TS EAMCET 2024 Chemistry Question Paper with Answer and Solution

(D) Given:
Internal diameter $d = (5.73 \pm 0.01) \text{ cm}$
External diameter $D = (6.01 \pm 0.01) \text{ cm}$
The thickness $t$ of the cylinder wall is given by the formula:
$t = \frac{D - d}{2}$
First,calculate the mean value:
$t_{\text{mean}} = \frac{6.01 - 5.73}{2} = \frac{0.28}{2} = 0.14 \text{ cm}$
Next,calculate the uncertainty in thickness:
Since the thickness involves subtraction,the absolute errors add up:
$\Delta t = \frac{\Delta D + \Delta d}{2} = \frac{0.01 + 0.01}{2} = \frac{0.02}{2} = 0.01 \text{ cm}$
Therefore,the thickness is $(0.14 \pm 0.01) \text{ cm}$.

(B) $1$. Identify the longest carbon chain containing the principal functional group $(-OH)$ and the double bond. The chain has $10$ carbons,so the parent alkane is decane.
$2$. Number the chain to give the lowest possible locants to the functional group $(-OH)$,then the double bond,and finally the substituents.
$3$. Starting from the right,the $-OH$ group is at $C-4$,the double bond starts at $C-6$,and the ethyl group is at $C-5$. However,the standard $IUPAC$ numbering gives the $-OH$ group the lowest locant $4$.
$4$. Following the numbering shown in the provided solution image: the $-OH$ group is at $C-7$,the double bond is at $C-4$,and the methyl group is at $C-2$.
$5$. The correct $IUPAC$ name is $2-$methyl$-6-$ethyldec$-4-$en$-7-$ol.

(A) $1$. Identify the principal functional group: The molecule contains an amino group $(-NH_2)$,a ketone group $(>C=O)$,and a hydroxyl group $(-OH)$. According to $IUPAC$ priority rules,the ketone group has higher priority than the alcohol and amine groups. Therefore,the suffix will be '-one'.
$2$. Select the longest carbon chain containing the principal functional group: The longest chain containing the ketone group has $6$ carbons.
$3$. Number the chain: Numbering should start from the end that gives the ketone group the lowest possible locant. Numbering from right to left,the ketone is at position $2$. The chain is numbered as follows: $C_1(H_2OH)-C_2(=O)-C_3(CH_3)=C_4(H)-C_5(H, NH_2)-C_6(H_3)$.
$4$. Identify substituents and their positions: There is an amino group at position $5$,a methyl group at position $3$,and a hydroxy group at position $1$.
$5$. Assemble the name: Combining these,we get $5-$amino$-1-$hydroxy$-3-$methylhex$-3-$en$-2-$one.

(B) Let us analyze each option:
$A$: The structure shows a phenol ring with methyl groups at positions $3$ and $4$. This is $3, 4-$dimethylphenol. This is correct.
$B$: The structure shows a benzene ring with a chlorine atom at position $1$ and nitro groups at positions $2$ and $4$. The $IUPAC$ name should be $1-$chloro$-2, 4-$dinitrobenzene. The given name $4-$chloro$-1, 3-$dinitrobenzene is incorrect.
$C$: The structure shows a benzene ring with a methyl group at position $1$,a chlorine atom at position $2$,and a nitro group at position $4$. This is $2-$chloro$-1-$methyl$-4-$nitrobenzene. This is correct.
$D$: The structure shows an aniline ring with a methyl group at position $2$ and an ethyl group at position $4$. This is $4-$ethyl$-2-$methylaniline. This is correct.
Therefore,the option that is not correctly matched is $B$.

(A) The stability of carbocations is determined by resonance,hyperconjugation,and inductive effects.
$I$ $(C_6H_5CH_2^+)$ is a benzylic carbocation,which is highly stabilized by resonance with the phenyl ring.
$III$ $(CH_3-CH^+(CH_3))$ is a secondary $(2^{\circ})$ carbocation,stabilized by hyperconjugation and inductive effects of two methyl groups.
$IV$ $(CH_3-CH_2^+)$ is a primary $(1^{\circ})$ carbocation,stabilized by hyperconjugation and inductive effects of one methyl group.
$II$ $(CH_2=CH^+)$ is a vinylic carbocation,and $V$ $(HC \equiv C^+)$ is an acetylenic carbocation. Both have the positive charge on an $sp^2$ and $sp$ hybridized carbon,respectively. Higher $s$-character increases electronegativity,making the positive charge less stable.
Since $sp$ hybridization ($50\% \ s$-character) is more electronegative than $sp^2$ hybridization ($33.3\% \ s$-character),$V$ is less stable than $II$.
Thus,the decreasing order of stability is $I > III > IV > II > V$.

(B) The $CHO$ (aldehyde) group attached to the benzene ring is an electron-withdrawing group.
In the given structures,the $\pi$-electrons of the benzene ring are delocalized towards the oxygen atom of the $CHO$ group.
This type of electron displacement,where the substituent withdraws electron density from the conjugated system through resonance,is known as the $-R$ effect (or $-M$ effect).

(C) The structural isomers for $C_4H_9Br$ are:
$1$. $CH_3CH_2CH_2CH_2Br$ ($1$-bromobutane)
$2$. $CH_3CH_2CH(Br)CH_3$ ($2$-bromobutane)
$3$. $(CH_3)_2CHCH_2Br$ ($1$-bromo-$2$-methylpropane)
$4$. $(CH_3)_3CBr$ ($2$-bromo-$2$-methylpropane)
Among these,$2$-bromobutane $(CH_3CH_2CH(Br)CH_3)$ has a chiral center at the $C_2$ position,meaning it exists as a pair of enantiomers ($R$ and $S$ configurations).
Therefore,the total number of isomers including stereoisomers is $4$ structural isomers + $1$ additional enantiomer = $5$ isomers.

(C) Given: Height $h = 20 ~m$,final velocity $v = 31.4 ~ms^{-1}$.
Using the equation of motion $v^2 = u^2 + 2gh$,where initial velocity $u = 0$:
$v^2 = 2gh$
$g = \frac{v^2}{2h} = \frac{31.4 \times 31.4}{2 \times 20} = \frac{985.96}{40} \approx 24.65 ~ms^{-2}$.
Note that $31.4 \approx 10\pi$,so $g = \frac{(10\pi)^2}{40} = \frac{100\pi^2}{40} = 2.5\pi^2 ~ms^{-2}$.
For a seconds pendulum,the time period $T = 2 ~s$.
The formula for the time period of a simple pendulum is $T = 2\pi \sqrt{\frac{l}{g}}$.
Squaring both sides: $T^2 = 4\pi^2 \frac{l}{g} \Rightarrow l = \frac{T^2 g}{4\pi^2}$.
Substituting $T = 2 ~s$ and $g = 2.5\pi^2 ~ms^{-2}$:
$l = \frac{2^2 \times 2.5\pi^2}{4\pi^2} = \frac{4 \times 2.5\pi^2}{4\pi^2} = 2.5 ~m$.

(D) Let the distances of the stars of masses $M$ and $2M$ from their center of mass be $r_1$ and $r_2$ respectively.
By the definition of center of mass,$M r_1 = (2M) r_2$ and $r_1 + r_2 = d$.
Solving these,we get $r_1 = \frac{2d}{3}$ and $r_2 = \frac{d}{3}$.
The gravitational force between the stars provides the necessary centripetal force for their circular motion.
For the star of mass $M$: $F_g = M \omega^2 r_1$.
Substituting the values: $\frac{G(M)(2M)}{d^2} = M \omega^2 \left(\frac{2d}{3}\right)$.
$\frac{2GM^2}{d^2} = M \omega^2 \left(\frac{2d}{3}\right)$.
$\omega^2 = \frac{3GM}{d^3}$.
Therefore,$\omega = \sqrt{\frac{3GM}{d^3}}$.

(D) The reaction sequence is the Kolbe-Schmitt reaction followed by acetylation to form aspirin $(C_9H_8O_4)$.
$1$. Phenol reacts with $NaOH$, $CO_2$, and $H^+$ to form salicylic acid.
$2$. Salicylic acid reacts with acetic anhydride $((CH_3CO)_2O)$ in the presence of $H^+$ to form aspirin $(Z)$ and acetic acid $(CH_3COOH)$.
$3$. The molecular formula of aspirin $(Z)$ is $C_9H_8O_4$.
$4$. The molar mass of $C_9H_8O_4 = (9 \times 12) + (8 \times 1) + (4 \times 16) = 108 + 8 + 64 = 180 \ g/mol$.
$5$. The mass of carbon in one mole of $C_9H_8O_4 = 9 \times 12 = 108 \ g$.
$6$. Percentage of carbon = $\frac{\text{Mass of Carbon}}{\text{Molar Mass of } Z} \times 100 = \frac{108}{180} \times 100 = 60 \%$.

(C) Decarboxylation of sodium salts of carboxylic acids with sodalime $(NaOH + CaO)$ removes one carbon atom as $Na_2CO_3$ to form an alkane with one less carbon atom.
To obtain propane ($CH_3-CH_2-CH_3$,$3$ carbons),the starting material must be a sodium salt of butanoic acid ($4$ carbons).
The reaction is: $CH_3-CH_2-CH_2-COONa + NaOH \xrightarrow[\Delta]{CaO} CH_3-CH_2-CH_3 + Na_2CO_3$.
Therefore,compound '$A$' is sodium butanoate $(CH_3-CH_2-CH_2-CO_2Na)$.

(A) The homologous series of alkanes is represented by the general formula $C_n H_{2n+2}$.
Methane is $CH_4$ $(n=1)$.
The next member in the homologous series is ethane,$C_2 H_6$ $(n=2)$.
$I$. Wurtz reaction: $2 CH_3 Br \xrightarrow[\text{dry ether}]{Na} CH_3-CH_3 + 2 NaBr$. This produces ethane.
$II$. Decarboxylation: $CH_3 COONa \xrightarrow[CaO, \Delta]{NaOH} CH_4 + Na_2 CO_3$. This produces methane.
$III$. Hydrogenation: $CH_3-CH=CH_2 \xrightarrow{H_2 / Pt} CH_3-CH_2-CH_3$. This produces propane.
$IV$. Reduction of alkyl halide: $CH_3-CH_2 Br \xrightarrow[H^{+}]{Zn} CH_3-CH_3 + HBr$. This produces ethane.
Therefore,reactions $I$ and $IV$ produce ethane.

(D) $1$. The alkene $X$ is $but-2-ene$ $(CH_3CH=CHCH_3)$,which exhibits geometrical isomerism.
$2$. Oxidation of $but-2-ene$ with acidic $KMnO_4$ yields acetic acid $(CH_3COOH)$ as $Y$.
$3$. The sodium salt of $Y$ is sodium acetate $(CH_3COONa)$.
$4$. Heating sodium acetate with soda lime $(NaOH + CaO)$ undergoes decarboxylation to produce methane $(CH_4)$ as $Z$.

(B) Acrolein is an unsaturated aldehyde with the chemical formula $CH_2=CH-CHO$.
It is formed as a byproduct of photochemical smog reactions involving unburnt hydrocarbons,nitrogen oxides $(NO_x)$,and ozone $(O_3)$ in the atmosphere.
The reaction products typically include formaldehyde $(CH_2=O)$,acrolein $(CH_2=CH-CHO)$,and peroxyacetyl nitrate $(CH_3CO(OO)NO_2)$.

(B) $1$. The reaction of $C_3H_4$ (propyne) with $H_2O$ in the presence of $Hg^{2+}/H^+$ at $333 \ K$ is a hydration reaction that follows Markovnikov's rule to form an enol intermediate,which then tautomerizes to form $X$ (acetone,$CH_3COCH_3$).
$2$. The second part of the reaction shows that $Y$ undergoes ozonolysis $((i) O_3, (ii) Zn/H_2O)$ to produce $X$ (acetone).
$3$. Ozonolysis of an alkene $R_2C=CR_2$ produces two carbonyl compounds. To obtain acetone $(CH_3COCH_3)$ as the product,the alkene $Y$ must be $2,3-$dimethyl$-2-$butene $((CH_3)_2C=C(CH_3)_2)$.
$4$. Looking at the options provided,the structure in option $B$ represents $2,3-$dimethyl$-2-$butene.

(C) The given reaction is the Friedel-Crafts alkylation of benzene with ethyl chloride $(C_2H_5Cl)$ in the presence of anhydrous $AlCl_3$ to form ethylbenzene $(X)$.
The chemical formula of ethylbenzene $(X)$ is $C_8H_{10}$.
Atomic mass of Carbon $(C)$ $= 12.01 \ g/mol$.
Atomic mass of Hydrogen $(H)$ $= 1.008 \ g/mol$.
Molar mass of $C_8H_{10} = (8 \times 12.01) + (10 \times 1.008) = 96.08 + 10.08 = 106.16 \ g/mol$.
Percentage of carbon in $C_8H_{10} = \frac{\text{Mass of Carbon}}{\text{Total Molar Mass}} \times 100$.
Percentage of carbon $= \frac{96.08}{106.16} \times 100 \approx 90.505 \% \approx 90.6 \%$.

(C) $1$. The reaction begins with propene $(C_3H_6)$.
$2$. The reagent $X$ is $HBr/ROOR$ (peroxide effect/anti-Markovnikov addition),which converts propene to $1$-bromopropane $(Y = CH_3CH_2CH_2Br)$.
$3$. In the presence of $AlCl_3$,$1$-bromopropane undergoes rearrangement to form a more stable isopropyl carbocation $(CH_3CH^+CH_3)$.
$4$. This isopropyl carbocation then undergoes Friedel-Crafts alkylation with benzene $(C_6H_6)$ to form isopropylbenzene (cumene) as the major product $(Z)$.
$5$. Thus,$X$ is $HBr/ROOR$ and $Z$ is isopropylbenzene.

(A) Activating groups are those that donate electron density to the benzene ring,typically via resonance (e.g.,lone pairs). Deactivating groups are those that withdraw electron density from the benzene ring,typically via inductive or resonance effects.
$1$. $-OCH_2CH_3$: Activating (due to lone pair on oxygen).
$2$. $-COCH_3$: Deactivating (electron-withdrawing carbonyl group).
$3$. $-NHCOCH_3$: Activating (due to lone pair on nitrogen).
$4$. $-COOCH_3$: Deactivating (electron-withdrawing ester group).
$5$. $-SO_3H$: Deactivating (electron-withdrawing sulfonic acid group).
Thus,there are $2$ activating groups $(-OCH_2CH_3, -NHCOCH_3)$ and $3$ deactivating groups $(-COCH_3, -COOCH_3, -SO_3H)$.
Therefore,the correct answer is $2, 3$.

(B) $1$. The reaction of $C_2H_4Br_2$ ($1$,$2$-dibromoethane) with $Alc. KOH$ followed by $NaNH_2$ leads to dehydrohalogenation to form ethyne $(X = HC \equiv CH)$.
$2$. Passing ethyne through a red-hot iron tube at $873 \ K$ results in cyclic polymerization to form benzene $(Y = C_6H_6)$.
$3$. Benzene reacts with acetyl chloride $(CH_3COCl)$ in the presence of anhydrous $AlCl_3$ via Friedel-Crafts acylation to produce acetophenone $(Z = C_6H_5COCH_3)$.

(B) The hydrides of alkaline earth metals,specifically $BeH_2$ and $MgH_2$,exhibit a polymeric structure due to the presence of electron-deficient three-center two-electron $(3c-2e)$ bonds.
These structures allow the metal atoms to achieve a stable coordination environment through bridging hydrogen atoms.

(C) The structure of $H_2O_2$ is non-planar.
In the gaseous phase,the dihedral angle is $111.5^{\circ}$.
In the solid phase,due to hydrogen bonding,the dihedral angle decreases to $90.2^{\circ}$.
Therefore,the dihedral angles in gaseous and solid phases are $111.5^{\circ}$ and $90.2^{\circ}$ respectively.

(A) $H_2O_2$ is unstable and decomposes into $H_2O$ and $O_2$ in the presence of light or rough surfaces like glass containing alkali oxides.
To prevent this,it is stored in wax-lined glass or plastic bottles to provide a smooth,inert surface.
Additionally,it must be kept in the dark to avoid photochemical decomposition.

(C) In reaction $I$,the oxidation state of $H$ in $H_2O$ changes from $+1$ to $0$ (in $H_2$),which is a decrease in oxidation number,hence water is reduced.
In reaction $II$,the oxidation state of $O$ in $H_2O$ changes from $-2$ to $0$ (in $O_2$),which is an increase in oxidation number,hence water is oxidized.
Therefore,in reaction $I$ water is reduced and in reaction $II$ water is oxidized.

(D) The relationship between volume strength and normality of $H_2O_2$ is given by the formula: $\text{Volume strength} = 5.6 \times \text{Normality}$.
Given,volume strength = $20$.
Therefore,$\text{Normality} = \frac{\text{Volume strength}}{5.6} = \frac{20}{5.6} \approx 3.571 \ N$.
Thus,the correct option is $D$.

(C) Lewis acid is an electron-pair acceptor.
$(i)$ $NH_3$: Has a lone pair on $N$,acts as a Lewis base.
$(ii)$ $AlCl_3$: $Al$ has an incomplete octet ($6$ electrons),acts as a Lewis acid.
$(iii)$ $SnCl_4$: $Sn$ has vacant $d$-orbitals and can expand its octet,acts as a Lewis acid.
$(iv)$ $CO_2$: The $C$ atom is electron-deficient due to the high electronegativity of oxygen atoms,acts as a Lewis acid.
$(v)$ $Ag^{+}$: $A$ metal cation with vacant orbitals,acts as a Lewis acid.
$(vi)$ $HSO_4^{-}$: Can act as a Lewis base (due to lone pairs on $O$) or a Brønsted-Lowry acid,but is not typically classified as a Lewis acid in this context.
Therefore,$AlCl_3$,$SnCl_4$,$CO_2$,and $Ag^{+}$ act as Lewis acids.
The total count is $4$.

(C) The average kinetic energy per molecule of an ideal gas is given by the formula $\langle KE \rangle = \frac{f}{2} k_B T$,where $f$ is the degrees of freedom,$k_B$ is the Boltzmann constant,and $T$ is the absolute temperature.
Since both hydrogen and nitrogen are diatomic gases at $30^{\circ} C$,they have the same number of degrees of freedom $(f=5)$.
Because the gases are in the same vessel,they are in thermal equilibrium,meaning they are at the same temperature $(T_{H_2} = T_{N_2} = T)$.
Therefore,the ratio of the average kinetic energies per molecule is $\frac{\langle KE \rangle_{H_2}}{\langle KE \rangle_{N_2}} = \frac{\frac{5}{2} k_B T}{\frac{5}{2} k_B T} = 1:1$.
The mass ratio is irrelevant because the average kinetic energy per molecule depends only on the temperature,not on the mass or the number of molecules.

(B) Given: Radius $r = 9 \ km = 9000 \ m$. Speed $v = 540 \ km/h = 540 \times \frac{5}{18} \ m/s = 150 \ m/s$. Acceleration due to gravity $g = 10 \ m/s^2$.
For a banked aircraft in a horizontal loop,the banking angle $\theta$ is given by the relation: $\tan \theta = \frac{v^2}{rg}$.
Substituting the values: $\tan \theta = \frac{150 \times 150}{9000 \times 10} = \frac{22500}{90000} = \frac{1}{4}$.
Since $\tan \theta = \frac{1}{4}$,we have $\cot \theta = 4$,which implies $\theta = \cot^{-1}(4)$.

(D) Given: Length of wire $l = 20 \ cm = 0.2 \ m$,Current $I = \frac{3}{\pi^2} \ A$.
When the wire is bent into a circle,its length becomes the circumference of the circle: $l = 2 \pi R$.
So,$0.2 = 2 \pi R \Rightarrow R = \frac{0.1}{\pi} \ m = \frac{10^{-1}}{\pi} \ m$.
The magnetic field $B$ at the centre of a circular loop is given by $B = \frac{\mu_0 I}{2R}$.
Substituting the values: $B = \frac{(4 \pi \times 10^{-7}) \times (\frac{3}{\pi^2})}{2 \times (\frac{10^{-1}}{\pi})}$.
Simplifying: $B = \frac{4 \pi \times 10^{-7} \times 3}{2 \times 10^{-1} \times \pi} = \frac{12 \pi \times 10^{-7}}{2 \pi \times 10^{-1}} = 6 \times 10^{-6} \ T$.

(D) Given: Diameter $D = 4 \ mm$,so radius $R = 2 \ mm$.
For a point inside the rod $(r < R)$,the magnetic field is given by $B_{in} = \frac{\mu_0 i r}{2 \pi R^2}$.
At $r_1 = 1 \ mm$,$B_1 = \frac{\mu_0 i (1)}{2 \pi (2)^2} = \frac{\mu_0 i}{8 \pi}$.
For a point outside the rod $(r > R)$,the magnetic field is given by $B_{out} = \frac{\mu_0 i}{2 \pi r}$.
At $r_2 = 4 \ mm$,$B_2 = \frac{\mu_0 i}{2 \pi (4)} = \frac{\mu_0 i}{8 \pi}$.
The ratio of the magnetic fields is $\frac{B_1}{B_2} = \frac{\mu_0 i / 8 \pi}{\mu_0 i / 8 \pi} = 1:1$.

(A) Physical laws are based on conserved quantities,which can be either scalars (like energy) or vectors (like linear momentum and angular momentum). Therefore,the statement that all conserved quantities are necessarily scalars is incorrect.

(D) Given: Side of the cube $a = 10 \ cm$, depth in water $x = 3 \ cm$, density of oil $\delta_1 = 0.9 \ g \ cm^{-3}$, density of water $\delta_2 = 1 \ g \ cm^{-3}$.
Since the cube is floating, the weight of the cube is equal to the total buoyant force exerted by the oil and water.
Weight of cube $W = mg$.
Buoyant force $F_B = (\text{Volume in oil} \times \delta_1 \times g) + (\text{Volume in water} \times \delta_2 \times g)$.
Volume of cube $V = a^3 = 10^3 = 1000 \ cm^3$.
Volume in water $V_2 = a^2 \times x = 10^2 \times 3 = 300 \ cm^3$.
Volume in oil $V_1 = a^2 \times (a - x) = 10^2 \times (10 - 3) = 700 \ cm^3$.
By the law of flotation, $mg = V_1 \delta_1 g + V_2 \delta_2 g$.
$m = V_1 \delta_1 + V_2 \delta_2$.
$m = (700 \times 0.9) + (300 \times 1)$.
$m = 630 + 300 = 930 \ g$.

(C) The volume of the liquid remains constant during the splitting process.
When a large drop of radius $R$ splits into $n$ small drops of radius $r$,the total volume is conserved: $\frac{4}{3} \pi R^3 = n \times \frac{4}{3} \pi r^3$,which implies $R^3 = nr^3$ or $r = R / n^{1/3}$.
The total surface area of the large drop is $A_1 = 4 \pi R^2$.
The total surface area of $n$ small drops is $A_2 = n \times 4 \pi r^2 = n \times 4 \pi (R / n^{1/3})^2 = n^{1/3} \times 4 \pi R^2$.
Since $n > 1$,$n^{1/3} > 1$,therefore $A_2 > A_1$. The surface area increases.
Surface energy is given by $U = T \times A$,where $T$ is surface tension.
Since the surface area increases,the surface energy of the system increases.
To increase the surface energy,work must be done on the system,meaning energy is absorbed from the surroundings.

(B) Given: Mass $m = 2 \ kg$,length $l = 2 \ m$,area $A = 1 \ mm^2 = 1 \times 10^{-6} \ m^2$,maximum elongation $\Delta l = 2 \ mm = 2 \times 10^{-3} \ m$,$g = 10 \ ms^{-2}$.
For the tension to be zero at the highest point in a vertical circle,the velocity at the bottom point must be $v = \sqrt{5gl}$.
$v = \sqrt{5 \times 10 \times 2} = \sqrt{100} = 10 \ ms^{-1}$.
The maximum tension $T_{\max}$ occurs at the lowest point of the vertical circle.
$T_{\max} = mg + \frac{mv^2}{l} = 2 \times 10 + \frac{2 \times (10)^2}{2} = 20 + 100 = 120 \ N$.
Using the formula for Young's modulus $Y = \frac{T_{\max} \cdot l}{A \cdot \Delta l}$:
$Y = \frac{120 \times 2}{1 \times 10^{-6} \times 2 \times 10^{-3}} = \frac{240}{2 \times 10^{-9}} = 120 \times 10^9 = 1.2 \times 10^{11} \ Nm^{-2}$.

(D) For uniformly accelerated motion,the distance covered in the $n^{th}$ second is given by $S_n = u + \frac{a}{2}(2n - 1)$.
For the fourth second $(n=4)$: $25 = u + \frac{a}{2}(2 \times 4 - 1) \implies 25 = u + 3.5a$ --- $(i)$
For the sixth second $(n=6)$: $37 = u + \frac{a}{2}(2 \times 6 - 1) \implies 37 = u + 5.5a$ --- (ii)
Subtracting $(i)$ from (ii): $(37 - 25) = (5.5a - 3.5a) \implies 12 = 2a \implies a = 6 \ m/s^2$.
Substituting $a=6$ in $(i)$: $25 = u + 3.5(6) \implies 25 = u + 21 \implies u = 4 \ m/s$.
The distance covered in the next two seconds (from $t=6 \ s$ to $t=8 \ s$) is calculated using $S = ut + \frac{1}{2}at^2$ where $u$ is the velocity at $t=6 \ s$.
Velocity at $t=6 \ s$: $v = u + at = 4 + (6 \times 6) = 40 \ m/s$.
Distance covered in $t=2 \ s$: $S = (40 \times 2) + \frac{1}{2} \times 6 \times (2)^2 = 80 + 12 = 92 \ m$.

(D) For projectile motion,the angle of projection is $\theta = \tan^{-1}(\sqrt{7})$,so $\tan \theta = \sqrt{7}$.
At any height $y$,the vertical component of velocity is $v_y^2 = u_y^2 - 2gy$.
At $y = \frac{H}{2}$,where $H = \frac{u^2 \sin^2 \theta}{2g}$ is the maximum height:
$v_y^2 = u^2 \sin^2 \theta - 2g \left( \frac{u^2 \sin^2 \theta}{4g} \right) = u^2 \sin^2 \theta - \frac{u^2 \sin^2 \theta}{2} = \frac{u^2 \sin^2 \theta}{2}$.
The horizontal component of velocity remains constant: $v_x = u \cos \theta$.
The speed $v$ at $y = \frac{H}{2}$ is $v = \sqrt{v_x^2 + v_y^2} = \sqrt{u^2 \cos^2 \theta + \frac{u^2 \sin^2 \theta}{2}}$.
Given $v = nu$,we have $n^2 u^2 = u^2 \cos^2 \theta + \frac{u^2 \sin^2 \theta}{2}$.
$n^2 = \cos^2 \theta + \frac{\sin^2 \theta}{2} = \cos^2 \theta + \frac{1 - \cos^2 \theta}{2} = \frac{1 + \cos^2 \theta}{2}$.
Since $\tan \theta = \sqrt{7}$,$\tan^2 \theta = 7$,so $\sec^2 \theta = 1 + 7 = 8$,which means $\cos^2 \theta = \frac{1}{8}$.
$n^2 = \frac{1 + 1/8}{2} = \frac{9/8}{2} = \frac{9}{16}$.
Therefore,$n = \frac{3}{4}$.

(C) The neutron multiplication factor $K$ (also denoted as $k$) is defined as the ratio of the number of neutrons produced in a generation to the number of neutrons produced in the preceding generation.
When $K = 1$,the rate of fission remains constant,and the reactor is said to be in a 'critical' state.
If $K > 1$,the reactor is 'supercritical' (power increases).
If $K < 1$,the reactor is 'subcritical' (power decreases).
Therefore,for a critical reactor,$K = 1$.

(D) In the decay of a nucleus with mass number $A=236$,the parent nucleus decays into an $\alpha$-particle (mass number $4$) and a daughter nucleus (mass number $A-4 = 232$).
By the law of conservation of linear momentum,the magnitude of momentum of the $\alpha$-particle $(P_{\alpha})$ must be equal to the magnitude of momentum of the daughter nucleus $(P_{d})$,so $P_{\alpha} = P_{d} = P$.
The kinetic energy $KE$ is given by $KE = \frac{P^2}{2m}$,which implies $KE \propto \frac{1}{m}$.
Therefore,the ratio of kinetic energies is $\frac{(KE)_{d}}{(KE)_{\alpha}} = \frac{m_{\alpha}}{m_{d}} = \frac{4}{232} = \frac{1}{58}$.
Given $(KE)_{\alpha} = E$,we find $(KE)_{d} = \frac{E}{58}$.
The total energy released $(Q)$ is the sum of the kinetic energies of the products:
$Q = (KE)_{\alpha} + (KE)_{d} = E + \frac{E}{58} = \frac{59E}{58}$.

(A) On balancing the equation $(1)$:
$2BF_3 + 6NaH \xrightarrow{450 \ K} B_2H_6 + 6NaF$
So,$X$ is $B_2H_6$ (diborane).
Balancing the equation $(2)$:
$B_2H_6 + 6H_2O \longrightarrow 2B(OH)_3 + 6H_2 \uparrow$
So,$Y$ is $B(OH)_3$ (boric acid).
$I$. $X$ $(B_2H_6)$ is an electron-deficient molecule (contains $3c-2e^-$ bonds). This is correct.
$II$. In $X$ $(B_2H_6)$,there is no direct $B-B$ bond; it has two bridging hydrogen atoms $(B-H-B)$. This is incorrect.
$III$. $Y$ $(B(OH)_3)$ is a weak monobasic Lewis acid,not a tribasic acid. This is incorrect.
$IV$. $Y$ $(B(OH)_3)$ acts as a Lewis acid as it accepts a lone pair of electrons from $OH^-$ ions. This is correct.
Therefore,statements $I$ and $IV$ are correct.

(A) The first reaction is: $4BF_3 + 3LiAlH_4 \xrightarrow{(C_2H_5)_2O} 2B_2H_6 + 3LiF + 3AlF_3$. Here,$X$ is $B_2H_6$ (diborane).
The second reaction is: $B_2H_6 + 2NaH \longrightarrow 2NaBH_4$. Here,$Y$ is $NaBH_4$ (sodium borohydride).
$NaBH_4$ contains the $BH_4^-$ ion,where hydrogen is in the $-1$ oxidation state.
$NaBH_4$ is a well-known reducing agent,not an oxidizing agent.
Therefore,the statement 'It is a good oxidizing agent' is incorrect.

(B) The reaction of diborane $(B_2 H_6)$ with carbon monoxide $(CO)$ under pressure yields the borane carbonyl adduct,$BH_3 \cdot CO$ $(X)$.
$B_2 H_6 + 2 CO \rightarrow 2 BH_3 \cdot CO$
The reaction of diborane $(B_2 H_6)$ with sodium hydride $(NaH)$ in the presence of diethyl ether $((C_2 H_5)_2 O)$ yields sodium borohydride,$NaBH_4$ $(Y)$.
$B_2 H_6 + 2 NaH \xrightarrow{(C_2 H_5)_2 O} 2 NaBH_4$
Therefore,$X = BH_3 \cdot CO$ and $Y = NaBH_4$.

(C) The chemical formula of borax is $Na_2[B_4O_5(OH)_4] \cdot 8H_2O$.
Comparing this with the given expression $Na_2[B_4O_5(OH)_x] \cdot yH_2O$,we get:
$x = 4$
$y = 8$
Therefore,the sum $x + y = 4 + 8 = 12$.

(B) $(i)$ $CCl_4$ does not undergo hydrolysis because it lacks vacant $d$-orbitals to accept lone pairs from water molecules.
$(ii)$ The structure of diamond consists of a $3-D$ network of carbon atoms with directional covalent bonds throughout the lattice.
$(iii)$ The thermodynamic stability order of carbon allotropes is $\text{Graphite} > \text{Diamond} > \text{Fullerene}$.
$(iv)$ Glass is a man-made amorphous silicate.
Therefore,statements $(ii)$ and $(iv)$ are correct.

(B) Silicones are organosilicon polymers which have the general empirical formula $(R_2SiO)_n$.
They are hydrophobic in nature due to the presence of organic groups $(R)$ attached to the silicon atoms.
Because of this hydrophobic nature,they are widely used for water proofing of fabrics.
Thus,both Assertion $(A)$ and Reason $(R)$ are correct,and $(R)$ is the correct explanation of $(A)$.

(D) The balanced chemical equations are:
$P_2O_3 + 3H_2O \rightarrow 2H_3PO_3$ $(X = H_3PO_3)$
$P_4O_{10} + 6H_2O \rightarrow 4H_3PO_4$ $(Y = H_3PO_4)$
In $H_3PO_3$ (phosphorous acid),the structure contains $1$ $P=O$ bond,$2$ $P-OH$ bonds,and $1$ $P-H$ bond.
In $H_3PO_4$ (phosphoric acid),the structure contains $1$ $P=O$ bond and $3$ $P-OH$ bonds.
Thus,the number of $P=O$ bonds in $X$ and $Y$ are $1$ and $1$ respectively.

(A) The chemical reaction is: $C + 2H_2SO_4 \rightarrow CO_2 + 2SO_2 + 2H_2O$.
Both $CO_2$ and $SO_2$ are non-metallic oxides,which are acidic in nature.

(B) The reaction of carbon with concentrated sulfuric acid is an oxidation-reduction reaction.
Carbon is oxidized to carbon dioxide $(CO_2)$ and sulfuric acid is reduced to sulfur dioxide $(SO_2)$.
The balanced chemical equation is:
$C + 2 H_2SO_4 \xrightarrow{\Delta} CO_2 + 2 SO_2 + 2 H_2O$
Thus,$X$ is $CO_2$ and $Y$ is $SO_2$.

(C) The boiling points of hydrogen halides follow the order: $HCl < HBr < HI < HF$.
$HF$ exhibits strong intermolecular hydrogen bonding,which significantly increases its boiling point compared to other hydrogen halides.
For the remaining hydrogen halides $(HCl, HBr, HI)$,the boiling point increases with an increase in molecular mass due to stronger van der Waals forces.

(C) For a plano-convex lens,the radius of curvature $R$ is related to the aperture radius $r$ and thickness $t$ by the geometric relation: $R^2 = (R-t)^2 + r^2$. Since $t$ is very small,$R^2 \approx R^2 - 2Rt + t^2 + r^2$,which simplifies to $2Rt \approx r^2$,or $R = \frac{r^2}{2t}$.
Given the diameter $d = 8.4 \text{ cm}$,the radius $r = \frac{d}{2} = 4.2 \text{ cm}$.
Using the Lens Maker's Formula: $\frac{1}{f} = (\mu - 1)(\frac{1}{R_1} - \frac{1}{R_2})$.
For a plano-convex lens,$R_1 = R$ and $R_2 = \infty$,so $\frac{1}{f} = (\mu - 1)(\frac{1}{R})$.
Thus,$R = f(\mu - 1)$.
Substituting $R = \frac{r^2}{2t}$,we get $\frac{r^2}{2t} = f(\mu - 1)$.
Rearranging for thickness $t$: $t = \frac{r^2}{2f(\mu - 1)}$.
Substituting the values: $r = 4.2 \text{ cm}$,$f = 73.5 \text{ cm}$,$\mu = \frac{5}{3}$.
$t = \frac{(4.2)^2}{2 \times 73.5 \times (\frac{5}{3} - 1)} = \frac{17.64}{147 \times \frac{2}{3}} = \frac{17.64}{98} = 0.18 \text{ cm} = 1.8 \text{ mm}$.

(C) Let the height of the object be $h_0 = 12 \ cm$.
When the lens is placed at the first position,the image height is $h_1 = 18 \ cm$.
When the lens is moved to the second position,the image height is $h_2$.
According to the displacement method for a convex lens,the relationship between the object height and the two image heights is given by the formula:
$h_0 = \sqrt{h_1 \cdot h_2}$
Substituting the given values:
$12 = \sqrt{18 \cdot h_2}$
Squaring both sides:
$144 = 18 \cdot h_2$
$h_2 = \frac{144}{18} = 8 \ cm$.
Thus,the height of the second image is $8 \ cm$.

(B) Assign oxidation states to each element:
In $KClO_3$,the oxidation state of $K$ is $+1$,$Cl$ is $+5$,and $O$ is $-2$.
In $KCl$,the oxidation state of $K$ is $+1$ and $Cl$ is $-1$.
In $O_2$,the oxidation state of $O$ is $0$.
Comparing the oxidation states:
$Cl$ changes from $+5$ to $-1$ (decrease in oxidation state,so $Cl$ is reduced).
$O$ changes from $-2$ to $0$ (increase in oxidation state,so $O$ is oxidized).
Therefore,$Cl$ is reduced and $O$ is oxidized.

(A) $Ti = [Ar] 3d^2 4s^2$
$Ti^{3+} = [Ar] 3d^1 4s^0$
$Ti^{4+} = [Ar] 3d^0 4s^0$
Since $Ti^{4+}$ has a vacant $d$-orbital and it acquires the nearest noble gas configuration,it is more stable compared to $Ti^{3+}$ and $Ti^{2+}$.
$Cu^{2+} / Cu = +0.34 \ V$ has an exceptionally positive reduction potential among $3d$ series elements.
$Sc$ shows $+3$ oxidation state and $Zn$ shows $+2$ oxidation state. They do not exhibit $+1$ oxidation state.
Therefore,only $(i)$ and $(ii)$ are correct.

(C) The standard electrode potential $(E^{\circ})$ for $M^{3+}/M^{2+}$ follows the order $Mn^{3+}/Mn^{2+} > Fe^{3+}/Fe^{2+} > Cr^{3+}/Cr^{2+}$. Thus,option $A$ is incorrect.
Magnetic moment is calculated as $\mu = \sqrt{n(n+2)}$,where $n$ is the number of unpaired electrons.
For $Mn^{2+}$ $(3d^5)$,$n=5$. For $Cr^{2+}$ $(3d^4)$,$n=4$. For $Fe^{2+}$ $(3d^6)$,$n=4$. The order is $Mn^{2+} > Cr^{2+} \approx Fe^{2+}$. Thus,option $B$ is incorrect.
Oxidizing power depends on the reduction potential and the stability of the lower oxidation state. $MnO_4^-$ $(Mn^{+7})$ is a stronger oxidizing agent than $Cr_2O_7^{2-}$ $(Cr^{+6})$ and $VO_2^+$ $(V^{+5})$. The correct order is $VO_2^+ < Cr_2O_7^{2-} < MnO_4^-$. Thus,option $C$ is correct.
First ionization enthalpy generally increases across the period but shows irregularities due to electronic configuration. The order is $Ti < V < Cr$ is incorrect; the correct order is $Ti < V < Cr$ is not followed due to stability of half-filled $d$-orbitals in $Cr$.

(A) The electronic configurations are:
$Eu (Z=63): [Xe] 4f^7 6s^2$ (Half-filled $f$-orbital)
$Pu (Z=94): [Rn] 5f^6 7s^2$
$Cf (Z=98): [Rn] 5f^{10} 7s^2$
$Sm (Z=62): [Xe] 4f^6 6s^2$
$Gd (Z=64): [Xe] 4f^7 5d^1 6s^2$ (Half-filled $f$-orbital)
$Cm (Z=96): [Rn] 5f^7 6d^1 7s^2$ (Half-filled $f$-orbital)
The elements with half-filled $f$-orbitals are $Eu, Gd,$ and $Cm$.
Total count = $3$.

(C) For a hydrogen electrode,$E = -0.0591 \times pH = -0.0591 \times 10 = -0.591 \ V$. Thus,statement $(A)$ is correct.
$(B)$ $\Lambda^{\circ}_{m}(CaCl_2) = \Lambda^{\circ}_{m}(Ca^{2+}) + 2 \times \Lambda^{\circ}_{m}(Cl^{-}) = 119 + 2(76) = 119 + 152 = 271 \ S \ cm^2 \ mol^{-1}$. Since $271 \neq 195$,statement $(B)$ is incorrect.
$(C)$ The Nernst equation at equilibrium is $E_{cell} = 0$,which gives $0 = E_{cell}^{0} - \frac{2.303 RT}{nF} \log K_{c}$,so $E_{cell}^{0} = \frac{2.303 RT}{nF} \log K_{c}$. Thus,statement $(C)$ is correct.
Therefore,statements $(A)$ and $(C)$ are correct.

(C) During the electrolysis of molten $NaCl$,$Na$ metal is deposited at the cathode,and $Cl_2$ gas is liberated at the anode.
At cathode: $Na^+ + e^- \rightarrow Na$
At anode: $2Cl^- \rightarrow Cl_2 + 2e^-$
Thus,Statement $I$ is correct.
When aqueous $CuSO_4$ is electrolyzed using $Pt$ electrodes,$Cu$ is deposited at the cathode and $O_2$ is liberated at the anode.
At cathode: $Cu^{2+} + 2e^- \rightarrow Cu$
At anode: $2H_2O \rightarrow O_2 + 4H^+ + 4e^-$
Since $O_2$ is liberated at the anode,not the cathode,Statement $II$ is incorrect.
Therefore,Statement $I$ is correct but Statement $II$ is incorrect.

(A) The standard electrode potential $(E^{\circ})$ represents the tendency of an electrode to undergo reduction.
For $Li^{+} / Li$,the standard electrode potential is $-3.04 \ V$.
For $Na^{+} / Na$,the standard electrode potential is $-2.714 \ V$.
Therefore,the values are $-3.04 \ V$ and $-2.714 \ V$ respectively.

(B) According to Faraday's law of electrolysis,$W = Z \times i \times t$,where $Z$ is the electrochemical equivalent.
Given:
$W = 0.592 \ g$
$i = 0.5 \ A$
$t = 60 \ minutes = 60 \times 60 \ s = 3600 \ s$
Substituting the values in the formula:
$0.592 = Z \times 0.5 \times 3600$
$0.592 = Z \times 1800$
$Z = \frac{0.592}{1800} \approx 3.288 \times 10^{-4} \ g \ C^{-1}$
Rounding to two significant figures,$Z \approx 3.3 \times 10^{-4} \ g \ C^{-1}$.

(D) Given: Concentration $C = 0.02 \ M = 0.02 \ mol \ L^{-1} = 0.02 \times 1000 \ mol \ m^{-3} = 20 \ mol \ m^{-3}$.
Cell constant $G^* = 129 \ m^{-1}$.
Molar conductivity $\Lambda_m = 124 \times 10^{-4} \ S \ m^2 \ mol^{-1}$.
Formula: $\Lambda_m = \frac{\kappa}{C}$,where $\kappa$ is conductivity.
$\kappa = \Lambda_m \times C = (124 \times 10^{-4} \ S \ m^2 \ mol^{-1}) \times (20 \ mol \ m^{-3}) = 2480 \times 10^{-4} \ S \ m^{-1} = 0.248 \ S \ m^{-1}$.
Resistance $R = \frac{G^*}{\kappa} = \frac{129 \ m^{-1}}{0.248 \ S \ m^{-1}} \approx 520.16 \ \Omega$.
Rounding to the nearest integer,$R = 520 \ \Omega$.

(B) An allylic halide is a compound in which the halogen atom is bonded to an $sp^3$ hybridized carbon atom that is adjacent to a carbon-carbon double bond $(C=C)$.
In option $B$,the chlorine atom is attached to a carbon atom that is directly adjacent to the $C=C$ double bond. This carbon is $sp^3$ hybridized,making it an allylic halide.
In option $A$,the chlorine is attached directly to the double-bonded carbon,which is a vinylic halide.
Therefore,the correct option is $B$.

(B) $1$. Identify the longest carbon chain attached to the oxygen atom. The chain has $5$ carbon atoms,so the parent alkane is pentane.
$2$. The phenoxy group $(-OC_6H_5)$ is attached to the first carbon of the pentane chain.
$3$. Number the chain starting from the carbon attached to the oxygen atom to give the lowest possible number to the substituent.
$4$. The methyl group is at the $4^{th}$ position.
$5$. Thus,the $IUPAC$ name is $1-$phenoxy$-4-$methylpentane.
$6$. However,based on the provided options and the structure,the compound is named as an ether derivative of benzene where the alkoxy group is $4-$methylpentoxy. Therefore,the name is $4-$methylpentoxybenzene.

(A) $1$. Bithionol: It contains two $-OH$ groups attached to the two benzene rings.
$2$. Terpineol: It contains one $-OH$ group in its structure.
$3$. Chloroxylenol: It contains one $-OH$ group attached to the benzene ring.
Therefore,the number of $-OH$ groups in bithionol,terpineol,and chloroxylenol is $2, 1, 1$ respectively.

(C) Kaolinite is a clay mineral with the chemical formula $Al_2Si_2O_5(OH)_4$,which contains the metal $Al$ (Aluminum).
Calamine is a zinc ore with the chemical formula $ZnCO_3$,which contains the metal $Zn$ (Zinc).
Therefore,$X$ is $Al$ and $Y$ is $Zn$.

(C) Copper matte is obtained during the extraction of copper from copper pyrites.
It is primarily a mixture of copper$(I)$ sulphide $(Cu_2S)$ and iron$(II)$ sulphide $(FeS)$.

(B) Distillation is the process used to refine metals that have low boiling points,such as $Zn$,$Cd$,and $Hg$. In this method,the impure metal is heated to vaporize the pure metal,leaving behind the high boiling point impurities as a residue. The vapors are then condensed to obtain the pure metal.

(D) Calcination is a process of heating an ore in a limited supply of air or in the absence of air to remove volatile impurities and moisture.
It is typically used for carbonate or hydrated oxide ores.
The reaction $Fe_2O_3 \cdot x H_2O \xrightarrow{\Delta} Fe_2O_{3(s)} + x H_2O_{(g)}$ represents the removal of water of hydration from limonite,which is a characteristic calcination process.

(A) The reaction sequence is as follows:
$1$. Hydrolysis of $C_4 H_9 Br$ $(X)$ gives an alcohol $Y$ $(C_4 H_9 OH)$.
$2$. The alcohol $Y$ undergoes dehydration with $20 \% H_3 PO_4$ at $358 \ K$ to form an alkene.
$3$. Tertiary butyl bromide,$(CH_3)_3 CBr$,on hydrolysis gives tertiary butyl alcohol,$(CH_3)_3 COH$.
$4$. Tertiary butyl alcohol undergoes easy dehydration with $20 \% H_3 PO_4$ at $358 \ K$ to form isobutylene ($2$-methylpropene),which is consistent with the provided reaction conditions.
$5$. Therefore,$X$ is $(CH_3)_3 CBr$.

(A) Step $1$: Dehydrohalogenation of tert-butyl bromide with alcoholic $KOH$ gives isobutylene ($2$-methylpropene).
$(CH_3)_3 CBr \xrightarrow{\text{alc. } KOH, \Delta} (CH_3)_2 C=CH_2 + KBr + H_2O$
Step $2$: Electrophilic addition of $HBr$ to isobutylene follows Markovnikov's rule to form tert-butyl bromide as the major product.
$(CH_3)_2 C=CH_2 + HBr \rightarrow (CH_3)_3 CBr$

(D) $S_{N}1$ mechanism proceeds via the formation of a carbocation intermediate. The stability of the carbocation determines the ease of the $S_{N}1$ reaction.
$(A)$ Benzyl chloride forms a benzyl carbocation $(C_6H_5CH_2^+)$,which is resonance-stabilized by the benzene ring.
$(B)$ $CH_2=CH-CH_2Cl$ forms an allyl carbocation $(CH_2=CH-CH_2^+)$,which is resonance-stabilized.
$(C)$ $(CH_3)_3C-Cl$ forms a tertiary carbocation $((CH_3)_3C^+)$,which is stabilized by the inductive effect $(+I)$ of three methyl groups.
$(D)$ $CH_3-CH=CH-Cl$ would form a vinylic carbocation $(CH_3-CH=CH^+)$. $A$ positive charge on a carbon atom involved in a double bond is highly unstable due to the high electronegativity of $sp$-hybridized carbon.
Therefore,$CH_3-CH=CH-Cl$ does not undergo hydrolysis by the $S_{N}1$ mechanism.

(C) The reaction sequence is as follows:
$1$. Aniline reacts with $NaNO_2/HCl$ at $0^{\circ}C$ followed by $Cu_2Cl_2/HCl$ to form chlorobenzene $(X)$. This is the Sandmeyer reaction.
$2$. Chlorobenzene $(X)$ reacts with methyl chloride $(CH_3Cl)$ in the presence of $Na$ and dry ether to form toluene $(Y)$.
$3$. The reaction between an aryl halide and an alkyl halide in the presence of sodium and dry ether is known as the Wurtz-Fittig reaction.

(B) $1$. The starting material is propene $(CH_3CH=CH_2)$.
$2$. Hydroboration-oxidation of propene with $(i) B_2H_6$ and $(ii) H_2O_2/NaOH$ gives propan$-1-$ol $(CH_3CH_2CH_2OH)$ as $X$.
$3$. Dehydrogenation of propan$-1-$ol $(X)$ using $Cu$ at $573 \ K$ gives propanal $(CH_3CH_2CHO)$ as $Y$.
$4$. Propanal $(Y)$ undergoes aldol condensation in the presence of dilute $NaOH$ followed by heating $(\Delta)$ to give the $\alpha,\beta$-unsaturated aldehyde,$2$-methylpent-$2$-enal $(CH_3CH_2CH=C(CH_3)CHO)$ as $Z$.

(C) In group $13$ elements,the stability of the $+1$ oxidation state increases down the group due to the inert pair effect.
For $Tl$ (Thallium),the $+1$ oxidation state is more stable than the $+3$ oxidation state because the $6s^2$ electrons are reluctant to participate in bonding.

(B) The thermal decomposition reactions are as follows:
$(NH_4)_2Cr_2O_7 \xrightarrow{\Delta} Cr_2O_3 + N_2 \uparrow + 4H_2O$
$Ba(N_3)_2 \xrightarrow{\Delta} Ba + 3N_2$
$2NaNO_3 \xrightarrow{\Delta} 2NaNO_2 + O_2$
From the above reactions,it is clear that ammonium dichromate $(ii)$ and barium azide $(iii)$ liberate dinitrogen gas upon heating,whereas sodium nitrate $(i)$ liberates dioxygen gas.
Therefore,the correct option is $ii$ and $iii$ only.

(C) The boiling point of the hydrides of the nitrogen family $(Group \ 15)$ follows the order: $PH_3 < AsH_3 < NH_3 < SbH_3 < BiH_3$.
$NH_3$ has an anomalously high boiling point due to intermolecular hydrogen bonding,but the boiling point increases down the group due to an increase in molecular mass and van der Waals forces.
Comparing the given hydrides: $PH_3$ has the lowest boiling point $(Y = PH_3)$.
$BiH_3$ has the highest boiling point $(X = BiH_3)$.
Therefore,$X = BiH_3$ and $Y = PH_3$.

(D) In Haber's process for the manufacture of ammonia,the catalyst used is finely divided $Fe$,the promoter used is $Mo$ (or $K_2O$ and $Al_2O_3$),and the poison for the catalyst is $CO$.

(A) The correct matches are as follows:
$A$. Ostwald's process uses $Pt$ gauge as a catalyst $(A-III)$.
$B$. Haber's process uses $Fe$ as a catalyst $(B-IV)$.
$C$. Deacon's process uses $CuCl_2$ as a catalyst $(C-I)$.
$D$. Cracking of hydrocarbons uses Zeolites as a catalyst $(D-II)$.
Therefore,the correct sequence is $A-III, B-IV, C-I, D-II$.

(A) To determine which oxoacids contain $P-O-P$ bonds,we examine their structures:
$I$. $H_4P_2O_5$ (Pyrophosphorous acid): Contains a $P-O-P$ linkage.
$II$. $H_4P_2O_6$ (Hypophosphoric acid): Contains a $P-P$ bond,not $P-O-P$.
$III$. $H_4P_2O_7$ (Pyrophosphoric acid): Contains a $P-O-P$ linkage.
$IV$. $(HPO_3)_3$ (Cyclotrimetaphosphoric acid): Contains a cyclic structure with $P-O-P$ linkages.
Therefore,$I$,$III$,and $IV$ all contain $P-O-P$ bonds. Given the options provided,both $A$ ($III$ and $IV$) and $C$ ($I$ and $III$) contain correct examples,but $I$,$III$,and $IV$ are the complete set.

(D) In the contact process for the manufacture of sulfuric acid $(H_2SO_4)$,the sulfur dioxide $(SO_2)$ gas obtained from burning sulfur or roasting sulfide ores contains impurities like arsenic compounds.
These impurities act as catalytic poisons for the platinum catalyst used in the converter.
To remove these arsenic impurities,an arsenic purifier is used.
The arsenic purifier contains ferric oxide $(Fe_2O_3 \cdot xH_2O)$,which effectively adsorbs the arsenic impurities from the gas stream.

(D) The given reaction is the industrial method for the preparation of chlorine gas,known as the Deacon's process.
In this process,$HCl$ is oxidized by atmospheric oxygen in the presence of $CuCl_2$ as a catalyst at $723 \ K$.
$4 HCl + O_2 \xrightarrow[723 \ K]{CuCl_2} 2 Cl_2 + 2 H_2 O$

(C) Neon $(Ne)$ is a noble gas,which is colorless,odorless,and chemically inert.
Option $C$ is incorrect because $Ne$ is not a fluorescent green gas and does not have a rotten egg smell.
$PH_3$ has a rotten fish smell,$Cl_2$ is a greenish-yellow gas with a pungent smell,and $SO_2$ is a colorless gas with a pungent smell.

(A) Bakelite is formed by the condensation polymerization of phenol and formaldehyde (methanal).
Melamine is formed by the condensation polymerization of melamine and formaldehyde (methanal).
Therefore,methanal is the common monomer present in both polymers.

(D) Condensation polymers are formed by repeated condensation reactions between two different bi-functional monomeric units,involving the elimination of small molecules such as water,alcohol,or hydrogen chloride.
$A$,$B$,and $C$ (Terylene,Nylon $6,6$,and Bakelite) are examples of condensation polymers.
Polystyrene is an addition polymer formed by the polymerization of styrene $(C_6H_5CH=CH_2)$ monomers without the elimination of any small molecules.

(B) The correct matches are:
$A$. $CF_2=CF_2$ is the monomer for Teflon $(III)$.
$B$. $NH_2(CH_2)_6NH_2$ (hexamethylenediamine) and $HO_2C(CH_2)_4CO_2H$ (adipic acid) are monomers for Nylon $6, 6$ $(IV)$.
$C$. $C_6H_5OH$ (phenol) and $HCHO$ (formaldehyde) are monomers for Bakelite $(II)$.
$D$. $CH_2=CH(Cl)-CH=CH_2$ (chloroprene) is the monomer for Neoprene $(I)$.
Thus,the correct matching is $A-III, B-IV, C-II, D-I$.

(C) Caprolactam on heating with water at $533-543 \ K$ undergoes ring-opening polymerization to form Nylon-$6$.
The reaction is:
$Caprolactam \xrightarrow[H_2O]{533-543 \ K} [CO-(CH_2)_5-NH]_n$
This polymer is known as Nylon-$6$. The structure of the repeating unit is $[CO-(CH_2)_5-NH]_n$.

(A) The reaction of ethylene glycol $(HO-CH_2-CH_2-OH)$ with phthalic acid produces Glyptal,which is used in the manufacture of paints. Thus,$X$ is phthalic acid.
The reaction of ethylene glycol with terephthalic acid produces Terylene (or Dacron),which is used in making safety helmets and synthetic fibers. Thus,$Y$ is terephthalic acid.
Therefore,$X$ is phthalic acid and $Y$ is terephthalic acid.

(B) The oxidation states of the transition elements in the given ions are $V$ in $VO_2^{+}$ is $+5$,$Cr$ in $Cr_2O_7^{2-}$ is $+6$,and $Mn$ in $MnO_4^{-}$ is $+7$.
Oxidizing power is directly related to the ease of reduction,which depends on the stability of the lower oxidation states formed.
For these transition metals,the stability of the reduced products follows the trend $Mn^{2+} > Cr^{3+} > V^{3+}$.
Consequently,the oxidizing power increases as the reduction potential increases,leading to the order $VO_2^{+} < Cr_2O_7^{2-} < MnO_4^{-}$.

(C) When $Pt$ reacts with a $3:1$ mixture of concentrated hydrochloric acid $(HCl)$ and concentrated nitric acid $(HNO_3)$,it forms a solution known as aqua regia. Aqua regia is capable of dissolving platinum by forming chlorocomplexes. The reaction typically forms $H_2[PtCl_6]$.
In this complex,the $Pt$ is in the $+4$ oxidation state. The balanced chemical equation for the reaction is:
$Pt + 4HNO_3 + 6HCl \rightarrow H_2[PtCl_6] + 4NO_2 + 4H_2O$
So,in $[PtCl_6]^{2-}$ the oxidation state of $Pt$ is:
$x + 6(-1) = -2$
$x - 6 = -2$
$x = +4$

(B) The reaction of white phosphorus with sodium hydroxide is a disproportionation reaction:
$P_4 + 3 NaOH + 3 H_2O \rightarrow PH_3 + 3 NaH_2PO_2$
Here,$X$ is phosphine $(PH_3)$.
Next,phosphine reacts with copper$(II)$ sulfate:
$2 PH_3 + 3 CuSO_4 \rightarrow Cu_3P_2 + 3 H_2SO_4$
Thus,$Y$ is copper phosphide $(Cu_3P_2)$.

(D) Cement is primarily composed of calcium oxide $(CaO)$ and silicon dioxide $(SiO_2)$.
Other significant components include aluminum oxide $(Al_2O_3)$,magnesium oxide $(MgO)$,iron oxide $(Fe_2O_3)$,and sulfur trioxide $(SO_3)$.
Among the given options,$SiO_2$ and $CaO$ constitute the largest percentage by mass in Portland cement.

(D) In a $ccp$ lattice,if the number of atoms is $N$,then the number of octahedral voids is $N$ and the number of tetrahedral voids is $2N$.
Given the formula $AB_2O_4$,let the number of $O$ atoms be $4$.
Since $O$ forms a $ccp$ lattice,$N = 4$.
Number of tetrahedral voids $= 2 \times 4 = 8$.
Number of octahedral voids $= 4$.
Atoms of $A$ occupy $\frac{1}{8}$ of tetrahedral voids,so number of $A$ atoms $= \frac{1}{8} \times 8 = 1$.
Atoms of $B$ occupy octahedral voids. From the formula $AB_2O_4$,there are $2$ atoms of $B$.
Fraction of occupied octahedral voids $= \frac{\text{Number of } B \text{ atoms}}{\text{Total octahedral voids}} = \frac{2}{4} = \frac{1}{2}$.
Fraction of vacant octahedral voids $= 1 - \frac{1}{2} = \frac{1}{2}$.

(B) The density of a unit cell is given by the formula: $\rho = \frac{Z \times M}{N_A \times a^3}$
Where $Z$ is the number of atoms per unit cell, $M$ is the molar mass, $N_A$ is Avogadro's number, and $a$ is the edge length.
Rearranging for $Z$: $Z = \frac{\rho \times N_A \times a^3}{M}$
Given:
$\rho = 2.7 \times 10^3 \ kg \ m^{-3}$
$M = 2.7 \times 10^{-2} \ kg \ mol^{-1}$
$a = 405 \ pm = 405 \times 10^{-12} \ m = 4.05 \times 10^{-10} \ m$
$N_A = 6.023 \times 10^{23} \ mol^{-1}$
Substituting the values:
$Z = \frac{(2.7 \times 10^3) \times (6.023 \times 10^{23}) \times (4.05 \times 10^{-10})^3}{2.7 \times 10^{-2}}$
$Z = \frac{2.7 \times 10^3 \times 6.023 \times 10^{23} \times 66.43 \times 10^{-30}}{2.7 \times 10^{-2}}$
$Z = \frac{1080.14 \times 10^{-4}}{2.7 \times 10^{-2}} \approx 4$
Thus, the number of atoms per unit cell is $4$.

(A) In a $ccp$ lattice,if the number of oxygen atoms is $4$,then:
Number of octahedral voids $= 4$
Number of tetrahedral voids $= 8$
Given the formula $AB_2O_4$,there are $1$ atom of $A$,$2$ atoms of $B$,and $4$ atoms of $O$.
For atom $A$ in tetrahedral voids: $x \% \text{ of } 8 = 1 \implies x = (1/8) \times 100 = 12.5 \%$.
For atom $B$ in octahedral voids: $y \% \text{ of } 4 = 2 \implies y = (2/4) \times 100 = 50 \%$.
Thus,$x = 12.5 \%$ and $y = 50 \%$.

(A) In a $BCC$ unit cell,there are $8$ corners and $1$ body center.
Number of $X$ atoms at corners = $8 - 3 = 5$.
Contribution of each corner atom = $1/8$.
Total $X$ atoms = $5 \times (1/8) = 5/8$.
Number of $Y$ atoms at the body center = $1$.
Contribution of body center atom = $1$.
Total $Y$ atoms = $1$.
Ratio $X : Y = 5/8 : 1 = 5 : 8$.
Therefore,the formula of the compound is $X_5 Y_8$.

(D) The correct statement is that the trapping of electrons in the lattice leads to the formation of $F$-centres.
$1-$ Frenkel defect is favoured when the difference between the size of the cation and anion is large.
$2-$ Frenkel defect is also known as a dislocation defect.
$3-$ Schottky defect affects the physical properties of solids,specifically by decreasing the density of the substance.

(C) Total pressure of air over water $= 1 \,atm$.
Partial pressure of $N_2$ and $O_2$ are:
$P_{N_2} = \frac{1 \times 79}{100} = 0.79 \,atm$
$P_{O_2} = \frac{1 \times 21}{100} = 0.21 \,atm$
Applying Henry's law $(P = K_H \times X)$:
$X_{N_2} = \frac{P_{N_2}}{K_{H, N_2}} = \frac{0.79}{8.57 \times 10^4} \approx 9.22 \times 10^{-6}$
$X_{O_2} = \frac{P_{O_2}}{K_{H, O_2}} = \frac{0.21}{4.56 \times 10^4} \approx 4.60 \times 10^{-6}$
Ratio of mole fractions of $N_2$ and $O_2$:
$\frac{X_{N_2}}{X_{O_2}} = \frac{9.22 \times 10^{-6}}{4.60 \times 10^{-6}} \approx 2$
Thus, the ratio is $2: 1$.

(C) For a dilute solution,the relative lowering of vapour pressure is given by: $\frac{P^0 - P_s}{P^0} = \frac{n_2}{n_1} = \frac{W_2 \times M_1}{M_2 \times W_1}$
Given: $W_2 = 0.06 \ kg = 60 \ g$,$W_1 = 1 \ kg = 1000 \ g$,$P^0 = 3.78 \ kPa$,$P_s = 3.768 \ kPa$,$M_1 = 18 \ g \ mol^{-1}$ (for water).
$\Delta P = P^0 - P_s = 3.78 - 3.768 = 0.012 \ kPa$.
Rearranging the formula for $M_2$: $M_2 = \frac{W_2 \times M_1 \times P^0}{\Delta P \times W_1}$
$M_2 = \frac{60 \times 18 \times 3.78}{0.012 \times 1000} = \frac{4082.4}{12} = 340.2 \ g \ mol^{-1}$
The closest integer value is $340 \ g \ mol^{-1}$.

(D) Given: $P_{T}^0 = 3.63 \ kPa$,$P_{B}^0 = 9.7 \ kPa$,$X_{T} = 0.4$.
Since $X_{T} + X_{B} = 1$,the mole fraction of benzene is $X_{B} = 1 - 0.4 = 0.6$.
The total vapour pressure of the solution is given by $P_{total} = P_{T}^0 X_{T} + P_{B}^0 X_{B}$.
$P_{total} = (3.63 \times 0.4) + (9.7 \times 0.6) = 1.452 + 5.82 = 7.272 \ kPa$.
The composition of toluene in the vapour phase $(y_{T})$ is calculated using Dalton's Law: $y_{T} = \frac{P_{T}}{P_{total}} = \frac{P_{T}^0 X_{T}}{P_{total}}$.
$y_{T} = \frac{1.452}{7.272} \approx 0.1996 \approx 0.20$.

(C) Given: $T_{b}^{\circ} = 373.15 \ K$,$T_{b} = 373.202 \ K$,$K_{b} = 0.52 \ K \ kg \ mol^{-1}$,$K_{f} = 1.86 \ K \ kg \ mol^{-1}$,$T_{f}^{\circ} = 273.15 \ K$.
Elevation in boiling point: $\Delta T_{b} = T_{b} - T_{b}^{\circ} = 373.202 - 373.15 = 0.052 \ K$.
Since $\Delta T_{b} = m \times K_{b}$,we have $0.052 = m \times 0.52$,which gives molality $m = 0.1 \ mol \ kg^{-1}$.
Depression in freezing point: $\Delta T_{f} = m \times K_{f} = 0.1 \times 1.86 = 0.186 \ K$.
Freezing point of the solution: $T_{f} = T_{f}^{\circ} - \Delta T_{f} = 273.15 - 0.186 = 272.964 \ K$.

(B) According to Raoult's law,the vapour pressure of a solution containing a non-volatile solute is given by $P_s = P_1^0 \times X_1$.
Given: $n_{\text{solute}} = 0.1 \ mol$,$n_{\text{solvent}} = 0.9 \ mol$,$P_1^0 = 0.9 \ bar$.
Mole fraction of solvent $(X_1)$ = $\frac{n_{\text{solvent}}}{n_{\text{solvent}} + n_{\text{solute}}} = \frac{0.9}{0.9 + 0.1} = 0.9$.
$P_s = 0.9 \times 0.9 = 0.81 \ bar$.

(A) According to the Freundlich adsorption isotherm:
$\frac{x}{m} = k \cdot p^{1/n}$
Taking logarithm on both sides:
$\log(\frac{x}{m}) = \log k + \frac{1}{n} \log p$
This equation represents a straight line of the form $y = mx + c$,where the slope is $\frac{1}{n}$.
From the given graph,the slope is calculated as:
$\text{Slope} = \frac{\text{change in } \log(\frac{x}{m})}{\text{change in } \log p} = \frac{3}{6} = \frac{1}{2}$
Therefore,$\frac{1}{n} = \frac{1}{2}$.
Substituting this back into the Freundlich equation:
$\frac{x}{m} = k \cdot p^{1/2}$
Thus,the extent of adsorption is proportional to $p^{1/2}$.

(C) The Freundlich adsorption isotherm equation is given by: $\log \frac{x}{m} = \log k + \frac{1}{n} \log p$.
Comparing this with the equation of a straight line $y = mx + c$,where $y = \log \frac{x}{m}$,$x = \log p$,slope $= \frac{1}{n}$,and intercept $= \log k$.
Given: Slope $(\frac{1}{n}) = 0.5$ and Intercept $(\log k) = 1.0$.
Pressure $(p) = 100 \ atm$.
Substituting the values into the equation: $\log \frac{x}{m} = 1.0 + 0.5 \times \log(100)$.
Since $\log(100) = 2$,we get: $\log \frac{x}{m} = 1.0 + 0.5 \times 2 = 1.0 + 1.0 = 2.0$.
Therefore,$\frac{x}{m} = 10^2 = 100$.