KCET 2017 Chemistry Question Paper with Answer and Solution

(D) Leaching is a process used when the ore is soluble in a suitable solvent. In the metallurgy of silver $(Ag)$ and gold $(Au)$,the metal is leached with a dilute solution of $NaCN$ or $KCN$ in the presence of air (for $O_2$). The metal is subsequently recovered by displacement using a more reactive metal like zinc $(Zn)$.
The chemical reaction is:
$4 M_{(s)} + 8 CN^{-}_{(aq)} + 2 H_2O_{(aq)} + O_{2(g)} \rightarrow 4[M(CN)_2]^{-}_{(aq)} + 4 OH^{-}_{(aq)}$ (where $M = Ag$ or $Au$)
Recovery step:
$2[M(CN)_2]^{-}_{(aq)} + Zn_{(s)} \rightarrow [Zn(CN)_4]^{2-}_{(aq)} + 2 M_{(s)}$

(A) Antibiotics are classified as bactericidal or bacteriostatic based on their action.
Bactericidal antibiotics kill the microorganisms in the body.
Examples of bactericidal antibiotics include $Penicillin$,$Ofloxacin$,and $Aminoglycosides$.
Bacteriostatic antibiotics inhibit the growth of microorganisms.
Examples of bacteriostatic antibiotics include $Chloramphenicol$,$Erythromycin$,and $Tetracycline$.
Therefore,$Ofloxacin$ is a bactericidal antibiotic.

(A) The force of attraction in real gases tends to show slightly lower pressure as compared to ideal gases.
$p_{ideal} = p_{real} + \frac{a n^2}{V^2}$ or $p_{real} = p_{ideal} - \frac{a n^2}{V^2}$
Here,'$a$' is the constant which measures the magnitude of attractive forces among the molecules of gas,'$n$' is the number of moles present,and '$V$' is the volume of the gas.

(B) The $N_{2}O$ molecule has a linear structure with the connectivity $N-N-O$.
Total valence electrons $= (5 \times 2) + 6 = 16 \ e^{-}$.
To satisfy the octet rule for all atoms,the most stable Lewis structure involves a triple bond between the two nitrogen atoms and a single bond between the central nitrogen and oxygen,or a double bond between both pairs.
The most significant resonance contributor is $:\ddot{N}^{-}-N^{+}\equiv O:$,where the central nitrogen has a formal charge of $+1$,the terminal nitrogen has a formal charge of $-1$,and the oxygen has a formal charge of $0$.

(C, D) The reaction quotient $Q_{C}$ is compared with the equilibrium constant $K_{C}$ to determine the direction of the reaction:
$1$. If $Q_{C} > K_{C}$,the system has excess products,so the reverse reaction is favoured to reach equilibrium.
$2$. If $Q_{C} < K_{C}$,the system has excess reactants,so the forward reaction is favoured to reach equilibrium.
$3$. If $Q_{C} = K_{C}$,the reaction is at equilibrium.
Therefore,statements $C$ and $D$ are incorrect.

(C) The given reaction is $3 ClO_{3}^{-}(aq) \rightarrow ClO^{-}(aq) + 2 ClO_{3}^{-}(aq)$.
Actually,the correct reaction for disproportionation of chlorate is $3 ClO_{3}^{-}(aq) \rightarrow ClO_{4}^{-}(aq) + 2 ClO_{2}(aq)$ or similar,but based on the provided image logic:
In the reaction $ClO_{3}^{-} \rightarrow ClO^{-} + Cl^{-}$,the oxidation state of $Cl$ changes from $+5$ to $+1$ and $-1$.
Since the same element $(Cl)$ is simultaneously reduced (from $+5$ to $+1$ and $-1$) and oxidized (if applicable,though here it is only reduction),this specific type of reaction where an element's oxidation state changes in both directions is called a disproportionation reaction.

(B) The molar mass of $H_2SO_4$ is $98 \ g/mol$.
Initial amount of $H_2SO_4 = 98 \ mg = 98 \times 10^{-3} \ g = 0.1 \times 10^{-2} \ mol = 1 \times 10^{-3} \ mol$.
Number of molecules in $1 \times 10^{-3} \ mol = 1 \times 10^{-3} \times 6.022 \times 10^{23} = 6.022 \times 10^{20}$ molecules.
Number of molecules removed $= 3.01 \times 10^{20}$.
Number of molecules left $= 6.022 \times 10^{20} - 3.01 \times 10^{20} = 3.012 \times 10^{20}$ molecules.
Number of moles left $= \frac{3.012 \times 10^{20}}{6.022 \times 10^{23}} \approx 0.5 \times 10^{-3} \ mol$.

(D) Photo-chemical smog is a mixture of air pollutants that have been chemically altered into further poisonous compounds upon exposure to sunlight.
The main components of photochemical smog include nitrogen oxides $(NO_x)$,volatile organic compounds $(VOCs)$,tropospheric ozone $(O_3)$,peroxyacetyl nitrate $(PAN)$,and acrolein.
Chlorofluorocarbons $(CFCs)$ are not components of photochemical smog; they are primarily responsible for ozone layer depletion in the stratosphere.

(B) For a reaction where $\Delta H < 0$ and $\Delta S < 0$,the spontaneity is governed by the Gibbs free energy equation: $\Delta G = \Delta H - T\Delta S$.
For the reaction to be spontaneous,$\Delta G$ must be negative.
Since $\Delta S$ is negative,the term $-T\Delta S$ is positive.
As temperature $(T)$ decreases,the magnitude of the positive term $-T\Delta S$ decreases,making $\Delta G$ more negative.
Thus,the reaction becomes more thermodynamically favorable at lower temperatures.
In the context of this question,the term "rate" refers to the thermodynamic favorability or the extent of the reaction.

(D) This is $Kolbe's$ electrolytic method which produces alkanes.
$2 RCOONa + 2 H_2O \xrightarrow{\text{Electrolysis}} R-R + 2 NaOH + H_2 + 2 CO_2$
Steps of reaction:
$2 CH_3COONa \rightleftharpoons 2 CH_3COO^{-} + 2 Na^{+}$
$2 H_2O \rightleftharpoons 2 OH^{-} + 2 H^{+}$
At anode:
$2 CH_3COO^{-} - 2 e^{-}$ $\rightarrow [2 CH_3COO]$ $\rightarrow CH_3-CH_3 + 2 CO_2$
At cathode:
$2 H^{+} + 2 e^{-}$ $\rightarrow [2 H]$ $\rightarrow H_2$

(C) The electronic configuration of a chlorine atom $(Z=17)$ is $1s^2 2s^2 2p^6 3s^2 3p^5$.
The unpaired electron is present in the $3p$ subshell.
For the $3p$ orbital: Principal quantum number $(n)$ = $3$,Azimuthal quantum number $(l)$ = $1$.
The $3p^5$ configuration fills the orbitals as follows: $m_l = -1$ ($2$ electrons),$m_l = 0$ ($2$ electrons),$m_l = +1$ ($1$ electron).
Thus,for the unpaired electron: $n=3$,$l=1$,$m_l=1$,and $m_s = \pm \frac{1}{2}$.

(D) $Al_{2}O_{3}$ is an amphoteric metallic oxide because it reacts with both acids and bases.
$Al_{2}O_{3} + 6HCl \rightarrow 2AlCl_{3} + 3H_{2}O$
$Al_{2}O_{3} + 2NaOH \rightarrow 2NaAlO_{2} + H_{2}O$

(B) The given curve is $y=x^{2}$ and the line is $y=16$.
To find the area,we integrate with respect to $y$ from $y=0$ to $y=16$.
The curve $y=x^{2}$ implies $x = \pm \sqrt{y}$.
The area is given by the integral:
$Area = \int_{-4}^{4} (16 - x^{2}) dx$
$= 2 \int_{0}^{4} (16 - x^{2}) dx$
$= 2 [16x - \frac{x^{3}}{3}]_{0}^{4}$
$= 2 [16(4) - \frac{4^{3}}{3}]$
$= 2 [64 - \frac{64}{3}]$
$= 2 [\frac{192 - 64}{3}]$
$= 2 [\frac{128}{3}] = \frac{256}{3} \text{ sq. units}$.
Alternatively,using integration with respect to $y$:
$Area = 2 \int_{0}^{16} \sqrt{y} dy$
$= 2 [\frac{y^{3/2}}{3/2}]_{0}^{16}$
$= 2 \times \frac{2}{3} [y^{3/2}]_{0}^{16}$
$= \frac{4}{3} [16^{3/2}]$
$= \frac{4}{3} [64] = \frac{256}{3} \text{ sq. units}$.

(A) In the manufacture of hydrogen from water gas $(CO + H_2)$,the process is known as the water-gas shift reaction.
The reaction is: $CO(g) + H_2O(g) \xrightarrow{\text{catalyst}} CO_2(g) + H_2(g)$.
$CO$ is oxidized to $CO_2$ with steam in the presence of an iron chromate catalyst.
The resulting $CO_2$ is then removed by scrubbing the mixture with a solution of sodium arsenite or alkali (like $KOH$ or $NaOH$) to obtain pure hydrogen.

(A) $Ofloxacin$ is a bactericidal antibiotic,which means it kills bacteria.
$Erythromycin$,$tetracycline$,and $chloramphenicol$ are bacteriostatic antibiotics,which inhibit the growth of microorganisms.

(C) Electronegativity generally increases across a period from left to right and decreases down a group from top to bottom.
In the periodic table,$C$ and $N$ belong to the $2^{nd}$ period,while $Si$ and $P$ belong to the $3^{rd}$ period.
Comparing the values: $Si$ $(1.90)$ < $P$ $(2.19)$ < $C$ $(2.55)$ < $N$ $(3.04)$.
Thus,the correct increasing order of electronegativity is $Si < P < C < N$.

(C) The correct answer is $(C)$.
Homolytic bond fission occurs when a covalent bond breaks such that each atom retains one electron,resulting in the formation of free radicals.
Free radical chlorination of methane involves the homolytic cleavage of the $Cl-Cl$ bond by $UV$ light or heat to produce chlorine radicals $(Cl^{\bullet})$.
$Cl-Cl \xrightarrow{h\nu} Cl^{\bullet} + Cl^{\bullet}$

(B) When a mineral acid (like $HCl$) is added to an aqueous solution of Borax $(Na_{2}B_{4}O_{7} \cdot 10H_{2}O)$,it reacts to form orthoboric acid $(H_{3}BO_{3})$.
The chemical reaction is:
$Na_{2}B_{4}O_{7} + 2HCl + 5H_{2}O \rightarrow 4H_{3}BO_{3} + 2NaCl$

(D) The equilibrium constant $(K_c)$ depends only on temperature and is independent of the presence of a catalyst. $A$ catalyst only increases the rate of both forward and backward reactions equally,allowing the system to reach equilibrium faster without changing the position of equilibrium or the value of the equilibrium constant. Therefore,the equilibrium constant remains $4 \times 10^{-4}$.

(A) Plaster of Paris is a hemihydrate of calcium sulfate.
It is chemically represented as $CaSO_4 \cdot \frac{1}{2} H_2O$.

(C) Option $(A)$: $n$-Butane and iso-butane are chain isomers,not functional isomers.
Option $(B)$: Dimethyl ether $(CH_3OCH_3)$ and ethanol $(CH_3CH_2OH)$ are functional isomers because they contain different functional groups (ether and alcohol).
Option $(C)$: Propan-$1$-ol $(CH_3CH_2CH_2OH)$ and propan-$2$-ol $(CH_3CH(OH)CH_3)$ are position isomers because the position of the hydroxyl $(-OH)$ group changes along the carbon chain.
Option $(D)$: Ethanoic acid $(CH_3COOH)$ and methyl methanoate $(HCOOCH_3)$ are functional isomers because they contain different functional groups (carboxylic acid and ester).

(C) According to $VSEPR$ theory:
For tetrahedral geometry: No. of electron pairs $= 4$,Bond pairs $= 4$,Lone pairs $= 0$.
For pyramidal geometry (e.g.,$NH_3$): No. of electron pairs $= 4$,Bond pairs $= 3$,Lone pairs $= 1$.
For trigonal planar geometry: No. of electron pairs $= 3$,Bond pairs $= 3$,Lone pairs $= 0$.
For octahedral geometry: No. of electron pairs $= 6$,Bond pairs $= 6$,Lone pairs $= 0$.
Therefore,the structure with three bond pairs and one lone pair is pyramidal.

(A) Antibiotics are classified as either bactericidal (kill bacteria) or bacteriostatic (inhibit the growth of bacteria).
$Ofloxacin$ is a bactericidal antibiotic.
$Tetracycline$,$Chloramphenicol$,and $Erythromycin$ are examples of bacteriostatic antibiotics.

(A) Lower members of aliphatic carboxylic acids are soluble in water. This is due to the formation of hydrogen bonds with water molecules.
Carboxylic acids are polar molecules; they tend to be soluble in water,but as the alkyl chain gets longer,their solubility decreases due to the increasing hydrophobic nature of the carbon chain.

(B) The reaction involved is an electrophilic aromatic substitution.
In the presence of $FeCl_{3}$ (a Lewis acid) and in the dark,the halogen $(X_{2})$ reacts to generate an electrophile $(X^{+})$.
This electrophile attacks the benzene ring of toluene.
Since the methyl group $(-CH_{3})$ is an ortho/para-directing group due to its $+I$ effect and hyperconjugation,the electrophilic substitution occurs primarily at the ortho and para positions to yield ortho and para-halotoluene.

(A) The reaction involved is:
$(CH_{3})_{3}C-O-CH_{3} + HI \rightarrow (CH_{3})_{3}C-I + CH_{3}OH$
This is because the reaction occurs by the $S_{N}1$ mechanism. The formation of products is governed by the stability of the carbocation formed from the cleavage of the $C-O$ bond in the protonated ether (oxonium ion).
Since the tert-butyl carbocation $[(CH_{3})_{3}C^{+}]$ is more stable than the methyl carbocation $[CH_{3}^{+}]$,the cleavage of the $C-O$ bond gives the more stable carbocation $[(CH_{3})_{3}C^{+}]$ and methanol as products.
Then,the iodide ion $(I^{-})$ attacks the tert-butyl carbocation to form tert-butyl iodide.

(A) reducing agent is a substance that loses electrons and gets oxidized.
The strength of a reducing agent is inversely proportional to its standard reduction potential $(E^{\circ})$.
The more negative the $E^{\circ}$ value,the greater the tendency to lose electrons,making it a stronger reducing agent.
Comparing the given values:
$E^{\circ} (Zn^{2+}/Zn) = -0.762 \ V$
$E^{\circ} (Cr^{3+}/Cr) = -0.740 \ V$
$E^{\circ} (H^{+}/H_2) = 0.0 \ V$
$E^{\circ} (F_2/F^-) = 2.87 \ V$
Since $-0.762 \ V$ is the most negative value,$Zn_{(s)}$ is the strongest reducing agent.

(C) The Cannizzaro reaction is a disproportionation reaction (auto-oxidation and reduction).
It occurs in aldehydes that do not possess an $\alpha$-hydrogen atom,such as formaldehyde $(HCHO)$ and benzaldehyde $(C_6H_5CHO)$.
When treated with concentrated alkali,one molecule of the aldehyde is oxidized to a carboxylic acid salt,and another molecule is reduced to an alcohol.

(C) In the reaction of $NaOH$ with $Cl_2$,hydrogen gas $(H_2)$ is not produced. The reaction depends on the concentration of $NaOH$.
For cold and dilute $NaOH$,the reaction is: $2NaOH + Cl_2 \rightarrow NaCl + NaClO + H_2O$.
For hot and concentrated $NaOH$,the reaction is: $6NaOH + 3Cl_2 \rightarrow 5NaCl + NaClO_3 + 3H_2O$.
Option $C$ is incorrect because it suggests the liberation of $H_2$ and $O_2$ gases,which does not occur.

(A) delta is formed where a river meets the sea due to the process of settling down of colloidal particles. The river water contains colloidal clay particles,while sea water contains various electrolytes. When these two meet,the electrolytes present in sea water cause the coagulation of the colloidal clay particles,leading to their deposition and the formation of a delta.

(B) The Van't Hoff factor '$i$' is defined as the ratio of the observed colligative property to the calculated colligative property.
It accounts for the extent of dissociation or association of solute particles in a solution.

(D) The reaction sequence is as follows:
$1$. $CH_{3}Br + KCN \rightarrow CH_{3}CN + KBr$ (Product $A$ is $CH_{3}CN$)
$2$. $CH_{3}CN + 2H_{2}O \xrightarrow{H^{+}} CH_{3}COOH + NH_{3}$ (Product $B$ is $CH_{3}COOH$)
$3$. $CH_{3}COOH \xrightarrow{LiAlH_{4}} CH_{3}CH_{2}OH$ (Product $C$ is $CH_{3}CH_{2}OH$)
Thus,the final product $C$ is ethyl alcohol.

(A) The condition $a \neq b \neq c$ and $\alpha \neq \beta \neq \gamma \neq 90^{\circ}$ represents the triclinic crystal system.
Among the given options,$K_2Cr_2O_7$ crystallizes in the triclinic system.
Therefore,the correct option is $A$.

(B) When an external pressure higher than the osmotic pressure is applied to the solution side,the solvent molecules move from the solution into the pure solvent through a semi-permeable membrane. This process is known as $Reverse \ osmosis$.

(C) The balanced chemical equation is: $NaClO_{3} + H_{2}O \rightarrow NaClO_{4} + H_{2}$.
The oxidation state of $Cl$ in $NaClO_{3}$ is $+5$ and in $NaClO_{4}$ is $+7$.
The change in oxidation state is $7 - 5 = 2$,which means $2$ moles of electrons are involved per mole of $NaClO_{3}$.
Thus,$2 \ F$ of charge is required to produce $1$ mole of $NaClO_{4}$.
Therefore,$1 \ F$ of charge produces $1/2$ mole of $NaClO_{4}$.
For $3 \ F$ of charge,the moles of $NaClO_{4}$ formed $= (1/2) \times 3 = 1.5 \ mol$.

(B) Acidified $KMnO_4$ is a strong oxidizing agent. It oxidizes primary alcohols directly to carboxylic acids rather than stopping at the aldehyde stage. Therefore,it cannot be used to prepare aldehydes from primary alcohols.

(B) Option $(B)$ is correct.
$1$. Frenkel defect occurs in compounds where there is a large difference in the size of ions,specifically where anions are much larger than cations.
$2$. Frenkel defect is known as a dislocation defect because an ion is dislocated from its normal lattice site to an interstitial site.
$3$. $F$-centers are formed by the trapping of electrons in anion vacancies,not protons.
$4$. Schottky defect leads to a decrease in the density of the crystal due to the presence of vacancies.

(A) The extraction of chlorine from brine solution ($NaCl$ solution) is based on the oxidation of chloride ions $(Cl^{-})$ to chlorine gas $(Cl_2)$ during electrolysis.
The reaction at the anode is: $2Cl^{-}_{(aq)} \rightarrow Cl_{2(g)} + 2e^{-}$.

(A) According to the Arrhenius equation,$k = A e^{\frac{-E_{a}}{RT}}$.
Taking the natural logarithm on both sides,we get $\ln k = \ln A - \frac{E_{a}}{R} \times \frac{1}{T}$.
From this equation,it is clear that as the temperature $(T)$ increases,the term $\frac{E_{a}}{RT}$ decreases,which causes the rate constant $(k)$ to increase exponentially.
Similarly,if the activation energy $(E_{a})$ decreases,the term $\frac{E_{a}}{RT}$ decreases,which also leads to an increase in the rate constant $(k)$.
Therefore,both statements $A$ and $B$ are technically correct based on the Arrhenius equation. However,in standard multiple-choice contexts,the dependence on temperature is the primary characteristic.

(A) The hydrogenation of vegetable oil involves the reaction of liquid oil with gaseous $H_2$ in the presence of solid $Ni$ catalyst.
Since the catalyst $(Ni)$ is in a solid phase and the reactants (oil and $H_2$) are in liquid/gas phases,the phases are different.
Therefore,this is an example of heterogeneous catalysis.

(C) The basicity of aliphatic amines in aqueous solution depends on the combined effect of inductive effect $(+I)$,solvation effect,and steric hindrance.
Alkyl groups are electron-releasing ($+I$ effect),which increases the electron density on the nitrogen atom,thereby increasing basicity.
However,in aqueous solutions,the stability of the conjugate acid (ammonium ion) formed by protonation is also crucial,which is influenced by hydrogen bonding (solvation) and steric hindrance.
For the given amines,the order of basicity is $NH_{3} < CH_{3}NH_{2} < (CH_{3})_{2}NH$.

(D) Option $A$: Cetyltrimethylammonium bromide is a popular cationic detergent used in hair conditioners,not air conditioners.
Option $B$: Non-ionic detergents are formed when polyethylene glycol reacts with stearic acid,not adipic acid.
Option $C$: Sodium dodecyl benzene sulphonate used in toothpaste is an anionic detergent,not a cationic detergent.
Option $D$: Since all statements $A$,$B$,and $C$ are incorrect,the correct choice is $D$.

(A) Novolac is a linear polymer formed by the condensation polymerization of phenol and formaldehyde in the presence of an acid or base catalyst.
The reaction involves the formation of $o-$ or $p-$hydroxymethylphenol intermediates,which then undergo further condensation to form the linear polymer Novolac.

(A) Sucrose is a disaccharide composed of one molecule of $\alpha$-$D$-glucose and one molecule of $\beta$-$D$-fructose.
These two monosaccharide units are joined together by a glycosidic linkage between $C-1$ of $\alpha$-$D$-glucose and $C-2$ of $\beta$-$D$-fructose.

(D) In an aqueous solution of sodium chloride,the following ions are present: $Na^{+}$,$Cl^{-}$,$H^{+}$,and $OH^{-}$.
At the anode,oxidation occurs. The two possible oxidation reactions are:
$1$) $2H_{2}O_{(l)} \rightarrow O_{2(g)} + 4H^{+}_{(aq)} + 4e^{-} ; E^{\circ} = 1.23 \ V$
$2$) $Cl^{-}_{(aq)} \rightarrow \frac{1}{2}Cl_{2(g)} + e^{-} ; E^{\circ} = 1.36 \ V$
Although the standard electrode potential for water oxidation is lower,the oxidation of $Cl^{-}$ is kinetically preferred due to the overpotential of oxygen evolution on the electrode surface. Thus,$Cl^{-}$ ions are preferentially oxidized at the anode.

(A) For a square planar complex of the type $M(abcd)$,there are $3$ possible geometric isomers.
In the case of $MAXBL$,we can fix one ligand (e.g.,$M$) and arrange the others.
There are $2$ isomers where two specific ligands are in the $cis$ position (adjacent) and $1$ isomer where they are in the $trans$ position (opposite).
Thus,the complex shows $2$ $cis$ and $1$ $trans$ isomers.

(B) The acidity of substituted phenols is significantly affected by the nature of the substituents.
Electron-withdrawing groups $(EWG)$ such as $-NO_2$ increase the acidic strength by stabilizing the phenoxide ion through the $-I$ and $-M$ effects.
Conversely,electron-donating groups $(EDG)$ such as $-CH_3$ (methyl group in cresol) decrease the acidic strength due to the $+I$ effect and hyperconjugation,which destabilizes the phenoxide ion.
Therefore,the acidic strength follows the order: $o-cresol < phenol < o-nitrophenol$.

(B) For a general reaction $aA \rightarrow bB$,the rate of reaction is given by:
Rate $= -\frac{1}{a} \frac{d[A]}{dt} = \frac{1}{b} \frac{d[B]}{dt}$
Given the reaction $\frac{1}{2} A \rightarrow 2 B$,we have $a = \frac{1}{2}$ and $b = 2$.
Substituting these values:
$-\frac{1}{1/2} \frac{d[A]}{dt} = \frac{1}{2} \frac{d[B]}{dt}$
$-2 \frac{d[A]}{dt} = \frac{1}{2} \frac{d[B]}{dt}$
Multiplying both sides by $\frac{1}{2}$:
$-\frac{d[A]}{dt} = \frac{1}{4} \frac{d[B]}{dt}$

(A) The fundamental assumption of $CFT$ is that $M-L$ interactions are purely electrostatic in nature.
In an octahedral metal complex,the electrostatic interaction of a positively charged metal ion with six negatively charged ligands or with the negative ends of dipoles associated with the six ligands is considered.

(B) The hormone that contains iodine is $Thyroxine$. It is secreted by the thyroid gland.

(D) Xerophthalmia is caused by the deficiency of Vitamin $A$. It is characterized by the hardening and dryness of the cornea of the eye. Deficiency of Vitamin $D$ causes rickets in children and osteomalacia in adults,not xerophthalmia. Therefore,the statement in option $D$ is incorrect.

(D) Ketones can be reduced to secondary alcohols using $NaBH_4$ or $LiAlH_4$.
They can also be reduced to alkanes using Clemmensen reduction $(Zn(Hg)/conc. HCl)$ or Wolff-Kishner reduction ($NH_2NH_2/KOH$ in ethylene glycol).
Hydrogen in the presence of palladium in Barium sulphate and quinoline is known as Lindlar's catalyst,which is specifically used for the partial reduction of alkynes to alkenes,not for the reduction of ketones.

(A) $N$ forms $p\pi-p\pi$ bond with itself because the $N$ atom has a small atomic size and high electronegativity,allowing it to form $1\sigma$ and $2\pi$ bonds (triple bonds) with another $N$ atom.
This results in the formation of a stable diatomic molecule,$N_2$.

(A) In a face-centred cubic $(FCC)$ unit cell,there are $8$ corners and $6$ face centres.
Number of $A$ atoms at corners = $7 \times \frac{1}{8} = \frac{7}{8}$ (since one corner is missing).
Number of $B$ atoms at face centres = $6 \times \frac{1}{2} = 3$.
The ratio of $A : B = \frac{7}{8} : 3$.
Multiplying both sides by $8$,we get the ratio $7 : 24$.
Therefore,the simplest formula of the compound is $A_{7} B_{24}$.

(B) The metal extracted by leaching with a cyanide is $Ag$ (or $Au$).
In the Mac-Arthur Forest process,silver $(Ag)$ or gold $(Au)$ ores are treated with a dilute solution of $NaCN$ in the presence of air $(O_2)$.
The chemical reaction is: $4M + 8CN^- + 2H_2O + O_2 \rightarrow 4[M(CN)_2]^- + 4OH^-$.
Subsequently,the metal is recovered by displacement using a more electropositive metal like zinc $(Zn)$: $2[M(CN)_2]^- + Zn \rightarrow [Zn(CN)_4]^{2-} + 2M$.

(D) $He$ (helium) has an unusual property of diffusing through materials such as rubber,glass,or plastic.
This is due to its very small atomic size.
The increasing order of diffusibility of noble gases is $Xe < Kr < Ar < Ne < He$.

(A) The complex is $[M(en)_{2}(C_{2}O_{4})]NO_{2}$.
In this complex,the counter ion is $NO_{2}^{-}$,which has a charge of $-1$. Therefore,the complex ion $[M(en)_{2}(C_{2}O_{4})]$ must have a charge of $+1$.
Let the oxidation state of $M$ be $x$.
$en$ (ethane-$1,2$-diamine) is a neutral bidentate ligand (charge $= 0$).
$C_{2}O_{4}^{2-}$ (oxalate) is a bidentate ligand with a charge of $-2$.
Setting up the equation for the charge of the complex ion:
$x + (2 \times 0) + (1 \times -2) = +1$
$x - 2 = +1$
$x = +3$
Thus,the oxidation state of $M$ is $+3$.
Coordination number calculation:
$en$ is bidentate ($2$ donor sites each,$2 \times 2 = 4$ sites).
$C_{2}O_{4}^{2-}$ is bidentate ($2$ donor sites).
Total coordination number $= 4 + 2 = 6$.
Therefore,the coordination number is $6$ and the oxidation state is $3$.

(D) Biodegradable polymers are polymers that decompose by bacterial action.
Polyhydroxy butyrate-co-$\beta$-hydroxyvalerate $(PHBV)$ and Nylon $2$-Nylon-$6$ are examples of biodegradable polymers.
Glyptal is a condensation polymer of ethylene glycol and phthalic acid,which is not biodegradable.

(D) The depression in freezing point is given by the formula $\Delta T_f = i \times K_f \times m$.
Since the molality $(m)$ and the cryoscopic constant $(K_f)$ are the same for all solutions,the depression in freezing point $(\Delta T_f)$ is directly proportional to the van't Hoff factor $(i)$.
The freezing point of the solution is $T_f = T_f^0 - \Delta T_f$. Therefore,the solution with the lowest value of $i$ will have the highest freezing point.

Compound	van't Hoff Factor $(i)$
$Al_2(SO_4)_3 \rightarrow 2Al^{3+} + 3SO_4^{2-}$	$5$
$BaCl_2 \rightarrow Ba^{2+} + 2Cl^{-}$	$3$
$AlCl_3 \rightarrow Al^{3+} + 3Cl^{-}$	$4$
$NH_4Cl \rightarrow NH_4^+ + Cl^{-}$	$2$

Comparing the values of $i$,$NH_4Cl$ has the lowest van't Hoff factor $(i = 2)$. Thus,it will have the minimum depression in freezing point and consequently the highest freezing point.

(B) The spin-only magnetic moment $(\mu_{spin})$ is calculated using the formula $\mu_{spin} = \sqrt{n(n+2)} \text{ BM}$,where $n$ is the number of unpaired electrons.
| Electronic Configuration | No. of Unpaired Electrons $(n)$ | Magnetic Moment $(\mu_{spin})$ |
| $3d^7$ | $3$ | $\sqrt{3(3+2)} = \sqrt{15} \text{ BM}$ |
| $3d^5$ | $5$ | $\sqrt{5(5+2)} = \sqrt{35} \text{ BM}$ |
| $3d^8$ | $2$ | $\sqrt{2(2+2)} = \sqrt{8} = 2\sqrt{2} \text{ BM}$ |
| $3d^2$ | $2$ | $\sqrt{2(2+2)} = \sqrt{8} = 2\sqrt{2} \text{ BM}$ |
Comparing the values,the $3d^5$ configuration has the highest number of unpaired electrons $(n=5)$,resulting in the highest magnetic moment of $\sqrt{35} \text{ BM}$.

(A) $Ln(III)$ ions are often colored due to $f-f$ transitions,which occur because of the presence of partly filled $4f$ orbitals,allowing them to absorb certain wavelengths from the visible region of the spectrum.
Therefore,the statement that "$Ln(III)$ compounds are generally colourless" is incorrect.

(A) Physical adsorption is an exothermic process. According to Le Chatelier's principle,an increase in temperature decreases the extent of adsorption. Therefore,high temperature is an unfavorable condition for physical adsorption.

(C) Gabriel phthalimide synthesis is a method used to prepare pure primary amines.
$1$. Phthalimide is reacted with alcoholic $KOH$ to form potassium phthalimide.
$2$. Potassium phthalimide is then reacted with an alkyl halide to form $N$-alkyl phthalimide.
$3$. Finally,$N$-alkyl phthalimide undergoes alkaline hydrolysis using aqueous $NaOH$ to yield the primary amine and phthalic acid (as a salt).
$HCl$ is not used in this process as the final step requires basic hydrolysis to release the amine.

(A) The rate law is determined experimentally by measuring the rate of reaction at different concentrations of reactants. Therefore,the statement that the rate law cannot be determined experimentally is incorrect. Complex reactions often exhibit fractional orders,bimolecular reactions involve the collision of two species,and molecularity is a concept defined specifically for elementary reactions.