JEE Main 2023 Chemistry Question Paper with Answer and Solution

(D) The reaction sequence is as follows:
$1$. Treatment of the carboxylic acid with $SOCl_2$ converts it into an acid chloride: $Ph-CH=CH-CH_2-COOH \xrightarrow{SOCl_2} Ph-CH=CH-CH_2-COCl$.
$2$. Reaction with $R-NH_2$ (amine) converts the acid chloride into an amide: $Ph-CH=CH-CH_2-COCl \xrightarrow{R-NH_2} Ph-CH=CH-CH_2-CONHR$.
$3$. Reduction with $LiAlH_4$ followed by $H_3O^+$ reduces the amide group $(-CONHR)$ to an amine group $(-CH_2NHR)$: $Ph-CH=CH-CH_2-CONHR \xrightarrow{LiAlH_4, H_3O^+} Ph-CH=CH-CH_2-CH_2NHR$.
Thus,the final product is $Ph-CH=CH-CH_2-CH_2NHR$.

(B) The chemical reaction for the oxidation of cyclohexanecarbaldehyde with Tollen's reagent is:
$C_6H_{11}CHO + 2[Ag(NH_3)_2]OH \rightarrow C_6H_{11}COONH_4 + 2Ag + 3NH_3 + H_2O$
Molar mass of cyclohexanecarbaldehyde $(C_7H_{12}O)$ = $(7 \times 12) + (12 \times 1) + 16 = 84 + 12 + 16 = 112 \ g/mol$.
From the stoichiometry,$1 \ mol$ of cyclohexanecarbaldehyde produces $3 \ mol$ of $NH_3$.
Moles of cyclohexanecarbaldehyde = $\frac{131.8 \times 10^3 \ g}{112 \ g/mol} \approx 1176.78 \ mol$.
Moles of $NH_3$ produced = $3 \times 1176.78 = 3530.34 \ mol$.
Mass of $NH_3$ = $3530.34 \ mol \times 17 \ g/mol = 60015.78 \ g \approx 60 \ kg$.

(B) The oligopeptide structure contains $4$ carbonyl groups $(C=O)$,each contributing $1$ $sp^2$ hybridized carbon. Total carbons from carbonyl groups = $4 \times 1 = 4$.
Additionally,the phenyl ring (side chain of phenylalanine) contains $6$ $sp^2$ hybridized carbons.
Total number of $sp^2$ hybridized carbons = $4 + 6 = 10$.

(B) $MgCl_2 \rightarrow Mg^{2+} + 2Cl^-$
Initial moles: $1$,$0$,$0$
At equilibrium: $1-\alpha$,$\alpha$,$2\alpha$
Total particles $i = 1 + 2\alpha$. Given $\alpha = 0.8$,so $i = 1 + 2(0.8) = 2.6$.
For a $1.0$ molal solution,$n_2 = 1$ mole of solute in $1000 \ g$ of water ($n_1 = 1000/18 = 55.55$ moles).
Using Raoult's Law: $\frac{p^{\circ} - p_s}{p^{\circ}} = \frac{i \times n_2}{n_1 + i \times n_2} \approx \frac{i \times n_2}{n_1}$.
$\frac{50 - p_s}{50} = \frac{2.6 \times 1}{55.55} = 0.0468$.
$50 - p_s = 2.34 \implies p_s = 47.66 \ mm \ Hg$.
Rounding to the nearest integer,$p_s \approx 48 \ mm \ Hg$.

(B) The starting material is $4$-ethoxyphenylacetaldehyde.
$1$. Strecker synthesis: Reaction with $NH_4Cl/KCN$ followed by hydrolysis gives $4$-ethoxyphenylalanine ('$A$').
$2$. Nitration: Treatment with $Conc. HNO_3-H_2SO_4$ $(2 \ eq)$ introduces two nitro groups at the ortho positions relative to the ethoxy group,yielding $3,5$-dinitro-$4$-ethoxyphenylalanine ('$B$').
$3$. Acetylation: Reaction with $(CH_3CO)_2O$ protects the amino group as an acetamide.
$4$. Esterification: Reaction with $EtOH/\Delta$ converts the carboxylic acid to an ethyl ester.
$5$. Reduction: $H_2, Pd/C$ reduces the two nitro groups to amino groups.
$6$. Diazotization: $HNO_2$ converts the amino groups to diazonium salts.
$7$. Sandmeyer-like reaction: $NaI$ replaces the diazonium groups with iodine atoms.
The final product '$D$' is $3,5$-diiodo-$4$-ethoxyphenylalanine ethyl ester derivative.
The molecular formula is $C_{15}H_{19}NO_4I_2$.
Comparing this with $C_xH_{19}NO_4I_2$,we find $x = 15$.

(B) The rate-determining step $(RDS)$ is the slow step: $NOBr_2 + NO \rightarrow 2 \ NOBr$.
The rate law is given by: $r = k [NOBr_2] [NO]$ ---- $(i)$
From the fast equilibrium step: $K_{eq} = \frac{[NOBr_2]}{[NO] [Br_2]}$,which implies $[NOBr_2] = K_{eq} [NO] [Br_2]$ ---- $(ii)$
Substituting $(ii)$ into $(i)$:
$r = k \cdot K_{eq} [NO] [Br_2] [NO]$
$r = k' [NO]^2 [Br_2]^1$
The overall order of the reaction is the sum of the powers of the concentration terms in the rate law: $2 + 1 = 3$.

(A) The given reactant is $N$-methylpyrrolidin$-2-$one,which is a cyclic amide (lactam).
Treatment with $NaOH$ and $\Delta$ (heating) causes the hydrolysis of the amide bond.
The hydroxide ion $(OH^-)$ attacks the carbonyl carbon,leading to the ring-opening of the lactam.
This results in the formation of the carboxylate salt of the corresponding amino acid,$CH_3NH(CH_2)_3COO^-Na^+$.
Subsequent acidification with $H^+$ protonates the carboxylate group to form the carboxylic acid,$CH_3NH(CH_2)_3COOH$.

(A) $A.$ Chlorophyll contains $Mg^{2+}$ ion,not $Co$. Thus,this is mismatched.
$B.$ $EDTA$ is used to estimate water hardness by forming a stable complex with $Ca^{2+}$ and $Mg^{2+}$ ions. This is a correct match.
$C.$ In black and white photography,the developed film is fixed by washing with hypo solution $(Na_2S_2O_3)$,which dissolves undecomposed $AgBr$ to form the complex ion $[Ag(S_2O_3)_2]^{3-}$,not $[Ag(CN)_2]^-$. Thus,this is mismatched.
$D.$ Wilkinson catalyst is $[(Ph_3P)_3RhCl]$. This is a correct match.
$E.$ $D^{-}$\text{Penicillamine} is a chelating ligand used to treat Wilson's disease by chelating excess copper. This is a correct match.
Therefore,the mismatched combinations are $A$ and $C$.

(D) $a.$ Nylon-$6$ is prepared from the monomer Caprolactam $(III)$.
$b.$ Vulcanized rubber is a cross-linked polymer formed by heating rubber with sulphur $(II)$.
$c.$ cis-$1,4$-polyisoprene is the chemical name for Natural Rubber $(I)$.
$d.$ Polychloroprene is commonly known as Neoprene $(IV)$.
Therefore,the correct matching is $a$ $\rightarrow III, b$ $\rightarrow II, c$ $\rightarrow I, d$ $\rightarrow IV$.

(A) Reaction with $C_2H_5O^-$: $C_2H_5O^-$ is a strong nucleophile. $2-$Methylpropyl bromide is a primary alkyl halide. Therefore,it undergoes an $S_N2$ reaction to form isobutyl ethyl ether $(A)$.
Reaction with $C_2H_5OH$: $C_2H_5OH$ is a weak nucleophile. The reaction proceeds via an $S_N1$ mechanism. The primary carbocation formed initially undergoes a $1,2-H$ shift to form a more stable tertiary carbocation,which then reacts with the nucleophile to form tert-butyl ethyl ether $(B)$.

(B) The reaction of $D-(+)-\text{Glyceraldehyde}$ with $HCN$ followed by hydrolysis and oxidation with $HNO_3$ is a Kiliani-Fischer synthesis sequence.
$1$. The addition of $HCN$ to the aldehyde group of $D-(+)-\text{Glyceraldehyde}$ creates a new chiral center,resulting in two diastereomeric cyanohydrins.
$2$. Hydrolysis of these cyanohydrins converts the $-CN$ group into a $-COOH$ group,forming two diastereomeric aldonic acids.
$3$. Oxidation with $HNO_3$ further oxidizes the terminal $-CH_2OH$ group to a $-COOH$ group,resulting in two dicarboxylic acids (aldaric acids).
$4$. One of the resulting products is $meso-\text{tartaric acid}$ (or a derivative),which is optically inactive due to an internal plane of symmetry.
$5$. The other product is an optically active isomer (e.g.,$L-(+)-\text{tartaric acid}$ derivative).
Thus,the final products are one optically inactive $(meso)$ product and one optically active product.

(A) The refining of metals depends on their properties.
$Zinc$ is refined by the distillation method because it has a low boiling point.
$Liquation$ is used for metals with low melting points like $Tin$ $(Sn)$.
$Titanium$ is refined by the $van \ Arkel$ method.
$Nickel$ is refined by the $Mond$ process.
$Copper$ is refined by $Electrolysis$.
Therefore,the mismatch is $Zinc - Liquation$.

(C) The central atom $Cl$ in $ClF_5$ has $7$ valence electrons. It forms $5$ bonds with $F$ atoms and has $1$ lone pair of electrons.
Total electron pairs = $5 + 1 = 6$,which corresponds to $sp^3d^2$ hybridization.
The geometry is octahedral,but due to the presence of one lone pair,the shape is square pyramidal.
$ClF_5$ exists as a colourless liquid at room temperature.

(C) The reaction proceeds as follows:
$1$. The alcohol group is protonated by $HCl$ and leaves as $H_2O$,forming a tertiary carbocation.
$2$. This carbocation undergoes a ring expansion (Wagner-Meerwein rearrangement) because the five-membered ring expands to a six-membered ring to form a more stable carbocation.
$3$. Finally,treatment with $KOH$ causes dehydrohalogenation (elimination of $H^+$) to form the most stable alkene,which is the major product '$B$'.
$4$. The final structure corresponds to the product shown in option $C$.

(D) Lyophilic sols act as protective colloids for lyophobic sols. When a lyophilic sol is added to a lyophobic sol,the lyophilic particles form a protective film or layer around the lyophobic particles,thereby preventing their coagulation by electrolytes.

(B) The third ionization energy $(IE_3)$ is high for elements that have stable electronic configurations in their $+2$ oxidation state,as removing the third electron disrupts this stability.
$Eu$ $(Z=63)$ has the configuration $[Xe] 4f^7 6s^2$. Its $Eu^{2+}$ ion is $[Xe] 4f^7$,which is a stable half-filled configuration.
$Yb$ $(Z=70)$ has the configuration $[Xe] 4f^{14} 6s^2$. Its $Yb^{2+}$ ion is $[Xe] 4f^{14}$,which is a stable fully-filled configuration.
Therefore,both $Eu$ and $Yb$ exhibit high third-ionization energy.

(C) For two solutions at the same temperature,if osmotic pressure $\pi_1 = \pi_2$,then their molar concentrations must be equal $(C_1 = C_2)$.
Given,$C_2 = 0.05 \ M$ (glucose).
For solution $A$,mass $w = 12 \ g$ and volume $V = 1000 \ mL = 1 \ L$.
Concentration $C_1 = \frac{n}{V} = \frac{w / M_A}{1 \ L} = \frac{12}{M_A} \ M$.
Equating the concentrations: $\frac{12}{M_A} = 0.05$.
$M_A = \frac{12}{0.05} = 240 \ g \ mol^{-1}$.
Thus,the molecular mass of $A$ is $240 \ g \ mol^{-1}$.

(C) Step $1$: Determine the concentration of $Ca(OH)_2$ using the titration with $HCl$.
Reaction: $Ca(OH)_2 + 2HCl \rightarrow CaCl_2 + 2H_2O$
Volume of $Ca(OH)_2$ used = $35.5 \, mL - 25.5 \, mL = 10 \, mL$.
Volume of $HCl = 20 \, mL$,Molarity of $HCl = 0.5 \, M$.
Millimoles of $HCl = 20 \times 0.5 = 10 \, mmol$.
From the stoichiometry,$1 \, mol$ of $Ca(OH)_2$ reacts with $2 \, mol$ of $HCl$.
Millimoles of $Ca(OH)_2 = \frac{10}{2} = 5 \, mmol$.
$M_{Ca(OH)_2} = \frac{5 \, mmol}{10 \, mL} = 0.5 \, M$.
Step $2$: Determine the concentration of $H_2SO_4$.
Reaction: $Ca(OH)_2 + H_2SO_4 \rightarrow CaSO_4 + 2H_2O$
Volume of $Ca(OH)_2 = 20 \, mL$,Molarity = $0.5 \, M$.
Millimoles of $Ca(OH)_2 = 20 \times 0.5 = 10 \, mmol$.
From the stoichiometry,$1 \, mol$ of $Ca(OH)_2$ reacts with $1 \, mol$ of $H_2SO_4$.
Millimoles of $H_2SO_4 = 10 \, mmol$.
Volume of $H_2SO_4 = 10 \, mL$.
$M_{H_2SO_4} = \frac{10 \, mmol}{10 \, mL} = 1 \, M$.

(C) For a first order reaction,the time taken for $50 \%$ completion is the half-life,$t_{50} = t_{1/2}$.
At $87.5 \%$ completion,the remaining amount of reactant is $A_t = A_0 - 0.875 A_0 = 0.125 A_0 = \frac{A_0}{8}$.
Using the half-life concept:
$A_0$ $\xrightarrow{t_{1/2}} \frac{A_0}{2}$ $\xrightarrow{t_{1/2}} \frac{A_0}{4}$ $\xrightarrow{t_{1/2}} \frac{A_0}{8}$.
This shows that $t_{87.5} = 3 \times t_{1/2}$.
Since $t_{50} = t_{1/2}$,we have $t_{87.5} = 3 \times t_{50}$.
Therefore,$x = 3$.

(C) The balanced redox reaction is given by:
$10[FeSO_4 \cdot (NH_4)_2SO_4 \cdot 6H_2O] + 2KMnO_4 + 8H_2SO_4$ $\rightarrow 5Fe_2(SO_4)_3 + 2MnSO_4 + 10(NH_4)_2SO_4 + K_2SO_4 + 68H_2O$
In this reaction,for every $2$ molecules of $KMnO_4$ consumed,$68$ molecules of water are produced.

(C) Step $1$: Calculate the volume of the nickel layer.
Volume = Area $\times$ thickness = $100 \, cm^2 \times 0.0001 \, cm = 0.01 \, cm^3$.
Step $2$: Calculate the mass of nickel required.
Mass = Density $\times$ Volume = $10 \, g \cdot cm^{-3} \times 0.01 \, cm^3 = 0.1 \, g$.
Step $3$: Use Faraday's law of electrolysis: $W = \frac{M \times I \times t}{n \times F}$.
Here,$W = 0.1 \, g$,$M = 60 \, g \cdot mol^{-1}$,$I = 2 \, A$,$n = 2$ (for $Ni^{2+} + 2e^- \rightarrow Ni$),$F = 96500 \, C \cdot mol^{-1}$.
Step $4$: Solve for $t$ (which is $x$):
$0.1 = \frac{60 \times 2 \times x}{2 \times 96500}$.
$0.1 = \frac{60 \times x}{96500}$.
$x = \frac{0.1 \times 96500}{60} = \frac{9650}{60} \approx 160.83 \, s$.
Rounding to the nearest integer,$x = 161$.

(B) In qualitative analysis,$(NH_4)_2CO_3$ is used as the group reagent for the $5^{th}$ group cations,which include $Ba^{2+}$,$Ca^{2+}$,and $Sr^{2+}$.
The chemical reaction for the precipitation of $Ba^{2+}$ is:
$Ba^{2+} + (NH_4)_2CO_3 \rightarrow BaCO_3 \downarrow + 2NH_4^{+}$
Thus,$Ba^{2+}$ is detected by obtaining a white precipitate of barium carbonate $(BaCO_3)$.

(C) The general structure of an amino acid contains one $-NH_2$ group and one $-COOH$ group.
$Arginine$,$Lysine$,and $Histidine$ are basic amino acids because they contain an additional basic functional group in their side chains.
$Asparagine$ $(Asn)$ has the side chain $-CH_2CONH_2$. The amide group $(-CONH_2)$ is neutral,not basic. Thus,$Asparagine$ contains only one basic amino group $(-NH_2)$ attached to the $\alpha$-carbon,making it the correct choice among the given options.

(D) Statement $I$ is correct,as Ellingham diagrams are graphical representations of the change in Gibbs free energy $(\Delta G^0)$ with temperature $(T)$ for the formation of oxides,sulfides,and halides of metals.
Statement $II$ is incorrect because Ellingham diagrams consist of plots of $\Delta G^0$ vs $T$,not $\Delta_f H^0$ vs $T$.

(C) The Tyndall effect is observed when the diameter of the dispersed particles is not much smaller than the wavelength of the light used.
When the particle size is large enough (comparable to the wavelength of light),the light hits the particles and scatters in all directions,making the path of the light visible.
Therefore,both Assertion $A$ and Reason $R$ are correct,and $R$ provides the correct explanation for $A$.

(B) The complex $[Cr(ox)_2ClBr]^{3-}$ is of the type $[M(AA)_2a_2]$ where $AA$ is a bidentate ligand (oxalate) and $a$ represents monodentate ligands ($Cl^-$ and $Br^-$).
$1$. Geometrical Isomerism: There are two geometrical isomers:
- $trans$-isomer: The two monodentate ligands ($Cl$ and $Br$) are at $180^{\circ}$ to each other. This isomer has a plane of symmetry and is optically inactive.
- $cis$-isomer: The two monodentate ligands ($Cl$ and $Br$) are at $90^{\circ}$ to each other. This isomer lacks a plane of symmetry and is chiral.
$2$. Optical Isomerism: The $cis$-isomer exists as a pair of enantiomers ($d$ and $l$ forms).
Total stereoisomers = $1$ $(trans)$ + $2$ ($cis$-$d$ and $cis$-$l$) = $3$.

(D) The reaction sequence is as follows:
$1$. $A$ is $Aniline$ $(C_6H_5NH_2)$.
$2$. When $Aniline$ reacts with $Br_2$ in water (or $Br_2/CS_2$ at low temperature),it undergoes electrophilic substitution to form $2,4,6-tribromoaniline$ (Compound $B$).
$3$. $2,4,6-tribromoaniline$ reacts with $NaNO_2/HCl$ at $0-5^{\circ}C$ to form the diazonium salt,$2,4,6-tribromobenzenediazonium$ chloride (Compound $C$).
$4$. Finally,reduction of the diazonium salt with $H_3PO_2$ and water removes the $-N_2^+Cl^-$ group,yielding $1,3,5-tribromobenzene$.
Thus,compound $A$ is $Aniline$.

(A) - $A$. Weak intermolecular forces of attraction: $2-$chloro$-1,3-$butadiene (chloroprene) is the monomer of Neoprene,which is an elastomer (rubber) characterized by weak intermolecular forces.
- $B$. Hydrogen bonding: Hexamethylenediamine reacting with adipic acid forms Nylon-$6,6$,which exhibits strong hydrogen bonding due to the presence of amide groups.
- $C$. Heavily branched polymer: Phenol and formaldehyde react to form Bakelite,which is a heavily branched (cross-linked) thermosetting polymer.
- $D$. High density polymer: $AlEt_3 + TiCl_4$ is the Ziegler-Natta catalyst used to prepare high-density polyethylene $(HDPE)$.
Therefore,the correct matching is $A-III, B-I, C-IV, D-II$.

(D) Among the halide ions,only $I^{-}$ can be oxidized to $I_2$ by oxygen in an acidic medium because it has the highest standard oxidation potential.
The balanced chemical equation is:
$4I^{-}_{(aq)} + 4H^{+}_{(aq)} + O_{2(g)} \rightarrow 2I_{2(s)} + 2H_2O_{(l)}$

(B) $1$. $CH_3-CH_2-CH_2-Br + KOH \text{ (alc)} \rightarrow CH_3-CH=CH_2$ (Alkene,$III$)
$2$. $CH_3-CH_2-CH_2-Br + KCN \text{ (alc)} \rightarrow CH_3-CH_2-CH_2-CN$ (Nitrile,$I$)
$3$. $CH_3-CH_2-CH_2-Br + AgNO_2 \rightarrow CH_3-CH_2-CH_2-NO_2 + AgBr$ (Nitroalkane,$IV$)
$4$. $CH_3-CH_2-CH_2-Br + CH_3COOAg \rightarrow CH_3COOCH_2-CH_2-CH_3 + AgBr$ (Ester,$II$)
Therefore,the correct matching is $A-III, B-I, C-IV, D-II$.

(B) The complex with the maximum number of unpaired electrons will exhibit the maximum attraction to an applied magnetic field (paramagnetism).
$1$. $[Zn(H_2O)_6]^{2+}$: $Zn^{2+}$ is a $d^{10}$ system,which has $0$ unpaired electrons.
$2$. $[Co(H_2O)_6]^{2+}$: $Co^{2+}$ is a $d^7$ system. In an octahedral field with a weak field ligand $(H_2O)$,the configuration is $t_{2g}^5 e_g^2$,resulting in $3$ unpaired electrons.
$3$. $[Co(en)_3]^{3+}$: $Co^{3+}$ is a $d^6$ system. Since $en$ is a strong field ligand,the configuration is $t_{2g}^6 e_g^0$,resulting in $0$ unpaired electrons.
$4$. $[Ni(H_2O)_6]^{2+}$: $Ni^{2+}$ is a $d^8$ system,which has $2$ unpaired electrons.
Since $[Co(H_2O)_6]^{2+}$ has the highest number of unpaired electrons $(3)$,it will exhibit the maximum attraction.

(D) The dissolution equilibrium is $Cu(OH)_{2(s)} \rightleftharpoons Cu^{2+}_{(aq)} + 2OH^{-}_{(aq)}$.
$K_{sp} = [Cu^{2+}][OH^{-}]^2 = 1 \times 10^{-20}$.
At $pH = 14$,$pOH = 14 - 14 = 0$,so $[OH^{-}] = 10^0 = 1 \, M$.
$[Cu^{2+}] = \frac{K_{sp}}{[OH^{-}]^2} = \frac{10^{-20}}{1^2} = 10^{-20} \, M$.
Using the Nernst equation for $Cu^{2+} + 2e^{-} \rightarrow Cu_{(s)}$:
$E = E^{\circ} - \frac{0.059}{2} \log_{10} \frac{1}{[Cu^{2+}]}$.
$E = 0.34 - \frac{0.059}{2} \log_{10} \frac{1}{10^{-20}}$.
$E = 0.34 - \frac{0.059}{2} \times 20 = 0.34 - 0.59 = -0.25 \, V$.
Given $E = -x \times 10^{-2} \, V$,we have $-0.25 = -x \times 10^{-2}$,so $x = 25$.

(D) For the reaction $A ( g ) \rightarrow 2 B ( g ) + C ( g )$,let the initial pressure of $A$ be $P_0 = 800 \ mm \ Hg$.
At $t = 10 \ min$,the total pressure $P_t = P_0 - x + 2x + x = P_0 + 2x = 1600 \ mm \ Hg$.
Substituting $P_0 = 800$,we get $800 + 2x = 1600$,so $2x = 800$,which means $x = 400 \ mm \ Hg$.
The pressure of $A$ remaining at $t = 10 \ min$ is $P_0 - x = 800 - 400 = 400 \ mm \ Hg$.
Since the initial pressure was $800 \ mm \ Hg$ and it became $400 \ mm \ Hg$ in $10 \ min$,the half-life $t_{1/2} = 10 \ min$.
After $30 \ min$ $(3 \times t_{1/2})$,the pressure of $A$ remaining is $P_A = P_0 \times (1/2)^3 = 800 \times (1/8) = 100 \ mm \ Hg$.
The amount of $A$ reacted is $800 - 100 = 700 \ mm \ Hg$.
According to stoichiometry,$A \rightarrow 2B + C$,the pressure of $B$ produced is $2 \times 700 = 1400 \ mm \ Hg$ and $C$ produced is $700 \ mm \ Hg$.
The total pressure at $t = 30 \ min$ is $P_A + P_B + P_C = 100 + 1400 + 700 = 2200 \ mm \ Hg$.

(D) For a body-centered cubic $(BCC)$ lattice,the relationship between the edge length $(a)$ and the atomic radius $(r)$ is given by: $\sqrt{3} \ a = 4 \ r$.
Given $a = 4 \ \mathring{A}$,we substitute the value into the equation:
$\sqrt{3} \times 4 = 4 \ r$.
$r = \sqrt{3} \ \mathring{A} \approx 1.732 \ \mathring{A}$.
To express this in the form $..... \times 10^{-1} \ \mathring{A}$,we multiply by $10/10$:
$r = 17.32 \times 10^{-1} \ \mathring{A}$.
Rounding to the nearest integer,we get $17 \times 10^{-1} \ \mathring{A}$.

(D) Mass of solution = $100 \, g$.
Mass of $NaCl = 29.25 \, g$,Mass of $MgCl_2 = 19 \, g$.
Mass of solvent $(H_2O)$ = $100 - (29.25 + 19) = 51.75 \, g = 0.05175 \, kg$.
For $NaCl$ $(i=2)$: Moles = $29.25 / 58.5 = 0.5 \, mol$.
For $MgCl_2$ $(i=3)$: Moles = $19 / 95 = 0.2 \, mol$.
Total moles of ions = $(i_{NaCl} \times n_{NaCl}) + (i_{MgCl_2} \times n_{MgCl_2}) = (2 \times 0.5) + (3 \times 0.2) = 1.0 + 0.6 = 1.6 \, mol$.
$\Delta T_b = i \times K_b \times m = K_b \times (n_{total_ions} / \text{mass of solvent in kg}) = 0.52 \times (1.6 / 0.05175) \approx 16.07 \, ^{\circ}C$.
Boiling point of solution = $100 + 16.07 = 116.07 \, ^{\circ}C$.
Nearest integer is $116$.

(A) The correct matches are as follows:
$A$. Tetrafluoroethene is the monomer of Teflon $(III)$.
$B$. Acrylonitrile is the monomer of Orlon $(I)$.
$C$. Caprolactam is the monomer of Nylon-$6$ $(IV)$.
$D$. Isoprene ($2$-methyl-$1,3$-butadiene) is the monomer of Natural rubber $(II)$.
Therefore,the correct matching is $A-III, B-I, C-IV, D-II$.

(B) Step $1$: Hydroboration-oxidation of propene $(CH_3-CH=CH_2)$ with $B_2H_6$ followed by $H_2O_2, NaOH$ gives propan$-1-$ol $(CH_3-CH_2-CH_2-OH)$.
Step $2$: Oxidation of propan$-1-$ol with $PCC$ (Pyridinium chlorochromate) gives propanal $(CH_3-CH_2-CHO)$.
Step $3$: Reaction of propanal with methylmagnesium bromide $(CH_3MgBr)$ followed by acidic workup gives butan$-2-$ol $(CH_3-CH_2-CH(OH)-CH_3)$.

(D) Calcination is the process of heating an ore in the absence or limited supply of air to remove moisture and volatile impurities.
Option $A$ is the dehydration of limonite (calcination).
Option $B$ is the calcination of limestone.
Option $C$ is the calcination of dolomite.
Option $D$ $(2PbS + 3O_2 \xrightarrow{\Delta} 2PbO + 2SO_2)$ involves the reaction of sulfide ore with oxygen,which is the definition of roasting.

(B) Step $1$: Aniline $(C_6H_5NH_2)$ reacts with $NaNO_2$ in the presence of $HCl$ at $0-5^{\circ}C$ to form benzenediazonium chloride $(C_6H_5N_2^+Cl^-)$,which is product '$A$'.
Step $2$: Benzenediazonium chloride undergoes a diazo coupling reaction with $N,N$-dimethylaniline in a weakly acidic medium. The diazonium group $( -N=N- )$ attacks the para-position of the $N,N$-dimethylaniline ring to form $N,N$-dimethyl$-4-$(phenylazo)aniline,which is product '$B$'.
The reaction is: $C_6H_5N_2^+Cl^- + C_6H_5N(CH_3)_2 \rightarrow C_6H_5-N=N-C_6H_4-N(CH_3)_2 + HCl$.

(A) $1$. In pyrophosphoric acid $(H_4P_2O_7)$,there is $1$ $P-O-P$ bond.
$2$. In cyclotrimetaphosphoric acid $((HPO_3)_3)$,there are $3$ $P-O-P$ bonds in the cyclic structure.
$3$. In phosphorus pentoxide $(P_4O_{10})$,there are $6$ $P-O-P$ bonds.
Therefore,the number of $P-O-P$ bonds are $1, 3, 6$ respectively.

(C) The conversion of $p$-nitrotoluene to $3$-bromo-$4$-chlorobenzoic acid involves the following steps:
$1$. Electrophilic aromatic substitution: Bromination of $p$-nitrotoluene using $Br_2 / Fe$ introduces a bromine atom at the ortho position relative to the methyl group.
$2$. Reduction: The nitro group $(-NO_2)$ is reduced to an amino group $(-NH_2)$ using $Fe / H^+$.
$3$. Diazotization: The amino group is converted to a diazonium salt $(-N_2^+Cl^-)$ using $HONO$ (nitrous acid) at $0-5 \ ^\circ C$.
$4$. Sandmeyer reaction: The diazonium group is replaced by a chlorine atom using $CuCl$.
$5$. Oxidation: The methyl group $(-CH_3)$ is oxidized to a carboxylic acid group $(-COOH)$ using $KMnO_4$.
Thus,the correct sequence is $(i) Br_2 / Fe, (ii) Fe / H^{+}, (iii) HONO, (iv) CuCl, (v) KMnO_4$.

(D) The starting material is ninhydrin ($1,2,3$-indantrione). When it reacts with water $(H_2O)$,the central carbonyl group,which is highly electrophilic due to the electron-withdrawing effect of the two adjacent carbonyl groups,undergoes nucleophilic addition by water to form a gem-diol. This gem-diol is stabilized by intramolecular hydrogen bonding between the hydroxyl groups and the adjacent carbonyl oxygen atoms. The product is $2,2$-dihydroxy-$1,3$-indanedione.

(D) Arginine is an amino acid that exists as a zwitterion in its solid state.
Due to the presence of ionic charges,it exhibits a high melting point and is a crystalline solid.
It is highly soluble in polar solvents like water but has low solubility in non-polar solvents like benzene.
Therefore,the statement that it has high solubility in benzene is incorrect.

(A) Chlorine is an ortho/para directing group due to the resonance effect.
In electrophilic aromatic substitution reactions,the para-isomer is generally the major product due to less steric hindrance compared to the ortho-isomer.
Therefore,the major product of the Friedel-Crafts acylation of chlorobenzene is $p-$chloroacetophenone,which corresponds to the structure shown in option $A$.

(A) The magnitude of crystal field splitting energy $\left(\Delta_0\right)$ depends on the oxidation state of the metal ion and the nature of the metal ion itself.
For a given ligand and oxidation state,$\Delta_0$ generally increases with the effective nuclear charge and the size of the $d$-orbitals.
Among the given $3d$ series metal ions in the $+3$ oxidation state,$Cr^{3+}$ has a $d^3$ configuration.
In octahedral complexes,$Cr^{3+}$ exhibits a high crystal field splitting energy due to the stability of the $t_{2g}^3$ configuration.
Comparing the given complexes,$[Cr(OH_2)_6]^{3+}$ has the highest $\Delta_0$ value among the choices provided.

(C) In a $CsCl$ crystal structure,the $Cs^{+}$ ion is located at the body center,and $Cl^{-}$ ions are located at the corners of the cube.
The body diagonal of the cube is given by $\sqrt{3} a$.
Since the $Cs^{+}$ ion at the center touches the $Cl^{-}$ ions at the corners along the body diagonal,the length of the body diagonal is equal to $2(r_{Cs^{+}} + r_{Cl^{-}})$.
Therefore,$2(r_{Cs^{+}} + r_{Cl^{-}}) = \sqrt{3} a$.
This simplifies to $r_{Cs^{+}} + r_{Cl^{-}} = \frac{\sqrt{3}}{2} a$.

(A) The complex is $[Co(H_2O)_6]^{2+}$.
$Co^{2+}$ has a $3d^7$ electronic configuration.
In an octahedral field,$H_2O$ is a weak field ligand,so the electrons fill the orbitals as $t_{2g}^5 e_g^2$.
The $t_{2g}$ orbitals are filled as $(t_{2g})^2 (t_{2g})^2 (t_{2g})^1$,which results in $1$ unpaired electron in the $t_{2g}$ set.

(A) The complex $[Cr(H_2O)_5Cl]Cl_2$ dissociates in water as: $[Cr(H_2O)_5Cl]Cl_2 \rightarrow [Cr(H_2O)_5Cl]^{2+} + 2Cl^-$.
Each mole of the complex provides $2$ moles of ionizable chloride ions $(Cl^-)$.
Number of millimoles of the complex $= Molarity \times Volume (mL) = 0.01 \, M \times 20 \, mL = 0.2 \, mmol$.
Since $1$ mole of complex gives $2$ moles of $Cl^-$,$0.2 \, mmol$ of complex gives $0.4 \, mmol$ of $Cl^-$.
The precipitation reaction is: $Ag^+ + Cl^- \rightarrow AgCl(s)$.
Thus,$0.4 \, mmol$ of $AgNO_3$ is required to precipitate $0.4 \, mmol$ of $Cl^-$.
Using $Molarity = \frac{millimoles}{Volume (mL)}$,we get $0.1 = \frac{0.4}{V}$.
$V = \frac{0.4}{0.1} = 4 \, mL$.

(A) The relative lowering of vapour pressure is given by $\frac{P^{\circ} - P_{s}}{P_{s}} = \frac{n_{solute}}{n_{solvent}}$.
Given: $30 \% \ (w/v)$ solution means $30 \ g$ of glucose in $100 \ mL$ of solution.
Mass of solution $= \text{density} \times \text{volume} = 1.2 \ g \ cm^{-3} \times 100 \ mL = 120 \ g$.
Mass of solvent (water) $= 120 \ g - 30 \ g = 90 \ g$.
Moles of glucose $(n_{solute})$ $= \frac{30 \ g}{180 \ g \ mol^{-1}} = \frac{1}{6} \ mol \approx 0.1667 \ mol$.
Moles of water $(n_{solvent})$ $= \frac{90 \ g}{18 \ g \ mol^{-1}} = 5 \ mol$.
Using the formula $\frac{24 - P_{s}}{P_{s}} = \frac{0.1667}{5} = 0.03334$.
$24 - P_{s} = 0.03334 \ P_{s} \implies 24 = 1.03334 \ P_{s}$.
$P_{s} = \frac{24}{1.03334} \approx 23.22 \ mm \ Hg$. The closest integer value is $23 \ mm \ Hg$.

(A) The coagulating value is defined as the minimum concentration of an electrolyte in millimoles per liter $(mmol/L)$ required to cause coagulation of a sol in $2$ hours.
Given:
Volume of $NaCl$ solution = $20 \ mL$
Molarity of $NaCl$ solution = $0.5 \ M$
Volume of $As_2S_3$ sol = $200 \ mL$
Number of millimoles of $NaCl$ = $Molarity \times Volume \ (in \ mL) = 0.5 \times 20 = 10 \ mmol$.
These $10 \ mmol$ of $NaCl$ are required for $200 \ mL$ of sol.
For $1000 \ mL$ $(1 \ L)$ of sol,the required millimoles = $(10 / 200) \times 1000 = 50 \ mmol/L$.
Therefore,the coagulating value of $NaCl$ is $50$.