WBJEE 2023 Chemistry Question Paper with Answer and Solution

(A) $1$. Compound $M$ reacts with ammoniacal $AgNO_3$ to give a white precipitate,which indicates that $M$ is a terminal alkyne $(R-C \equiv CH)$.
$2$. $M$ undergoes hydrogenation with $H_2$ and Lindlar catalyst to form $N$,which is an alkene.
$3$. $N$ undergoes ozonolysis to give $O$ and $P$. $O$ (benzaldehyde,$Ph-CHO$) reacts with $(CH_3CO)_2O$ and $CH_3COONa$ (Perkin's condensation) to form cinnamic acid $(Ph-CH=CH-COOH)$.
$4$. Based on the reaction sequence,$M$ must be phenylacetylene $(Ph-C \equiv CH)$.
$5$. $Ph-C \equiv CH + H_2 \xrightarrow{Lindlar} Ph-CH=CH_2$ $(N)$.
$6$. $Ph-CH=CH_2 \xrightarrow{O_3} Ph-CHO$ $(O)$ + $HCHO$ $(P)$.
$7$. Thus,$M$ is $Ph-C \equiv CH$.

(D) To determine the molecular shapes and lone pairs ($L$.$P$.) on the central atoms:
$1$. For $SF_4$: The central atom $S$ has $6$ valence electrons. It forms $4$ bonds with $F$ atoms,leaving $1$ lone pair. The shape is see-saw.
$2$. For $CF_4$: The central atom $C$ has $4$ valence electrons. It forms $4$ bonds with $F$ atoms,leaving $0$ lone pairs. The shape is tetrahedral.
$3$. For $XeF_4$: The central atom $Xe$ has $8$ valence electrons. It forms $4$ bonds with $F$ atoms,leaving $2$ lone pairs. The shape is square planar.
Thus,the shapes are different with $1, 0$ and $2$ lone pairs of electrons on the central atoms,respectively.

(C) The hybridisation of an atom can be determined by calculating the steric number $(SN)$:
$SN = \frac{1}{2} [V + M - C + A]$
Where $V$ is the number of valence electrons of the central atom,$M$ is the number of monovalent atoms,$C$ is the cationic charge,and $A$ is the anionic charge.
For $NO_2^{+}$:
$V = 5$ (nitrogen),$M = 0$,$C = 1$,$A = 0$.
$SN = \frac{1}{2} [5 + 0 - 1 + 0] = \frac{4}{2} = 2$.
$A$ steric number of $2$ corresponds to $sp$ hybridisation.
Therefore,the nitrogen atom in $NO_2^{+}$ is $sp$ hybridised.

(B) The bond length of the $C=O$ bond depends on the extent of resonance. Greater resonance character increases the single bond character of the $C=O$ bond,thereby increasing its bond length.
$I$: Ethyl propanoate has no conjugation with the $C=O$ group.
$II$: Ethyl propenoate has conjugation between the $C=C$ double bond and the $C=O$ group,which increases the single bond character of $C=O$.
$III$: Ethenyl propanoate has conjugation between the lone pair of the oxygen atom and the $C=O$ group (back-bonding),which significantly increases the single bond character of $C=O$.
Comparing the resonance effects:
In $III$,the lone pair on the oxygen atom is directly conjugated with the $C=O$ group,leading to a strong resonance effect that makes the $C=O$ bond more like a single bond.
In $II$,the $C=C$ double bond is conjugated with the $C=O$ group.
In $I$,there is no such conjugation.
Thus,the extent of single bond character is $III > II > I$.
Therefore,the order of $C=O$ bond length is $III > II > I$.

(B) The ionization energy order for the given noble gases is $He > Ne > Ar > Kr$.
Since the energy of the photon is sufficient to ionize $Ar$,it will also be sufficient to ionize $Kr$ because the ionization energy of $Kr$ is lower than that of $Ar$.
However,the energy is not sufficient to ionize $He$ or $Ne$ as they have higher ionization energies than $Ar$.
Therefore,the ions present in the mixture will be $Ar^{+}$ and $Kr^{+}$.

(B) $1$. Identify the principal functional group: The aldehyde group $(-CHO)$ has the highest priority and must be assigned the lowest possible number,so it is at position $1$.
$2$. Select the longest carbon chain containing the principal functional group and the double bond: The chain starting from the aldehyde carbon $(C-1)$ through the double bond gives a $4-$carbon chain (but$-2-$enal).
$3$. Number the chain: The aldehyde carbon is $C-1$,the carbon with the double bond is $C-2$,the next is $C-3$,and the terminal methyl is $C-4$.
$4$. Identify substituents: There is an ethyl group $(-CH_2CH_3)$ attached to $C-2$.
$5$. Combine: The $IUPAC$ name is $2-$ethylbut$-2-$enal.

(C) The stability of carbocations is determined by the electron-donating or electron-withdrawing nature of the substituents attached to the conjugated system.
$1$. In $III$ and $IV$,the groups $-NMe_2$ and $-OMe$ show a strong $+R$ (resonance) effect,which stabilizes the carbocation significantly.
$2$. Since nitrogen is less electronegative than oxygen,the lone pair on $N$ is more easily donated,making $-NMe_2$ a stronger $+R$ group than $-OMe$. Thus,$III > IV$.
$3$. In $I$,the $-CH_3$ group shows a weak $+I$ (inductive) effect and hyperconjugation,providing moderate stability.
$4$. In $II$,the $-BMe_2$ group has an empty $p$-orbital and shows a $-R$ effect,which destabilizes the carbocation.
$5$. Therefore,the correct stability order is $III > IV > I > II$.

(A) For all atoms to lie in a single plane,all carbon atoms and the atoms attached to the benzene ring must be $sp^2$ hybridized to maintain planarity through conjugation.
In $4-$Nitrobenzaldehyde $(O_2N-C_6H_4-CHO)$,the benzene ring,the nitro group $(-NO_2)$,and the aldehyde group $(-CHO)$ are all $sp^2$ hybridized.
The conjugation extends throughout the molecule,allowing all atoms (including the hydrogen of the aldehyde group) to lie in the same plane.
In other options,groups like $-OCH_3$ (in $4-$Methoxybenzaldehyde),$-CH_3$ (in $4-$Methylnitrobenzene),and $-COCH_3$ (in $4-$Nitroacetophenone) contain $sp^3$ hybridized carbon atoms,which possess tetrahedral geometry and prevent all atoms from lying in a single plane.

(B) The given pair of compounds are non-superimposable mirror images of each other.
By assigning $R/S$ configurations using the Cahn-Ingold-Prelog $(CIP)$ priority rules:
$1$. The priority order is $-OH > -CO_2H > -CH=CH-CH_3 > -H$.
$2$. In the first structure,the configuration is $R$.
$3$. In the second structure,the configuration is $S$.
Since they are mirror images and have opposite configurations at the chiral center,they are enantiomers.

(A, D) The reaction of $Br_2$ in $CCl_4$ is typically used for the electrophilic addition to alkenes.
$(A)$ $PhCH=CHCH_3 + Br_2 \xrightarrow{CCl_4} PhCHBrCHBrCH_3$. This is a standard addition reaction.
$(B)$ Bromination of benzoic acid requires a Lewis acid catalyst like $FeBr_3$ and is not achieved by $Br_2/CCl_4$ alone.
$(C)$ $CH_3CH_2COOH \rightarrow CH_3CHBrCOOH$ is the Hell-Volhard-Zelinsky $(HVZ)$ reaction,which requires $Br_2$ and red phosphorus $(P)$.
$(D)$ The conversion of silver benzoate to bromobenzene is the Borodine-Hunsdiecker reaction,which uses $Br_2$ in $CCl_4$ under reflux conditions.
Therefore,the conversions that can be carried out by $Br_2$ in $CCl_4$ are $(A)$ and $(D)$.

(B, C) buffer solution is formed by a mixture of a weak acid and its salt with a strong base,or a weak base and its salt with a strong acid.
Option $B$: $NH_4OH$ (weak base) + $HCl$ (strong acid) in $2:1$ mole ratio results in $NH_4Cl$ (salt) and remaining $NH_4OH$. This forms a basic buffer.
Option $C$: $CH_3COOH$ (weak acid) + $NaOH$ (strong base) in $2:1$ mole ratio results in $CH_3COONa$ (salt) and remaining $CH_3COOH$. This forms an acidic buffer.
Both $B$ and $C$ act as buffer solutions.

(B) The self-ionization of interhalogen compounds like $BrF_3$ occurs through the transfer of a fluoride ion $(F^-)$.
The reaction is represented as: $2 BrF_3 \rightleftharpoons BrF_2^+ + BrF_4^-$.
Here,$BrF_3$ acts as both an acid and a base,forming the $BrF_2^+$ cation and the $BrF_4^-$ anion.

(C) In the given reaction: $2 Cr(OH)_3 + 4 OH^{-} + KIO_3 \rightarrow 2 CrO_4^{2-} + 5 H_2O + KI$
The oxidation state of iodine in $KIO_3$ changes from $+5$ to $-1$ in $KI$.
The change in oxidation state of iodine is $|5 - (-1)| = 6$.
The equivalent weight of an oxidizing agent is given by the formula: $\text{Equivalent weight} = \frac{\text{Molecular mass}}{\text{Change in oxidation state}}$.
Therefore,the equivalent weight of $KIO_3 = M / 6$.

(C) In the Solvay process,$NaHCO_3$ is precipitated because it is sparingly soluble in water and can be easily separated by filtration.
However,in the case of potassium,the reaction is: $(NH_4)HCO_3 + KCl \longrightarrow KHCO_3(aq) + NH_4Cl(aq)$.
Since $KHCO_3$ is appreciably soluble in water,it does not precipitate out of the solution.
Therefore,it cannot be isolated from the reaction medium to be converted into $K_2CO_3$ by heating.

(A) Calculate the mass of each substance:
$I$. $1$ mole of $N_2 = 1 \times 28 \ g = 28 \ g$
$II$. $0.5$ mole of $O_3 = 0.5 \times 48 \ g = 24 \ g$
$III$. $3.011 \times 10^{23}$ molecules of $O_2 = 0.5$ mole of $O_2 = 0.5 \times 32 \ g = 16 \ g$
$IV$. $0.5$ gram atom of $O = 0.5 \times 16 \ g = 8 \ g$
Comparing the masses: $8 \ g < 16 \ g < 24 \ g < 28 \ g$.
Thus,the order of increasing mass is $IV < III < II < I$.

(C) The formula for root mean square speed is $C_{rms} = \sqrt{\frac{3RT}{M}}$.
Initially,for $X_2$ gas at temperature $T$,$C_1 = x = \sqrt{\frac{3RT}{M_{X_2}}}$.
When temperature is doubled $(T_2 = 2T)$ and $X_2$ dissociates into $2X$,the molar mass becomes $M_2 = \frac{M_{X_2}}{2}$.
The new rms speed $C_2$ is given by $C_2 = \sqrt{\frac{3R(2T)}{M_{X_2}/2}} = \sqrt{4 \times \frac{3RT}{M_{X_2}}} = 2 \times \sqrt{\frac{3RT}{M_{X_2}}}$.
Substituting $x$ for the initial speed,we get $C_2 = 2x \ m/s$.

(B) The average kinetic energy $(KE_{avg})$ of an ideal gas is given by $\frac{3}{2}kT$. Since the temperature $T$ is halved in Case $-2$,the average kinetic energy is also halved.
The average speed $(C_{avg})$ is given by $\sqrt{\frac{8RT}{\pi M}}$.
For Case $-1$: $C_1 \propto \sqrt{\frac{T}{M}}$.
For Case $-2$: $C_2 \propto \sqrt{\frac{T/2}{2M}} = \sqrt{\frac{T}{4M}} = \frac{1}{2} \sqrt{\frac{T}{M}}$.
Thus,$C_2 = \frac{1}{2} C_1$.
Therefore,both the average kinetic energy and the average speed are halved in the second case.

(A, C, D) Above the critical temperature $(T_c)$,a gas cannot be liquefied by the application of pressure alone.
At this temperature,the distinction between the liquid and gas phases disappears,meaning the liquid phase cannot be distinguished from the gas phase.
Furthermore,the density of the substance changes continuously with changes in pressure $(P)$ or volume $(V)$ as it exists as a supercritical fluid.
Therefore,statements $A$,$C$,and $D$ are applicable.

(B) The de Broglie wavelength is given by $\lambda = \frac{h}{mv}$.
Given: $m_1 = 100 \ g$,$m_2 = 50 \ g$,and $v_2 = 1.5 \ v_1$.
The ratio of wavelengths is $\frac{\lambda_1}{\lambda_2} = \frac{h / (m_1 v_1)}{h / (m_2 v_2)} = \frac{m_2 v_2}{m_1 v_1}$.
Substituting the values: $\frac{\lambda_1}{\lambda_2} = \frac{50 \times 1.5 \ v_1}{100 \times v_1} = \frac{75}{100} = \frac{3}{4}$.
Thus,the ratio is $3 : 4$.

(A) For an electron in the $5d$ orbital,the principal quantum number $n = 5$ and the azimuthal quantum number $l = 2$ (since $s=0, p=1, d=2, f=3$).
The magnetic quantum number $m_l$ can take any integer value from $-l$ to $+l$,which is $-2, -1, 0, 1, 2$.
Both $(5, 2, 1)$ and $(5, 2, -1)$ are valid sets of quantum numbers for a $5d$ electron.
However,in multiple-choice questions of this type,if multiple options are mathematically valid,we select the one provided in the set. Since both $A$ and $D$ are valid,but $D$ is often cited in standard examples,we identify the valid range.

(B) The dissociation constant of water at $STP$ $(298 \ K)$ is $K_w = [H^{+}][OH^{-}] = 10^{-7} \times 10^{-7} = 10^{-14}$.
The relation between standard free energy change and equilibrium constant is $\Delta G^{\circ} = -RT \ln K_w = -2.303 RT \log K_w$.
Given $R = 1.987 \ cal \ K^{-1} mol^{-1}$,$T = 298 \ K$,and $K_w = 10^{-14}$.
$\Delta G^{\circ} = -2.303 \times 1.987 \times 298 \times \log(10^{-14})$.
$\Delta G^{\circ} = -2.303 \times 1.987 \times 298 \times (-14)$.
$\Delta G^{\circ} = 2.303 \times 1.987 \times 298 \times 14 \approx 19091 \ cal / mol$.

(C) The reaction sequence involves the alkaline hydrolysis (saponification) of an ester $G$ $(C_6H_{12}O_2)$ to form an alcohol $(H)$ and a carboxylate salt,which upon acidification gives a carboxylic acid $(I)$.
$H$ is then oxidized by $CrO_3/H^+$ to form the same carboxylic acid $(I)$.
This indicates that the alcohol $H$ is a primary alcohol,which oxidizes to a carboxylic acid with the same number of carbon atoms.
Looking at the options,we need an ester that hydrolyzes into a primary alcohol and a carboxylic acid such that the alcohol can be oxidized to the acid.
Option $C$ is $CH_3CH_2COOCH_2CH_2CH_3$ (propyl propionate).
Hydrolysis: $CH_3CH_2COOCH_2CH_2CH_3 + OH^- \rightarrow CH_3CH_2COO^- + CH_3CH_2CH_2OH$.
Acidification: $CH_3CH_2COO^- + H^+ \rightarrow CH_3CH_2COOH$ (propanoic acid).
Oxidation of $H$ $(CH_3CH_2CH_2OH)$: $CH_3CH_2CH_2OH \xrightarrow{CrO_3/H^+} CH_3CH_2COOH$.
Since the acid formed from hydrolysis and the acid formed from oxidation of the alcohol must be the same for the cycle to work as shown,the ester must be $CH_3CH_2COOCH_2CH_2CH_3$.

(A) The final product is $Ph_2C=CH_2$ ($1$,$1$-diphenylethene).
The last step is a Wittig reaction: $N + Ph_3P=CH_2 \rightarrow Ph_2C=CH_2$. This implies $N$ must be benzophenone $(Ph_2C=O)$.
The step $M \xrightarrow{CrO_3 / H^{\oplus}} N$ is an oxidation of a secondary alcohol to a ketone. Thus,$M$ is diphenylmethanol $(Ph_2CHOH)$.
The first step is the Grignard reaction: $L \xrightarrow[(ii) H_3O^{\oplus}]{(i) PhMgBr} Ph_2CHOH$.
Benzaldehyde $(PhCHO)$ reacts with $PhMgBr$ followed by hydrolysis to give diphenylmethanol $(Ph_2CHOH)$.
Therefore,$L$ is benzaldehyde.

(C) The boiling point depends on the strength of intermolecular forces.
$1$. Ethoxyethane $(II)$ is an ether,which exhibits only weak dipole-dipole interactions.
$2$. $N$-ethylethanamine $(I)$ is a secondary amine,which exhibits hydrogen bonding,but it is weaker than the hydrogen bonding in alcohols due to the lower electronegativity of $N$ compared to $O$.
$3$. Butan$-2-$ol $(III)$ is an alcohol,which exhibits strong intermolecular hydrogen bonding.
Therefore,the order of boiling points is $II < I < III$.

(C) Let the rate law be $Rate = k[A]^x[B]^y$.
From experiments $2$ and $3$ (where $[B]$ is constant):
$\frac{10}{1} = \frac{k(10)^x(1)^y}{k(1)^x(1)^y}$ $\Rightarrow 10 = 10^x$ $\Rightarrow x = 1$.
From experiments $1$ and $2$ (where $[A]$ is constant):
$\frac{100}{1} = \frac{k(1)^x(10)^y}{k(1)^x(1)^y}$ $\Rightarrow 100 = 10^y$ $\Rightarrow 10^2 = 10^y$ $\Rightarrow y = 2$.
Thus,the order with respect to $A$ is $1$ and with respect to $B$ is $2$.

(B) The relationship between half-life $(T_{1/2})$ and decay constant $(\lambda)$ is given by: $T_{1/2} = \frac{0.693}{\lambda}$.
Given that the magnitudes are identical,let $T_{1/2} = \lambda = x$.
Substituting this into the equation: $x = \frac{0.693}{x}$.
This simplifies to: $x^2 = 0.693$.
Therefore,$x = \sqrt{0.693} = (0.693)^{1/2}$.

(D) The Arrhenius equation is given by:
$k = A e^{-E_a / RT}$
Taking the natural logarithm on both sides:
$\ln k = \ln A - \frac{E_a}{R} \left( \frac{1}{T} \right)$
This equation is in the form of a straight line equation $y = mx + c$,where:
$y = \ln k$
$x = \frac{1}{T}$
$m = -\frac{E_a}{R}$ (slope)
$c = \ln A$ (intercept)
Therefore,a plot of $\ln k$ versus $1 / T$ gives a linear plot with a slope of $-E_a / R$.

(A) In both complexes,the central metal ion is $Fe^{3+}$. The electronic configuration of $Fe^{3+}$ is $[Ar] 3d^5$.
For $[Fe(CN)_6]^{3-}$: $CN^-$ is a strong field ligand,which causes pairing of electrons in the $3d$ orbitals. This results in one unpaired electron $(t_{2g}^5 e_g^0)$,making it paramagnetic.
For $[FeF_6]^{3-}$: $F^-$ is a weak field ligand,which does not cause pairing of electrons. This results in five unpaired electrons $(t_{2g}^3 e_g^2)$,making it paramagnetic.
Therefore,both complexes are paramagnetic.

(D) Let us analyze each complex:
$1$. In $[VF_6]^{3-}$,$V$ is in $+3$ oxidation state. $V^{3+}$ is $3d^2$. It has $2$ unpaired electrons,so it is paramagnetic. This statement is correct.
$2$. In $[CuCl_4]^{2-}$,$Cu$ is in $+2$ oxidation state. $Cu^{2+}$ is $3d^9$. It has $1$ unpaired electron,so it is paramagnetic. This statement is correct.
$3$. In $[Co(NH_3)_6]^{3+}$,$Co$ is in $+3$ oxidation state. $Co^{3+}$ is $3d^6$. $NH_3$ is a strong field ligand,causing pairing of electrons. All $6$ electrons are paired,so it is diamagnetic. This statement is correct.
$4$. In $[CoF_6]^{3-}$,$Co$ is in $+3$ oxidation state. $Co^{3+}$ is $3d^6$. $F^-$ is a weak field ligand,so no pairing occurs. The $3d$ orbitals have $4$ unpaired electrons. Thus,it is paramagnetic with $4$ unpaired electrons. The statement claiming $2$ unpaired electrons is incorrect.

(A) The central metal ion is $Ni^{2+}$,which has an electronic configuration of $[Ar] 3d^8$.
Since the complex $[NiX_4]^{2-}$ is paramagnetic and has a coordination number of $4$,it adopts a tetrahedral geometry.
In a tetrahedral geometry,the hybridisation is $sp^3$.
For $Ni^{2+}$ $(3d^8)$,the $3d$ orbitals have two unpaired electrons.
Therefore,the hybridisation is $sp^3$ and the number of unpaired electrons is $2$.

(B) The atomic number of $Pr$ (Praseodymium) is $59$.
The electronic configuration of $Pr$ is $[Xe] 4f^3 6s^2$.
When $Pr$ forms a $Pr^{3+}$ ion,it loses three electrons (two from $6s$ and one from $4f$).
Therefore,the electronic configuration of $Pr^{3+}$ is $[Xe] 4f^2$.

(B) According to Kohlrausch's law of independent migration of ions,the equivalent conductance at infinite dilution for acetic acid is given by:
$\wedge^{0}_{CH_{3}COOH} = \wedge^{0}_{CH_{3}COONa} + \wedge^{0}_{HCl} - \wedge^{0}_{NaCl}$
Substituting the given values:
$\wedge^{0}_{CH_{3}COOH} = 91 + 426.16 - 126.45$
$\wedge^{0}_{CH_{3}COOH} = 390.71 \ \Omega^{-1} \ cm^{2} \ eq^{-1}$

(C) The acidity of phenols depends on the stability of the phenoxide ion formed after the loss of a proton. Electron-withdrawing groups $(EWG)$ stabilize the phenoxide ion and increase acidity,while electron-donating groups $(EDG)$ destabilize it and decrease acidity.
$1$. Compound $(IV)$ has a $-NO_2$ group,which is a strong electron-withdrawing group ($-R$ and $-I$ effect),making it the most acidic.
$2$. Compound $(III)$ has a $-CH=CH_2$ group,which is an electron-withdrawing group ($-R$ effect),making it more acidic than $(I)$ and $(II)$.
$3$. Compound $(I)$ has a $-CH_3$ group,which is an electron-donating group ($+I$ and hyperconjugation),making it less acidic than $(III)$ but more acidic than $(II)$.
$4$. Compound $(II)$ has a $-NMe_2$ group,which is a very strong electron-donating group ($+R$ effect),making it the least acidic.
Therefore,the correct order of acidity is $(IV) > (III) > (I) > (II)$.

(A) In the first step,acetone $(CH_3COCH_3)$ reacts with $Br_2$ in an acidic medium $(AcOH)$. Acidic halogenation of ketones typically results in monohalogenation at the $\alpha$-carbon,yielding $X = CH_3COCH_2Br$.
In the second step,the monobromoacetone $(CH_3COCH_2Br)$ reacts with $Br_2$ in a basic medium $(NaOH)$. This is a haloform reaction. The base promotes further halogenation of the $\alpha$-carbon followed by cleavage to form bromoform $(CHBr_3)$ and the corresponding carboxylate salt,which is sodium acetate $(CH_3CO_2Na)$.

(D) The reaction sequence involves two $S_N2$ steps,each resulting in an inversion of configuration at the chiral center.
$1$. The starting material is $sec$-butyl chloride. Reaction with aqueous $NaOH$ proceeds via $S_N2$ mechanism,leading to the formation of $sec$-butyl alcohol with inversion of configuration.
$2$. The alcohol then reacts with $p$-toluenesulfonyl chloride $(TsCl)$ in pyridine to form a tosylate intermediate,which is a good leaving group.
$3$. Finally,the tosylate reacts with $NaI$ in acetone via another $S_N2$ reaction,resulting in a second inversion of configuration.
$4$. Since there are two inversions,the final product $C_4H_9I$ will have the same configuration as the starting material $C_4H_9Cl$.
$5$. Comparing the structures,option $D$ correctly represents the intermediate alcohol (inverted) and the final iodide (retained configuration relative to the starting material).

(A, C, D) To synthesize benzoic acid from benzene:
Option $(A)$: Benzene reacts with $Br_2/Fe$ to form bromobenzene,which forms a Grignard reagent $(PhMgBr)$ with $Mg/\text{dry ether}$. This reacts with $CO_2$ followed by acid hydrolysis $(H_3O^{\oplus})$ to yield benzoic acid.
Option $(C)$: Friedel-Crafts alkylation of benzene with $CH_3Cl/AlCl_3$ gives toluene. Oxidation of toluene with alkaline $KMnO_4$ followed by acidification gives benzoic acid.
Option $(D)$: Friedel-Crafts acylation of benzene with $CH_3COCl/AlCl_3$ gives acetophenone. The haloform reaction with $Br_2/NaOH$ followed by acidification gives benzoic acid.
Option $(B)$ is incorrect as the reaction sequence does not lead to benzoic acid.
Therefore,the correct sets are $(A), (C),$ and $(D)$.

(D) For a body-centred cubic $(BCC)$ unit cell, the relationship between the edge length $(a)$ and the atomic radius $(r)$ is given by the formula: $4r = \sqrt{3}a$.
Given that the atomic radius $r = 75 \ pm$.
Substituting the value of $r$ into the formula: $a = \frac{4r}{\sqrt{3}}$.
$a = \frac{4 \times 75}{\sqrt{3}} = \frac{300}{\sqrt{3}}$.
Rationalizing the denominator: $a = \frac{300 \times \sqrt{3}}{3} = 100 \times \sqrt{3}$.
Since $\sqrt{3} \approx 1.732$, we get $a = 100 \times 1.732 = 173.2 \ pm$.

(B) The osmotic pressure formula is $\pi = CRT$,where $C$ is the molarity,$R$ is the gas constant $(0.0821 \ L \ atm \ K^{-1} \ mol^{-1})$,and $T$ is the temperature in Kelvin.
Given: $\pi = 3.0 \times 10^{-4} \ atm$,$T = 27 + 273 = 300 \ K$,$V = 4.0 \ L$,and mass of solute $w = 4 \ g$.
$C = \frac{\pi}{RT} = \frac{3.0 \times 10^{-4}}{0.0821 \times 300} = 1.218 \times 10^{-5} \ mol \ L^{-1}$.
Since $C = \frac{w}{M \times V}$,where $M$ is the molar mass:
$1.218 \times 10^{-5} = \frac{4}{M \times 4.0}$.
$M = \frac{4}{1.218 \times 10^{-5} \times 4.0} = \frac{1}{1.218 \times 10^{-5}} \approx 82101 \ g \ mol^{-1}$.
Rounding to the nearest given option,the molar mass is $82000 \ g \ mol^{-1}$.

(B) Step $1$: Calculate the number of moles of the solute.
Moles $= \frac{\text{Mass}}{\text{Molar Mass}} = \frac{63 \ g}{126 \ g/mol} = 0.5 \ mol$.
Step $2$: Calculate the total mass of the solution.
Mass of solution $= \text{Mass of solute} + \text{Mass of solvent} = 63 \ g + 500 \ g = 563 \ g$.
Step $3$: Calculate the volume of the solution using density.
Volume $= \frac{\text{Mass}}{\text{Density}} = \frac{563 \ g}{1.126 \ g/mL} = 500 \ mL = 0.5 \ L$.
Step $4$: Calculate the molarity.
Molarity $(M) = \frac{\text{Moles of solute}}{\text{Volume of solution in } L} = \frac{0.5 \ mol}{0.5 \ L} = 1.0 \ M$.

(A) For both complexes,the central metal ion is $Fe^{3+}$.
Electronic configuration of $Fe^{3+}$ is $[Ar] 3d^5$.
In $[FeCl_4]^-$,$Cl^-$ is a weak field ligand,so no pairing of electrons occurs. The number of unpaired electrons $(n)$ is $5$.
The spin-only magnetic moment is $\mu = \sqrt{n(n+2)} = \sqrt{5(5+2)} = \sqrt{35} \approx 5.9 \ BM$.
In $[Fe(CN)_6]^{3-}$,$CN^-$ is a strong field ligand,causing pairing of electrons. The configuration becomes $t_{2g}^5 e_g^0$,leaving $n = 1$ unpaired electron.
The spin-only magnetic moment is $\mu = \sqrt{1(1+2)} = \sqrt{3} \approx 1.732 \ BM$.
Thus,the values are $5.9 \ BM$ and $1.732 \ BM$.