AP EAMCET 2021 Chemistry Question Paper with Answer and Solution

(C) The hydrocarbon '$X$' reacts with sodamide $(NaNH_2)$,which indicates that '$X$' is a terminal alkyne.
Ozonolysis of a terminal alkyne followed by oxidative workup ($O_3$ followed by $Zn/H_2O_2$ or $H_2O_2$) leads to the cleavage of the triple bond to form a carboxylic acid and formic acid $(HCOOH)$.
For one of the products to be optically active,the carboxylic acid formed from the alkyl group must contain a chiral center.
Starting with $3-Methylpent-1-yne$ $(CH_3CH_2CH(CH_3)C \equiv CH)$:
$CH_3CH_2CH(CH_3)C \equiv CH \xrightarrow{O_3, H_2O_2} CH_3CH_2CH(CH_3)COOH + HCOOH$.
$2-Methylbutanoic acid$ $(CH_3CH_2CH(CH_3)COOH)$ contains a chiral carbon at the $C-2$ position,making it optically active.

(C) Step $1$: $2, 3-$dibromobutane reacts with $Zn / Cu$ in the presence of heat to undergo debromination,forming $trans-$but$-2-$ene.
Step $2$: The $trans-$but$-2-$ene undergoes reductive ozonolysis with $O_3$ followed by $Zn / H_2O$.
Step $3$: The double bond breaks to form two molecules of acetaldehyde $(CH_3CHO)$.
Thus,the final product is $2 \ CH_3CHO$.

(A) Molar mass of benzene $= 78 \ g/mol$.
Mole of benzene $= \frac{7.8 \ g}{78 \ g/mol} = 0.1 \ mol$.
The reaction is $C_6H_6 + CH_3COCl \xrightarrow{AlCl_3} C_8H_8O + HCl$.
According to the stoichiometry,$1 \ mol$ of benzene produces $1 \ mol$ of $C_8H_8O$.
Therefore,$0.1 \ mol$ of benzene should theoretically produce $0.1 \ mol$ of $C_8H_8O$.
Molar mass of $C_8H_8O = (8 \times 12) + (8 \times 1) + (1 \times 16) = 96 + 8 + 16 = 120 \ g/mol$.
Theoretical yield $= 0.1 \ mol \times 120 \ g/mol = 12 \ g$.
Percentage yield $= \frac{\text{Experimental yield}}{\text{Theoretical yield}} \times 100 = \frac{8.4 \ g}{12 \ g} \times 100 = 70\%$.

(C) The molecular formula of benzene is $C_6H_6$.
The balanced chemical equation for the complete combustion of benzene is:
$C_6H_6(l) + \frac{15}{2} O_2(g) \rightarrow 6 CO_2(g) + 3 H_2O(l)$
From the stoichiometry of the reaction,$1 \text{ mole}$ of benzene produces $6 \text{ moles}$ of carbon dioxide $(CO_2)$.
The molar mass of $CO_2$ is $12 + (2 \times 16) = 44 \ g/mol$.
Therefore,the mass of $6 \text{ moles}$ of $CO_2$ is $6 \times 44 \ g = 264 \ g$.
Thus,the complete combustion of one mole of benzene produces $264 \ g$ of carbon dioxide.

(A) When benzene reacts with excess chlorine in the presence of ultraviolet $(UV)$ light at $500 \ K$,addition reaction occurs.
The delocalized $\pi$-electron system of the benzene ring is broken,and one chlorine atom adds to each of the six carbon atoms.
This results in the formation of $1,2,3,4,5,6-\text{hexachlorocyclohexane}$,commonly known as benzene hexachloride $(BHC)$ or gammaxene.
Therefore,$(P) = \text{Benzene hexachloride}$.

(C) The acidity is directly proportional to the tendency of the hydrogen donor.
It depends upon the bond dissociation enthalpy of the $H-X$ bond,where $X$ is $O, S, Se, Te$.
As we move down the group,the bond dissociation energy of hydrides decreases due to an increase in bond length,which makes the release of $H^+$ ions easier.
Therefore,the acidic strength increases down the group.
Hence,the correct order of acidic character of the given hydrides is $H_2Te > H_2Se > H_2S > H_2O$.

(A) The metal oxide that yields $H_2O_2$ upon treatment with dilute acid is $BaO_2$ because it contains the peroxide ion $(O_2^{2-})$.
When barium peroxide reacts with dilute sulfuric acid,it produces barium sulfate and hydrogen peroxide:
$BaO_2 + H_2SO_4 \rightarrow BaSO_4 + H_2O_2$

(B) $H_2 O_2$ is obtained by the electrolysis of a cold $50 \%$ solution of $H_2 SO_4$.
The electrolysis of $HSO_4^-$ ions at high current density leads to the formation of peroxodisulphate $(H_2 S_2 O_8)$,which upon hydrolysis yields hydrogen peroxide.
The reaction is:
$2 HSO_4^- \xrightarrow{\text{Electrolysis}} HO_3 SOOSO_3 H + 2 e^-$.
$HO_3 SOOSO_3 H + 2 H_2 O \rightarrow 2 H_2 SO_4 + H_2 O_2$.

(C) Old lead paintings often turn black due to the formation of lead sulfide $(PbS)$ by the reaction of lead pigments with atmospheric $H_2S$.
The colour is restored by washing with $H_2O_2$,which oxidizes the black $PbS$ to white lead sulfate $(PbSO_4)$.
The reaction is: $PbS(s) + 4H_2O_2(aq) \rightarrow PbSO_4(s) + 4H_2O(l)$.
In this reaction,$H_2O_2$ acts as an oxidizing agent,not a reducing agent.
Therefore,Assertion $(A)$ is true,but Reason $(R)$ is false.

(A) conjugate acid is formed when a base accepts a proton $(H^{+})$.
$NH_{3} + H^{+} \rightarrow NH_{4}^{+}$
$HCO_{3}^{-} + H^{+} \rightarrow H_{2}CO_{3}$
$H_{2}O + H^{+} \rightarrow H_{3}O^{+}$
$HSO_{4}^{-} + H^{+} \rightarrow H_{2}SO_{4}$
Therefore,the correct matching is $(a$ $\rightarrow iii), (b$ $\rightarrow i), (c$ $\rightarrow ii), (d$ $\rightarrow iv)$.

(D) An acid loses a proton to form an anion known as its conjugate base.
According to the Bronsted-Lowry theory,a strong acid forms a weak conjugate base,and a weak acid forms a strong conjugate base.
Among the given halide ions $(F^{-}, Cl^{-}, Br^{-}, I^{-})$,the corresponding acids are $HF, HCl, HBr,$ and $HI$.
$HI$ is the strongest acid and $HF$ is the weakest acid among these hydrohalic acids.
Since $HF$ is the weakest acid,its conjugate base $F^{-}$ is the strongest conjugate base.

(D) Bronsted-Lowry acid is a proton ($H^{+}$ ion) donor,while a Bronsted-Lowry base is a proton ($H^{+}$ ion) acceptor.
$HSO_4^{-} + H_2O \rightleftharpoons H_2SO_4 + OH^{-}$ (acts as a Bronsted base)
$HSO_4^{-} + H_2O \rightleftharpoons SO_4^{2-} + H_3O^{+}$ (acts as a Bronsted acid)
Since $HSO_4^{-}$ can both donate a proton to form $SO_4^{2-}$ and accept a proton to form $H_2SO_4$,it acts as both a Bronsted acid and a Bronsted base.

(A) For a strong acid like $HCl$,it dissociates completely in water: $HCl \rightarrow H^{+} + Cl^{-}$.
Given concentration of $HCl = 0.10 \ M$,so $[H^{+}] = 0.10 \ M = 10^{-1} \ M$.
The $pH$ of the solution is calculated as: $pH = -\log[H^{+}] = -\log(10^{-1}) = 1$.
Using the relation at $25^{\circ}C$: $pH + pOH = 14$.
Therefore,$pOH = 14 - pH = 14 - 1 = 13$.

(D) The dissociation reaction of acetic acid is: $CH_3COOH \rightleftharpoons CH_3COO^{-} + H^{+}$
Given concentration $c = 0.1 \ M$ and degree of dissociation $\alpha = 0.0132$.
The concentration of hydrogen ions is given by $[H^{+}] = c \times \alpha$.
$[H^{+}] = 0.1 \times 0.0132 = 0.00132 \ M$.
The $pH$ is calculated as $pH = -\log[H^{+}]$.
$pH = -\log(0.00132) = -\log(1.32 \times 10^{-3}) = 3 - \log(1.32) \approx 3 - 0.12 = 2.88$.
Thus,the $pH$ of the solution is $2.88$.

(C) The $pH$ scale ranges from $0$ to $14$. Substances with $pH > 7$ are basic,and higher $pH$ values indicate stronger basicity.
$1 \ M$ $HCl$ is a strong acid $(pH \approx 0)$.
Distilled $H_2O$ is neutral $(pH = 7)$.
$1 \ M$ $aq$ $NH_3$ is a weak base $(pH \approx 11.6)$.
$1 \ M$ $NaOH$ is a strong base $(pH = 14)$.
Therefore,$1 \ M$ $NaOH$ has the highest $pH$.

(B) An acidic buffer is a solution that resists changes in $pH$ and is prepared by mixing a weak acid with its salt of a strong base.
$HCOOH$ (weak acid) and $HCOOK$ (salt) form an acidic buffer.
$C_6H_5COOH$ (weak acid) and $C_6H_5COONa$ (salt) form an acidic buffer.
$HCN$ (weak acid) and $KCN$ (salt) form an acidic buffer.
$HClO_4$ is a strong acid,and $NaClO_4$ is its corresponding salt. $A$ mixture of a strong acid and its salt does not act as a buffer solution because a buffer requires a weak acid to maintain the equilibrium $HA \rightleftharpoons H^+ + A^-$. Therefore,$HClO_4 \& NaClO_4$ is not an acidic buffer.

(C) buffer solution consists of a mixture of a weak acid and its salt with a strong base,or a weak base and its salt with a strong acid.
$LiOH$ is a strong base,while $HNO_3$ and $HBr$ are strong acids.
$HNO_2$ is a weak acid and $NaNO_2$ is its salt formed by the combination of the weak acid $(HNO_2)$ and a strong base $(NaOH)$.
Therefore,the pair $(HNO_2 + NaNO_2)$ constitutes an acidic buffer.

(B) The solubility product $(K_{sp})$ of $AgBr$ is $5 \times 10^{-10}$.
For the dissociation reaction: $AgBr_{(s)} \rightleftharpoons Ag^{+}_{(aq)} + Br^{-}_{(aq)}$.
The expression for $K_{sp}$ is $K_{sp} = [Ag^{+}][Br^{-}]$.
In $0.2 \ M$ $NaBr$ solution,the concentration of $Br^{-}$ ions is $0.2 \ M$ due to the common ion effect.
Let the solubility of $AgBr$ be $s$.
Then $[Ag^{+}] = s$ and $[Br^{-}] = 0.2 + s \approx 0.2$ (since $s$ is very small).
$5 \times 10^{-10} = s \times 0.2$.
$s = \frac{5 \times 10^{-10}}{0.2} = 25 \times 10^{-10} \ M$.

(D) The dissociation of $Ag_2CO_3$ is given by: $Ag_2CO_3(s) \rightleftharpoons 2Ag^{+}(aq) + CO_3^{2-}(aq)$.
Given that $[Ag^{+}] = 1.20 \times 10^{-4} \ mol \ L^{-1}$.
From the stoichiometry of the reaction,$[CO_3^{2-}] = \frac{1}{2} [Ag^{+}] = \frac{1}{2} \times 1.20 \times 10^{-4} = 0.60 \times 10^{-4} \ mol \ L^{-1}$.
The solubility product expression is $K_{sp} = [Ag^{+}]^2 [CO_3^{2-}]$.
Substituting the values: $K_{sp} = (1.20 \times 10^{-4})^2 \times (0.60 \times 10^{-4})$.
$K_{sp} = (1.44 \times 10^{-8}) \times (0.60 \times 10^{-4}) = 0.864 \times 10^{-12} = 8.64 \times 10^{-13}$.
Wait,re-evaluating based on standard solubility $S$: If $[Ag^{+}] = 2S = 1.20 \times 10^{-4}$,then $S = 0.60 \times 10^{-4}$.
$K_{sp} = 4S^3 = 4 \times (0.60 \times 10^{-4})^3 = 4 \times 0.216 \times 10^{-12} = 0.864 \times 10^{-12} = 8.64 \times 10^{-13}$.
Given the options provided,there is a discrepancy. If we assume the question implies $[Ag^{+}] = 2S$ and the result matches option $D$ $(6.90 \times 10^{-12})$,it suggests $S = 1.20 \times 10^{-4}$ was intended as the solubility $S$ itself,not the concentration of $Ag^{+}$.
Assuming $S = 1.20 \times 10^{-4} \ mol \ L^{-1}$,then $K_{sp} = 4S^3 = 4 \times (1.20 \times 10^{-4})^3 = 4 \times 1.728 \times 10^{-12} = 6.912 \times 10^{-12} \approx 6.90 \times 10^{-12}$.

(B) For the salt $AB_4$,the dissociation is:
$AB_{4(s)} \rightleftharpoons A^{4+}_{(aq)} + 4B^{-}_{(aq)}$
If $S$ is the molar solubility,then:
$[A^{4+}] = S$
$[B^{-}] = 4S$
The solubility product $K_{sp}$ is given by:
$K_{sp} = [A^{4+}][B^{-}]^4$
$K_{sp} = (S)(4S)^4$
$K_{sp} = S \cdot 256S^4$
$K_{sp} = 256S^5$
$S^5 = \frac{K_{sp}}{256}$
$S = \left(\frac{K_{sp}}{256}\right)^{1/5}$

(C) From the ideal gas equation,$PV = nRT$.
Since $n = \frac{m}{M}$,we have $PV = \frac{m}{M}RT$,where $m$ is the mass and $M$ is the molar mass.
This implies $m = \frac{PVM}{RT}$,so $m \propto \frac{P}{T}$.
Let $m_1 = 6 \ g$,$P_1 = P$,and $T_1 = 400 \ K$.
Let $m_2$ be the final mass,$P_2 = \frac{P}{2}$,and $T_2 = 300 \ K$.
Using the ratio $\frac{m_2}{m_1} = \frac{P_2}{P_1} \times \frac{T_1}{T_2}$,we get:
$\frac{m_2}{6} = \frac{P/2}{P} \times \frac{400}{300} = \frac{1}{2} \times \frac{4}{3} = \frac{2}{3}$.
Therefore,$m_2 = 6 \times \frac{2}{3} = 4 \ g$.
The mass of oxygen that leaked out is $\Delta m = m_1 - m_2 = 6 \ g - 4 \ g = 2 \ g$.

(C) Given: Mass of the car $m = 500 \ kg$,radius of the turn $r = 50 \ m$,and velocity $v = 36 \ km/h$.
First,convert the velocity into $SI$ units $(m/s)$:
$v = 36 \times \frac{5}{18} = 10 \ m/s$.
The formula for centripetal force is $F = \frac{mv^2}{r}$.
Substituting the values into the formula:
$F = \frac{500 \times (10)^2}{50}$
$F = \frac{500 \times 100}{50}$
$F = 10 \times 100 = 1000 \ N$.
Therefore,the centripetal force acting on the car is $1000 \ N$.

(A) According to Ampere's Circuital Law,the magnetic field $\vec{B}$ at a distance $r$ from the axis is given by $\oint \vec{B} \cdot d\vec{l} = \mu_0 I_{\text{enclosed}}$.
For a point outside the cable $(r > R_{\text{outer}})$,the total current enclosed is $I_{\text{enclosed}} = I_{\text{inner}} + I_{\text{outer}}$.
Since the currents are equal and in opposite directions,$I_{\text{inner}} = I$ and $I_{\text{outer}} = -I$.
Therefore,$I_{\text{enclosed}} = I + (-I) = 0$.
Since the enclosed current is zero,the magnetic field $\vec{B}$ outside the cable is zero.

(C) Given that,the length of the wire is $L$. The current passing through the wire is $I$. Let $R$ be the radius of the circular loop formed by the wire.
Then,the circumference $2 \pi R = L$,which gives $R = \frac{L}{2 \pi}$.
Consider a small element of the wire $AB$ of length $\Delta l$ which subtends an angle $\Delta \theta$ at the center $O$. Thus,$\Delta l = R \Delta \theta$.
The magnetic force acting on this small element $AB$ is $F = I B \Delta l = I B R \Delta \theta$,directed radially outward.
Let $T$ be the tension in the wire. The components of tension $T \cos(\Delta \theta / 2)$ at the ends of the element cancel each other out.
The radial component of the tension force is $2 T \sin(\Delta \theta / 2)$.
For equilibrium,the magnetic force must be balanced by the radial component of the tension:
$2 T \sin(\Delta \theta / 2) = F = I B R \Delta \theta$.
Since $\Delta \theta$ is very small,$\sin(\Delta \theta / 2) \approx \Delta \theta / 2$.
Therefore,$2 T (\Delta \theta / 2) = I B R \Delta \theta$,which simplifies to $T = I B R$.
Substituting $R = \frac{L}{2 \pi}$ into the expression for tension,we get $T = I B (\frac{L}{2 \pi}) = \frac{I B L}{2 \pi}$.

(B) When a body moves with a constant speed $v$ along a circle,it experiences a centripetal acceleration directed towards the center,although its linear speed remains constant.
Since the centripetal force $F$ acting on the body is always directed towards the center and the displacement $s$ is always along the tangent to the circle,the angle $\theta$ between the force and the displacement is $90^{\circ}$.
The work done $W$ is given by the formula:
$W = F \cdot s = F s \cos \theta$
Since $\theta = 90^{\circ}$,we have $\cos 90^{\circ} = 0$.
Therefore,$W = F s (0) = 0$.
Hence,no work is done on the body.

(A) Time-period in spring-block system is given by, $T = 2 \pi \sqrt{\frac{m}{k}} $
If mass is doubled, $T^{\prime} = 2 \pi \sqrt{\frac{2m}{k}} $ $ \Rightarrow \quad T^{\prime} = \sqrt{2} T $
Hence, time period becomes $ \sqrt{2} $ times, if the mass of the block is doubled.
$ (A \rightarrow 4) $
(B) For spring, $ \mathrm{PE} = \frac{1}{2} k x^2 $
If spring constant is increased 4 times, PE becomes 4 times.
$ (B \rightarrow 3) $
(C) Speed of the block, $ v = \omega A $
Hence, velocity of the block becomes twice when amplitude of the oscillation is doubled.
$ (C \rightarrow 2) $
(D) Energy of the oscillation, $ E = \frac{1}{2} m \omega^2 a^2 $
$ \begin{aligned} & & E^{\prime} = \frac{1}{2} m \omega^{\prime 2} a^2 \\ \Rightarrow & E^{\prime} & = \frac{1}{2} m(2 \omega)^2 a^2 = \frac{4}{2} m \omega^2 a^2 \quad \left( \because \omega^{\prime} = 2 \omega \right) \\ \Rightarrow & E^{\prime} & = 4 E \end{aligned} $
Energy of the oscillation becomes 4 times when angular frequency is doubled.
$ (D \rightarrow 1) $

(A) When borax $(Na_2B_4O_7 \cdot 10H_2O)$ is dissolved in water,it undergoes hydrolysis to form orthoboric acid and sodium hydroxide.
The reaction is as follows:
$Na_2B_4O_7 + 7H_2O \rightarrow 2NaOH + 4H_3BO_3$
Since $NaOH$ is a strong base,the resulting solution is alkaline in nature.

(A) Diborane $(B_2H_6)$ is an electron-deficient molecule.
In its structure,the four terminal hydrogen atoms and two boron atoms lie in the same plane.
The two bridging hydrogen atoms lie above and below this plane.
The terminal $B-H$ bonds are normal $2-$centered$-2-$electron $(2c-2e)$ bonds with a bond angle of $120^{\circ}$ $(x)$.
The bridging $B-H-B$ bonds are $3-$centered$-2-$electron $(3c-2e)$ bonds,which are commonly known as banana-bonds or tau-bonds.
The bridging $H-B-H$ bond angle $(y)$ is $97^{\circ}$.

(B) $H_3BO_3$ is a weak monobasic Lewis acid. It does not dissociate to give $H^{+}$ ions directly in water.
Instead,it acts as a Lewis acid by accepting an $OH^{-}$ ion from the water molecule.
This process releases a proton $(H^{+})$ from the water molecule,thereby increasing the acidity of the solution.
The reaction is: $B(OH)_3 + 2H_2O \rightarrow [B(OH)_4]^{-} + H_3O^{+}$.
Therefore,$H_3BO_3$ is considered an acid because its molecule accepts $OH^{-}$ from water,releasing a proton.

(A) The molecular geometry of $BCl_3$ is trigonal planar because boron is $sp^2$ hybridized with three bond pairs and no lone pairs.
In the adduct $BCl_3 \cdot NH_3$,the boron atom forms a coordinate bond with the nitrogen atom of $NH_3$.
After the formation of the coordinate bond,the boron atom is surrounded by four atoms (three $Cl$ and one $N$),resulting in $sp^3$ hybridization.
Therefore,the geometry of $BCl_3 \cdot NH_3$ is tetrahedral.

(B) Boron compounds are often employed as Lewis acids due to their strong electrophilic nature,which is caused by a vacant $p$-orbital that can accept electrons.
Boron behaves as a Lewis acid because it has only $3$ valence electrons,leaving its octet incomplete.
For example,$B_2H_6$ is an electron-deficient compound.
The electronic configuration of Boron is $B (Z=5) = [He] 2s^2 2p^1$.
This results in a total of $6$ electrons in the valence shell of the central atom in compounds like $BF_3$,which is less than the $8$ required for a complete octet.

(C) $NaBH_4$ is a mild reducing agent. When sodium borohydride reacts with iodine,it produces diborane,sodium iodide,and hydrogen gas. This reaction involves the oxidation of sodium borohydride with iodine in diglyme to yield diborane. This approach is common in the industrial production of $B_2H_6$.
The balanced chemical equation is:
$2NaBH_4 + I_2 \xrightarrow{\text{Diglyme}} B_2H_6 + 2NaI + H_2$

(C) $AlF_3$ is insoluble in anhydrous $HF$ because it lacks free $F^-$ ions to interact with the $Al^{3+}$ center.
In the presence of $KF$,$KF$ dissociates to provide $F^-$ ions,which react with $AlF_3$ to form a soluble complex.
The reaction is:
$AlF_3 + 3KF \rightarrow K_3[AlF_6]$
This complex dissociates into $3K^+$ and $[AlF_6]^{3-}$.
Therefore,the correct option is $[AlF_6]^{3-}$.

(D) The reaction for the formation of baking soda $(NaHCO_3)$ from sodium carbonate $(Na_2CO_3)$ is given by the equation:
$Na_2CO_3 + H_2O + CO_2 \rightarrow 2NaHCO_3$
From the balanced chemical equation,it is clear that $1 \text{ mole}$ of $H_2O$ reacts with $1 \text{ mole}$ of $CO_2$.
Therefore,the molar ratio of $H_2O : CO_2$ is $1:1$.

(C) The coordination number of an element depends on the availability of vacant orbitals in the valence shell.
Carbon is in the $2^{nd}$ period with the electronic configuration $1s^2 2s^2 2p^2$. It lacks vacant $d$-orbitals,limiting its maximum covalency to $4$.
Germanium is in the $4^{th}$ period with the electronic configuration $[Ar] 3d^{10} 4s^2 4p^2$.
Due to the presence of vacant $d$-orbitals,germanium can expand its coordination number beyond $4$.

(B) Fullerenes are allotropes of carbon consisting of clusters of carbon atoms,such as $C_{60}$,$C_{70}$,$C_{76}$,and $C_{84}$.
Fullerenes are synthesized by heating graphite in an electric arc in the presence of an inert gas like helium or argon at low pressure.
During this process,the carbon vaporizes and condenses to form a soot containing various fullerenes.
$C_{60}$ (Buckminsterfullerene) is the most abundant product,while $C_{70}$ is obtained in smaller quantities.

(B) When silicon tetrachloride $(SiCl_4)$ reacts with an excess of water,it undergoes complete hydrolysis.
Silicon has vacant $d$-orbitals,which allow the water molecule to attack the silicon atom,leading to the substitution of chlorine atoms by hydroxyl groups.
The reaction proceeds as follows:
$SiCl_4 + 4H_2O \rightarrow Si(OH)_4 + 4HCl$
Therefore,the major product is silicic acid,$Si(OH)_4$.

(A) $SiO_2$ is a covalent three-dimensional network solid.
Silica (commonly known as quartz) is $SiO_2$,which is usually used to manufacture glass,ceramics,and piezoelectric materials.
The most common use of silica is in water filtration; because of its uniform shape and size,it is an effective filtration bed that removes contaminants from water.
$SiO_2$ is less reactive due to high $Si-O$ bond enthalpy. Therefore,the statement that it is highly reactive is incorrect.

(B) The two most abundant elements in the Earth's crust are oxygen and silicon. The compound $(A)$ formed by these elements is silicon dioxide $(SiO_2)$,which is widely used in building construction (e.g.,in sand and concrete).
When silicon dioxide $(SiO_2)$ reacts with carbon at high temperatures,it undergoes a reduction reaction:
$SiO_2 + 3C \xrightarrow{\Delta} SiC + 2CO$
Here,$(B)$ is carbon monoxide $(CO)$,which is a poisonous,stable diatomic gas.

(A) Silicones are used as greases (lubricants) and in surgical implants due to their inert nature and biocompatibility.
Zeolites are used as catalysts (e.g.,$ZSM-5$) and as ion exchangers in water softening.
Therefore,the correct matches are:
$A$ (Silicones) $\rightarrow$ $(ii)$ Greases,$(iii)$ In surgical implants.
$B$ (Zeolites) $\rightarrow$ $(i)$ As a catalyst,$(iv)$ As ion exchangers.
The correct code is $A-(ii, iii), B-(i, iv)$.

(C) Nitric acid $(HNO_3)$ is a strong oxidizing agent. It usually oxidizes the $H_2$ gas produced to water and itself gets reduced to nitrogen oxides.
However,magnesium $(Mg)$ and manganese $(Mn)$ react with very dilute nitric acid to evolve hydrogen gas.
$Mg + 2HNO_3 \text{ (very dilute)} \rightarrow Mg(NO_3)_2 + H_2 \uparrow$

(B) The ozone molecule $(O_3)$ exhibits resonance between two canonical structures.
In each resonance structure,the central oxygen atom is bonded to the two terminal oxygen atoms by one single bond and one double bond.
$A$ single bond consists of $1 \sigma$ bond,and a double bond consists of $1 \sigma$ and $1 \pi$ bond.
Therefore,the total number of $\sigma$ bonds is $2$ and the total number of $\pi$ bonds is $1$.
The bond angle in the angular ozone molecule is approximately $117^{\circ}$.

(B) The reaction of calcium metal with water is: $Ca + 2H_2O \longrightarrow Ca(OH)_2 + H_2(g)$.
The reaction of calcium hydride with water is: $CaH_2 + 2H_2O \longrightarrow Ca(OH)_2 + 2H_2(g)$.
Both reactions produce $H_2$ gas as a product upon hydrolysis.

(D) The correct option is $(d)$. Chlorine reacts with water to form hypochlorous acid $(HOCl)$,which is unstable and decomposes to give nascent oxygen $([O])$.
$Cl_2 + H_2O \longrightarrow HCl + HOCl$
$HOCl \longrightarrow HCl + [O]$
This nascent oxygen is responsible for the bleaching and oxidizing properties of chlorine water.

(C) The reaction of iodine with chlorine in the presence of water is an oxidation reaction where iodine is oxidized to iodic acid $(HIO_3)$ and chlorine is reduced to hydrochloric acid $(HCl)$.
The balanced chemical equation is:
$I_2 + 5Cl_2 + 6H_2O \longrightarrow 2HIO_3 + 10HCl$
Comparing this with the given reaction $I_2 + 5Cl_2 + 6H_2O \longrightarrow X + Y$,we identify $X$ as $HIO_3$ and $Y$ as $HCl$.

(B) The acidity of oxoacids depends on the oxidation state and electronegativity of the central atom.
$1$. The oxidation state of $Cl$ in $HClO_4$ is $+7$,while that of $S$ in $H_2SO_4$ is $+6$. $A$ higher oxidation state leads to stronger electron-withdrawing effects,making the $O-H$ bond more polar and easier to break.
$2$. $Cl$ is more electronegative than $S$,which further increases the electron-withdrawing effect on the $O-H$ bond.
$4$. Due to the higher oxidation state and electronegativity of $Cl$,the $ClO_3$ group exerts a stronger inductive effect than the $SO_3$ group,facilitating the release of $H^+$.
Statement $3$ is a property of these acids but does not explain why $HClO_4$ is a stronger acid than $H_2SO_4$.
Therefore,statements $1$,$2$ and $4$ are the correct reasons.

(B) Molecular bromine $(Br_2)$ reacts with $3$ moles of molecular fluorine $(F_2)$ to produce $2$ moles of $BrF_3$.
$BrF_3$ is an interhalogen compound.
The chemical reaction is:
$Br_{2(g)} + 3F_{2(g)} \longrightarrow 2BrF_{3(g)}$
In $BrF_3$,the central bromine atom has $7$ valence electrons. It forms $3$ covalent bonds with $3$ fluorine atoms,using $3$ electrons.
Remaining electrons = $7 - 3 = 4$ electrons,which form $2$ lone pairs.
Thus,the total number of lone pairs at the central bromine atom is $2$.

(A) First,arrange the $6$ men around a round table. The number of ways to arrange $n$ objects in a circle is $(n-1)!$. So,$6$ men can be seated in $(6-1)! = 5!$ ways.
There are $6$ gaps created between the $6$ men where the $5$ women can be seated so that no two women sit together.
The number of ways to arrange $5$ women in these $6$ gaps is given by the permutation formula $P(6, 5) = \frac{6!}{(6-5)!} = 6!$.
Therefore,the total number of ways is $5! \times 6!$.

(B) The centroid $C$ of a triangle with vertices $(x_1, y_1)$,$(x_2, y_2)$,and $(x_3, y_3)$ is given by $\left(\frac{x_1 + x_2 + x_3}{3}, \frac{y_1 + y_2 + y_3}{3}\right)$.
For vertices $(3, -1)$,$(1, 3)$,and $(2, 4)$,the centroid is $C = \left(\frac{3 + 1 + 2}{3}, \frac{-1 + 3 + 4}{3}\right) = (2, 2)$.
To find the intersection point $P$ of $x + 3y - 1 = 0$ and $3x - y + 1 = 0$,we solve the system:
$x + 3y = 1$ and $3x - y = -1$.
Multiplying the second equation by $3$: $9x - 3y = -3$.
Adding the two equations: $(x + 3y) + (9x - 3y) = 1 - 3$ $\Rightarrow 10x = -2$ $\Rightarrow x = -\frac{1}{5}$.
Substituting $x = -\frac{1}{5}$ into $x + 3y = 1$: $-\frac{1}{5} + 3y = 1$ $\Rightarrow 3y = \frac{6}{5}$ $\Rightarrow y = \frac{2}{5}$.
So,$P = \left(-\frac{1}{5}, \frac{2}{5}\right)$.
The slope $m$ of the line passing through $C(2, 2)$ and $P\left(-\frac{1}{5}, \frac{2}{5}\right)$ is $m = \frac{2 - 2/5}{2 - (-1/5)} = \frac{8/5}{11/5} = \frac{8}{11}$.
The equation of the line is $y - 2 = \frac{8}{11}(x - 2)$ $\Rightarrow 11y - 22 = 8x - 16$ $\Rightarrow 8x - 11y + 6 = 0$.
Checking the options,for $(-9, -6)$: $8(-9) - 11(-6) + 6 = -72 + 66 + 6 = 0$.
Thus,the line passes through $(-9, -6)$.

(B) Since the axes are rotated through an angle $\theta = 45^{\circ}$,we replace $(x, y)$ by $(x \cos 45^{\circ} - y \sin 45^{\circ}, x \sin 45^{\circ} + y \cos 45^{\circ})$,which is $(\frac{x-y}{\sqrt{2}}, \frac{x+y}{\sqrt{2}})$. Substituting these into the equation $3x^2 + 3y^2 + 2xy = 2$:
$3(\frac{x-y}{\sqrt{2}})^2 + 3(\frac{x+y}{\sqrt{2}})^2 + 2(\frac{x-y}{\sqrt{2}})(\frac{x+y}{\sqrt{2}}) = 2$
$\frac{3}{2}(x^2 + y^2 - 2xy) + \frac{3}{2}(x^2 + y^2 + 2xy) + \frac{2}{2}(x^2 - y^2) = 2$
$\frac{3}{2}(2x^2 + 2y^2) + (x^2 - y^2) = 2$
$3x^2 + 3y^2 + x^2 - y^2 = 2$
$4x^2 + 2y^2 = 2$
Dividing by $2$,we get $2x^2 + y^2 = 1$.

(B) The reaction sequence is as follows:
$1$. Acetylation of aniline with $(CH_3CO)_2O$ and pyridine protects the $-NH_2$ group,forming acetanilide.
$2$. Electrophilic aromatic substitution with $Br_2$ in $CH_3CO_2H$ occurs at the para-position due to the steric hindrance of the $-NHCOCH_3$ group,yielding $p$-bromoacetanilide.
$3$. Hydrolysis with $NaOH(aq)$ removes the acetyl group to regenerate $p$-bromoaniline.
$4$. Diazotization with $HNO_2$ at $0-5 \ ^\circ C$ converts the $-NH_2$ group into a diazonium salt,$-N_2^+Cl^-$.
$5$. The Sandmeyer reaction with $CuCl/HCl$ replaces the diazonium group with a chlorine atom,resulting in $1$-bromo-$4$-chlorobenzene.

(C) The reaction of phenol with $CHCl_3$ in the presence of aqueous $NaOH$ is known as the $Reimer-Tiemann$ reaction.
In this reaction,$CHCl_3$ reacts with $NaOH$ to generate a dichlorocarbene intermediate $(:CCl_2)$.
This electrophilic carbene attacks the phenoxide ion at the ortho position.
The intermediate formed undergoes hydrolysis to yield salicylaldehyde ($2$-hydroxybenzaldehyde) in its salt form,which is sodium $2-$formylphenolate.

(D) Step $(i)$ and $(ii)$: The oxidation of styrene $(C_6H_5CH=CH_2)$ with alkaline $KMnO_4$ followed by acidic workup $(H_3O^+)$ results in the formation of benzoic acid $(C_6H_5COOH)$.
Step $(iii)$: Benzoic acid contains a $-COOH$ group,which is a deactivating and meta-directing group. Therefore,electrophilic aromatic substitution with $Br_2/FeBr_3$ will direct the bromine atom to the meta position.
The final product is $m$-bromobenzoic acid.

(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) 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.

(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).