KCET 2011 Chemistry Question Paper with Answer and Solution

(A) The extension $\Delta \ell$ of a wire is given by the formula $\Delta \ell = \frac{F L}{A Y}$,where $F$ is the tension,$L$ is the length,$A$ is the cross-sectional area,and $Y$ is Young's modulus.
Since $A = \pi r^2$,we have $\Delta \ell = \frac{F L}{\pi r^2 Y}$.
Given that $F$ and $Y$ are constant for all wires,the extension $\Delta \ell$ is proportional to $\frac{L}{r^2}$.
Calculating the ratio $\frac{L}{r^2}$ for each option:
$A: \frac{100}{(0.2)^2} = \frac{100}{0.04} = 2500 \ cm/mm^2$
$B: \frac{200}{(0.4)^2} = \frac{200}{0.16} = 1250 \ cm/mm^2$
$C: \frac{300}{(0.6)^2} = \frac{300}{0.36} \approx 833.33 \ cm/mm^2$
$D: \frac{400}{(0.8)^2} = \frac{400}{0.64} = 625 \ cm/mm^2$
Comparing these values,the ratio is largest for option $A$.

(C) $1$. The circuit consists of two $OR$ gates followed by an $AND$ gate.
$2$. The inputs to the first $OR$ gate are $A$ and $B$. Therefore,its output is $(A + B)$.
$3$. The inputs to the second $OR$ gate are $A$ and $C$. Therefore,its output is $(A + C)$.
$4$. These two outputs are fed into an $AND$ gate.
$5$. The final output $Y$ of the $AND$ gate is the product of its inputs: $Y = (A + B) \cdot (A + C)$.

(A) The condition for the first minima in a single-slit diffraction pattern is given by $a \sin \theta = n \lambda$,where $n = 1$ for the first minima.
Given:
Wavelength $\lambda = 6000 \, \mathring{A} = 6000 \times 10^{-10} \, \text{m} = 6 \times 10^{-7} \, \text{m}$.
Slit width $a = 0.3 \, \text{mm} = 0.3 \times 10^{-3} \, \text{m} = 3 \times 10^{-4} \, \text{m}$.
For small angles,$\sin \theta \approx \theta$.
Therefore,$\theta = \frac{\lambda}{a} = \frac{6 \times 10^{-7}}{3 \times 10^{-4}} = 2 \times 10^{-3} \, \text{rad}$.

(D) In the life cycle of plants,the dominant phase refers to the stage that is independent,photosynthetic,and longer-lived.
$A$. $Hibiscus$ is an Angiosperm,where the sporophyte is the dominant phase.
$B$. $Nephrolepis$ is a Pteridophyte,where the sporophyte is the dominant phase.
$C$. $Cycas$ is a Gymnosperm,where the sporophyte is the dominant phase.
$D$. $Riccia$ is a Bryophyte. In Bryophytes,the main plant body is a haploid gametophyte,which is the dominant,independent,and photosynthetic phase,while the sporophyte is dependent on the gametophyte for nutrition.

(C) Oxidative decarboxylation is a process where a carboxyl group is removed from a molecule as $CO_2$ along with the oxidation of the substrate.
During aerobic respiration, the pyruvic acid formed during glycolysis enters the mitochondrial matrix.
Here, it undergoes oxidative decarboxylation to form acetyl $Co-A$.
This reaction is catalyzed by the pyruvate dehydrogenase complex in the presence of $NAD^+$ and coenzyme-$A$.
The chemical equation is: $\text{Pyruvic acid} + NAD^+ + CoA \xrightarrow{\text{Pyruvate dehydrogenase}} \text{Acetyl } Co-A + NADH + H^+ + CO_2$.

(B) Gibberellic acids $(GA)$ induce the sub-apical meristem to develop faster.
This process causes the elongation of the reduced stem,a phenomenon known as bolting,which is observed in rosette plants like cabbage (e.g.,$Brassica$ $oleracea$ var. $capitata$) and root crops like radish.

(C) transformer consists of two coils,the primary coil and the secondary coil,wound on a common magnetic core. When an alternating current flows through the primary coil,it produces a time-varying magnetic flux in the core. This changing flux is linked with the secondary coil,inducing an electromotive force $(EMF)$ in it. This phenomenon,where a change in current in one coil induces an $EMF$ in a neighboring coil,is known as mutual induction. Therefore,a transformer works on the principle of mutual induction.

(B) The wavelength of spectral lines in the hydrogen atom is given by the Rydberg formula: $\frac{1}{\lambda} = R \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right)$.
For the minimum wavelength,the transition must be from $n_2 = \infty$ to $n_1$.
For the Lyman series,$n_1 = 1$,so $\frac{1}{\lambda_L} = R \left( \frac{1}{1^2} - \frac{1}{\infty^2} \right) = R$,which gives $\lambda_L = \frac{1}{R}$.
For the Balmer series,$n_1 = 2$,so $\frac{1}{\lambda_B} = R \left( \frac{1}{2^2} - \frac{1}{\infty^2} \right) = \frac{R}{4}$,which gives $\lambda_B = \frac{4}{R}$.
The ratio of the minimum wavelengths is $\frac{\lambda_L}{\lambda_B} = \frac{1/R}{4/R} = \frac{1}{4} = 0.25$.

(D) The molecular orbital configuration of $O_{2}$ is $(\sigma 1s)^{2}(\sigma^{*} 1s)^{2}(\sigma 2s)^{2}(\sigma^{*} 2s)^{2}(\sigma 2p_{z})^{2}(\pi 2p_{x})^{2}(\pi 2p_{y})^{2}(\pi^{*} 2p_{x})^{1}(\pi^{*} 2p_{y})^{1}$.
Bond order $= \frac{N_{b} - N_{a}}{2} = \frac{10 - 6}{2} = 2$. It has $2$ unpaired electrons.
For $O_{2}^{+}$,the configuration is $(\sigma 1s)^{2}(\sigma^{*} 1s)^{2}(\sigma 2s)^{2}(\sigma^{*} 2s)^{2}(\sigma 2p_{z})^{2}(\pi 2p_{x})^{2}(\pi 2p_{y})^{2}(\pi^{*} 2p_{x})^{1}$.
Bond order $= \frac{10 - 5}{2} = 2.5$. It has $1$ unpaired electron.
Thus,when $O_{2}$ is converted into $O_{2}^{+}$,the paramagnetic character decreases and the bond order increases.

(D) The bond length depends on the bond order; higher bond order results in shorter bond length.
$A: C_2H_4$ (double bond,$1.34 \ \mathring{A}$)
$B: C_2H_2$ (triple bond,$1.20 \ \mathring{A}$)
$C: C_6H_6$ (partial double bond,$1.39 \ \mathring{A}$)
$D: C_2H_6$ (single bond,$1.54 \ \mathring{A}$)
Comparing the values: $1.20 \ \mathring{A} (B) < 1.34 \ \mathring{A} (A) < 1.39 \ \mathring{A} (C) < 1.54 \ \mathring{A} (D)$.
Thus,the increasing order is $B < A < C < D$.

(B) For the first reaction: $SO_{2(g)} + \frac{1}{2} O_{2(g)} \rightleftharpoons SO_{3(g)}$; $K_{1} = \frac{[SO_{3}]}{[SO_{2}][O_{2}]^{1/2}}$
For the second reaction: $2 SO_{3(g)} \rightleftharpoons 2 SO_{2(g)} + O_{2(g)}$; $K_{2} = \frac{[SO_{2}]^{2}[O_{2}]}{[SO_{3}]^{2}}$
Comparing the two expressions,we can see that $K_{2} = \left( \frac{[SO_{2}][O_{2}]^{1/2}}{[SO_{3}]} \right)^{2} = \left( \frac{1}{K_{1}} \right)^{2} = \frac{1}{K_{1}^{2}}$
Therefore,$K_{1}^{2} = \frac{1}{K_{2}}$.

(A) The given compound is $CH_2=CH-CH_2-COOH$.
$1$. Identify the longest carbon chain containing the principal functional group (carboxylic acid). The chain has $4$ carbon atoms,so the root word is $but$.
$2$. Number the chain starting from the carboxylic acid carbon as $C-1$. Thus,the double bond starts at $C-3$.
$3$. The suffix for the carboxylic acid is $-oic \ acid$ and for the double bond is $-ene$.
$4$. Combining these,the $IUPAC$ name is $but-3-enoic \ acid$.

(C) In acetic acid $(CH_3COOH)$,the methyl group $(-CH_3)$ releases electrons due to the $+I$ effect,which increases the electron density on the carboxylate group and destabilizes the conjugate base,thereby decreasing its acidic character.
Formic acid $(HCOOH)$ does not contain any such electron-donating group attached to the carboxyl group.
Hence,due to the $+I$ effect of the methyl group in acetic acid,formic acid is a stronger acid than acetic acid.

(B) The nucleophilicity of a nitrogen atom depends on the availability of its lone pair of electrons for donation.
In the given molecule,$H_2N-NH-CO-NH_2$ (semicarbazide/urea derivative structure):
- The nitrogen atom labeled $III$ has its lone pair involved in resonance with the adjacent carbonyl group $(C=O)$,making it less available.
- The nitrogen atom labeled $II$ has its lone pair involved in resonance with the carbonyl group,and it is also attached to another nitrogen atom,which further reduces its availability.
- The nitrogen atom labeled $I$ has its lone pair available for donation because it is not directly involved in resonance with the carbonyl group. Although it is attached to another nitrogen atom,its lone pair is the most available among the three for acting as a nucleophile.
Therefore,the nitrogen atom labeled $I$ is the most nucleophilic.

(A) The molecule $1-$bromo$-2-$methylcyclobutane contains two chiral centers at $C1$ and $C2$.
Since the ring is not symmetric,the number of stereoisomers is given by $2^n$,where $n$ is the number of chiral centers.
Here,$n = 2$,so the total number of stereoisomers is $2^2 = 4$.
These $4$ stereoisomers consist of two pairs of enantiomers (cis-isomer pair and trans-isomer pair),all of which are optically active.
Therefore,the maximum number of possible optical isomers is $4$.

(D) The stability of cyclohexane conformations is determined by their potential energy. The order of stability is: $Chair > Twisted \ boat > Boat > Half \ chair$.
Conversely,the order of energy is: $Half \ chair > Boat > Twisted \ boat > Chair$.
The $Half \ chair$ conformation is the most energetic (least stable) due to significant torsional strain and steric interactions.

(B) The formula for the percentage of nitrogen in the Kjeldahl method is:
$\text{Percentage of Nitrogen} = \frac{1.4 \times N \times V}{W}$
Where:
$N = \text{Normality of } HCl = 1 \ N$
$V = \text{Volume of } HCl \text{ consumed} = 30 \ cm^3$
$W = \text{Weight of organic compound} = 1.2 \ g$
Substituting the values:
$\text{Percentage of Nitrogen} = \frac{1.4 \times 1 \times 30}{1.2} = \frac{42}{1.2} = 35 \%$

(D) Chromite ore is $FeCr_2O_4$,which can be represented as $FeO \cdot Cr_2O_3$.
In this compound,the oxidation number of $Fe$ is $+2$ and the oxidation number of $Cr$ is $+3$.

(A) $Ethyl$ benzene cannot be prepared by the $Wurtz$ reaction.
The $Wurtz$ reaction involves the coupling of two alkyl halides to form a higher symmetric alkane.
It is not suitable for the synthesis of alkyl-substituted aromatic compounds like $Ethyl$ benzene.
$Wurtz-Fittig$ reaction,$Friedel-Crafts$ alkylation,and $Clemmensen$ reduction are all valid methods for preparing $Ethyl$ benzene.

(C) The reaction of benzene with $CH_3Cl$ in the presence of anhydrous $AlCl_3$ is a Friedel-Crafts alkylation reaction.
In the first step,the Lewis acid $AlCl_3$ reacts with $CH_3Cl$ to generate the electrophile,which is the methyl carbocation $(CH_3^{+})$.
The reaction is as follows:
$CH_3Cl + AlCl_3 \rightarrow CH_3^{+} + [AlCl_4]^{-}$
Thus,$CH_3^{+}$ is the electrophilic intermediate formed during the reaction.

(B) Benzene does not contain $3$ distinct single bonds and $3$ distinct double bonds. Due to resonance,all carbon-carbon bonds in benzene are equivalent,having a bond order of $1.5$.

(A) For a solution to have $pH = 1$,the concentration of $[H^+]$ must be $0.1 \ M$.
$(A)$ $0.1 \ M \ CH_3COOH$ is a weak acid and does not dissociate completely,so $[H^+] < 0.1 \ M$,hence $pH > 1$.
$(B)$ $0.1 \ M \ HNO_3$ is a strong acid,$[H^+] = 0.1 \ M$,so $pH = -\log(0.1) = 1$.
$(C)$ $0.05 \ M \ H_2SO_4$ is a strong acid,$[H^+] = 2 \times 0.05 = 0.1 \ M$,so $pH = -\log(0.1) = 1$.
$(D)$ For the mixture: $n(H^+) = 50 \times 0.4 = 20 \ mmol$,$n(OH^-) = 50 \times 0.2 = 10 \ mmol$. Remaining $n(H^+) = 20 - 10 = 10 \ mmol$. Total volume = $100 \ cm^3$. $[H^+] = 10 / 100 = 0.1 \ M$,so $pH = 1$.

(D) The chemical reaction is: $2Na + 2H_{2}O \longrightarrow 2NaOH + H_{2}$
Molar mass of $Na = 23 \ g/mol$.
Moles of $Na = \frac{0.023 \ g}{23 \ g/mol} = 0.001 \ mol$.
According to the stoichiometry,$2 \ mol$ of $Na$ produces $2 \ mol$ of $NaOH$.
Therefore,$0.001 \ mol$ of $Na$ produces $0.001 \ mol$ of $NaOH$.
The volume of the solution is $100 \ cm^{3} = 0.1 \ L$.
Concentration of $[OH^{-}] = \frac{0.001 \ mol}{0.1 \ L} = 0.01 \ M = 10^{-2} \ M$.
$pOH = -\log[OH^{-}] = -\log(10^{-2}) = 2$.
Since $pH + pOH = 14$,then $pH = 14 - 2 = 12$.

(D) The $pH$ of a buffer solution is given by the Henderson-Hasselbalch equation: $pH = p K_{a} + \log \frac{[salt]}{[acid]}$.
Initially,the solution contains $0.1 \ mol$ of sodium acetate in $1000 \ cm^{3}$ $(1 \ L)$ of $0.1 \ M$ acetic acid.
After adding an additional $0.1 \ mol$ of sodium acetate,the total amount of salt becomes $0.1 + 0.1 = 0.2 \ mol$.
The concentration of salt $[salt] = \frac{0.2 \ mol}{1 \ L} = 0.2 \ M$.
The concentration of acid $[acid] = 0.1 \ M$.
Substituting these values into the equation: $pH = p K_{a} + \log \frac{0.2}{0.1}$.
$pH = p K_{a} + \log 2$.

(B) The ionic product $(IP)$ for $ZnS$ is $[Zn^{2+}][S^{2-}] = 0.1 \times 8.1 \times 10^{-19} = 8.1 \times 10^{-20}$. Since $8.1 \times 10^{-20} > 3 \times 10^{-22}$ ($K_{sp}$ of $ZnS$),$ZnS$ precipitates.
The ionic product $(IP)$ for $CuS$ is $[Cu^{2+}][S^{2-}] = 0.01 \times 8.1 \times 10^{-19} = 8.1 \times 10^{-21}$. Since $8.1 \times 10^{-21} > 8 \times 10^{-36}$ ($K_{sp}$ of $CuS$),$CuS$ precipitates.
Therefore,both $CuS$ and $ZnS$ precipitate.

(B) Compound $A$ is $o$-nitrophenol and compound $B$ is $p$-nitrophenol.
$o$-Nitrophenol exhibits intramolecular hydrogen bonding,which reduces its intermolecular attraction,making it more volatile.
$p$-Nitrophenol exhibits intermolecular hydrogen bonding,which leads to association of molecules,making it less volatile.
Therefore,the vapour pressure of $o$-nitrophenol $(A)$ is higher than that of $p$-nitrophenol $(B)$.
Thus,the vapour pressure of $B$ is lower than that of $A$.

(A) The chemical reactions involved are:
$CaCl_2 + Na_2CO_3 \rightarrow CaCO_3 + 2NaCl$
$CaCO_3 \xrightarrow{\Delta} CaO + CO_2$
From the stoichiometry,$1 \ mol$ of $CaO$ is obtained from $1 \ mol$ of $CaCO_3$,which in turn comes from $1 \ mol$ of $CaCl_2$.
Molar mass of $CaO = 40 + 16 = 56 \ g/mol$.
Molar mass of $CaCl_2 = 40 + 2 \times 35.5 = 111 \ g/mol$.
Given mass of $CaO = 0.56 \ g$,which is $0.56 / 56 = 0.01 \ mol$.
Therefore,moles of $CaCl_2 = 0.01 \ mol$.
Mass of $CaCl_2 = 0.01 \times 111 = 1.11 \ g$.
Mass of $NaCl$ in the mixture = Total mass - Mass of $CaCl_2 = 4.44 \ g - 1.11 \ g = 3.33 \ g$.
Percentage of $NaCl = (3.33 / 4.44) \times 100 = 75\%$.

(D) The formula for rms velocity is $v_{rms} = \sqrt{\frac{3RT}{M}}$.
Given that $v_{rms(H_{2})} = \sqrt{7} v_{rms(N_{2})}$.
Substituting the formula: $\sqrt{\frac{3RT_{H_{2}}}{M_{H_{2}}}} = \sqrt{7} \sqrt{\frac{3RT_{N_{2}}}{M_{N_{2}}}}$.
Squaring both sides: $\frac{T_{H_{2}}}{2} = 7 \times \frac{T_{N_{2}}}{28}$.
Simplifying: $\frac{T_{H_{2}}}{2} = \frac{T_{N_{2}}}{4}$.
Therefore,$T_{N_{2}} = 2 T_{H_{2}}$.

(C) According to the ideal gas equation,$PV = nRT$,where $n = \frac{m}{M}$.
Since $P$,$V$,$T$,and $m$ are given,$V = \frac{mRT}{PM}$.
For a fixed mass $(m = 25 \ g)$,temperature $(T)$,and pressure $(P)$,the volume $V$ is inversely proportional to the molar mass $M$ $(V \propto \frac{1}{M})$.
The molar masses of the gases are: $M(HF) = 20 \ g/mol$,$M(HCl) = 36.5 \ g/mol$,$M(HBr) = 81 \ g/mol$,and $M(HI) = 128 \ g/mol$.
Since $HI$ has the highest molar mass,it will have the least number of moles and consequently the least volume.

(A) The energy of a photon is given by the formula $E = \frac{hc}{\lambda}$,where $h$ is Planck's constant,$c$ is the speed of light,and $\lambda$ is the wavelength.
Since $h$ and $c$ are constants,$E \propto \frac{1}{\lambda}$.
Therefore,the ratio of energies is inversely proportional to the ratio of their wavelengths:
$\frac{E_1}{E_2} = \frac{\lambda_2}{\lambda_1}$
Given $\frac{E_1}{E_2} = \frac{3}{2}$,we have $\frac{3}{2} = \frac{\lambda_2}{\lambda_1}$.
Thus,the ratio of wavelengths $\lambda_1 : \lambda_2 = 2:3$.

(C) According to the $(n+l)$ rule,the energy of an orbital increases as the value of $(n+l)$ increases.
For option $A$: $n+l = 4+0 = 4$.
For option $B$: $n+l = 3+1 = 4$.
For option $C$: $n+l = 3+2 = 5$.
For option $D$: $n+l = 3+0 = 3$.
Since the value of $(n+l)$ is maximum for option $C$,it represents the highest energy level.

(D) The first law of thermodynamics is given by the equation: $\Delta U = Q + W$.
For a cyclic process,the system returns to its initial state,so the change in internal energy is zero: $\Delta U = 0$.
Substituting this into the first law equation: $0 = Q + W$,which simplifies to $Q = -W$.
Therefore,the correct statement is that for a cyclic process,$Q = -W$.

(B) The first ionisation energy generally increases across a period due to an increase in effective nuclear charge.
Exceptions occur when a more stable electronic configuration (like fully filled or half-filled orbitals) is present in the preceding element.
$Be$ $(2s^2)$ has a higher ionisation energy than $B$ $(2s^2 2p^1)$.
$Mg$ $(3s^2)$ has a higher ionisation energy than $Al$ $(3s^2 3p^1)$.
$N$ $(2s^2 2p^3)$ has a higher ionisation energy than $O$ $(2s^2 2p^4)$.
$Na$ $(3s^1)$ has a lower ionisation energy than $Mg$ $(3s^2)$,which follows the general trend.
Therefore,$Na$ and $Mg$ is not an exception.

(A) The reaction between $HCl$ and $NaOH$ is a neutralization reaction: $HCl + NaOH \rightarrow NaCl + H_{2}O$.
The heat of neutralization for a strong acid and a strong base is $-57.3 \ kJ \ mol^{-1}$.
Number of moles of $HCl = M \times V(L) = 0.1 \times 0.5 = 0.05 \ mol$.
Number of moles of $NaOH = M \times V(L) = 0.2 \times 0.2 = 0.04 \ mol$.
Since $NaOH$ is the limiting reagent,the amount of heat evolved depends on the moles of $NaOH$ reacted.
Heat evolved $= 0.04 \ mol \times 57.3 \ kJ \ mol^{-1} = 2.292 \ kJ$.

(A) The enthalpy of vaporization $(\Delta H_{vap})$ is $+35.3 \ kJ/mol = +35300 \ J/mol$.
The boiling point $(T)$ is $80^{\circ} C = 80 + 273 = 353 \ K$.
For the transition of liquid to vapour,the entropy change is $\Delta S_{vap} = \frac{\Delta H_{vap}}{T} = \frac{35300 \ J/mol}{353 \ K} = +100 \ J/mol \cdot K$.
The question asks for the entropy change in the transition of vapour to liquid (condensation),which is the reverse process.
Therefore,$\Delta S_{cond} = -\Delta S_{vap} = -100 \ J/mol \cdot K$.

(C) For the reaction $A_{(s)} + B_{(g)} \rightleftharpoons C_{(g)} + D_{(g)}$,the standard Gibbs free energy change is $\Delta G^{\circ} = -350 \ kJ$.
The relationship between $\Delta G^{\circ}$ and the equilibrium constant $K_{eq}$ is given by the equation: $\Delta G^{\circ} = -RT \ln K_{eq}$.
Since $\Delta G^{\circ}$ is negative $(-350 \ kJ)$,the value of $\ln K_{eq}$ must be positive,which implies that $K_{eq} > 1$.
Therefore,the equilibrium constant is greater than one.

(C) transformer works on the principle of mutual induction.
Mutual induction is a phenomenon where a change in current in one coil induces an electromotive force $(emf)$ in an adjacent coil due to the change in magnetic flux linked with it.

(D) transformer is a device that transfers electrical energy between two or more circuits through electromagnetic induction.
It consists of two coils,the primary and the secondary,wound on a common magnetic core.
When an alternating current flows through the primary coil,it creates a changing magnetic flux in the core.
This changing flux links with the secondary coil and induces an electromotive force $(EMF)$ in it,which is the phenomenon of mutual induction.
Therefore,a transformer works on the principle of mutual induction.

(C) Lucas reagent is a mixture of concentrated $HCl$ and anhydrous $ZnCl_2$.
Tertiary $(3^{\circ})$ alcohols react immediately with Lucas reagent at room temperature to form alkyl chlorides,which appear as turbidity.
Secondary $(2^{\circ})$ alcohols react within $5-10$ minutes.
Primary $(1^{\circ})$ alcohols do not react at room temperature.
Among the given options,$2-$methyl propan$-2-$ol is a tertiary alcohol,so it gives turbidity immediately.

(B) Both acetaldehyde $(CH_{3}CHO)$ and acetone $(CH_{3}COCH_{3})$ contain the methyl ketone group $(CH_{3}CO-)$,which makes them undergo the iodoform reaction.
When treated with iodine $(I_{2})$ in the presence of sodium hydroxide $(NaOH)$,they form a yellow precipitate of iodoform $(CHI_{3})$.
Acetaldehyde reaction:
$CH_{3}CHO + 3I_{2} + 4NaOH \longrightarrow CHI_{3} + HCOONa + 3NaI + 3H_{2}O$
Acetone reaction:
$CH_{3}COCH_{3} + 3I_{2} + 4NaOH \longrightarrow CHI_{3} + CH_{3}COONa + 3NaI + 3H_{2}O$
Fehling's solution and Tollen's reagent only react with aldehydes (like acetaldehyde) and not with ketones (like acetone).

(D) Maltose is a disaccharide composed of two $\alpha$-$D$-glucopyranose units.
These two units are joined together by an $\alpha$-glycosidic linkage between the $C1$ of one glucose unit and the $C4$ of the other glucose unit,known as a $1-4$ glycosidic linkage.

(B) In an alkaline medium (high $pH$),the amino acid glycine exists as a negatively charged particle because the carboxyl group $(-COOH)$ loses a proton to form a carboxylate ion $(-COO^-)$.
$H_2N-CH_2-COOH + OH^- \rightarrow H_2N-CH_2-COO^- + H_2O$
Thus,it exists as an $anion$.

(C) For a zero order reaction,the integrated rate equation is $[A] = [A]_0 - kt$.
At $100 \%$ completion,the final concentration $[A] = 0$.
Given initial concentration $[A]_0 = a$.
Substituting these values: $0 = a - kt$.
Therefore,$kt = a$,which gives $t = \frac{a}{k}$.

(B) The Arrhenius equation is given by $k = A e^{-E_{a} / RT}$.
Taking the ratio of rate constant $k$ to the Arrhenius factor $A$,we get $k/A = e^{-E_{a} / RT}$.
Given the activation energy $E_{a} = 2.303 \ RT \ J \ mol^{-1}$.
Substituting $E_{a}$ into the equation: $k/A = e^{-(2.303 \ RT) / RT} = e^{-2.303}$.
Taking the natural logarithm on both sides: $\ln(k/A) = -2.303$.
Since $\ln(x) = 2.303 \log_{10}(x)$,we have $2.303 \log_{10}(k/A) = -2.303$.
Dividing by $2.303$,we get $\log_{10}(k/A) = -1$.
Therefore,$k/A = 10^{-1} = 0.1$.

(D) Oxytocin is a peptide hormone,not an enzyme.
Silk is a natural polyamide (protein).
Lipase is an enzyme.
Butter is a fat (lipid).
Therefore,the incorrect match is $D$.

(C) The group $18$ elements (noble gases) have completely filled valence subshells ($ns^2 np^6$,except $He$ which is $1s^2$).
Due to the presence of paired electrons in all orbitals,these elements are diamagnetic.
The naturally occurring group $18$ elements are $He, Ne, Ar, Kr, Xe,$ and $Rn$.
Since $Rn$ is radioactive,the number of naturally occurring stable $p$-block elements that are diamagnetic is $5$ $(He, Ne, Ar, Kr, Xe)$.

(B) In the complex $[Co(NH_3)_4 Cl_2] Cl$,the ligands are $4$ ammine $(NH_3)$ and $2$ chloro $(Cl^-)$ groups.
According to $IUPAC$ nomenclature,ligands are named in alphabetical order: ammine before chloro.
Thus,the name is tetraammine dichloro.
The oxidation state of cobalt $(Co)$ is calculated as: $x + 4(0) + 2(-1) + (-1) = 0$,which gives $x = +3$.
Therefore,the $IUPAC$ name is tetraammine dichloro cobalt $(III)$ chloride.

(C) The electronic configuration of $Fe^{2+}$ is $[Ar] 3d^{6}$.
It has $4$ unpaired electrons as shown in the $d$-orbital diagram:
$d^{6} = [\uparrow\downarrow] [\uparrow] [\uparrow] [\uparrow] [\uparrow]$
The spin-only magnetic moment is calculated using the formula:
$\mu = \sqrt{n(n+2)} \ BM$
Where $n$ is the number of unpaired electrons.
$\mu = \sqrt{4(4+2)} = \sqrt{24} \approx 4.89 \ BM$
Rounding to the nearest integer,the value is $5 \ BM$.

(B) The chromyl chloride test is used to detect the presence of chloride ions.
When a chloride salt is heated with potassium dichromate $(K_2Cr_2O_7)$ and concentrated sulfuric acid $(H_2SO_4)$,red vapors of chromyl chloride $(CrO_2Cl_2)$ are evolved.
When these vapors are passed into a solution of sodium hydroxide $(NaOH)$,a yellow solution of sodium chromate $(Na_2CrO_4)$ is formed.
Upon adding lead acetate $(CH_3COO)_2Pb$ to this yellow solution,a yellow precipitate of lead chromate $(PbCrO_4)$ is obtained.

(B) The emf of a galvanic cell is given by the Nernst equation: $E_{cell} = E^{\circ}_{cell} - \frac{0.0591}{n} \log \frac{[Product]}{[Reactant]}$.
For the reaction $Zn(s) + Cu^{2+}(aq) \rightarrow Zn^{2+}(aq) + Cu(s)$,the equation is $E = E^{\circ} - \frac{0.0591}{2} \log \frac{[Zn^{2+}]}{[Cu^{2+}]} = E^{\circ} + \frac{0.0591}{2} \log \frac{[Cu^{2+}]}{[Zn^{2+}]}$.
As the ratio $\frac{[Cu^{2+}]}{[Zn^{2+}]}$ increases,the value of $E$ increases.
Based on the standard cell configurations,the order of emf values is $E_{1} > E_{2} > E_{3}$.

(A) The relationship between the standard $emf$ $(E^{\circ})$ and the equilibrium constant $(K_{eq})$ is given by the Nernst equation at equilibrium:
$E^{\circ} = \frac{0.0591}{n} \log K_{eq}$ at $298 \ K$.
Given:
$n = 2$
$E^{\circ} = 0.59 \ V$
Substituting the values:
$0.59 = \frac{0.059}{2} \log K_{eq}$
$\log K_{eq} = \frac{0.59 \times 2}{0.059}$
$\log K_{eq} = 10 \times 2 = 20$
$K_{eq} = 10^{20}$

(C) The reaction for the deposition of magnesium is: $Mg^{2+} + 2e^- \rightarrow Mg(s)$.
$2 \times 96500 \ C$ of charge is required to deposit $1 \text{ mole}$ $(24 \ g)$ of $Mg$.
Charge passed $= 9.65 \ C$.
Moles of $Mg$ deposited $= \frac{9.65}{2 \times 96500} = \frac{1}{2 \times 10000} = 0.5 \times 10^{-4} = 5 \times 10^{-5} \text{ moles}$.
Since $1 \text{ mole}$ of $Mg$ produces $1 \text{ mole}$ of Grignard reagent $(R-Mg-X)$,the moles of Grignard reagent obtained is $5 \times 10^{-5}$.

(D) The conversion of $m$-nitrophenol to resorcinol proceeds through the following steps:
$1$. Reduction: $m$-nitrophenol is reduced to $m$-aminophenol using a reducing agent like $[H]$.
$2$. Diazotization: $m$-aminophenol is treated with $HNO_2$ at $0-5 \ ^{\circ}C$ to form a diazonium salt.
$3$. Hydrolysis: The diazonium salt is hydrolyzed with $H_2O$ to yield resorcinol,releasing $N_2$ and $HCl$ as byproducts.
Thus,the sequence is reduction,diazotization,and hydrolysis.

(A) Dry distillation of a mixture of calcium formate and another calcium salt of a carboxylic acid yields an aldehyde.
Specifically,the reaction between calcium formate $(HCOO)_2Ca$ and calcium acetate $(CH_3COO)_2Ca$ produces acetaldehyde $(CH_3CHO)$ along with calcium carbonate $(CaCO_3)$.
The reaction is: $(HCOO)_2Ca + (CH_3COO)_2Ca \xrightarrow{400^{\circ}C} 2CH_3CHO + 2CaCO_3$.

(A) In the Bayer's process,$NaOH$ is used to leach bauxite ore $(Al_2O_3 \cdot 2H_2O)$ to separate it from impurities like $Fe_2O_3$ and $SiO_2$.
$NaOH$ is not a primary standard because it is hygroscopic and absorbs $CO_2$ from the air.
Manganous hydroxide,$Mn(OH)_2$,is insoluble in excess $NaOH$.
$NaOH$ reacts with $Cl_2$ to form sodium chloride and sodium hypochlorite or chlorate depending on the temperature.

(D) According to the Ellingham diagram,the line for the formation of $Fe_2O_3$ lies below the line for the formation of $CO$ at temperatures below $983 \ K$.
This indicates that the formation of $Fe_2O_3$ is more spontaneous (more negative $\Delta G^\circ$) than the formation of $CO$ in this temperature range.
Therefore,iron has a higher affinity for oxygen than carbon below $983 \ K$,making the reduction of $Fe_2O_3$ by carbon non-spontaneous.

(D) Roasting is the process of heating an ore in a regular supply of air in a furnace at a temperature below the melting point of the metal. $ZnS$ (zinc blende) is a sulfide ore,and sulfide ores are always subjected to roasting,not calcination,to convert them into their respective oxides. Therefore,the statement that zinc blende is subjected to calcination is false.

(C) The hydrolysis of $t$-butyl bromide $((CH_3)_3CBr)$ with aqueous $NaOH$ follows the $S_{N}1$ mechanism.
In the $S_{N}1$ mechanism,the rate-determining step is the formation of a carbocation intermediate,which depends only on the concentration of the substrate ($t$-butyl bromide).
Rate $= k[(CH_3)_3CBr]$.
Since the rate is independent of the concentration of the nucleophile $(OH^-)$,the rate of the reaction does not change when the concentration of alkali is doubled.
Therefore,the statement that the rate of the reaction doubles when the concentration of alkali is doubled is false.

(A) The general reaction is $R-Cl + KCN \rightarrow R-CN + KCl$.
Propanenitrile is $CH_3CH_2CN$.
For this to form,the alkyl group $R$ must be $CH_3CH_2-$.
Therefore,$R-Cl$ is $CH_3CH_2Cl$,which is chloroethane.

(D) In Ramsay and Rayleigh's isolation of noble gases from air,the nitrogen of the air is finally converted into $NaNO_{2}$ and $NaNO_{3}$.
$N_{2} + O_{2} \xrightarrow{\text{Electric discharge}} 2 NO$
$2 NO + O_{2} \longrightarrow 2 NO_{2(g)}$
$2 NO_{2} + 2 NaOH \longrightarrow NaNO_{2} + NaNO_{3} + H_{2}O$

(B) Initial equivalents of $HCl = N_1 V_1 = 0.2 \ N \times 50 \ cm^{3} = 10 \ meq$.
Equivalents of $NaOH$ added $= N_2 V_2 = 0.1 \ N \times 50 \ cm^{3} = 5 \ meq$.
Equivalents of $HCl$ remaining $= 10 \ meq - 5 \ meq = 5 \ meq$.
To neutralize the remaining $5 \ meq$ of $HCl$,we use $0.5 \ N \ KOH$.
Volume of $KOH$ required $= \frac{\text{Equivalents}}{\text{Normality}} = \frac{5 \ meq}{0.5 \ N} = 10 \ cm^{3}$.

(D) $CH_{2}=CH_{2} \xrightarrow{\text{Conc. } H_{2}SO_{4}} CH_{3}CH_{2}HSO_{4} \text{ (A)}$
$CH_{3}CH_{2}HSO_{4} \xrightarrow{\Delta, H_{2}O} CH_{3}CH_{2}OH \text{ (B)}$
$CH_{3}CH_{2}OH \xrightarrow{PBr_{3}} CH_{3}CH_{2}Br \text{ (C)}$
$CH_{3}CH_{2}Br \xrightarrow{\text{Alc. } KOH, \Delta} CH_{2}=CH_{2} \text{ (D)}$
Thus,$B$ is ethanol $(CH_{3}CH_{2}OH)$ and $D$ is alcoholic $KOH$.

(C) In quartz $(SiO_2)$,the lattice points are occupied by silicon and oxygen atoms,which are held together by a continuous network of covalent bonds.
Therefore,quartz is classified as a covalent or network crystal.

(A) When a solution boils at a temperature higher than the boiling points of its individual components,it indicates that the vapour pressure of the solution is lower than what is expected from Raoult's law. This decrease in vapour pressure is characteristic of a negative deviation from Raoult's law.

(B) For isotonic solutions,the molar concentration is the same,so $C_1 = C_2$.
Given that $0.05 \ M$ glucose solution has the same osmotic pressure as the compound solution,the molarity of the compound solution is $0.05 \ M$.
Assuming the volume of the solution is $V \ L$,the number of moles of the compound is $n = M \times V = 0.05 \times V$.
Also,$n = \frac{\text{mass}}{\text{molar mass}} = \frac{3}{M_X}$.
Equating these,$\frac{3}{M_X} = 0.05 \times V$.
Assuming the volume is $1 \ L$ for standard comparison,$M_X = \frac{3}{0.05} = 60 \ g/mol$.
The empirical formula mass of $CH_2O$ is $12 + 2(1) + 16 = 30 \ g/mol$.
The value of $n$ is $\frac{\text{molecular mass}}{\text{empirical formula mass}} = \frac{60}{30} = 2$.
Therefore,the molecular formula is $2 \times CH_2O = C_2H_4O_2$.

(A) The formula of the complex is $[Cr(H_2O)_5Cl]Cl_2$.
Each mole of the complex releases $2 \ mol$ of ionizable $Cl^-$ ions.
Number of moles of complex = $Molarity \times Volume (L) = 0.01 \ M \times 0.1 \ L = 0.001 \ mol$.
Number of moles of $Cl^-$ ions = $2 \times 0.001 \ mol = 0.002 \ mol$.
Reaction: $Ag^+(aq) + Cl^-(aq) \rightarrow AgCl(s)$.
Moles of $AgCl$ formed = $0.002 \ mol$.
Molar mass of $AgCl = 108 + 35.5 = 143.5 \ g/mol$.
Mass of $AgCl = 0.002 \ mol \times 143.5 \ g/mol = 0.287 \ g = 287 \times 10^{-3} \ g$.

(A) The reaction is $2 A_{(g)} \rightarrow B_{(g)} + C_{(s)}$.
Let the initial pressure of $A$ be $P_0$.
At $t = \infty$ (completion),only $B_{(g)}$ is present,so $P_B = P_0 / 2 = 200 \ Pa$,which means $P_0 = 400 \ Pa$.
At $t = 10 \ min$,let the pressure of $A$ reacted be $2x$.
Initial: $P_A = 400, P_B = 0, P_C = 0$.
At $t = 10$: $P_A = 400 - 2x, P_B = x, P_C = 0$ (since $C$ is solid).
Total pressure $P_t = (400 - 2x) + x = 400 - x = 300 \ Pa$.
Therefore,$x = 100 \ Pa$.
The partial pressure of $A$ at $t = 10 \ min$ is $P_A = 400 - 2(100) = 200 \ Pa$.
For a first-order reaction,$k = \frac{2.303}{t} \log \left( \frac{P_0}{P_A} \right)$.
$k = \frac{2.303}{10} \log \left( \frac{400}{200} \right) = \frac{2.303}{10} \log 2$.
$k = \frac{2.303 \times 0.3010}{10} \approx 0.0693 \ min^{-1}$.

(C) The partition coefficient $(K)$ for silver between molten zinc and molten lead is $300$.
Let $x$ be the mass of $Ag$ in $Zn$ and $(1 - x)$ be the mass of $Ag$ in $Pb$.
Partition coefficient $= \frac{\text{Conc. of } Ag \text{ in molten } Zn}{\text{Conc. of } Ag \text{ in molten } Pb} = \frac{x / 10}{(1 - x) / 100} = 300$.
Solving for $x$: $\frac{10x}{1 - x} = 300 \implies 10x = 300 - 300x \implies 310x = 300 \implies x = \frac{30}{31}$.
Amount of silver in molten lead $= 1 - x = 1 - \frac{30}{31} = \frac{1}{31} \ g$.
Percentage of silver in lead $= \frac{1/31}{1} \times 100 \approx 3.22 \% \approx 3 \%$.

(D) Coagulation is the process of aggregating colloidal particles to form a precipitate.
$A$,$B$,and $C$ involve the aggregation of particles (coagulation).
Peptization is the reverse process,where a freshly prepared precipitate is converted into a colloidal sol by adding a suitable electrolyte (peptizing agent).
Therefore,peptization does not involve coagulation.

(A) Adsorption is a spontaneous exothermic process.
Since it is exothermic,the enthalpy change $(\Delta H)$ is negative $(\Delta H < 0)$.
As gas molecules are adsorbed on the solid surface,their freedom of movement decreases,leading to a decrease in entropy $(\Delta S < 0)$.