JEE Main 2022 Chemistry Question Paper with Answer and Solution

(B) The borax bead test involves the following reactions:
$Na_{2}B_{4}O_{7} \cdot 10 H_{2}O \xrightarrow{\Delta} Na_{2}B_{4}O_{7} + 10 H_{2}O$
$Na_{2}B_{4}O_{7} \xrightarrow{\Delta} 2 NaBO_{2} + B_{2}O_{3}$
$B_{2}O_{3} + CoO \rightarrow Co(BO_{2})_{2} \text{ (cobalt (II) metaborate)}$
The blue coloured bead is due to the formation of cobalt $(II)$ metaborate,$Co(BO_{2})_{2}$.

(C) At high altitudes,$N_2$ and $O_2$ react to form $NO$:
$N_{2(g)} + O_{2(g)} \rightarrow 2NO_{(g)}$
$NO$ further reacts with oxygen to form $NO_2$:
$2NO_{(g)} + O_{2(g)} \rightarrow 2NO_{2(g)}$
$NO_2$ is a toxic gas that can damage plant leaves and retard the process of photosynthesis.

(A) In $\gamma-$methylcyclohexanecarbaldehyde,the parent ring is cyclohexane with a carbaldehyde $(-CHO)$ group attached at position $1$.
The carbon atom directly attached to the $-CHO$ group is the $\alpha-$carbon.
The next carbon is the $\beta-$carbon,and the carbon after that is the $\gamma-$carbon.
Therefore,the methyl group is attached at the $3-$position relative to the $-CHO$ group on the cyclohexane ring.
This corresponds to the structure where the methyl group is at the meta position relative to the $-CHO$ group.

(B) The reaction of $2$-bromo-$3$-methylbutane with $Mg$ in $Et_2O$ forms a Grignard reagent,$3$-methylbutylmagnesium bromide $(CH_3-CH(CH_3)-CH(MgBr)-CH_3)$.
When this Grignard reagent is treated with $D_2O$,the $MgBr$ group is replaced by a deuterium atom $(D)$,resulting in $2$-deuterio-$3$-methylbutane $(CH_3-CH(CH_3)-CH(D)-CH_3)$.
In $2$-deuterio-$3$-methylbutane,the carbon atom at position $2$ is bonded to four different groups: a hydrogen atom $(H)$,a deuterium atom $(D)$,a methyl group $(-CH_3)$,and an isopropyl group $(-CH(CH_3)_2)$.
Since the carbon at position $2$ is bonded to four distinct groups,it is a chiral center,making the molecule chiral.

(D) Statement $I$: The compound has a chiral center bonded to four different groups: $-H$,$-NO_2$,a $cis-but-2-enyl$ group,and another $cis-but-2-enyl$ group. However,since the two alkenyl groups are identical,the molecule possesses a plane of symmetry or is achiral if the groups are arranged such that they superimpose. Upon closer inspection of the structure,the chiral center is bonded to two identical groups,making it achiral. Thus,Statement $I$ is incorrect.
Statement $II$: The second structure is indeed the mirror image of the first structure. Since the first structure is achiral,its mirror image is identical to it. Thus,Statement $II$ is correct.
Therefore,Statement $I$ is incorrect but Statement $II$ is correct.

(C) $1$. Ethanol $(CH_{3}CH_{2}OH)$ on heating with concentrated $H_{2}SO_{4}$ undergoes dehydration to produce ethene gas $(CH_{2}=CH_{2})$.
$2$. Ethene reacts with cold dilute aqueous solution of alkaline $KMnO_{4}$ (Baeyer's reagent) to undergo hydroxylation,resulting in the formation of ethane$-1,2-$diol,commonly known as glycol $(HOCH_{2}-CH_{2}OH)$.

(C) The neutralization of a weak acid $(CH_3COOH)$ with a strong base $(NaOH)$ results in a basic equivalence point $(pH > 7)$.
Phenolphthalein is a weak organic acid $(HIn)$ that is colorless in acidic medium and turns pink in basic medium due to the formation of its ionized form $(In^-)$.
Its $pH$ range is approximately $8.2 - 10.0$,which matches the equivalence point of the titration between a weak acid and a strong base.

(B) According to Dalton's Law of Partial Pressures,$P_{total} = P_{O_2} + P_{Ne}$.
Given $P_{total} = 25 \ bar$ and $P_{Ne} = 20 \ bar$,so $P_{O_2} = 25 - 20 = 5 \ bar$.
Partial pressure is related to mole fraction $(X)$ by $P_i = X_i \times P_{total}$.
Moles of $Ne = \frac{200 \ g}{20 \ g \ mol^{-1}} = 10 \ mol$.
Moles of $O_2 = \frac{x \ g}{32 \ g \ mol^{-1}} = \frac{x}{32} \ mol$.
Mole fraction of $O_2$ is $X_{O_2} = \frac{n_{O_2}}{n_{O_2} + n_{Ne}} = \frac{x/32}{x/32 + 10}$.
Using $P_{O_2} = X_{O_2} \times P_{total}$,we get $5 = \left( \frac{x/32}{x/32 + 10} \right) \times 25$.
Dividing by $5$,we get $1 = \left( \frac{x/32}{x/32 + 10} \right) \times 5$.
$1 = \frac{5x/32}{(x + 320)/32} = \frac{5x}{x + 320}$.
$x + 320 = 5x \implies 4x = 320 \implies x = 80 \ g$.

(B) To determine the hybridisation,we use the formula: $\text{Steric Number} = \frac{1}{2} (V + M - C + A)$,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.
$1$. $PF_{5}$: $\text{Steric Number} = \frac{1}{2}(5 + 5) = 5$ $(sp^{3}d)$
$2$. $BrF_{5}$: $\text{Steric Number} = \frac{1}{2}(7 + 5) = 6$ $(sp^{3}d^{2})$
$3$. $PCl_{3}$: $\text{Steric Number} = \frac{1}{2}(5 + 3) = 4$ $(sp^{3})$
$4$. $SF_{6}$: $\text{Steric Number} = \frac{1}{2}(6 + 6) = 6$ $(sp^{3}d^{2})$
$5$. $[ICl_{4}]^{-}$: $\text{Steric Number} = \frac{1}{2}(7 + 4 + 1) = 6$ $(sp^{3}d^{2})$
$6$. $ClF_{3}$: $\text{Steric Number} = \frac{1}{2}(7 + 3) = 5$ $(sp^{3}d)$
$7$. $IF_{5}$: $\text{Steric Number} = \frac{1}{2}(7 + 5) = 6$ $(sp^{3}d^{2})$
The molecules/ions with $sp^{3}d^{2}$ hybridisation are $BrF_{5}$,$SF_{6}$,$[ICl_{4}]^{-}$,and $IF_{5}$.
Total count = $4$.

(A) The reaction of a polyhydric alcohol with sodium or similar reagents releases $H_2$ gas based on the number of hydroxyl groups $(x)$:
$R(OH)_x + xNa \rightarrow R(ONa)_x + \frac{x}{2} H_2$
Using the Principle of Atom Conservation $(PoAC)$ for hydrogen atoms in the $OH$ groups:
$x \times \text{moles of } X = 2 \times \text{moles of } H_2$
Given:
Mass of $X = 1.84 \, mg = 1.84 \times 10^{-3} \, g$
Molar mass of $X = 92.0 \, g/mol$
Volume of $H_2$ at $STP = 1.344 \, mL = 1.344 \times 10^{-3} \, L$
Moles of $X = \frac{1.84 \times 10^{-3}}{92} = 2 \times 10^{-5} \, mol$
Moles of $H_2 = \frac{1.344 \times 10^{-3}}{22.4} = 6 \times 10^{-5} \, mol$
Substituting into the equation:
$x \times (2 \times 10^{-5}) = 2 \times (6 \times 10^{-5})$
$x = \frac{12 \times 10^{-5}}{2 \times 10^{-5}} = 6$
Thus,the number of alcoholic hydrogens is $6$.

(D) The starting material is a racemic mixture of $(\pm) Ph(C=O)C(OH)(CN)Ph$. This means it contains two enantiomers,$(+)$ and $(-)$.
When this reacts with $HCN$,the carbonyl group $(C=O)$ is converted into a cyanohydrin group $(C(OH)(CN))$.
The original chiral center at the second carbon remains unchanged,while a new chiral center is created at the carbonyl carbon.
Each enantiomer of the starting material will produce two diastereomers because the new chiral center can have either $R$ or $S$ configuration.
Since we start with a racemic mixture (a $1:1$ mixture of two enantiomers),we get two pairs of diastereomers.
Specifically,$(+) \text{ isomer} \rightarrow (R,R) + (R,S)$ and $(-) \text{ isomer} \rightarrow (S,S) + (S,R)$.
Thus,a total of $4$ stereoisomers are formed. However,looking at the options provided,the question likely asks for the number of stereoisomers formed from one enantiomer or is based on a specific interpretation. Given the standard nature of this problem,the correct answer is $4$. Since $4$ is not an option,and based on common textbook problems of this type,if the question implies the number of diastereomeric products from the racemic mixture,the answer is $4$. If the question is flawed,we select the most logical outcome. Re-evaluating: the product is $Ph-C(OH)(CN)-C(OH)(CN)-Ph$. This molecule has two chiral centers. The $(R,R)$,$(S,S)$,$(R,S)$,and $(S,R)$ forms are possible. The $(R,S)$ and $(S,R)$ are identical (meso form). Thus,there are $3$ stereoisomers: $(R,R)$,$(S,S)$,and the meso $(R,S)$. Therefore,the correct option is $D$.

(D) The argument of the sine function must be dimensionless,so the dimensions of $\frac{\alpha x}{kt}$ must be $[M^0L^0T^0]$.
Since $[kt] = [\text{Energy}] = [ML^2T^{-2}]$,we have $[\alpha] = \frac{[kt]}{[x]} = \frac{[ML^2T^{-2}]}{[L]} = [MLT^{-2}]$.
The energy density $u$ has dimensions of energy per unit volume,so $[u] = \frac{[ML^2T^{-2}]}{[L^3]} = [ML^{-1}T^{-2}]$.
From the given equation $u = \frac{\alpha}{\beta} \sin(\dots)$,the dimensions of $\beta$ are given by $[\beta] = \frac{[\alpha]}{[u]}$.
Substituting the dimensions: $[\beta] = \frac{[MLT^{-2}]}{[ML^{-1}T^{-2}]} = [M^0L^2T^0]$.
Therefore,the correct option is $D$.

(D) For a body-centred cubic $(BCC)$ unit cell,the number of atoms per unit cell is $Z = 2$.
The density formula is given by $\rho = \frac{Z \times M_{atomic}}{N_A \times a^3}$.
Given: $\rho = 6.0 \, g \, cm^{-3}$,$a = 300 \, pm = 300 \times 10^{-10} \, cm = 3 \times 10^{-8} \, cm$.
$6.0 = \frac{2 \times M_{atomic}}{6.022 \times 10^{23} \times (3 \times 10^{-8})^3}$.
$6.0 = \frac{2 \times M_{atomic}}{6.022 \times 10^{23} \times 27 \times 10^{-24}} = \frac{2 \times M_{atomic}}{16.2594}$.
$M_{atomic} = \frac{6.0 \times 16.2594}{2} = 48.7782 \, g \, mol^{-1}$.
Number of moles in $180 \, g = \frac{180}{48.7782} \approx 3.6902 \, mol$.
Number of atoms $= \text{moles} \times N_A = 3.6902 \times 6.022 \times 10^{23} \approx 22.22 \times 10^{23}$.
The nearest integer is $22$.

(C) Step $1$: Calculate the moles of ascorbic acid $(C_{6}H_{8}O_{6})$.
$n = \frac{\text{mass}}{\text{molar mass}} = \frac{10.2 \ g}{176 \ g \ mol^{-1}} \approx 0.05795 \ mol$.
Step $2$: Calculate the molality $(m)$ of the solution.
$m = \frac{\text{moles of solute}}{\text{mass of solvent in kg}} = \frac{0.05795 \ mol}{0.150 \ kg} \approx 0.3863 \ mol \ kg^{-1}$.
Step $3$: Calculate the depression in freezing point $(\Delta T_{f})$.
$\Delta T_{f} = K_{f} \times m = 3.9 \ K \ kg \ mol^{-1} \times 0.3863 \ mol \ kg^{-1} \approx 1.5066^{\circ} C$.
Step $4$: Express $\Delta T_{f}$ in the form $(x \times 10^{-1})^{\circ} C$.
$1.5066 \approx 15.066 \times 10^{-1}$.
Rounding to the nearest integer,$x = 15$.

(C) For a first-order reaction,the rate constant $K$ is given by $K = \frac{0.693}{t_{1/2}}$.
Given $t_{1/2} = 0.3010 \ min$,so $K = \frac{0.693}{0.3010} \approx 2.303 \ min^{-1}$.
The integrated rate equation is $\ln \frac{[A]_0}{[A]_t} = Kt$,which can be written as $\log \frac{[A]_0}{[A]_t} = \frac{Kt}{2.303}$.
Substituting the values: $\log \frac{[A]_0}{[A]_t} = \frac{2.303 \times 2.0}{2.303} = 2.0$.
Therefore,$\frac{[A]_0}{[A]_t} = 10^2 = 100$.

(B) The interhalogens with a square pyramidal structure are those of the type $AX_{5}$,which undergo $sp^{3}d^{2}$ hybridization.
From the given list,the molecules with square pyramidal geometry are $BrF_{5}$,$IF_{5}$,and $ClF_{5}$.
Therefore,the total number of such interhalogens is $3$.

(A) The complex $[Cu(en)_2(SCN)_2]$ exhibits linkage isomerism due to the ambidentate ligand $SCN^-$.
For the trans-isomer,the possible linkage isomers are:
$1$. $[Cu(en)_2(SCN)_2]$ (trans-dithiocyanato)
$2$. $[Cu(en)_2(NCS)_2]$ (trans-diisothiocyanato)
$3$. $[Cu(en)_2(SCN)(NCS)]$ (trans-thiocyanato-isothiocyanato)
These three forms represent distinct linkage isomers. Since the question asks for the number of relatively more stable isomers,and these three linkage isomers are chemically distinct and stable,the total number is $3$.

(A) The Gibbs free energy change is given by the equation $\Delta G = \Delta H - T \Delta S$.
Since the entropy of a liquid is higher than that of a solid,the entropy change $(\Delta S)$ for the formation of liquid metal is more positive compared to solid metal.
As $\Delta S$ increases,the term $-T \Delta S$ becomes more negative,which makes the overall $\Delta G^{\ominus}$ value more negative.
$A$ more negative $\Delta G^{\ominus}$ indicates a more spontaneous reaction,making the reduction of metal oxide easier.
Therefore,both Assertion $A$ and Reason $R$ are correct,and $R$ is the correct explanation of $A$.

(B) The reaction of white phosphorus $(P_{4})$ with thionyl chloride $(SOCl_{2})$ is a standard chemical reaction used to produce phosphorus trichloride $(PCl_{3})$,sulfur dioxide $(SO_{2})$,and disulfur dichloride $(S_{2}Cl_{2})$.
The balanced chemical equation is:
$P_{4} + 8 SOCl_{2} \rightarrow 4 PCl_{3} + 4 SO_{2} + 2 S_{2}Cl_{2}$

(C) Iodine $(I_{2})$ is a non-metal that acts as a reducing agent when it reacts with strong oxidizing agents like concentrated nitric acid $(HNO_{3})$.
In this reaction,$I_{2}$ is oxidized to iodic acid $(HIO_{3})$,while $HNO_{3}$ is reduced to nitrogen dioxide $(NO_{2})$.
The balanced chemical equation is:
$I_{2} + 10 HNO_{3(conc)} \rightarrow 2 HIO_{3} + 10 NO_{2} + 4 H_{2}O$

(D) Isoelectronic species are those that have the same number of electrons.
$1. Sm^{2+}: 62 - 2 = 60 \text{ electrons}; Er^{3+}: 68 - 3 = 65 \text{ electrons}$. Since $60 \neq 65$,they are not isoelectronic.
$2. Yb^{2+}: 70 - 2 = 68 \text{ electrons}; Lu^{3+}: 71 - 3 = 68 \text{ electrons}$. Since $68 = 68$,they are isoelectronic.
$3. Tb^{2+}: 65 - 2 = 63 \text{ electrons}; Tm^{4+}: 69 - 4 = 65 \text{ electrons}$. Since $63 \neq 65$,they are not isoelectronic.
Therefore,both pairs $(A)$ and $(C)$ are not isoelectronic.

(A) $KMnO_4$ is a strong oxidizing agent. When $HCl$ is used in permanganate titrations,$KMnO_4$ oxidizes $HCl$ to $Cl_2$ gas according to the following reaction:
$2KMnO_4 + 16HCl \rightarrow 2MnCl_2 + 2KCl + 8H_2O + 5Cl_2$
Because $HCl$ is consumed in this side reaction,it interferes with the titration,leading to inaccurate results. Therefore,$HCl$ is not used. Both the assertion and the reason are correct,and the reason explains the assertion.

(B) $Ni(CO)_4$: The oxidation state of $Ni$ is $0$. The configuration is $3d^8 4s^2$. Due to the strong field ligand $CO$,electrons pair up,resulting in $sp^3$ hybridization.
$[Ni(CN)_4]^{2-}$: The oxidation state of $Ni$ is $+2$. The configuration is $3d^8$. Due to the strong field ligand $CN^-$,the electrons pair up,resulting in $dsp^2$ hybridization.
$[Co(CN)_6]^{3-}$: The oxidation state of $Co$ is $+3$. The configuration is $3d^6$. Due to the strong field ligand $CN^-$,the electrons pair up,resulting in $d^2sp^3$ hybridization.
$[CoF_6]^{3-}$: The oxidation state of $Co$ is $+3$. The configuration is $3d^6$. Due to the weak field ligand $F^-$,the electrons do not pair up,resulting in $sp^3d^2$ hybridization.
Therefore,the correct matching is $A-I, B-IV, C-III, D-II$.

(D) The reaction between $N_{2}$ and $O_{2}$ is highly endothermic,requiring a very high temperature (approximately $1483-2000 \ K$) to proceed:
$N_{2}(g) + O_{2}(g) \xrightarrow{1483-2000 \ K} 2NO(g)$
Since such high temperatures are not available in the atmosphere,they do not react under normal conditions.

(C) The reaction proceeds through the following steps:
$1$. Protonation of the hydroxyl group by $H^+$ followed by the loss of a water molecule to form a tertiary carbocation.
$2$. The double bond in the chain attacks the carbocation,leading to a cyclization reaction to form a more stable five-membered ring carbocation.
$3$. The bromide ion $(Br^-)$ then attacks the carbocation to form the final product.
$4$. The structure corresponding to this mechanism is option $C$.

(D) The reaction of an alkyl aryl ether with concentrated $HI$ involves the cleavage of the $C-O$ bond between the alkyl group and the oxygen atom. This is because the $C(aryl)-O$ bond has partial double bond character due to resonance and is stronger.
Thus,the reaction of phenyl isopropyl ether with $HI$ yields phenol $(A)$ and isopropyl iodide.
However,in the context of the given options,the reaction proceeds to form phenol $(A)$.
When phenol $(A)$ is heated with $Zn$ dust,it undergoes reduction to form benzene $(B)$.

(A) Direct nitration of aniline with concentrated $HNO_3$ and $H_2SO_4$ (nitrating mixture) yields a mixture of ortho,meta,and para nitroanilines.
Assertion $A$ is true because the formation of the meta-isomer occurs due to the protonation of the $-NH_2$ group in the strongly acidic medium,forming the anilinium ion $(-NH_3^+)$,which is meta-directing.
Reason $R$ is also true,as the nitrating mixture is indeed a strong acidic mixture.
Since the formation of the meta-product is specifically due to the acidic nature of the medium (which protonates the amino group),Reason $R$ correctly explains Assertion $A$.

(A) The given polymers are identified as follows:
$A$. $(-CH_2-C(Cl)=CH-CH_2-)_n$ is Neoprene,which is an elastomer $(III)$.
$B$. $(-NH-(CH_2)_6-NH-CO-(CH_2)_4-CO-)_n$ is Nylon-$6,6$,which is a fiber $(II)$.
$C$. $(-CH_2-CH(Cl)-)_n$ is Polyvinyl chloride $(PVC)$,which is a thermoplastic polymer $(IV)$.
$D$. $(-C_6H_3(OH)-CH_2-)_n$ is Novolac,which is a thermosetting polymer $(I)$.
Therefore,the correct matching is $A-III, B-II, C-IV, D-I$.

(D) Statement $I$ is false because inhibitors can also bind to allosteric sites,which are different from the active site,to block enzyme action.
Statement $II$ is true because some inhibitors (irreversible inhibitors) form strong covalent bonds with the enzyme,permanently deactivating it.
Therefore,Statement $I$ is false and Statement $II$ is true.

(A) $1$. Reaction with $HCN$ followed by hydrolysis $(H_{3}O^+)$ adds a carbon atom to the carbonyl group of Fructose $(C_{6}H_{12}O_{6})$,resulting in a heptonic acid derivative $(A)$ with the formula $C_{7}H_{14}O_{8}$.
$2$. Reduction of Fructose with $NaBH_{4}$ followed by heating with $HI/P$ reduces all hydroxyl and carbonyl groups to a hydrocarbon chain. Since Fructose is a hexose,the final product $B$ is $n$-hexane $(C_{6}H_{14})$.
$3$. Therefore,$A = C_{7}H_{14}O_{8}$ and $B = C_{6}H_{14}$.

(D) For an $FCC$ lattice,the number of atoms per unit cell,$Z = 4$.
Given edge length,$a = 4.0 \times 10^{-8} \ cm$.
Density,$d = 9.03 \ g \ cm^{-3}$.
Avogadro's number,$N_{A} = 6.02 \times 10^{23} \ mol^{-1}$.
The formula for density is $d = \frac{Z \times M}{N_{A} \times a^{3}}$.
Rearranging for molar mass $M$: $M = \frac{d \times N_{A} \times a^{3}}{Z}$.
Substituting the values: $M = \frac{9.03 \times 6.02 \times 10^{23} \times (4.0 \times 10^{-8})^{3}}{4}$.
$M = \frac{9.03 \times 6.02 \times 10^{23} \times 64 \times 10^{-24}}{4}$.
$M = \frac{9.03 \times 6.02 \times 6.4}{4} = 86.97 \ g/mol$.
Rounding to the nearest integer,$M \approx 87 \ g/mol$.

(C) According to Dalton's Law of partial pressures,the partial pressure of $A$ is given by $P_A = Y_A \times P_{total}$.
Given $Y_A = 0.5$ and $P_{total} = 0.8 \ atm$,we have $P_A = 0.5 \times 0.8 = 0.4 \ atm$.
According to Raoult's Law,$P_A = P_A^0 \times X_A$,where $X_A$ is the mole fraction in the liquid phase.
Given $X_A = 0.2$,we have $0.4 = P_A^0 \times 0.2$.
Therefore,$P_A^0 = \frac{0.4}{0.2} = 2 \ atm$.

(A) The Arrhenius equation is given by $k = Ae^{-E_a/RT}$.
Taking the natural logarithm on both sides,we get $\ln k = -\frac{E_a}{R}(\frac{1}{T}) + \ln A$.
Comparing this with the equation of a straight line $y = mx + c$,the slope $m = -\frac{E_a}{R}$.
From the given graph,the slope is calculated as $\frac{y_2 - y_1}{x_2 - x_1} = \frac{0 - 20}{5 - 0} = -4$.
Equating the slopes: $-\frac{E_a}{R} = -4$,which gives $E_a = 4 \times R$.
Given $R = 2 \ cal \ K^{-1} \ mol^{-1}$,we have $E_a = 4 \times 2 = 8 \ cal \ mol^{-1}$.

(B) The Freundlich adsorption isotherm is given by the equation: $\frac{x}{m} = K P^{\frac{1}{n}}$
Taking the logarithm on both sides,we get: $\log \frac{x}{m} = \frac{1}{n} \log P + \log K$
This is a linear equation of the form $y = mx + c$,where the plot of $\log \frac{x}{m}$ versus $\log P$ should be a straight line with a slope of $\frac{1}{n}$ and an intercept of $\log K$.
Analyzing the given curves:
$1$. The first curve $(a)$ is a non-linear curve of $\log \frac{x}{m}$ vs $\log P$,which is incorrect.
$2$. The second curve $(b)$ is a linear plot of $\log \frac{x}{m}$ vs $P$,which is incorrect.
$3$. The third curve $(c)$ is a linear plot of $\log \frac{x}{m}$ vs $P$,which is incorrect.
$4$. The fourth curve $(d)$ is a linear plot of $\log \frac{x}{m}$ vs $\log P$ with a positive intercept,which is correct.
Therefore,there are $3$ curves that are not in accordance with the Freundlich adsorption isotherm.

(A) Albumin is a protein,which is a macromolecule that forms a lyophilic sol when dispersed in a suitable solvent like water or dilute salt solutions.
Since albumin has a high affinity for the dispersion medium,it forms a lyophilic sol.

(A) Earthy and undesired materials present in the ore,other than the desired metal,are known as gangue.

(D) The basicity of an amine depends on the availability of the lone pair of electrons on the nitrogen atom.
In $N$-methylpyrrole,the lone pair is involved in aromaticity,making it very weakly basic.
In acyclic amines like triethylamine and diethylamine,the nitrogen atom can undergo amine inversion,which affects its basicity.
Quinuclidine is a bicyclic tertiary amine where the nitrogen atom is held in a rigid structure.
Due to this rigid bicyclic structure,the lone pair on the nitrogen is highly available for protonation,and the molecule cannot undergo amine inversion,which would otherwise decrease the availability of the lone pair.
Therefore,quinuclidine is the strongest Bronsted base among the given options.

(A) $NaCN$ is an ionic compound,so it provides $CN^-$ ions in solution. The carbon atom is more nucleophilic than the nitrogen atom,leading to the formation of alkyl cyanide $(R-CN)$ as the major product.
$AgCN$ is a covalent compound. The nitrogen atom has a lone pair of electrons available for nucleophilic attack,while the carbon atom is involved in a covalent bond with silver. This leads to the formation of alkyl isocyanide $(R-NC)$ as the major product.
Therefore,for the reaction with $AgCN$,the product $A$ is $CH_3CH_2NC$ (ethyl isocyanide),and for the reaction with $NaCN$,the product $B$ is $CH_3CH_2CN$ (ethyl cyanide).

(D) Step $1$: The reaction of $4$-hydroxybenzaldehyde with $Br_2$ leads to electrophilic aromatic substitution at the ortho positions relative to the $-OH$ group,forming $3,5$-dibromo-$4$-hydroxybenzaldehyde (Product $A$).
Step $2$: The reaction of $A$ with $NH_2OH$ (hydroxylamine) converts the aldehyde group $(-CHO)$ into an oxime group $(-CH=N-OH)$,forming $3,5$-dibromo-$4$-hydroxybenzaldehyde oxime (Product $B$).
Step $3$: The reaction of $B$ with $P_2O_5$ (a dehydrating agent) causes the dehydration of the oxime group to a nitrile group $(-CN)$,yielding $3,5$-dibromo-$4$-hydroxybenzonitrile (Product $C$).

(A) The reaction sequence is as follows:
$1$. The starting material $A$ reacts with $NH_3$ to form an amide $(C_8H_8ClNO)$.
$2$. The amide then undergoes the Hofmann bromamide degradation reaction with $Br_2$ and $NaOH$ to form an amine.
$3$. Looking at the final product,which is $3-chloro-4-methylaniline$ (or a similar substituted aniline derivative),we can trace back the structure.
$4$. The final product has a $Cl$ group at the ortho position and a $CH_3$ group at the meta position relative to the $NH_2$ group.
$5$. The starting material $A$ must be $4-chloro-2-methylbenzoyl$ chloride,which upon reaction with $NH_3$ gives $4-chloro-2-methylbenzamide$,and subsequent Hofmann degradation yields $3-chloro-4-methylaniline$.

(A) The given reaction is a coupling reaction of benzene diazonium chloride with a phenol or amine in a basic medium to form an azo dye.
The reaction of benzene diazonium chloride with $\beta$-naphthol in the presence of $NaOH$ (basic medium) yields an orange-red azo dye.
The chemical equation is:
$C_6H_5N_2^+Cl^- + \beta\text{-Naphthol} \xrightarrow{NaOH} \text{Orange-red dye}$

(B) Step $1$: The reduction of $4\text{-aminobenzonitrile}$ with $DIBAL-H$ $(AlH(i-Bu)_2)$ followed by hydrolysis yields $4\text{-aminobenzaldehyde}$ as product '$A$'.
Step $2$: The reaction of $4\text{-aminobenzaldehyde}$ with acetaldehyde $(CH_3CHO)$ in the presence of dilute $NaOH$ and heat is a cross-aldol condensation reaction.
Step $3$: The aldehyde group of $4\text{-aminobenzaldehyde}$ reacts with the $\alpha\text{-hydrogen}$ of acetaldehyde to form $4\text{-aminocinnamaldehyde}$ as the major product '$B$'.

(B) $Amytal$ is a hypnotic drug used to treat sleeping disorders. It belongs to the class of barbiturates,which act as central nervous system depressants.

(B) $1$. The compound $'X'$ is acidic and soluble in $NaOH$ but insoluble in $NaHCO_3$. This indicates that the compound is more acidic than water but less acidic than carbonic acid $(H_2CO_3)$. Phenols typically satisfy this condition.
$2$. The compound gives a violet colour with neutral $FeCl_3$ solution. This is a characteristic test for the presence of a phenolic group ($-OH$ group attached directly to a benzene ring).
$3$. Among the given options,Phenol $(C_6H_5OH)$ is the only compound that contains a phenolic group and satisfies all the given chemical properties.
$4$. The reaction with $FeCl_3$ is: $6C_6H_5OH + FeCl_3 \rightarrow [Fe(C_6H_5O)_6]^{3-} + 6H^+ + 3HCl$ (forming a violet complex).

(A) Cell constant $G^* = \frac{\ell}{A} = 129 \; m^{-1} = 1.29 \; cm^{-1}$.
Conductivity $\kappa = \frac{G^*}{R}$.
For solution $1$ $(74.5 \; ppm)$: $\kappa_1 = \frac{1.29}{100} \; S \; cm^{-1}$.
Concentration $C_1 \propto 74.5 \; ppm$.
For solution $2$ $(149 \; ppm)$: $\kappa_2 = \frac{1.29}{50} \; S \; cm^{-1}$.
Concentration $C_2 \propto 149 \; ppm$.
Since $ppm$ is proportional to molarity $(M)$,$\frac{C_1}{C_2} = \frac{74.5}{149} = \frac{1}{2}$.
Molar conductivity $\wedge_m = \frac{1000 \kappa}{C}$.
$\frac{\wedge_1}{\wedge_2} = \frac{\kappa_1}{\kappa_2} \times \frac{C_2}{C_1} = \frac{1.29/100}{1.29/50} \times \frac{149}{74.5} = \frac{50}{100} \times 2 = 1$.
Given $\frac{\wedge_1}{\wedge_2} = x \times 10^{-3} = 1$.
Therefore,$x = 1000$.

(A) In a crystal where the anion forms a cubic close packing $(CCP)$ and the cation occupies all octahedral voids,the structure is of the $NaCl$ type.
For an $NaCl$ type structure,the edge length $a$ is related to the ionic radii of the cation $(r_{+})$ and the anion $(r_{-})$ by the formula: $a = 2(r_{+} + r_{-})$.
Given: $r_{+} = 102 \ pm$ and $r_{-} = 181 \ pm$.
Substituting the values: $a = 2(102 \ pm + 181 \ pm)$.
$a = 2(283 \ pm) = 566 \ pm$.

(B) According to Henry's Law,$p = K_H \times x$,where $x$ is the mole fraction of the gas in the solution.
Given $p = 0.920 \; bar$ and $K_H = 46.82 \; kbar = 46820 \; bar$.
$x = \frac{p}{K_H} = \frac{0.920}{46820} \approx 1.965 \times 10^{-5}$.
Since the solubility is very small,the number of moles of $H_2O$ in $1 \; L$ $(1000 \; g)$ is $n_{H_2O} = \frac{1000}{18} \approx 55.55 \; mol$.
$x = \frac{n_{O_2}}{n_{O_2} + n_{H_2O}} \approx \frac{n_{O_2}}{n_{H_2O}}$.
$n_{O_2} = x \times n_{H_2O} = (1.965 \times 10^{-5}) \times 55.55 \approx 1.09 \times 10^{-3} \; mol$.
$n_{O_2} \approx 1.09 \; millimoles$.
The nearest integer is $1$.

(C) The rate law for the reaction is given by $r = k[X]^1[Y]^0 = k[X]$.
From experiment $I$,$2 \times 10^{-3} = k(0.1) \Rightarrow k = 2 \times 10^{-2} \ min^{-1}$.
For experiment $III$,the concentration of $X$ is $0.4 \ mol \ L^{-1}$.
Therefore,the initial rate is $r = k[X] = (2 \times 10^{-2})(0.4) = 0.8 \times 10^{-2} = 8 \times 10^{-3} \ mol \ L^{-1} \ min^{-1}$.
Comparing this with $M \times 10^{-3}$,we get $M = 8$.
The ratio of the numerical value of $M$ to $0.2$ is $\frac{8}{0.2} = 40$.

(C) tetrapeptide is formed by the linkage of $4$ amino acids.
In a linear polypeptide chain,the number of peptide bonds is always $(n - 1)$,where $n$ is the number of amino acids.
For a tetrapeptide,$n = 4$.
Number of peptide bonds $= 4 - 1 = 3$.
Therefore,(number of amino acids) - (number of peptide bonds) $= 4 - 3 = 1$.

(D) The complex is $[Fe(CN)_{6}]^{3-}$.
$Fe$ is in the $+3$ oxidation state,so its electronic configuration is $3d^{5}$.
$CN^{-}$ is a strong field ligand,which causes pairing of electrons in the $d$-orbitals.
For an octahedral complex,the $d$-orbitals split into $t_{2g}$ and $e_{g}$ sets.
The $5$ electrons occupy the $t_{2g}$ orbitals as $(t_{2g}^{5} e_{g}^{0})$.
The crystal field stabilization energy $(CFSE)$ is calculated as:
$CFSE = n(t_{2g}) \times (-0.4 \Delta_{o}) + n(e_{g}) \times (0.6 \Delta_{o})$
$CFSE = 5 \times (-0.4 \Delta_{o}) + 0 \times (0.6 \Delta_{o}) = -2.0 \Delta_{o}$.
Thus,the value is $2$.

(A) The extent of adsorption of a gas on a solid adsorbent like charcoal depends on the ease of liquefaction of the gas.
Easily liquefiable gases have higher critical temperatures $(T_c)$ and are adsorbed to a greater extent.
Conversely,gases with lower critical temperatures are difficult to liquefy and show the least adsorption.
Comparing the given values: $T_c(He) = 5.2 \ K$,$T_c(CH_4) = 190 \ K$,$T_c(CO_2) = 304.2 \ K$,and $T_c(NH_3) = 405.5 \ K$.
Since $He$ has the lowest critical temperature $(5.2 \ K)$,it shows the least adsorption on charcoal.

(C) The liquation process is used for metals having low melting points,such as $Sn$ (tin).
In this process,the metal is heated to a molten state and made to flow down a sloping hearth.
The metal flows down,while impurities with higher melting points remain behind on the top.

(B) The enthalpy of hydration $(\Delta_{hyd}H)$ is directly proportional to the charge density of the ion.
Charge density is defined as $\frac{\text{charge}}{\text{size}}$.
For the given $3d$-metal ions $(Cr^{2+}, Mn^{2+}, Fe^{2+}, Co^{2+})$,the charge is the same $(+2)$.
Therefore,the ion with the largest ionic radius will have the lowest charge density and consequently the lowest enthalpy of hydration.
According to periodic trends,the ionic radius decreases across the $3d$ series due to increasing effective nuclear charge.
Among the given ions,$Mn^{2+}$ has the largest ionic radius.
Thus,$Mn^{2+}$ has the lowest enthalpy of hydration.

(D) $Cu^{2+}$ has a $d^9$ electronic configuration,which leads to an unsymmetrical filling of the $e_g$ orbitals $(t_{2g}^6 e_g^3)$.
This causes Jahn-Teller distortion in octahedral complexes.
The magnitude of structural distortion is directly proportional to the difference in the field strength of the ligands present in the coordination sphere.
In $[Cu(en)_2Cl_2]$,the ligands are $en$ (a strong field ligand) and $Cl^-$ (a weak field ligand).
Because $en$ and $Cl^-$ have the largest difference in their ligand field strengths compared to the other options,$[Cu(en)_2Cl_2]$ exhibits the maximum structural distortion.
Specifically,for $trans-[Cu(en)_2Cl_2]$,the axial ligands $(Cl^-)$ are significantly different from the equatorial ligands $(en)$,leading to maximum elongation.

(A) The Hinsberg reagent is $Benzene$ $sulphonyl$ $chloride$ $(C_6H_5SO_2Cl)$.
It is used for the separation and identification of primary,secondary,and tertiary amines.

(D) Natural polymers are those that occur naturally in plants and animals. Examples include $Protein$,$Starch$,and $Rubber$.
$Rayon$ is a regenerated cellulose fiber,which is classified as a semisynthetic polymer because it is derived from natural cellulose through chemical processing.

(D) Assertion $A$ states that Amylose is insoluble in water,which is incorrect because Amylose is a water-soluble component of starch.
Reason $R$ states that Amylose is a long linear molecule with more than $200$ glucose units,which is correct as it consists of $200$ to $1000$ $\alpha-D-(+)$-glucose units held by $\alpha$-glycosidic linkages.
Therefore,$A$ is incorrect but $R$ is correct.

(B) Mass of ethanol = $62.5 \ cm^{3} \times 0.80 \ g \ cm^{-3} = 50 \ g = 0.05 \ kg$.
Freezing point depression,$\Delta T_{f} = T_{f}^{\circ} - T_{f} = 156.0 \ K - 155.1 \ K = 0.9 \ K$.
Using the formula $\Delta T_{f} = K_{f} \times m$,where $m$ is molality:
$0.9 = 2.00 \times \frac{1.80 \ g / M_{w}}{0.05 \ kg}$.
$M_{w} = \frac{2.00 \times 1.80}{0.9 \times 0.05} = \frac{3.6}{0.045} = 80 \ g \ mol^{-1}$.

(C) The cell reaction is: $Cu_{(s)} + 2Ag^{+}_{(aq)} \rightarrow Cu^{2+}_{(aq)} + 2Ag_{(s)}$.
Using the Nernst equation: $E_{\text{cell}} = E^{\Theta}_{\text{cell}} - \frac{0.06}{n} \log \frac{[Cu^{2+}]}{[Ag^{+}]^2}$.
Here,$n = 2$,$[Cu^{2+}] = 10^{-3} \, M$,and $[Ag^{+}] = 10^{-2} \, M$.
$0.43 = E^{\Theta}_{\text{cell}} - \frac{0.06}{2} \log \frac{10^{-3}}{(10^{-2})^2}$.
$0.43 = E^{\Theta}_{\text{cell}} - 0.03 \log \frac{10^{-3}}{10^{-4}} = E^{\Theta}_{\text{cell}} - 0.03 \log(10) = E^{\Theta}_{\text{cell}} - 0.03$.
$E^{\Theta}_{\text{cell}} = 0.43 + 0.03 = 0.46 \, V$.
We know $E^{\Theta}_{\text{cell}} = E^{\Theta}_{\text{cathode}} - E^{\Theta}_{\text{anode}} = E^{\Theta}_{Ag^{+}/Ag} - E^{\Theta}_{Cu^{2+}/Cu}$.
$0.46 = 0.80 - E^{\Theta}_{Cu^{2+}/Cu}$.
$E^{\Theta}_{Cu^{2+}/Cu} = 0.80 - 0.46 = 0.34 \, V$.
Thus,$E^{\Theta}_{Cu^{2+}/Cu} = 34 \times 10^{-2} \, V$.

(C) Radioactive decay follows first-order kinetics: $N_t = N_0 \times (1/2)^{t/t_{1/2}}$.
Given: $N_0 = 1 \, \mu g$,$t_{1/2} = 30 \ years$,$t = 100 \ years$.
$N_t = 1 \times (1/2)^{100/30} = (1/2)^{10/3}$.
Taking $\log$ on both sides: $\log N_t = \frac{10}{3} \log(0.5) = \frac{10}{3} \times (-0.30) = -1$.
$N_t = 10^{-1} \, \mu g$.
Comparing with $n \times 10^{-1} \, \mu g$,we get $n = 1$.

(C) The complex is $Na[Co(bpy)Cl_4]$.
$bpy$ $(2,2'-bipyridine)$ is a bidentate ligand,and $Cl^-$ is a monodentate ligand.
Coordination number of $Co = (1 \times 2) + (4 \times 1) = 6$.
Let the oxidation state of $Co$ be $x$.
The charge on $Na$ is $+1$,$bpy$ is $0$,and $Cl$ is $-1$.
$1 + x + 0 + 4(-1) = 0$
$1 + x - 4 = 0$
$x - 3 = 0$
$x = +3$.
Magnitude of oxidation state $= 3$.
Sum $= 6 + 3 = 9$.

(A) To determine the number of oxoacids with a peroxo $(O-O)$ bond,we examine their structures:
$1$. $H_2SO_3$ (Sulphurous acid): Contains only $S=O$ and $S-OH$ bonds. No peroxo bond.
$2$. $H_2SO_4$ (Sulphuric acid): Contains $S=O$ and $S-OH$ bonds. No peroxo bond.
$3$. $H_2S_2O_8$ (Peroxodisulphuric acid): Contains a central $O-O$ peroxo linkage between two sulphur atoms.
$4$. $H_2S_2O_7$ (Pyrosulphuric acid): Contains an $S-O-S$ linkage. No peroxo bond.
Thus,only $H_2S_2O_8$ contains a peroxo bond.
The total count is $1$.