NEET 2021 Chemistry Question Paper with Answer and Solution

(B) According to Newton's Law of Cooling,the rate of cooling is given by $\frac{dT}{dt} = K(T_{avg} - T_0)$,where $T_{avg}$ is the average temperature of the body and $T_0$ is the surrounding temperature.
For the first case,cooling from $90^{\circ} C$ to $80^{\circ} C$ in time $t$:
$\frac{90 - 80}{t} = K \left( \frac{90 + 80}{2} - 20 \right)$
$\frac{10}{t} = K(85 - 20) = 65K$ --- $(1)$
For the second case,cooling from $80^{\circ} C$ to $60^{\circ} C$ in time $t'$:
$\frac{80 - 60}{t'} = K \left( \frac{80 + 60}{2} - 20 \right)$
$\frac{20}{t'} = K(70 - 20) = 50K$ --- $(2)$
Dividing equation $(1)$ by $(2)$:
$\frac{10/t}{20/t'} = \frac{65K}{50K}$
$\frac{10}{t} \times \frac{t'}{20} = \frac{13}{10}$
$\frac{t'}{2t} = \frac{13}{10}$
$t' = \frac{13}{5} t$

(A) Let the particle be at height $h$ from the surface of the Earth.
By the law of conservation of mechanical energy, the total energy at height $S$ is equal to the total energy at height $h$.
Total energy $E = mgS$.
At height $h$, the potential energy is $PE = mgh$ and kinetic energy is $KE = mg(S-h)$.
Given that $KE = 3 PE$, we have $mg(S-h) = 3 mgh$.
Dividing by $mg$, we get $S-h = 3h$, which implies $S = 4h$ or $h = S/4$.
Now, to find the speed $V$ at this height, we use $KE = \frac{1}{2} mV^2$.
Since $KE = 3 PE = 3 mgh = 3 mg(S/4) = \frac{3 mgS}{4}$.
Equating the two expressions for $KE$: $\frac{1}{2} mV^2 = \frac{3 mgS}{4}$.
Solving for $V$, we get $V^2 = \frac{3 gS}{2}$, so $V = \sqrt{\frac{3 gS}{2}}$.
Thus, the height is $S/4$ and the speed is $\sqrt{\frac{3 gS}{2}}$.

(D) According to Fajan's rule,covalent character is directly proportional to the polarising power of the cation.
Polarising power is inversely proportional to the size of the cation $(Polarising \ power \propto \frac{1}{\text{size of cation}})$.
Among the given alkaline earth metal ions,the size order is $Be^{2+} < Mg^{2+} < Ca^{2+} < Sr^{2+}$.
Since $Be^{2+}$ has the smallest size,it has the highest polarising power,making $BeCl_2$ the most covalent compound.
Due to its covalent nature,$BeCl_2$ is soluble in organic solvents.

(A) The acidic strength of hydrohalic acids $(HX)$ depends on the bond dissociation enthalpy of the $H-X$ bond.
As we move down the group from $F$ to $I$,the atomic size increases,which leads to an increase in the bond length and a decrease in the bond dissociation enthalpy.
Consequently,the $H-X$ bond becomes weaker,making it easier to release $H^+$ ions.
Therefore,the acidic strength increases in the order $HF < HCl < HBr < HI$.
Both Statement $I$ and Statement $II$ are true.

(A) In the presence of peroxides like benzoyl peroxide $((C_6H_5CO)_2O_2)$,the addition of $HBr$ to unsymmetrical alkenes follows the Anti-Markovnikov rule (also known as the peroxide effect or Kharasch effect).
In this mechanism,the bromine radical attacks the less substituted carbon atom of the double bond to form a more stable radical intermediate.
Reaction: $CH_3-CH(CH_3)-CH=CH_2 + HBr \xrightarrow{\text{Peroxide}} CH_3-CH(CH_3)-CH_2-CH_2-Br$ ($1$-bromo$-3-$methylbutane).

(A) In the solid state,$BeCl_2$ exists as a polymeric chain structure where each $Be$ atom is $sp^3$ hybridized and bonded to four $Cl$ atoms.
In the vapour phase,at high temperatures $(1200 \ K)$,$BeCl_2$ exists as a monomeric linear molecule,while at lower temperatures $(1000 \ K)$,it exists as a dimeric structure $(Be_2Cl_4)$.

(C) $1$. Calculate the percentage of hydrogen: $100\% - 78\% = 22\%$.
$2$. Determine the mole ratio of elements:
For $C$: $\frac{78}{12} = 6.5$.
For $H$: $\frac{22}{1} = 22$.
$3$. Divide by the smallest value $(6.5)$:
For $C$: $\frac{6.5}{6.5} = 1$.
For $H$: $\frac{22}{6.5} \approx 3.38 \approx 3$.
$4$. The empirical formula is $CH_{3}$.

(C) To determine the structure of $2,6-$Dimethyldec$-4-$ene,follow these steps:
$1$. The parent chain is $decene$,which contains $10$ carbon atoms.
$2$. The double bond is located at the $4^{th}$ position (between $C4$ and $C5$).
$3$. There are methyl groups at the $2^{nd}$ and $6^{th}$ positions.
$4$. Analyzing the options,we look for a $10$-carbon chain where the double bond starts at $C4$ and methyl substituents are at $C2$ and $C6$.
$5$. Structure $C$ represents a $10$-carbon chain with a double bond at the $4^{th}$ position and methyl groups at the $2^{nd}$ and $6^{th}$ positions,which matches the $IUPAC$ name $2,6-$Dimethyldec$-4-$ene.

(B) metal displacement reaction involves a metal in a compound being displaced by another metal in the uncombined state.
In the reaction $Cr_{2}O_{3} + 2 Al \stackrel{\Delta}{\longrightarrow} Al_{2}O_{3} + 2 Cr$,the metal $Al$ displaces the metal $Cr$ from its oxide $Cr_{2}O_{3}$.
Therefore,this is a metal displacement reaction.

(D) Metamerism is shown by compounds having a polyvalent functional group (like ether,thioether,or amine) where the alkyl groups attached to the functional group differ.
$C_4H_{10}O$ can represent diethyl ether $(CH_3CH_2-O-CH_2CH_3)$ and methyl propyl ether $(CH_3-O-CH_2CH_2CH_3)$.
Since the alkyl groups attached to the oxygen atom are different in these two isomers,they are metamers.
Therefore,the correct option is $D$.

(A) The least stable conformer of ethane is the eclipsed form.
In the eclipsed conformation,the hydrogen atoms on the front carbon are directly aligned with the hydrogen atoms on the back carbon,resulting in maximum torsional strain.
This corresponds to a dihedral angle of $0^{\circ}$.

(D) According to the ideal gas equation,$PV = nRT$.
For a fixed amount of gas,$P = \frac{nRT}{V}$.
At a constant volume $V$,$P \propto T$.
This means that for a given volume,the pressure is higher at a higher temperature.
Therefore,if $T_1 > T_2 > T_3$,then the corresponding pressures at a constant volume will be $P_1 > P_2 > P_3$.
Looking at the graphs,the curve for the highest temperature $(600 \ K)$ must be the uppermost curve,followed by $400 \ K$ and then $200 \ K$ at the bottom.
This matches the representation where the curve for $600 \ K$ is above the curve for $400 \ K$,which is above the curve for $200 \ K$.

(C) Dimethylammonium acetate is a salt of a weak acid (acetic acid) and a weak base (dimethylamine).
The formula for the $pH$ of a salt of a weak acid and a weak base is given by:
$pH = 7 + \frac{1}{2} pK_{a} - \frac{1}{2} pK_{b}$
Given:
$pK_{a} = 4.77$
$pK_{b} = 3.27$
Substituting the values into the formula:
$pH = 7 + \frac{1}{2} (4.77) - \frac{1}{2} (3.27)$
$pH = 7 + \frac{1}{2} (4.77 - 3.27)$
$pH = 7 + \frac{1}{2} (1.50)$
$pH = 7 + 0.75$
$pH = 7.75$

(A) To determine the geometry,we calculate the Steric Number $(SN)$ using the formula: $SN = \text{Number of } \sigma \text{ bonds} + \text{Number of lone pairs}$.
$1$. $PCl_{5}$: $SN = 5 + 0 = 5$. Geometry: Trigonal bipyramidal $(iv)$.
$2$. $SF_{6}$: $SN = 6 + 0 = 6$. Geometry: Octahedral $(iii)$.
$3$. $BrF_{5}$: $SN = 5 + 1 = 6$. Geometry: Square pyramidal $(i)$.
$4$. $BF_{3}$: $SN = 3 + 0 = 3$. Geometry: Trigonal planar $(ii)$.
Thus,the correct matching is $(a)-(iv), (b)-(iii), (c)-(i), (d)-(ii)$.

(C) In $BF_3$,the central atom is Boron $(B)$.
Boron has $3$ valence electrons and forms $3$ $\sigma$-bonds with $3$ Fluorine atoms.
Number of lone pairs $(lp)$ on central atom = $0$.
Steric number = (Number of $\sigma$-bonds) + (Number of lone pairs) = $3 + 0 = 3$.
$A$ steric number of $3$ corresponds to $sp^2$ hybridization.
The total number of electrons around the central Boron atom is $3 \times 2 = 6$ electrons (due to $3$ covalent bonds).
Thus,the hybridization is $sp^2$ and the number of electrons is $6$.

(A) The relationship between frequency $(\nu)$,wavelength $(\lambda)$,and speed of light $(c)$ is given by the formula: $\nu = \frac{c}{\lambda}$.
Rearranging the formula to solve for wavelength: $\lambda = \frac{c}{\nu}$.
Given: $c = 3.0 \times 10^{8} \, m \, s^{-1}$ and $\nu = 1,368 \, kHz = 1,368 \times 10^{3} \, Hz$.
Substituting the values: $\lambda = \frac{3.0 \times 10^{8} \, m \, s^{-1}}{1,368 \times 10^{3} \, s^{-1}}$.
$\lambda = \frac{300,000,000}{1,368,000} \, m \approx 219.29 \, m$.
Rounding to one decimal place,we get $\lambda \approx 219.3 \, m$.

(B) For one mole of an ideal gas,the relationship between molar heat capacity at constant pressure $(C_{P})$ and molar heat capacity at constant volume $(C_{V})$ is given by Mayer's relation:
$C_{P} - C_{V} = R$
where $R$ is the universal gas constant.

(C) For an ideal gas,the internal energy $ \Delta U $ depends only on temperature. Since the process is isothermal $( \Delta T = 0 )$,$ \Delta U = 0 $.
For any irreversible process,the total entropy change of the universe is always positive,i.e.,$ \Delta S_{total} > 0 $.
Therefore,$ \Delta S_{total} \neq 0 $.
Thus,the correct condition is $ \Delta U = 0 $ and $ \Delta S_{total} \neq 0 $.

(C) The total pressure is calculated using the ideal gas equation: $PV = n_{total}RT$.
First,calculate the number of moles of each gas:
$n_{O_2} = \frac{4 \ g}{32 \ g/mol} = 0.125 \ mol$
$n_{H_2} = \frac{2 \ g}{2 \ g/mol} = 1.0 \ mol$
Total moles $n_{total} = 0.125 + 1.0 = 1.125 \ mol$.
Using $PV = nRT$ with $V = 1 \ L$,$R = 0.082 \ L \ atm \ mol^{-1} \ K^{-1}$,and $T = 273 \ K$:
$P \times 1 = 1.125 \times 0.082 \times 273$
$P = 25.18 \ atm$.

(C) molecule is non-polar if its net dipole moment is zero.
$SbCl_5$ has a trigonal bipyramidal geometry where the three equatorial $Sb-Cl$ bonds cancel each other out,and the two axial $Sb-Cl$ bonds also cancel each other out,resulting in a net dipole moment of zero.
$POCl_3$,$CH_2O$,and $NO_2$ are polar molecules due to their asymmetric structures and the presence of different atoms or lone pairs leading to a non-zero net dipole moment.

(D) Iso-electronic species are those that have the same number of electrons.
$O^{2-}$ has $8 + 2 = 10$ electrons; $F^{-}$ has $9 + 1 = 10$ electrons. (Iso-electronic)
$Na^{+}$ has $11 - 1 = 10$ electrons; $Mg^{2+}$ has $12 - 2 = 10$ electrons. (Iso-electronic)
$Mn^{2+}$ has $25 - 2 = 23$ electrons; $Fe^{3+}$ has $26 - 3 = 23$ electrons. (Iso-electronic)
$Fe^{2+}$ has $26 - 2 = 24$ electrons; $Mn^{2+}$ has $25 - 2 = 23$ electrons. (Not iso-electronic)
Therefore,the pair $Fe^{2+}, Mn^{2+}$ is not iso-electronic.

(C) The correct matching is as follows:
$(a)$ $2 SO_{2(g)} + O_{2(g)} \rightarrow 2 SO_{3(g)}$ represents the formation of $SO_3$ which leads to acid rain,so $(a)-(iv)$ (as it is a part of tropospheric pollution).
$(b)$ $HOCl_{(g)} \stackrel{h \nu}{\longrightarrow} \dot{O}H + \dot{Cl}$ is a reaction involved in the depletion of the ozone layer,so $(b)-(iii)$.
$(c)$ $CaCO_{3} + H_{2}SO_{4} \rightarrow CaSO_{4} + H_{2}O + CO_{2}$ is the reaction of acid rain with marble/limestone,so $(c)-(i)$.
$(d)$ $NO_{2(g)} \stackrel{h \nu}{\longrightarrow} NO_{(g)} + O_{(g)}$ is a key step in the formation of photochemical smog,so $(d)-(ii)$.
Thus,the correct sequence is $(a)-(iv), (b)-(iii), (c)-(i), (d)-(ii)$.

(C) The given reaction is the decarboxylation of sodium propionate to form ethane.
This reaction occurs in the presence of sodalime,which is a mixture of $NaOH$ and $CaO$.
Therefore,the missing reagent is $CaO$.

(B) For option $A$: The acidic strength of hydrohalic acids increases down the group due to the decrease in bond dissociation enthalpy. Thus,$HF < HCl < HBr < HI$ is correct.
For option $B$: The acidic strength of hydrides of group $16$ increases down the group $(H_{2}O < H_{2}S < H_{2}Se < H_{2}Te)$. Since $pK_{a} = -\log(K_{a})$,higher acidic strength implies a lower $K_{a}$ value,which means a lower $pK_{a}$ value. Therefore,the correct order of $pK_{a}$ values should be $H_{2}O > H_{2}S > H_{2}Se > H_{2}Te$. Thus,the given order is incorrect.
For option $C$: The acidic character of hydrides of group $15$ increases down the group due to the increase in bond length and decrease in bond dissociation energy. Thus,$NH_{3} < PH_{3} < AsH_{3} < SbH_{3}$ is correct.
For option $D$: The oxidising power of group $14$ oxides increases down the group due to the inert pair effect,making the $+2$ oxidation state more stable than the $+4$ state for heavier elements. Thus,$CO_{2} < SiO_{2} < SnO_{2} < PbO_{2}$ is correct.

(B) The direction of propagation of an electromagnetic wave is given by the direction of the Poynting vector,which is $\vec{E} \times \vec{B}$.
Given the wave propagates in the $x$-direction,we must have $\vec{E} \times \vec{B} \parallel \hat{i}$.
For option $B$: $\vec{E} = -\hat{j} + \hat{k}$ and $\vec{B} = -\hat{j} - \hat{k}$.
Calculating the cross product: $\vec{E} \times \vec{B} = (-\hat{j} + \hat{k}) \times (-\hat{j} - \hat{k}) = (-\hat{j} \times -\hat{j}) + (-\hat{j} \times -\hat{k}) + (\hat{k} \times -\hat{j}) + (\hat{k} \times -\hat{k}) = 0 + \hat{i} + \hat{i} + 0 = 2\hat{i}$.
Since $2\hat{i}$ is in the $x$-direction,option $B$ is correct.

(B) The bond enthalpy of the $C-X$ bond depends on the bond length.
As we move down the group in the periodic table,the atomic size of the halogen $(X)$ increases.
This leads to an increase in the $C-X$ bond length.
As the bond length increases,the bond strength (and thus bond enthalpy) decreases.
Therefore,the correct order of bond enthalpy is $CH_{3}-F > CH_{3}-Cl > CH_{3}-Br > CH_{3}-I$.

(C) Metals that are liquid at room temperature,such as $Hg$ (mercury),are purified by the process of distillation.
In distillation,the impure metal is heated to its boiling point,and the vapors are condensed to obtain the pure metal.
Since the boiling point of the metal is significantly different from the boiling points of the impurities,this method is highly effective for volatile metals.

(D) In the $14$ Bravais lattices,the body-centred $(BCC)$ unit cells are found in the following crystal systems:
$1$. Cubic system: $1$ $(BCC)$
$2$. Tetragonal system: $1$ $(BCC)$
$3$. Orthorhombic system: $1$ $(BCC)$
Therefore,the total number of body-centred unit cells is $1 + 1 + 1 = 3$.

(C) The atomic and ionic radii of $Zr$ $(Z=40)$ and $Hf$ $(Z=72)$ are very similar.
This phenomenon is known as the lanthanoid contraction.
As we move from $4d$ series to $5d$ series elements in the same group,the filling of $4f$ orbitals occurs.
The poor shielding effect of $4f$ electrons leads to a decrease in atomic size,which almost perfectly compensates for the expected increase in size due to the addition of a new shell.

(B) In a blast furnace,the temperature varies at different zones.
The combustion zone,where coke burns to produce heat,reaches the maximum temperature of approximately $2200 \ K$.

(D) The reaction of acetone $(CH_3COCH_3)$ with ethylmagnesium bromide $(C_2H_5MgBr)$ in the presence of dry ether is a nucleophilic addition reaction.
The nucleophilic ethyl group $(C_2H_5^-)$ attacks the electrophilic carbonyl carbon of acetone to form an intermediate magnesium alkoxide.
Subsequent hydrolysis with $H_3O^+$ yields the final alcohol product.
The reaction sequence is:
$CH_3COCH_3 + C_2H_5MgBr \rightarrow CH_3-C(OMgBr)(CH_3)-C_2H_5$
$CH_3-C(OMgBr)(CH_3)-C_2H_5 + H_2O/H^+ \rightarrow CH_3-C(OH)(CH_3)-CH_2CH_3 + Mg(OH)Br$
The structure of the product is $CH_3-C(OH)(CH_3)-CH_2-CH_3$.
Selecting the longest carbon chain containing the $-OH$ group gives a butane chain. Numbering from the end closest to the $-OH$ group,the $-OH$ is at position $2$ and a methyl group is also at position $2$.
Thus,the $IUPAC$ name is $2-$methylbutan$-2-$ol.

(A) Teflon is an addition polymer formed by the addition polymerisation of tetrafluoroethene $(CF_2=CF_2)$.
The reaction is as follows:
$n \ CF_2=CF_2 \xrightarrow{\text{Polymerisation}} [-CF_2-CF_2-]_n$
Nylon-$6,6$,Novolac,and Dacron are condensation polymers.

(D) In a hexagonal close-packed $(HCP)$ unit cell,the number of atoms per unit cell $(Z)$ is $6$.
The number of octahedral voids $(OHV)$ is equal to the number of atoms per unit cell,so $OHV = Z = 6$.
The number of tetrahedral voids $(THV)$ is equal to twice the number of atoms per unit cell,so $THV = 2 \times Z = 2 \times 6 = 12$.
Therefore,the number of tetrahedral and octahedral voids are $12$ and $6$ respectively.

(B) The correct answer is $B$.
Most trivalent Lanthanoid ions $(Ln^{3+})$ are colored in both solid and aqueous states due to $f-f$ transitions.
Actinoid contraction is indeed greater than Lanthanoid contraction because the $5f$ electrons provide poorer shielding than $4f$ electrons.
Lanthanoids are typical metallic elements and are good conductors of heat and electricity.
Actinoids are highly reactive metals,especially when finely divided,as they form a protective oxide layer slowly.

(B) Statement $I$ is incorrect because Aspirin and Paracetamol are non-narcotic analgesics.
Statement $II$ is incorrect because Morphine and Heroin are narcotic analgesics.
Therefore,both statements are false.

(A) The correct answer is $(A)$.
According to $Saytzeff's \ Rule$,in dehydrohalogenation reactions,the major product is the more highly substituted alkene,which is the one with more alkyl groups attached to the double-bonded carbons.
For $2-Bromopentane$,the elimination of $HBr$ can yield either $Pent-1-ene$ or $Pent-2-ene$.
$Pent-2-ene$ is a disubstituted alkene,whereas $Pent-1-ene$ is a monosubstituted alkene.
Therefore,$Pent-2-ene$ is the more stable and major product.

(B) The enthalpy of reaction $\Delta H$ is given by the difference between the activation energy of the forward reaction $(E_a)_f$ and the activation energy of the backward reaction $(E_a)_b$:
$\Delta H = (E_a)_f - (E_a)_b$
Given $\Delta H = -4.2 \ kJ \ mol^{-1}$ and $(E_a)_f = 9.6 \ kJ \ mol^{-1}$:
$-4.2 = 9.6 - (E_a)_b$
$(E_a)_b = 9.6 + 4.2 = 13.8 \ kJ \ mol^{-1}$
Since $\Delta H$ is negative,the reaction is exothermic,meaning the potential energy of the product $B$ must be lower than the potential energy of the reactant $A$. Among the given options,the graph where the product $B$ is at a lower energy level than reactant $A$ is represented by option $B$.

(A) The structure of the ethylene diaminetetraacetate $(EDTA^{4-})$ ion consists of two nitrogen atoms and four oxygen atoms (from the four carboxylate groups) that can act as donor sites.
Since it has six donor atoms capable of binding to a central metal ion,it is classified as a hexadentate ligand.
Therefore,the correct option is $A$.

(B) Noble gases have stable electronic configurations ($ns^2 np^6$,except $He$ which is $1s^2$).
Due to their stable configuration,they have very weak dispersion forces between their atoms,which leads to very low melting and boiling points.
Therefore,the statement that they have very high melting and boiling points is incorrect.

(A) The deficiency of Vitamin $B_{12}$ (cyanocobalamin) leads to pernicious anemia,which is characterized by a decrease in the number of red blood cells ($RBC$s) and the production of abnormally large,immature $RBC$s.

(A) Tritium $(^{3}_{1}H)$ is a radioactive isotope of hydrogen. It undergoes radioactive decay by emitting a beta particle $(\beta^{-})$ to form Helium-$3$ $(^{3}_{2}He)$. The nuclear reaction is: $^{3}_{1}H \rightarrow ^{3}_{2}He + ^{0}_{-1}e + \bar{\nu}_{e}$.

(B) According to Kohlrausch's law of independent migration of ions,the molar conductivity at infinite dilution for $CH_{3}COOH$ is given by:
$\Lambda^{\circ}_{m}(CH_{3}COOH) = \Lambda^{\circ}_{m}(CH_{3}COONa) + \Lambda^{\circ}_{m}(HCl) - \Lambda^{\circ}_{m}(NaCl)$
Substituting the given values:
$\Lambda^{\circ}_{m}(CH_{3}COOH) = 91.0 + 426.16 - 126.45$
$\Lambda^{\circ}_{m}(CH_{3}COOH) = 517.16 - 126.45$
$\Lambda^{\circ}_{m}(CH_{3}COOH) = 390.71 \ S \ cm^{2} \ mol^{-1}$

(C) Hinsberg's reagent is benzenesulfonyl chloride $(C_6H_5SO_2Cl)$.
$1^{\circ}$ amines $(R-NH_2)$ react with Hinsberg's reagent to form $N$-alkylbenzenesulfonamide,which contains an acidic hydrogen atom attached to the nitrogen. This makes the product soluble in alkali (e.g.,$KOH$ or $NaOH$).
$2^{\circ}$ amines form $N,N$-dialkylbenzenesulfonamides,which do not have an acidic hydrogen and are insoluble in alkali.
$3^{\circ}$ amines do not react with Hinsberg's reagent.
Among the given options,$CH_3-CH_2-NH_2$ is a $1^{\circ}$ amine,so it will react to form a product that is soluble in alkali.

(C) The Tyndall effect is the scattering of light by particles in a colloid or a very fine suspension.
Among the given options,$NaCl$,glucose,and urea form true solutions,which do not exhibit the Tyndall effect.
Starch forms a colloidal solution,which scatters light and therefore exhibits the Tyndall effect.

(A) The osmotic pressure $\Pi$ is given by the formula $\Pi = CRT$,where $C$ is the molar concentration,$R$ is the gas constant,and $T$ is the temperature.
Since the mass of solute $(10 \ g)$ and the volume of solvent $(250 \ mL)$ are the same for all solutions,the molar concentration $C = \frac{n}{V} = \frac{mass}{M_{w} \times V}$ is inversely proportional to the molar mass $(M_{w})$ of the solute.
Therefore,$\Pi \propto \frac{1}{M_{w}}$.
Calculating molar masses:
Glucose $(C_{6}H_{12}O_{6})$: $M_{w1} = 180 \ g/mol$
Urea $(CH_{4}N_{2}O)$: $M_{w2} = 60 \ g/mol$
Sucrose $(C_{12}H_{22}O_{11})$: $M_{w3} = 342 \ g/mol$
Since $M_{w2} < M_{w1} < M_{w3}$,the osmotic pressure follows the order $P_{2} > P_{1} > P_{3}$.

(A) The given reaction is the $Etard$ reaction,which is used for the oxidation of toluene to benzaldehyde using chromyl chloride $(CrO_2Cl_2)$ in the presence of a non-polar solvent like carbon disulfide $(CS_2)$.
In this reaction,the methyl group of toluene is oxidized to a chromium complex intermediate,which is $C_6H_5CH(OCrOHCl_2)_2$.
This intermediate is then hydrolyzed with water $(H_3O^+)$ to yield benzaldehyde.

(C) According to Raoult's law for an ideal solution,the total vapour pressure $P_s$ is given by:
$P_s = P_A^0 x_A + P_B^0 x_B$
Given:
$P_A^0$ (benzene) $= 280 \ mm \ Hg$
$P_B^0$ (octane) $= 420 \ mm \ Hg$
Molar ratio $= 3 : 2$,so mole fractions are:
$x_A = \frac{3}{3+2} = \frac{3}{5} = 0.6$
$x_B = \frac{2}{3+2} = \frac{2}{5} = 0.4$
Substituting the values:
$P_s = (280 \times 0.6) + (420 \times 0.4)$
$P_s = 168 + 168$
$P_s = 336 \ mm \ Hg$

(D) $NaBH_4$ (Sodium borohydride) is a selective reducing agent that reduces aldehydes and ketones to their corresponding alcohols but does not reduce esters,carboxylic acids,or amides under normal conditions. In the given substrate,there is a ketone group and an ester group $(-COOCH_3)$. Therefore,$NaBH_4$ will selectively reduce the ketone group to a secondary alcohol $(-CHOH)$ while leaving the ester group unaffected. The final product is a cyclohexane ring with an $-OH$ group,a $-CH_2-COOCH_3$ group,and a $-CH_3$ group.

(C) First,calculate the limiting molar conductivity of acetic acid:
$\Lambda_{m}^{\circ}(CH_{3}COOH) = \Lambda_{H^{+}}^{\circ} + \Lambda_{CH_{3}COO^{-}}^{\circ} = 350 + 50 = 400 \ S \ cm^{2} \ mol^{-1}$
Next,calculate the degree of dissociation $(\alpha)$:
$\alpha = \frac{\Lambda_{m}}{\Lambda_{m}^{\circ}} = \frac{20}{400} = 0.05$
Finally,calculate the dissociation constant $(K_{a})$ using the formula $K_{a} = C \alpha^{2} / (1 - \alpha)$. Since $\alpha$ is very small,we use $K_{a} \approx C \alpha^{2}$:
$K_{a} = 0.007 \times (0.05)^{2}$
$K_{a} = 7 \times 10^{-3} \times 25 \times 10^{-4}$
$K_{a} = 175 \times 10^{-7} = 1.75 \times 10^{-5} \ mol \ L^{-1}$
Thus,the value is $1.75$.

(A) The Arrhenius equation is given by $\ln k = \ln A - \frac{E_{a}}{RT}$.
Comparing this with the equation of a straight line $y = mx + c$,where $y = \ln k$ and $x = \frac{1}{T}$,the slope $m = -\frac{E_{a}}{R}$.
Given that the slope is $-5 \times 10^{3} \, K$,we have $-\frac{E_{a}}{R} = -5 \times 10^{3} \, K$.
Therefore,$E_{a} = 5 \times 10^{3} \times 8.314 \, J \, mol^{-1} = 41570 \, J \, mol^{-1}$.
Converting to $kJ \, mol^{-1}$,we get $E_{a} = 41.57 \, kJ \, mol^{-1} \approx 41.5 \, kJ \, mol^{-1}$.

(B) The given reaction sequence involves the conversion of $2,4,6$-tribromoaniline to $2,4,6$-tribromobenzenediazonium chloride using $NaNO_{2} / HCl$ at $0-5^{\circ}C$.
To replace the diazonium group $(-N_{2}^{+}Cl^{-})$ with a hydrogen atom,a reducing agent is required.
Ethanol $(CH_{3}CH_{2}OH)$ acts as a mild reducing agent that reduces the diazonium salt to the corresponding aromatic hydrocarbon (in this case,$1,3,5$-tribromobenzene) while being oxidized to acetaldehyde.
Therefore,the reagent '$R$' is $CH_{3}CH_{2}OH$.

(D) The correct matches are:
$(a)$ The reaction of benzene with $CO$ and $HCl$ in the presence of anhydrous $AlCl_3/CuCl$ is the $\text{Gattermann-Koch reaction}$. Thus,$(a)-(ii)$.
$(b)$ The reaction of a methyl ketone with $NaOX$ $(X = Cl, Br, I)$ is the $\text{Haloform reaction}$. Thus,$(b)-(iii)$.
$(c)$ The reaction of an alcohol with a carboxylic acid in the presence of concentrated $H_2SO_4$ is $\text{Esterification}$. Thus,$(c)-(iv)$.
$(d)$ The reaction of a carboxylic acid containing $\alpha$-hydrogen with $X_2/\text{Red P}$ followed by water is the $\text{Hell-Volhard-Zelinsky (HVZ) reaction}$. Thus,$(d)-(i)$.
Therefore,the correct sequence is $(a)-(ii), (b)-(iii), (c)-(iv), (d)-(i)$.

(D) To find the magnetic moment,we calculate the number of unpaired electrons $(n)$ in each complex:
$a. [Fe(CN)_{6}]^{3-}$: $Fe^{3+}$ is $3d^{5}$. $CN^-$ is a strong field ligand,causing pairing. $n=1$,$\mu = \sqrt{1(1+2)} = 1.73 \ BM$.
$b. [Fe(H_{2}O)_{6}]^{3+}$: $Fe^{3+}$ is $3d^{5}$. $H_2O$ is a weak field ligand,no pairing. $n=5$,$\mu = \sqrt{5(5+2)} = 5.92 \ BM$.
$c. [Fe(CN)_{6}]^{4-}$: $Fe^{2+}$ is $3d^{6}$. $CN^-$ is a strong field ligand,causing pairing. $n=0$,$\mu = 0 \ BM$.
$d. [Fe(H_{2}O)_{6}]^{2+}$: $Fe^{2+}$ is $3d^{6}$. $H_2O$ is a weak field ligand,no pairing. $n=4$,$\mu = \sqrt{4(4+2)} = 4.90 \ BM$.
Matching these values: $a-iv, b-i, c-ii, d-iii$.