ mL \times 0.2 \ M = 2 \ mmol)$ are exactly half the initial moles of $HX$ $(20 \ mL \times 0.2 \ M = 4 \ mmol)$. At the half-equivalence point,the concentration of the acid $[HX]$ equals the concentration of its conjugate base $[X^-]$. According to the Henderson-Hasselbalch equation,$pH = pK_a + \log([X^-]/[HX])$. Since $[X^-] = [HX]$,$pH = pK_a = 3.3$.

(B) $1$. Calculate the initial moles of acetic acid $(CH_3COOH)$: Since $20 \text{ mL}$ of acetic acid is neutralized by $28.4 \text{ mL}$ of $0.1 \text{ M } NaOH$,the moles of acid = moles of base = $0.1 \text{ M} \times 28.4 \text{ mL} = 2.84 \text{ mmol}$.
$2$. Calculate the composition of solution $(X)$: Solution $(X)$ is formed by mixing $2.84 \text{ mmol}$ of $CH_3COOH$ and $1.42 \text{ mmol}$ of $NaOH$ $(0.1 \text{ M} \times 14.2 \text{ mL} = 1.42 \text{ mmol})$.
$3$. Reaction: $CH_3COOH + NaOH \rightarrow CH_3COONa + H_2O$.
$4$. After reaction: Remaining $CH_3COOH = 2.84 - 1.42 = 1.42 \text{ mmol}$. Formed $CH_3COONa$ (salt) = $1.42 \text{ mmol}$.
$5$. Apply Henderson-Hasselbalch equation: $pH = pK_a + \log \left( \frac{[Salt]}{[Acid]} \right)$.
$6$. Since $[Salt] = [Acid] = 1.42 \text{ mmol}$,the ratio is $1$. Thus,$pH = 4.75 + \log(1) = 4.75 + 0 = 4.75$.

(C) To determine the pH order,we calculate the millimoles (mmol) of $H^+$ and $OH^-$ ions in each mixture:
$A$. $Ca(OH)_2$ is a strong base. $OH^-$ mmol $= 10 \times 0.2 \times 2 = 4$. $HCl$ is a strong acid. $H^+$ mmol $= 25 \times 0.1 = 2.5$. Excess $OH^- = 4 - 2.5 = 1.5 \text{ mmol}$. The solution is basic,so pH $> 7$.
$B$. $H_2SO_4$ is a strong acid. $H^+$ mmol $= 10 \times 0.01 \times 2 = 0.2$. $Ca(OH)_2$ is a strong base. $OH^-$ mmol $= 10 \times 0.01 \times 2 = 0.2$. Since $H^+$ mmol $= OH^-$ mmol,the solution is neutral,so pH $= 7$.
$C$. $H_2SO_4$ is a strong acid. $H^+$ mmol $= 10 \times 0.1 \times 2 = 2$. $KOH$ is a strong base. $OH^-$ mmol $= 10 \times 0.1 = 1$. Excess $H^+ = 2 - 1 = 1 \text{ mmol}$. The solution is acidic,so pH $< 7$.
Comparing the pH values: $C$ (acidic) $< B$ (neutral) $< A$ (basic). Therefore,the increasing order is $C < B < A$.

(C) The target reaction is: $\frac{1}{2}x_2 + \frac{1}{2}y_2 + \frac{1}{2}z_2 \rightleftharpoons xyz$.
We are given:
$(1)$ $2xy \rightleftharpoons x_2 + y_2$ with $K_1 = 2.5 \times 10^5$
$(2)$ $xy + \frac{1}{2}z_2 \rightleftharpoons xyz$ with $K_2 = 5 \times 10^{-3}$
To obtain the target reaction,we perform: (Reaction $2$) - $\frac{1}{2} \times$ (Reaction $1$).
This corresponds to: $(xy + \frac{1}{2}z_2) - \frac{1}{2}(2xy) \rightleftharpoons xyz - \frac{1}{2}(x_2 + y_2)$.
Rearranging gives: $\frac{1}{2}x_2 + \frac{1}{2}y_2 + \frac{1}{2}z_2 \rightleftharpoons xyz$.
The equilibrium constant $K_3$ is given by $K_3 = \frac{K_2}{(K_1)^{1/2}}$.
$K_3 = \frac{5 \times 10^{-3}}{\sqrt{2.5 \times 10^5}} = \frac{5 \times 10^{-3}}{\sqrt{25 \times 10^4}} = \frac{5 \times 10^{-3}}{500} = \frac{5 \times 10^{-3}}{5 \times 10^2} = 1.0 \times 10^{-5}$.

(D) We need to find the enthalpy of formation for the reaction: $2Al(s) + 3Cl_2(g) \to Al_2Cl_6(s)$.
Given equations:
$(i)$ $2Al(s) + 6HCl(aq) \to Al_2Cl_6(aq) + 3H_2(g)$,$\Delta H_1 = -1200 \text{ kJ/mol}$
(ii) $H_2(g) + Cl_2(g) \to 2HCl(g)$,$\Delta H_2 = -164 \text{ kJ/mol}$
(iii) $HCl(g) + aq \to HCl(aq)$,$\Delta H_3 = -83 \text{ kJ/mol}$
(iv) $Al_2Cl_6(s) + aq \to Al_2Cl_6(aq)$,$\Delta H_4 = -663 \text{ kJ/mol}$
To get the target reaction,we perform the following operation:
Target = $(i)$ + $3$ $\times$ (ii) + $6$ $\times$ (iii) - (iv)
$\Delta H_f = (-1200) + 3(-164) + 6(-83) - (-663)$
$\Delta H_f = -1200 - 492 - 498 + 663$
$\Delta H_f = -2190 + 663 = -1527 \text{ kJ/mol}$.

(B) The standard enthalpy change of the reaction is calculated as:
$\Delta H^\circ = \Sigma \Delta_f H^\circ(\text{products}) - \Sigma \Delta_f H^\circ(\text{reactants})$
$\Delta H^\circ = 2 \times \Delta_f H^\circ(XY) - [\Delta_f H^\circ(X_2) + \Delta_f H^\circ(Y_2)]$
$\Delta H^\circ = 2(42) - (8 + 80) = 84 - 88 = -4 \ kJ \ mol^{-1} = -4000 \ J \ mol^{-1}$.
The standard entropy change of the reaction is:
$\Delta S^\circ = \Sigma S^\circ(\text{products}) - \Sigma S^\circ(\text{reactants})$
$\Delta S^\circ = 2 \times S^\circ(XY) - [S^\circ(X_2) + S^\circ(Y_2)]$
$\Delta S^\circ = 2(200) - (140 + 250) = 400 - 390 = 10 \ J \ K^{-1} \ mol^{-1}$.
Using the Gibbs-Helmholtz equation:
$\Delta G^\circ = \Delta H^\circ - T\Delta S^\circ$
$\Delta G^\circ = -4000 \ J \ mol^{-1} - (600 \ K \times 10 \ J \ K^{-1} \ mol^{-1})$
$\Delta G^\circ = -4000 - 6000 = -10000 \ J \ mol^{-1} = -10 \ kJ \ mol^{-1}$.

(C) The molar heat capacity $(C_p)$ at constant pressure for gases is generally higher than for liquids and solids because a portion of the heat supplied is used to perform work against the external pressure during expansion.
For monatomic gases like Helium $(He)$,$C_p = \frac{5}{2}R \approx 20.8 \ J \cdot K^{-1} \cdot mol^{-1}$.
For liquids like Bromine $(Br_2)$,the molar heat capacity is approximately $75.7 \ J \cdot K^{-1} \cdot mol^{-1}$ (Note: While $C_p$ for liquids can be high,in the context of standard chemistry problems comparing states of matter,gases are often considered to have higher degrees of freedom or specific heat characteristics relative to solids).
However,based on standard thermodynamic data,the molar heat capacity of liquids like $Br_2$ is often higher than that of monatomic gases. Re-evaluating the standard trend: $C_p$ (liquid) > $C_p$ (gas) > $C_p$ (solid).
Given the specific options provided in competitive chemistry contexts,the intended order is $Helium(g) > Bromine(l) > Copper(s)$ based on the degrees of freedom and expansion work.

(C) For an isothermal process,the magnitude of work done during compression is greater than the magnitude of work done during expansion between the same initial and final states.
In the case of expansion,the work done in multi-stage expansion $(w_B)$ is greater than in single-stage expansion $(w_A)$,so $|w_B| > |w_A|$.
In the case of compression,the work done in single-stage compression $(w_C)$ is greater than in multi-stage compression $(w_D)$,so $|w_C| > |w_D|$.
Comparing the two,the magnitude of work required for compression is always greater than the magnitude of work obtained during expansion between the same two states.
Therefore,the correct order of magnitude is $|w_C| > |w_D| > |w_B| > |w_A|$.

(B) $SF_4$ has a see-saw structure (steric number $5$,$4$ bond pairs,$1$ lone pair).
$1$. $BrF_4^{\ominus}$: Steric number = $\frac{1}{2}(7+4+1) = 6$ (Square planar).
$2$. $CH_4$: Steric number = $4$ (Tetrahedral).
$3$. $IF_4^{\oplus}$: Steric number = $\frac{1}{2}(7+4-1) = 5$ ($4$ bond pairs,$1$ lone pair,see-saw).
$4$. $XeF_4$: Steric number = $\frac{1}{2}(8+4) = 6$ (Square planar).
$5$. $XeO_2F_2$: Steric number = $\frac{1}{2}(8+2+2) = 5$ ($4$ bond pairs,$1$ lone pair,see-saw).
Both $IF_4^{\oplus}$ and $XeO_2F_2$ are isostructural with $SF_4$.

(C) $\pi^*$ antibonding molecular orbital is formed by the out-of-phase overlap of two atomic orbitals (e.g.,$p_x$ or $p_y$ orbitals).
This results in the presence of a nodal plane between the two nuclei where the electron probability density is zero,in addition to the nodal plane containing the internuclear axis.
In the $\pi^*$ orbital,the lobes of the atomic orbitals have opposite signs (represented by shading) facing each other,leading to destructive interference.
Diagram $C$ correctly depicts this out-of-phase combination,showing the nodal plane between the nuclei.

(D) Statement $I$: In $SO_2$ (valence electrons around $S = 10$),$SO_3$ (valence electrons around $S = 12$),$SF_4$ (valence electrons around $S = 10$),and $SF_6$ (valence electrons around $S = 12$),the octet rule is not obeyed (expanded octet).
In $H_2S$,sulphur has $8$ electrons in its valence shell,thus it follows the octet rule.
Therefore,there are $4$ compounds that do not obey the octet rule. Statement $I$ is false.
Statement $II$: Let us count the lone pairs $(LP)$ on the central atom:
$1$. $[H_2O(2 LP), ClF_3(2 LP), SF_4(1 LP)]$
$2$. $[NH_3(1 LP), BrF_5(1 LP), SF_4(1 LP)]$
$3$. $[BrF_5(1 LP), ClF_3(2 LP), XeF_4(2 LP)]$
$4$. $[XeF_4(2 LP), ClF_3(2 LP), H_2O(2 LP)]$
Only the second set contains molecules where all have exactly $1$ lone pair on the central atom.
Thus,there is only $1$ such set. Statement $II$ is true.

(D) Statement $(I)$: The first ionisation enthalpy generally increases across a period from left to right. The correct order for these elements is $Ar > Cl > Mg > Na$. Therefore,the given order $Na > Mg > Cl > Ar$ is incorrect. Thus,Statement $(I)$ is false.
Statement $(II)$: The electronic configuration of $Ca$ $(Z=20)$ is $[Ar] 4s^2$. The first two electrons are removed from the $4s$ orbital. The third electron must be removed from the $3p^6$ orbital,which is a stable noble gas configuration (Argon core). Removing an electron from a stable,fully-filled shell requires a very high amount of energy. Thus,Statement $(II)$ is true.

(C) The electronic configuration of $Mg$ is $[Ne]3s^2$.
Removing the first electron requires $737 \text{ kJ/mol}$ to form the $Mg^+$ ion with configuration $[Ne]3s^1$.
The second ionization enthalpy involves removing the second electron from the $Mg^+$ ion.
Since the electron is being removed from a positively charged species,the electrostatic attraction is stronger,requiring significantly more energy than the first ionization.
Therefore,the $2^{nd}$ ionization enthalpy must be greater than the $1^{st}$ ionization enthalpy $(737 \text{ kJ/mol})$ and must be positive.
Among the given options,$+1450 \text{ kJ/mol}$ is the only value that is greater than $737 \text{ kJ/mol}$ and positive,making it the most probable experimental value.

(D) Statement $I$: $[Al_2O_3, Cr_2O_3]$ are both amphoteric. $[Cl_2O_7, Mn_2O_7]$ are both acidic. $[Na_2O]$ is basic,while $[V_2O_3]$ is amphoteric. $[CO, N_2O]$ are both neutral. Thus,only $3$ pairs have the same nature. Therefore,Statement $I$ is false.
Statement $II$: $Na_2O$ is a highly basic metallic oxide and $Cl_2O_7$ is a strongly acidic non-metallic oxide. Therefore,Statement $II$ is true.

(A) Statement $I$: The standard electrode potential $E^\circ_{Al^{3+}/Al}$ is approximately $-1.66 \ V$,while $E^\circ_{Tl^{3+}/Tl}$ is approximately $+1.26 \ V$. $A$ more negative electrode potential indicates a stronger reducing agent and a more electropositive nature. Therefore,Statement $I$ is true.
Statement $II$: The sum of the first three ionization enthalpies of Boron is extremely high,which prevents the formation of $B^{3+}$ ions,leading it to form only covalent compounds. In contrast,Aluminium has a significantly lower sum of ionization enthalpies,which allows it to form $Al^{3+}$ ions under specific conditions. Therefore,Statement $II$ is true.

(B) primary standard is a reagent that is pure,stable,and has a high molar mass.
Statement $A$ is incorrect because hydrated salts often lose or gain water of crystallization,making them unstable.
Statement $B$ is correct; a primary standard must be stable and not react with air (e.g.,not hygroscopic or efflorescent).
Statement $C$ is correct; the reaction must be rapid and follow a definite stoichiometry.
Statement $D$ is incorrect; a primary standard must be readily soluble in water to prepare a standard solution.
Statement $E$ is incorrect; a primary standard should have a high relative molar mass to minimize weighing errors.
Therefore,only statement $B$ and $C$ are correct,but based on standard criteria,$B$ and $C$ are the most accurate. Given the options,$B$ is the best fit.

(B) The reactions of glucose are as follows:
$1$. Glucose reacts with hydroxylamine $(NH_2OH)$ to form an oxime,known as Glucoxime $(A-IV)$.
$2$. Glucose reacts with bromine water ($Br_2$ water) to undergo mild oxidation of the aldehyde group to a carboxylic acid,forming Gluconic acid $(B-I)$.
$3$. Glucose reacts with excess acetic anhydride to form Glucose pentaacetate,indicating the presence of five hydroxyl groups $(C-II)$.
$4$. Glucose reacts with concentrated nitric acid $(HNO_3)$ to undergo strong oxidation of both the aldehyde and the primary alcohol group,forming Saccharic acid $(D-III)$.
Thus,the correct matching is $A-IV, B-I, C-II, D-III$.

(B) Statement-$I$: The electronegativity order for the given elements is $N > P > As > Sb$. Thus,$X = N$ (most electronegative) and $Y = Sb$ (least electronegative). The oxide $N_{2}O_{3}$ is acidic,and $Sb_{2}O_{3}$ is amphoteric. Therefore,Statement-$I$ is true.
Statement-$II$: $BCl_{3}$ is covalent and undergoes hydrolysis in water to form boric acid,$B(OH)_{3}$ (or $H_{3}BO_{3}$),and $HCl$. The reaction is $BCl_{3} + 3H_{2}O \rightarrow B(OH)_{3} + 3HCl$. The statement claims it produces $[B(OH)_{4}]^{-}$ and $[B(H_{2}O)_{6}]^{3+}$,which is incorrect. Therefore,Statement-$II$ is false.

(D) Statement $(A)$ is incorrect because the factor $e^{-Ea/RT}$ represents the fraction of molecules having kinetic energy equal to or greater than the activation energy $(Ea)$.
Statement $(B)$ is correct because a lower activation energy $(Ea)$ means a larger fraction of molecules can cross the energy barrier,leading to a faster reaction rate.
Statement $(C)$ is correct as a general rule of thumb for many reactions,an increase in temperature by $10^{\circ}C$ approximately doubles the rate constant.
Statement $(D)$ is incorrect because the plot of $\log k$ vs $\frac{1}{T}$ gives a straight line with a slope of $-\frac{Ea}{2.303R}$,not $-\frac{Ea}{R}$.
Therefore,statements $(B)$ and $(C)$ are correct.

(C) The reaction proceeds in two steps:
$1$. Cyclohexylmethanamine reacts with benzoyl chloride $(C_6H_5COCl)$ in the presence of a base $(NaOH)$ to form an amide,$N$-(cyclohexylmethyl)benzamide,which is product $[A]$.
$2$. The amide $[A]$ is then reduced using lithium aluminium hydride $(LiAlH_4)$ followed by hydrolysis $(H_2O)$ to yield the corresponding amine,$N$-benzylcyclohexylmethanamine,which is the final product $[B]$.

(B) The electronic configurations are:
$Cr = [Ar] 3d^{5} 4s^{1}$
$Mn = [Ar] 3d^{5} 4s^{2}$
Statement-$I$: The first ionization enthalpy $(IE_{1})$ of $Cr$ is lower than that of $Mn$ because $Mn$ has a stable half-filled $d$-subshell and a full $s$-subshell,making it harder to remove an electron. Thus,Statement-$I$ is true.
Statement-$II$: For $IE_{2}$,$Cr$ loses its $4s^{1}$ electron to reach a stable $3d^{5}$ configuration,while $Mn$ loses its $4s^{1}$ electron from a $3d^{5} 4s^{2}$ configuration. However,$IE_{2}$ of $Cr$ is higher than $Mn$ because removing the second electron from $Cr$ disrupts the stable $d^{5}$ configuration. For $IE_{3}$,$Mn$ has a higher value because it involves removing an electron from the stable $d^{5}$ configuration of $Mn^{2+}$. Therefore,Statement-$II$ is false.

(C) Total mass of $H_2SO_4 = (100 \times 0.98) + (100 \times 0.49) = 98 + 49 = 147 \ g$.
Total mass of solution $= 100 + 100 = 200 \ g$.
Total mass of $H_2O = 200 - 147 = 53 \ g$.
Moles of $H_2SO_4 = \frac{147}{98} = 1.5 \ mol$.
Moles of $H_2O = \frac{53}{18} \approx 2.944 \ mol$.
Mole fraction of $H_2SO_4 = \frac{n_{H_2SO_4}}{n_{H_2SO_4} + n_{H_2O}} = \frac{1.5}{1.5 + 2.944} = \frac{1.5}{4.444} \approx 0.337$.

(C) Equating the two rate constants: $k_1 = k_2$
$10^6 e^{\frac{-30000}{T}} = 10^4 e^{\frac{-24000}{T}}$
Divide both sides by $10^4 e^{\frac{-30000}{T}}$:
$10^2 = e^{\frac{-24000}{T} - (\frac{-30000}{T})}$
$100 = e^{\frac{6000}{T}}$
Taking natural logarithm on both sides:
$\ln(100) = \frac{6000}{T}$
$2 \ln(10) = \frac{6000}{T}$
$2 \times 2.303 = \frac{6000}{T}$
$T = \frac{6000}{4.606} \approx 1302.64 \ K$
Rounding to the nearest integer,we get $T = 1303 \ K$.

(C) The cell reaction is: $3HSnO_2^- + Bi_2O_3 + 3H_2O + 6OH^- \rightarrow 3Sn(OH)_6^{2-} + 2Bi$.
The number of electrons involved in the balanced equation is $n = 6$.
The standard cell potential is $E^0_{\text{cell}} = E^0_{\text{cathode}} - E^0_{\text{anode}} = -0.44 - (-0.90) = +0.46 \ V$.
Applying the Nernst equation at $298 \ K$: $E_{\text{cell}} = E^0_{\text{cell}} - \frac{0.0591}{n} \log Q$.
Using $0.06$ as an approximation for $\frac{0.0591}{1}$: $E_{\text{cell}} = 0.46 - \frac{0.06}{6} \log(10^6)$.
$E_{\text{cell}} = 0.46 - 0.01 \times 6 = 0.46 - 0.06 = 0.40 \ V$.
Thus,$E_{\text{cell}} = 4 \times 10^{-1} \ V$.
The value is $4$.

(B) Statement $I$ is true because Henry's law states that $p = K_{H}x$,where $K_{H}$ is a constant for a given gas-solvent system at a constant temperature,especially in the range where the solution behaves as an ideally dilute solution.
Statement $II$ is false because $K_{H}$ depends on the nature of the gas and the nature of the solvent. Therefore,$K_{H}$ will differ for the same solute in different solvents.

(D) In chlorocyclohexane,the carbon atom attached to chlorine is $sp^3$ hybridized,while in chlorobenzene,it is $sp^2$ hybridized.
Due to the resonance effect ($+M$ effect) in chlorobenzene,the $C-Cl$ bond acquires partial double bond character ($B$ is correct).
This resonance also reduces the polarity of the $C-Cl$ bond compared to chlorocyclohexane ($C$ is correct).
The $C-Cl$ bond in chlorobenzene is formed using an $sp^2$ hybridized orbital of carbon,which is shorter and stronger than the $sp^3$ bond in chlorocyclohexane ($E$ is correct).
Statement $A$ is incorrect because the resonance effect decreases the negative charge on chlorine. Statement $D$ is incorrect because the $C-Cl$ bond is actually shorter in chlorobenzene due to the partial double bond character.

(C) The rate of $S_{N}1$ reaction is directly proportional to the stability of the carbocation $(C^{\oplus})$ formed.
$I$ is $1-$bromobicyclo[$2.2$.$1$]heptane and $II$ is $1-$bromobicyclo[$2.2$.$2$]octane. Both are bridgehead halides and are extremely unreactive in $S_{N}1$ reactions due to Bredt's rule,which prevents the formation of a planar carbocation at the bridgehead position.
Between $I$ and $II$,$II$ is slightly more reactive than $I$ because the bridgehead carbocation in $II$ is less strained than in $I$.
$III$ is a tert-butyl carbocation,which is more stable than the bridgehead carbocations.
$IV$ is a triphenylmethyl carbocation,which is the most stable due to extensive resonance stabilization by three phenyl rings.
Thus,the correct order of reactivity is $II < I < III < IV$.

(C) $1$. The molecular formula $C_8H_9ON$ and the reaction with $Br_2 / HO^{(-)}$ (Hofmann bromamide degradation) suggest that $A$ is an amide.
$2$. The product $B$ is an amine,which is soluble in dilute acid.
$3$. The reaction sequence $B$ $\xrightarrow{NaNO_2/HCl} C$ $\xrightarrow{CuCN} D$ $\xrightarrow{H_3O^+} E$ is the standard conversion of an amine to a carboxylic acid via a diazonium salt and nitrile.
$4$. $E$ is also obtained from the hydrolysis of $A$,confirming $A$ is an amide.
$5$. $E$ on oxidation with $KMnO_4$ gives $F$. If $E$ is $p$-toluic acid,oxidation gives terephthalic acid $(F)$,which has two types of hydrogen atoms (aromatic and carboxylic).
$6$. Based on the reaction path,$A$ must be $p$-methylbenzamide.

(D) Sucrose is dextrorotatory $(+66.5^\circ)$.
Upon hydrolysis,it yields an equimolar mixture of $D-(+)$-glucose $(+52.5^\circ)$ and $D-(-)$-fructose $(-92.4^\circ)$.
Since the magnitude of the laevorotation of fructose $(-92.4^\circ)$ is greater than the dextrorotation of glucose $(+52.5^\circ)$,the resulting mixture is laevorotatory.
This process is known as the inversion of sugar.
Statement $I$ is correct.
Statement $II$ is incorrect because it incorrectly states that glucose is laevorotatory and fructose is dextrorotatory,whereas the opposite is true.

(C) Statement $I$: The bond length in $HX$ increases as the size of the halogen increases $(HF < HCl < HBr < HI)$. The halogen with the longest bond is $I$,which has the largest covalent radius,not the smallest. Thus,Statement $I$ is false.
Statement $II$: The boiling point order for group $15$ hydrides is $PH_3 < AsH_3 < NH_3 < SbH_3 < BiH_3$. The hydride with the lowest boiling point is $PH_3$. The element $E$ is $P$ (phosphorus). The maximum covalency of phosphorus is $6$ (e.g.,in $[PF_6]^-$),not $4$. Thus,Statement $II$ is false.

(A) The correct matches are as follows:
$A$. $NH_2-NH_2, KOH$ is used in the Wolff-Kishner reduction $(III)$.
$B$. $[Ag(NH_3)_2]OH$ is the reagent for Tollen's test $(I)$.
$C$. Aqueous $CuSO_4$,Sodium Potassium tartarate,and $KOH$ constitute the Fehling's solution used in Fehling's test $(IV)$.
$D$. $Zn-Hg, HCl$ is used in the Clemmensen reduction $(II)$.
Thus,the correct sequence is $A-III, B-I, C-IV, D-II$.

(B) For a $1^{st}$ order reaction,the rate is given by $Rate = k[A]$.
In Run $1$,the concentration of $A$ is $10 \ M$.
In Run $2$,the volume is doubled,but the concentration of $A$ remains $10 \ M$,so the rate is the same as Run $1$.
In Run $3$,$100 \ mL$ of $H_2O$ is added to $100 \ mL$ of $10 \ M$ solution of $A$,which dilutes the solution to $5 \ M$.
Since the rate depends on the concentration of $A$,the rate in Run $3$ will be lower than in Run $1$ and Run $2$.
Therefore,the correct order is $Run \ 3 < Run \ 1 = Run \ 2$.

(C) Statement $I$ is incorrect. Nitrobenzene is a highly deactivated ring due to the strong electron-withdrawing $-NO_{2}$ group. Therefore,it does not undergo Friedel-Crafts acylation reaction with $CH_{3}COCl / AlCl_{3}$.
Statement $II$ is correct. The $-NO_{2}$ group is indeed a meta-directing and deactivating group because it withdraws electron density from the benzene ring via both inductive and resonance effects.

(B) In a polar protic solvent like methanol,nucleophilicity is determined by the ability of the nucleophile to donate an electron pair.
The order of nucleophilicity is governed by basicity and polarizability.
$C_2H_5O^{-}$ is a strong base and a good nucleophile.
$C_6H_5O^{-}$ (phenoxide) is less basic than $C_2H_5O^{-}$ due to resonance stabilization of the negative charge on the oxygen atom.
$I^{-}$ is a very good nucleophile in polar protic solvents due to its high polarizability,despite being a weak base.
$F^{-}$ is a small,highly solvated ion in methanol,making it a poor nucleophile.
Thus,the correct order is $I^{-} > C_2H_5O^{-} > C_6H_5O^{-} > F^{-}$.

(D) The metal $M$ that does not liberate $H_{2}$ gas from dilute $HCl$ is $Cu$ (Copper),as it has a positive standard reduction potential $(E^{\circ}_{Cu^{2+}/Cu} = +0.34 \ V)$.
When $CuSO_{4}$ is treated with excess $KCN$,it forms a very stable complex $[Cu(CN)_{4}]^{3-}$.
The reaction is: $2CuSO_{4} + 10KCN \rightarrow 2K_{3}[Cu(CN)_{4}] + (CN)_{2} + 2K_{2}SO_{4}$.
Because the complex $[Cu(CN)_{4}]^{3-}$ is extremely stable (a 'perfect' complex),the concentration of free $Cu^{2+}$ ions in the solution is negligible.
Therefore,when $H_{2}S_{(g)}$ is passed through this solution,the ionic product does not exceed the solubility product $(K_{sp})$ of $CuS$,and no precipitate of $CuS$ is formed.
Thus,the amount of $MS$ formed is $0 \ mol$.

(C) According to Henry's law,the partial pressure of a gas $(P_g)$ is directly proportional to its mole fraction $(X_g)$ in the solution: $P_g = K_H \cdot X_g$.
Taking the logarithm on both sides: $log(P_g) = log(K_H \cdot X_g) = log(K_H) + log(X_g)$.
This equation is of the form $y = mx + c$,where $y = log(P_g)$,$x = log(X_g)$,the slope $m = 1$,and the intercept $c = log(K_H)$.
Since the slope is $1$ (positive) and there is a positive intercept $log(K_H)$,the plot is a straight line with a positive slope that does not pass through the origin. This corresponds to Plot $C$.

(C) The reaction of $EH_{3}$ with $K_{2}HgI_{4}$ (Nessler's reagent) in an alkaline medium to form a brown precipitate of basic mercury$(II)$amido-iodine is a characteristic test for ammonia $(NH_{3})$.
Therefore,the element $E$ is nitrogen $(N)$.
The first ionization enthalpy values for the first elements of groups $13, 14, 15$,and $16$ are $B$ $(801 \ kJ \ mol^{-1})$,$C$ $(1086 \ kJ \ mol^{-1})$,$N$ $(1402 \ kJ \ mol^{-1})$,and $O$ $(1314 \ kJ \ mol^{-1})$.
Since $E$ is $N$,its first ionization enthalpy is $1402 \ kJ \ mol^{-1}$.

(D) The reaction of phenol with $CHCl_3$ and $aq. KOH$ is the Reimer-Tiemann reaction.
In this reaction,$o$-hydroxybenzaldehyde (salicylaldehyde) is the major product due to intramolecular hydrogen bonding,while $p$-hydroxybenzaldehyde is the minor product.
Therefore,Statement $I$ is false.
$o$-hydroxybenzaldehyde exhibits intramolecular hydrogen bonding,making it volatile with steam,whereas $p$-hydroxybenzaldehyde exhibits intermolecular hydrogen bonding,making it less volatile.
Thus,they can be separated by steam distillation.
Therefore,Statement $II$ is true.

(A) For low spin octahedral complexes,electrons fill the $t_{2g}$ orbitals first.
$Co^{3+} (3d^6): t_{2g}^6 e_g^0$,unpaired electrons $= 0$.
$Fe^{3+} (3d^5): t_{2g}^5 e_g^0$,unpaired electrons $= 1$.
$Mn^{3+} (3d^4): t_{2g}^4 e_g^0$,unpaired electrons $= 2$.
$Cr^{3+} (3d^3): t_{2g}^3 e_g^0$,unpaired electrons $= 3$.
Thus,the decreasing order is $Cr^{3+} > Mn^{3+} > Fe^{3+} > Co^{3+}$.

(B) $1$. An apple green flame test is a characteristic feature of $Ba^{2+}$ ions.
$2$. $Ba^{2+}$ ions react with $K_{2}CrO_{4}$ in an acetic acid medium to form a yellow precipitate of barium chromate $(BaCrO_{4})$.
$3$. The formation of a canary yellow precipitate upon heating the sodium carbonate extract with conc. $HNO_{3}$ and ammonium molybdate is a confirmatory test for the presence of phosphate ions $(PO_{4}^{3-})$.
$4$. Therefore,the salt contains $Ba^{2+}$ as the cation and $PO_{4}^{3-}$ as the anion.

(A) The chemical equation for the Hoffmann bromamide degradation reaction is: $RCONH_{2} + Br_{2} + 4NaOH \rightarrow RNH_{2} + Na_{2}CO_{3} + 2NaBr + 2H_{2}O$.
$(A)$ is correct because the stoichiometry of the reaction requires $4$ moles of $NaOH$ and $1$ mole of $Br_{2}$ for every mole of primary amide.
$(B)$ is false because alkyl amides (primary amides) readily undergo this reaction.
$(C)$ is correct because this reaction is a standard laboratory method for the synthesis of primary amines.
$(D)$ is false because secondary amides $(RCONHR')$ lack the necessary $-NH_{2}$ group required for the rearrangement mechanism.
$(E)$ is correct because the by-products formed are indeed $Na_{2}CO_{3}$,$NaBr$,and $H_{2}O$.

(C) Let the empirical formula be $(COH)_{n}$.
The molar mass is $12n + 16n + n = 29n = 58 \Rightarrow n = 2$.
Thus,the molecular formula is $C_{2}H_{2}O_{2}$,which corresponds to glyoxal $(CHO-CHO)$.
Glyoxal contains two aldehyde groups and lacks $\alpha$-hydrogen atoms,so it undergoes the Cannizzaro reaction in the presence of $50\% \ KOH$.
In this reaction,one aldehyde group is oxidized to a carboxylate group and the other is reduced to a primary alcohol group.
$CHO-CHO + KOH \rightarrow HO-CH_{2}-COOK$.
Upon acidification,the salt converts to glycolic acid: $HO-CH_{2}-COOH$.
Therefore,the correct structure of $y$ is $HO-CH_{2}-COOH$.

(A) Statement $I$: Acetaldehyde $(CH_{3}CHO)$ reacts with semicarbazide $(H_{2}NNHCONH_{2})$ via a nucleophilic addition-elimination reaction to form a semicarbazone. The product $CH_{3} - CH = N - NH - C(=O) - NH_{2}$ is correctly represented.
Statement $II$: The molecule $Ph - CH(OH)(OCH_{3})$ is a hemiacetal. In the presence of dilute acid,hemiacetals undergo hydrolysis to regenerate the corresponding aldehyde $(Ph-CHO)$ and alcohol $(CH_{3}OH)$. Thus,both statements are correct.

(B) Step $1$: Formation of 'x'. Hydrolysis of prop$-1-$yne $(CH_{3}C \equiv CH)$ in the presence of $HgSO_{4}/H^{+}$ at $333 \ K$ follows Kucherov's reaction to yield acetone $(CH_{3}COCH_{3})$.
Step $2$: Formation of 'y'. Reaction of ethanenitrile $(CH_{3}CN)$ with methylmagnesium bromide $(CH_{3}MgBr)$ in dry ether followed by hydrolysis yields acetone $(CH_{3}COCH_{3})$.
Step $3$: Aldol condensation. Both 'x' and 'y' are acetone. The reaction of two molecules of acetone in the presence of $Ba(OH)_{2}$ undergoes self-aldol condensation to form $4-$hydroxy$-4-$methylpentan$-2-$one $(CH_{3}COCH_{2}C(OH)(CH_{3})_{2})$.
Step $4$: Dehydration. Heating the aldol product causes dehydration (loss of $H_{2}O$) to form the $\alpha,\beta$-unsaturated ketone,$4$-methylpent$-3-$en$-2-$one $(CH_{3}COCH=C(CH_{3})_{2})$.

(A) Step $(i)$: $NaBH_{4}$ in $MeOH$ selectively reduces the aldehyde group $(-CHO)$ to a primary alcohol $(-CH_{2}OH)$. The cyclic amide (lactam) remains unaffected.
Step (ii): $NaOH(aq.), \Delta$ causes the alkaline hydrolysis of the cyclic amide (lactam). The ring opens to form the corresponding amino acid salt (carboxylate).
Step (iii): $H_{3}O^{+}$ protonates the carboxylate to form a carboxylic acid $(-COOH)$ and the amine to form an ammonium salt $(-NH_{2}^{+}-)$.
The final product is a linear amino acid derivative with a primary alcohol group: $HOOC-(CH_{2})_{4}-NH-CH(CH_{3})CH_{2}OH$.

(D) The reaction involves the cleavage of an ether,benzyl phenyl ether $(C_6H_5-O-CH_2-C_6H_5)$,with hydrogen iodide $(HI)$.
In the reaction of an alkyl aryl ether with $HI$,the cleavage occurs such that the alkyl group forms an alkyl iodide and the aryl group forms a phenol.
This is because the $O-C_{aryl}$ bond has partial double bond character due to resonance and is stronger,while the $O-C_{alkyl}$ bond is weaker and undergoes nucleophilic substitution.
Here,the $O-CH_2C_6H_5$ bond breaks to form benzyl iodide $(C_6H_5CH_2I)$ and phenol $(C_6H_5OH)$.
Therefore,Statement $I$ is false because the products are benzyl iodide and phenol,not benzyl alcohol and iodobenzene.
Statement $II$ is true because the $O-CH_2$ bond is indeed the one that undergoes cleavage.

(C) The starting material is $tert$-butyl alcohol (or $2$-methylpropan-$2$-ol).
$1$. Treatment with $Cu/573 \ K$ is a dehydrogenation reaction. However,$tert$-butyl alcohol does not undergo dehydrogenation to form an aldehyde or ketone because it lacks an $\alpha$-hydrogen. Instead,it undergoes dehydration to form isobutylene ($2$-methylpropene).
$2$. The second step involves the reaction of the alkene with benzoic acid $(PhCOOH)$ in the presence of an acid catalyst $(H^+)$. This is an electrophilic addition reaction where the alkene adds to the carboxylic acid to form an ester.
$3$. The reaction of isobutylene with benzoic acid yields $tert$-butyl benzoate as the major product.
Therefore,the correct option is $(C)$.

(D) The reaction of an ether with $HI$ involves the protonation of the ether oxygen atom,followed by a nucleophilic attack by the iodide ion $(I^-)$.
In the given ether,benzyl phenyl ether $(C_6H_5-O-CH_2-C_6H_5)$,the oxygen atom is attached to a phenyl group and a benzyl group.
The cleavage occurs such that the more stable carbocation is formed. The benzyl carbocation $(C_6H_5CH_2^+)$ is resonance-stabilized and more stable than the phenyl carbocation.
Therefore,the $C-O$ bond on the benzylic side is cleaved,leading to the formation of benzyl iodide $(C_6H_5CH_2I)$ and phenol $(C_6H_5OH)$.
Statement $I$ claims the products are benzyl alcohol and iodobenzene,which is incorrect.
Statement $II$ correctly states that the $-O-CH_2-$ bond is cleaved.
Thus,Statement $I$ is false and Statement $II$ is true.

A is true: alcoholic $KOH$ is a stronger base and promotes elimination. E is true: $SOCl_2$ works well for alcohols, but phenols are not effectively converted to aryl chlorides this way. C is false: aryl halides are very different (less reactive). D is false: allyl chloride is a haloalkene, not a haloalkyne.

(B) . Finkelstein reaction involves the exchange of halogen atoms,typically using $NaI$ in acetone to prepare alkyl iodides.
$B$. Swarts reaction is used to prepare alkyl fluorides by treating alkyl chlorides or bromides with metallic fluorides like $SbF_3$ or $AgF$.
$C$. Sandmeyer's reaction involves the conversion of diazonium salts into aryl halides using copper$(I)$ salts like $Cu_2Cl_2$.
$D$. Fittig reaction involves the coupling of two aryl halide molecules in the presence of $Na$ metal and dry ether to form diaryls.
Therefore,the correct matching is $A-III, B-I, C-IV, D-II$.

(C) . Williamson synthesis involves the reaction of an alkoxide ion with an alkyl halide,which is a nucleophilic substitution $(III)$ reaction.
$B$. Friedel-Crafts reaction involves the substitution of an aromatic hydrogen by an alkyl or acyl group,which is an electrophilic aromatic substitution $(IV)$ reaction.
$C$. Bromination of vinyl benzene (an alkene) involves the addition of bromine across the double bond,which is an electrophilic addition $(I)$ reaction.
$D$. Chlorination of toluene in the presence of light involves the substitution of a hydrogen atom on the side chain via a free radical mechanism,which is a free radical substitution $(II)$ reaction.
Therefore,the correct matching is $A-III, B-IV, C-I, D-II$.

(A) $1$. Reaction $A$: Benzyl bromide $(C_6H_5CH_2Br)$ reacts with $AgCN$ to form benzyl isocyanide $(C_6H_5CH_2NC)$. This is because $AgCN$ is a covalent compound,and the nucleophilic attack occurs through the nitrogen atom of the ambident $CN^-$ ion.
$2$. Reaction $B$: Benzylamine $(C_6H_5CH_2NH_2)$ undergoes the carbylamine reaction with $CHCl_3$ and $KOH$ (or $NaOH$) to form benzyl isocyanide $(C_6H_5CH_2NC)$.
$3$. Reaction $C$: Bromobenzene does not undergo nucleophilic substitution with $AgCN$ under normal conditions due to the partial double bond character of the $C-Br$ bond.
$4$. Reaction $D$: Aniline $(C_6H_5NH_2)$ undergoes the carbylamine reaction to form phenyl isocyanide $(C_6H_5NC)$,not benzyl isocyanide.
$5$. Reaction $E$: $2-$Phenylethyl bromide reacts with $KCN$ to form $3-$phenylpropanenitrile $(C_6H_5CH_2CH_2CN)$.
Therefore,benzyl isocyanide is obtained from reactions $A$ and $B$.

(A) Statement $I$ is correct: Transition metals have high enthalpies of atomisation due to strong interatomic metallic bonding,which is facilitated by the presence of a large number of unpaired electrons in their $(n-1)d$ orbitals.
Statement $II$ is correct: $[Fe(H_2O)_6]^{3+}$ is an octahedral complex where the $d$-orbitals split into $t_{2g}$ $(d_{xy}, d_{xz}, d_{yz})$ and $e_g$ $(d_{x^2-y^2}, d_{z^2})$ sets,with $t_{2g} < e_g$. $[Ni(Cl)_4]^{2-}$ is a tetrahedral complex where the $d$-orbitals split into $e$ $(d_{x^2-y^2}, d_{z^2})$ and $t_2$ $(d_{xy}, d_{xz}, d_{yz})$ sets,with $e < t_2$. Thus,the splitting patterns provided are correct.

(B) Interstitial compounds are formed when small atoms like $H, C, N$ are trapped inside the crystal lattice of transition metals.
Their key characteristics include:
$1$. They have high melting points,higher than those of pure metals.
$2$. They are very hard (some approach diamond in hardness).
$3$. They retain metallic conductivity.
$4$. They are chemically inert and usually non-stoichiometric.
Since they are extremely hard and possess metallic properties,the statement that they are 'very soft and ionic' is incorrect. Therefore,option $B$ is the correct answer.

(C) The highest manganese fluoride is $MnF_4$ (oxidation state of Mn is $+4$).
The highest manganese oxide is $Mn_2O_7$ (oxidation state of Mn is $+7$).
The difference $|x| = |7 - 4| = 3$.
We need to identify ions with $3$ unpaired electrons:
$Sc^{3+}$: $[Ar] 3d^0$,unpaired electrons = $0$.
$Zn^{2+}$: $[Ar] 3d^{10}$,unpaired electrons = $0$.
$V^{2+}$: $[Ar] 3d^3$,unpaired electrons = $3$.
$Fe^{2+}$: $[Ar] 3d^6$,unpaired electrons = $4$.
$Co^{2+}$: $[Ar] 3d^7$,unpaired electrons = $3$.
Thus,$V^{2+}$ and $Co^{2+}$ have $3$ unpaired electrons.

(D) The electronic configurations of the given elements are:
$Eu (63): [Xe] 4f^7 6s^2$
$Gd (64): [Xe] 4f^7 5d^1 6s^2$
$Dy (66): [Xe] 4f^{10} 6s^2$
$Ho (67): [Xe] 4f^{11} 6s^2$
$Yb (70): [Xe] 4f^{14} 6s^2$
$Lu (71): [Xe] 4f^{14} 5d^1 6s^2$
$Tm (69): [Xe] 4f^{13} 6s^2$
$Hf (72): [Xe] 4f^{14} 5d^2 6s^2$
Comparing the number of electrons in the $4f$ orbital:
Pair $A$ ($Eu$ and $Gd$): Both have $7$ electrons in the $4f$ orbital. (Correct)
Pair $B$ ($Dy$ and $Ho$): $Dy$ has $10$ and $Ho$ has $11$ electrons. (Incorrect)
Pair $C$ ($Yb$ and $Hf$): Both have $14$ electrons in the $4f$ orbital. (Correct)
Pair $D$ ($Lu$ and $Tm$): $Lu$ has $14$ and $Tm$ has $13$ electrons. (Incorrect)
Therefore,pairs $A$ and $C$ are correct.

$(D)$ $A: [NiCl_4]^{2-}, Ni^{2+}(3d^8)$,weak field ligand,$sp^3$ hybridization,tetrahedral geometry,paramagnetic (two unpaired electrons). Correct.
$B: [Ni(NH_3)_6]^{2+}, Ni^{2+}(3d^8)$,weak field ligand,$sp^3d^2$ hybridization,octahedral geometry,paramagnetic (two unpaired electrons). Correct.
$C: [Ni(CO)_4], Ni^0(3d^8 4s^2)$,strong field ligand $(CO)$,$sp^3$ hybridization,tetrahedral geometry,diamagnetic (all electrons paired). Incorrect (given as paramagnetic).
$D: [Ni(CN)_4]^{2-}, Ni^{2+}(3d^8)$,strong field ligand $(CN^-)$,$dsp^2$ hybridization,square planar geometry,diamagnetic (all electrons paired). Correct.
Thus,the correct sequences are $A, B$,and $D$.

(C) is correct: The chelate effect makes $[Fe(C_2O_4)_3]^{3-}$ more stable than complexes with monodentate ligands like $OH^-$ or $SCN^-$.
$B$ is incorrect: $[Cu(en)_2]^{2+}$ is more stable than $[Cu(NH_3)_4]^{2+}$ because $en$ (ethylenediamine) is a bidentate ligand that forms stable $5-$membered chelate rings.
$C$ is correct: In $K_4[Fe(CN)_6]$,$Fe$ is in the $+2$ oxidation state ($d^6$ configuration). $CN^-$ is a strong field ligand,causing pairing of electrons,resulting in $d^2sp^3$ hybridization.
$D$ is correct: $NO_2^-$ is an ambidentate ligand,so $[Fe(NO_2)_3Cl_3]^{3-}$ can exhibit linkage isomerism.
$E$ is incorrect: $NO_2^-$ and $SCN^-$ are classic examples of ambidentate ligands.
Therefore,statements $A, C,$ and $D$ are correct.

(A) According to Crystal Field Theory $(CFT)$,in an octahedral complex,the five $d$-orbitals split into two sets due to the approach of ligands.
$1$. The $t_{2g}$ set $(d_{xy}, d_{yz}, d_{zx})$ is lower in energy by $0.4\Delta_o$ relative to the barycenter,providing stabilization.
$2$. The $e_g$ set $(d_{x^2-y^2}, d_{z^2})$ is higher in energy by $0.6\Delta_o$ relative to the barycenter,causing destabilization.
$3$. In a free metal ion,all five $d$-orbitals are degenerate (same energy). The presence of ligands creates an asymmetric electrostatic field,which lifts this degeneracy.
Both statements are scientifically accurate.

(D) The spin-only magnetic moment $(mu)$ is calculated using the formula $\mu = \sqrt{n(n+2)} \text{ BM}$,where $n$ is the number of unpaired electrons.
$A$. $[Cr(H_2O)_6]^{2+}$: $Cr^{2+}$ has a $d^4$ configuration. Number of unpaired electrons $(n)$ = $4$. $\mu = \sqrt{4(4+2)} = \sqrt{24} \approx 4.90 \text{ BM}$ $(III)$.
$B$. $[Co(H_2O)_6]^{2+}$: $Co^{2+}$ has a $d^7$ configuration. Number of unpaired electrons $(n)$ = $3$. $\mu = \sqrt{3(3+2)} = \sqrt{15} \approx 3.87 \text{ BM}$ $(I)$.
$C$. $[Cu(H_2O)_6]^{2+}$: $Cu^{2+}$ has a $d^9$ configuration. Number of unpaired electrons $(n)$ = $1$. $\mu = \sqrt{1(1+2)} = \sqrt{3} \approx 1.73 \text{ BM}$ $(IV)$.
$D$. $[Mn(H_2O)_6]^{2+}$: $Mn^{2+}$ has a $d^5$ configuration. Number of unpaired electrons $(n)$ = $5$. $\mu = \sqrt{5(5+2)} = \sqrt{35} \approx 5.92 \text{ BM}$ $(II)$.
Therefore,the correct matching is $A-III, B-I, C-IV, D-II$.

(D) The crystal field splitting energy ($\Delta_o$) depends on the strength of the ligand according to the spectrochemical series.
The spectrochemical series order for the given ligands is: $F^- < H_2O < en < CN^-$.
As the ligand field strength increases, the value of $\Delta_o$ increases.
Therefore, the order of $\Delta_o$ values is: $[CrF_6]^{3-} < [Cr(H_2O)_6]^{3+} < [Cr(en)_3]^{3+} < [Cr(CN)_6]^{3-}$.
Matching the values:
$A$. $[Cr(CN)_6]^{3-}$ corresponds to $IV$ ($26,600$ $cm^{-1}$).
$B$. $[CrF_6]^{3-}$ corresponds to $I$ ($15,060$ $cm^{-1}$).
$C$. $[Cr(H_2O)_6]^{3+}$ corresponds to $II$ ($17,400$ $cm^{-1}$).
$D$. $[Cr(en)_3]^{3+}$ corresponds to $III$ ($22,300$ $cm^{-1}$).
Thus, the correct match is $A-IV, B-I, C-II, D-III$.

(C) -aldotetrose has the general structure $CHO-(CHOH)_2-CH_2OH$.
Upon oxidation with concentrated $HNO_3$,the terminal aldehyde $(-CHO)$ and primary alcohol $(-CH_2OH)$ groups are oxidized to carboxylic acid $(-COOH)$ groups,resulting in a tartaric acid derivative $(HOOC-(CHOH)_2-COOH)$.
For this dicarboxylic acid to be optically inactive,it must contain a plane of symmetry,making it a meso compound.
In the Fischer projection of the $D$-aldotetrose,the $D$-configuration implies that the hydroxyl group at the chiral carbon furthest from the aldehyde group (i.e.,$C-3$) is on the right side.
For the resulting dicarboxylic acid to be meso,the hydroxyl groups must be on the same side (erythro configuration) so that a plane of symmetry exists.
Therefore,the $D$-aldotetrose must have both hydroxyl groups on the right side,which corresponds to $D$-erythrose.

(D) . Glucose exists in $\alpha$ and $\beta$ anomeric forms. (Correct)
$B$. Anomers differ in configuration only at the hemiacetal carbon $(C_1)$. (Correct)
$C$. Melting points: $\alpha$-$D$-glucose is $419 \text{ K}$ and $\beta$-$D$-glucose is $423 \text{ K}$. Thus,the melting point of $\beta$-anomer is greater than $\alpha$-anomer. (Incorrect)
$D$. Specific rotation: $\alpha$-anomer is $+112^\circ$ and $\beta$-anomer is $+19^\circ$. The values are swapped in the statement. (Incorrect)
$E$. Crystallization of glucose from hot saturated aqueous solution at $371 \text{ K}$ yields $\alpha$-$D$-glucose,and below $303 \text{ K}$ yields $\beta$-$D$-glucose. (Correct)
Therefore,statements $A, B$,and $E$ are correct.

(A) . Glutamine: Contains an amide group $(-CONH_2)$ in the side chain,which can undergo Hoffman bromamide degradation $(IV)$.
$B$. Lysine: Contains a primary amino group $(-NH_2)$ in the side chain,which reacts with benzenesulfonyl chloride in Hinsberg's test $(I)$.
$C$. Tyrosine: Contains a phenolic group ($-OH$ attached to a benzene ring),which gives a positive violet color with neutral $FeCl_3$ solution $(II)$.
$D$. Serine: Contains a primary alcohol group $(-CH_2OH)$ in the side chain,which gives a positive red color with Ceric ammonium nitrate test $(III)$.
Therefore,the correct matching is: $A-IV, B-I, C-II, D-III$.

(D) The reaction sequence is as follows:
$1$. Glucose reacts with $HI/\Delta$ to form $n$-hexane.
$2$. $n$-hexane undergoes aromatization in the presence of $V_2O_5$ at $773 \ K$ and high pressure to form benzene.
$3$. Benzene reacts with benzoyl chloride in the presence of anhydrous $AlCl_3$ (Friedel-Crafts acylation) to form benzophenone $(Ph-CO-Ph)$.
In the product,benzophenone $(C_{13}H_{10}O)$:
- Each phenyl ring has $3$ $\pi$ bonds,so $2$ rings have $6$ $\pi$ bonds.
- The $C=O$ group has $1$ $\pi$ bond.
- Total $\pi$ bonds = $6 + 1 = 7$. Each $\pi$ bond contains $2$ electrons,so total $\pi$ electrons = $7 \times 2 = 14$.
- The oxygen atom in the $C=O$ group has $2$ lone pairs of electrons,which equals $2 \times 2 = 4$ electrons.
- Total number of electrons = $14 + 4 = 18$.