AIEEE 2012 Chemistry Question Paper with Answer and Solution

(A) Let the solubility of $AgCl$ be $x \ mol \ L^{-1}$.
The dissociation equilibrium is: $AgCl(s) \rightleftharpoons Ag^{+}(aq) + Cl^{-}(aq)$.
The solubility product expression is: $K_{sp} = [Ag^{+}][Cl^{-}]$.
In a $0.1 \ M \ KCl$ solution,$KCl$ dissociates completely: $KCl(aq) \rightarrow K^{+}(aq) + Cl^{-}(aq)$.
Thus,the concentration of $Cl^{-}$ from $KCl$ is $0.1 \ M$.
The total concentration of $Cl^{-}$ is $[Cl^{-}] = (x + 0.1) \ M \approx 0.1 \ M$ (since $x$ is very small).
Substituting the values into the $K_{sp}$ expression: $1.0 \times 10^{-10} = (x)(0.1)$.
Solving for $x$: $x = \frac{1.0 \times 10^{-10}}{0.1} = 1.0 \times 10^{-9} \ mol \ L^{-1}$.

(B) Water in the solid phase (ice) is best described as a molecular solid.
In ice,$H_2O$ molecules are held together by hydrogen bonding,which is a type of intermolecular force characteristic of molecular solids.

(B) For reaction $(I): N_2 + 2O_2 \rightleftharpoons 2NO_2$,the equilibrium constant is $K_1 = \frac{[NO_2]^2}{[N_2][O_2]^2} \dots (i)$
For reaction $(II): 2NO_2 \rightleftharpoons N_2 + 2O_2$,which is the reverse of $(I)$,the equilibrium constant is $K_2 = \frac{[N_2][O_2]^2}{[NO_2]^2} = \frac{1}{K_1} \dots (ii)$
For reaction $(III): NO_2 \rightleftharpoons \frac{1}{2} N_2 + O_2$,the equilibrium constant is $K_3 = \frac{[N_2]^{1/2}[O_2]}{[NO_2]}$
Squaring $K_3$,we get $(K_3)^2 = \frac{[N_2][O_2]^2}{[NO_2]^2} = K_2 = \frac{1}{K_1} \dots (iii)$
Thus,$K_1 = \frac{1}{K_2} = \frac{1}{(K_3)^2}$.

(A) $95\%\, H_2SO_4$ by weight means $100\, g$ of $H_2SO_4$ solution contains $95\, g$ of $H_2SO_4$ by mass.
Molar mass of $H_2SO_4 = 98\, g\, mol^{-1}$.
Moles of $H_2SO_4 = \frac{95\, g}{98\, g\, mol^{-1}} = 0.969\, mol$.
Volume of $100\, g$ of $H_2SO_4$ solution $= \frac{\text{mass}}{\text{density}} = \frac{100\, g}{1.834\, g\, cm^{-3}} = 54.52\, cm^3 = 0.05452\, L$.
Molarity $(M) = \frac{\text{Moles of solute}}{\text{Volume of solution in } L} = \frac{0.969\, mol}{0.05452\, L} \approx 17.8\, M$.

(B) The molecular formula $C_4H_6$ corresponds to a degree of unsaturation (double bond equivalent) of $2$.
For cyclic structures,this means the molecule must contain either two double bonds,one triple bond,or two rings,or one ring and one double bond.
The possible cyclic structures for $C_4H_6$ are:
$1$. Cyclobutene
$2$. $1$-methylcyclopropene
$3$. $3$-methylcyclopropene (often referred to as $2$-methylcyclopropene in some contexts)
$4$. Methylenecyclopropane
$5$. Bicyclo$[1.1.0]$butane
Thus,there are $5$ possible cyclic structures.

(D) The given compound is a three-membered cyclic ether containing an oxygen atom.
According to $IUPAC$ nomenclature rules for epoxides,the oxygen atom is treated as an epoxy substituent on the alkane chain.
The parent chain is propane ($3$ carbons).
The oxygen atom is bonded to carbon $1$ and carbon $2$.
Therefore,the correct $IUPAC$ name is $1, 2-$ epoxy propane.

(B) The Global Warming Potential $(GWP)$ is a measure of how much heat a greenhouse gas traps in the atmosphere relative to carbon dioxide $(CO_2)$.
Comparing the $GWP$ values (over a $100$-year horizon):
$CFCs$ (Chlorofluorocarbons) have a very high $GWP$ (ranging from thousands to tens of thousands).
$N_2O$ (Nitrous oxide) has a $GWP$ of approximately $273$.
$CH_4$ (Methane) has a $GWP$ of approximately $28$.
$CO_2$ (Carbon dioxide) is the reference gas with a $GWP$ of $1$.
Therefore,the correct sequence is $CFC > N_2O > CH_4 > CO_2$.

(D) For $16.0 \ g$ $O_3$: Molar mass $= 48 \ g/mol$. Moles $= 16/48 = 1/3 \ mol$. Number of $O$ atoms $= 3 \times (1/3) \times N_A = N_A$.
For $28.0 \ g$ $CO$: Molar mass $= 28 \ g/mol$. Moles $= 28/28 = 1 \ mol$. Number of $O$ atoms $= 1 \times 1 \times N_A = N_A$.
For $16.0 \ g$ $O_2$: Molar mass $= 32 \ g/mol$. Moles $= 16/32 = 0.5 \ mol$. Number of $O$ atoms $= 2 \times 0.5 \times N_A = N_A$.
Thus,the ratio of oxygen atoms is $N_A : N_A : N_A$,which simplifies to $1 : 1 : 1$.

(B) In $BF_3$,the boron atom is $sp^2$ hybridized and possesses an empty $p_z$ orbital.
When $BF_3$ reacts with $NH_3$,the lone pair of electrons from the nitrogen atom in $NH_3$ is donated into the empty $p_z$ orbital of boron.
This coordinate covalent bond formation changes the geometry of boron from trigonal planar to tetrahedral.
Consequently,the hybridization of boron changes from $sp^2$ to $sp^3$.

(C) The reaction of metal $M$ (specifically $Mg$) with nitrogen gas is: $3Mg + N_2 \to Mg_3N_2$ $(Y)$.
$Mg_3N_2$ $(Y)$ reacts with water to produce ammonia gas: $Mg_3N_2 + 6H_2O \to 3Mg(OH)_2 + 2NH_3 \uparrow$ (colourless gas).
When $NH_3$ is passed through $CuSO_4$ solution,it forms a deep blue complex: $CuSO_4 + 4NH_3 \to [Cu(NH_3)_4]SO_4$ (Blue complex).
Therefore,$Y$ is $Mg_3N_2$.

(A) The $BOD$ (Biological Oxygen Demand) value is a measure of the amount of dissolved oxygen required by bacteria to decompose organic matter in a water sample.
Clean water has a $BOD$ value of less than $5 \ ppm$,whereas highly polluted water has a $BOD$ value of more than $17 \ ppm$.

(A) The first ionization enthalpy increases across a period from left to right. However,Nitrogen $(N)$ has a stable half-filled $2p^3$ electronic configuration $(1s^2, 2s^2, 2p^3)$.
Due to this extra stability,it requires more energy to remove an electron from Nitrogen compared to its neighbors like Carbon and Oxygen.

(C) The $N_2$ molecule consists of two nitrogen atoms with identical electronegativity,resulting in a non-polar covalent bond.
Additionally,the $N \equiv N$ triple bond has a very high bond dissociation energy $(941 \ kJ/mol)$,making it chemically inert compared to the $CN^{-}$ ion.

(C) The radius of the $n^{th}$ orbit is given by $r_n = a_0 n^2$.
For the third orbit $(n = 3)$, the radius is $r = a_0 \times (3)^2 = 9 a_0$.
According to Bohr's postulate, the angular momentum is $mvr = \frac{nh}{2\pi}$.
Substituting the values, $mv = \frac{nh}{2\pi r} = \frac{3h}{2\pi \times 9 a_0} = \frac{h}{6\pi a_0}$.
The de-Broglie wavelength is given by $\lambda = \frac{h}{mv}$.
Substituting $mv$, we get $\lambda = \frac{h}{h / (6\pi a_0)} = 6\pi a_0$.

(C) The reaction of an alkyne with $H^{+}/Hg^{2+}$ (mercuric sulfate in dilute sulfuric acid) is a hydration reaction that follows Markovnikov's rule.
$Ph-C \equiv C-CH_3 + H_2O \xrightarrow{H^{+}/Hg^{2+}} Ph-C(OH)=CH-CH_3$ (enol intermediate).
The enol intermediate undergoes tautomerization to form the more stable ketone.
$Ph-C(OH)=CH-CH_3 \rightleftharpoons Ph-CO-CH_2CH_3$ (propiophenone).
Thus,the final product $A$ is $Ph-CO-CH_2CH_3$.

(C) Reaction: $2AB_{3(g)} \rightleftharpoons A_{2(g)} + 3B_{2(g)}$
Initial moles: $8 \ mol$ of $AB_3$,$0 \ mol$ of $A_2$,$0 \ mol$ of $B_2$
At equilibrium,$2 \ mol$ of $A_2$ are formed. According to the stoichiometry,$2 \ mol$ of $A_2$ are produced from $4 \ mol$ of $AB_3$ and produce $6 \ mol$ of $B_2$.
Equilibrium moles: $AB_3 = 8 - 4 = 4 \ mol$,$A_2 = 2 \ mol$,$B_2 = 6 \ mol$
Equilibrium concentrations (in $1 \ dm^3$): $[AB_3] = 4 \ M$,$[A_2] = 2 \ M$,$[B_2] = 6 \ M$
$K_C = \frac{[A_2][B_2]^3}{[AB_3]^2} = \frac{2 \times (6)^3}{(4)^2} = \frac{2 \times 216}{16} = \frac{432}{16} = 27$

(C) The chemical equation for the nitration of benzene is:
$C_6H_6 + HNO_3 \to C_6H_5NO_2 + H_2O$
Molar mass of benzene $(C_6H_6)$ = $(6 \times 12) + (6 \times 1) = 78\, g/mol$.
Molar mass of nitrobenzene $(C_6H_5NO_2)$ = $(6 \times 12) + (5 \times 1) + 14 + (2 \times 16) = 123\, g/mol$.
According to the stoichiometry,$78\, g$ of benzene produces $123\, g$ of nitrobenzene.
Therefore,$5\, g$ of benzene will produce:
$\text{Theoretical yield} = \frac{123}{78} \times 5 = 7.88\, g$.
Rounding to the nearest provided option,the value is $8.09\, g$ (noting that $7.88$ is closest to $8.09$ among the choices).

(B) The strength of a base is determined by its ability to accept a proton $(H^{+})$.
In reaction $(ii)$,$CN^{-}$ reacts with $H_2O$ to form $OH^{-}$. The equilibrium constant $K_b = 1.6 \times 10^{-5}$ indicates the extent of this reaction.
Since $OH^{-}$ is the product of the base dissociation of $CN^{-}$,and $K_b$ is relatively large,$OH^{-}$ is a stronger base than $CN^{-}$.
Comparing the conjugate bases: $OH^{-}$ is the conjugate base of $H_2O$,and $CN^{-}$ is the conjugate base of $HCN$.
Since $HCN$ is a weak acid $(K_a = 6.2 \times 10^{-10})$ and $H_2O$ is an extremely weak acid,the conjugate base of the weaker acid is stronger.
Thus,the order of relative base strength is $OH^{-} > CN^{-} > H_2O$.

(A) The reactivity of benzene derivatives towards electrophilic substitution depends on the electron density of the benzene ring. Electron-donating groups increase the electron density (activating groups),while electron-withdrawing groups decrease it (deactivating groups).
$(i)$ $-OCH_3$: This group exerts a strong $+M$ (mesomeric) effect,which significantly increases the electron density on the benzene ring,making it highly reactive.
(ii) $-CH_3$: This group exerts a $+I$ (inductive) effect and hyperconjugation,which increases the electron density,but less effectively than the $+M$ effect of $-OCH_3$.
(iii) Benzene: This is the reference compound with no substituents.
(iv) $-CF_3$: This group exerts a strong $-I$ effect,which significantly decreases the electron density on the benzene ring,making it the least reactive.
Therefore,the decreasing order of reactivity is $(i) > (ii) > (iii) > (iv)$.

(D) The relationship between enthalpy change and internal energy change is given by $\Delta H = \Delta U + \Delta n_g RT$.
Therefore,the difference is $\Delta H - \Delta U = \Delta n_g RT$.
For the given reaction $2C_6H_{6(l)} + 15O_{2(g)} \longrightarrow 12CO_{2(g)} + 6H_2O_{(l)}$,the change in the number of gaseous moles is $\Delta n_g = (n_{products, g}) - (n_{reactants, g}) = 12 - 15 = -3$.
Substituting the values: $\Delta H - \Delta U = -3 \times 8.314 \times 300 = -7482 \ J \ mol^{-1}$.

(A) The expressions for the velocities are:
$u = \sqrt{\frac{3RT}{M}}$ (root mean square velocity)
$v = \sqrt{\frac{8RT}{\pi M}}$ (average velocity)
$\alpha = \sqrt{\frac{2RT}{M}}$ (most probable velocity)
Comparing their ratios:
$u : v : \alpha = \sqrt{3} : \sqrt{\frac{8}{\pi}} : \sqrt{2}$
Using numerical values $(\pi \approx 3.14)$:
$u : v : \alpha = 1.732 : 1.596 : 1.414$
Therefore, the correct order is $u > v > \alpha$.

(A) The given reaction is a redox reaction involving the reduction of permanganate and oxidation of oxalate.
Reduction half-reaction:
$MnO_4^- + 8H^+ + 5e^- \to Mn^{2+} + 4H_2O$ ... $(i)$
Oxidation half-reaction:
$C_2O_4^{2-} \to 2CO_2 + 2e^-$ ... $(ii)$
To balance the electrons,multiply equation $(i)$ by $2$ and equation $(ii)$ by $5$:
$2MnO_4^- + 16H^+ + 10e^- \to 2Mn^{2+} + 8H_2O$
$5C_2O_4^{2-} \to 10CO_2 + 10e^-$
Adding these two equations gives:
$2MnO_4^- + 5C_2O_4^{2-} + 16H^+ \to 2Mn^{2+} + 10CO_2 + 8H_2O$
Comparing this with the given equation $X\,MnO_4^- + Y\,C_2O_4^{2-} + Z\,H^+ \rightleftharpoons X\,Mn^{2+} + 2Y\,CO_2 + \frac{Z}{2}\,H_2O$,we get:
$X = 2$,$Y = 5$,$Z = 16$.

(D) $CCl_4$ has the lowest (zero) dipole moment.
This is due to its symmetrical tetrahedral structure.
In this molecule,the individual bond dipole moments cancel each other out due to the symmetrical arrangement of the four $C-Cl$ bonds.

(A) precipitate forms when the ionic product of the ions in the solution exceeds the solubility product constant $(K_{sp})$.
For the dissociation of $CaF_2$: $CaF_2 \rightleftharpoons Ca^{2+} + 2F^{-}$.
The ionic product is given by: $Q_{sp} = [Ca^{2+}][F^{-}]^2$.
For option $(A)$: $[Ca^{2+}] = 1 \times 10^{-2}\,M$ and $[F^{-}] = 1 \times 10^{-3}\,M$.
$Q_{sp} = (1 \times 10^{-2}) \times (1 \times 10^{-3})^2 = (1 \times 10^{-2}) \times (1 \times 10^{-6}) = 1 \times 10^{-8}$.
Since $1 \times 10^{-8} > 1.7 \times 10^{-10}$,the ionic product exceeds $K_{sp}$,leading to the formation of a precipitate.
Therefore,option $(A)$ is correct.

(D) For the reaction: ${N_2}{O_4}_{(g)} \rightleftharpoons 2N{O_2}_{(g)}$
Let the initial moles of ${N_2}{O_4}$ be $1$ and degree of dissociation be $\alpha$.
At equilibrium,moles of ${N_2}{O_4} = (1 - \alpha)$ and moles of $N{O_2} = 2\alpha$.
Total moles at equilibrium $= (1 - \alpha) + 2\alpha = (1 + \alpha)$.
Partial pressure of ${N_2}{O_4} = \frac{(1 - \alpha)}{(1 + \alpha)} \times P$.
Partial pressure of $N{O_2} = \frac{2\alpha}{(1 + \alpha)} \times P$.
$K_P = \frac{(P_{N{O_2}})^2}{P_{{N_2}{O_4}}} = \frac{[\frac{2\alpha}{1 + \alpha} \times P]^2}{\frac{1 - \alpha}{1 + \alpha} \times P} = \frac{4\alpha^2 P}{1 - \alpha^2}$.
Given $K_P = 2$ and $P = 0.5 \ atm$:
$2 = \frac{4 \times \alpha^2 \times 0.5}{1 - \alpha^2} = \frac{2\alpha^2}{1 - \alpha^2}$.
$1 - \alpha^2 = \alpha^2 \implies 2\alpha^2 = 1 \implies \alpha^2 = 0.5$.
$\alpha = \sqrt{0.5} \approx 0.707$.
Percentage dissociation $= \alpha \times 100 = 70.7 \% \approx 71 \%$.

(A) Elements with atomic numbers $40$ $(Zr)$ and $72$ $(Hf)$ belong to the same group $(Group \ 4)$ in the periodic table.
Elements in the same group exhibit similar chemical properties due to having the same valence shell electronic configuration.
Therefore,the pair $(40, 72)$ is expected to have similar properties.

(A) $1.$ The longest carbon chain containing both multiple bonds has $7$ carbon atoms,so the parent name is hept.
$2.$ Numbering the chain from left to right gives the multiple bonds the lowest locants ($2$ and $4$). The double bond is at $C2$ and the triple bond is at $C4$.
$3.$ Since the higher priority groups ($CH_3$ and the alkyne group) are on opposite sides of the double bond,the configuration is $(E)$.
$4.$ Therefore,the $IUPAC$ name is $(E)-2-\text{hepten}-4-\text{yne}$.

(C) The reactivity of alkali metals with water increases as we move down the group from $Li$ to $Cs$.
This trend is due to the increase in the electropositive character and the decrease in ionization enthalpy down the group.
The order of reactivity is $Li < Na < K < Rb < Cs$.
Therefore,$Rb$ (Rubidium) reacts most vigorously with water among the given options.

(A) The photobromination of $2-$methylbutane involves the formation of a free radical intermediate.
$CH_3-CH(CH_3)-CH_2-CH_3 \xrightarrow[Br_2]{hv} CH_3-C(Br)(CH_3)-CH_2-CH_3$ ($2-$bromo$-2-$methylbutane).
The reactivity of hydrogen atoms towards free radical substitution follows the order: $3^{\circ} > 2^{\circ} > 1^{\circ}$.
Since $2-$methylbutane has one tertiary $(3^{\circ})$ hydrogen atom at the $C-2$ position,the abstraction of this hydrogen leads to the most stable tertiary free radical.
Therefore,$2-$bromo$-2-$methylbutane is the major product.

(B) The de-Broglie wavelength is given by the relation: $\lambda = \frac{h}{p} = \frac{h}{\sqrt{2m(KE)}}$.
From this relation,we can see that $\lambda \propto \frac{1}{\sqrt{KE}}$.
Let the initial kinetic energy be $KE_1$ and the final kinetic energy be $KE_2 = 4 \times KE_1$.
Let the initial wavelength be $\lambda_1$ and the final wavelength be $\lambda_2$.
Then,$\frac{\lambda_2}{\lambda_1} = \sqrt{\frac{KE_1}{KE_2}} = \sqrt{\frac{KE_1}{4 \times KE_1}} = \sqrt{\frac{1}{4}} = \frac{1}{2}$.
Therefore,the new wavelength becomes half of the original wavelength.

(A) The chemical reaction is: $CO_2(g) + C(s) \to 2CO(g)$
Let the volume of $CO_2$ reacted be $x\,mL$.
According to the stoichiometry,$x\,mL$ of $CO_2$ produces $2x\,mL$ of $CO$.
Initial volume of $CO_2 = 500\,mL$.
Remaining volume of $CO_2 = (500 - x)\,mL$.
Volume of $CO$ produced = $2x\,mL$.
Total final volume = $(500 - x) + 2x = 500 + x$.
Given that the total final volume is $700\,mL$,so $500 + x = 700$,which gives $x = 200\,mL$.
Volume of $CO_2$ remaining = $500 - 200 = 300\,mL$.
Volume of $CO$ produced = $2 \times 200 = 400\,mL$.
Thus,the composition is $CO_2 = 300\,mL$ and $CO = 400\,mL$.

(C) Let the volume of the vessel be $V$. Initially,the volume of air at $T_1 = 300 \ K$ is $V$.
Since $2/5$ of the air is expelled,the remaining volume of air at $300 \ K$ is $V - (2/5)V = (3/5)V$.
When the vessel is heated to a new temperature $T_2$,this remaining air expands to fill the entire volume $V$ of the vessel.
Since the pressure remains constant in an open vessel,we use Charles's Law: $\frac{V_1}{T_1} = \frac{V_2}{T_2}$.
Substituting the values: $\frac{(3/5)V}{300} = \frac{V}{T_2}$.
Solving for $T_2$: $T_2 = 300 \times \frac{5}{3} = 500 \ K$.

(B) $XeF_4$ has a zero dipole moment.
It has a square planar structure due to which the bond moments of $Xe-F$ bonds cancel each other out.

(D) The enthalpy of neutralisation of a strong acid with a strong base is the enthalpy of formation of water from $H^+$ and $OH^-$ ions,which is $-55.90 \ kJ \ mol^{-1}$.
$NH_4OH$ is a weak base,so it requires energy for ionisation.
Let the enthalpy of ionisation of $NH_4OH$ be $x \ kJ \ mol^{-1}$.
The total enthalpy of neutralisation is the sum of the enthalpy of ionisation of the weak base and the enthalpy of neutralisation of $H^+$ and $OH^-$.
$\Delta H_{neutralisation} = \Delta H_{ionisation} + \Delta H_{H^+ + OH^- \rightarrow H_2O}$
$-51.46 = x + (-55.90)$
$x = -51.46 + 55.90$
$x = +4.44 \ kJ \ mol^{-1}$

(B) The Beilstein test is a simple chemical test used for the detection of halogens ($Cl$,$Br$,$I$) in organic compounds.
In this test,a copper wire is heated in a flame until it no longer imparts any color to the flame.
The wire is then dipped into the organic compound and heated again in the flame.
$A$ green or blue-green flame indicates the presence of a halogen due to the formation of volatile copper halides.

(A) Given $\vec{a} + \vec{b} + \vec{c} = 0$,we can write $\vec{a} + \vec{b} = -\vec{c}$.
Taking the dot product of both sides with themselves:
$(\vec{a} + \vec{b}) \cdot (\vec{a} + \vec{b}) = (-\vec{c}) \cdot (-\vec{c})$
$|\vec{a}|^2 + |\vec{b}|^2 + 2(\vec{a} \cdot \vec{b}) = |\vec{c}|^2$
Since $\vec{a} \cdot \vec{b} = |\vec{a}| |\vec{b}| \cos \theta$,where $\theta$ is the angle between $\vec{a}$ and $\vec{b}$:
$3^2 + 5^2 + 2(3)(5) \cos \theta = 7^2$
$9 + 25 + 30 \cos \theta = 49$
$34 + 30 \cos \theta = 49$
$30 \cos \theta = 15$
$\cos \theta = \frac{15}{30} = \frac{1}{2}$
Therefore,$\theta = \frac{\pi}{3}$.

(B) First,identify the orbitals based on the given quantum numbers:
$(i)$ $n = 4, l = 1$ corresponds to $4p$ orbital. $(n+l) = 4 + 1 = 5$.
$(ii)$ $n = 4, l = 0$ corresponds to $4s$ orbital. $(n+l) = 4 + 0 = 4$.
$(iii)$ $n = 3, l = 2$ corresponds to $3d$ orbital. $(n+l) = 3 + 2 = 5$.
$(iv)$ $n = 3, l = 1$ corresponds to $3p$ orbital. $(n+l) = 3 + 1 = 4$.
According to the $(n+l)$ rule,energy increases as the $(n+l)$ value increases.
For $(iv)$ and $(ii)$,both have $(n+l) = 4$. Since $(iv)$ has a lower $n$ value $(n=3)$ than $(ii)$ $(n=4)$,$(iv)$ has lower energy.
For $(iii)$ and $(i)$,both have $(n+l) = 5$. Since $(iii)$ has a lower $n$ value $(n=3)$ than $(i)$ $(n=4)$,$(iii)$ has lower energy.
Thus,the increasing order of energy is: $(iv) < (ii) < (iii) < (i)$.

(D) Maleic acid is the $cis$-form and fumaric acid is the $trans$-form of butenedioic acid $(HOOC-CH=CH-COOH)$.
Since they differ in the spatial arrangement of groups around the double bond,they are classified as geometrical isomers.

(D) compound is antiaromatic if it is cyclic,planar,fully conjugated,and contains $4n \pi$ electrons (where $n = 1, 2, ...$).
$(I)$ Furan: $6 \pi$ electrons (aromatic).
$(II)$ Cycloheptatriene: Not fully conjugated (non-aromatic).
$(III)$ Cyclooctatetraene: Non-planar tub-shaped (non-aromatic).
$(IV)$ Cyclopentadienyl anion: $6 \pi$ electrons (aromatic).
$(V)$ Cyclopentadienyl cation: $4 \pi$ electrons,planar,fully conjugated (antiaromatic).
$(VI)$ Cyclobutadiene: $4 \pi$ electrons,planar,fully conjugated (antiaromatic).
Therefore,compounds $(V)$ and $(VI)$ are antiaromatic.

(D) The second ionization enthalpy corresponds to the energy required to remove an electron from a unipositive ion $(M^+ \rightarrow M^{2+} + e^-)$.
Electronic configurations of the ions are:
$C^+: 1s^2 2s^2 2p^1$
$N^+: 1s^2 2s^2 2p^2$
$O^+: 1s^2 2s^2 2p^3$
$F^+: 1s^2 2s^2 2p^4$
$O^+$ has a half-filled $2p$ subshell $(2p^3)$,which is exceptionally stable.
Therefore,the energy required to remove an electron from $O^+$ is higher than that for $F^+$.
The general trend of ionization enthalpy increases across a period,but due to the stability of the half-filled $p$-orbital in $O^+$,the order is $O > F > N > C$.

(C) Given,vapour density $(V.D.)$ $= 94.8$.
Molecular weight of metal chloride $= 2 \times V.D. = 2 \times 94.8 = 189.6 \ g/mol$.
Mass of chlorine in $189.6 \ g$ of metal chloride $= 74.75\% \text{ of } 189.6 = 0.7475 \times 189.6 \approx 141.7 \ g$.
Number of chlorine atoms $= \frac{141.7}{35.5} \approx 4$.
Mass of metal $= 189.6 - 141.7 = 47.9 \ g$.
Since the metal chloride is $MCl_4$,the valency of the metal $M$ is $4$.
Therefore,the formula of the metal chloride is $MCl_4$.

(A) trigonal pyramidal molecular shape occurs when the central atom has three bond pairs and one lone pair of electrons.
$1$. $NI_3$: Nitrogen has $5$ valence electrons. It forms $3$ bonds with $I$ atoms and has $1$ lone pair. Shape: Trigonal pyramidal.
$2$. $I_3^-$: The central $I$ atom has $3$ lone pairs and $2$ bond pairs. Shape: Linear.
$3$. $SO_3^{2-}$: Sulfur has $6$ valence electrons. It forms $3$ bonds with $O$ atoms (one double,two single) and has $1$ lone pair. Shape: Trigonal pyramidal.
$4$. $NO_3^-$: Nitrogen has $5$ valence electrons. It forms $3$ bonds with $O$ atoms and has no lone pairs. Shape: Trigonal planar.
Thus,$NI_3$ $(I)$ and $SO_3^{2-}$ $(III)$ have a trigonal pyramidal shape.

(A) The reaction sequence is as follows:
$1.$ Allylic chlorination: $CH_3-CH=CH_2 + Cl_2 \xrightarrow{700\,K} Cl-CH_2-CH=CH_2 (A) + HCl$
$2.$ Hydrolysis: $Cl-CH_2-CH=CH_2 \xrightarrow[420\,K, 12\,atm]{Na_2CO_3, H_2O} HO-CH_2-CH=CH_2 (B)$
$3.$ Chlorohydrination and hydrolysis: $HO-CH_2-CH=CH_2$ $\xrightarrow{HOCl} HO-CH_2-CH(OH)-CH_2Cl$ $\xrightarrow{NaOH} HO-CH_2-CH(OH)-CH_2OH (C)$.
Compound $C$ is Glycerol.

(C) Soda-acid fire extinguishers work on the principle of the reaction between an acid and a carbonate or bicarbonate salt.
They contain a solution of $NaHCO_3$ (sodium bicarbonate) and a separate bottle of $H_2SO_4$ (sulfuric acid).
When the extinguisher is activated,the acid reacts with the bicarbonate to release $CO_2$ gas,which helps in extinguishing the fire.
Therefore,the correct component is $NaHCO_3$.

(B) $NO$ (nitric oxide) and freons (chlorofluorocarbons) are responsible for the depletion of the ozone layer in the stratosphere.

(A) molecule exhibits a dipole moment if its net dipole moment is non-zero.
In $1, 2-$ dichlorobenzene,the two $C-Cl$ bonds are at an angle of $60^\circ$ to each other,making the molecule unsymmetrical and resulting in a net dipole moment.
In contrast,$1, 4-$ dichlorobenzene,trans-$2, 3-$ dichloro$-2-$butene,and trans$-1, 2-$ dinitroethene are centrosymmetric molecules where the individual bond dipoles cancel each other out,resulting in a net dipole moment of zero.

(D) The dissociation of $PbI_2$ is given by: $PbI_2(s) \rightleftharpoons Pb^{2+}(aq) + 2I^{-}(aq)$.
First,calculate the molar solubility $(s)$ in $mol\, L^{-1}$:
$s = \frac{\text{solubility in } g\, L^{-1}}{\text{molar mass}} = \frac{0.7}{461.2} \approx 1.5178 \times 10^{-3} \, mol\, L^{-1}$.
The solubility product expression is $K_{sp} = [Pb^{2+}][I^{-}]^2 = (s)(2s)^2 = 4s^3$.
Substituting the value of $s$:
$K_{sp} = 4 \times (1.5178 \times 10^{-3})^3 \approx 4 \times 3.5 \times 10^{-9} = 14.0 \times 10^{-9}$.

(A) The hydration of propyne $(CH_3-C\equiv CH)$ in the presence of $H_2SO_4$ and $HgSO_4$ follows Markovnikov's addition.
First,an unstable enol intermediate,$CH_3-C(OH)=CH_2$ (prop-$1$-en-$2$-ol),is formed.
This enol undergoes tautomerization to form the stable ketone,$CH_3-CO-CH_3$ (propan-$2$-one or acetone).
Reaction: $CH_3-C\equiv CH + H_2O$ $\xrightarrow[HgSO_4]{H_2SO_4} [CH_3-C(OH)=CH_2]$ $\rightarrow CH_3-CO-CH_3$.

(B) For an isothermal process,the change in entropy $\Delta S$ is given by the formula: $\Delta S = nR \ ln \ (P_1 / P_2)$.
Given that the gas is expanded to half of its initial pressure,$P_2 = P_1 / 2$,so $P_1 / P_2 = 2$.
For $n = 1 \ mol$,the expression becomes: $\Delta S = 1 \times R \times ln \ (2)$.
Substituting the values: $\Delta S = 8.314 \times 0.693 = 5.76 \ J \ K^{-1} \ mol^{-1}$.

(A) $1$. In $SO_3$,there is no $S-S$ bond. Number of $S-S$ bonds = $0$.
$2$. In $S_2O_3^{2-}$ (thiosulfate ion),there is one $S-S$ bond. Number of $S-S$ bonds = $1$.
$3$. In $S_2O_6^{2-}$ (dithionate ion),there is one $S-S$ bond. Number of $S-S$ bonds = $1$.
$4$. In $S_2O_8^{2-}$ (peroxydisulfate ion),there is no $S-S$ bond; it contains an $O-O$ peroxide linkage. Number of $S-S$ bonds = $0$.
Thus,the number of $S-S$ bonds are $0, 1, 1, 0$ respectively.

(D) The magnetic moment (spin only) is calculated using the formula $\mu = \sqrt{n(n+2)} \ BM$,where $n$ is the number of unpaired electrons.
$1$. For $Mn^{2+}$ $(d^5)$: $n = 5$,$\mu = \sqrt{5(5+2)} = \sqrt{35} \approx 5.92 \ BM$.
$2$. For $Ti^{2+}$ $(d^2)$: $n = 2$,$\mu = \sqrt{2(2+2)} = \sqrt{8} \approx 2.83 \ BM$.
$3$. For $V^{2+}$ $(d^3)$: $n = 3$,$\mu = \sqrt{3(3+2)} = \sqrt{15} \approx 3.87 \ BM$.
$4$. For $Cr^{2+}$ $(d^4)$: $n = 4$,$\mu = \sqrt{4(4+2)} = \sqrt{24} \approx 4.90 \ BM$.
The value $4.90 \ BM$ is of the order of $5 \ BM$. Thus,$Cr^{2+}$ is the correct answer.

(A) $XeF_4$ has a square planar structure because it has $4$ bond pairs and $2$ lone pairs on the central $Xe$ atom,resulting in $sp^3d^2$ hybridization.
$NH_4^+$,$BF_4^-$,and $CCl_4$ all have $4$ bond pairs and $0$ lone pairs,resulting in $sp^3$ hybridization and a tetrahedral structure.

(B) For a reaction of order $n$,the half-life is given by $t_{1/2} \propto (a_0)^{1-n}$.
Taking logarithm on both sides,we get $\log\,t_{1/2} = (1-n)\log\,a_0 + \text{constant}$.
Comparing this with the equation of a straight line $y = mx + c$,the slope $m = 1-n$.
From the given figure,the slope is $\tan(45^{\circ}) = 1$.
Therefore,$1-n = 1$,which gives $n = 0$.

(C) Polyamides are polymers containing amide linkages $(-CONH-)$ in their backbone.
$Nylon$ (e.g.,$Nylon-6,6$ or $Nylon-6$) is a well-known example of a synthetic polyamide.
$Teflon$ is a fluoropolymer,$Orlon$ is a polyacrylonitrile,and $Terylene$ is a polyester.

(C) The deposition of metals at the cathode during electrolysis depends on their standard reduction potentials $(E^o)$.
Metals with higher reduction potentials are deposited more easily at the cathode.
The given standard reduction potentials are: $E^o(Ag^+/Ag) = 0.80 \ V$,$E^o(Hg_2^{2+}/2Hg) = 0.79 \ V$,$E^o(Cu^{2+}/Cu) = 0.34 \ V$,and $E^o(Mg^{2+}/Mg) = -2.37 \ V$.
In an aqueous solution,water can also be reduced at the cathode: $2H_2O + 2e^- \rightarrow H_2 + 2OH^-$,with $E^o = -0.83 \ V$.
Since $Mg^{2+}$ has a reduction potential of $-2.37 \ V$,which is much lower than that of water $(-0.83 \ V)$,$Mg$ will not deposit from an aqueous solution; instead,$H_2$ gas will be evolved.
Therefore,the metals that will deposit in the order of increasing voltage are $Ag$,$Hg$,and $Cu$.
The correct sequence is $Ag > Hg > Cu$.

(B) $N$-bromosuccinimide $(NBS)$ is a reagent used for allylic or benzylic bromination via a free radical mechanism.
In ethylbenzene $(C_6H_5CH_2CH_3)$,the benzylic position is the carbon atom directly attached to the benzene ring.
The hydrogen atoms at the benzylic position are more reactive due to the stability of the resulting benzylic radical.
Therefore,$NBS$ selectively replaces a hydrogen atom at the benzylic position with a bromine atom,yielding $1$-bromo-$1$-phenylethane $(C_6H_5CH(Br)CH_3)$.

(B) Gold number is defined as the minimum amount of protective colloid in milligrams that prevents the coagulation of $10 \, mL$ of a standard gold sol when $1 \, mL$ of $10 \% \, NaCl$ solution is added to it.
Given mass of starch $= 0.025 \, g$.
Converting mass to milligrams: $0.025 \, g = 0.025 \times 1000 \, mg = 25 \, mg$.
Since this amount of starch just prevents the coagulation of $10 \, mL$ of gold sol,the gold number of starch is $25$.

(C) Sucrose is a non-reducing sugar because it does not contain a free hemiacetal or hemiketal group.
It does not reduce Fehling's reagent or Tollen's reagent.

(C) Fog is a type of aerosol where the dispersed phase is a liquid and the dispersion medium is a gas.

(C) Aspirin,also known as acetylsalicylic acid,is prepared by the acetylation of the phenolic $-OH$ group of salicylic acid using acetic anhydride in the presence of an acid catalyst like $H_2SO_4$.
The chemical reaction is as follows:
$C_6H_4(OH)COOH + (CH_3CO)_2O \xrightarrow{H_2SO_4} C_6H_4(OCOCH_3)COOH + CH_3COOH$
Thus,the correct option is $C$.

(A) The complex $[PtCl_2(NH_3)_2]$ involves a $Pt^{2+}$ central metal ion with a $d^8$ electronic configuration.
It undergoes $dsp^2$ hybridization,which results in a square-planar geometry.
In contrast,$[NiCl_4]^{2-}$ has a tetrahedral geometry due to $sp^3$ hybridization,while $MnO_4^-$ and $CrO_4^{2-}$ exhibit tetrahedral geometry.

(A) $MnO_2$ is a selective oxidizing agent that specifically oxidizes allylic and benzylic alcohols to their corresponding aldehydes or ketones without affecting the double bond.
Allyl alcohol $(CH_2=CH-CH_2OH)$ is oxidized to acrolein $(CH_2=CH-CHO)$ using $MnO_2$.

(B) In the froth-floatation process,froth stabilizers such as cresols and aniline are added to stabilize the froth and enhance the non-wettability of the mineral particles.

(C) The activation energy $(E_a)$ can be calculated using the Arrhenius equation: $\log(\frac{K_2}{K_1}) = \frac{E_a}{2.303 \ R} \times \frac{T_2 - T_1}{T_1 \times T_2}$.
Given: $\frac{K_2}{K_1} = 2$,$T_1 = 298 \ K$,$T_2 = 308 \ K$,and $R = 8.314 \ J \ K^{-1} \ mol^{-1}$.
Substituting the values: $\log(2) = \frac{E_a}{2.303 \times 8.314} \times \frac{308 - 298}{308 \times 298}$.
$0.3010 = \frac{E_a}{19.147} \times \frac{10}{91784}$.
$E_a = \frac{0.3010 \times 19.147 \times 91784}{10} \approx 52897 \ J \ mol^{-1} = 52.9 \ kJ \ mol^{-1}$.

(C) All aldehydes show reaction with Tollen's reagent and Fehling solutions,but ketones do not show this reaction.
$Note$ :- Benzaldehyde does not give reaction with Fehling solution.

(A) The basicity of amines depends on the availability of the lone pair of electrons on the nitrogen atom.
$I$ (Piperidine): The nitrogen atom is $sp^3$ hybridized,and the lone pair is not involved in resonance. It is the most basic.
$II$ (Pyridine): The nitrogen atom is $sp^2$ hybridized. The lone pair is in an $sp^2$ orbital,which is more electronegative than an $sp^3$ orbital,making it less basic than piperidine.
$III$ (Morpholine): Similar to piperidine,the nitrogen is $sp^3$ hybridized,but the presence of an electronegative oxygen atom exerts an inductive effect ($-I$ effect),which reduces the electron density on the nitrogen atom,making it less basic than piperidine but more basic than pyridine.
$IV$ (Pyrrole): The nitrogen atom is $sp^2$ hybridized,and its lone pair is involved in the aromatic sextet (delocalized). Thus,it is the least basic.
Therefore,the order of basicity is $I > III > II > IV$.

(D) The depression in freezing point is given by $\Delta T_f = T_f^{\circ} - T_f = 0 - (-1.91) = 1.91 \ K$.
Using the formula $\Delta T_f = i \times K_f \times m$,where $i$ is the van't Hoff factor,$K_f = 1.86 \ K \ kg \ mol^{-1}$,and $m = 1.00 \ m$:
$i = \frac{\Delta T_f}{K_f \times m} = \frac{1.91}{1.86 \times 1.00} \approx 1.027$.
For the dissociation of $HF$: $HF \rightleftharpoons H^+ + F^-$.
If $\alpha$ is the degree of dissociation,then $i = 1 + \alpha$.
$1 + \alpha = 1.027 \implies \alpha = 0.027$.
Therefore,the percentage dissociation is $0.027 \times 100 = 2.7 \%$.

(A) $1$. $RNA$ (Ribonucleic acid) is primarily responsible for protein synthesis in the cell.
$2$. The sugar present in $DNA$ is $2$-deoxy-$D$-ribose,not $D-(-)$-ribose (which is found in $RNA$).
$3$. $DNA$ typically has a double-stranded helix structure,whereas $RNA$ is usually single-stranded.
$4$. $DNA$ is primarily found in the nucleus of the cell,while $RNA$ is found in the cytoplasm and nucleus.
Therefore,the correct statement is that $RNA$ controls the synthesis of proteins.

(C) Bakelite is a thermosetting phenol-formaldehyde resin.
It is obtained by the condensation polymerization of phenol and formaldehyde in the presence of an acid or base catalyst.

(D) The correct answer is $D$.
In metals,electrical conductivity decreases with an increase in temperature due to the increased vibrations of metal ions,which scatter electrons.
In semiconductors,conductivity increases with temperature due to the excitation of more electrons from the valence band to the conduction band.
Therefore,the statement that conductivity of both metals and semiconductors increases with temperature is incorrect.

(C) The electronic configuration of $Ce$ $(Z = 58)$ is $[Xe] 4f^1 5d^1 6s^2$.
By losing four electrons,it attains the stable noble gas configuration of $Xe$ $([Xe])$,making the $+4$ oxidation state stable for $Ce$.

(B) The process of removing impurities from a colloidal solution is known as purification.
Electrodialysis is a technique used to accelerate the process of dialysis by applying an electric field,which helps in the faster removal of ionic impurities from the colloidal solution.
Therefore,colloidal solutions can be purified by electrodialysis.

(B) For the reaction $Cu^{2+} + e^- \to Cu^{+}$,$\Delta G_1^o = -n_1 E_1^o F = -1 \times 0.15 \times F = -0.15F$.
For the reaction $Cu^{2+} + 2e^- \to Cu$,$\Delta G_2^o = -n_2 E_2^o F = -2 \times 0.34 \times F = -0.68F$.
To find the potential for $Cu^{+} + e^- \to Cu$,we subtract the first reaction from the second:
$(Cu^{2+} + 2e^- \to Cu) - (Cu^{2+} + e^- \to Cu^{+}) \implies Cu^{+} + e^- \to Cu$.
$\Delta G^o = \Delta G_2^o - \Delta G_1^o = -0.68F - (-0.15F) = -0.53F$.
Since $\Delta G^o = -n E^o F$,where $n=1$:
$-0.53F = -1 \times E^o \times F
\implies E^o = 0.53 \, V$.

(B) The given complex is of the type $[MABCD]$,where $M = Pt^{2+}$,$A = NO_2^-$,$B = Py$,$C = NH_3$,and $D = NH_2OH$.
Square planar complexes of the type $[MABCD]$ exhibit $3$ geometrical isomers.
These isomers are formed by fixing one ligand and varying the positions of the other three ligands relative to it.
As shown in the structures $(I)$,$(II)$,and $(III)$,there are $3$ possible arrangements for the ligands around the central metal atom.

(C) Sulphonamides are a group of synthetic drugs that contain the sulphonamide group. They are widely used as antimicrobial agents because they inhibit the growth of bacteria by interfering with the synthesis of folic acid in them.

(B) The electronic configuration of Gadolinium $[Z = 64]$ is $[Xe] 4f^7 5d^1 6s^2$.
In the $4f$ subshell,there are $7$ unpaired electrons.
In the $5d$ subshell,there is $1$ unpaired electron.
Therefore,the total number of unpaired electrons is $7 + 1 = 8$.

(C) Except for glycine,all other naturally occurring amino acids contain a chiral center at the $\alpha$-carbon atom.
Glycine has the structure $H_2N-CH_2-COOH$,where the $\alpha$-carbon is bonded to two identical hydrogen atoms,making it achiral.
Therefore,all amino acids except glycine are optically active.