TS EAMCET 2022 Chemistry Question Paper with Answer and Solution

(D) In the $Wurtz$ reaction,alkanes are prepared by the reaction of alkyl halides with sodium metal $(Na)$.
Since sodium metal is highly reactive,the solvent used must be inert towards it.
An aprotic solvent is required as the reaction medium to prevent side reactions.
Dry ether is a commonly used,suitable aprotic solvent for this reaction.

(C) $1$. The reaction of an alkyne with $H_2$ in the presence of Lindlar's catalyst (poisoned $Pd/CaCO_3$) is a syn-addition,which yields a (cis)-alkene as the major product $P$.
$2$. The reaction of an alkyne with $Na$ in liquid $NH_3$ (Birch reduction) is an anti-addition,which yields a (trans)-alkene as the major product $Q$.
$3$. Therefore,$P$ is (cis)-alkene and $Q$ is (trans)-alkene.

(A) The addition of $Br_2/\text{water}$ to an alkene is an electrophilic addition reaction. The rate-determining step involves the formation of a carbocation intermediate.
Electron-donating groups (like alkyl groups) increase the electron density of the double bond and stabilize the carbocation intermediate,thereby increasing the rate of reaction.
Electron-withdrawing groups (like $-NO_2$) decrease the electron density of the double bond and destabilize the carbocation intermediate,thereby decreasing the rate of reaction.
Comparing the substituents:
$1$. $CH_3-CH=CH_2$ $(D)$: Has one methyl group (+$I$ effect),stabilizing the carbocation.
$2$. $CH_3-CH_2-CH=CH_2$ $(C)$: Has an ethyl group (+$I$ effect),which is slightly more effective than methyl at stabilizing,but steric factors and hyperconjugation make $D$ and $C$ very reactive.
$3$. $CH_2=CH_2$ $(A)$: No substituents.
$4$. $CH_2=CH-NO_2$ $(B)$: The $-NO_2$ group is a strong electron-withdrawing group (-$I$ and -$M$ effect),making it the least reactive.
Thus,the correct order of reactivity is $C > D > A > B$.

(D) The reaction of an alkene with cold,dilute,aqueous $KMnO_4$ (Bayer's reagent) results in syn-dihydroxylation,where a double bond is converted into a vicinal diol (two $-OH$ groups are added across the double bond).
In the given compound,$5-$allylcyclohex$-3-$ene$-1-$ol,there are two double bonds: one in the cyclohexene ring and one in the allyl side chain.
$1$. The double bond in the cyclohexene ring undergoes dihydroxylation,adding two $-OH$ groups.
$2$. The double bond in the allyl side chain also undergoes dihydroxylation,adding two more $-OH$ groups.
$3$. The original molecule already contains one $-OH$ group at the $1-$position of the ring.
Total $-OH$ groups = $1$ (original) $+ 2$ (from ring double bond) $+ 2$ (from side chain double bond) = $5$ hydroxyl groups.

(A) In reaction $(A)$,Friedel-Crafts alkylation of benzene with $n$-propyl chloride in the presence of $AlCl_3$ leads to the formation of a carbocation intermediate. The primary propyl carbocation rearranges to a more stable secondary isopropyl carbocation,which then attacks the benzene ring to form isopropylbenzene as the major product $P$.
In reaction $(B)$:
$(i)$ Friedel-Crafts alkylation of benzene with $CH_3Cl$ in the presence of $AlCl_3$ gives toluene.
$(ii)$ Free radical halogenation of toluene with $Br_2$ in the presence of $hv$ (light) gives benzyl bromide $(C_6H_5CH_2Br)$.
$(iii)$ Nucleophilic substitution of benzyl bromide with aqueous $KOH$ gives benzyl alcohol $(C_6H_5CH_2OH)$ as the major product $Q$.

(C) Step $1$: Dehydrohalogenation of $C_6H_5-CHBr-CH_2Br$ with $alc. KOH$ followed by $NaNH_2$ leads to the formation of phenylacetylene $(C_6H_5-C \equiv CH)$.
Step $2$: The cyclic trimerization of phenylacetylene occurs when passed through a red hot iron tube at $873 \ K$.
Step $3$: The trimerization of $C_6H_5-C \equiv CH$ yields $1,3,5-triphenylbenzene$ as the major product due to steric hindrance,which favors the meta-substituted product.

(D) When benzene $(C_6H_6)$ reacts with excess chlorine $(Cl_2)$ in the presence of ultraviolet light,an addition reaction occurs.
This results in the formation of benzene hexachloride $(C_6H_6Cl_6)$,also known as $BHC$ or Gammaxene.
The reaction is: $C_6H_6 + 3Cl_2 \xrightarrow{uv/hv} C_6H_6Cl_6$.

(C) Boiling removes only temporary hardness of water.
During boiling,the soluble $Mg(HCO_3)_2$ is converted into insoluble magnesium hydroxide and $Ca(HCO_3)_2$ is changed to insoluble calcium carbonate.
In the permutit process,hydrated sodium aluminium silicate is used. For simplicity,sodium aluminium silicate $(NaAlSiO_4)$ can be written as $NaZ$. When this is added to hard water,exchange reactions take place:
$2 NaZ_{(s)} + M^{2+}_{(aq)} \rightarrow MZ_{2(s)} + 2 Na^{+}_{(aq)}$ (where $M = Mg, Ca$).
In the permutit process,only cations causing hardness are removed.
In the ion-exchange process,both the cations and anions causing permanent hardness of water can be removed by using cation and anion exchange resins. Demineralized water is obtained after this process.
Hence,the most effective method for the removal of hardness of water is the ion-exchange process.

(C) $1$. Compound '$A$' is $Ca(OH)_2$ (slaked lime),which is used for water softening.
$2$. $Ca(OH)_2$ reacts with $Na_2CO_3$ to form $NaOH$ (compound '$B$') and $CaCO_3$ (precipitate). $NaOH$ is a strong base with $pH=14$.
$Ca(OH)_2 + Na_2CO_3 \rightarrow CaCO_3 + 2NaOH$
$3$. $Ca(OH)_2$ reacts with $CO_2$ to form $CaCO_3$,which appears as a cloudy precipitate.
$Ca(OH)_2 + CO_2 \rightarrow CaCO_3 + H_2O$
$4$. The mixture of $NaOH$ and $CaO$ is known as soda lime. It performs decarboxylation on organic acids. When $C_6H_5COOH$ (benzoic acid,compound '$C$') reacts with soda lime,it produces benzene $(C_6H_6)$.
$C_6H_5COOH + NaOH + CaO \xrightarrow{\Delta} C_6H_6 + Na_2CO_3$
Therefore,'$A$' is $Ca(OH)_2$,'$B$' is $NaOH$,and '$C$' is $C_6H_5COOH$.

(D) Deionized water is obtained by passing hard water through a series of ion exchange resins. First,it passes through a cation exchange resin (which removes $Ca^{2+}$,$Mg^{2+}$,etc.) and then through an anion exchange resin (which removes $Cl^-$,$SO_4^{2-}$,etc.). Thus,both cationic and anionic exchangers are used one after the other.

(B) Heavy water,denoted as $D_2O$,has physical properties slightly different from ordinary water $(H_2O)$.
At $1 \ atm$ pressure,the freezing point of heavy water is $3.8 \ ^{\circ}C$.

(D) The reaction of calcium carbide $(CaC_2)$ with heavy water $(D_2O)$ is an acid-base type reaction where the carbide ion $(C_2^{2-})$ reacts with $D^+$ ions from $D_2O$.
$CaC_2 + 2D_2O \rightarrow Ca(OD)_2 + C_2D_2$
Thus,the product "$P$" is $C_2D_2$ (deuteroacetylene).

(A) According to the Lewis concept,a base is a substance that can donate an electron pair to form a coordinate covalent bond.
In the $NH_3$ molecule,the nitrogen atom has one lone pair of electrons.
Because $NH_3$ can donate this lone pair of electrons to an electron-deficient species,it acts as a Lewis base.

(D) At $25^{\circ} C$,the ionic product of water $(K_w)$ is $1.0 \times 10^{-14}$.
For pure water,$[H^{ }] = [OH^{-}] = \sqrt{K_w} = 10^{-7} \ M$,which gives $pH = 7$.
The auto-ionization of water is an endothermic process $(\Delta H > 0)$.
According to Le Chatelier's principle,an increase in temperature shifts the equilibrium to the right,increasing the value of $K_w$.
At $80^{\circ} C$,$K_w > 10^{-14}$,which implies $[H^{ }] > 10^{-7} \ M$.
Since $pH = -\log[H^{ }]$,an increase in $[H^{ }]$ results in a decrease in $pH$.
Therefore,at $80^{\circ} C$,the $pH$ of pure water is $< 7.0$.

(A) $AlCl_3$ is a salt of a weak base and a strong acid; it undergoes cationic hydrolysis:
$Al^{3+} + 3H_2O \rightarrow Al(OH)_3 + 3H^+$
$H^+$ ions are released into the solution,making the aqueous solution of $AlCl_3$ acidic $(A-II)$.
$CH_3COONa$ is a salt of a weak acid and a strong base; it undergoes anionic hydrolysis:
$CH_3COO^- + H_2O \rightarrow CH_3COOH + OH^-$
$OH^-$ ions are released,making the solution basic $(B-I)$.
$KCl$ is a strong electrolyte and an ionic compound,making its aqueous solution highly conductive $(C-III)$.
$Al_2O_3$ is an amphoteric oxide as it reacts with both acids and bases $(D-V)$.
Therefore,the correct match is $A-II, B-I, C-III, D-V$.

(B) The solubility product $(K_{sp})$ values for the given metal sulfides are:
$A$. $PbS$: $8.0 \times 10^{-28}$ $(II)$
$B$. $HgS$: $4.0 \times 10^{-53}$ $(I)$
$C$. $MnS$: $2.5 \times 10^{-13}$ $(IV)$
$D$. $ZnS$: $1.6 \times 10^{-24}$ $(III)$
Therefore,the correct matching sequence is $A-II, B-I, C-IV, D-III$.

(A) Let $I = \int \sqrt{4 \cos ^2 x - 5 \sin ^2 x} \cos x \, dx$.
Using the identity $\cos ^2 x = 1 - \sin ^2 x$,we get:
$I = \int \sqrt{4(1 - \sin ^2 x) - 5 \sin ^2 x} \cos x \, dx = \int \sqrt{4 - 9 \sin ^2 x} \cos x \, dx$.
Let $3 \sin x = t$. Then $3 \cos x \, dx = dt$,which implies $\cos x \, dx = \frac{1}{3} dt$.
Substituting these into the integral:
$I = \int \sqrt{4 - t^2} \cdot \frac{1}{3} dt = \frac{1}{3} \int \sqrt{2^2 - t^2} dt$.
Using the standard formula $\int \sqrt{a^2 - x^2} dx = \frac{x}{2} \sqrt{a^2 - x^2} + \frac{a^2}{2} \sin ^{-1}\left(\frac{x}{a}\right) + c$:
$I = \frac{1}{3} \left[ \frac{t}{2} \sqrt{4 - t^2} + \frac{4}{2} \sin ^{-1}\left(\frac{t}{2}\right) \right] + c$.
Substituting $t = 3 \sin x$ back:
$I = \frac{1}{3} \left[ \frac{3 \sin x}{2} \sqrt{4 - 9 \sin ^2 x} + 2 \sin ^{-1}\left(\frac{3 \sin x}{2}\right) \right] + c = \frac{1}{2} \sin x \sqrt{4 - 9 \sin ^2 x} + \frac{2}{3} \sin ^{-1}\left(\frac{3 \sin x}{2}\right) + c$.

(C) The Lewis acidity of boron trihalides depends on the extent of $p\pi-p\pi$ back-bonding between the halogen atom and the boron atom.
In $BF_3$,the small size of the $F$ atom allows for effective $2p-2p$ back-bonding,which reduces the electron deficiency of the boron atom.
As we move from $F$ to $I$,the size of the halogen atom increases,making the $p\pi-p\pi$ back-bonding less effective.
Therefore,the electron deficiency of the boron atom increases,leading to an increase in Lewis acidic character.
The correct order is $BI_3 > BBr_3 > BCl_3 > BF_3$.

(D) The assertion is false because boron has a small atomic size and lacks $d$-orbitals in its valence shell,which prevents it from expanding its coordination number beyond $4$. Therefore,the formation of an octahedral $[B(OH_2)_6]^{3+}$ complex is impossible.
However,the reason is true because boron is electron-deficient ($sextet$ of electrons) and acts as a Lewis acid,reacting with Lewis bases like $OH^{-}$ to form the stable tetrahedral $[B(OH)_4]^{-}$ ion.
Thus,$A$ is false but $R$ is true.

(A) The element with $ns^2 np^1$ configuration is a member of Group $13$.
Boron $(B)$ is a black-coloured non-metal that exists in crystalline and amorphous forms.
Crystalline boron is very hard and unreactive towards air at room temperature.
Amorphous boron reacts with oxygen in the air to form boron trioxide $(B_2O_3)$,which is acidic in nature:
$4B(s) + 3O_2(g) \rightarrow 2B_2O_3(s)$
$B_2O_3$ further reacts with water to form boric acid:
$B_2O_3(s) + 3H_2O(l) \rightarrow 2B(OH)_3(aq)$

(A) Aluminium metal reacts with aqueous sodium hydroxide to produce sodium aluminate and hydrogen gas.
The balanced chemical equation is:
$2 Al_{(s)} + 2 NaOH_{(aq)} + 6 H_2O_{(l)} \rightarrow 2 Na[Al(OH)_4]_{(aq)} + 3 H_{2(g)}$
The evolution of hydrogen gas can be tested by bringing a burning candle near the reaction,which produces a characteristic pop sound.

(C) The reaction of aluminum $(Al)$ with an excess of aqueous sodium hydroxide $(NaOH)$ is a characteristic reaction where aluminum acts as an amphoteric metal.
The balanced chemical equation for this reaction is:
$2Al(s) + 2NaOH(aq) + 6H_2O(l) \longrightarrow 2Na^{+}[Al(OH)_4]^-(aq) + 3H_2(g)$
In this reaction,$P$ is the complex ion sodium tetrahydroxoaluminate$(III)$,$Na^{+}[Al(OH)_4]^-$,and $Q$ is hydrogen gas,$H_2$.
Therefore,the correct option is $C$.

(A) Carbon $(C)$ has the smallest atomic size and high bond dissociation energy,which allows it to form stable $C-C$ bonds,leading to the property of maximum catenation.
Germanium $(Ge)$ is the least abundant element among the group $14$ elements on the earth's crust.
Therefore,$X = C$ and $Y = Ge$.

(C) The standard enthalpy of formation $\Delta_{f} H^{\circ}$ for the most stable allotrope of an element is defined as $0 \ kJ/mol$.
Graphite $(II)$ is the most stable allotrope of carbon,so its $\Delta_{f} H^{\circ} = 0 \ kJ/mol$.
Diamond $(I)$ is less stable than graphite,with $\Delta_{f} H^{\circ} \approx 1.90 \ kJ/mol$.
Fullerene $(III)$ is the least stable among these,with $\Delta_{f} H^{\circ} \approx 38.1 \ kJ/mol$.
Therefore,the correct order of $\Delta_{f} H^{\circ}$ values is $III > I > II$.

(B) In Group $14$,the acidic character of oxides decreases as we move down the group from $Si$ to $Pb$.
$SiO_2$ is a well-known acidic oxide.
$SnO_2$ and $PbO_2$ are amphoteric in nature.
$SnO$ is also amphoteric.
Therefore,$SiO_2$ is the correct acidic oxide.

(D) Catenation is the ability of an element to form stable bonds with other atoms of the same element to form long chains or rings.
This property decreases down the group in Group $14$ because the bond energy of $M-M$ bonds decreases as the atomic size increases.
$C$ shows the highest catenation.
$Pb$ (Lead) is a metal and does not exhibit catenation due to its large atomic size and weak $Pb-Pb$ bond energy.

(D) All the elements of Group $16$ form hydrides of the type $H_2E$ (where $E = O, S, Se, Te, Po$).
Boiling point depends on the molecular mass and intermolecular forces.
As the size of the central atom increases,the van der Waals forces increase,which generally leads to an increase in boiling point from $H_2S$ to $H_2Te$.
However,$H_2O$ exhibits exceptionally high boiling point due to the presence of strong intermolecular hydrogen bonding.
Therefore,the correct order of boiling points is $H_2O > H_2Te > H_2Se > H_2S$.

(A) disproportionation reaction is a redox reaction in which the same element is simultaneously oxidized and reduced.
$1. 2 H_2\stackrel{-1}{O_2} \rightarrow 2 H_2\stackrel{-2}{O} + \stackrel{0}{O_2}$
Here,the oxidation state of oxygen changes from $-1$ to $-2$ (reduction) and from $-1$ to $0$ (oxidation).
$2. \stackrel{0}{Cl_2} + 2 OH^{-} \rightarrow \stackrel{-1}{Cl^{-}} + \stackrel{+1}{ClO^{-}} + H_2O$
Here,the oxidation state of chlorine changes from $0$ to $-1$ (reduction) and from $0$ to $+1$ (oxidation).
Both reactions involve the same element undergoing both oxidation and reduction,hence they are disproportionation reactions.

(C) In the disproportionation reaction of $Se_2Cl_2$,selenium in the $+1$ oxidation state is converted to $+4$ and $0$ oxidation states.
The balanced chemical equation is:
$2Se_2Cl_2 \rightarrow SeCl_4 + 3Se$
Thus,$P$ is $SeCl_4$ and $Q$ is $Se$.

(A) disproportionation reaction is a redox reaction in which the same element is simultaneously oxidized and reduced.
$(A)$ $Cl_2 + 2 NaOH \rightarrow NaCl + NaOCl + H_2O$: The oxidation state of $Cl$ changes from $0$ to $-1$ (reduction) and from $0$ to $+1$ (oxidation). This is a disproportionation reaction.
$(B)$ $2 H_2O_2 \rightarrow 2 H_2O + O_2$: The oxidation state of $O$ in $H_2O_2$ is $-1$. It changes to $-2$ in $H_2O$ (reduction) and to $0$ in $O_2$ (oxidation). This is a disproportionation reaction.
$(C)$ $2 KMnO_4 \rightarrow K_2MnO_4 + MnO_2 + O_2$: The oxidation state of $Mn$ changes from $+7$ to $+6$ and $+4$ (both reduction). This is not a disproportionation reaction.
$(D)$ $3 MnO_4^{2-} + 4 H^+ \rightarrow 2 MnO_4^- + MnO_2 + 2 H_2O$: The oxidation state of $Mn$ changes from $+6$ to $+7$ (oxidation) and from $+6$ to $+4$ (reduction). This is a disproportionation reaction.
Thus,reactions $(A)$,$(B)$,and $(D)$ are disproportionation reactions.

(D) The given unbalanced reaction is: $aKNO_3 + 5C_{12}H_{22}O_{11} \rightarrow bN_2 + cCO_2 + dH_2O + eK_2CO_3$.
To balance the equation,we equate the number of atoms of each element on both sides:
For $K$: $a = 2e$
For $N$: $a = 2b$
For $C$: $5 \times 12 = c + e \Rightarrow 60 = c + e$
For $H$: $5 \times 22 = 2d$ $\Rightarrow 110 = 2d$ $\Rightarrow d = 55$
For $O$: $3a + 5 \times 11 = 2c + d + 3e$ $\Rightarrow 3a + 55 = 2c + 55 + 3e$ $\Rightarrow 3a = 2c + 3e$
Setting $e = 24$,we get $a = 48$.
Since $a = 2b$,$b = 24$.
Since $c + e = 60$,$c = 60 - 24 = 36$.
Thus,the balanced equation is: $48KNO_3 + 5C_{12}H_{22}O_{11} \rightarrow 24N_2 + 36CO_2 + 55H_2O + 24K_2CO_3$.
The values are $A=48, B=24, C=36, D=55, E=24$.

(C) diprotic acid is an acid that can donate two protons ($H^+$ ions) per molecule.
$1$. Citric acid $(C_6H_8O_7)$ is a triprotic acid.
$2$. Chromic acid $(H_2CrO_4)$ is a diprotic acid.
$3$. Oxalic acid $(H_2C_2O_4)$ is a diprotic acid.
$4$. Pyrosulfuric acid $(H_2S_2O_7)$ is a diprotic acid.
$5$. Sulfurous acid $(H_2SO_3)$ is a diprotic acid.
Thus,there are $4$ diprotic acids in the given list.

(D) The balanced redox reaction between potassium permanganate $(KMnO_4)$ and ferrous oxalate $(FeC_2O_4)$ in an acidic medium is:
$3MnO_4^- + 5FeC_2O_4 + 24H^+ \rightarrow 3Mn^{2+} + 5Fe^{3+} + 10CO_2 + 12H_2O$
From the stoichiometry of the balanced equation,$3$ moles of $MnO_4^-$ react with $5$ moles of $FeC_2O_4$.
Therefore,$1$ mole of $MnO_4^-$ will react with $\frac{5}{3}$ moles of $FeC_2O_4$.

(B) Lithium nitrate $(LiNO_3)$ decomposes on heating to form lithium oxide $(Li_2O)$,nitrogen dioxide $(NO_2)$,and oxygen gas $(O_2)$.
The balanced chemical equation is:
$4LiNO_3(s) \xrightarrow{\Delta} 2Li_2O(s) + 4NO_2(g) + O_2(g)$

(A) The melting point of ionic compounds depends on the lattice energy of the crystal lattice.
Lattice energy is inversely proportional to the interionic distance.
As the size of the anion increases from $F^{-}$ to $Cl^{-}$ to $I^{-}$,the interionic distance increases,leading to a decrease in lattice energy.
Therefore,the melting point decreases in the order $LiF > LiCl > LiI$.
This corresponds to the order $I > II > III$.

(C) In the same period,alkaline earth metals $(Group \ 2)$ have a higher nuclear charge than alkali metals $(Group \ 1)$.
Due to the higher nuclear charge,the electrons in alkaline earth metals are pulled more strongly towards the nucleus,resulting in smaller ionic radii compared to alkali metals.
Therefore,the Assertion is true.
However,the Reason states that alkali metals have a higher nuclear charge than alkaline earth metals,which is incorrect; alkali metals have a lower nuclear charge.
Thus,$(A)$ is true but $(R)$ is false.

(C) Potassium superoxide $(KO_{2})$ reacts with water to give potassium hydroxide,hydrogen peroxide,and oxygen gas.
The balanced chemical equation for the hydrolysis is:
$2KO_{2} + 2H_{2}O \rightarrow 2K^{+} + 2OH^{-} + H_{2}O_{2} + O_{2}$

(D) $BeH_2$ and $MgH_2$ are electron-deficient compounds and exist as polymeric structures in the solid state. This statement is correct.
$(B)$ $LiH$ is unreactive to oxygen at moderate temperatures,whereas it reacts at higher temperatures. This statement is correct.
$(C)$ Due to the small size and high polarizing power of $Be^{2+}$ and $Mg^{2+}$ ions,$BeH_2$ and $MgH_2$ possess significant covalent character. This statement is correct.
$(D)$ The stability of alkali metal hydrides decreases as the size of the alkali metal cation increases down the group. Therefore,the correct order is $LiH > NaH > KH > RbH > CsH$. This statement is incorrect.
Thus,the correct statements are $(A)$,$(B)$,and $(C)$.

(B) The circuit consists of two $NOT$ gates (formed by $NOR$ gates with shorted inputs) followed by a $NOR$ gate.
$1$. The input $A$ passes through a $NOR$ gate with shorted inputs,which acts as a $NOT$ gate,producing output $\bar{A}$.
$2$. Similarly,the input $B$ passes through a $NOR$ gate with shorted inputs,producing output $\bar{B}$.
$3$. These two outputs $\bar{A}$ and $\bar{B}$ are then fed as inputs to a final $NOR$ gate.
$4$. The output $Y$ of the final $NOR$ gate is given by $Y = \overline{\bar{A} + \bar{B}}$.
$5$. Using De Morgan's theorem,$\overline{\bar{A} + \bar{B}} = \overline{\bar{A}} \cdot \overline{\bar{B}} = A \cdot B$.
$6$. Thus,$Y = A \cdot B$,which is the Boolean expression for an $AND$ gate.

(A) The percent labeling of oleum is defined as $100 + x$,where $x$ is the mass of water required to convert all $SO_3$ in $100 \ g$ of oleum into $H_2SO_4$.
For $118\%$ oleum,$x = 18 \ g$ of $H_2O$.
The reaction is $SO_3 + H_2O \rightarrow H_2SO_4$.
Since $18 \ g$ of $H_2O$ is $1 \ mole$,it reacts with $1 \ mole$ of $SO_3$ $(80 \ g)$.
Thus,in $100 \ g$ of oleum,there is $80 \ g$ of $SO_3$ and $20 \ g$ of $H_2SO_4$.
Moles of $SO_3 = \frac{80 \ g}{80 \ g/mol} = 1 \ mol$.
Moles of $H_2SO_4 = \frac{20 \ g}{98 \ g/mol} \approx 0.204 \ mol$.
Neutralization reactions are:
$SO_3 + 2NaOH \rightarrow Na_2SO_4 + H_2O$ (requires $2 \ mol$ $NaOH$ per $mol$ $SO_3$)
$H_2SO_4 + 2NaOH \rightarrow Na_2SO_4 + 2H_2O$ (requires $2 \ mol$ $NaOH$ per $mol$ $H_2SO_4$)
Total $NaOH$ moles $= 2(1) + 2(0.204) = 2 + 0.408 = 2.408 \ mol \approx 2.4 \ mol$.

(D) According to Avogadro's Law,the volume ratio of gases is equal to the mole ratio under the same conditions of temperature and pressure.
Let the hydrocarbon be $C_xH_y$.
The combustion reaction is: $C_xH_y + (x + y/4) O_2 \rightarrow x CO_2 + (y/2) H_2O$.
Given volume ratio: $V(C_xH_y) : V(CO_2) : V(H_2O) = 10 : 40 : 50 = 1 : 4 : 5$.
From the stoichiometry,$x = 4$ (moles of $CO_2$ per mole of hydrocarbon).
Also,$y/2 = 5$,which implies $y = 10$.
Therefore,the molecular formula is $C_4H_{10}$.

(D) The balanced chemical equation for the reduction of magnetite $(Fe_3O_4)$ is:
$Fe_3O_4 + 4CO \rightarrow 3Fe + 4CO_2$
Molar mass of $Fe_3O_4 = (3 \times 56) + (4 \times 16) = 168 + 64 = 232 \ g/mol$.
From the stoichiometry,$232 \ g$ of $Fe_3O_4$ produces $3 \times 56 = 168 \ g$ of $Fe$ at $100 \%$ efficiency.
Given the process is $80 \%$ efficient,the actual yield of $Fe$ from $232 \ g$ of $Fe_3O_4$ is $168 \times 0.8 = 134.4 \ g$.
To obtain $4 \ kg$ $(4000 \ g)$ of $Fe$,the mass of $Fe_3O_4$ required is:
$\text{Mass} = \frac{232 \times 4000}{134.4} \approx 6904.76 \ g \approx 6.9 \ kg$.

(B) The balanced chemical equation is:
$CaCO_3 + 2HCl \rightarrow CaCl_2 + CO_2 + H_2O$
Molar mass of $CaCO_3 = 40 + 12 + (3 \times 16) = 100 \ g/mol$.
Moles of $CaCO_3 = \frac{200 \ g}{100 \ g/mol} = 2 \ mol$.
According to the stoichiometry,$1 \ mol$ of $CaCO_3$ reacts with $2 \ mol$ of $HCl$.
Therefore,$2 \ mol$ of $CaCO_3$ requires $4 \ mol$ of $HCl$.
Molar mass of $HCl = 1 + 35.5 = 36.5 \ g/mol$.
Mass of pure $HCl$ required $= 4 \ mol \times 36.5 \ g/mol = 146 \ g$.
Given that the solution is $50\% \ (w/w)$,it means $50 \ g$ of $HCl$ is present in $100 \ g$ of solution.
Mass of solution $= \frac{146 \ g}{0.50} = 292 \ g$.

(D) The average molecular weight is calculated by taking the weighted average of the molar masses of the components based on their volume percentages.
Average molecular weight $= (\% \text{ of } O_2 \times \text{Molar mass of } O_2) + (\% \text{ of } N_2 \times \text{Molar mass of } N_2)$
Average molecular weight $= (\frac{21}{100} \times 32) + (\frac{79}{100} \times 28)$
Average molecular weight $= 6.72 + 22.12 = 28.8 \text{ g/mol}$

(D) According to the rules for significant figures:
$1$. All non-zero digits are significant.
$2$. Trailing zeros in a number containing a decimal point are significant.
In the number $2.0400$,the digits $2, 0, 4, 0, 0$ are all significant.
Therefore,the total number of significant figures is $5$.

(C) To round $234555359$ to $3$ significant figures,we look at the first three digits: $2, 3, 4$.
The next digit is $5$.
According to the rules of rounding,if the digit following the last significant figure is $5$ followed by non-zero digits,we round up.
Therefore,$234555359$ rounded to $3$ significant figures is $235000000$.

(A) Step $1$: Calculate the number of moles of each gas.
$n_{CH_4} = \frac{4 \ g}{16 \ g/mol} = 0.25 \ mol$
$n_{CO_2} = \frac{4.4 \ g}{44 \ g/mol} = 0.1 \ mol$
Step $2$: Calculate the total number of moles $(n_T)$.
$n_T = 0.25 \ mol + 0.1 \ mol = 0.35 \ mol$
Step $3$: Use the ideal gas equation $PV = nRT$ to find the pressure.
Given: $V = 1 \ L$,$T = 27 + 273 = 300 \ K$,$R = 0.082 \ L \ atm \ K^{-1} \ mol^{-1}$.
$P = \frac{n_T RT}{V} = \frac{0.35 \times 0.082 \times 300}{1} = 8.61 \ atm \approx 8.6 \ atm$.

(B) According to Dalton's Law of Partial Pressures,the partial pressure of a gas is given by the product of its mole fraction and the total pressure of the mixture.
$P_C = \chi_C \times P_{\text{total}}$
Where $\chi_C$ is the mole fraction of gas $C$.
Total moles of gas $= 2 + 3 + 5 + 10 = 20 \ \text{moles}$.
Moles of $C = 5 \ \text{moles}$.
Mole fraction of $C$ $(\chi_C)$ $= \frac{5}{20} = 0.25$.
Given $P_C = 1.5 \ atm$.
$1.5 = 0.25 \times P_{\text{total}}$.
$P_{\text{total}} = \frac{1.5}{0.25} = 6 \ atm$.

(A) According to Graham's law of diffusion,the rate of diffusion $(r)$ is inversely proportional to the square root of the density $(\rho)$ of the gas: $r \propto \frac{1}{\sqrt{\rho}}$.
From the ideal gas equation,$PV = nRT = \frac{m}{MW} RT$,where $m$ is the mass and $MW$ is the molecular weight.
Rearranging gives $\frac{m}{V} = \rho = \frac{P \times MW}{RT}$.
Since $T$ is constant,$\rho \propto P \times MW$.
Therefore,the rate of diffusion is given by $r \propto \frac{1}{\sqrt{P \times MW}}$.
For two gases,$\frac{r_{CH_4}}{r_X} = \sqrt{\frac{P_X \times (MW)_X}{P_{CH_4} \times (MW)_{CH_4}}}$.
Given $r_{CH_4} = 2r_X$,$P_{CH_4} = 1.0 \ atm$,$P_X = 1.45 \ atm$,and $(MW)_{CH_4} = 16 \ g/mol$:
$2 = \sqrt{\frac{1.45 \times (MW)_X}{1.0 \times 16}}$.
Squaring both sides: $4 = \frac{1.45 \times (MW)_X}{16}$.
$(MW)_X = \frac{4 \times 16}{1.45} = \frac{64}{1.45} \approx 44.13 \approx 44$.

(C) According to the kinetic molecular theory of gases:
$a$) The actual volume of the molecules is negligible in comparison to the empty space between them. This means the particle size is considered a point mass.
$b$) Collisions of gas molecules are perfectly elastic,not inelastic. Thus,statement $b$ is incorrect.
$c$) At any particular time,different particles in the gas have different speeds and different kinetic energies. Only the average kinetic energy is constant at a given temperature. Thus,statement $c$ is incorrect.
$d$) Pressure is exerted by the gas as a result of the collision of the particles with the walls of the container. This is a correct statement.
Therefore,statements $a$ and $d$ are correct.

(C) The unit of the rate constant $k$ is $s^{-1}$,which indicates that the reaction is a $1^{st}$ order reaction.
For a $1^{st}$ order reaction,the integrated rate equation is given by $\ln [A] = \ln [A]_0 - kt$.
Comparing this with the equation of a straight line $y = mx + c$,where $y = \ln [A]$,$x = t$,$m = -k$ (slope),and $c = \ln [A]_0$ (intercept).
Therefore,a plot of $\ln [A]$ versus $t$ will be a straight line with a negative slope equal to $-k$.

(D) The reaction $H_2 + Cl_2 \xrightarrow{hv} 2 \ HCl$ is a photochemical reaction.
Photochemical reactions involving the absorption of light are typically zero order reactions because the rate depends on the intensity of light absorbed rather than the concentration of the reactants.

(B) Given,$T_1 = 27 + 273 = 300 \ K$ and $T_2 = 37 + 273 = 310 \ K$.
The temperature coefficient of a reaction is defined as the ratio of rate constants at temperatures differing by $10 \ K$,which is typically $2$.
Let $K_1 = k$ and $K_2 = 2k$.
Using the Arrhenius equation: $\log \frac{K_2}{K_1} = \frac{E_a}{2.303 \ R} \left( \frac{T_2 - T_1}{T_1 \ T_2} \right)$.
Substituting the values: $\log 2 = \frac{E_a}{2.303 \times 8.314 \times 10^{-3}} \times \left( \frac{310 - 300}{310 \times 300} \right)$.
$0.3010 = \frac{E_a}{0.019147} \times \frac{10}{93000}$.
$E_a = \frac{0.3010 \times 0.019147 \times 93000}{10} \approx 53.5 \ kJ \cdot mol^{-1}$.

(C) According to the Arrhenius equation,$k = Ae^{-E_a / RT}$.
Let $E_{a_1}$ be the initial activation energy and $E_{a_2}$ be the activation energy after adding the catalyst.
$k_1 = Ae^{-E_{a_1} / RT} \dots (1)$
$k_2 = 4k_1 = Ae^{-E_{a_2} / RT} \dots (2)$
Dividing $(2)$ by $(1)$:
$4 = e^{(E_{a_1} - E_{a_2}) / RT}$
Taking natural logarithm on both sides:
$\ln 4 = \frac{E_{a_1} - E_{a_2}}{RT}$
$E_{a_2} - E_{a_1} = -RT \ln 4$
Given $T = 27 + 273 = 300 \ K$ and $\ln 4 = 1.386$.
$\Delta E_a = E_{a_2} - E_{a_1} = -(8.314 \ J \ K^{-1} \ mol^{-1} \times 300 \ K \times 1.386)$
$\Delta E_a = -3457.3 \ J / mol \approx -3.45 \ kJ / mol$.

(A) The relationship between threshold energy $(E_T)$,activation energy $(E_a)$,and the energy of reactants $(E_R)$ is given by the formula:
$E_T = E_a + E_R$
Given:
$E_T = 75 \ kJ/mol$
$E_R = 20 \ kJ/mol$
Substituting the values:
$75 = E_a + 20$
$E_a = 75 - 20 = 55 \ kJ/mol$
Therefore,the activation energy is $55 \ kJ/mol$.

(D) Group $16$ elements are known as chalcogens because they are ore-forming elements.

(C) The structures of the given metal carbonyls are as follows:
$I. [Co_2(CO)_8]$: It exists in a bridged form in the solid state,containing two $CO$ bridges.
$II. [Fe_3(CO)_{12}]$: It contains two $CO$ bridges in its structure.
$III. [Mn_2(CO)_{10}]$: It has a direct $Mn-Mn$ bond and no $CO$ bridges.
$IV. [Fe_2(CO)_9]$: It contains three $CO$ bridges.
Thus,the complexes with $CO$ bridges are $I, II,$ and $IV$.

(C) The magnetic moment (spin only) is given by $\mu = \sqrt{n(n+2)} \text{ BM}$,where $n$ is the number of unpaired electrons.
$1. [CoCl_4]^{2-}: Co^{2+} (d^7)$ is a tetrahedral complex. The configuration is $e^4 t_2^3$,so $n = 3$.
$2. [Fe(H_2O)_6]^{2+}: Fe^{2+} (d^6)$ is an octahedral complex. The configuration is $t_{2g}^4 e_g^2$,so $n = 4$.
$3. [Mn(H_2O)_6]^{2+}: Mn^{2+} (d^5)$ is an octahedral complex. The configuration is $t_{2g}^3 e_g^2$,so $n = 5$.
$4. [Cr(H_2O)_6]^{2+}: Cr^{2+} (d^4)$ is an octahedral complex. The configuration is $t_{2g}^3 e_g^1$,so $n = 4$.
Comparing the number of unpaired electrons,both $[Fe(H_2O)_6]^{2+}$ and $[Cr(H_2O)_6]^{2+}$ have $n = 4$,thus they have the same magnetic moment.

(D) The secondary valence of a metal complex is equal to its coordination number ($C$.$N$.),which is the total number of ligand donor atoms directly bonded to the central metal ion.
$1$. For $(I) \ CoCl_3 \cdot 6 H_2O$:
Since $3$ moles of $AgCl$ are precipitated,all $3$ $Cl^-$ ions are outside the coordination sphere. The formula is $[Co(H_2O)_6]Cl_3$. The coordination number is $6$ (six $H_2O$ ligands).
$2$. For $(II) \ NiCl_3 \cdot 6 H_2O$:
Since $2$ moles of $AgCl$ are precipitated,$2$ $Cl^-$ ions are outside the coordination sphere. The formula is $[Ni(H_2O)_5Cl]Cl_2 \cdot H_2O$. The coordination number is $6$ (five $H_2O$ and one $Cl^-$ ligand).
$3$. For $(III) \ Co(SO_4)Br \cdot 5 NH_3$:
Since $1$ mole of $AgBr$ is precipitated,the $Br^-$ ion is outside the coordination sphere. The formula is $[Co(NH_3)_5(SO_4)]Br$. The coordination number is $6$ (five $NH_3$ and one $SO_4^{2-}$ ligand).
Thus,the secondary valences for $(I), (II),$ and $(III)$ are $6, 6, 6$ respectively.

(C) The crystal field splitting,$\Delta_0$,depends upon the field produced by the ligand and the charge on the metal ion. Ligands are arranged in a spectrochemical series based on their field strength. The order of increasing field strength is:
$I^{-} < Br^{-} < SCN^{-} < Cl^{-} < S^{2-} < F^{-} < OH^{-} < C_2O_4^{2-} < H_2O < NCS^{-} < edta^{4-} < NH_3 < en < CN^{-} < CO$
Comparing the given ligands:
$III$ $(en)$ is a strong field ligand.
$I$ $(NCS^{-})$ is stronger than $IV$ $(SCN^{-})$.
$II$ $(S^{2-})$ is a weak field ligand.
Thus,the decreasing order of field strength is $III > I > IV > II$.

(D) According to the colour wheel,the colour observed is the complementary colour of the light absorbed.
When a metal complex absorbs orange light,it appears in its complementary colour,which is blue.

(A) The transition metals (with the exception of $Zn$,$Cd$,and $Hg$) are very hard and have low volatility.
Their melting and boiling points are high.
The high melting points of these metals are attributed to the involvement of a greater number of electrons from $(n-1)d$ orbitals in addition to the $ns$ electrons in the interatomic metallic bonding.
Since more electrons participate in metallic bonding,the strength of the metallic bond increases,leading to higher melting points.
Thus,the Assertion is true and the Reason is the correct explanation for the Assertion.

(C) The ionic radii of trivalent lanthanide ions ($La^{3+}$ to $Lu^{3+}$) decrease due to lanthanide contraction.
$La^{3+}$ $(Z=57)$ has the largest radius,while $Lu^{3+}$ $(Z=71)$ has the smallest radius among the lanthanides.
$Y^{3+}$ $(Z=39)$ is a $4d$ transition metal ion and is significantly smaller than the $4f$ lanthanide ions.
Therefore,the order of ionic radii is $Y^{3+} < Lu^{3+} < Eu^{3+} < La^{3+}$.

(D) The general electronic configuration of group $12$ elements is $(n-1)d^{10} ns^2$.
When these elements form $+2$ oxidation state ions,they lose two electrons from the $ns$ orbital.
Therefore,the electronic configuration of $Zn^{2+}$,$Cd^{2+}$,and $Hg^{2+}$ becomes $(n-1)d^{10} ns^0$,which represents a fully filled $d$-subshell.

(B) The electronic configuration of $Fe^{2+}$ is $[Ar] 3d^6$ and that of $Fe^{3+}$ is $[Ar] 3d^5$.
In $Fe^{3+}$,the $d$-orbital is exactly half-filled $(d^5)$.
According to Hund's rule,half-filled and fully-filled orbitals possess extra stability due to maximum exchange energy and symmetrical distribution of electrons.
Therefore,$Fe^{3+}$ is more stable than $Fe^{2+}$.

(C) The reaction of $FeCl_3$ with $NH_3$ produces $Fe(OH)_3$ which is a brown precipitate. Thus,$(A)$ matches with $(III)$.
The reaction of $AgCl$ with $NH_3$ produces $[Ag(NH_3)_2]Cl$ which is a colourless solution. Thus,$(B)$ matches with $(IV)$.
Therefore,the correct sequence is $(A-III, B-IV)$.

(C) The thermal decomposition of potassium permanganate $(KMnO_4)$ at $513 \ K$ is given by the following reaction:
$2KMnO_4 \xrightarrow{\Delta} K_2MnO_4 + MnO_2 + O_2 \uparrow$
In the products,the oxidation states of manganese are:
$K_2MnO_4$: $Mn$ is in $+6$ oxidation state.
$MnO_2$: $Mn$ is in $+4$ oxidation state.
The product with the higher oxidation state is $K_2MnO_4$ $(Mn^{+6})$.
The electronic configuration of $Mn^{+6}$ is $[Ar] 3d^1 4s^0$.
Since it has one unpaired electron,it is paramagnetic.
$K_2MnO_4$ is known to have a characteristic green colour.
Therefore,the product with the higher oxidation state $(Mn^{+6})$ is paramagnetic and green.

(B) Due to Lanthanide contraction,the ionic radii decrease from $La^{3+}$ to $Lu^{3+}$.
As the size decreases,the polarizing power of the metal ion increases,which increases the covalent character of the $M-OH$ bond. Consequently,the basic strength decreases from $La(OH)_3$ to $Lu(OH)_3$.
Chemical reactivity decreases from $La$ to $Lu$ due to the increase in ionization energy and decrease in size.
$Zr$ and $Hf$ exhibit similar atomic radii due to the Lanthanide contraction,which makes their chemical properties very similar.
Separation of Lanthanides is difficult because their chemical properties are very similar due to the small change in ionic radii.
Therefore,statements $A, B, C,$ and $D$ are correct.

(B) According to Faraday's first law of electrolysis,the mass of the substance deposited $(W)$ is given by the formula: $W = \frac{E \times i \times t}{96500}$.
Here,$W = 0.96 \ g$,$i = 1.2 \ A$,and $t = 40 \ min = 40 \times 60 \ s = 2400 \ s$.
Substituting the values into the formula:
$0.96 = \frac{E \times 1.2 \times 2400}{96500}$.
$E = \frac{0.96 \times 96500}{1.2 \times 2400}$.
$E = \frac{92640}{2880} = 32.16 \ g$ (approximately $31.75 \ g$ based on standard atomic mass).
Thus,the equivalent weight of copper is $31.75 \ g$.

(C) The half-cell reactions are:
$(1)$ $Mn^{7+} + 5e^- \rightarrow Mn^{2+}$,$\Delta G^{\circ}_1 = -5FE^{\circ}_{Mn^{7+}/Mn^{2+}}$
$(2)$ $Mn^{4+} + 2e^- \rightarrow Mn^{2+}$,$\Delta G^{\circ}_2 = -2FE^{\circ}_{Mn^{4+}/Mn^{2+}}$
$(3)$ $Mn^{7+} + 3e^- \rightarrow Mn^{4+}$,$\Delta G^{\circ}_3 = -3FE^{\circ}_{Mn^{7+}/Mn^{4+}}$
Since reaction $(3)$ = $(1)$ - $(2)$,we have $\Delta G^{\circ}_3 = \Delta G^{\circ}_1 - \Delta G^{\circ}_2$.
Substituting the values: $-3FE^{\circ}_3 = -5F(1.51) - (-2F(1.23))$.
$3E^{\circ}_3 = 5(1.51) - 2(1.23) = 7.55 - 2.46 = 5.09$.
$E^{\circ}_3 = 5.09 / 3 = 1.696 \ V \approx 1.70 \ V$.

(C) According to Faraday's second law of electrolysis,when the same quantity of electricity is passed through different electrolytes,the masses of substances deposited are proportional to their equivalent weights: $\frac{m_{Ag}}{m_{Au}} = \frac{E_{Ag}}{E_{Au}}$.
Given $m_{Ag} = 2.15 \ g$,$m_{Au} = 1.31 \ g$,and $E_{Ag} = \frac{107.9}{1} = 107.9$.
Substituting the values: $\frac{2.15}{1.31} = \frac{107.9}{E_{Au}}$.
$E_{Au} = \frac{107.9 \times 1.31}{2.15} \approx 65.7$.
Since $\text{Equivalent weight} = \frac{\text{Atomic mass}}{\text{Valency}}$,we have $\text{Valency} = \frac{197}{65.7} \approx 3$.

(C) The chemical reaction at the anode is:
$2H_2O(l) \rightarrow O_2(g) + 4H^+(aq) + 4e^-$
Total charge passed $Q = I \times t = 3 \ A \times (10 \times 60 \ s) = 1800 \ C$.
According to the stoichiometry,$4 \ mol$ of electrons produce $1 \ mol$ of $O_2$ gas.
$4 \times 96500 \ C$ of charge produces $22.4 \ L$ of $O_2$ at $STP$.
Therefore,$1800 \ C$ of charge produces volume $V = \frac{22.4 \times 1800}{4 \times 96500} \approx 0.1045 \ L$.
The closest value is $0.1 \ L$.

(B) The reaction is: $Fe^{3+} + Ce^{3+} \rightarrow Fe^{2+} + Ce^{4+}$
Here,$Fe^{3+}$ is reduced to $Fe^{2+}$ (cathode) and $Ce^{3+}$ is oxidised to $Ce^{4+}$ (anode).
$E^{\circ}_{cell} = E^{\circ}_{cathode} - E^{\circ}_{anode}$
$E^{\circ}_{cell} = E^{\circ}_{Fe^{3+}/Fe^{2+}} - E^{\circ}_{Ce^{4+}/Ce^{3+}}$
$E^{\circ}_{cell} = 0.77 \ V - 1.6 \ V = -0.83 \ V$

(D) The electric charge required for the deposition of $1$ equivalent of a substance is equal to $1 \ Faraday$.
According to Faraday's laws of electrolysis,$1 \ mol$ of electrons is required to deposit $1$ equivalent of any substance.
The charge carried by $1 \ mol$ of electrons is calculated as:
Charge $= N_A \times e^-$
Charge $= (6.022 \times 10^{23} \ mol^{-1}) \times (1.602 \times 10^{-19} \ C)$
Charge $\approx 96500 \ C \ mol^{-1} = 1 \ Faraday$.
Therefore,the charge is equal to the charge on $1 \ mol$ of electrons.

(B) Acetic acid $(CH_3COOH)$ is a weak electrolyte.
For weak electrolytes,the degree of dissociation increases with dilution (decrease in concentration).
As a result,the molar conductivity $(\lambda_{m})$ increases sharply as the concentration $(C)$ approaches zero.
The plot of $\lambda_{m}$ versus $\sqrt{C}$ for a weak electrolyte shows a steep increase as $\sqrt{C}$ decreases,and it does not intersect the y-axis at finite concentration.
Graph $B$ correctly represents this behavior,where $\lambda_{m}$ increases as $\sqrt{C}$ decreases.

(C) The given compound is a tertiary amine where a nitrogen atom is attached to a benzene ring,a methyl group $(-CH_3)$,and an ethyl group $(-CH_2CH_3)$.
According to $IUPAC$ nomenclature rules for $N$-substituted amines,the parent chain is the benzene ring (benzenamine).
The substituents on the nitrogen atom are listed in alphabetical order.
Since 'ethyl' comes before 'methyl' alphabetically,the name is $N$-Ethyl-$N$-methylbenzenamine.
Therefore,the correct option is $C$.

(C) $NaHCO_3$ is a mild base and only compounds more acidic than carbonic acid $(H_2CO_3)$ can react with it to evolve $CO_2$ gas.
Carboxylic acids are significantly more acidic than phenols and amines.
Compound $(III)$ is benzoic acid and compound $(IV)$ is $p$-nitrobenzoic acid.
Both $(III)$ and $(IV)$ are carboxylic acids and are strong enough to react with $NaHCO_3$.
Compound $(I)$ is an amine-substituted phenol and compound $(II)$ is phenol; both are less acidic than $H_2CO_3$ and do not react with $NaHCO_3$.
Therefore,only $(III)$ and $(IV)$ react with $NaHCO_3$.

(A) The acidity of phenols is inversely proportional to their $pK_{a}$ values $(Acidity \propto \frac{1}{pK_{a}})$.
Electron-withdrawing groups (like $-NO_{2}$) increase the acidity of phenol by stabilizing the phenoxide ion through $-I$ and $-M$ effects.
$1$. Phenol $(A)$ has no electron-withdrawing group,so it is the least acidic (highest $pK_{a}$).
$2$. meta-Nitrophenol $(C)$ exhibits only the $-I$ effect,making it more acidic than phenol but less acidic than ortho and para isomers.
$3$. ortho-Nitrophenol $(B)$ and para-Nitrophenol $(D)$ exhibit both $-I$ and $-M$ effects. para-Nitrophenol $(D)$ is the most acidic due to the strong $-M$ effect of the $-NO_{2}$ group at the para position.
ortho-Nitrophenol $(B)$ is less acidic than para-Nitrophenol $(D)$ due to intramolecular hydrogen bonding,which stabilizes the undissociated molecule.
Thus,the order of acidity is: $D > B > C > A$.
Since $pK_{a}$ is inversely proportional to acidity,the decreasing order of $pK_{a}$ is: $A > C > B > D$.

(D) The acidity of carboxylic acids is directly proportional to the stability of their conjugate bases,which is enhanced by electron-withdrawing groups ($-I$ effect).
$pK_a$ is inversely proportional to acidity.
Comparing the compounds:
$II. CF_3COOH$ (Strongest acid due to three highly electronegative $F$ atoms)
$I. CCl_3COOH$ (Strong acid,but $Cl$ is less electronegative than $F$)
$III. NO_2CH_2COOH$ (Stronger acid than $IV$ because $-NO_2$ is a stronger electron-withdrawing group than $-CN$)
$IV. NCCH_2COOH$
Acidity order: $II > I > III > IV$
Since $pK_a = -\log(K_a)$,the order of $pK_a$ is the reverse of the acidity order:
$pK_a$ order: $IV > III > I > II$
This corresponds to $(II) < (I) < (III) < (IV)$.

(D) The $SN_1$ reaction rate depends on the stability of the carbocation formed in the rate-determining step.
The carbocations formed are:
$I$. $CH_3CH_2CH_2CH_2^+$ (Primary)
$II$. $CH_3CH_2CH^+CH_3$ (Secondary)
$III$. $(CH_3)_2CHCH_2^+$ (Primary,but branched)
$IV$. $(CH_3)_3C^+$ (Tertiary)
Tertiary carbocations are more stable than secondary,which are more stable than primary carbocations.
Among the primary carbocations,$III$ is slightly more stable than $I$ due to the inductive effect of the methyl group.
Therefore,the stability order is $I < III < II < IV$.
Since the rate of $SN_1$ reaction follows the stability order of carbocations,the reactivity order is $I < III < II < IV$.

(B) Reaction $1$: Sodium $t-butoxide$ $((CH_3)_3CONa)$ is a strong base and a bulky nucleophile. It reacts with methyl bromide $(CH_3Br)$ via an $S_N2$ mechanism to produce $t-butyl$ methyl ether $(CH_3-O-C(CH_3)_3)$ as the major product $(P)$.
Reaction $2$: Sodium methoxide $(CH_3ONa)$ is a strong base. When it reacts with a tertiary substrate like $t-butyl$ bromide $((CH_3)_3CBr)$,the steric hindrance prevents $S_N2$ substitution. Instead,an $E2$ elimination reaction occurs,producing $2-methylpropene$ $((CH_3)_2C=CH_2)$ as the major product $(Q)$.

(C) The reaction proceeds as follows:
$1$. The first step is the anti-Markovnikov addition of $HBr$ to methylenecyclohexane in the presence of a peroxide,$(C_6H_5CO)_2O_2$. This yields (cyclohexylmethyl) bromide,$C_6H_{11}CH_2Br$.
$2$. The second step is a nucleophilic substitution reaction with $KCN$,where the bromide ion is replaced by a cyanide group,yielding cyclohexylacetonitrile,$C_6H_{11}CH_2CN$.
$3$. The third step is the acid-catalyzed hydrolysis of the nitrile group followed by heating,which converts the $-CN$ group into a carboxylic acid group,$-COOH$. The final product is cyclohexylacetic acid,$C_6H_{11}CH_2COOH$.

(A) Reaction $(A)$: $CH_3CH_2CH_2I$ is a primary alkyl halide. In the presence of aqueous $KOH$ (a strong nucleophile),it undergoes a bimolecular nucleophilic substitution $(S_N2)$ reaction to form propan$-1-$ol.
Reaction $(B)$: $(CH_3)_3CCl$ is a tertiary alkyl halide. In the presence of water (a polar protic solvent),it undergoes a unimolecular nucleophilic substitution $(S_N1)$ reaction to form tert-butyl alcohol.
Therefore,the correct matching is $A-III$ and $B-I$.

(A) The reaction of an alkyl halide with $Zn$ and dilute $HCl$ is a reduction reaction.
In this reaction,the halogen atom is replaced by a hydrogen atom to form the corresponding alkane.
The chemical equation is:
$(CH_3)_2CH-CH_2-CH_2Cl + [H] \xrightarrow{Zn/dil.HCl} (CH_3)_2CH-CH_2-CH_3 + HCl$
Here,$1-$chloro$-3-$methylbutane is reduced to $2-$methylbutane.

(A) The reaction proceeds as follows:
$(i)$ $CH_3OH$ reacts with $KI$ to form $CH_3I$.
(ii) $CH_3I$ reacts with $Mg$ in dry ether to form the Grignard reagent $CH_3MgI$.
(iii) $CH_3MgI$ undergoes nucleophilic addition to the carbonyl group of $1-$indanone ($2$,$3$-dihydro-1H-inden$-1-$one).
(iv) Subsequent hydrolysis with $H_2O$ yields $1-$methyl$-2,3-$dihydro-1H-inden$-1-$ol.
$(v)$ Dehydration of the alcohol using $20\% H_3PO_4$ at $358 \ K$ results in the formation of the more stable conjugated alkene,$1$-methyl-1H-indene,as the major product.

(C) The reaction sequence is as follows:
$1$. Toluene reacts with $Br_2$ in the presence of $UV$ light to undergo free-radical substitution at the benzylic position,forming benzyl bromide $(C_6H_5CH_2Br)$.
$2$. Benzyl bromide reacts with $Mg$ in dry ether to form the Grignard reagent,benzylmagnesium bromide $(C_6H_5CH_2MgBr)$.
$3$. The Grignard reagent reacts with formaldehyde $(CH_2O)$ followed by acidic hydrolysis $(H_3O^+)$ to form $2-$phenylethanol $(C_6H_5CH_2CH_2OH)$.
$4$. Finally,$2$-phenylethanol reacts with $PBr_3$ to replace the hydroxyl group with a bromine atom,yielding ($2$-bromoethyl)benzene $(C_6H_5CH_2CH_2Br)$.

(B) The reaction proceeds as follows:
$1$. Bromocyclohexane reacts with $Mg$ in dry ether to form cyclohexylmagnesium bromide (a Grignard reagent).
$2$. The Grignard reagent reacts with $CO_2$ followed by acidic workup $(H_3O^+)$ to form cyclohexanecarboxylic acid.
$3$. Cyclohexanecarboxylic acid reacts with $SOCl_2$ to form cyclohexanecarbonyl chloride.
$4$. Finally,the reaction of cyclohexanecarbonyl chloride with dimethylcadmium,$(CH_3)_2Cd$,yields cyclohexyl methyl ketone as the major product. This is a standard method for the preparation of ketones from acid chlorides.

(A) The given reaction is an example of the $Wolff-Kishner$ reduction.
This reaction involves the reduction of a carbonyl group $(C=O)$ to a methylene group $(CH_2)$.
First,the ketone reacts with hydrazine $(NH_2NH_2)$ to form a hydrazone intermediate.
Then,in the presence of a strong base like $KOH$ and a high-boiling solvent like ethylene glycol,the hydrazone undergoes base-catalyzed decomposition to release nitrogen gas $(N_2)$ and form the corresponding alkane.
In this specific case,$3,3-dimethylcyclopentanone$ is reduced to $3,3-dimethylcyclopentane$.

(A) $1$. Propene undergoes anti-Markovnikov addition with $HBr$ in the presence of peroxide to form $1-$bromopropane: $CH_3-CH=CH_2 + HBr \xrightarrow{\text{peroxide}} CH_3-CH_2-CH_2Br$.
$2$. $1-$bromopropane reacts with $Mg$ in the presence of dry ether to form the Grignard reagent,propylmagnesium bromide: $CH_3-CH_2-CH_2Br + Mg \xrightarrow{\text{dry ether}} CH_3-CH_2-CH_2MgBr$.
$3$. Propanal $(CH_3-CH_2-CHO)$ undergoes nucleophilic addition with propylmagnesium bromide followed by hydrolysis to form hexan$-3-$ol: $CH_3-CH_2-CHO + CH_3-CH_2-CH_2MgBr \rightarrow CH_3-CH_2-CH(OH)-CH_2-CH_2-CH_3$.
$4$. Hexan$-3-$ol on oxidation with $CrO_3$ forms hexan$-3-$one: $CH_3-CH_2-CH(OH)-CH_2-CH_2-CH_3 \xrightarrow{CrO_3} CH_3-CH_2-CO-CH_2-CH_2-CH_3$.

(B) Step $(i)$: Benzene reacts with $Br_2$ in the presence of anhydrous $FeBr_3$ to form bromobenzene via electrophilic aromatic substitution.
Step (ii): Bromobenzene undergoes Friedel-Crafts alkylation with $CH_3Cl$ and anhydrous $AlCl_3$. Since the $-Br$ group is ortho/para directing,the major product is $p$-bromotoluene.
Step (iii): $p$-Bromotoluene reacts with $Mg$ in dry ether to form the Grignard reagent,$p$-tolylmagnesium bromide $(CH_3-C_6H_4-MgBr)$.
Step (iv) and $(v)$: The Grignard reagent reacts with acetaldehyde $(CH_3CHO)$ followed by hydrolysis $(H_2O)$ to form a secondary alcohol. The nucleophilic carbon of the Grignard reagent attacks the carbonyl carbon of acetaldehyde,resulting in $1-(4-methylphenyl)ethanol$.

(D) Step $1$: Phenol reacts with zinc dust $(Zn)$ under heating $(\Delta)$ to undergo reduction,resulting in the formation of benzene.
$C_6H_5OH Zn \xrightarrow{\Delta} C_6H_6 ZnO$
Step $2$: Benzene then undergoes electrophilic aromatic substitution (chlorination) in the presence of anhydrous ferric chloride $(FeCl_3)$ and chlorine $(Cl_2)$ to form chlorobenzene.
$C_6H_6 Cl_2 \xrightarrow{\text{anhyd. } FeCl_3} C_6H_5Cl HCl$
Therefore,the major product is chlorobenzene.

(A) The reaction proceeds as follows:
$(i)$ Phenol reacts with $NaOH$ and $CH_3I$ to form anisole (methoxybenzene) via Williamson ether synthesis.
(ii) Anisole undergoes Friedel-Crafts acylation with propionyl chloride $(CH_3CH_2COCl)$ in the presence of anhydrous $AlCl_3$. Since the $-OCH_3$ group is ortho/para-directing,and the para-position is sterically less hindered,the major product is $p$-methoxypropiophenone.
(iii) Reduction of the ketone group with $NaBH_4$ yields the corresponding secondary alcohol,$1$-($4$-methoxyphenyl)propan$-1-$ol.
Therefore,the correct option is $A$.

(C) The reaction proceeds as follows:
$(i)$ $Sn/HCl$ reduces the $-NO_2$ group to an $-NH_2$ group,forming $p$-toluidine.
(ii) $Br_2$ $(1 \ eq.)$ in the presence of the strongly activating $-NH_2$ group leads to ortho-bromination,yielding $2$-bromo-$4$-methylaniline.
(iii) $NaNO_2/HCl$ at $273-278 \ K$ converts the $-NH_2$ group into a diazonium salt,$-N_2^+Cl^-$.
(iv) $Cu_2Br_2/HBr$ (Sandmeyer reaction) replaces the diazonium group with a $-Br$ atom,resulting in $1,2$-dibromo-$4$-methylbenzene (also known as $3,4$-dibromotoluene).

(C) When aluminium chloride $(AlCl_3)$ is dissolved in water,it undergoes hydration to form the octahedral complex ion $[Al(H_2O)_6]^{3+}$ and releases chloride ions $(Cl^-)$.
The reaction is: $AlCl_3 + 6H_2O \rightarrow [Al(H_2O)_6]^{3+} + 3Cl^-$.
In this complex,the $Al^{3+}$ ion is coordinated to six water molecules,and the hybridization of $Al$ is $sp^3d^2$.

(A) Noble gases are monoatomic and possess the largest atomic radius in their respective periods.
The atomic size of noble gases increases down the group from $He$ to $Rn$ due to the addition of new shells.
Among the noble gases,$He$ has the smallest atomic size.
Due to its extremely small size,$He$ atoms can easily diffuse through the pores of silica glass.
The atomic radius of noble gases is referred to as the van der Waals radius.

(C) $(A)\ NO_2^-(aq) + NH_4^+(s) \xrightarrow{\Delta} N_{2(g)} + 2H_2O$. This reaction liberates $N_2$ gas.
$(B)\ CO(NH_2)_{2(s)} + 2HNO_{2(l)} \longrightarrow 2N_{2(g)} + CO_2 + 3H_2O$. This reaction liberates $N_2$ gas.
$(C)\ 2NH_{3(g)} + 3NaOCl_{(aq)} \xrightarrow{\text{gelatine}} N_2H_4 + 3NaCl + H_2O$. This reaction produces hydrazine $(N_2H_4)$ and does not liberate $N_2$ gas.
$(D)\ (NH_4)_2Cr_2O_{7(s)} \xrightarrow{\Delta} Cr_2O_3 + N_{2(g)} + 4H_2O$. This reaction liberates $N_2$ gas.

(D) Reaction $(A): NH_4 NO_3 \xrightarrow{\Delta} N_2 O + 2 H_2 O$. $N_2 O$ is diamagnetic.
Reaction $(B): 2 NaNO_2 + H_2 SO_4 \rightarrow Na_2 SO_4 + H_2 O + NO_2 + NO$. (Note: The reaction produces $NO$ and $NO_2$,both are paramagnetic).
Reaction $(C): 2 Pb(NO_3)_2 \xrightarrow{673 \ K} 2 PbO + 4 NO_2 + O_2$. $NO_2$ and $O_2$ are paramagnetic.
The paramagnetic gaseous products are $NO$ (from $B$),$NO_2$ (from $B$ and $C$),and $O_2$ (from $C$).
Total number of unique paramagnetic gaseous products is $3$.

(B) The reaction described is the Brown Ring Test,which is used to detect the presence of nitrate ions $(NO_3^-)$ in a solution.
When dilute $FeSO_4$ is added to a solution containing nitrate ions,followed by the careful addition of concentrated $H_2SO_4$,the following reaction occurs:
$2NO_3^- + 3Fe^{2+} + 4H^+ \rightarrow 2NO + 3Fe^{3+} + 2H_2O$
The nitrogen oxide formed "in situ" is nitric oxide $(NO)$.
This $NO$ then reacts with the hydrated ferrous sulfate complex to form the brown-colored complex $[Fe(H_2O)_5(NO)]SO_4$.

(A) The reaction between $N_2$ and $O_2$ is highly endothermic and requires a very high temperature to proceed.
This reaction is part of the Birkeland-Eyde process,which uses an electric arc furnace to provide temperatures greater than $3000 \ K$ to facilitate the formation of $NO$ from $N_2$ and $O_2$.

(A) Phosphorous acid $(H_3PO_3)$ has two $P-OH$ bonds,making it a dibasic acid.
Phosphoric acid $(H_3PO_4)$ has three $P-OH$ bonds,making it a tribasic acid.
Therefore,they are respectively dibasic and tribasic acids.