KCET 2015 Chemistry Question Paper with Answer and Solution

(C) According to Curie's Law,the magnetic susceptibility $\chi_m$ of a paramagnetic material is inversely proportional to its absolute temperature $T$ (in Kelvin).
$\chi_m \propto \frac{1}{T}$ or $\chi_m T = \text{constant}$.
Given:
$T_1 = -73\,^{\circ}C = -73 + 273 = 200\,K$
$\chi_{m1} = 0.0075$
$T_2 = -173\,^{\circ}C = -173 + 273 = 100\,K$
Using the relation $\chi_{m1} T_1 = \chi_{m2} T_2$:
$0.0075 \times 200 = \chi_{m2} \times 100$
$\chi_{m2} = \frac{0.0075 \times 200}{100}$
$\chi_{m2} = 0.0075 \times 2 = 0.015$.

(A) The pair of compounds which cannot exist together in solution is $NaHCO_3$ and $NaOH$.
$NaHCO_3$ is an acidic salt (amphoteric in nature) and $NaOH$ is a strong base.
When they are present in the same solution,they undergo an acid-base neutralization reaction as follows:
$NaHCO_3 + NaOH \longrightarrow Na_2CO_3 + H_2O$
Since they react with each other,they cannot coexist in the same solution.

(D) The correct statement is $(d)$.
Prokaryotic cells are characterized by the absence of a well-defined nucleus and membrane-bound organelles such as mitochondria,chloroplasts,endoplasmic reticulum,and Golgi apparatus.
Option $(a)$ is incorrect because prokaryotic cells lack a true nucleus.
Option $(b)$ is incorrect because animal cells do not possess a cell wall.
Option $(c)$ is incorrect because,according to the cell theory (Omnis cellula-e cellula),new cells arise from pre-existing cells,not from abiotic materials.

(C) The reaction sequence is as follows:
$1$. The reaction of a ketone with $PCl_5$ replaces the carbonyl oxygen with two chlorine atoms,forming a gem-dichloride: $\text{Cyclohexyl-CO-CH}_3 + PCl_5 \rightarrow \text{Cyclohexyl-CCl}_2\text{-CH}_3 (P) + POCl_3$.
$2$. The treatment of the gem-dichloride $(P)$ with excess $\text{NaNH}_2$ (a strong base) leads to double dehydrohalogenation,resulting in the formation of an alkyne: $\text{Cyclohexyl-CCl}_2\text{-CH}_3 + \text{excess NaNH}_2$ $\rightarrow \text{Cyclohexyl-C}\equiv\text{CH} (Q) + 2\text{NaCl} + 2\text{NH}_3$.
Therefore,the final product $Q$ is cyclohexylacetylene,which corresponds to option $C$.

(C) For the conversion $CH_4 \rightarrow C_2H_6$,there is no change in the hybridization of the carbon atom. Both $CH_4$ and $C_2H_6$ have $sp^3$ hybridization and tetrahedral geometry.
For the conversion $NH_3 \rightarrow NH_4^+$,nitrogen in $NH_3$ is $sp^3$ hybridized with a pyramidal shape. In $NH_4^+$,it remains $sp^3$ hybridized but the geometry becomes tetrahedral.
For the conversion $H_2O \rightarrow H_3O^+$,oxygen in $H_2O$ is $sp^3$ hybridized with a bent shape. In $H_3O^+$,it remains $sp^3$ hybridized but the shape becomes pyramidal.
For the conversion $BF_3 \rightarrow BF_4^-$,boron in $BF_3$ is $sp^2$ hybridized with a trigonal planar geometry. In $BF_4^-$,the hybridization changes to $sp^3$ and the geometry changes to tetrahedral.

(C) Molecular orbital configurations:
$O_{2}^{+} (15 \ e^{-}): \sigma 1s^{2} \sigma^{*} 1s^{2} \sigma 2s^{2} \sigma^{*} 2s^{2} \sigma 2p_{z}^{2} \pi 2p_{x}^{2} = \pi 2p_{y}^{2} \pi^{*} 2p_{x}^{1}$
Bond order $= \frac{1}{2}(10-5) = 2.5$,and it is paramagnetic due to one unpaired electron.
$O_{2}^{-} (17 \ e^{-}): \sigma 1s^{2} \sigma^{*} 1s^{2} \sigma 2s^{2} \sigma^{*} 2s^{2} \sigma 2p_{z}^{2} \pi 2p_{x}^{2} = \pi 2p_{y}^{2} \pi^{*} 2p_{x}^{2} = \pi^{*} 2p_{y}^{1}$
Bond order $= \frac{1}{2}(10-7) = 1.5$,and it is paramagnetic due to one unpaired electron.
Since both species have unpaired electrons,both are paramagnetic.
Option $C$ is incorrect because $O_{2}^{+}$ is paramagnetic,not diamagnetic.

(C) The dissociation of $H_2S$ is given by the equilibrium: $H_2S \rightleftharpoons 2H^+ + S^{2-}$.
In the presence of $HCl$,the concentration of $H^+$ ions increases significantly.
According to the common ion effect,this shift in equilibrium suppresses the dissociation of $H_2S$,resulting in a lower concentration of $S^{2-}$ ions.
Group-$2$ metal sulfides have very low solubility products $(K_{sp})$,so they precipitate even at this low $S^{2-}$ concentration.
Group-$4$ metal sulfides have higher $K_{sp}$ values and require a higher $S^{2-}$ concentration to precipitate,which is not achieved in the presence of $HCl$.

(C) The jump between $IE_3$ and $IE_4$ is very large ($2750$ to $11580 \ kJ \cdot mol^{-1}$),indicating that the fourth electron is removed from a stable inner shell (noble gas configuration).
This means the element has $3$ valence electrons.
Among the given options,Aluminum $(Al)$ has $3$ valence electrons $([Ne] 3s^2 3p^1)$.

(C) According to Curie's Law,the magnetic susceptibility $\chi$ of a paramagnetic material is inversely proportional to its absolute temperature $T$,i.e.,$\chi \propto \frac{1}{T}$.
Given:
$\chi_{1} = 0.0075$
$T_{1} = -73^{\circ} C = (-73 + 273) K = 200 K$
$T_{2} = -173^{\circ} C = (-173 + 273) K = 100 K$
Using the relation $\frac{\chi_{1}}{\chi_{2}} = \frac{T_{2}}{T_{1}}$:
$\frac{0.0075}{\chi_{2}} = \frac{100 K}{200 K}$
$\frac{0.0075}{\chi_{2}} = \frac{1}{2}$
$\chi_{2} = 2 \times 0.0075 = 0.015$
Thus,the magnetic susceptibility at $-173^{\circ} C$ is $0.015$.

(C) $H_2O_2$ acts as an oxidizing agent in many reactions.
$H_2O_2$ oxidizes $PbS$ to $PbSO_4$,$Na_2SO_3$ to $Na_2SO_4$,and $KI$ to $I_2$.
However,$H_2O_2$ reacts with $O_3$ to form $H_2O$ and $O_2$ $(H_2O_2 + O_3 \rightarrow H_2O + 2O_2)$.
In this reaction,$O_3$ is reduced to $O_2$,and $H_2O_2$ acts as a reducing agent,not an oxidizing agent.

(B) The general formula for alkynes is $C_{n}H_{2n-2}$. For $n=5$,the formula is $C_{5}H_{8}$.
There are three possible isomeric alkynes for this formula:
$(I)$ $CH \equiv C-CH_{2}-CH_{2}-CH_{3}$ (Pent-$1$-yne)
$(II)$ $CH_{3}-C \equiv C-CH_{2}-CH_{3}$ (Pent-$2$-yne)
$(III)$ $CH \equiv C-CH(CH_{3})_{2}$ ($3$-Methylbut-$1$-yne)
Thus,the total number of possible alkynes is $3$.

(D) Clark's process is a commercial method for the removal of temporary hardness from water.
In this process,a calculated amount of lime $(Ca(OH)_2)$ is added to the hard water.
The lime reacts with the soluble bicarbonates of calcium and magnesium to form insoluble carbonates,which are then removed by filtration.
The chemical reactions are as follows:
$CaO + H_2O \rightarrow Ca(OH)_2$
$Ca(HCO_3)_2 + Ca(OH)_2 \rightarrow 2 CaCO_3 \downarrow + 2 H_2O$
$Mg(HCO_3)_2 + 2 Ca(OH)_2 \rightarrow Mg(OH)_2 \downarrow + 2 CaCO_3 \downarrow + 2 H_2O$

(D) The $pH$ of the aqueous solution of these salts depends on the strength of the corresponding conjugate acid formed upon hydrolysis.
When these salts react with water,they form $NaOH$ (a strong base) and the corresponding oxoacid of chlorine.
The reactions are:
$NaClO + H_{2}O \rightarrow NaOH + HClO$
$NaClO_{2} + H_{2}O \rightarrow NaOH + HClO_{2}$
$NaClO_{3} + H_{2}O \rightarrow NaOH + HClO_{3}$
$NaClO_{4} + H_{2}O \rightarrow NaOH + HClO_{4}$
Since $NaOH$ is common to all,the $pH$ is determined by the acidity of the resulting acid.
$HClO_{4}$ is the strongest acid among the given options $(HClO < HClO_{2} < HClO_{3} < HClO_{4})$.
Therefore,the solution of $NaClO_{4}$ will have the highest concentration of $H^{+}$ ions,resulting in the lowest $pH$ value.

(A) In the given circuit,the battery of $ 12 \ V $ is connected in series with a $ 4 \ \Omega $ resistor. The two branches containing diodes $ D_{1} $ and $ D_{2} $ are in parallel with each other.
Diode $ D_{1} $ is forward-biased,allowing current to flow through the $ 3 \ \Omega $ resistor.
Diode $ D_{2} $ is reverse-biased,acting as an open circuit,so no current flows through the $ 2 \ \Omega $ resistor.
The total resistance of the circuit is the sum of the series resistor and the resistor in the active branch: $ R_{eq} = 4 \ \Omega + 3 \ \Omega = 7 \ \Omega $.
Using Ohm's law,the current $ I $ flowing in the circuit is:
$ I = \frac{V}{R_{eq}} = \frac{12 \ V}{7 \ \Omega} \approx 1.71 \ A $.

(C) In boron halides $(BX_{3})$,the boron atom has an incomplete octet and acts as a Lewis acid.
Back bonding occurs due to the donation of electron density from the filled $p$-orbital of the halogen to the empty $p$-orbital of boron.
In $BF_{3}$,the $2p$ orbital of $F$ and the $2p$ orbital of $B$ have similar sizes,leading to effective $p\pi-p\pi$ overlap.
Therefore,there is maximum $p\pi-p\pi$ back bonding in $BF_{3}$,which reduces its Lewis acidity.

(D) The reaction between sodium hydride $(NaH)$ and diborane $(B_{2}H_{6})$ occurs in an ether solvent,typically diethyl ether,$(C_{2}H_{5})_{2}O$,to produce sodium borohydride $(NaBH_{4})$.
The chemical equation is: $2NaH + B_{2}H_{6} \xrightarrow{(C_{2}H_{5})_{2}O} 2NaBH_{4}$.
Here,'$A$' is $(C_{2}H_{5})_{2}O$ and '$B$' is $NaBH_{4}$,which is a well-known reducing agent.

(A) $NaHCO_{3}$ is an acidic salt (amphoteric) and $NaOH$ is a strong base.
They react with each other to form sodium carbonate and water.
The reaction is:
$NaHCO_{3} + NaOH \rightarrow Na_{2}CO_{3} + H_{2}O$.
Therefore,they cannot coexist in a solution.

(B) $1$ mole of $O_2 = 6.022 \times 10^{23}$ molecules $= 32 \ g$ of $O_2$.
$1$ molecule of $O_2 = \frac{32}{6.022 \times 10^{23}} \ g$ of $O_2$.
Therefore,$1.8 \times 10^{22}$ molecules of $O_2 = \frac{32}{6.022 \times 10^{23}} \times 1.8 \times 10^{22} \ g$.
$= 0.960 \ g$ of $O_2$.

(D) $1$. Calculate the mass of $C$ and $H$ in the compound:
Mass of $C = \frac{12}{44} \times 0.44 \ g = 0.12 \ g$
Mass of $H = \frac{2}{18} \times 0.18 \ g = 0.02 \ g$
Mass of $O = 0.30 - (0.12 + 0.02) = 0.16 \ g$
$2$. Calculate the molar ratio:
Moles of $C = \frac{0.12}{12} = 0.01$
Moles of $H = \frac{0.02}{1} = 0.02$
Moles of $O = \frac{0.16}{16} = 0.01$
$3$. Determine the empirical formula:
The ratio $C:H:O = 0.01:0.02:0.01 = 1:2:1$. Thus,the empirical formula is $CH_2O$.
$4$. Determine the molecular formula:
Empirical formula mass $= 12 + 2(1) + 16 = 30 \ g \ mol^{-1}$.
$n = \frac{\text{Molecular mass}}{\text{Empirical formula mass}} = \frac{60}{30} = 2$.
Molecular formula $= 2 \times (CH_2O) = C_2H_4O_2$.

(C) In Maxwell's distribution of velocities,as the temperature increases,the peak of the curve shifts to the right (higher velocity) and the height of the peak decreases.
From the given plot,the peak corresponding to $T_{1}$ is at a higher velocity $(V_{1})$ and has a lower height $(f_{1})$ compared to the peak corresponding to $T_{2}$ ($V_{2}$ and $f_{2}$).
Therefore,$T_{1} > T_{2}$,$V_{1} > V_{2}$,and $f_{2} > f_{1}$.
Thus,the correct statement is $T_{1} > T_{2}$.

(B) For electron $P$,the quantum numbers are $n = 3$ and $l = 2$.
Thus,the value of $(n + l) = 3 + 2 = 5$.
For electron $Q$,the quantum numbers are $n = 3$ and $l = 0$.
Thus,the value of $(n + l) = 3 + 0 = 3$.
According to the $(n + l)$ rule (Aufbau principle),the orbital with a higher $(n + l)$ value has higher energy.
Therefore,$P$ has greater energy than $Q$.

(C) Given equations:
$1) \ S_{(s)} + \frac{3}{2} O_{2(g)} \rightarrow SO_{3(g)}; \Delta H_{1} = 2x \ kJ$
$2) \ SO_{2(g)} + \frac{1}{2} O_{2(g)} \rightarrow SO_{3(g)}; \Delta H_{2} = y \ kJ$
We need to find the heat of formation of $SO_{2}$,which is the enthalpy change for the reaction:
$S_{(s)} + O_{2(g)} \rightarrow SO_{2(g)}; \Delta H_{3} = ?$
Reverse equation $(2)$:
$SO_{3(g)} \rightarrow SO_{2(g)} + \frac{1}{2} O_{2(g)}; \Delta H_{4} = -y \ kJ$ $(3)$
Add equation $(1)$ and equation $(3)$:
$(S_{(s)} + \frac{3}{2} O_{2(g)}) + (SO_{3(g)}) \rightarrow (SO_{3(g)}) + (SO_{2(g)} + \frac{1}{2} O_{2(g)})$
$S_{(s)} + O_{2(g)} \rightarrow SO_{2(g)}$
The enthalpy change is $\Delta H_{3} = \Delta H_{1} + \Delta H_{4} = 2x - y \ kJ$.

(A) The bond dissociation energy of a $C-H$ bond depends on the stability of the radical formed after the homolytic cleavage of the bond.
In toluene $(C_6H_5CH_3)$,the removal of a hydrogen atom from the methyl group results in the formation of a benzyl radical $(C_6H_5CH_2^{\bullet})$.
This benzyl radical is highly stabilized by resonance with the benzene ring,as shown in the provided resonance structures.
Since the resulting radical is very stable,the energy required to break the $C-H$ bond in the methyl group of toluene is the lowest among the given options.

(C) The reaction of bromocyclohexane with magnesium in the presence of dry ether forms the Grignard reagent,cyclohexylmagnesium bromide $(A)$.
$C_6H_{11}Br + Mg \xrightarrow{\text{dry ether}} C_6H_{11}MgBr (A)$
Subsequent hydrolysis of the Grignard reagent with water in an acidic medium yields cyclohexane $(B)$ as the final product.
$C_6H_{11}MgBr + H_2O \xrightarrow{H^+} C_6H_{12} (B) + Mg(OH)Br$
Therefore,the product '$B$' is cyclohexane.

(A) The acidic strength of phenols depends on the stability of the phenoxide ion formed after the loss of a proton. Electron-withdrawing groups $(EWG)$ stabilize the phenoxide ion and increase acidity,while electron-donating groups $(EDG)$ destabilize it and decrease acidity.
$1$. $m-$cresol $(ii)$: The $-CH_3$ group is an $EDG$ due to $+I$ effect and hyperconjugation,making it less acidic than phenol.
$2$. Phenol $(iii)$: The reference compound.
$3$. $m-$chlorophenol $(iv)$: The $-Cl$ group is an $EWG$ due to its strong $-I$ effect,which outweighs its $+R$ effect at the meta position,making it more acidic than phenol.
$4$. $m-$nitrophenol $(i)$: The $-NO_2$ group is a strong $EWG$ due to both $-I$ and $-R$ effects,making it the most acidic among the given compounds.
Thus,the increasing order of acidic strength is: $m-$cresol $(ii)$ < phenol $(iii)$ < $m-$chlorophenol $(iv)$ < $m-$nitrophenol $(i)$.
The correct option is $A$.

(B) The formation of cyanohydrin from a ketone by reaction with $HCN$ is an example of a nucleophilic addition reaction.
In this reaction,the nucleophile $CN^-$ attacks the electrophilic carbonyl carbon of the ketone,leading to the formation of an intermediate alkoxide ion,which is then protonated to form the cyanohydrin.

(B) Methanoic acid $(HCOOH)$ contains a hydrogen atom attached to the carbonyl carbon,similar to an aldehyde group $(H-C(=O)-)$.
Therefore,like aldehydes,it reduces Tollen's reagent to metallic silver.
Ethanoic acid $(CH_3COOH)$ is a typical carboxylic acid that lacks this specific $H-C(=O)-$ group and does not reduce Tollen's reagent.
Thus,Tollen's test is used to distinguish between them.

(C) Iodoform test is given by compounds containing the $-COCH_{3}$ group or the $-CH(OH)CH_{3}$ group.
Propan$-2-$ol contains the $-CH(OH)CH_{3}$ group.
Butan$-2-$one contains the $-COCH_{3}$ group.
Acetophenone contains the $-COCH_{3}$ group attached to a benzene ring.
Propan$-1-$ol is a primary alcohol $(CH_{3}CH_{2}CH_{2}OH)$ and does not contain the $-CH(OH)CH_{3}$ group. Ethanol is the only primary alcohol that gives the iodoform reaction.

(B) The reaction of white phosphorus with concentrated $NaOH$ is: $P_{4} + 3NaOH + 3H_{2}O \xrightarrow{\Delta, CO_{2}} 3NaH_{2}PO_{2} + PH_{3}\uparrow$ (phosphine).
$PH_{3}$ is less basic than $NH_{3}$ because the lone pair of electrons on the phosphorus atom is less available for donation compared to nitrogen,due to the larger size of the phosphorus atom and the involvement of $s$-character in the bonding.
Therefore,the statement that $PH_{3}$ is more basic than $NH_{3}$ is incorrect.

(B) Only primary amides $(R-CONH_2)$ undergo the Hoffmann bromamide reaction to yield primary amines.
$R-CONH_2 + Br_2 + 4NaOH \xrightarrow{\Delta} R-NH_2 + Na_2CO_3 + 2NaBr + 2H_2O$
In the given options,$CH_3CONHCH_3$ is a secondary amide ($N$-methylacetamide) because the nitrogen atom is attached to one alkyl group in addition to the carbonyl group.
Since secondary amides lack the two hydrogen atoms on the nitrogen required for the rearrangement mechanism,it does not undergo the Hoffmann bromamide reaction.

(D) Glycogen is often referred to as animal starch because its structure is similar to amylopectin.
Both are polymers of $\alpha-D$-glucose units.
The primary difference is that glycogen is more highly branched than amylopectin.
While amylopectin has branching every $20-25$ glucose units,glycogen has branching every $10-14$ glucose units,making it more compact and extensively branched.

(C) Essential amino acids are those that the human body cannot synthesize on its own and must be obtained through the diet.
Out of the $20$ standard amino acids,$9$ are classified as essential.
Isoleucine is one of these $9$ essential amino acids.
Therefore,the correct option is $C$.

(C) The deficiency of $Vitamin \ B_2$ (Riboflavin) leads to cheilosis (fissuring at corners of mouth and lips),digestive disorders,and burning sensation of the skin.

(D) The reaction sequence is as follows:
$1$. Starting with $p$-nitro toluene $(P)$,bromination gives $2-$bromo$-4-$nitrotoluene.
$2$. Reduction with $Sn/HCl$ converts the $-NO_2$ group to $-NH_2$,yielding $2-$bromo$-4-$aminotoluene $(Q)$.
$3$. Diazotization with $NaNO_2/HCl$ at $273-278 \ K$ followed by reduction with $H_3PO_2/H_2O$ removes the $-NH_2$ group,resulting in $o$-bromotoluene $(R)$.
$4$. Oxidation of the methyl group with $KMnO_4/OH^-$ gives $2-$bromobenzoic acid.
Thus,the starting compound $(P)$ is $p$-nitro toluene.

(A) The boiling point of haloalkanes increases with an increase in the molecular mass and the number of halogen atoms due to stronger van der Waals forces of attraction.
Comparing the compounds:
$iii$. Chloromethane ($CH_{3}Cl$,molar mass $\approx 50.5 \ g/mol$)
$i$. Bromomethane ($CH_{3}Br$,molar mass $\approx 95 \ g/mol$)
$iv$. Dibromomethane ($CH_{2}Br_{2}$,molar mass $\approx 173.8 \ g/mol$)
$ii$. Bromoform ($CHBr_{3}$,molar mass $\approx 252.7 \ g/mol$)
Thus,the increasing order of boiling points is $iii < i < iv < ii$.

(D) The rate law for the esterification reaction is $Rate = k[CH_3COOH][CH_3OH]$.
In the first case,the total volume is $100 \text{ cm}^3 + 100 \text{ cm}^3 = 200 \text{ cm}^3$.
In the second case,each solution is diluted with an equal volume of water before mixing. Thus,$100 \text{ cm}^3$ of acid is diluted to $200 \text{ cm}^3$ and $100 \text{ cm}^3$ of alcohol is diluted to $200 \text{ cm}^3$. When these are mixed,the total volume becomes $400 \text{ cm}^3$.
Since the total volume of the mixture is doubled,the concentration of each reactant ($CH_3COOH$ and $CH_3OH$) becomes half of its initial concentration in the mixture.
Let the initial concentrations be $[C_1]$ and $[C_2]$. Then $Rate_1 = k[C_1][C_2]$.
In the second case,the new concentrations are $[C_1/2]$ and $[C_2/2]$.
$Rate_2 = k[C_1/2][C_2/2] = \frac{1}{4} k[C_1][C_2] = 0.25 \times Rate_1$.
Thus,the rate becomes $0.25$ times the initial rate.

(B) For a first order reaction,the rate constant $k$ is given by $k = \frac{2.303}{t} \log \frac{[A]_0}{[A]}$.
Given that the concentration is reduced to $12.5 \%$ in $1 \ hr$ $(60 \ min)$,we have $[A]_0 = 100$ and $[A] = 12.5$.
$k = \frac{2.303}{60} \log \frac{100}{12.5} = \frac{2.303}{60} \log 8 = \frac{2.303}{60} \times 0.903 = 0.0346 \ min^{-1}$.
The half-life $t_{1/2}$ is calculated as $t_{1/2} = \frac{0.693}{k} = \frac{0.693}{0.0346} \approx 20 \ min$.
Alternatively,$12.5 \% = (1/2)^3$,which means $3$ half-lives have passed in $60 \ min$. Thus,$t_{1/2} = 60/3 = 20 \ min$.

(C) For a first-order reaction,the rate constant $k$ is given by $k = \frac{0.693}{t_{1/2}}$.
Given $t_{1/2} = 10 \ min$,so $k = \frac{0.693}{10} = 0.0693 \ min^{-1}$.
After $20 \ min$ (which is $2 \times t_{1/2}$),the concentration $[A]$ will be reduced to $\frac{1}{4}$ of the initial concentration.
$[A] = \frac{12}{4} = 3 \ M$.
The rate of reaction is given by $Rate = k[A]$.
$Rate = 0.0693 \times 3 \ M \ min^{-1}$.

(C) $Prontosil$ is converted into $sulphanilamide$ in the body,which is the real active compound. Therefore,the statement that $Prontosil$ is not converted into $sulphanilamide$ is false.

(B) Mohr salt is a double salt with the chemical formula $FeSO_{4} \cdot (NH_{4})_{2}SO_{4} \cdot 6H_{2}O$.
When dissolved in excess water,it dissociates completely into its constituent ions:
$FeSO_{4} \cdot (NH_{4})_{2}SO_{4} \cdot 6H_{2}O \rightarrow Fe^{2+} + 2NH_{4}^{+} + 2SO_{4}^{2-} + 6H_{2}O$.
The total number of ions produced per molecule is $1 (Fe^{2+}) + 2 (NH_{4}^{+}) + 2 (SO_{4}^{2-}) = 5$ ions.

(C) Tetrahedral complexes do not show geometrical isomerism because,due to their symmetrical structure,the relative positions of the ligands are the same with respect to each other.
Square planar complexes of the type $M A_{3} B$ do not show geometrical isomerism because all possible spatial arrangements are equivalent.
Square planar complexes of the type $M A B C D$ show three geometrical isomers. These can be obtained by fixing the position of one ligand and placing any of the remaining three ligands at the trans-position one by one.

(C) The magnitude of $\Delta_{0}$ depends on the nature of the ligand and the oxidation state of the central metal ion.
According to the spectrochemical series,the field strength of the ligands is $Cl^{-} < H_{2}O < NH_{3} < CN^{-}$.
Ligands with lower field strength result in a smaller crystal field splitting energy $(\Delta_{0})$.
Since $Cl^{-}$ is the weakest field ligand among the given options,the complex $[Co(Cl)_{6}]^{3-}$ will have the minimum magnitude of $\Delta_{0}$.

(D) The atomic number of $Gd$ is $64$.
The electronic configuration of neutral $Gd$ is $[Xe] 4f^7 5d^1 6s^2$.
When forming the $Gd^{2+}$ ion,electrons are removed from the outermost shell first,which is the $6s$ orbital.
Removing two electrons from the $6s$ orbital results in the configuration $[Xe] 4f^7 5d^1$.

(B) The thermal decomposition of potassium permanganate $(KMnO_4)$ at $513 \ K$ is given by the following reaction:
$2KMnO_4 \rightarrow K_2MnO_4 + MnO_2 + O_2$
From the reaction,it is clear that $K_2MnO_4$,$MnO_2$,and $O_2$ are produced.
Therefore,$MnO$ is not obtained upon heating potassium permanganate.

(C) The oxidation reaction of water is: $H_2O \rightarrow \frac{1}{2}O_2 + 2H^{+} + 2e^-$.
For the oxidation of $1 \ mol$ of $H_2O$,$2 \ mol$ of electrons are required.
Quantity of electricity required $(Q)$ = $nF$.
$Q = 2 \times 96500 \ C = 193000 \ C = 1.93 \times 10^5 \ C$.

(B) During the charging process,the lead storage battery acts as an electrolytic cell.
At the anode (positive terminal during discharge,now connected to positive terminal of external source),$PbSO_{4}$ is oxidized to $PbO_{2}$:
$PbSO_{4(s)} + 2H_{2}O_{(l)} \rightarrow PbO_{2(s)} + S{O_{4}}^{2-}_{(aq)} + 4H^{+}_{(aq)} + 2e^{-}$
At the cathode (negative terminal during discharge,now connected to negative terminal of external source),$PbSO_{4}$ is reduced to $Pb$:
$PbSO_{4(s)} + 2e^{-} \rightarrow Pb_{(s)} + S{O_{4}}^{2-}_{(aq)}$
Therefore,$PbSO_{4}$ on the cathode is reduced to $Pb$.

(B) In a hydrogen-oxygen fuel cell,the reduction of oxygen occurs at the cathode.
The reaction is:
$O_{2(g)} + 2H_2O_{(l)} + 4e^- \rightarrow 4OH^{-}_{(aq)}$

(C) In the Bessemer converter,copper is extracted through a self-reduction process. The reactions involved are:
$2Cu_2S + 3O_2 \rightarrow 2Cu_2O + 2SO_2$
$2Cu_2O + Cu_2S \rightarrow 6Cu + SO_2$
These reactions occur because,at high temperatures,sulphur has a greater affinity for oxygen than copper does,allowing sulphur to remove oxygen from $Cu_2O$ to form $SO_2$ gas. Thus,copper has less affinity for oxygen than sulphur.

(B) The reaction sequence is as follows:
$ZnSO_4$ $\xrightarrow{NH_4OH} Zn(OH)_2 \downarrow \text{ (White ppt., } X)$ $\xrightarrow{NH_4OH \text{ (excess)}} [Zn(NH_3)_4]^{2+} \text{ (Clear solution, } Y)$ $\xrightarrow{H_2S} ZnS \downarrow \text{ (Ppt., } Z)$
Therefore,the metal $M$ is $Zn$ and the precipitate $Z$ is $ZnS$.

(C) Cryolite is $Na_{3}AlF_{6}$.
In the Hall-Heroult process,pure alumina $(Al_{2}O_{3})$ has a very high melting point $(2323 \ K)$ and is a poor conductor of electricity.
Cryolite $(Na_{3}AlF_{6})$ is added to the electrolyte to lower the melting point of the mixture to about $1240 \ K$ and to increase its electrical conductivity,making the electrolysis process more efficient.

(D) $2-$Bromobutane is a secondary alkyl halide. In the presence of aqueous $NaOH$,the reaction proceeds primarily through the $S_N1$ mechanism.
This mechanism involves the formation of a planar carbocation intermediate.
The nucleophile $(OH^-)$ can attack the carbocation from either side with equal probability.
This leads to the formation of both enantiomers in equal amounts,resulting in a racemic mixture of $(\pm)$ butan$-2-$ol.

(C) When acetic acid reacts with slaked lime,calcium acetate is formed,which on dry distillation produces a ketone.
$2 CH_3COOH + Ca(OH)_2 \rightarrow (CH_3COO)_2Ca + 2 H_2O$
$(CH_3COO)_2Ca \xrightarrow{\Delta} CH_3COCH_3 + CaCO_3$
The final product obtained is propanone $(CH_3COCH_3)$.

(D) $AgCN$ is a covalent compound,therefore,the nucleophilic attack occurs through the nitrogen atom of the $CN^-$ group,leading to the formation of an isocyanide $(R-NC)$.
Step $1$: $CH_3-CH(Br)-CH_3 + AgCN \xrightarrow{\Delta} CH_3-CH(NC)-CH_3$ (Product $X$ is isopropyl isocyanide).
Step $2$: Reduction of isocyanide with $LiAlH_4$ yields a secondary amine.
$CH_3-CH(NC)-CH_3 \xrightarrow{LiAlH_4} CH_3-CH(NHCH_3)-CH_3$ (Product $Y$ is $N-\text{methylpropan-2-amine}$).

(A) The $AgI/Ag^{+}$ sol is a positively charged sol.
According to the Hardy-Schulze rule,the flocculation power of an electrolyte depends on the valency of the coagulating ion (the ion with a charge opposite to that of the sol).
For a positively charged sol,the coagulating power increases with the increase in the negative charge of the anion: $Cl^{-} < SO_{4}^{2-} < PO_{4}^{3-}$.
Flocculation value is inversely proportional to flocculation power.
Since $Cl^{-}$ has the minimum flocculation power,$NaCl$ will have the maximum flocculation value.

(C) The hybridization of $Xe$ in $XeO_3$ is $sp^3$,which involves three bond pairs and one lone pair.
Due to the presence of one lone pair,$XeO_3$ adopts a pyramidal geometry.
$XeOF_4$ has a square pyramidal geometry,$XeF_2$ is linear,and $XeF_4$ is square planar.

(C) Orlon is a polymer of acrylonitrile,which is also known as vinyl cyanide.
The polymerization reaction is as follows:
$nCH_2=CH(CN) \xrightarrow{\text{Peroxide}} -(CH_2-CH(CN))_n-$
Thus,the monomeric unit of Orlon is vinyl cyanide.

(B) The reaction of $Ba(NO_3)_2$ with $H_2SO_4$ is a double displacement reaction that results in the formation of a white precipitate of $BaSO_4$.
Because $BaSO_4$ is highly insoluble,the reaction proceeds in the forward direction even with dilute $H_2SO_4$.
The chemical equation is: $Ba(NO_3)_2 + H_2SO_4 \rightarrow BaSO_4(s) + 2HNO_3$.

(B) The crystal system with dimensions $a \neq b \neq c$,$\alpha = \gamma = 90^{\circ}$ and $\beta \neq 90^{\circ}$ is known as the Monoclinic system.

(D) For a $B.C.C.$ (Body-Centered Cubic) lattice,the relationship between the edge length $(a)$ and the atomic radius $(r)$ is given by the formula: $r = \frac{\sqrt{3}}{4} a$.
Given the edge length $a = 4.29 \ \mathring{A}$.
Substituting the value of $a$ into the formula:
$r = \frac{\sqrt{3}}{4} \times 4.29$
$r = 0.433 \times 4.29$
$r \approx 1.857 \ \mathring{A}$.

(C) Isotonic solutions have the same molar concentration at the same temperature.
The molarity of the urea solution is calculated as:
$M = \frac{w \times 1000}{M_w \times V(mL)} = \frac{0.06 \times 1000}{60 \times 100} = 0.01 \ M$.
Since the molarity of the urea solution is $0.01 \ M$,it is isotonic with a $0.01 \ M$ glucose solution.

(B) The freezing point depression is given by $\Delta T_f = i \cdot K_f \cdot m$.
The freezing point of the solution is $T_f = T_f^\circ - \Delta T_f$.
For the highest freezing point,$\Delta T_f$ must be minimum,which implies that the product $i \cdot M$ must be minimum.
$(A)$ For $0.1 \ M$ Sucrose,$i = 1$,$i \cdot M = 1 \times 0.1 = 0.1$.
$(B)$ For $0.01 \ M \ NaCl$,$i = 2$,$i \cdot M = 2 \times 0.01 = 0.02$.
$(C)$ For $0.1 \ M \ NaCl$,$i = 2$,$i \cdot M = 2 \times 0.1 = 0.2$.
$(D)$ For $0.01 \ M \ Na_2SO_4$,$i = 3$,$i \cdot M = 3 \times 0.01 = 0.03$.
Since $0.01 \ M \ NaCl$ has the lowest $i \cdot M$ value,it has the highest freezing point.

(B) Given:
Freezing point of solution $(T_f)$ = $-0.186^{\circ} C$
Freezing point of pure water $(T_f^{\circ})$ = $0^{\circ} C$
$\Delta T_f = T_f^{\circ} - T_f = 0 - (-0.186) = 0.186 \text{ K}$
We know that $\Delta T_f = K_f \cdot m$ and $\Delta T_b = K_b \cdot m$
Dividing the two equations: $\frac{\Delta T_b}{\Delta T_f} = \frac{K_b}{K_f}$
$\Delta T_b = \Delta T_f \times \frac{K_b}{K_f}$
$\Delta T_b = 0.186 \times \frac{0.521}{1.86} = 0.1 \times 0.521 = 0.0521 \text{ K}$.

(C) The Freundlich adsorption isotherm is given by the equation:
$\frac{x}{m} = k p^{\frac{1}{n}}$
Taking the logarithm on both sides:
$\log \left(\frac{x}{m}\right) = \log k + \frac{1}{n} \log p$
This equation is of the form $y = mx + c$,where:
$y = \log \left(\frac{x}{m}\right)$
$x = \log p$
Slope $= \frac{1}{n}$
Intercept $= \log k$
Therefore,plotting $\log \left(\frac{x}{m}\right)$ against $\log p$ gives a straight line with a positive intercept $\log k$ and a slope of $\frac{1}{n}$. This corresponds to the curve shown in option $C$.

(C) Macromolecular colloids are quite stable and resemble true solutions in many respects.
Due to their high stability and the presence of a solvation layer,they cannot be easily coagulated compared to lyophobic colloids.

(D) nucleoside is formed by the combination of a pentose sugar and a nitrogenous base.
Adenosine consists of the sugar ribose and the nitrogenous base adenine.
Since it lacks a phosphate group,it is classified as a nucleoside.