KCET 2026 Chemistry Question Paper with Answer and Solution

(A) Step $1$: Kjeldahl's method is used for the estimation of nitrogen in organic compounds. However, it is not applicable to compounds containing nitrogen in a ring system, such as pyridine $(C_5H_5N)$, because the nitrogen atom in the ring is not easily converted to ammonium sulphate $( (NH_4)_2SO_4 )$ by heating with conc. $H_2SO_4$.
Step $2$: Since the nitrogen in the pyridine ring is highly stable and does not readily form ammonium sulphate, Statement $I$ is true.
Step $3$: Statement $II$ claims that nitrogen in pyridine changes to ammonium sulphate, which is incorrect as pyridine resists this reaction under standard Kjeldahl conditions. Therefore, Statement $II$ is false.

(B) Step $1$: The given molecular formula is $C_5H_{12}$, which follows the general formula $C_nH_{2n+2}$, indicating it is an alkane.
Step $2$: The chain isomers are structural isomers that differ in the arrangement of the carbon chain.
Step $3$: The possible isomers for $C_5H_{12}$ are:
$(i)$ $CH_3-CH_2-CH_2-CH_2-CH_3$ (n-pentane)
(ii) $CH_3-CH(CH_3)-CH_2-CH_3$ (isopentane or $2-$methylbutane)
(iii) $CH_3-C(CH_3)_2-CH_3$ (neopentane or $2,2-$dimethylpropane)
Step $4$: There are $3$ possible chain isomers.

(C) The general formula for an alkane is $C_nH_{2n+2}$.
Given the molecular formula $C_{20}H_{42}$, we have $n = 20$.
Checking the hydrogen count: $2(20) + 2 = 40 + 2 = 42$.
Since it fits the general formula, it is an alkane with $20$ carbon atoms.
The $IUPAC$ name for an alkane with $20$ carbon atoms is $Eicosane$.

(C) $1$. Identify the longest carbon chain attached to the oxygen atom. The chain containing the ether group is a $3$-carbon chain, which is propane.
$2$. Number the chain starting from the end that gives the substituent the lowest possible locant. Numbering from either end, the substituent is at position $2$.
$3$. The substituents are an ethoxy group $(-OC_2H_5)$ and a methyl group $(-CH_3)$ both at position $2$.
$4$. Combining these, the $IUPAC$ name is $2$-ethoxy-$2$-methylpropane.

(C) Step $1$: The staggered conformation of ethane has the hydrogen atoms on adjacent carbons as far apart as possible, resulting in minimum torsional strain and maximum stability.
Step $2$: The eclipsed conformation has the hydrogen atoms as close as possible, resulting in maximum torsional strain and minimum stability.
Step $3$: Therefore, Statement $I$ is true because staggered is more stable, and Statement $II$ is false because the torsional strain in staggered conformation is actually less than in the eclipsed conformation.

(C) Step $1$: Observe that the percentage composition of elements ($Cu$, $C$, and $O$) in both the natural and synthetic samples of $Cupric \ Carbonate$ is identical.
Step $2$: The Law of Definite Proportions states that a given chemical compound always contains its constituent elements in a fixed ratio by mass, regardless of its source or method of preparation.
Step $3$: Since the composition remains constant for both samples, this data illustrates the Law of Definite Proportions.

(A) Step $1$: $(a)$ $2 \text{ moles of ethene } (C_2H_4) = 2 \times 28 \text{ g} = 56 \text{ g}$. Matches $(ii)$.
Step $2$: $(b)$ Molar mass of $CO_2 = 44 \text{ g/mol}$. $1.5 \text{ moles} = 1.5 \times 44 \text{ g} = 66 \text{ g}$. Matches $(iv)$.
Step $3$: $(c)$ $1 \text{ g of } H_2 = \frac{1}{2} \text{ mole} = 0.5 \times 6.022 \times 10^{23} \text{ molecules} = 3.011 \times 10^{23} \text{ molecules}$. Note: The provided options suggest a mismatch in the original question's $(c)$ and $(d)$ mapping. Based on standard calculations, $(c)$ corresponds to $0.5 \text{ mole}$ and $(d)$ $2 \text{ moles of } H_2O$ at $STP$ is $44.8 \text{ L}$. Given the options, $(a-ii, b-iv, c-i, d-iii)$ is the intended logical set.

(A) Step $1$: $Ra$ $(Z=88)$ belongs to group $2$ (alkaline earth metals) and period $7$. Thus, $a-iv$.
Step $2$: $Ga$ $(Z=31)$ has electronic configuration $[Ar] 3d^{10} 4s^2 4p^1$, placing it in period $4$ and group $13$. Thus, $b-i$.
Step $3$: $W$ $(Z=74)$ is a transition metal in period $6$ and group $6$. Thus, $c-ii$.
Step $4$: $Pd$ $(Z=46)$ is a transition metal in period $5$ and group $10$. Thus, $d-iii$.
Conclusion: The correct matching is $a-iv, b-i, c-ii, d-iii$.

(B) $1$. For $NO_2^+$: Steric number = $1/2 \times (5 + 0 - 1 + 0) = 2$. Hybridization is $sp$.
$2$. For $NO_3^-$: Steric number = $1/2 \times (5 + 0 + 1 - 0) = 3$. Hybridization is $sp^2$.
$3$. For $NH_4^+$: Steric number = $1/2 \times (5 + 4 - 1 + 0) = 4$. Hybridization is $sp^3$.
Thus, the correct order is $sp, sp^2$ and $sp^3$.

(D) Step $1$: Analyze $(a)$. The dipole moment of $BF_3$ is $0$ (non-polar). In $NH_3$ and $NF_3$, both have lone pairs, but in $NH_3$, the bond dipoles and lone pair dipole are in the same direction, whereas in $NF_3$, they oppose each other. Thus, the order is $NH_3 > NF_3 > BF_3$. The given order $BF_3 < NF_3 < NH_3$ is correct.
Step $2$: Analyze $(b)$. $HgCl_2$ is linear $(180^\circ)$, $NH_4^+$ is tetrahedral $(109.5^\circ)$, and $SF_6$ is octahedral $(90^\circ)$. The order $HgCl_2 > NH_4^+ > SF_6$ is correct.
Step $3$: Since both orders are correct, the statement 'does not agree' is false for both. Therefore, the correct option is $(d)$.

(D) Step $1$: In resonance structures, only the positions of electrons change, not the positions of atoms. Thus, statement $(c)$ is incorrect.
Step $2$: The $CO_3^{2-}$ ion has three equivalent resonance structures, making all $C-O$ bonds equivalent due to resonance hybrid. Thus, $(a)$ and $(b)$ are correct.
Step $3$: The formal charge on an atom is calculated as $FC = V - L - \frac{1}{2}S$. For carbon in $CO_3^{2-}$, $V=4$, $L=0$, $S=8$, so $FC = 4 - 0 - 4 = 0$. Thus, $(d)$ is correct.
Step $4$: Combining these, statements $(a)$, $(b)$, and $(d)$ are correct.

(C) Step $1$: Analyze Statement $I$. In $H_2O_2$ (hydrogen peroxide), oxygen is in the peroxide state, so its oxidation number is $-1$. In $RbO_2$ (rubidium superoxide), oxygen is in the superoxide state, so its oxidation number is $-\frac{1}{2}$. Thus, Statement $I$ is correct.
Step $2$: Analyze Statement $II$. In $HAuCl_4$, the oxidation state of $Au$ is $+3$, so it is represented as $HAu(III)Cl_4$. In $MnO_2$, the oxidation state of $Mn$ is $+4$ (since $x + 2(-2) = 0 \implies x = +4$), so it should be represented as $Mn(IV)O_2$. The statement claims $Mn(II)O_2$, which is incorrect. Thus, Statement $II$ is incorrect.

(A) Step $1$: Write the half-reactions.
Oxidation: $C_2O_4^{2-} \rightarrow 2 CO_2 + 2 e^-$
Reduction: $MnO_4^- + 8 H^+ + 5 e^- \rightarrow Mn^{2+} + 4 H_2O$
Step $2$: Balance the electrons by multiplying the oxidation half-reaction by $5$ and the reduction half-reaction by $2$.
$5 C_2O_4^{2-} \rightarrow 10 CO_2 + 10 e^-$
$2 MnO_4^- + 16 H^+ + 10 e^- \rightarrow 2 Mn^{2+} + 8 H_2O$
Step $3$: Add the two equations to get the balanced net ionic equation:
$5 C_2O_4^{2-} + 2 MnO_4^- + 16 H^+ \rightarrow 2 Mn^{2+} + 8 H_2O + 10 CO_2$
Comparing this with the given equation, we get $a = 5$ and $x = 2$.

(B) The de Broglie wavelength $\lambda$ is given by the relation $\lambda = \frac{h}{p}$.
Since momentum $p = mv$, substituting this into the equation gives $\lambda = \frac{h}{mv}$.
Therefore, the correct representation is $\lambda = \frac{h}{mv}$.

(B) $1$. The wave function $\psi$ represents the amplitude of the electron wave.
$2$. The square of the wave function, $\psi^2$, represents the probability density of finding an electron at a specific point in space.
$3$. Therefore, $\psi^2$ is known as the probability density.

(A) Step $1$: Statement $(A)$ is the Third Law of Thermodynamics, which states that the entropy of a perfect crystalline substance at absolute zero $(0 \text{ K})$ is zero. Thus, $(A)$ is true.
Step $2$: Statement $(B)$ refers to the Gibbs free energy equation: $\Delta G = \Delta H - T\Delta S$. For a reaction to be spontaneous, $\Delta G < 0$. Therefore, $\Delta H - T\Delta S < 0$, which implies $\Delta H < T\Delta S$ or $T\Delta S > \Delta H$. Thus, $(B)$ is true.
Step $3$: Since both statements are correct, the correct option is $(A)$.

(B) $1$. Internal energy $(U)$ and entropy $(S)$ depend only on the state of the system, not on the path taken to reach that state, hence they are state functions.
$2$. Work $(w)$ and heat $(q)$ are path functions, not state functions.
$3$. Internal energy does not change in an isothermal process for an ideal gas $(\Delta U = 0)$.
$4$. Work done in an adiabatic process is not necessarily zero; it is given by $w = \Delta U$.

(D) According to the first law of thermodynamics, $\Delta U = q + w$.
For path $ACB$:
Heat flow into the system, $q_{ACB} = 60 \text{ J}$.
Work done by the system, $w_{ACB} = -30 \text{ J}$ (since work is done by the system).
Change in internal energy, $\Delta U = q_{ACB} + w_{ACB} = 60 \text{ J} - 30 \text{ J} = 30 \text{ J}$.
Since internal energy $\Delta U$ is a state function, it remains the same for any path between $A$ and $B$.
Therefore, for path $ADB$, $\Delta U = 30 \text{ J}$.
Given for path $ADB$, work done by the system, $w_{ADB} = -10 \text{ J}$.
Using $\Delta U = q_{ADB} + w_{ADB}$:
$30 \text{ J} = q_{ADB} - 10 \text{ J}$.
$q_{ADB} = 30 \text{ J} + 10 \text{ J} = 40 \text{ J}$.

(D) The relationship between $K_p$ and $K_c$ is given by $K_p = K_c(RT)^{\Delta n}$.
For the reaction $N_{2(g)} + 3H_{2(g)} \rightleftharpoons 2NH_{3(g)}$, the change in moles of gas is $\Delta n = n_p - n_r = 2 - (1 + 3) = -2$.
The temperature $T = 500 + 273 = 773 \text{ K}$.
The gas constant $R = 0.082 \text{ L atm K}^{-1} \text{ mol}^{-1}$.
Substituting these values into the formula: $K_p = K_c(RT)^{-2}$.
Rearranging for $K_c$, we get $K_c = \frac{K_p}{(RT)^{-2}}$.
Substituting the given $K_p$ value: $K_c = \frac{1.44 \times 10^{-5}}{(0.082 \times 773)^{-2}}$.

(C) $1$. The reaction involves a change in the total number of moles of gaseous species: $4 \text{ moles}$ on the reactant side and $1 \text{ mole}$ on the product side. According to Le Chatelier's principle, a change in pressure will shift the equilibrium.
$2$. Since the reaction is exothermic ($\Delta H = -q \text{ kJ}$, where $q > 0$), a change in temperature will also shift the equilibrium.
$3$. $A$ catalyst only increases the rate of both forward and backward reactions equally and does not affect the equilibrium position or the amount of product formed.
$4$. Therefore, the amount of product is affected by both temperature and pressure.

(A) $1$. The solution is a basic buffer consisting of a weak base (pyridine) and its salt (pyridinium chloride).
$2$. The concentration of the base $[Base] = 0.2 \text{ M}$.
$3$. The concentration of the salt $[Salt] = \frac{0.15 \text{ mol}}{0.5 \text{ L}} = 0.3 \text{ M}$.
$4$. Using the Henderson-Hasselbalch equation for a basic buffer: $pOH = pK_b + \log\left(\frac{[Salt]}{[Base]}\right)$.
$5$. $pOH = 8.75 + \log\left(\frac{0.3}{0.2}\right) = 8.75 + \log(1.5) = 8.75 + 0.176 = 8.926$.
$6$. Since $pH + pOH = 14$, $pH = 14 - 8.926 = 5.074 \approx 5.07$.

(D) $1$. Heteropolar reactions, also known as ionic reactions, involve the heterolytic cleavage of covalent bonds.
$2$. In heterolytic cleavage, the shared pair of electrons is transferred to one of the bonded atoms.
$3$. This process results in the formation of charged species, specifically cations and anions, as reaction intermediates.
$4$. Therefore, both anions and cations serve as intermediates in heteropolar reactions.

(A) The electrolysis reaction of water is: $2H_2O(l) \rightarrow 2H_2(g) + O_2(g)$.
According to Faraday's law of electrolysis, the number of equivalents of $H_2$ and $O_2$ produced are equal.
Equivalents of $O_2 = \frac{\text{mass}}{\text{equivalent mass}} = \frac{16 \text{ g}}{8 \text{ g/eq}} = 2 \text{ eq}$.
Therefore, equivalents of $H_2 = 2 \text{ eq}$.
Since the equivalent mass of $H_2$ is $1 \text{ g/eq}$, the mass of $H_2 = 2 \text{ eq} \times 1 \text{ g/eq} = 2 \text{ g}$.
Number of moles of $H_2 = \frac{\text{mass}}{\text{molar mass}} = \frac{2 \text{ g}}{2 \text{ g/mol}} = 1 \text{ mol}$.
At $STP$, $1 \text{ mole}$ of any gas occupies $22.4 \text{ L}$.
Thus, the volume of $H_2$ gas is $22.4 \text{ L}$.

(B) According to Kohlrausch's law of independent migration of ions:
$\Lambda_m^0(NH_4OH) = \lambda^0(NH_4^+) + \lambda^0(OH^-)$
To obtain this, we use strong electrolytes:
$\Lambda_m^0(NH_4Cl) = \lambda^0(NH_4^+) + \lambda^0(Cl^-)$
$\Lambda_m^0(NaOH) = \lambda^0(Na^+) + \lambda^0(OH^-)$
$\Lambda_m^0(NaCl) = \lambda^0(Na^+) + \lambda^0(Cl^-)$
Adding the first two and subtracting the third:
$\Lambda_m^0(NH_4Cl) + \Lambda_m^0(NaOH) - \Lambda_m^0(NaCl) = (\lambda^0(NH_4^+) + \lambda^0(Cl^-)) + (\lambda^0(Na^+) + \lambda^0(OH^-)) - (\lambda^0(Na^+) + \lambda^0(Cl^-))$
$= \lambda^0(NH_4^+) + \lambda^0(OH^-) = \Lambda_m^0(NH_4OH)$.

(A) Step $1$: Identify the oxidation and reduction half-reactions from the overall reaction $3Mn^{2+} \rightarrow Mn + 2Mn^{3+}$.
Step $2$: Oxidation: $Mn^{2+} \rightarrow Mn^{3+} + e^-$ $(E^0_{ox} = -1.51 \text{ V})$.
Step $3$: Reduction: $Mn^{2+} + 2e^- \rightarrow Mn$ $(E^0_{red} = -1.18 \text{ V})$.
Step $4$: Calculate $E^0_{cell} = E^0_{red} + E^0_{ox} = -1.18 \text{ V} + (-1.51 \text{ V}) = -2.69 \text{ V}$.
Step $5$: Since $E^0_{cell} < 0$, the reaction is non-spontaneous.

(B) Given:
Conductivity $(\kappa) = 0.021\text{ }\Omega^{-1}\text{ cm}^{-1}$
Resistance $(R) = 60\text{ }\Omega$
Cell constant $(G^*) = \kappa \times R$
$G^* = 0.021\text{ }\Omega^{-1}\text{ cm}^{-1} \times 60\text{ }\Omega = 1.26\text{ cm}^{-1}$.

(A) For a $1^{st}$ order reaction, the integrated rate equation is $[R] = [R]_0 e^{-kt}$.
$(1)$ The plot of $[R]$ vs $t$ is an exponential decay curve, not a straight line.
$(2)$ The plot of $\ln[R]$ vs $t$ is a straight line with slope $-k$ and intercept $\ln[R]_0$.
$(3)$ The plot of $\log_{10}[R]$ vs $t$ is a straight line with slope $-\frac{k}{2.303}$ and intercept $\log_{10}[R]_0$.
$(4)$ The plot of $\log_{10}\frac{[R]_0}{[R]}$ vs $t$ is a straight line with slope $\frac{k}{2.303}$ passing through the origin.
Since the graph in option $A$ is shown as a horizontal straight line (which represents a zero-order reaction), it is not applicable for a $1^{st}$ order reaction.

(A) The overall rate constant is given by $K = \frac{k_1 k_2}{k_3}$.
Taking the natural logarithm on both sides: $\ln K = \ln k_1 + \ln k_2 - \ln k_3$.
Differentiating with respect to temperature $T$: $\frac{d(\ln K)}{dT} = \frac{d(\ln k_1)}{dT} + \frac{d(\ln k_2)}{dT} - \frac{d(\ln k_3)}{dT}$.
Using the Arrhenius equation $\frac{d(\ln k)}{dT} = \frac{E_a}{RT^2}$, we get: $\frac{E_a}{RT^2} = \frac{E_{a1}}{RT^2} + \frac{E_{a2}}{RT^2} - \frac{E_{a3}}{RT^2}$.
Thus, $E_a = E_{a1} + E_{a2} - E_{a3}$.
Substituting the given values: $E_a = 40 + 50 - 60 = 30 \text{ kJ mol}^{-1}$.

(D) $1$. In methyl chloride $(CH_3Cl)$, the carbon atom is $sp^3$ hybridized. The $C - Cl$ bond is a pure single bond.
$2$. In chlorobenzene $(C_6H_5Cl)$, the lone pair of electrons on the chlorine atom participates in resonance with the benzene ring. This imparts partial double bond character to the $C - Cl$ bond.
$3$. Due to resonance, the $C - Cl$ bond in chlorobenzene is shorter and stronger than in methyl chloride.
$4$. Conversely, the $C - Cl$ bond in methyl chloride is longer and weaker than in chlorobenzene.

(C) Step $1$: Chlorobenzene can be prepared from aniline via the Sandmeyer reaction using $NaNO_2/HCl$ followed by $CuCl/HCl$.
Step $2$: It can be prepared from benzene via electrophilic substitution using $Cl_2/FeCl_3$.
Step $3$: It can be prepared from benzene diazonium chloride via the Sandmeyer reaction.
Step $4$: In phenol, the $C-O$ bond has partial double bond character due to resonance, making it very strong and difficult to break to replace the $-OH$ group with a $-Cl$ atom. Thus, chlorobenzene is not easily prepared from phenol.

(B) $S_N1$ reactions proceed via a planar carbocation intermediate.
For a racemic mixture to be formed, the starting alkyl halide must be chiral.
In $2$-bromobutane, the carbon atom attached to the bromine is chiral (bonded to four different groups: $-H, -CH_3, -C_2H_5, -Br$).
Upon hydrolysis, the leaving group departs to form a planar carbocation, which can be attacked by the nucleophile from either side with equal probability, resulting in a racemic mixture.

(B) $Chloramphenicol$ is an antibiotic used for the treatment of $Typhoid \text{ } fever$ $(a-iv)$.
$Thyroxine$ deficiency leads to $Goiter$ $(b-iii)$.
$Chloroquine$ is an antimalarial drug used for $Malaria$ $(c-i)$.
$Chloroform$ acts as an $Anaesthetic$ $(d-ii)$.
Therefore, the correct matching is $a-iv, b-iii, c-i, d-ii$.

(D) $1$. $R - CH_2OH$ is a primary alcohol.
$2$. Alkaline $KMnO_4$ is a strong oxidizing agent that oxidizes primary alcohols directly to carboxylic acids.
$3$. $LiAlH_4$ and $Na/C_2H_5OH$ are reducing agents.
$4$. $PCC$ (Pyridinium Chlorochromate) is a mild oxidizing agent that selectively oxidizes primary alcohols to aldehydes without further oxidation to carboxylic acids.

(B) The chemical structure of glycerol is $CH_2(OH)CH(OH)CH_2(OH)$.
In this structure, the two terminal carbon atoms are bonded to one $-OH$ group each, making them primary alcoholic groups.
The central carbon atom is bonded to one $-OH$ group and two other carbon atoms, making it a secondary alcoholic group.
Therefore, glycerol contains two primary and one secondary alcoholic groups.

(B) $1$. Hydroboration-oxidation of propene $(CH_3-CH=CH_2)$ follows anti-Markovnikov addition to yield propan$-1-$ol $(P = CH_3-CH_2-CH_2-OH)$.
$2$. Oxidation of primary alcohol $(P)$ with $CrO_3$ in anhydrous medium stops at the aldehyde stage, yielding propanal $(Q = CH_3-CH_2-CHO)$.
$3$. Reaction of propanal $(Q)$ with methylmagnesium bromide $(CH_3MgBr)$ followed by acid hydrolysis yields butan$-2-$ol $(R = CH_3-CH_2-CH(OH)-CH_3)$.

(A) $1$. Reaction of carbonyl compounds with hydroxylamine $(NH_2OH)$ yields an oxime: $(a) \rightarrow (ii)$.
$2$. Reaction of carbonyl compounds with primary amines $(R-NH_2)$ yields a Schiff base: $(b) \rightarrow (iii)$.
$3$. Reaction of carbonyl compounds with alcohols $(R-OH)$ in the presence of dry $HCl$ yields an acetal: $(c) \rightarrow (iv)$.
$4$. Reaction of carbonyl compounds with hydrogen cyanide $(HCN)$ yields a cyanohydrin: $(d) \rightarrow (i)$.
Therefore, the correct matching is $a-ii, b-iii, c-iv, d-i$.

(D) The reaction between benzaldehyde and acetophenone in the presence of a dilute base $(OH^-)$ is a Claisen-Schmidt condensation reaction.
$1$. Acetophenone acts as the nucleophile because it has $\alpha$-hydrogens, which are removed by the base to form an enolate ion.
$2$. The enolate ion attacks the carbonyl carbon of benzaldehyde to form a $\beta$-hydroxy ketone.
$3$. The $\beta$-hydroxy ketone undergoes dehydration (loss of $H_2O$) to form an $\alpha, \beta$-unsaturated ketone, which is the stable major product.
$4$. The product is $1,3-\text{diphenylprop-2-en-1-one}$ (also known as chalcone), represented by option $D$.

(D) Step $1$: The acidity of a compound depends on the stability of its conjugate base.
Step $2$: The conjugate base of a carboxylic acid is the carboxylate ion $(RCOO^-)$, where the negative charge is delocalized over two highly electronegative oxygen atoms.
Step $3$: The conjugate base of a phenol is the phenoxide ion $(C_6H_5O^-)$, where the negative charge is delocalized over the carbon atoms of the benzene ring and one oxygen atom.
Step $4$: Since oxygen is more electronegative than carbon, the negative charge is more effectively stabilized in the carboxylate ion than in the phenoxide ion. Therefore, carboxylic acids are more acidic.

(C) The iodoform test is given by compounds containing the $CH_3CO-$ group or the $CH_3CH(OH)-$ group attached to a carbon or hydrogen atom.
$(1)$ Ethanal $(CH_3CHO)$ contains the $CH_3CO-$ group.
$(2)$ Acetone $(CH_3COCH_3)$ contains the $CH_3CO-$ group.
$(3)$ Acetophenone $(C_6H_5COCH_3)$ contains the $CH_3CO-$ group.
$(4)$ Ethanoic acid $(CH_3COOH)$ does not contain the $CH_3CO-$ group attached to a carbon or hydrogen atom, as the carbonyl carbon is bonded to an $-OH$ group. Therefore, it does not give the iodoform test.

(B) $1$. In a strong acidic medium, the lone pair of electrons on the nitrogen atom of aniline is protonated by the acid.
$2$. This results in the formation of the anilinium ion $(-NH_3^+)$.
$3$. The $-NH_3^+$ group is strongly electron-withdrawing due to its positive charge and exerts a $-I$ effect.
$4$. Consequently, it deactivates the benzene ring and directs the incoming electrophile to the meta position.

(D) Step $1$: The basic strength of alkylamines in the aqueous phase depends on three main factors: inductive effect, solvation effect, and steric hindrance.
Step $2$: The inductive effect ($+I$ effect) increases the electron density on the nitrogen atom, making it more basic.
Step $3$: The solvation effect (stabilization of the conjugate acid by hydrogen bonding with water) stabilizes the cation; more substituted amines have less hydrogen bonding.
Step $4$: Steric hindrance (crowding around the nitrogen atom) hinders the approach of the proton.
Step $5$: Hyperconjugation effect does not play a significant role in determining the basicity of alkylamines in the aqueous phase. Therefore, the correct option is $D$.

(C) Step $1$: Nitrobenzene reacts with conc. $HNO_3$ and conc. $H_2SO_4$ at $323-333 \text{ K}$ to form $m$-dinitrobenzene $(A)$.
Step $2$: $m$-dinitrobenzene is reduced by $Fe/HCl$ to form $m$-nitroaniline $(B)$.
Step $3$: $m$-nitroaniline reacts with $NaNO_2/HCl$ at $273 \text{ K}$ to form $m$-nitrobenzenediazonium chloride $(C)$.
Step $4$: $m$-nitrobenzenediazonium chloride reacts with ethanol $(C_2H_5OH)$ to form nitrobenzene $(D)$, acetaldehyde $(CH_3CHO)$, $N_2$ gas, and $HCl$. Thus, $D$ is nitrobenzene.

(B) $1$. An oxidising agent is a substance that undergoes reduction by gaining electrons.
$2$. In $CrO_3$, $CrO_4^{2-}$, and $Cr_2O_7^{2-}$, chromium is in the $+6$ oxidation state, which is a common state for chromium to be reduced to lower oxidation states.
$3$. In $MoO_3$, molybdenum is in the $+6$ oxidation state. Due to the stability of the $+6$ oxidation state in group $6$ elements, $Mo(VI)$ is much more stable than $Cr(VI)$.
$4$. Therefore, $MoO_3$ does not readily undergo reduction and acts as a very poor oxidising agent compared to the chromium species.

(D) Step $1$: Manganese can exhibit a maximum oxidation state of $+7$ by using all its valence electrons $(3d^5 4s^2)$.
Step $2$: Oxygen has the unique ability to form multiple bonds (specifically $p\pi - d\pi$ multiple bonds) with transition metals, which stabilizes the higher oxidation states of the metal.
Step $3$: Fluorine, being highly electronegative and small, can only form single covalent bonds $(Mn-F)$. It cannot form multiple bonds with manganese.
Step $4$: Therefore, oxygen can satisfy the high valency of manganese in $Mn_2O_7$ through multiple bonding, whereas fluorine is restricted to $MnF_4$.

(B) The electronic configuration of $Cr$ is $[Ar] 3d^5 4s^1$.
The electronic configuration of $Cr^{2+}$ ion is $[Ar] 3d^4 4s^0$.
Number of unpaired electrons $(n)$ = $4$.
The spin-only magnetic moment formula is $\mu = \sqrt{n(n+2)} \text{ BM}$.
Substituting $n = 4$ into the formula: $\mu = \sqrt{4(4+2)} = \sqrt{4 \times 6} = \sqrt{24} \approx 4.90 \text{ BM}$.

(C) $1$. The stability of a coordination complex is significantly enhanced by the chelate effect.
$2$. The ligand $C_2O_4^{2-}$ (oxalate) is a bidentate chelating ligand that forms five-membered stable rings with the central metal ion.
$3$. Among the given options, $[Fe(C_2O_4)_3]^{3-}$ forms a stable chelated structure, making it the most stable complex due to the chelate effect.

(C) The complex $[Co(NH_3)_6]Cl_2$ dissociates in solution as follows:
$[Co(NH_3)_6]Cl_2 \rightarrow [Co(NH_3)_6]^{2+} + 2Cl^-$
Number of $[Co(NH_3)_6]^{2+}$ ions = $1$
Number of $Cl^-$ ions = $2$
Total number of ions per molecule = $1 + 2 = 3$.

(A) Step $1$: In metal carbonyls, the $M - C \sigma$ bond is formed by the donation of a lone pair of electrons from the carbonyl carbon atom into a vacant orbital of the metal atom.
Step $2$: The $M - C \pi$ bond is formed by the back-donation of a pair of electrons from a filled $d$-orbital of the metal into the vacant antibonding $\pi^*$ molecular orbital of the carbon monoxide ligand.
Step $3$: Both statements accurately describe the synergistic bonding mechanism in metal carbonyls.
Conclusion: Both Statement $I$ and Statement $II$ are correct.

(A) Step $1$: Determine the hybridization and geometry of each complex.
Step $2$: $[Co(NH_3)_6]^{3+}$ has $d^2sp^3$ hybridization, which corresponds to an octahedral geometry (ii).
Step $3$: $[NiCl_4]^{2-}$ has $sp^3$ hybridization, which corresponds to a tetrahedral geometry (iii).
Step $4$: $[Ni(CN)_4]^{2-}$ has $dsp^2$ hybridization, which corresponds to a square planar geometry (iv).
Step $5$: $[Fe(CO)_5]$ has $dsp^3$ hybridization, which corresponds to a trigonal bipyramidal geometry $(i)$.
Therefore, the correct match is $a-ii, b-iii, c-iv, d-i$.

(B) Step $1$: Identify the deficiency diseases for each vitamin.
- Vitamin $B_1$ (Thiamine) deficiency leads to Beri-Beri, but among the given options, it is associated with retarded growth in some contexts or general metabolic issues.
- Vitamin $B_2$ (Riboflavin) deficiency causes cheilosis, glossitis, and burning sensation of the skin.
- Vitamin $B_6$ (Pyridoxine) deficiency leads to convulsions.
- Vitamin $B_{12}$ (Cyanocobalamin) deficiency leads to pernicious anaemia ($RBC$ deficiency in Haemoglobin).
Step $2$: Match the items:
$(a)$ $B_1$ - (iii) Retarded growth
$(b)$ $B_2$ - (iv) Burning sensation of the skin
$(c)$ $B_6$ - $(i)$ Convulsions
$(d)$ $B_{12}$ - (ii) $RBC$ deficiency in Haemoglobin
Step $3$: The correct sequence is $a-iii, b-iv, c-i, d-ii$.

(C) Step $1$: Monosaccharides contain a free aldehyde or ketone group, which makes them reducing sugars. Thus, Statement $I$ is correct.
Step $2$: Sucrose is a disaccharide composed of glucose and fructose linked by a glycosidic bond between their anomeric carbons. It lacks a free aldehyde or ketone group, making it a non-reducing sugar. Therefore, it cannot reduce ammoniacal silver nitrate solution (Tollens' reagent). Thus, Statement $II$ is incorrect.

(C) $1$. Methionine is an essential amino acid, which is correct.
$2$. Glycine $(H_2N-CH_2-COOH)$ has no chiral carbon atom, so it does not exhibit enantiomerism. This is correct.
$3$. Glycylalanylglutamine is a tripeptide formed by three amino acids (Glycine, Alanine, Glutamine). $A$ tripeptide contains only two peptide (amide) linkages. Thus, the statement that it has three amide linkages is incorrect.
$4$. Zwitterions of amino acids contain both acidic $(-COO^-)$ and basic $(-NH_3^+)$ groups, allowing them to act as amphoteric substances. This is correct.

(B) Step $1$: Aliphatic alcohols react with ceric ammonium nitrate to give a red color, so $(a)$ matches (iii).
Step $2$: $C_6H_5NH_2$ (aniline) is a primary aromatic amine which gives the azo dye test, so $(b)$ matches (ii).
Step $3$: $CH_3CH_2CHO$ (propanal) is an aldehyde which gives a silver mirror with Tollen’s reagent, so $(c)$ matches (iv).
Step $4$: Phenol reacts with neutral ferric chloride to give a violet color, so $(d)$ matches $(i)$.
Therefore, the correct matching is $a-iii, b-ii, c-iv, d-i$.

(D) Step $1$: Reaction with concentrated $H_2SO_4$: Bromide salts $(Br^-)$ react with concentrated $H_2SO_4$ to produce hydrogen bromide gas, which is further oxidized to reddish-brown bromine $(Br_2)$ gas.
Step $2$: Reaction with $AgNO_3$: Bromide ions $(Br^-)$ react with silver nitrate $(AgNO_3)$ to form silver bromide $(AgBr)$, which appears as a pale yellow precipitate.
Step $3$: Conclusion: Since both observations match the characteristics of the bromide ion, the anion $A^-$ is $Br^-$.

(D) $1$. Osmotic pressure $(\pi = iCRT)$: For $0.1 \text{ M NaCl}$, $i=2$, so $\pi$ is higher than $0.1 \text{ M Urea}$ $(i=1)$. Option $A$ is incorrect.
$2$. Concentration: $2 \text{ M}$ is much higher than $2 \text{ ppm}$. Option $B$ is incorrect.
$3$. Elevation in boiling point $(\Delta T_b = i K_b m)$: For $0.02 \text{ M NaCl}$, $i=2$, so $\Delta T_b$ is higher than $0.02 \text{ M Urea}$ $(i=1)$. Option $C$ is incorrect.
$4$. Vapour pressure: Adding a non-volatile solute to a solvent lowers the vapour pressure. Thus, salt water has lower vapour pressure than pure water. Option $D$ is correct.

(C) Step $1$: Henry's law states that the partial pressure of a gas in vapor phase is proportional to the mole fraction of the gas in the solution, given by $p = K_H x$. Thus, $(a) \rightarrow (ii)$.
Step $2$: Raoult's law states that for a solution of volatile liquids, the partial vapor pressure of each component is $p_1 = x_1 p_1^o$. Thus, $(b) \rightarrow (i)$.
Step $3$: The first law of thermodynamics states that the change in internal energy is the sum of heat and work, given by $\Delta U = q + w$. Thus, $(c) \rightarrow (iv)$.
Step $4$: Kohlrausch's law of independent migration of ions states that the limiting molar conductivity of an electrolyte is the sum of the individual contributions of its ions, given by $\Lambda_m^o = \nu_+ \lambda_+^o + \nu_- \lambda_-^o$. Thus, $(d) \rightarrow (iii)$.
Conclusion: The correct matching is $a – ii, b – i, c – iv, d – iii$.

(A) Step $1$: Calculate the Van't Hoff factor $(i)$ using the formula: $i = \frac{\text{observed freezing point depression}}{\text{calculated freezing point depression}} = \frac{0.0205}{0.0197} \approx 1.041$.
Step $2$: For the dissociation of acetic acid $(CH_3COOH \rightleftharpoons CH_3COO^- + H^+)$, the number of ions produced per molecule is $n = 2$.
Step $3$: Calculate the degree of dissociation $(\alpha)$ using the formula: $\alpha = \frac{i-1}{n-1} = \frac{1.041-1}{2-1} = 0.041$.

(C) Step $1$: Calculate moles of urea $(n_2)$: $n_2 = \frac{18 \text{ g}}{60 \text{ g mol}^{-1}} = 0.3 \text{ mol}$.
Step $2$: Calculate moles of water $(n_1)$: $n_1 = \frac{100 \text{ g}}{18 \text{ g mol}^{-1}} \approx 5.55 \text{ mol}$.
Step $3$: Calculate relative lowering of vapour pressure using the formula $\frac{P_A^o - P_A}{P_A^o} = \frac{n_2}{n_1 + n_2}$.
Step $4$: Substitute the values: $\frac{0.3}{5.55 + 0.3} = \frac{0.3}{5.85} \approx 0.051 \approx 0.05$.

(B) For a $1^{st}$ order reaction, the rate constant $k$ is given by $k = \frac{2.303}{t} \log \frac{[R]_0}{[R]_t}$.
Substituting the values: $k = \frac{2.303}{40} \log \frac{0.1}{0.025} = \frac{2.303}{40} \log 4 = \frac{2.303 \times 0.6021}{40} \approx 0.03466 \text{ min}^{-1}$.
The rate of reaction is given by $\text{Rate} = k[R]$.
For $[R] = 0.01 \text{ M}$, $\text{Rate} = 0.03466 \times 0.01 = 3.466 \times 10^{-4} \text{ M min}^{-1} \approx 3.47 \times 10^{-4} \text{ M min}^{-1}$.

(B) The relationship between enthalpy change $\Delta H$, activation energy of the forward reaction $(E_a)_f$, and activation energy of the backward reaction $(E_a)_b$ is given by: $\Delta H = (E_a)_f - (E_a)_b$.
Given: $(E_a)_f = 150 \text{ kJ mol}^{-1}$ and $\Delta H = -135 \text{ kJ mol}^{-1}$.
Substituting the values: $-135 = 150 - (E_a)_b$.
Rearranging for $(E_a)_b$: $(E_a)_b = 150 + 135 = 285 \text{ kJ mol}^{-1}$.