KCET 2012 Chemistry Question Paper with Answer and Solution

(B) The chemiosmotic theory,proposed by Peter Mitchell,explains the mechanism of $ATP$ synthesis in mitochondria and chloroplasts.
In mitochondria,the electron transport system $(ETS)$ pumps protons ($H^+$ ions) from the mitochondrial matrix into the intermembrane space.
This creates a proton gradient (or $H^+$ gradient) across the inner mitochondrial membrane.
The potential energy stored in this electrochemical gradient is used by the enzyme $ATP$ synthase to phosphorylate $ADP$ into $ATP$ as protons flow back into the matrix through the $F_0-F_1$ particle.
Therefore,the synthesis of $ATP$ is directly dependent on the $H^+$ gradient.

(A) Power alcohol is defined as a mixture of $80 \% \text{ petrol}$ and $20 \% \text{ ethanol}$ with a small amount of benzene added to act as a co-solvent to prevent phase separation.
It is primarily used as a fuel in internal combustion engines.

(C) The compound $NaNC$ (sodium isocyanide) contains the following bonds:
$1$. Ionic bond: Between $Na^+$ and $[NC]^-$ ions.
$2$. Covalent bond: Between $N$ and $C$ atoms (triple bond).
$3$. Coordinate bond: The lone pair of electrons on $N$ is donated to $C$ to form the triple bond structure in the isocyanide ion $[N \equiv C]^-$.
Thus,$NaNC$ contains all three types of bonds.

(A) Covalent bonds are directional in nature.
Therefore,covalent compounds possess a definite geometry,such as $CH_4$ being tetrahedral and $C_2H_2$ being linear.
Thus,the statement 'No definite geometry' is incorrect for covalent compounds.

(D) In ethane $(C_2H_6)$,each carbon atom is $sp^3$ hybridized.
The $C-H$ bond is formed by the overlap of the $sp^3$ hybrid orbital of carbon and the $1s$ orbital of hydrogen,which is an $sp^3-s$ overlap.
The $C-C$ bond is formed by the overlap of the $sp^3$ hybrid orbital of one carbon atom with the $sp^3$ hybrid orbital of the other carbon atom,which is an $sp^3-sp^3$ overlap.
Therefore,the correct types of overlap are $sp^3-s$ and $sp^3-sp^3$.

(B) In graphite,each $C$ atom is $sp^{2}$ hybridized.
$\therefore$ $p$-character $= \frac{2}{3} \times 100 \approx 67 \%$.
In diamond,each $C$ atom is $sp^{3}$ hybridized.
$\therefore$ $p$-character $= \frac{3}{4} \times 100 = 75 \%$.
Thus,the values are $67 \%$ and $75 \%$ respectively.

(C) According to Molecular Orbital Theory $(MOT)$,the electronic configuration of $O_2$ ($16$ electrons) is: $KK, \sigma 2s^2, \sigma^{*} 2s^2, \sigma 2p_z^2, \pi 2p_x^2 = \pi 2p_y^2, \pi^{*} 2p_x^1 = \pi^{*} 2p_y^1$. It has $2$ unpaired electrons.
$(a)$ $O_{2}^{-}$ ($17$ electrons): Configuration ends with $\pi^{*} 2p_x^2, \pi^{*} 2p_y^1$. It has $1$ unpaired electron.
$(b)$ $O_{2}^{+}$ ($15$ electrons): Configuration ends with $\pi^{*} 2p_x^1$. It has $1$ unpaired electron.
$(c)$ $O_{2}^{2-}$ ($18$ electrons): Configuration ends with $\pi^{*} 2p_x^2, \pi^{*} 2p_y^2$. All electrons are paired.
$(d)$ $O_{2}$ ($16$ electrons): As shown above,it has $2$ unpaired electrons.
Therefore,$O_{2}^{2-}$ has no unpaired electrons.

(B) Electronegativity increases across a period from left to right and decreases down a group.
Comparing the elements:
$F$ $(Z=9)$ is in period $2$,group $17$.
$O$ $(Z=8)$ is in period $2$,group $16$.
$N$ $(Z=7)$ is in period $2$,group $15$.
$P$ $(Z=15)$ is in period $3$,group $15$.
Since $F$,$O$,and $N$ are in the same period $(2)$,their electronegativity follows the order $F > O > N$.
$P$ is below $N$ in group $15$,so its electronegativity is lower than $N$.
Thus,the correct order is $F > O > N > P$.

(B) On moving from left to right in a period,ionisation energy generally increases due to an increase in effective nuclear charge.
Thus,the expected order of ionisation energy for $C$,$N$,$O$,and $F$ is $C < N < O < F$.
However,the ionisation energy of $N$ $(1s^2 2s^2 2p^3)$ is greater than that of $O$ $(1s^2 2s^2 2p^4)$.
This is because $N$ has a stable half-filled $p$-orbital configuration,which requires more energy to remove an electron.
Therefore,the correct order is $C < O < N < F$.

(D) Given that $p=\frac{b \times c}{[a b c]}, q=\frac{c \times a}{[a b c]}, r=\frac{a \times b}{[a b c]}$.
We know that $[a b c] = a \cdot (b \times c) = b \cdot (c \times a) = c \cdot (a \times b)$.
Let $T_{1} = (a+b) \cdot p = a \cdot p + b \cdot p$.
$T_{1} = a \cdot \frac{b \times c}{[a b c]} + b \cdot \frac{b \times c}{[a b c]} = \frac{[a b c]}{[a b c]} + \frac{[b b c]}{[a b c]} = 1 + 0 = 1$.
Similarly,$T_{2} = (b+c) \cdot q = b \cdot q + c \cdot q = b \cdot \frac{c \times a}{[a b c]} + c \cdot \frac{c \times a}{[a b c]} = \frac{[b c a]}{[a b c]} + 0 = 1$.
Similarly,$T_{3} = (c+a) \cdot r = c \cdot r + a \cdot r = c \cdot \frac{a \times b}{[a b c]} + a \cdot \frac{a \times b}{[a b c]} = \frac{[c a b]}{[a b c]} + 0 = 1$.
Therefore,$(a+b) \cdot p + (b+c) \cdot q + (c+a) \cdot r = T_{1} + T_{2} + T_{3} = 1 + 1 + 1 = 3$.

(D) The standard free energy change is given by the formula: $\Delta G^{\circ} = -RT \ln K = -2.303 RT \log K$.
Given: $R = 8.314 \ J \ K^{-1} \ mol^{-1}$,$T = 298 \ K$,$K = 0.008$.
Substituting the values: $\Delta G^{\circ} = -2.303 \times 8.314 \times 298 \times \log(0.008)$.
Since $\log(0.008) = \log(8 \times 10^{-3}) = \log 8 - 3 = 0.903 - 3 = -2.097$.
$\Delta G^{\circ} = -2.303 \times 8.314 \times 298 \times (-2.097) \approx 11965 \ J \ mol^{-1} = +11.96 \ kJ \ mol^{-1}$.

(D) The neutralization reaction is: $CH_3COOH + NaOH \rightarrow CH_3COONa + H_2O$.
Moles of $CH_3COOH = 0.1 \ M \times 0.1 \ L = 0.01 \ mol$.
Thus,$0.01 \ mol$ of $CH_3COONa$ is formed.
Kolbe's electrolysis reaction is: $2CH_3COONa + 2H_2O \rightarrow C_2H_6 + 2CO_2 + H_2 + 2NaOH$.
According to the stoichiometry,$2 \ mol$ of $CH_3COONa$ produces $1 \ mol$ of $C_2H_6$.
Therefore,$0.01 \ mol$ of $CH_3COONa$ produces $0.005 \ mol$ of $C_2H_6$.
Volume of $C_2H_6$ at $STP = 0.005 \ mol \times 22400 \ mL/mol = 112 \ mL$.

$(A)$ A Ruby laser uses a synthetic ruby crystal ($Al_2O_3$) doped with chromium ($Cr^{3+}$) ions. These chromium ions are the active centers that undergo population inversion. The specific red color (wavelength of $694.3$ nm) is emitted when these chromium ions transition between energy levels.

(A) The given compound is $CH_3COCH_2CH(OH)CH_3$.
$1$. Identify the principal functional group: The ketone group $(-CO-)$ has higher priority than the hydroxyl group $(-OH)$.
$2$. Number the carbon chain starting from the end that gives the ketone group the lowest possible number: The ketone carbon gets position $2$.
$3$. The chain has $5$ carbons,so the parent alkane is pentane. With the ketone group,it is pentan$-2-$one.
$4$. The hydroxyl group is at position $4$,so it is named as a hydroxy substituent.
$5$. Combining these,the $IUPAC$ name is $4-$hydroxy$2-$pentanone.

(D) In the electromeric effect,there is a complete transfer of $\pi$-electrons from one atom to another to produce temporary full charges on the atoms joined by multiple bonds.
Therefore,the statement that it results in the appearance of partial charges is incorrect,as it produces full charges (formal charges) on the atoms.

(B) The percentage of oxygen $= 100 - (52 + 13) = 35 \%$.
Calculating the empirical formula:
$C: 52/12 = 4.33 \rightarrow 2$
$H: 13/1 = 13.00 \rightarrow 6$
$O: 35/16 = 2.18 \rightarrow 1$
Empirical formula $= C_2H_6O$.
Empirical formula weight $= 2 \times 12 + 6 \times 1 + 16 = 46$.
Molecular weight $= 2 \times \text{vapour density} = 2 \times 23 = 46$.
Since molecular weight equals empirical formula weight,the molecular formula is $C_2H_6O$.
Since the compound reacts with sodium metal to liberate $H_2$ gas,it must be an alcohol,which is $C_2H_5OH$ (ethanol).
The functional isomer of ethanol $(C_2H_5OH)$ is methoxy methane $(CH_3OCH_3)$.

(A) Calculate the milliequivalents $(meq)$ of $NaOH$ and $HCl$ present in the solution.
$meq \ of \ NaOH = N_1 \times V_1 = 0.4 \times 5 = 2.0 \ meq$.
$meq \ of \ HCl = N_2 \times V_2 = 0.1 \times 20 = 2.0 \ meq$.
Since the number of milliequivalents of the strong base $(NaOH)$ is equal to the number of milliequivalents of the strong acid $(HCl)$,the neutralization reaction is complete.
The resulting solution contains only the salt $(NaCl)$ and water,making the solution neutral.
Therefore,the $pH$ of the resulting solution is $7$.

(C) The $pH$ of a solution is determined by the concentration of $H^+$ ions. Adding a solution with the same $pH$ (i.e.,same concentration of $H^+$ ions) will not change the $pH$ of the original solution.
For $20 \ mL$ of $0.1 \ N \ HCl$,the concentration of $H^+$ is $0.1 \ M$,which corresponds to $pH = -\log(0.1) = 1$.
Option $(c)$ provides $500 \ mL$ of $HCl$ with $pH = 1$. Since the concentration of $H^+$ ions in this solution is the same as the original solution $(0.1 \ M)$,adding it will not change the $pH$ of the original $20 \ mL$ of $0.1 \ N \ HCl$ solution.

(B) For solution $I, pH=3$
$\Rightarrow [H^{+}] = 10^{-3} \ M$
$V_{1} = 100 \ mL$
For solution $II, pH = 4$
$\Rightarrow [H^{+}] = 10^{-4} \ M$
$V_{2} = 400 \ mL$
Concentration of resulting solution $M = \frac{M_{1}V_{1} + M_{2}V_{2}}{V_{1} + V_{2}}$
$M = \frac{10^{-3} \times 100 + 10^{-4} \times 400}{100 + 400}$
$M = \frac{0.1 + 0.04}{500} = \frac{0.14}{500} = 2.8 \times 10^{-4} \ M$
$[H^{+}] = 2.8 \times 10^{-4} \ M$
$pH = -\log [H^{+}] = -\log (2.8 \times 10^{-4})$
$pH = 4 - \log 2.8$

(A) The atomic weight of a metal is calculated as: $\text{Atomic weight} = \text{Equivalent weight} \times \text{Valency}$.
Given that the metal is bivalent,its valency is $2$.
$\text{Atomic weight} = 20 \times 2 = 40 \ g \ mol^{-1}$.
The molecular formula of the anhydrous chloride of a bivalent metal $M$ is $MCl_2$.
The molecular mass of $MCl_2 = \text{Atomic mass of } M + 2 \times \text{Atomic mass of } Cl$.
$\text{Molecular mass} = 40 + 2 \times 35.5 = 40 + 71 = 111 \ g \ mol^{-1}$.

(B) The balanced chemical equation for the reaction is: $2KMnO_4 + 3H_2SO_4 + 5H_2C_2O_4 \rightarrow K_2SO_4 + 2MnSO_4 + 8H_2O + 10CO_2$.
From the stoichiometry,$2 \text{ moles of } KMnO_4$ react with $5 \text{ moles of } H_2C_2O_4$.
Using the equivalence concept: $n_{factor} \times M \times V = \text{constant}$.
For $KMnO_4$ in acidic medium,$n_{factor} = 5$.
For $H_2C_2O_4$,$n_{factor} = 2$.
Equating equivalents: $n_{factor,1} \times M_1 \times V_1 = n_{factor,2} \times M_2 \times V_2$.
$5 \times 0.025 \times 20 = 2 \times 0.1 \times V$.
$2.5 = 0.2 \times V$.
$V = \frac{2.5}{0.2} = 12.5 \ mL$.

(B) The reaction of sodium with water is highly exothermic and vigorous.
By forming an amalgam of sodium with mercury $(Na-Hg)$,the activity of sodium is reduced,which makes the reaction with water less vigorous.

(B) Given volume of water $= 18 \ mL$ and density $= 1 \ g \ mL^{-1}$.
Mass of water $= \text{density} \times \text{volume} = 1 \ g \ mL^{-1} \times 18 \ mL = 18 \ g$.
Molar mass of $H_2O = (2 \times 1) + 16 = 18 \ g \ mol^{-1}$.
Number of moles of $H_2O = \frac{18 \ g}{18 \ g \ mol^{-1}} = 1 \ mol$.
One molecule of $H_2O$ contains $2 \times 1 (\text{from } H) + 8 (\text{from } O) = 10$ electrons.
Total number of electrons in $1 \ mol$ of $H_2O = 1 \ mol \times 6.022 \times 10^{23} \text{ molecules/mol} \times 10 \text{ electrons/molecule}$.
Total number of electrons $= 6.022 \times 10^{24}$.

(D) The reaction is $N_2O_{4(g)} \rightleftharpoons 2NO_2(g)$.
Initially,$n_1 = 2 \ mol$ at $T_1 = 298 \ K$ and $P_1 = 1 \ atm$.
$20 \%$ by mass of $N_2O_4$ decomposes. Since molar mass is constant,this corresponds to $20 \%$ of moles decomposing.
Moles of $N_2O_4$ reacted $= 2 \times 0.2 = 0.4 \ mol$.
Moles of $N_2O_4$ remaining $= 2 - 0.4 = 1.6 \ mol$.
Moles of $NO_2$ produced $= 2 \times 0.4 = 0.8 \ mol$.
Total moles at equilibrium,$n_2 = 1.6 + 0.8 = 2.4 \ mol$.
Using the ideal gas law for a constant volume container: $\frac{P_1}{n_1 T_1} = \frac{P_2}{n_2 T_2}$.
$\frac{1}{2 \times 298} = \frac{P_2}{2.4 \times 596}$.
$P_2 = \frac{2.4 \times 596}{2 \times 298} = \frac{2.4 \times 596}{596} = 2.4 \ atm$.

(C) The combustion reaction of methane is: $CH_{4(g)} + 2O_{2(g)} \rightarrow CO_{2(g)} + 2H_2O_{(l)}$
Initially: $20 \ mL$ of $CH_4$ and $50 \ mL$ of $O_2$ are taken.
According to the stoichiometry,$1 \ volume$ of $CH_4$ requires $2 \ volumes$ of $O_2$.
So,$20 \ mL$ of $CH_4$ will react with $2 \times 20 = 40 \ mL$ of $O_2$.
Remaining $O_2 = 50 \ mL - 40 \ mL = 10 \ mL$.
Volume of $CO_2$ produced = $20 \ mL$.
Since $H_2O$ is produced as a liquid at room temperature,its volume is negligible.
Total volume of gas left = $Volume \ of \ remaining \ O_2 + Volume \ of \ CO_2 = 10 \ mL + 20 \ mL = 30 \ mL$.

(A) For any orbital,the principal quantum number $n$ and the azimuthal quantum number $l$ must satisfy the condition $l < n$.
For the $3f$ orbital,$n = 3$ and for an $f$-subshell,$l = 3$.
Since $l$ cannot be equal to or greater than $n$ ($l < n$ is required),the $3f$ orbital is impossible.
In the other options:
$2p$: $n=2, l=1$ ($1 < 2$ is valid).
$4d$: $n=4, l=2$ ($2 < 4$ is valid).
$2s$: $n=2, l=0$ ($0 < 2$ is valid).

(C) The electronic configuration of sodium $(Z=11)$ is $1s^2, 2s^2, 2p^6, 3s^1$.
For the outermost electron,which is in the $3s$ orbital:
Principal quantum number $(n) = 3$.
Azimuthal quantum number $(l)$ for $s$-orbital $= 0$.
Magnetic quantum number $(m_l) = 0$.
Spin quantum number $(m_s) = +\frac{1}{2}$ (or $-\frac{1}{2}$).
Thus,the set of quantum numbers is $(3, 0, 0, \frac{1}{2})$.

(A) The work done during expansion is given by the formula $W = -P_{ext} \Delta V$.
Since the gas is expanding against an external pressure,the work done by the gas is $W = P_{ext} \Delta V$.
Given $P_{ext} = 10^{5} \ Nm^{-2}$,$V_{1} = 1 \ m^{3}$,and $V_{2} = 2 \ m^{3}$.
$\Delta V = V_{2} - V_{1} = 2 \ m^{3} - 1 \ m^{3} = 1 \ m^{3}$.
$W = 10^{5} \ Nm^{-2} \times 1 \ m^{3} = 10^{5} \ J$.
Converting to kilojoules,$10^{5} \ J = 100 \ kJ = 10^{2} \ kJ$.

(C) According to the $1^{st}$ Law of Thermodynamics,energy can neither be created nor destroyed,meaning the total energy of the universe remains constant.
According to the $2^{nd}$ Law of Thermodynamics,the entropy of the universe is continuously increasing for any spontaneous process.
Therefore,the statement that the total energy of the universe remains constant is true.

(C) The reaction of formaldehyde $(HCHO)$ with a Grignard reagent $(RMgX)$ followed by acid hydrolysis yields a primary alcohol.
In this case,the product is ethanol $(C_{2}H_{5}OH)$,which contains two carbon atoms.
Since formaldehyde provides one carbon atom,the Grignard reagent must provide the other methyl group $(CH_{3})$.
Therefore,the reagent $X$ is methylmagnesium iodide $(CH_{3}MgI)$.
The reaction is: $HCHO + CH_{3}MgI$ $\rightarrow CH_{3}CH_{2}OMgI$ $\xrightarrow{H_{3}O^{+}} CH_{3}CH_{2}OH + Mg(OH)I$.

(A) In aniline $(C_6H_5NH_2)$,the lone pair of electrons on the nitrogen atom is in conjugation with the benzene ring and is delocalised due to resonance. This makes the lone pair less available for protonation,resulting in lower basicity.
In benzylamine $(C_6H_5CH_2NH_2)$,the lone pair of electrons on the nitrogen atom is not involved in resonance with the benzene ring because the nitrogen atom is separated from the ring by a $CH_2$ group. Thus,the lone pair is more available for protonation,making benzylamine a stronger base than aniline.

(A) Carbohydrates that can reduce Tollen's reagent or Fehling's solution are called reducing sugars. All monosaccharides and most disaccharides (except sucrose) are reducing sugars.
In sucrose,the glycosidic linkage is formed between the $C1$ of glucose and $C2$ of fructose. Since both the anomeric carbons are involved in the linkage,there is no free aldehyde or keto group available to act as a reducing agent. Therefore,it fails to react with Tollen's reagent and Fehling's solution.

(C) The acidic strength of substituted benzoic acids is influenced by electronic effects and the ortho effect.
$1$. The $CH_3$ group is an electron-donating group ($+I$ effect),which decreases the acidity of benzoic acid when present at the $meta$ or $para$ positions.
$2$. $p-$toluic acid has a $CH_3$ group at the $para$ position,which destabilizes the carboxylate anion via the $+I$ effect,making it less acidic than benzoic acid.
$3$. $o-$toluic acid experiences the 'ortho effect',which significantly increases its acidity compared to benzoic acid,regardless of the electronic nature of the substituent.
$4$. Therefore,the decreasing order of acidic strength is: $o-$toluic acid $>$ benzoic acid $>$ $p-$toluic acid.

(D) $Methyl$ salicylate is also known as oil of wintergreen.
It is an organic ester with the chemical structure $C_6H_4(OH)(COOCH_3)$.

(B) For a first order reaction,the rate constant is given by $k = \frac{2.303}{t} \log \frac{a}{a-x}$.
Case $I$: Given $x = 60 \%$,$t = 20 \ min$,so $a-x = 40$.
$k = \frac{2.303}{20} \log \frac{100}{40} = \frac{2.303}{20} \log 2.5$.
$k = \frac{2.303}{20} \times 0.3979 \approx 0.0458 \ min^{-1}$.
Case $II$: For $84 \%$ completion,$x = 84$,so $a-x = 16$.
$t = \frac{2.303}{k} \log \frac{100}{16} = \frac{2.303}{0.0458} \log 6.25$.
$t = \frac{2.303}{0.0458} \times 0.7959 \approx 40 \ min$.

(C) The reaction is $A(g) \longrightarrow B(g) + C(g)$.
Let the initial pressure of $A$ be $P_i$.
At $t = 0$: $P_A = P_i$,$P_B = 0$,$P_C = 0$. Total pressure $P_{total} = P_i = 180 \ mm \ of \ Hg$ (at completion).
At $t = 20 \ min$: $P_A = P_i - x$,$P_B = x$,$P_C = x$.
Total pressure $P_t = (P_i - x) + x + x = P_i + x = 100 \ mm \ of \ Hg$.
Since $P_i = 180 \ mm \ of \ Hg$,we have $180 + x = 100$,which is not possible as $x$ must be positive. Re-evaluating the data: The total pressure at completion is $2P_i = 180 \ mm \ of \ Hg$,so $P_i = 90 \ mm \ of \ Hg$.
At $t = 20 \ min$: $P_t = P_i + x = 100 \ mm \ of \ Hg$.
$90 + x = 100 \implies x = 10 \ mm \ of \ Hg$.
The partial pressure of $A$ at $20 \ min$ is $P_A = P_i - x = 90 - 10 = 80 \ mm \ of \ Hg$.

(D) Let the order of reaction with respect to $A$ and $B$ be $m$ and $n$ respectively.
Then,$rate = k[A]^{m}[B]^{n}$
From the table:
$2 = k[0.2]^{m}[0.2]^{n}$ $(i)$
$4 = k[0.2]^{m}[0.4]^{n}$ $(ii)$
$36 = k[0.6]^{m}[0.4]^{n}$ $(iii)$
On comparing Eqs. $(i)$ and $(ii)$:
$\frac{4}{2} = \frac{k[0.2]^{m}[0.4]^{n}}{k[0.2]^{m}[0.2]^{n}}$
$2 = (\frac{0.4}{0.2})^{n} = 2^{n}$
$n = 1$
On comparing Eqs. $(ii)$ and $(iii)$:
$\frac{36}{4} = \frac{k[0.6]^{m}[0.4]^{n}}{k[0.2]^{m}[0.4]^{n}}$
$9 = (\frac{0.6}{0.2})^{m} = 3^{m}$
$3^{2} = 3^{m}$
$m = 2$
Therefore,the rate law is $r = k[A]^{2}[B]$.

(C) The rate constant $k$ is inversely proportional to the time $t$ taken for the reaction to occur $(k \propto 1/t)$.
Given $t_1 = 5 \ h$ at $T_1 = 300 \ K$ and $t_2 = 50 \ h$ at $T_2 = 270 \ K$.
Thus,$k_1/k_2 = t_2/t_1 = 50/5 = 10$,which means $k_2/k_1 = 1/10$.
Using the Arrhenius equation: $\log(k_2/k_1) = \frac{E_a}{2.303 R} \left[ \frac{T_2 - T_1}{T_1 T_2} \right]$.
Substituting the values: $\log(1/10) = \frac{E_a}{2.303 R} \left[ \frac{270 - 300}{270 \times 300} \right]$.
$-1 = \frac{E_a}{2.303 R} \left[ \frac{-30}{81000} \right]$.
$-1 = \frac{E_a}{2.303 R} \left[ \frac{-1}{2700} \right]$.
$E_a = 2.303 \times 2700 \times R \ J \cdot mol^{-1} = 2.303 \times 2.7 \times R \ kJ \cdot mol^{-1}$.

(A) Saccharin is manufactured from $toluene$ through a series of chemical reactions as shown below:
$1$. $Toluene$ is reacted with $ClSO_3H$ to form $o$-toluenesulphonyl chloride and $p$-toluenesulphonyl chloride.
$2$. The $o$-isomer is treated with $NH_3$ to form $o$-toluenesulphonamide.
$3$. This is then oxidized using alkaline $KMnO_4$ to form $o$-sulphamoylbenzoic acid.
$4$. Finally,upon heating,it undergoes dehydration to form $Saccharin$.

(B) The oxidation state of $Co$ in $Na_{3}[Co(NO_{2})_{4} Cl_{2}]$ is calculated as follows:
$3(+1) + x + 4(-1) + 2(-1) = 0$
$3 + x - 4 - 2 = 0$
$x - 3 = 0$
$x = +3$
The $EAN$ (Effective Atomic Number) is given by the formula:
$EAN = (\text{Atomic number of metal}) - (\text{Oxidation state of metal}) + 2 \times (\text{Coordination number})$
$EAN = (27 - 3) + 2 \times 6$
$EAN = 24 + 12 = 36$.

(C) Ligands are species that donate at least one lone pair of electrons to the central metal atom or ion to form a coordinate covalent bond.
According to the Lewis theory,an electron pair donor is defined as a Lewis base.
Therefore,a ligand acts as a Lewis base.

(D) Bidentate ligands are those that have two donor atoms to coordinate with the central metal ion.
$1$. Cyano $(CN^-)$ is a negatively charged monodentate ligand.
$2$. Ethylene diamine ($en$ or $NH_2CH_2CH_2NH_2$) is a neutral bidentate ligand.
$3$. Acetato $(CH_3COO^-)$ is a negatively charged monodentate ligand.
$4$. Dimethyl glyoximato $(dmg^-)$ is a negatively charged bidentate ligand.
Therefore,the correct option is $D$.

(C) Secondary valency corresponds to the coordination number of the central metal atom.
The formula of the given complex is $[Pt(NH_{3})_{4}Cl_{2}]Cl_{2}$.
In this complex,the central metal atom $Pt$ is bonded to $4$ ammonia ligands and $2$ chloride ligands.
Therefore,the coordination number of $Pt = 4 + 2 = 6$.
$\therefore$ The secondary valency of $Pt$ is $6$.

(C) The electronic configuration of $Ni$ $(Z=28)$ is $[Ar] 3d^8 4s^2$.
$Ni^{2+}$ ion has the configuration $[Ar] 3d^8$.
In an aqueous solution,$Ni^{2+}$ forms the octahedral complex $[Ni(H_2O)_6]^{2+}$.
In the $3d$ subshell,the $8$ electrons are arranged as follows: three orbitals are doubly occupied and two orbitals are singly occupied.
Therefore,the number of unpaired electrons $(n)$ $= 2$.
The spin-only magnetic moment is calculated using the formula $\mu = \sqrt{n(n + 2)} \ BM$.
Substituting $n = 2$,we get $\mu = \sqrt{2(2 + 2)} = \sqrt{8} \ BM$.

(D) The magnetic moment $(\mu)$ is calculated using the formula: $\mu = \sqrt{n(n+2)} \ BM$,where $n$ is the number of unpaired electrons.
For $Ti^{3+}$ $(3d^1)$,$n = 1$. Thus,$\mu = \sqrt{1(1+2)} = \sqrt{3} \approx 1.73 \ BM$.
For $V^{3+}$ $(3d^2)$,$n = 2$. Thus,$\mu = \sqrt{2(2+2)} = \sqrt{8} \approx 2.82 \ BM$.
For $Cr^{3+}$ $(3d^3)$,$n = 3$. Thus,$\mu = \sqrt{3(3+2)} = \sqrt{15} \approx 3.87 \ BM$.
For $Fe^{3+}$ $(3d^5)$,$n = 5$. Thus,$\mu = \sqrt{5(5+2)} = \sqrt{35} \approx 5.91 \ BM$.
Therefore,$Ti^{3+}$ has a magnetic moment of approximately $1.73-1.75 \ BM$.

(B) Silver iodide $(AgI)$ has a crystal structure similar to that of ice.
Due to this structural similarity,it acts as an effective nucleating agent for water droplets in clouds.
This procedure is known as 'Cloud seeding' and is used to induce artificial rain.

(A) For the hydrogen electrode $Pt, \frac{1}{2} H_{2}(1 \ atm) \mid H^{+}(aq)$,the reduction potential is given by the Nernst equation:
$E = E^{\circ} - \frac{0.0591}{1} \log \frac{1}{[H^{+}]}$
Since $E^{\circ} = 0 \ V$ for the Standard Hydrogen Electrode $(SHE)$,the equation simplifies to:
$E = 0.0591 \log [H^{+}]$
For $HCl$ solutions,$[H^{+}] = [HCl]$.
$(A)$ For $[HCl] = 2 \ M$:
$E = 0.0591 \log(2) \approx 0.0591 \times 0.3010 = 0.0178 \ V$
$(B)$ For $[HCl] = 0.1 \ M$:
$E = 0.0591 \log(0.1) = 0.0591 \times (-1) = -0.0591 \ V$
$(C)$ For $[HCl] = 0.5 \ M$:
$E = 0.0591 \log(0.5) \approx 0.0591 \times (-0.3010) = -0.0178 \ V$
$(D)$ For $[HCl] = 1 \ M$:
$E = 0.0591 \log(1) = 0 \ V$
Thus,only the electrode in option $(A)$ has a potential greater than zero.

(D) The reduction reaction at the cathode is: $Ca^{2+} + 2e^{-} \longrightarrow Ca$.
From the stoichiometry,$1 \ mol$ of $Ca$ requires $2 \ F$ of charge.
Number of moles of $Ca$ $(n)$ = $\frac{\text{mass}}{\text{molar mass}} = \frac{30 \ g}{40 \ g/mol} = 0.75 \ mol$.
Total charge required $(Q)$ = $n \times 2 \ F = 0.75 \times 2 \times 96500 \ C = 144750 \ C$.
Using the relation $Q = i \times t$,where $i = 5 \ A$:
$t = \frac{Q}{i} = \frac{144750 \ C}{5 \ A} = 28950 \ s$.
To convert time into hours: $t = \frac{28950}{3600} \ h \approx 8.04 \ h$.
Thus,the approximate time is $8 \ h$.

(D) The given reaction is $A_{(s)} + 2B^{2+}_{(aq)} \rightleftharpoons A^{2+}_{(aq)} + 2B_{(s)}$.
Here,the number of electrons transferred,$n = 2$.
Given $E^{\circ}_{\text{cell}} = 0.0295 \ V$.
The relationship between equilibrium constant $K_c$ and $E^{\circ}_{\text{cell}}$ is given by the Nernst equation at equilibrium:
$E^{\circ}_{\text{cell}} = \frac{0.059}{n} \log K_c$.
Substituting the values:
$0.0295 = \frac{0.059}{2} \log K_c$.
$0.0295 = 0.0295 \log K_c$.
$\log K_c = 1$.
$K_c = 10^1 = 10$.

(A) In the froth floatation process,potassium ethyl xanthate acts as a collector.
It attaches itself by its polar group to the grains of minerals,making the mineral surface hydrophobic.
This increases the non-wettability of the mineral particles,effectively making the ore water repellant.

(B) In the Ellingham diagram,the line for the formation of $CO$ from $C$ and $O_{2}$ intersects the line for the formation of $CO_{2}$ at $983 \ K$.
Below $983 \ K$,the formation of $CO_{2}$ is more spontaneous (more negative $\Delta_{f}G^{\circ}$),meaning $CO_{2}$ is more stable than $CO$.
Above $983 \ K$,the formation of $CO$ becomes more spontaneous (more negative $\Delta_{f}G^{\circ}$),meaning $CO$ is more stable than $CO_{2}$.
Therefore,the statement that '$CO$ is less stable than $CO_{2}$ at more than $983 \ K$' is false.

(D) The froth floatation process is primarily used for the concentration of sulphide ores.
Among the given options,chalcopyrites $(CuFeS_{2})$ is a sulphide ore,whereas cryolite $(Na_{3}AlF_{6})$,cuprite $(Cu_{2}O)$,and calamine $(ZnCO_{3})$ are not.
Therefore,chalcopyrites is concentrated by the froth floatation process.

(C) $S_{N}1$ reactions proceed via the formation of a carbocation intermediate.
The rate of the reaction depends only on the concentration of the substrate,not on the concentration of the nucleophile.
$3^{\circ}$-alkyl halides are most reactive towards $S_{N}1$ due to the stability of the resulting carbocation.
$1^{\circ}$-alkyl halides generally undergo $S_{N}2$ reactions because they form unstable carbocations.
Polar solvents stabilize the transition state and the carbocation intermediate,thus favouring $S_{N}1$ reactions.
Therefore,the statement that $1^{\circ}$-alkyl halides generally react through $S_{N}1$ reaction is incorrect.

(B) The reaction of a mixture of methyl bromide $(CH_3Br)$ and bromobenzene $(C_6H_5Br)$ with sodium metal in dry ether is a combination of Wurtz and Fittig reactions,known as the Wurtz-Fittig reaction.
Possible products include:
$1$. Wurtz reaction: $2CH_3Br + 2Na \rightarrow CH_3-CH_3 + 2NaBr$ (Ethane is formed).
$2$. Wurtz-Fittig reaction: $C_6H_5Br + CH_3Br + 2Na \rightarrow C_6H_5-CH_3 + 2NaBr$ (Toluene is formed).
$3$. Fittig reaction: $2C_6H_5Br + 2Na \rightarrow C_6H_5-C_6H_5 + 2NaBr$ (Diphenyl is formed).
Propane $(C_3H_8)$ cannot be formed in this reaction as there is no source of a three-carbon chain.

(B) The reaction sequence is the industrial preparation of phenol from cumene:
$1$. Benzene reacts with isopropyl chloride in the presence of anhydrous $AlCl_3$ to form cumene $(A)$.
$2$. Cumene $(A)$ is oxidized by $O_2$ at $130^{\circ}C$ to form cumene hydroperoxide $(B)$.
$3$. Cumene hydroperoxide $(B)$ is treated with dilute $H_2SO_4$ at $100^{\circ}C$ to yield phenol and propanone $(C)$.
Therefore,$C$ is propanone.

(D) $1$ mole of linseed oil has the maximum number of moles of hydrogen added because it contains the highest degree of unsaturation.
Linseed oil is a typical drying oil.
It consists primarily of linolenic acid esters (polyunsaturated fatty acids),which have multiple double bonds,allowing for the addition of more hydrogen compared to other oils listed.

(A) Since the given compound '$A$' burns with a sooty flame,it must be an aromatic compound.
When it reacts with Borsche's reagent (acidic $2,4-$dinitrophenylhydrazine solution),it yields an orange precipitate,which indicates that it is a carbonyl compound (aldehyde or ketone).
Since '$A$' gives a negative result for Tollen's reagent test,it cannot be an aldehyde.
Therefore,it must be an aromatic ketone.
Hence,'$A$' is acetophenone.

(C) $Buna-S$ rubber is a copolymer of $1,3-butadiene$ and styrene.
It is prepared by the polymerization of $1,3-butadiene$ $(75 \%)$ and styrene $(25 \%)$ in the presence of sodium $(Na)$ as a catalyst.
Thus,the $s$-block element used is $Na$.

(A) Given: $\frac{P_A^0}{P_B^0} = \frac{1}{2}$ and $\frac{n_A}{n_B} = \frac{1}{2}$.
Mole fraction of $A$ in liquid phase,$x_A = \frac{n_A}{n_A + n_B} = \frac{1}{1+2} = \frac{1}{3}$.
Mole fraction of $B$ in liquid phase,$x_B = \frac{n_B}{n_A + n_B} = \frac{2}{1+2} = \frac{2}{3}$.
According to Raoult's law,partial pressures are:
$P_A = x_A P_A^0 = \frac{1}{3} P_A^0$
$P_B = x_B P_B^0 = \frac{2}{3} P_B^0 = \frac{2}{3} (2 P_A^0) = \frac{4}{3} P_A^0$
Total pressure $P_{total} = P_A + P_B = \frac{1}{3} P_A^0 + \frac{4}{3} P_A^0 = \frac{5}{3} P_A^0$.
Mole fraction of $A$ in vapour phase,$y_A = \frac{P_A}{P_{total}} = \frac{\frac{1}{3} P_A^0}{\frac{5}{3} P_A^0} = \frac{1}{5} = 0.2$.

(D) According to Raoult's law for relative lowering of vapour pressure:
$\frac{p^{\circ} - p}{p^{\circ}} = \frac{n}{n + N} \approx \frac{n}{N} = \frac{w \times M}{m \times W}$
Given:
$p^{\circ} = 100, \quad p = 100 - 20 = 80$
$m = 40 \ g \ mol^{-1}, \quad M(C_{8}H_{18}) = 8 \times 12 + 18 \times 1 = 114 \ g \ mol^{-1}$
$W = 114 \ g, \quad w = ?$
Substituting the values:
$\frac{100 - 80}{100} = \frac{w \times 114}{40 \times 114}$
$\frac{20}{100} = \frac{w}{40}$
$w = \frac{0.2 \times 40} = 8 \ g$
Wait,checking the formula $\frac{p^{\circ} - p}{p} = \frac{w \times M}{m \times W}$:
$\frac{20}{80} = \frac{w \times 114}{40 \times 114}$
$0.25 = \frac{w}{40}$
$w = 0.25 \times 40 = 10 \ g$

(A) Colloidal particles are larger aggregates compared to solute particles in a true solution.
Since the number of particles in a colloidal dispersion is significantly smaller than in a true solution at the same concentration,the colligative properties,such as osmotic pressure,are of a low order.
Therefore,colloidal dispersions exhibit low osmotic pressure.

(B) Dialysis is a process used to separate colloidal particles from crystalloids through a semi-permeable membrane.
Colloidal particles (like proteins) cannot pass through the membrane,while crystalloids (like glucose) can pass through.
Therefore,dialysis is used to separate $glucose$ and $protein$.

(B) During the adsorption of a gas on the surface of a solid,there is a decrease in surface energy,which means it is an exothermic process. Therefore,$\Delta H < 0$.
When a gas is adsorbed,the freedom of movement of its molecules becomes restricted,leading to a decrease in the entropy of the system. Therefore,$\Delta S < 0$.
According to the Gibbs free energy equation,$\Delta G = \Delta H - T \Delta S$. For a spontaneous process,$\Delta G$ must be negative. Since adsorption is a spontaneous process,$\Delta G < 0$.