JEE Main 2025 Chemistry Question Paper with Answer and Solution

(C) According to the Born-Haber cycle for the dissolution of an ionic solid,the heat of solution $(\Delta H_{sol})$ is the sum of the lattice energy $(L.E.)$ and the hydration energies $(H.E.)$ of the constituent ions.
$\Delta H_{sol} = L.E. + (H.E.)_{K^{+}} + (H.E.)_{Cl^{-}}$
Given:
$L.E. = -z \ kJ / mol$
$(H.E.)_{K^{+}} = -x \ kJ / mol$
$(H.E.)_{Cl^{-}} = -y \ kJ / mol$
Substituting these values into the equation:
$\Delta H_{sol} = (-z) + (-x) + (-y)$
$\Delta H_{sol} = -z - x - y$
$\Delta H_{sol} = -(z + x + y)$
Wait,let us re-evaluate the standard definition. Usually,$L.E.$ is defined as the energy released when gaseous ions form a solid,so $L.E. = +z$ (energy required to break the lattice). If $L.E.$ is given as $-z$ (energy released),then the energy to break the lattice is $+z$.
Thus,$\Delta H_{sol} = z + (-x) + (-y) = z - (x + y)$.

(B) In a period,the atomic radius decreases from left to right due to an increase in effective nuclear charge. In a group,the atomic radius increases from top to bottom due to the addition of new shells.
$(A)$ $r_{Br} < r_{K}$ is correct because $K$ is in Group $1$ and $Br$ is in Group $17$ of the same period.
$(B)$ $r_{Mg} < r_{Al}$ is incorrect because $Mg$ (Group $2$) has a larger atomic radius than $Al$ (Group $13$) in the same period.
$(C)$ $r_{Rb} < r_{Na}$ is incorrect because $Rb$ is in the $5th$ period and $Na$ is in the $3rd$ period,so $r_{Rb} > r_{Na}$.
$(D)$ $r_{At} < r_{Cs}$ is correct because $Cs$ is in Group $1$ and $At$ is in Group $17$ of the same period.
Since the question asks for the incorrect trend,both $(B)$ and $(C)$ represent incorrect trends. However,in standard competitive chemistry,$r_{Mg} > r_{Al}$ is the classic example of a periodic trend error.

(D) The standard state of a substance is defined as its pure form at a pressure of $1 \ bar$ and a specified temperature (usually $298 \ K$).
$\Delta_{f} H^{\theta}$ (standard enthalpy of formation) is defined as zero for elements in their most stable state at the specified temperature.
For $O_{2(g)}$,which is the most stable form of oxygen,$\Delta_{f} H^{\theta} = 0$ at any specified temperature.
Therefore,$\Delta_{f} H_{500}^{\theta}$ is zero for $O_{2(g)}$ is the correct statement.

(A) To determine the hybridization of the central atom,we use the formula: $\text{Steric Number} = (\text{Number of sigma bonds}) + (\text{Number of lone pairs})$.
$1$. For $SO_2$: The central atom $S$ has $2$ sigma bonds and $1$ lone pair. $\text{Steric Number} = 2 + 1 = 3$,which corresponds to $sp^2$ hybridization.
$2$. For $NO_2^-$: The central atom $N$ has $2$ sigma bonds and $1$ lone pair. $\text{Steric Number} = 2 + 1 = 3$,which corresponds to $sp^2$ hybridization.
$3$. For $N_3^-$: The central atom $N$ has $2$ sigma bonds and $0$ lone pairs. $\text{Steric Number} = 2 + 0 = 2$,which corresponds to $sp$ hybridization.
Therefore,the hybridizations are $sp^2, sp^2$ and $sp$ respectively.

(B) The given compound is $CH_3-CH=CH-CH(CH_3)-CH=CH-CH(CH_3)-CH=CH-CH_3$. Upon complete ozonolysis $(O_3/Zn, H_2O)$,the double bonds are cleaved to form carbonyl compounds.
Specifically,the structure $CH_3-CH=CH-CH(CH_3)-CH=CH-CH(CH_3)-CH=CH-CH_3$ breaks down into:
$1$. $CH_3CHO$ (Acetaldehyde) - $2$ molecules.
$2$. $OHC-CH(CH_3)-CHO$ (Methylmalonaldehyde) - $2$ molecules.
Acetaldehyde $(CH_3CHO)$ is achiral and thus optically inactive.
Methylmalonaldehyde $(OHC-CH(CH_3)-CHO)$ has a chiral center at the carbon atom attached to the methyl group. However,in this specific molecule,the two aldehyde groups are identical,making the molecule achiral due to the presence of a plane of symmetry or because it is a meso-like structure in its conformation. Thus,it is optically inactive.
Therefore,the number of optically active products is $0$.

(B) The extra stability of half-filled subshells is attributed to the following factors:
$(I)$ Symmetrical distribution of electrons: Electrons are distributed symmetrically in the orbitals,which leads to lower energy.
$(II)$ Larger exchange energy: The number of possible exchanges for electrons with parallel spins is maximum in half-filled and fully-filled configurations,resulting in higher exchange energy and greater stability.
$(III)$ Smaller coulombic repulsion: Due to the symmetrical arrangement,the inter-electronic repulsion is minimized.
$(IV)$ Smaller shielding of electrons by one another: The effective nuclear charge experienced by the electrons is optimized in these configurations.
Therefore,statements $(A), (B), (D),$ and $(E)$ are correct.

(B) $(i)$ Statement $(A)$ is true: Due to the inert pair effect,$Tl^{+}$ is more stable than $Tl^{3+}$. Therefore,$Tl^{3+}$ acts as a powerful oxidising agent to gain electrons and form the more stable $Tl^{+}$ state.
$(ii)$ Statement $(B)$ is true: The standard reduction potential $E^0_{Al^{3+}/Al} = -1.66 \ V$ is very low,meaning $Al^{3+}$ is not easily reduced.
$(iii)$ Statement $(C)$ is false: $Tl^{3+}$ is unstable in solution due to the inert pair effect.
$(iv)$ Statement $(D)$ is true: $Tl^{+}$ is more stable than $Tl^{3+}$ due to the inert pair effect.
$(v)$ Statement $(E)$ is true: $Al^{3+}$ is the most stable oxidation state for Aluminum,and $Tl^{+}$ is the most stable oxidation state for Thallium.
Thus,statements $(A), (B), (D), \text{ and } (E)$ are correct.

(A) Initial $pOH = pK_b + \log \frac{[NH_4^+]}{[NH_3]} = 4.745 + \log \frac{0.10}{0.10} = 4.745$.
Since $pH + pOH = 14$,initial $pH = 14 - 4.745 = 9.255$.
On adding $0.05 \ mol$ of $HCl$,the reaction is: $NH_3 + H^+ \rightarrow NH_4^+$.
New moles: $NH_3 = 0.10 - 0.05 = 0.05 \ mol$,$NH_4^+ = 0.10 + 0.05 = 0.15 \ mol$.
New $pOH' = 4.745 + \log \frac{0.15}{0.05} = 4.745 + \log 3 = 4.745 + 0.477 = 5.222$.
New $pH' = 14 - 5.222 = 8.778$.
Change in $pH = |pH' - pH| = |8.778 - 9.255| = 0.477$.
$0.477 = 47.7 \times 10^{-2} \approx 48 \times 10^{-2}$.

(B) $1$. Calculate the pressure of dry $N_2$ gas: $P_{N_2} = P_{total} - P_{aqueous \ tension} = 715 \ mm \ Hg - 15 \ mm \ Hg = 700 \ mm \ Hg = \frac{700}{760} \ atm$.
$2$. Calculate the number of moles of $N_2$ using the ideal gas equation $PV = nRT$: $n_{N_2} = \frac{P_{N_2} V}{RT} = \frac{700}{760} \times \frac{50 \times 10^{-3}}{0.0821 \times 300} \approx 0.001826 \ mol$.
$3$. Calculate the mass of $N$ atoms: $Mass \ of \ N = 2 \times n_{N_2} \times 14 \ g/mol = 2 \times 0.001826 \times 14 \approx 0.05113 \ g = 51.13 \ mg$.
$4$. Calculate the percentage of $N$: $\% \ N = \frac{Mass \ of \ N}{Mass \ of \ compound} \times 100 = \frac{51.13 \ mg}{292 \ mg} \times 100 \approx 17.51 \%$.
$5$. The nearest integer is $18 \%$.

(C) The balanced chemical equation is: $C_4H_{10} + \frac{13}{2} O_2 \rightarrow 4 CO_2 + 5 H_2O$.
Molar mass of $C_4H_{10} = 4 \times 12 + 10 \times 1 = 58 \ g \ mol^{-1}$.
Molar mass of $O_2 = 2 \times 16 = 32 \ g \ mol^{-1}$.
Moles of $C_4H_{10} = \frac{174 \times 10^3 \ g}{58 \ g \ mol^{-1}} = 3000 \ mol$.
Moles of $O_2 = \frac{320 \times 10^3 \ g}{32 \ g \ mol^{-1}} = 10000 \ mol$.
According to the stoichiometry,$1 \ mol$ of $C_4H_{10}$ requires $6.5 \ mol$ of $O_2$.
For $3000 \ mol$ of $C_4H_{10}$,$O_2$ required $= 3000 \times 6.5 = 19500 \ mol$.
Since only $10000 \ mol$ of $O_2$ is available,$O_2$ is the limiting reagent.
Moles of $H_2O$ formed $= 5 \times (\frac{10000}{6.5}) = 7692.3 \ mol$.
Mass of $H_2O = 7692.3 \ mol \times 18 \ g \ mol^{-1} = 138461.5 \ g$.
Since density $= 1 \ g \ mL^{-1}$,volume $= 138461.5 \ mL = 138.46 \ L$.
The nearest integer is $138$.

(A) Step $1$: $Ph-CO-CH_3$ reacts with $PCl_5$ to form $Ph-CCl_2-CH_3$ $(A)$.
Step $2$: $A$ reacts with $3 \text{ eq. } NaNH_2/NH_3$ to undergo dehydrohalogenation to form the acetylide ion $Ph-C \equiv C^- Na^+$ $(B)$.
Step $3$: Acidification of $B$ gives phenylacetylene $Ph-C \equiv CH$ $(C)$.
Step $4$: Hydroboration-oxidation of $C$ $(1. B_2H_6, 2. H_2O_2/OH^-)$ gives an enol $Ph-CH=CH-OH$,which tautomerises to phenylacetaldehyde $Ph-CH_2-CHO$ $(D)$.
Product $D$ is $Ph-CH_2-CHO$.
The phenyl ring contains $6$ $sp^2$ carbon atoms,and the carbonyl carbon in the aldehyde group is also $sp^2$ hybridised.
Total $sp^2$ carbon atoms $= 6 + 1 = 7$.

(D) . Carbocation $\rightarrow sp^2$ hybridized carbon with empty $p$-orbital.
$B$. Carbon free radical $\rightarrow sp^2/sp^3$ hybridized carbon with one unpaired electron.
$C$. Nucleophile $\rightarrow$ Species that can supply a pair of electrons.
$D$. Electrophile $\rightarrow$ Species that can receive a pair of electrons.
Therefore,the correct matching is $A-III, B-IV, C-I, D-II$.

(B) The reaction sequence is as follows:
$1$. Dehydrohalogenation of $1, 2-$dibromocyclooctane with alc. $KOH$ followed by $NaNH_2$ yields cyclooctyne.
$2$. Hydration of cyclooctyne using $Hg^{2+} / H^{+}$ (Kucherov reaction) gives an enol intermediate,which tautomerizes to form cyclooctanone.
$3$. Clemmensen reduction of cyclooctanone using $Zn-Hg / H^{+}$ reduces the carbonyl group to a methylene group,resulting in cyclooctane as the final major product $P$.

(C) The given atomic numbers correspond to the following elements:
$32 \Rightarrow Ge$ (Germanium)
$35 \Rightarrow Br$ (Bromine)
$87 \Rightarrow Fr$ (Francium)
$19 \Rightarrow K$ (Potassium)
Ionisation enthalpy generally decreases down a group and increases across a period.
$Fr$ is an alkali metal located in the $7^{th}$ period and $1^{st}$ group.
Due to its large atomic size and the shielding effect of inner shells,$Fr$ has the lowest $1^{st}$ ionisation enthalpy among the given elements.
Therefore,the correct answer is $87$.

(D) $1$. According to $IUPAC$ nomenclature rules,the principal functional group $(-OH)$ is given the lowest possible locant.
$2$. The ring is numbered starting from the carbon attached to the $-OH$ group as $1$.
$3$. The double bond is then given the lowest possible locant,which is $2$.
$4$. The ethyl substituent is at position $4$.
$5$. Thus,the correct name is $4-$ethylcyclopent$-2-$en$-1-$ol.

(B) In the combustion of an organic compound,all $C$ in $CO_2$ and all $H$ in $H_2O$ comes from the organic compound.
Mass of $C$ in $CO_2 = \frac{12}{44} \times 0.307 \ g = 0.0837 \ g$.
Mass of $H$ in $H_2O = \frac{2}{18} \times 0.127 \ g = 0.0141 \ g$.
Percentage of $H = \frac{0.0141}{0.210} \times 100 = 6.714 \% \approx 6.72 \%$.
Mass of $O$ in the compound = Total mass - (Mass of $C$ + Mass of $H$) = $0.210 - (0.0837 + 0.0141) = 0.1122 \ g$.
Percentage of $O = \frac{0.1122}{0.210} \times 100 = 53.428 \% \approx 53.41 \%$.
Therefore,the percentages of $H$ and $O$ are $6.72 \%$ and $53.41 \%$ respectively.

(B) The element with atomic number $9$ is Fluorine $(F)$,with electronic configuration $1s^2 2s^2 2p^5$.
$(A)$ In $1s^2 2s^2 2p^5$,there are $9$ electrons. Filling them: $1s$ $(2)$,$2s$ $(2)$,$2p$ $(5)$. Total up-spins $(m_s = +\frac{1}{2})$ are $1+1+3 = 5$ and down-spins $(m_s = -\frac{1}{2})$ are $1+1+2 = 4$. Statement $A$ is correct.
$(B)$ The $2p^5$ configuration means $p_x^2 p_y^2 p_z^1$ or similar. The unpaired electron can be in any of the three $p$ orbitals,so $B$ is not necessarily true. Statement $B$ is incorrect.
$(C)$ The last electron enters the $2p$ subshell where $n=2$ and $l=1$. Statement $C$ is correct.
$(D)$ Angular nodes = $l$. For $1s$ $(l=0)$,$2s$ $(l=0)$,$2p$ $(l=1)$. Total angular nodes = $0+0+1+1+1 = 3$. Statement $D$ is incorrect.
Therefore,only $A$ and $C$ are correct.

(A) $(1)$ Carboxylic acid reacts with sodium bicarbonate solution to release $CO_2$ gas,causing effervescence.
$(2)$ Phenolic $-OH$ group reacts with neutral $FeCl_3$ to form a violet-coloured complex.
$(3)$ Alcoholic $-OH$ group reacts with ceric ammonium nitrate to produce a red-coloured complex.
$(4)$ Alkaline $KMnO_4$ (Baeyer's reagent) reacts with unsaturated compounds (alkenes or alkynes) to cause the disappearance of the purple colour,indicating unsaturation.

(A) The resonance energy is calculated as $\Delta H_{R.E.} = \Delta H_{f(exp)} - \Delta H_{f(theo)}$.
Given $\Delta H_{f(exp)} = 80 \ kJ \ mol^{-1}$.
For the theoretical enthalpy of formation,we consider the reaction: $X \equiv X(g) + \frac{1}{2} Y = Y(g) \rightarrow X=X=Y(g)$.
$\Delta H_{f(theo)} = (BE_{X \equiv X} + \frac{1}{2} BE_{Y=Y}) - (BE_{X=X} + BE_{X=Y})$.
Substituting the given values: $\Delta H_{f(theo)} = (940 + \frac{1}{2} \times 500) - (410 + 602)$.
$\Delta H_{f(theo)} = (940 + 250) - (1012) = 1190 - 1012 = 178 \ kJ \ mol^{-1}$.
Now,$\Delta H_{R.E.} = 80 - 178 = -98 \ kJ \ mol^{-1}$.
The magnitude of resonance energy is $|-98| = 98 \ kJ \ mol^{-1}$.

(B) The energy of an electron in a hydrogen-like species is given by the formula: $E_n = -13.6 \times \frac{Z^2}{n^2} \ eV$.
For the $H$ atom,the first Bohr orbit $(n=1, Z=1)$ has energy $E_1 = -13.6 \ eV$.
For the $Be^{3+}$ ion $(Z=4)$,the first excited state corresponds to $n=2$.
Substituting these values into the formula: $E = -13.6 \times \frac{4^2}{2^2} \ eV$.
$E = -13.6 \times \frac{16}{4} \ eV = -13.6 \times 4 \ eV = -54.4 \ eV$.
The magnitude of the energy is $|E| = 54.4 \ eV$.
The nearest integer value is $54$.

(C) The chemical reaction is: $NaI_{(aq)} + AgNO_{3(aq)} \rightarrow AgI_{(s)} + NaNO_{3(aq)}$
$1$. Calculate the molar mass of $AgI$: $108 + 127 = 235 \ g \ mol^{-1}$.
$2$. Calculate the moles of $AgI$ formed: $n(AgI) = \frac{4.74 \ g}{235 \ g \ mol^{-1}} \approx 0.02017 \ mol$.
$3$. From the stoichiometry,$1 \ mol$ of $NaI$ produces $1 \ mol$ of $AgI$. Therefore,moles of $NaI = 0.02017 \ mol$.
$4$. Calculate the molarity $(M)$: $M = \frac{n}{V(L)} = \frac{0.02017 \ mol}{0.020 \ L} = 1.0085 \ M$.
$5$. The nearest integer value is $1$.

(D) The reaction is $H_2O_{(g)} \rightleftharpoons H_{2(g)} + \frac{1}{2} O_{2(g)}$.
At $t = 0$,we have $1 \ mole$ of $H_2O$.
At equilibrium,the moles are: $H_2O = 1-\alpha$,$H_2 = \alpha$,$O_2 = \frac{\alpha}{2}$.
Total moles $n_T = 1 + \frac{\alpha}{2} \approx 1$ (since $\alpha \ll 1$).
The equilibrium constant $K_p$ is given by:
$K_p = \frac{P_{H_2} \cdot P_{O_2}^{1/2}}{P_{H_2O}} = \frac{(\alpha \cdot P) \cdot (\frac{\alpha}{2} \cdot P)^{1/2}}{(1-\alpha) \cdot P}$.
Given $P = 1 \ bar$,$K_p = \frac{\alpha \cdot (\alpha/2)^{1/2}}{1} = \frac{\alpha^{3/2}}{\sqrt{2}}$.
$8.0 \times 10^{-3} = \frac{\alpha^{3/2}}{\sqrt{2}}$.
$\alpha^{3/2} = 8.0 \times 10^{-3} \times \sqrt{2} \approx 11.31 \times 10^{-3}$.
$\alpha = (11.31 \times 10^{-3})^{2/3} \approx 5.04 \times 10^{-2}$.
The nearest integer value is $5$.

(B) $1$. Hydrolysis of dipeptide $X$ yields two amino acids,$Y$ and $Z$.
$2$. $Y$ reacts with aqueous $HNO_2$ to form lactic acid $(CH_3CH(OH)COOH)$. This indicates that $Y$ is alanine $(CH_3CH(NH_2)COOH)$.
$3$. $Z$ on heating forms a cyclic molecule. The structure provided is $2,5-$diketopiperazine,which is formed by the dimerization of glycine $(NH_2CH_2COOH)$. Thus,$Z$ is glycine.
$4$. Therefore,the dipeptide $X$ is $alanine-glycine$.

(B) Statement $(I) :$ When alkyl chlorides $(R-Cl)$ are treated with aqueous $KOH$,they undergo nucleophilic substitution $(S_N)$ reaction to form alcohols $(R-OH)$. This is not an elimination reaction. Thus,Statement $(I)$ is incorrect.
Statement $(II) :$ When alkyl chlorides are treated with alcoholic $KOH$,they undergo dehydrohalogenation (an elimination reaction) where the base abstracts a hydrogen atom from the $\beta-$carbon to form an alkene. Thus,Statement $(II)$ is correct.

(D) Statement $-I$: The molal depression constant $K_f$ is defined as $K_f = \frac{M_1 R T_f^2}{\Delta H_{\text {fus }}}$. Since $\Delta S_{\text {fus }} = \frac{\Delta H_{\text {fus }}}{T_f}$,we can write $K_f = \frac{M_1 R T_f}{\Delta S_{\text {fus }}}$. Thus,Statement $-I$ is correct.
Statement $-II$: The value of $K_f$ for benzene is $5.12 \ \text{K kg mol}^{-1}$ and for water is $1.86 \ \text{K kg mol}^{-1}$. Since $5.12 > 1.86$,the $K_f$ for benzene is greater than that for water. Thus,Statement $-II$ is incorrect.

(C) The reaction sequence is as follows:
$1$. $A$ is Methanol $(CH_3OH)$. In acidic medium,it forms $CH_3-OH_2^+$,which reacts with $NaCN$ followed by hydrolysis to yield Acetic acid $(CH_3COOH)$ as $B$. Acetic acid is a common food preservative (vinegar).
$2$. $B$ (Acetic acid) is reduced by diborane $(B_2H_6)$ to form Ethanol $(C_2H_5OH)$ as $C$. Ethanol is used as a fuel additive.
$3$. $C$ (Ethanol) reacts with oleum $(H_2SO_4 + SO_3)$ at $140^{\circ} C$ to undergo intermolecular dehydration,yielding Diethyl ether $(C_2H_5-O-C_2H_5)$ as $D$,which is an inhalable anesthetic.
Therefore,the correct sequence is Methanol,Acetic acid,Ethanol,Diethyl ether.

(A) To determine the number of unpaired electrons,we analyze the electronic configuration of the metal ion in each complex:
$1$. $[FeF_6]^{3-}$: $Fe^{3+}$ is $3d^5$. Since $F^-$ is a weak field ligand,electrons remain unpaired. Number of unpaired electrons = $5$.
$2$. $[CoF_6]^{3-}$: $Co^{3+}$ is $3d^6$. Since $F^-$ is a weak field ligand,it forms a high-spin complex. Number of unpaired electrons = $4$.
$3$. $[Ni(CN)_4]^{2-}$: $Ni^{2+}$ is $3d^8$. $CN^-$ is a strong field ligand,causing pairing of electrons. Number of unpaired electrons = $0$.
$4$. $[Ni(CO)_4]$: $Ni$ is $3d^8 4s^2$. $CO$ is a strong field ligand,causing pairing of all electrons in $3d$ orbital. Number of unpaired electrons = $0$.
Thus,the order is $[FeF_6]^{3-} (5) > [CoF_6]^{3-} (4) > [Ni(CN)_4]^{2-} (0) = [Ni(CO)_4] (0)$.

(A) According to the Arrhenius equation: $k = A e^{-Ea / RT}$
Taking log on both sides: $\log k = \log A - \frac{Ea}{2.303 RT}$
For a plot of $\log k$ versus $\frac{1}{T}$,the slope of the line is given by: $\text{Slope} = -\frac{Ea}{2.303 R}$
Since the slope is negative,the magnitude of the slope is directly proportional to the activation energy $(Ea)$: $|\text{Slope}| = \frac{Ea}{2.303 R}$
From the given graph,the steepness (magnitude of slope) of the lines follows the order: $(2) > (1) > (3)$
Therefore,the correct order of activation energies is: $Ea_2 > Ea_1 > Ea_3$

(C) The most stable complex ion among the given options is $[Fe(C_2O_4)_3]^{3-}$ due to the chelation effect of the oxalate ligand.
In $[Fe(C_2O_4)_3]^{3-}$,the oxidation state of $Fe$ is $+3$,which corresponds to a $d^5$ configuration.
Since $C_2O_4^{2-}$ is a weak field ligand,the $d^5$ configuration remains high-spin: $t_{2g}^3 e_g^2$.
The number of electrons in the $t_{2g}$ orbitals is $X = 3$.
Therefore,the oxide is $V_2O_3$.
$V_2O_3$ is a basic oxide.

(C) For a zero order reaction,the half-life is given by $t_{1/2} = \frac{[A]_0}{2k}$.
Given $t_{1/2} = 1 \ \text{hour} = 60 \ \text{min}$ and $[A]_0 = 2.0 \ \text{mol L}^{-1}$.
$60 \ \text{min} = \frac{2.0}{2k} \implies k = \frac{2.0}{2 \times 60} = \frac{1}{60} \ \text{mol L}^{-1} \ \text{min}^{-1}$.
For a zero order reaction,the integrated rate law is $[A]_t = [A]_0 - kt$.
To find the time $t$ for concentration to decrease from $0.50 \ \text{mol L}^{-1}$ to $0.25 \ \text{mol L}^{-1}$:
$t = \frac{[A]_0 - [A]_t}{k} = \frac{0.50 - 0.25}{1/60} = 0.25 \times 60 = 15 \ \text{min}$.

(A) Sea water is $6 \ M$ in $NaCl$,so $1000 \ mL$ of sea water contains $6 \ mol$ of $NaCl$.
$\text{Mass of solution} = \text{Volume} \times \text{density} = 1000 \times 2 = 2000 \ g$.
$\text{ppm} = \frac{\text{mass of } O_2}{\text{mass of solution}} \times 10^6$.
$5.8 = \frac{\text{mass of } O_2}{2000} \times 10^6$ $\Rightarrow \text{mass of } O_2 = 5.8 \times 2 \times 10^{-3} = 1.16 \times 10^{-2} \ g$.
$\text{Molality of } O_2 = \frac{\text{moles of } O_2}{\text{mass of solvent in } kg}$.
$\text{Mass of solvent} = \text{Mass of solution} - \text{Mass of } NaCl = 2000 - (6 \times 58.5) = 2000 - 351 = 1649 \ g = 1.649 \ kg$.
$\text{Moles of } O_2 = \frac{1.16 \times 10^{-2}}{32} = 0.03625 \times 10^{-2} = 3.625 \times 10^{-4} \ mol$.
$\text{Molality} = \frac{3.625 \times 10^{-4}}{1.649} \approx 2.197 \times 10^{-4} \ m$.
Thus,$x \approx 2$.

(C) $1$. The central atom $Br$ in $BrF_5$ has $7$ valence electrons. It forms $5$ bonds with $F$ atoms and has $1$ lone pair. Thus,$x = 1$.
$2$. Since $x = 1$,the complex $MCl_4 \cdot 3NH_3$ releases $1$ mole of $AgCl$ per mole of complex,indicating the formula is $[M(NH_3)_3Cl_3]Cl$.
$3$. The coordination sphere is $[M(NH_3)_3Cl_3]$,which is of the type $Ma_3b_3$.
$4$. Complexes of the type $Ma_3b_3$ exhibit $2$ geometrical isomers: facial (fac) and meridional (mer).

(D) Given: $\lambda_{m}^{\circ}(NH_4Cl) = 185 \ S \ cm^2 \ mol^{-1}$,$\lambda_{m}^{\circ}(OH^-) = 170 \ S \ cm^2 \ mol^{-1}$,$\lambda_{m}^{\circ}(Cl^-) = 70 \ S \ cm^2 \ mol^{-1}$.
Using Kohlrausch's law: $\lambda_{m}^{\circ}(NH_4Cl) = \lambda_{m}^{\circ}(NH_4^+) + \lambda_{m}^{\circ}(Cl^-) = 185$.
Therefore,$\lambda_{m}^{\circ}(NH_4^+) = 185 - 70 = 115 \ S \ cm^2 \ mol^{-1}$.
Now,$\lambda_{m}^{\circ}(NH_4OH) = \lambda_{m}^{\circ}(NH_4^+) + \lambda_{m}^{\circ}(OH^-) = 115 + 170 = 285 \ S \ cm^2 \ mol^{-1}$.
The degree of dissociation $(\alpha)$ is given by $\alpha = \frac{\lambda_{m}}{\lambda_{m}^{\circ}} = \frac{85.5}{285} = 0.3$.
Since $\alpha = x \times 10^{-1}$,we have $0.3 = x \times 10^{-1}$,which gives $x = 3$.

(C) The carbylamine test is a characteristic reaction for primary $(1^{\circ})$ amines.
Aliphatic or aromatic primary amines react with chloroform $(CHCl_3)$ and alcoholic potassium hydroxide $(KOH)$ to form isocyanides (carbylamines),which have a foul,pungent smell.
Secondary $(2^{\circ})$ and tertiary $(3^{\circ})$ amines do not give this test.
In the given options:
$A. CH_3CH_2NH_2$ is a primary amine.
$B. (CH_3)_2NH$ is a secondary amine.
$C. CH_3NH_2$ is a primary amine.
$D. (CH_3)_3N$ is a tertiary amine.
$E. C_6H_5NH_2$ is a primary amine (aniline).
Therefore,$A, C,$ and $E$ are primary amines and will give a positive carbylamine test.

(C) For the reaction $A_{(g)} \rightarrow 2B_{(g)} + C_{(g)}$:
At $t=0$,$P_A = P_0$ and $P_{total} = P_0$.
At $t=\infty$,all $A$ is consumed,so $P_{\infty} = 2P_0 + P_0 = 3P_0 = 240 \ mm \ Hg$,which gives $P_0 = 80 \ mm \ Hg$. Thus,option $(a)$ is correct.
At $t=10 \ min$,$P_{total} = (P_0 - x) + 2x + x = P_0 + 2x = 160 \ mm \ Hg$.
Substituting $P_0 = 80$,we get $80 + 2x = 160$,so $x = 40 \ mm \ Hg$.
Partial pressure of $A$ at $10 \ min = P_0 - x = 80 - 40 = 40 \ mm \ Hg$. Thus,option $(d)$ is correct.
Rate constant $k = \frac{2.303}{t} \log \frac{P_0}{P_A} = \frac{2.303}{10} \log \frac{80}{40} = \frac{2.303 \times 0.3010}{10} = 0.0693 \ min^{-1}$. Thus,option $(c)$ is incorrect.
First order reactions theoretically never go to completion,so option $(b)$ is correct.

(A) Statement $I$ is false because dimethyl ether is a gas at room temperature and is soluble in water,but diethyl ether is also soluble to a limited extent ($7.5 \ g$ per $100 \ mL$ of water) due to hydrogen bonding. The statement implies a contrast that is not strictly accurate in the context of solubility trends.
Statement $II$ is true because sodium metal reacts with ethyl alcohol $(C_2H_5OH)$ to produce sodium ethoxide and hydrogen gas $(H_2)$,making it unsuitable for drying. Sodium does not react with diethyl ether $(C_2H_5OC_2H_5)$,so it can be used to remove traces of moisture from it.

(C) Mohr's salt is $FeSO_4 \cdot (NH_4)_2SO_4 \cdot 6H_2O$.
It dissociates into $Fe^{2+}$,$2NH_4^+$,and $2SO_4^{2-}$ ions. Thus,Statement $I$ is true.
According to Kohlrausch's law of independent migration of ions,the molar conductance at infinite dilution is the sum of the molar conductances of the constituent ions multiplied by their respective stoichiometric coefficients.
$\lambda_{m}^{\infty} (FeSO_4 \cdot (NH_4)_2SO_4 \cdot 6H_2O) = \lambda^{\infty}(Fe^{2+}) + 2\lambda^{\infty}(NH_4^+) + 2\lambda^{\infty}(SO_4^{2-}) = x_1 + 2x_2 + 2x_3$.
Since the given expression in Statement $II$ is $x_1 + x_2 + 2x_3$,Statement $II$ is false.

(C) The enthalpy of atomisation depends on the number of unpaired electrons available for metallic bonding. Among the given transition metals ($Cr$,$Co$,$Fe$,$Ni$),$Cr$ has the lowest enthalpy of atomisation because it has a stable $d^5$ configuration,which limits the number of electrons participating in metallic bonding compared to others.
For $Cr$ $(Z=24)$,the electronic configuration is $[Ar] 3d^5 4s^1$.
The total number of valence electrons is the sum of electrons in the outermost $s$ and $d$ orbitals.
Valence electrons = $5 (d) + 1 (s) = 6$.

(B) When an ammonium salt is treated with sodium hydroxide $(NaOH)$,it releases ammonia gas $(NH_3)$:
$NH_4^{+} + OH^{-} \longrightarrow NH_3 \uparrow + H_2O$
Ammonia gas $(NH_3)$ reacts with Nessler's reagent $(K_2[HgI_4] + KOH)$ to form a brown-coloured precipitate known as the iodide of Millon's base:
$2[HgI_4]^{2-} + NH_3 + 3OH^{-} \longrightarrow HgO \cdot Hg(NH_2)I + 7I^{-} + 2H_2O$
Thus,$X = NH_3$ and $Y = K_2HgI_4 + KOH$.

(C) In the $3d$ transition series,Vanadium $(V)$ has the highest enthalpy of atomisation.
The aquated ion $V^{3+}$ exists in a green colour.
The oxide formed by $V^{3+}$ is $V_2O_3$.
$V_2O_3$ is a basic oxide.

(D) The reaction with aqueous $NaHCO_3$ to release $CO_2$ gas occurs only if the compound is more acidic than carbonic acid ($H_2CO_3$,$pK_a \approx 6.35$).
$1$. $2,4,6$-Trinitrophenol (Picric acid) is a very strong acid $(pK_a \approx 0.38)$ and readily reacts with $NaHCO_3$.
$2$. $4$-Nitrobenzoic acid is a strong organic acid $(pK_a \approx 3.4)$ and reacts with $NaHCO_3$.
$3$. $PhNH_3^+ Cl^-$ (Anilinium chloride) is the salt of a weak base and a strong acid. In aqueous solution,it exists as $PhNH_3^+$ and $Cl^-$. $PhNH_3^+$ is a weak acid $(pK_a \approx 4.6)$,which is stronger than $H_2CO_3$,so it can react with $NaHCO_3$ to release $CO_2$.
$4$. $3$-Nitrophenol is a weak acid $(pK_a \approx 8.39)$,which is less acidic than $H_2CO_3$ $(pK_a \approx 6.35)$. Therefore,it does not react with $NaHCO_3$ to give effervescence of $CO_2$.

(B) *Before applying medicine:
$\frac{dN}{dt} = KN$ (First order growth kinetics)
Integrating this gives $\ln(\frac{N}{N_0}) = Kt$, which implies $\frac{N}{N_0} = e^{Kt}$. This is an exponential growth curve starting from $1$ at $t=0$.
*After applying medicine:
The rate of decay $r$ is proportional to the square of the number of bacteria:
$r = -\frac{dN}{dt} = KN^2$
This represents a parabolic relationship where $r$ increases quadratically with $N$ $(r \propto N^2)$.
Comparing the options, option $B$ shows the correct exponential growth for the 'before' case and a parabolic curve $(r \propto N^2)$ for the 'after' case.

(B) Sucrose on hydrolysis yields $D-(+)-\text{glucose}$ and $D-(-)-\text{fructose}$.
Statement $I$ is incorrect because it incorrectly lists $D-(+)-\text{fructose}$ instead of $D-(-)-\text{fructose}$.
Statement $II$ is correct because the equimolar mixture of $D-(+)-\text{glucose}$ and $D-(-)-\text{fructose}$ obtained from sucrose hydrolysis is known as invert sugar.

(C) The complex $A$ reacts with $AgNO_3$ to give a white precipitate,which indicates the presence of free $Cl^-$ ions outside the coordination sphere. Thus,$A$ is $[Co(NH_3)_5(SO_4)]Cl$.
The complex $B$ reacts with $BaCl_2$ to give a white precipitate,which indicates the presence of free $SO_4^{2-}$ ions outside the coordination sphere. Thus,$B$ is $[Co(NH_3)_5Cl]SO_4$.
Since the isomers differ in the ions they provide in solution,this is an example of Ionisation isomerism.

(D) Let us analyze each reaction:
$A.$ $2$-bromomethylcyclohexane with $NaOEt$ undergoes $E2$ elimination to form an alkene.
$B.$ $2$-bromobutane with aqueous $KOH$ undergoes nucleophilic substitution ($SN^2$ or $SN^1$) to form butan-$2$-ol,not an alkene.
$C.$ $2$-bromo-$2$-methylpropane with $NaOMe$ undergoes $E2$ elimination to form $2$-methylpropene (an alkene).
$D.$ Phenol with $Na_2Cr_2O_7 / H_2SO_4$ undergoes oxidation to form $p$-benzoquinone,not an alkene.
$E.$ $2$-methylpropan-$2$-ol with $Cu$ at $573 \ K$ undergoes dehydration to form $2$-methylpropene (an alkene).
Thus,reactions $B$ and $D$ do not produce an alkene.

(B) The molecular formula of thyroxine is $C_{15}H_{11}O_4NI_4$.
Calculate the molecular mass of thyroxine:
$C: 15 \times 12 = 180$
$H: 11 \times 1 = 11$
$O: 16 \times 4 = 64$
$N: 14 \times 1 = 14$
$I: 127 \times 4 = 508$
Total molecular mass $= 180 + 11 + 64 + 14 + 508 = 777 \ g \ mol^{-1}$.
Percentage of iodine $= \frac{\text{Mass of Iodine}}{\text{Total Molecular Mass}} \times 100$
Percentage of iodine $= \frac{508}{777} \times 100 \approx 65.38 \%$.
The nearest integer is $65$.

(A) $1$. For $Cu^{2+} + 2e^{-} \longrightarrow Cu$:
Initial moles of $Cu^{2+} = 1.5 \ M \times 1 \ L = 1.5 \ mol$.
$1 \ F$ electricity corresponds to $1 \ mol$ of electrons.
Since $2 \ mol$ of electrons reduce $1 \ mol$ of $Cu^{2+}$,$1 \ mol$ of electrons reduces $0.5 \ mol$ of $Cu^{2+}$.
Remaining moles of $Cu^{2+} = 1.5 - 0.5 = 1.0 \ mol$.
$[Cu^{2+}] = 1.0 \ M$.
$2$. For $Ag^{+} + e^{-} \longrightarrow Ag$:
Initial moles of $Ag^{+} = 0.2 \ M \times 1 \ L = 0.2 \ mol$.
$0.1 \ F$ electricity corresponds to $0.1 \ mol$ of electrons.
$1 \ mol$ of electrons reduces $1 \ mol$ of $Ag^{+}$,so $0.1 \ mol$ of electrons reduces $0.1 \ mol$ of $Ag^{+}$.
Remaining moles of $Ag^{+} = 0.2 - 0.1 = 0.1 \ mol$.
$[Ag^{+}] = 0.1 \ M$.
$3$. Cell reaction: $Cu(s) + 2Ag^{+}(aq) \longrightarrow Cu^{2+}(aq) + 2Ag(s)$.
$E^{\circ}_{cell} = E^{\circ}_{cathode} - E^{\circ}_{anode} = 0.8 - 0.34 = 0.46 \ V$.
Using Nernst equation: $E = E^{\circ}_{cell} - \frac{0.06}{n} \log \frac{[Cu^{2+}]}{[Ag^{+}]^2}$.
$E = 0.46 - \frac{0.06}{2} \log \frac{1}{(0.1)^2} = 0.46 - 0.03 \log(100) = 0.46 - 0.03 \times 2 = 0.46 - 0.06 = 0.40 \ V$.

(D) The dissociation reaction for the salt is: $MX_3 \rightarrow M^{3+} + 3X^{-}$.
Here,the number of ions produced per formula unit is $n = 1 + 3 = 4$.
The relationship between the van't Hoff factor $(i)$ and the degree of dissociation $(\alpha)$ is given by: $i = 1 + (n - 1)\alpha$.
Substituting the given values: $2 = 1 + (4 - 1)\alpha$.
$2 = 1 + 3\alpha$.
$3\alpha = 1$.
$\alpha = \frac{1}{3} \approx 0.3333$.
Therefore,the percentage dissociation is $0.3333 \times 100 = 33.33\%$.
The nearest integer is $33\%$.

(A) To determine the hybridization and magnetic nature,we analyze the oxidation state of the metal and the nature of the ligand:

Complex	Hybridization	Magnetic Nature
$[FeF_6]^{3-}$	$sp^3d^2$	Paramagnetic
$[Fe(CN)_6]^{3-}$	$d^2sp^3$	Paramagnetic
$[Mn(CN)_6]^{3-}$	$d^2sp^3$	Paramagnetic
$[Co(C_2O_4)_3]^{3-}$	$d^2sp^3$	Diamagnetic
$[MnCl_6]^{3-}$	$sp^3d^2$	Paramagnetic
$[CoF_6]^{3-}$	$sp^3d^2$	Paramagnetic

From the table,the complexes that involve $d^2sp^3$ hybridization are $[Fe(CN)_6]^{3-}, [Mn(CN)_6]^{3-}$,and $[Co(C_2O_4)_3]^{3-}$.
Among these,$[Fe(CN)_6]^{3-}$ and $[Mn(CN)_6]^{3-}$ are paramagnetic.
Therefore,the number of such complexes is $2$.

(C) $(i)$ Proteins on hydrolysis yield a large number of $\alpha$-amino acids,not $\beta$-amino acids. Thus,Statement $(I)$ is incorrect.
$(ii)$ Fibrous proteins are generally insoluble in water. Denaturation involves the loss of biological activity and changes in the secondary and tertiary structures,but it does not involve the conversion of water-soluble fibrous proteins to water-insoluble forms in the manner described. Thus,Statement $(II)$ is also incorrect.

(D) When $NaOH$ is added to the mixture,benzoic acid reacts with $NaOH$ to form sodium benzoate,which is a salt and is soluble in the aqueous layer.
$C_6H_5COOH + NaOH \rightarrow C_6H_5COONa + H_2O$
Chlorobenzene and aniline do not react with $NaOH$ and remain in the organic ethyl acetate layer.
Therefore,the ethyl acetate layer contains chlorobenzene and aniline.

(C) For the first order reaction $A_{(g)} \rightarrow B_{(g)} + C_{(g)}$:
At $t = 0$,$P_A = P_0$,$P_B = 0$,$P_C = 0$
At $t = t$,$P_A = P_0 - x$,$P_B = x$,$P_C = x$
Total pressure $P_t = (P_0 - x) + x + x = P_0 + x \Rightarrow x = P_t - P_0$
At $t = \infty$,$P_A = 0$,$P_B = P_0$,$P_C = P_0$
Total pressure $P_{\infty} = 2P_0 \Rightarrow P_0 = \frac{P_{\infty}}{2}$
Pressure of $A$ at time $t$: $P_A = P_0 - x = P_0 - (P_t - P_0) = 2P_0 - P_t$
Substituting $P_0 = \frac{P_{\infty}}{2}$: $P_A = 2(\frac{P_{\infty}}{2}) - P_t = P_{\infty} - P_t$
Rate constant $k = \frac{1}{t} \ln \frac{P_0}{P_A} = \frac{1}{t} \ln \frac{P_{\infty}/2}{P_{\infty} - P_t} = \frac{1}{t} \ln \frac{P_{\infty}}{2(P_{\infty} - P_t)}$.

(B) $1$. The molecular formula $C_6H_{12}O$ corresponds to the general formula $C_nH_{2n}O$,which indicates that the compound $P$ is either an aldehyde or a ketone.
$2$. $P$ gives a positive test with $2,4-$dinitrophenylhydrazine $(2,4-DNP)$,which confirms the presence of a carbonyl group $(C=O)$.
$3$. $P$ gives a negative test with Tollens reagent,which indicates that $P$ is a ketone,not an aldehyde.
$4$. $P$ is optically active,meaning it must contain at least one chiral carbon atom.
$5$. Let's examine the options:
- Option $A$: $CH_3-CO-CH_2-CH_2-CH_2-CH_3$ (Hexan-$2$-one) is not optically active.
- Option $B$: $CH_3-CO-CH(CH_3)-CH_2-CH_3$ ($3-$methylpentan-$2$-one) has a chiral carbon at the $C-3$ position. It is a ketone and gives a positive $2,4-DNP$ test and a negative Tollens test.
- Option $C$: This is an aldehyde,which would give a positive Tollens test.
- Option $D$: $CH_3-CO-CH_2-CH(CH_3)_2$ ($4-$methylpentan-$2$-one) is not optically active.
$6$. Therefore,the correct structure is $3-$methylpentan-$2$-one.

(A) The nucleophilic aromatic substitution reaction of aryl halides with $NaOH$ depends on the presence of electron-withdrawing groups $(-NO_2)$ on the ring.
$A$. Chlorobenzene requires harsh conditions: $(a)$ $NaOH, 623 \ K, 300 \ atm$; $(b)$ $H_3O^+$ $(IV)$.
$B$. $1$-Chloro-$4$-nitrobenzene has one $-NO_2$ group,requiring milder conditions: $(a)$ $NaOH, 443 \ K$; $(b)$ $H_3O^+$ $(III)$.
$C$. $1$-Chloro-$2,4$-dinitrobenzene has two $-NO_2$ groups,requiring even milder conditions: $(a)$ $NaOH, 368 \ K$; $(b)$ $H_3O^+$ $(II)$.
$D$. $1$-Chloro-$2,4,6$-trinitrobenzene has three $-NO_2$ groups,which activate the ring significantly,allowing substitution with just warm $H_2O$ $(I)$.
Thus,the correct match is $A-IV, B-III, C-II, D-I$.

(C) According to Raoult's law,the partial pressures are $P_A = x_A P_A^o$ and $P_B = x_B P_B^o$.
In the vapour phase,$P_A = y_A P_{total}$ and $P_B = y_B P_{total}$.
Therefore,$\frac{y_A}{y_B} = \frac{P_A}{P_B} = \frac{x_A P_A^o}{x_B P_B^o} = \left(\frac{P_A^o}{P_B^o}\right) \left(\frac{x_A}{x_B}\right)$.
Given $P_A^o = 350 \ mm \ Hg$ and $P_B^o = 750 \ mm \ Hg$,we have $\frac{P_A^o}{P_B^o} = \frac{350}{750} < 1$.
Since $\frac{P_A^o}{P_B^o} < 1$,it follows that $\frac{y_A}{y_B} < \frac{x_A}{x_B}$,which is equivalent to $\frac{x_A}{x_B} > \frac{y_A}{y_B}$.

(A) $1$. Identification of acidic oxides:
$CrO_3$ and $Mn_2O_7$ are acidic oxides.
$VO_2$ is amphoteric,$V_2O_3$ is basic,and $V_2O_5$ is amphoteric.
Therefore,$X = 2$.
$2$. Determination of primary valency:
In the complex $[Co(en)_3]_2(SO_4)_3$,the coordination entity is $[Co(en)_3]^{3+}$.
The oxidation state of $Co$ is $+3$,which represents its primary valency.
Therefore,$Y = 3$.
$3$. Calculation:
$X + Y = 2 + 3 = 5$.

(B) To determine the basicity,we consider the availability of the lone pair on the nitrogen atom.
$1$. Aliphatic amines are generally more basic than aromatic amines because the lone pair on the nitrogen in aromatic amines is delocalized into the benzene ring.
$2$. Among aliphatic amines,$(CH_3)_2NH$ $(E)$ is more basic than $CH_3NH_2$ $(D)$ due to the greater $+I$ effect of two methyl groups.
$3$. Among aromatic amines,the basicity depends on the substituents on the benzene ring. An electron-donating group (like $-OCH_3$) increases basicity,while an electron-withdrawing group (like $-NO_2$) decreases it.
$4$. Thus,$p$-methoxyaniline $(B)$ > aniline $(A)$ > $p$-nitroaniline $(C)$.
$5$. Combining these,the order is: $(CH_3)_2NH$ $(E)$ > $CH_3NH_2$ $(D)$ > $p$-methoxyaniline $(B)$ > aniline $(A)$ > $p$-nitroaniline $(C)$.
Therefore,the correct order is $E > D > B > A > C$.

(A) Primary valency is equal to the oxidation state of the central metal atom.
Secondary valency is equal to the coordination number of the central metal atom.
$1$. $[Co(en)_2 Cl_2] Cl$: $Co$ is in $+3$ state,$CN = 2(2) + 2 = 6$. So,$(A-I)$.
$2$. $[Pt(NH_3)_2 Cl(NO_2)]$: $Pt$ is in $+2$ state,$CN = 2 + 1 + 1 = 4$. So,$(B-II)$.
$3$. $Hg[Co(SCN)_4]$: $Co$ is in $+3$ state (considering $Hg^{2+}$),$CN = 4$. So,$(C-III)$.
$4$. $[Mg(EDTA)]^{2-}$: $Mg$ is in $+2$ state,$EDTA$ is hexadentate,$CN = 6$. So,$(D-IV)$.
Thus,the correct match is $A-I, B-II, C-III, D-IV$.

(A) . The solution of chloroform and acetone shows negative deviation from Raoult's law,resulting in a maximum boiling azeotrope $(A-III)$.
$B$. The solution of ethanol and water shows positive deviation from Raoult's law,resulting in a minimum boiling azeotrope $(B-I)$.
$C$. The solution of benzene and toluene forms an ideal solution,which follows $\Delta V_{mix} = 0$ $(C-IV)$.
$D$. Acetic acid in benzene undergoes hydrogen bonding to form a dimer $(D-II)$.
Therefore,the correct matching is $A-III, B-I, C-IV, D-II$.

(A) $[Fe(CN)_6]^{3-}$: $Fe^{3+}$ is $3d^5$. $CN^-$ is a strong field ligand,so pairing occurs. $t_{2g}^5, e_g^0$. Unpaired electrons = $1$.
$[FeF_6]^{3-}$: $Fe^{3+}$ is $3d^5$. $F^-$ is a weak field ligand,so no pairing. $t_{2g}^3, e_g^2$. Unpaired electrons = $5$.
$[CoF_6]^{3-}$: $Co^{3+}$ is $3d^6$. $F^-$ is a weak field ligand. $t_{2g}^4, e_g^2$. Unpaired electrons = $4$.
$[Mn(CN)_6]^{3-}$: $Mn^{3+}$ is $3d^4$. $CN^-$ is a strong field ligand. $t_{2g}^4, e_g^0$. Unpaired electrons = $2$.

(A) Statement $(I)$ is correct. $cis-1,2-dichloroethene$ exhibits greater polarity than $cis-1,2-dibromoethene$ due to the stronger $p\pi-d\pi$ back-bonding in the $C-Cl$ bond compared to the $C-Br$ bond,which enhances the dipole moment.
Statement $(II)$ is incorrect. While the boiling point of $trans-1,2-dibromoethene$ is indeed lower than that of $cis-1,2-dibromoethene$ (because the $cis$ isomer has a net dipole moment $\mu \neq 0$ while the $trans$ isomer has $\mu = 0$),the $trans$ isomer is $NOT$ more polar than the $cis$ isomer. The $trans$ isomer is non-polar $(\mu = 0)$,whereas the $cis$ isomer is polar $(\mu \neq 0)$.

(B) Statement $(I)$ is correct: Metals with higher standard reduction potential $(E^{\theta})$ are deposited first at the cathode. The order of $E^{\theta}$ values is $Ag^{+} (0.80 \ V) > Hg_2^{2+} (0.79 \ V) > Cu^{2+} (0.24 \ V) > Mg^{2+} (-2.37 \ V)$. Thus,the sequence of deposition is $Ag, Hg, Cu$.
Statement $(II)$ is incorrect: At the cathode,the reduction of water $(2H_2O + 2e^- \rightarrow H_2 + 2OH^-)$ occurs instead of $Mg^{2+}$ reduction because $E^{\theta}_{H_2O/H_2} > E^{\theta}_{Mg^{2+}/Mg}$. Therefore,$H_2$ gas is evolved at the cathode,not oxygen gas (oxygen gas is evolved at the anode).

(D) Let us analyze the number of unpaired electrons $(n)$ for each complex:
$1.$ $[Co(NH_3)_6]^{3+}: Co^{3+} (3d^6)$,$\text{NH}_3$ is a strong field ligand $(SFL)$ $\rightarrow t_{2g}^6 e_g^0, n=0$ (Diamagnetic)
$2.$ $[Co(C_2O_4)_3]^{3-}: Co^{3+} (3d^6)$,$\text{C}_2\text{O}_4^{2-}$ is a $SFL$ with $Co^{3+} \rightarrow t_{2g}^6 e_g^0, n=0$ (Diamagnetic)
$3.$ $[MnCl_6]^{3-}: Mn^{3+} (3d^4)$,$\text{Cl}^{-}$ is a weak field ligand $(WFL)$ $\rightarrow t_{2g}^3 e_g^1, n=4$ (Paramagnetic)
$4.$ $[Mn(CN)_6]^{3-}: Mn^{3+} (3d^4)$,$\text{CN}^{-}$ is a $SFL$ $\rightarrow t_{2g}^4 e_g^0, n=2$ (Paramagnetic)
$5.$ $[CoF_6]^{3-}: Co^{3+} (3d^6)$,$\text{F}^{-}$ is a $WFL$ $\rightarrow t_{2g}^4 e_g^2, n=4$ (Paramagnetic)
$6.$ $[Fe(CN)_6]^{3-}: Fe^{3+} (3d^5)$,$\text{CN}^{-}$ is a $SFL$ $\rightarrow t_{2g}^5 e_g^0, n=1$ (Paramagnetic)
$7.$ $[FeF_6]^{3-}: Fe^{3+} (3d^5)$,$\text{F}^{-}$ is a $WFL$ $\rightarrow t_{2g}^3 e_g^2, n=5$ (Paramagnetic)
The paramagnetic species with the same number of unpaired electrons $(n=4)$ are $[MnCl_6]^{3-}$ and $[CoF_6]^{3-}$.
Therefore,the number of such species is $2$.

(D) For a first order reaction,the time taken $t$ to reach a concentration $C_t$ from initial concentration $C_0$ is given by $t = \frac{2.303}{k} \log(\frac{C_0}{C_t})$.
For $C_t = C_0 / 4$,$t_1 = \frac{2.303}{k} \log(\frac{C_0}{C_0/4}) = \frac{2.303}{k} \log(4) = \frac{2.303}{k} \times 2 \log(2)$.
For $C_t = C_0 / 8$,$t_2 = \frac{2.303}{k} \log(\frac{C_0}{C_0/8}) = \frac{2.303}{k} \log(8) = \frac{2.303}{k} \times 3 \log(2)$.
Therefore,the ratio $\frac{t_1}{t_2} = \frac{2 \log(2)}{3 \log(2)} = \frac{2}{3}$.

(A) $1$. Molar mass of $A$ is $121 \ g/mol$. Benzamide $(C_6H_5CONH_2)$ has a molar mass of $121 \ g/mol$ and contains nitrogen,thus giving a positive Lassaigne's test.
$2$. The reaction of benzamide with $NaOH$ followed by $H_3O^+$ is an alkaline hydrolysis,which converts the amide group $(-CONH_2)$ into a carboxylic acid group $(-COOH)$,yielding benzoic acid $(B)$.
$3$. Benzoic acid $(B)$ reacts with $NaHCO_3$ to release $CO_2$ gas,causing effervescence.
$4$. The reaction of benzoic acid $(B)$ with ethanol $(EtOH)$ in the presence of $H_2SO_4$ and heat is an esterification reaction,which produces ethyl benzoate $(C)$,characterized by a fruity smell.

(A) The conversion of ethylbenzene to $p$-bromostyrene involves three steps:
$1$. Electrophilic aromatic substitution: Treatment of ethylbenzene with $Br_2 / Fe$ (or $FeBr_3$) leads to bromination at the para position due to the ortho/para-directing nature of the ethyl group.
$2$. Free radical halogenation: Treatment with $Cl_2, \Delta$ (or $h\nu$) results in the chlorination of the benzylic position of the ethyl group.
$3$. Dehydrohalogenation: Treatment with alcoholic $KOH$ (alc. $KOH$) causes the elimination of $HCl$ from the side chain to form the vinyl group (styrene derivative).
Thus,the correct sequence is $Br_2 / Fe ; Cl_2, \Delta ; \text{alc. } KOH$.

(D) Statement $I$ is true: $A$ homoleptic octahedral complex with only one type of monodentate ligand (e.g.,$[Ma_6]$) does not exhibit stereoisomerism.
Statement $II$ is false: $cis-$platin and $trans-$platin have the formula $[Pt(NH_3)_2Cl_2]$. These are heteroleptic complexes of $Pt$ (Platinum),not $Pd$ (Palladium).
Therefore,Statement $I$ is true but Statement $II$ is false.

(B) binary mixture of $C_6H_5OH$ and $C_6H_5NH_2$ shows a negative deviation from Raoult's law.
This occurs because the intermolecular hydrogen bonding between phenol and aniline is stronger than the interactions in the pure components.
As a result,the vapour pressure of the solution is lower than that of the pure components,which leads to a higher boiling point for the solution compared to the pure components.
Therefore,this mixture forms a maximum boiling azeotrope,not a minimum boiling azeotrope.

(A) The formula for freezing point depression is $\Delta T_{f} = i K_{f} m$.
Given $\Delta T_{f} = 0.20^{\circ} C$,$K_{f} = 1.8 \ K \ kg \ mol^{-1}$,and $m = 0.1 \ m$.
$0.20 = i \times 1.8 \times 0.1$
$i = \frac{0.20}{0.18} = \frac{10}{9}$.
For the dissociation $HA \rightleftharpoons H^{+} + A^{-}$,the van't Hoff factor $i = 1 + \alpha$,where $\alpha$ is the degree of dissociation.
$1 + \alpha = \frac{10}{9} \implies \alpha = \frac{1}{9}$.
The dissociation constant $K_{a} = \frac{C \alpha^2}{1 - \alpha}$.
Substituting $C = 0.1$ and $\alpha = \frac{1}{9}$:
$K_{a} = \frac{0.1 \times (1/9)^2}{1 - 1/9} = \frac{0.1 \times (1/81)}{8/9} = \frac{0.1}{81} \times \frac{9}{8} = \frac{0.1}{72} = \frac{1}{720} \approx 1.38 \times 10^{-3}$.

(C) The spin only magnetic moment is calculated using the formula $\mu = \sqrt{n(n+2)} \ BM$,where $n$ is the number of unpaired electrons.
$Cu^{+} : [Ar] 3d^{10}$,$n = 0$,$\mu = 0 \ BM$.
$Cu^{2+} : [Ar] 3d^{9}$,$n = 1$,$\mu = \sqrt{1(1+2)} = \sqrt{3} \approx 1.73 \ BM$.
$Cr^{2+} : [Ar] 3d^{4}$,$n = 4$,$\mu = \sqrt{4(4+2)} = \sqrt{24} \approx 4.90 \ BM$.
$Cr^{3+} : [Ar] 3d^{3}$,$n = 3$,$\mu = \sqrt{3(3+2)} = \sqrt{15} \approx 3.87 \ BM$.
Comparing the values: $4.90 > 3.87 > 1.73 > 0$.
Therefore,the decreasing order is $Cr^{2+} > Cr^{3+} > Cu^{2+} > Cu^{+}$.

(D) The Williamson ether synthesis involves the reaction of an alkoxide ion with an alkyl halide. For the reaction to proceed via $S_N2$ mechanism to form an ether,the alkyl halide should ideally be primary. If the alkyl halide is secondary or tertiary,the strong base (alkoxide) will prefer to undergo an $E2$ elimination reaction rather than $S_N2$ substitution.
In option $A$,${}^{t}BuO^{\ominus}$ is a bulky base and $EtBr$ is a primary halide,so $S_N2$ is possible,but often elimination competes. However,in option $D$,$iso-BuO^{\ominus}$ is a base reacting with $sec-BuBr$ (a secondary alkyl halide). Due to steric hindrance and the nature of the secondary halide,the $E2$ elimination reaction is highly favored over $S_N2$ substitution,leading to alkene formation as the major product instead of the desired ether.

(A) Amino acids are amphoteric in nature. In an acidic medium (low $pH$),the amino group $(-NH_2)$ gets protonated to form $-NH_3^+$.
Thus,at $pH = 2$,the structure $A$ is $H_3N^+-CH(CH(CH_3)_2)-COOH$.
In a basic medium (high $pH$),the carboxylic acid group $(-COOH)$ gets deprotonated to form $-COO^-$.
Thus,at $pH = 10$,the structure $B$ is $H_2N-CH(CH(CH_3)_2)-COO^-$.
Therefore,the correct option is $A$.