TS EAMCET 2023 Chemistry Question Paper with Answer and Solution

(B) Acidic strength $\propto \ \% \ s$-character.
The acidity of hydrogen atoms attached to carbon depends on the hybridization of the carbon atom. The order of electronegativity is $sp > sp^2 > sp^3$,which corresponds to $50\% \ s$,$33.3\% \ s$,and $25\% \ s$ character respectively.
$1$. $CH \equiv CH$ $(II)$: $sp$ hybridized carbon ($50\% \ s$-character),most acidic.
$2$. $CH_3-C \equiv CH$ $(III)$: Terminal alkyne,but the $+I$ effect of the $-CH_3$ group decreases the acidity compared to ethyne.
$3$. $H_2C=CH_2$ $(I)$: $sp^2$ hybridized carbon ($33.3\% \ s$-character).
$4$. $CH_3-CH_3$ $(IV)$: $sp^3$ hybridized carbon ($25\% \ s$-character),least acidic.
Thus,the correct order of acidic strength is $IV < I < III < II$.

(A) Hyperconjugation requires the presence of at least one $\alpha$-hydrogen atom on a carbon atom adjacent to an unsaturated system (like a double bond or a benzene ring) or a carbocation.
In the given options:
$A$) $C_6H_5-C(CH_3)_3$: The $\alpha$-carbon is attached to three methyl groups and the benzene ring. It has no $\alpha$-hydrogen atoms,so hyperconjugation is not possible.
$B$) $C_6H_5-CH(CH_3)_2$: The $\alpha$-carbon has one $\alpha$-hydrogen atom.
$C$) $C_6H_5-CH_2-CH_3$: The $\alpha$-carbon has two $\alpha$-hydrogen atoms.
$D$) $C_6H_5-CH_3$: The $\alpha$-carbon has three $\alpha$-hydrogen atoms.
Therefore,the correct option is $(A)$.

(A) The stability of resonance structures is determined by the following rules:
$1$. Neutral structures are more stable than charged structures.
$2$. Structures with a complete octet for all atoms are more stable.
$3$. Negative charge on a more electronegative atom is more stable.
Structure $II$ is a neutral molecule where all atoms have complete octets,making it the most stable.
Structure $I$ has a positive charge on carbon and a negative charge on oxygen. Oxygen is more electronegative than carbon,so having a negative charge on oxygen is relatively stable.
Structure $III$ has a negative charge on carbon and a positive charge on oxygen. Having a positive charge on a highly electronegative atom like oxygen is highly unstable.
Therefore,the stability order is $II > I > III$.

(B) The phenomenon involving the delocalization of $\sigma$ electrons of a $C-H$ bond of an alkyl group attached to an unsaturated system (like benzene) or a vacant $p$-orbital is known as hyperconjugation.
This effect is also referred to as 'no-bond resonance' or the Baker-Nathan effect.

(A) To determine the number of $\pi$-electrons in each species:
$1$. Phenanthrene: It consists of three fused benzene rings. It has $14$ $\pi$-electrons.
$2$. Naphthalene: It consists of two fused benzene rings. It has $10$ $\pi$-electrons.
$3$. Cyclopentadienyl anion: It has two double bonds ($4$ $\pi$-electrons) and one lone pair on the carbon atom involved in resonance,contributing $2$ $\pi$-electrons,totaling $6$ $\pi$-electrons.
$4$. Cycloheptatrienyl cation: It has three double bonds ($6$ $\pi$-electrons) and a vacant $p$-orbital on the positively charged carbon,totaling $6$ $\pi$-electrons.
Comparing these,Phenanthrene has the maximum number of $\pi$-electrons $(14)$.
Therefore,the correct option is $A$.

(B) Kernite is a mineral of Boron $(B)$ with the formula $Na_2B_4O_6(OH)_2 \cdot 3H_2O$.
Cryolite is a mineral of Aluminum $(Al)$ with the formula $Na_3AlF_6$.
Therefore,$X$ is $B$ and $Z$ is $Al$.

(C) Zeolite is a hydrated sodium aluminium silicate.
Its general chemical formula is $Na_2Al_2Si_2O_8 \cdot xH_2O$.
Thus,the metal ions present are $Na^{+}$ and $Al^{3+}$.

(A) Step $1$: The hydrogenation of propyne $(CH_3-C \equiv CH)$ with $H_2 / Pd-C$ (Lindlar's catalyst or similar) reduces the alkyne to an alkene,specifically propene $(CH_3-CH=CH_2)$,which is product $A$.
Step $2$: The reaction of propene with $HBr$ in the presence of benzoyl peroxide $((C_6H_5CO)_2O_2)$ follows the anti-Markovnikov addition (peroxide effect).
Step $3$: According to the anti-Markovnikov rule,the bromine atom attaches to the carbon atom with more hydrogen atoms,resulting in $1$-bromopropane $(CH_3-CH_2-CH_2Br)$ as the major product $B$.

(A) The oxidative cleavage of alkenes using hot acidic $KMnO_4$ results in the breaking of the $C=C$ double bond.
For reaction $I$: $(CH_3)_2 C=CH_2$ undergoes cleavage to form $(CH_3)_2 C=O$ (acetone,a ketone) and $CO_2 + H_2 O$. Thus,$X$ is a ketone.
For reaction $II$: $CH_3-CH=CH-CH_3$ undergoes cleavage to form $2CH_3COOH$ (acetic acid,a carboxylic acid). However,looking at the provided options and the structure of $Y$ in the image,the product is $CH_3COOH$. Given the options,the correct functional groups are Ketone $(X)$ and Carboxylic acid $(Y)$.

(A) Ozonolysis of an alkene involves the cleavage of the double bond to form two carbonyl compounds.
Given products are $CH_3-CH_2-CHO$ (propanal) and $CH_3-CHO$ (ethanal).
By removing the oxygen atoms from the carbonyl groups and joining the carbon atoms with a double bond,we get:
$CH_3-CH_2-CH=CH-CH_3$.
The structure corresponds to $Pent-2-ene$.
Therefore,the correct option is $A$.

(C) $1$. The starting material is $2$-methylhexane. Upon heating with $Cr_2O_3$ at $773 \ K$ and $10-20 \ atm$,it undergoes aromatization to form $A$ (Toluene).
$2$. Toluene $(A)$ reacts with alkaline $KMnO_4$ followed by acidic workup $(H_3O^+)$ to undergo oxidation of the methyl group to a carboxylic acid group,forming $B$ (Benzoic acid).
$3$. Benzoic acid $(B)$ reacts with concentrated $HNO_3$ and $H_2SO_4$ (nitration). Since the $-COOH$ group is meta-directing,the $-NO_2$ group attaches at the meta position,forming $C$ ($m$-nitrobenzoic acid).
$4$. Compound $C$ has substituents $D = -COOH$ and $E = -NO_2$.

(A) $1$. Ozonolysis of alkene $X$ $(C_6H_{12})$ yields acetaldehyde $(CH_3CHO)$ and ethyl methyl ketone $(CH_3COCH_2CH_3)$.
$2$. By reversing the ozonolysis process,we can determine the structure of $X$ by joining the carbonyl carbons with a double bond: $CH_3CH=C(CH_3)CH_2CH_3$.
$3$. The structure of $X$ is $3-methylpent-2-ene$.
$4$. When $X$ reacts with $HBr$,the reaction follows Markovnikov's rule,where the $H^+$ adds to the carbon with more hydrogen atoms and $Br^-$ adds to the more substituted carbon.
$5$. The reaction is: $CH_3CH=C(CH_3)CH_2CH_3 + HBr \rightarrow CH_3CH_2-C(Br)(CH_3)CH_2CH_3$.
$6$. The product formed is $3-bromo-3-methylpentane$.

(A) Ozonolysis of an alkene involves the cleavage of the $C=C$ double bond to form carbonyl compounds.
To find the original alkene,we join the two carbonyl products by removing the oxygen atoms and forming a double bond between the carbon atoms.
The products are $Propan-2-one$ $(CH_3-CO-CH_3)$ and $methanal$ $(H-CHO)$.
Joining these: $(CH_3)_2C=O + O=CH_2 \rightarrow (CH_3)_2C=CH_2$.
The resulting alkene is $2-Methylpropene$.

(B) The addition of $HBr$ to propene in the presence of a peroxide occurs contrary to Markovnikov's rule,which is known as the peroxide effect or Kharasch effect.
This reaction proceeds via a free radical chain mechanism.
The mechanism is initiated by the homolysis of benzoyl peroxide as shown below:
$(i) \ C_6H_5-CO-O-O-CO-C_6H_5$ $\xrightarrow{\text{Homolysis}} 2C_6H_5-CO-O^{\bullet}$ $\rightarrow 2C_6H_5^{\bullet} + 2CO_2$

(C) $1$. The reaction of $2CH_3CH_2CH_2Br$ with $Na/Ether$ is a Wurtz reaction,which produces $n$-hexane $(X = CH_3CH_2CH_2CH_2CH_2CH_3)$.
$2$. The aromatization of $n$-hexane in the presence of $Mo_2O_3$ at $773K$ and $10-20 \ atm$ yields benzene $(Y = C_6H_6)$.
$3$. The reaction of benzene $(Y)$ with $CH_3Cl$ in the presence of anhydrous $AlCl_3$ is a Friedel-Crafts alkylation,which produces toluene $(Z = C_6H_5CH_3)$.

(A) The reactions are identified as follows:
$A$. $\text{Benzene} + Cl_2 \xrightarrow{AlCl_3} \text{Chlorobenzene}$ (Electrophilic substitution),so $A-III$.
$B$. $\text{Benzene} + CH_3Cl \xrightarrow{\text{anhyd. } AlCl_3} \text{Toluene}$ (Friedel-Crafts alkylation),so $B-IV$.
$C$. $\text{Benzene} + R-CO-Cl \xrightarrow{\text{anhyd. } AlCl_3} \text{Alkyl phenyl ketone}$ (Friedel-Crafts acylation),so $C-II$.
$D$. $\text{Toluene} \xrightarrow{(i) KMnO_4/KOH, (ii) H_3O^+} \text{Benzoic acid}$ (Oxidation of side chain),so $D-I$.
Thus,the correct matching is $A-III, B-IV, C-II, D-I$.

(C) The reaction shown is a Friedel-Crafts acylation of benzene $(A)$ to form acetophenone,which is then further substituted to form $B$. The electrophile in the Friedel-Crafts acylation reaction is the acylium ion,$CH_3-C\equiv O^{\oplus}$ (or $CH_3CO^{\oplus}$),which is generated by the reaction of acetyl chloride $(CH_3COCl)$ with the Lewis acid catalyst $AlCl_3$.

(A) Electrophilic aromatic substitution is favored by electron-donating groups (activating groups) and disfavored by electron-withdrawing groups (deactivating groups).
$-OH$,$-NH_2$,and $-CH_3$ are electron-donating groups that increase the electron density of the benzene ring,thereby facilitating the electrophilic attack.
$-NO_2$ is a strong electron-withdrawing group due to both its $-I$ (inductive) and $-M$ (mesomeric) effects.
It significantly reduces the electron density of the benzene ring,making it the least reactive towards electrophilic attack.

(C) Statement $(I)$ is incorrect because the reaction between $NaH$ and $H_2O$ releases hydrogen gas,not oxygen gas: $NaH + H_2O \rightarrow NaOH + H_2 \uparrow$.
Statement $(II)$ is correct; $CH_4$ and $SiH_4$ have the required number of electrons to form normal covalent bonds,making them electron-precise hydrides.
Statement $(III)$ is incorrect; $NH_3$ and $H_2O$ are electron-rich hydrides because they possess lone pairs of electrons on the central atom.
Statement $(IV)$ is correct; hydrides of $Be$ and $Mg$ ($BeH_2$ and $MgH_2$) are electron-deficient and exist in polymeric structures.
Thus,statements $(II)$ and $(IV)$ are correct.

(D) Statement $(I)$ is incorrect because in saline (ionic) hydrides,hydrogen exists as the hydride ion $(H^-)$,meaning its oxidation state is $-1$.
Statement $(II)$ is correct because lanthanum hydride $(LaH_{2.87})$ is a non-stoichiometric interstitial hydride.
Statement $(III)$ is correct because $NH_3$ and $H_2O$ contain highly electronegative atoms ($N$ and $O$) bonded to hydrogen,allowing for hydrogen bonding.
Statement $(IV)$ is correct because during the electrolysis of molten $NaH$,the $H^-$ ions migrate to the anode and are oxidized to $H_2$ gas $(2H^- \rightarrow H_2 + 2e^-)$.
Therefore,statements $(II)$ and $(III)$ are correct.

(D) In photosynthesis,$H_2O$ is oxidized to $O_2$ in photosystem $II$ at the oxygen-evolving complex.
Statement $A$ is correct.
$MgH_2$ is a saline (ionic) hydride,not an interstitial hydride. Interstitial hydrides are formed by $d$-block and $f$-block elements. Statement $B$ is incorrect.
Sodium hexametaphosphate $(Na_6P_6O_{18})$,also known as Calgon,is used to remove permanent hardness of water by sequestering $Ca^{2+}$ and $Mg^{2+}$ ions. Statement $C$ is correct.
Therefore,statements $A$ and $C$ are correct.

(C) Electron-precise hydrides are formed by group $14$ elements.
These elements have exactly the number of electrons required to form normal covalent bonds,such as in $CH_4$,$SiH_4$,$GeH_4$,and $SnH_4$.

(D) Calgon is the commercial name for sodium hexametaphosphate,which has the chemical formula $Na_6P_6O_{18}$.
It is commonly used in water treatment to soften water by sequestering calcium and magnesium ions.

(A) Given that the side length $a = BC = 1$ unit. The perimeter of the triangle is $a + b + c$. The arithmetic mean of the sines of its angles is $\frac{\sin A + \sin B + \sin C}{3}$.
According to the problem,$a + b + c = 6 \times \frac{\sin A + \sin B + \sin C}{3}$.
This simplifies to $a + b + c = 2(\sin A + \sin B + \sin C)$.
Using the sine rule,$a = 2R \sin A$,$b = 2R \sin B$,and $c = 2R \sin C$,where $R$ is the circumradius.
Substituting these into the equation: $2R(\sin A + \sin B + \sin C) = 2(\sin A + \sin B + \sin C)$.
Since $\sin A + \sin B + \sin C \neq 0$ for a triangle,we get $2R = 2$,which implies $R = 1$.
From the sine rule,$\frac{a}{\sin A} = 2R$.
Substituting $a = 1$ and $R = 1$,we get $\frac{1}{\sin A} = 2(1) = 2$.
Therefore,$\sin A = \frac{1}{2}$,which gives $A = \frac{\pi}{6}$ (since $A$ is an angle of a triangle).

(C) The function is $y = |\sin 2x|$. We need to find the area bounded by this curve and the $X$-axis in the interval $[0, 2\pi]$.
Since the period of $|\sin 2x|$ is $\frac{\pi}{2}$,the function repeats its shape every $\frac{\pi}{2}$ units.
In the interval $[0, 2\pi]$,there are $4$ such identical humps (from $0$ to $\frac{\pi}{2}$,$\frac{\pi}{2}$ to $\pi$,$\pi$ to $\frac{3\pi}{2}$,and $\frac{3\pi}{2}$ to $2\pi$).
Thus,the total area $A$ is given by:
$A = \int_0^{2\pi} |\sin 2x| \, dx = 4 \int_0^{\frac{\pi}{2}} \sin 2x \, dx$
$A = 4 \left[ \frac{-\cos 2x}{2} \right]_0^{\frac{\pi}{2}}$
$A = -2 [\cos(2 \cdot \frac{\pi}{2}) - \cos(0)]$
$A = -2 [\cos(\pi) - \cos(0)]$
$A = -2 [-1 - 1] = -2 [-2] = 4$
Therefore,the area is $4$ square units.

(A) The solution is a buffer solution consisting of a weak acid $(CH_3COOH)$ and its conjugate base $(CH_3COO^-)$.
The Henderson-Hasselbalch equation is: $pH = pK_{a} + \log \frac{[\text{Salt}]}{[\text{Acid}]}$
Calculate the concentration of acid: $[\text{Acid}] = \frac{10 \ mL \times 1.0 \ M}{100 \ mL} = 0.1 \ M$
Calculate the concentration of salt: $[\text{Salt}] = \frac{20 \ mL \times 0.5 \ M}{100 \ mL} = 0.1 \ M$
Substitute the values into the equation: $pH = 4.76 + \log \frac{0.1}{0.1}$
Since $\log(1) = 0$,we get $pH = 4.76 + 0 = 4.76$.

(C) buffer solution is formed by a mixture of a weak acid and its conjugate base,or a weak base and its conjugate acid.
In option $A$,$NH_3$ (weak base) and $HCl$ (strong acid) in a $2:1$ ratio result in $NH_4Cl$ and $NH_3$,forming a basic buffer.
In option $B$,$CH_3COOH$ (weak acid) and $NaOH$ (strong base) in a $2:1$ ratio result in $CH_3COONa$ and $CH_3COOH$,forming an acidic buffer.
In option $C$,$NaOH$ and $CH_3COOH$ in a $1:1$ ratio undergo complete neutralization to form $CH_3COONa$ (a salt of a weak acid and strong base),which does not act as a buffer.
In option $D$,$NH_4Cl$ and $NH_3$ form a standard basic buffer.

(C) The equation of a parabola with the origin as its focus and the $X$-axis as its axis is given by $y^2 = 4a(x+a)$,where $a$ is a parameter.
Differentiating with respect to $x$,we get $2y \frac{dy}{dx} = 4a$,which implies $a = \frac{y}{2} \frac{dy}{dx}$.
Substituting the value of $a$ back into the original equation:
$y^2 = 4 \left( \frac{y}{2} \frac{dy}{dx} \right) \left( x + \frac{y}{2} \frac{dy}{dx} \right)$
$y^2 = 2y \frac{dy}{dx} \left( x + \frac{y}{2} \frac{dy}{dx} \right)$
$y^2 = 2xy \frac{dy}{dx} + y^2 \left( \frac{dy}{dx} \right)^2$.
Here,the highest order derivative is $\frac{dy}{dx}$,so the order $m = 1$.
The power of the highest order derivative is $2$,so the degree $n = 2$.
Therefore,$mn - m + n = (1)(2) - 1 + 2 = 2 - 1 + 2 = 3$.

(B) Let the origin be $O(0,0,0)$ and the foot of the perpendicular from the origin to the plane be $P(2,-1,3)$.
Since $OP$ is perpendicular to the plane,the vector $\vec{OP}$ is the normal vector to the plane.
The direction ratios of the normal vector $\vec{OP}$ are $(2-0, -1-0, 3-0) = (2, -1, 3)$.
The equation of a plane passing through a point $(x_1, y_1, z_1)$ with normal vector $(a, b, c)$ is given by $a(x-x_1) + b(y-y_1) + c(z-z_1) = 0$.
Substituting the point $P(2, -1, 3)$ and the normal vector $(2, -1, 3)$ into the equation:
$2(x-2) - 1(y-(-1)) + 3(z-3) = 0$
$2(x-2) - 1(y+1) + 3(z-3) = 0$
$2x - 4 - y - 1 + 3z - 9 = 0$
$2x - y + 3z - 14 = 0$
Thus,the equation of the plane is $2x - y + 3z - 14 = 0$.

(B) Let the equation of the plane be $ax + by + cz + d = 0$.
Since the line segment joining the origin $(0, 0, 0)$ and the point $(2, -1, 3)$ is perpendicular to the plane,the direction ratios of the normal to the plane are given by the coordinates of the foot of the perpendicular:
$a = 2 - 0 = 2$
$b = -1 - 0 = -1$
$c = 3 - 0 = 3$
Thus,the equation of the plane is $2x - y + 3z + d = 0$.
Since the plane passes through the point $(2, -1, 3)$,we substitute these coordinates into the equation:
$2(2) - (-1) + 3(3) + d = 0$
$4 + 1 + 9 + d = 0$
$14 + d = 0$
$d = -14$
Therefore,the equation of the plane is $2x - y + 3z - 14 = 0$.

(B) For the quadratic equation $x^2+ax+b=0$ to have real roots,the discriminant $D$ must be greater than or equal to $0$.
$D = a^2 - 4b \geq 0 \implies a^2 \geq 4b$.
Given $a \in \{3, 4, 5\}$ and $b \in \{1, 2, 3, 4\}$,the total number of possible pairs $(a, b)$ is $3 \times 4 = 12$.
We check the condition $a^2 \geq 4b$ for each $a$:
$1$. If $a=3$,$a^2=9$. $9 \geq 4b \implies b \leq 2.25$. Possible $b$ values are $\{1, 2\}$ ($2$ pairs).
$2$. If $a=4$,$a^2=16$. $16 \geq 4b \implies b \leq 4$. Possible $b$ values are $\{1, 2, 3, 4\}$ ($4$ pairs).
$3$. If $a=5$,$a^2=25$. $25 \geq 4b \implies b \leq 6.25$. Possible $b$ values are $\{1, 2, 3, 4\}$ ($4$ pairs).
Total favorable outcomes = $2 + 4 + 4 = 10$.
Probability = $\frac{\text{Favorable outcomes}}{\text{Total outcomes}} = \frac{10}{12} = \frac{5}{6}$.

(C) The elements belonging to group $14$ are $C$,$Si$,$Ge$,$Sn$,and $Pb$.
$C$ and $Si$ are non-metals,$Ge$ is a metalloid,while $Sn$ and $Pb$ are metals.
These elements form covalent networks in their solid state.
The melting point depends on the strength of the interatomic forces (covalent or metallic bonds).
As we move down the group,the atomic size increases,which leads to a decrease in the strength of the interatomic forces.
Therefore,the melting point decreases from $C$ to $Pb$.
Thus,the correct order of melting temperature is $C > Si > Ge$.

(A) Statement $A$ is correct: $AlCl_3$ forms a dimer $(Al_2Cl_6)$ to complete its octet.
Statement $B$ is correct: $BCl_3$ has only $6$ electrons in the valence shell of boron,making it electron-deficient.
Statement $C$ is incorrect: The standard reduction potential $E^0_{Al^{3+}/Al}$ is $-1.66 \ V$,not $+1.26 \ V$.
Statement $D$ is incorrect: Due to the inert pair effect,the $+1$ oxidation state of thallium $(Tl^+)$ is more stable than the $+3$ oxidation state $(Tl^{3+})$.
Therefore,statements $C$ and $D$ are incorrect.

(A) Statement $(I)$ is correct due to the inert pair effect,where the stability of the $+1$ oxidation state increases down the group $(Ga < In < Tl)$.
Statement $(II)$ is incorrect; Boron has a very high melting point due to its giant covalent structure.
Statement $(III)$ is correct; Boron lacks $d$-orbitals and can accommodate a maximum of $4$ electron pairs in its valence shell.
Statement $(IV)$ is incorrect; the correct order of atomic radius is $Al > Ga < In$ (due to poor shielding of $d$-electrons in $Ga$).
Statement $(V)$ is correct; Aluminium forms a protective oxide layer on its surface when treated with concentrated nitric acid,making it passive.
Thus,statements $(I, III, V)$ are correct.

(B) To find the element $X$ for which $T_2 - T_1$ is maximum,we calculate the difference for each element given in the table:
For $B$: $3923 - 2453 = 1470 \ K$
For $Al$: $2740 - 933 = 1807 \ K$
For $Ga$: $2676 - 303 = 2373 \ K$
For $In$: $2353 - 430 = 1923 \ K$
Comparing these values,the maximum difference is $2373 \ K$,which corresponds to $Ga$.

(D) When aluminium reacts with dilute hydrochloric acid,it forms aluminium chloride and releases hydrogen gas:
$2 Al_{(s)} + 6 HCl_{(aq)} \longrightarrow 2 AlCl_{3(aq)} + 3 H_2 \uparrow$
Similarly,aluminium reacts with aqueous alkali to release hydrogen gas:
$2 Al_{(s)} + 2 NaOH_{(aq)} + 6 H_2O_{(l)} \longrightarrow 2 Na[Al(OH)_4]_{(aq)} + 3 H_2 \uparrow$
Thus,both gases $A$ and $B$ are $H_2$.

(C) Graphite is an allotrope of carbon that has a layered structure,making it soft and slippery,which allows it to be used as a dry lubricant at high temperatures.
Thus,Assertion $(A)$ is true.
The reason for this soft and slippery nature is that the hexagonal layers of carbon atoms are held together by weak van der Waals forces,allowing them to slip over one another.
Thus,Reason $(R)$ is true and it correctly explains why graphite acts as a lubricant.

(C) The oxides of $Sn$ and $Pb$ (Group $14$) exhibit amphoteric character.
Specifically,$SnO_2, SnO, PbO,$ and $PbO_2$ are amphoteric because they can react with both acids and bases.
This property arises due to the stability of different oxidation states ($+2$ and $+4$) influenced by the inert pair effect.

(C) In Group $14$,the melting point generally decreases as we move down the group due to the weakening of metallic bonding.
However,there is an anomaly where the melting point of $Pb$ is slightly higher than that of $Sn$.
The order of melting points for the elements is $C > Si > Ge > Pb > Sn$.
Therefore,$Sn$ has the lowest melting point.

(D) Nitrates of alkali metals (except $LiNO_3$) decompose on heating to form metal nitrites and oxygen gas,without releasing $NO_2$.
For potassium nitrate: $2 KNO_3 \xrightarrow{\Delta} 2 KNO_2 + O_2$.
Nitrates of heavy metals (like $Pb$) and $LiNO_3$ decompose to form metal oxides,$NO_2$,and $O_2$.

(C) The chemical formula of Peroxydisulfuric acid (Marshall's acid) is $H_2S_2O_8$.
Its structure is $HO-S(=O)_2-O-O-S(=O)_2-OH$.
From the structure,each sulfur atom is bonded to two double-bonded oxygen atoms,one oxygen atom of the peroxide linkage $(-O-O-)$,and one hydroxyl group $(-OH)$.
The oxidation state of each sulfur atom is $+6$.
There are two $S-OH$ bonds in the molecule (one on each sulfur atom).

(D) disproportionation reaction is a type of redox reaction where the same element in a single oxidation state is simultaneously oxidized and reduced.
In option $A$,$P$ in $H_3 PO_3$ $(+3)$ is oxidized to $H_3 PO_4$ $(+5)$ and reduced to $PH_3$ $(-3)$.
In option $B$,$O$ in $H_2 O_2$ $(-1)$ is oxidized to $O_2$ $(0)$ and reduced to $H_2 O$ $(-2)$.
In option $C$,$P$ in $P_4$ $(0)$ is oxidized to $NaH_2 PO_2$ $(+1)$ and reduced to $PH_3$ $(-3)$.
In option $D$,$P$ in $P_4$ $(0)$ is oxidized to $PCl_3$ $(+3)$ and $S$ in $SOCl_2$ $(+4)$ is reduced to $S_2 Cl_2$ $(+1)$. Since different elements undergo oxidation and reduction,it is a redox reaction but not a disproportionation reaction.

(C) In group $14$,the stability of the $+2$ oxidation state increases down the group due to the inert pair effect. Therefore,$Pb^{2+}$ is the most stable state for lead. Consequently,$Pb^{2+}$ acts as an oxidizing agent because it tends to gain electrons to reach the more stable $Pb^0$ state or remain in the $Pb^{2+}$ state,whereas $Sn^{2+}$ acts as a reducing agent because it tends to lose electrons to reach the more stable $Sn^{4+}$ state. Thus,the statement that lead in $+2$ oxidation state is a reducing agent is incorrect.

(A) The structure of carbon suboxide $(C_3O_2)$ is $O=C=C=C=O$.
In this structure,the two terminal carbon atoms are bonded to oxygen atoms.
Since the oxidation state of oxygen is $-2$,each terminal carbon atom must have an oxidation state of $+2$ to balance the bond.
The central carbon atom is bonded to two other carbon atoms,and since there is no electronegativity difference between identical atoms,its oxidation state is $0$.
Thus,the oxidation states are $+2, 0, +2$.

(A) The equivalent weight is calculated as: $E = \frac{\text{Molar mass}}{n\text{-factor}}$.
$1$. For $Na_2C_2O_4 \cdot 2H_2O$: Molar mass $= 2(23) + 2(12) + 4(16) + 2(18) = 46 + 24 + 64 + 36 = 170 \ g/mol$. Here,$C_2O_4^{2-} \rightarrow 2CO_2 + 2e^-$,so $n$-factor $= 2$. $E = \frac{170}{2} = 85$.
$2$. For $KMnO_4$ in acidic medium: Molar mass $= 39 + 55 + 4(16) = 158 \ g/mol$. $MnO_4^- + 8H^+ + 5e^- \rightarrow Mn^{2+} + 4H_2O$,so $n$-factor $= 5$. $E = \frac{158}{5} = 31.6$.
$3$. For $KMnO_4$ in neutral medium: $MnO_4^- + 2H_2O + 3e^- \rightarrow MnO_2 + 4OH^-$,so $n$-factor $= 3$. $E = \frac{158}{3} \approx 52.67$.
$4$. For $K_2Cr_2O_7$ in acidic medium: Molar mass $= 2(39) + 2(52) + 7(16) = 78 + 104 + 112 = 294 \ g/mol$. $Cr_2O_7^{2-} + 14H^+ + 6e^- \rightarrow 2Cr^{3+} + 7H_2O$,so $n$-factor $= 6$. $E = \frac{294}{6} = 49$.
Comparing the values $(85, 31.6, 52.67, 49)$,the maximum equivalent weight is $85$.

(A) Assertion $(A)$ is true because alkali metals have low ionization enthalpies,meaning their valence electrons are loosely held.
When these metals or their salts are heated in a flame,the energy from the flame is sufficient to excite these valence electrons to higher energy levels.
When these excited electrons return to their ground state,they emit energy in the form of visible light,which gives characteristic flame colours.
Therefore,the reason $(R)$ correctly explains the assertion $(A)$.

(D) Lithium nitrate $(LiNO_3)$ decomposes upon heating to form lithium oxide $(Li_2O)$,nitrogen dioxide $(NO_2)$,and oxygen $(O_2)$.
The balanced chemical equation is:
$4LiNO_3 \rightarrow 2Li_2O + 4NO_2 + O_2$

(B) Lithium nitrate $(LiNO_3)$ is unique among alkali metal nitrates because it decomposes upon heating to form lithium oxide $(Li_2O)$,nitrogen dioxide $(NO_2)$,and oxygen $(O_2)$.
Other alkali metal nitrates like sodium nitrate $(NaNO_3)$ decompose to form the corresponding nitrite $(NaNO_2)$ and oxygen $(O_2)$.
Therefore,the reaction $2 LiNO_3 \rightarrow 2 LiNO_2 + O_2$ is incorrect.

(C) The alkali metals react with oxygen to form different types of oxides depending on their size.
$Li$ forms only the monoxide $(Li_2O)$.
$Na$ forms the peroxide $(Na_2O_2)$.
$K, Rb,$ and $Cs$ have a strong tendency to form superoxides $(MO_2)$ due to the stabilization of the large superoxide anion $(O_2^-)$ by the large alkali metal cations.

(A) Assertion $(A)$: $Be$ and $Mg$ have very high ionization enthalpies due to their small atomic size.
Reason $(R)$: Because of their high ionization enthalpies,the electrons in $Be$ and $Mg$ are too strongly bound to be excited to higher energy levels by the energy available in the Bunsen flame.
Consequently,they do not impart any characteristic colour to the flame.
Therefore,both $(A)$ and $(R)$ are true,and $(R)$ is the correct explanation of $(A)$.

(B) Tranquilizers are chemical compounds used for the treatment of stress,and mild or even severe mental diseases. Examples include $Valium$ and $Serotonin$.
$Heroin$ and $Codeine$ are analgesics.
$Dimetapp$ and $Seldane$ are antihistamines.
$Cimetidine$ and $Ranitidine$ are antacids.

(A) The given structure is of $Chloramphenicol$,which is a broad-spectrum antibiotic.
It is used to treat various bacterial infections.

(B) Dimetapp (brompheniramine) is a synthetic drug that acts as an antihistamine.
Heroin is a narcotic analgesic.
Nardil is an antidepressant drug.
Veronal is a tranquilizer.

(B) Sodium rosinate is added to soap during its manufacture to improve its lathering property. It acts as a lathering agent.

(C) For $K_3[Cr(C_2O_4)_3]$: Oxidation state $= +3$,Coordination number $= 6$,Sum $= 3 + 6 = 9$
For $[Cr(CO)_6]$: Oxidation state $= 0$,Coordination number $= 6$,Sum $= 0 + 6 = 6$
For $K_2[PtCl_6]$: Oxidation state $= +4$,Coordination number $= 6$,Sum $= 4 + 6 = 10$
For $K_4[Fe(CN)_6]$: Oxidation state $= +2$,Coordination number $= 6$,Sum $= 2 + 6 = 8$
Thus,the sum is maximum for $K_2[PtCl_6]$.

(C) In $[Cr(H_2O)_6]Br_2$,$Cr$ is in $+2$ oxidation state $(3d^4)$,which has $4$ unpaired electrons,so it is paramagnetic.
In $Na_4[Fe(CN)_6]$,$Fe$ is in $+2$ oxidation state $(3d^6)$. $CN^-$ is a strong field ligand,causing pairing of electrons,resulting in $0$ unpaired electrons,so it is diamagnetic.
In $[Ni(CO)_4]$,$Ni$ is in $0$ oxidation state $(3d^8 4s^2)$. $CO$ is a strong field ligand,causing pairing of all electrons,resulting in $0$ unpaired electrons,so it is diamagnetic.
In $Na_2[NiCl_4]$,$Ni$ is in $+2$ oxidation state $(3d^8)$. $Cl^-$ is a weak field ligand,resulting in $2$ unpaired electrons,so it is paramagnetic.
Therefore,the incorrect match is $[Ni(CO)_4]$ - Paramagnetic.

(C) According to Werner's theory,the secondary valency of a central metal atom or ion in a coordination complex represents the number of ligands (neutral or negatively charged groups) directly bonded to it.
This is also known as the coordination number.
The primary valency corresponds to the oxidation state of the metal atom or ion.

(A) $[Co(NH_3)_6]^{3+}$: $NH_3$ is a strong field ligand. $Co^{3+}$ is $3d^6$. In octahedral field,it becomes $t_{2g}^6 e_g^0$ $(II)$.
$[CoF_6]^{3-}$: $F^-$ is a weak field ligand. $Co^{3+}$ is $3d^6$. In octahedral field,it becomes $t_{2g}^4 e_g^2$ $(III)$.
$[Ni(CO)_4]$: $CO$ is a strong field ligand. $Ni$ is in $0$ oxidation state $(3d^8 4s^2)$. It forms a tetrahedral complex with configuration $e^4 t_2^4$ (often represented as $t^4 e^6$ in some contexts for tetrahedral splitting) $(IV)$.
$[Fe(CN)_6]^{3-}$: $CN^-$ is a strong field ligand. $Fe^{3+}$ is $3d^5$. In octahedral field,it becomes $t_{2g}^5 e_g^0$ $(I)$.
Therefore,the correct match is $A-II, B-III, C-IV, D-I$.

(C) The melting points of transition metals in a group generally increase as we move down the group due to the increase in the number of unpaired electrons and the strength of metallic bonding.
For the group $6$ elements ($Cr$,$Mo$,$W$),the melting points follow the order $W > Mo > Cr$.

(A) $Mn$ is stable in the $+2$ oxidation state. Thus,$Mn^{3+}$ acts as an oxidizing agent by getting reduced to $Mn^{2+}$ and attaining the stable $d^5$ configuration.
$Cr$ is stable in the $+3$ oxidation state. Thus,$Cr^{2+}$ acts as a reducing agent by getting oxidized to $Cr^{3+}$.

(C) $Mn$ has the electronic configuration $[Ar] 3d^5 4s^2$.
$Mn^{2+}$ has the configuration $[Ar] 3d^5$,which is a half-filled $d$-orbital configuration,making it extra stable.
$Mn^{3+}$ has the configuration $[Ar] 3d^4$.
Due to the extra stability of the half-filled $d^5$ subshell,$Mn^{2+}$ is more stable than $Mn^{3+}$.
Therefore,the statement that $Mn^{3+}$ is more stable than $Mn^{2+}$ is incorrect.

(A) The reaction of copper with dilute nitric acid is given by: $3Cu + 8HNO_3 (\text{dilute}) \rightarrow 3Cu(NO_3)_2 + 2NO + 4H_2O$.
The reaction of copper with concentrated nitric acid is: $Cu + 4HNO_3 (\text{conc.}) \rightarrow Cu(NO_3)_2 + 2NO_2 + 2H_2O$.
The reaction of zinc with dilute nitric acid is: $4Zn + 10HNO_3 (\text{dilute}) \rightarrow 4Zn(NO_3)_2 + 5H_2O + N_2O$.
The reaction of zinc with concentrated nitric acid is: $Zn + 4HNO_3 (\text{conc.}) \rightarrow Zn(NO_3)_2 + 2H_2O + 2NO_2$.
Therefore,the reaction that produces nitrogen $(II)$ oxide $(NO)$ is the reaction between copper and dilute nitric acid.

(C) The thermal decomposition of ammonium dichromate produces nitrogen gas,chromium $(III)$ oxide,and water vapour.
$(NH_4)_2Cr_2O_7 \stackrel{\Delta}{\longrightarrow} N_2 + Cr_2O_3 + 4H_2O$

(D) The ground state electronic configurations are:
$Gd (Z=64) = [Xe] 4f^7 5d^1 6s^2$
$Yb (Z=70) = [Xe] 4f^{14} 6s^2$
For $Gd^{3+}$,three electrons are removed ($1$ from $5d$ and $2$ from $6s$),resulting in $4f^7$. Thus,$x = 7$.
For $Yb^{2+}$,two electrons are removed from $6s$,resulting in $4f^{14}$. Thus,$y = 14$.
The sum $x + y = 7 + 14 = 21$.

(C) Lanthanoids in the $+2$ oxidation state,such as $Eu^{+2}$ and $Yb^{+2}$,act as strong reducing agents.
This is because they tend to lose one electron to acquire the more stable $+3$ oxidation state,which is the most common and stable oxidation state for lanthanoids.

(A) The reduction half-reaction for a chlorine electrode is: $Cl_2(g) + 2e^- \rightarrow 2Cl^-(aq)$.
According to the Nernst equation,the electrode potential $E$ is given by: $E = E^{\circ} - \frac{0.0591}{2} \log \frac{[Cl^-]^2}{P_{Cl_2}}$.
Assuming standard pressure $P_{Cl_2} = 1 \ bar$,the equation becomes: $E = E^{\circ} - 0.0591 \log [Cl^-]$.
To maximize the electrode potential $E$,the term $-0.0591 \log [Cl^-]$ must be as large as possible,which means $\log [Cl^-]$ must be as small as possible (most negative).
This occurs when the concentration $[Cl^-]$ is the lowest among the given options.
Comparing the options,the lowest concentration is $2.5 \times 10^{-3} \ mol \ L^{-1}$.
Therefore,$X = 2.5 \times 10^{-3}$.

(B) The cell reaction is: $A_{(s)} + B^{+}_{(aq)} \rightleftharpoons A^{+}_{(aq)} + B_{(s)}$
Here,$A$ is oxidized to $A^{+}$ $(n=1)$ and $B^{+}$ is reduced to $B$.
$E^{\circ}_{cell} = E^{\circ}_{cathode} - E^{\circ}_{anode} = E^{\circ}_{B^{+}/B} - E^{\circ}_{A^{+}/A} = 0.5 - (-0.5) = 1.0 \ V$
Using the relation: $E^{\circ}_{cell} = \frac{0.06}{n} \log K_C$
Substituting the values: $1.0 = \frac{0.06}{1} \log K_C$
Therefore,$\log K_C = \frac{1.0}{0.06} = \frac{100}{6}$

(A) The anodic reaction is: $2Cl^{-} \rightarrow Cl_2 + 2e^{-}$
The total charge passed is $Q = I \times t = 15.0 \ A \times (45 \times 60) \ s = 40500 \ C$.
The number of moles of electrons passed is $n_{e^-} = \frac{Q}{F} = \frac{40500}{96500} \approx 0.4197 \ mol$.
From the stoichiometry of the reaction,$2 \ mol$ of $e^-$ produce $1 \ mol$ of $Cl_2$. Therefore,$n_{Cl_2} = \frac{n_{e^-}}{2} = \frac{0.4197}{2} \approx 0.20985 \ mol$.
Using the ideal gas law $PV = nRT$ at $1 \ atm$ and $273 \ K$ ($STP$ conditions),the volume is $V = \frac{nRT}{P} = \frac{0.20985 \times 0.082 \times 273}{1} \approx 4.7 \ L$.

(B) The reduction half-reaction is: $Fe^{3+} + e^{-} \rightarrow Fe^{2+}$
Using the Nernst equation:
$E_{Fe^{3+} \mid Fe^{2+}} = E^0_{Fe^{3+} \mid Fe^{2+}} - \frac{2.303 \ RT}{nF} \log \frac{[Fe^{2+}]}{[Fe^{3+}]}$
Here,$n = 1$,$[Fe^{2+}] = 2.0 \ M$,$[Fe^{3+}] = 0.02 \ M$,$\frac{2.303 \ RT}{F} = 0.059 \ V$,and $E^0 = 0.771 \ V$.
Substituting the values:
$E = 0.771 - 0.059 \log \left(\frac{2.0}{0.02}\right)$
$E = 0.771 - 0.059 \log(100)$
$E = 0.771 - 0.059 \times 2$
$E = 0.771 - 0.118 = 0.653 \ V$

(B) The degree of dissociation $(\alpha)$ is given by the formula: $\alpha = \frac{\Lambda_m}{\Lambda_m^\circ}$
First,calculate the molar conductivity $(\Lambda_m)$ using the formula: $\Lambda_m = \frac{\kappa \times 1000}{C}$
Given: $\kappa = 19.5 \times 10^{-5} \ S \ cm^{-1}$ and $C = 0.01 \ mol \ L^{-1}$
$\Lambda_m = \frac{19.5 \times 10^{-5} \times 1000}{0.01} = 19.5 \ S \ cm^2 \ mol^{-1}$
Now,calculate $\alpha$:
$\alpha = \frac{19.5}{390} = 0.05 = 5.0 \times 10^{-2}$

(D) The molar conductivity $\Lambda_{m}$ is related to the degree of dissociation $\alpha$ and limiting molar conductivity $\Lambda_{m}^{\circ}$ by the formula: $\Lambda_{m} = \alpha \times \Lambda_{m}^{\circ}$.
Substituting the given values: $\Lambda_{m} = 0.02 \times 400 = 8.0 \ S \ cm^2 \ mol^{-1}$.
The conductivity $\kappa$ is related to molar conductivity $\Lambda_{m}$ and concentration $C$ (in $mol \ L^{-1}$) by: $\kappa = \frac{\Lambda_{m} \times C}{1000}$.
Substituting the values: $\kappa = \frac{8.0 \times 0.1}{1000} = 0.8 \times 10^{-3} = 8 \times 10^{-4} \ S \ cm^{-1}$.

(A) Benzene-$1,3$-diol consists of a benzene ring with two hydroxyl $(-OH)$ groups attached at the $1$ and $3$ positions.
Common names for dihydroxybenzenes are:
$1,2$-dihydroxybenzene is known as Catechol.
$1,3$-dihydroxybenzene is known as Resorcinol.
$1,4$-dihydroxybenzene is known as Hydroquinol (or Quinol).
Therefore,the correct common name for benzene-$1,3$-diol is Resorcinol.

(B) The reactivity towards $S_{N}1$ reaction depends on the stability of the carbocation formed in the rate-determining step.
In $C_6H_5Cl$,the $C-Cl$ bond has partial double bond character due to resonance,making it extremely difficult to break.
Furthermore,the resulting phenyl cation $(C_6H_5^+)$ is highly unstable because the positive charge is on an $sp$-hybridized carbon atom.
In contrast,the other options form relatively stable carbocations: $C_6H_5CH_2^+$ (benzyl),$CH_2=CH-CH_2^+$ (allyl),and $(C_6H_5)_2CH^+$ (diphenylmethyl).
Therefore,$C_6H_5Cl$ is the least reactive towards $S_{N}1$ reaction.

(D) $pK_a = -\log K_a$. As $K_a$ increases,$pK_a$ decreases.
Benzoic acid $(pK_a \approx 4.2)$ is a stronger acid than phenol $(pK_a \approx 10)$,so the $pK_a$ of benzoic acid is smaller than that of phenol.
Therefore,Assertion $(A)$ is false.
The phenoxide ion is stabilized by non-equivalent resonance structures,while the benzoate ion is stabilized by two equivalent resonance structures,which makes the benzoate ion more stable and benzoic acid a stronger acid.
Thus,Reason $(R)$ is true.

(A) Sphalerite is the sulphide ore of zinc $(Zn)$,which has an atomic number of $30$.
Siderite is the carbonate ore of iron $(Fe)$,which has an atomic number of $26$.
Malachite is the carbonate hydroxide ore of copper $(Cu)$,which has an atomic number of $29$.
Therefore,the atomic numbers of $X$,$Y$,and $Z$ are $30, 26, 29$ respectively.

(A) The given reaction is an example of cyanide leaching,which is used for the extraction of metals like $Ag$ and $Au$ from their ores.
In this process,the ore is treated with a suitable reagent (like $NaCN$ or $KCN$) to dissolve the metal in the form of a complex.
Levigation is a gravity separation process based on the difference in densities of the ore and gangue.
Froth flotation is a physical process used primarily for sulphide ores.
Liquation is a refining process used for metals with low melting points.

(B) $Bauxite$ is the chief ore of $Al$ that is concentrated by the leaching process (Baeyer's process).
$Haematite$ is concentrated by magnetic separation.
$Zinc$ $blende$ and $Calamine$ are concentrated by the froth floatation process.

(C) The correct matches are as follows:
$(A)$ Zone refining is used for the purification of semiconductors like $Ga, Si, Ge$. Thus,$A \rightarrow III$.
$(B)$ Poling is used for the purification of copper $(Cu)$ containing $Cu_2O$ as an impurity. Thus,$B \rightarrow IV$.
$(C)$ Liquation is used for metals with low melting points like tin $(Sn)$. Thus,$C \rightarrow II$.
$(D)$ Vapour phase refining is used for metals like $Ti$ and $Zr$ (Van Arkel method) or $Ni$ (Mond process). Thus,$D \rightarrow I$.
Therefore,the correct matching is $A-III, B-IV, C-II, D-I$.

(A) The Mond process is specifically used for the refining of nickel $(Ni)$.
The van Arkel method is used for the refining of metals like zirconium $(Zr)$ or titanium $(Ti)$.
Zone refining is used for producing semiconductors and other metals of very high purity,such as germanium $(Ge)$,silicon $(Si)$,boron $(B)$,gallium $(Ga)$,and indium $(In)$.
Therefore,$(X) = Ni$,$(Y) = Zr$,and $(Z) = Ga$.

(D) $S_N2$ mechanism involves a single-step attack of the nucleophile and departure of the leaving group.
Reactivity towards $S_N2$ reaction depends on steric hindrance at the $\alpha$-carbon.
As the number of alkyl groups attached to the $\alpha$-carbon increases,steric hindrance increases,making the nucleophilic attack more difficult.
The order of reactivity towards $S_N2$ is: $\text{Primary} > \text{Secondary} > \text{Tertiary}$.
In $(CH_3)_3C-Br$,the carbon attached to the bromine is a tertiary carbon,which is highly sterically hindered,making it the least reactive towards $S_N2$ reaction.

(D) Sodium metal reacts with compounds containing acidic hydrogen atoms (like $-OH$ or $-COOH$ groups) to evolve hydrogen gas.
Phenol $(C_6H_5OH)$,ethanol $(C_2H_5OH)$,and benzoic acid $(C_6H_5COOH)$ all contain acidic hydrogen atoms.
Anisole $(C_6H_5OCH_3)$ is an ether and does not contain any acidic hydrogen atom,therefore it does not react with sodium metal.

(A) The Swarts reaction is a method used for the preparation of alkyl fluorides from alkyl chlorides or alkyl bromides by heating them in the presence of metallic fluorides such as $AgF$,$Hg_2F_2$,$CoF_2$,or $SbF_3$.
The reaction is represented as:
$CH_3-CH_2-Br \xrightarrow{CoF_2} CH_3-CH_2-F$
Therefore,option $A$ represents the Swarts reaction.

(C) The reactivity of alkyl halides towards nucleophilic $S_{N}1$ reaction depends on the stability of the intermediate carbocation formed in the rate-determining step.
Since the stability of carbocations follows the order $3^{\circ} > 2^{\circ} > 1^{\circ}$,the reactivity towards $S_{N}1$ follows the same order.
Therefore,the correct order is $3^{\circ} \text{ halide} > 2^{\circ} \text{ halide} > 1^{\circ} \text{ halide}$.

(B) The alkyl halide $X$ $(C_4H_9Br)$ undergoes $S_N2$ reaction,which implies it must be a primary alkyl halide to minimize steric hindrance. Thus,$X$ is $n$-butyl bromide $(CH_3CH_2CH_2CH_2Br)$.
When $n$-butyl bromide reacts with $Mg$ in dry ether,it forms a Grignard reagent,$n$-butylmagnesium bromide $(CH_3CH_2CH_2CH_2MgBr)$.
Subsequent reaction with $D_2O$ replaces the $MgBr$ group with a deuterium atom $(D)$,resulting in $n$-butane$-1-$d $(CH_3CH_2CH_2CH_2D)$.

(B) The Swarts reaction is a method used for the synthesis of alkyl fluorides by heating alkyl chlorides or alkyl bromides in the presence of a metallic fluoride such as $AgF$,$Hg_2F_2$,$CoF_2$,or $SbF_3$.
This is a classic example of a halogen exchange reaction.
The general reaction is: $H_3C-Br + AgF \rightarrow H_3C-F + AgBr$.

(B) The reaction sequence is as follows:
$1$. Chlorobenzene reacts with $HNO_3$ and $Conc. H_2SO_4$ (nitration) to form a mixture of $o$-nitrochlorobenzene and $p$-nitrochlorobenzene. The major product '$X$' is $p$-nitrochlorobenzene.
$2$. $p$-Nitrochlorobenzene then undergoes nucleophilic aromatic substitution with $NaOH$ at $443 \ K$,followed by acidification with $H^+$,to replace the $-Cl$ group with an $-OH$ group,yielding $p$-nitrophenol ($4$-nitrophenol) as the final product '$Y$'.
Therefore,the correct option is $B$.

(A) $1$. The reaction of acetone $(CH_3)_2C=O$ with $CH_3MgBr$ followed by acidic hydrolysis gives tert-butyl alcohol $(A)$,which is $(CH_3)_3C-OH$.
$2$. Treatment of $(A)$ with $SOCl_2$ leads to the formation of tert-butyl chloride $(B)$,which is $(CH_3)_3C-Cl$.
$3$. The reaction of $(B)$ with $CH_3ONa$ (a strong base) under heating conditions leads to dehydrohalogenation via an $E2$ mechanism,resulting in the formation of isobutylene ($2$-methylpropene) as the major product $(C)$,which is $(CH_3)_2C=CH_2$.

(A) Step $1$: Ozonolysis of $but-2-ene$ $(CH_3-CH=CH-CH_3)$ yields two molecules of acetaldehyde $(CH_3CHO)$. Thus,$X = CH_3CHO$.
Step $2$: Aldol condensation of $2$ molecules of acetaldehyde $(2CH_3CHO)$ in the presence of dilute $NaOH$ followed by heating $(\Delta)$ gives $but-2-enal$ $(CH_3-CH=CH-CHO)$,which is $Z$.
Step $3$: Ozonolysis of $but-2-enal$ $(CH_3-CH=CH-CHO)$ breaks the double bond to form acetaldehyde $(CH_3CHO)$ and glyoxal $(CHO-CHO)$.
Therefore,$A$ and $B$ are $CH_3CHO$ and $CHO-CHO$.

(A) The reaction of aniline with acetic anhydride $(CH_3CO)_2O$ leads to the formation of acetanilide $(A)$. This process is known as acetylation. The $-NHCOCH_3$ group is less activating than the $-NH_2$ group due to the resonance of the lone pair of nitrogen with the carbonyl group. This reduces the electron density on the ring,preventing polybromination. When acetanilide $(A)$ is treated with $Br_2$ in $CH_3COOH$,the bulky acetamido group directs the incoming bromine atom primarily to the para position due to steric hindrance at the ortho position. Thus,the major product $B$ is $p$-bromoacetanilide.

(B) The reaction of $m$-chlorobenzaldehyde with concentrated $NaOH$ is a Cannizzaro reaction.
Aldehydes that do not contain any $\alpha$-hydrogen atoms undergo self-oxidation and reduction (disproportionation) in the presence of a concentrated base.
In this reaction,one molecule of $m$-chlorobenzaldehyde is oxidized to the corresponding salt of the carboxylic acid (sodium $3$-chlorobenzoate),and another molecule is reduced to the corresponding alcohol ($3$-chlorobenzyl alcohol).
The reaction is: $2 \text{ } m\text{-Cl-C}_6\text{H}_4\text{CHO} + \text{NaOH (conc.)}$ $\rightarrow m\text{-Cl-C}_6\text{H}_4\text{CH}_2\text{OH} + m\text{-Cl-C}_6\text{H}_4\text{COONa}$.

(D) The reaction sequence is as follows:
$1$. The first step is the Etard reaction,where toluene is oxidized to benzaldehyde using chromyl chloride $(CrO_2Cl_2)$ in the presence of carbon disulfide $(CS_2)$,followed by hydrolysis with $H_3O^+$.
$2$. The second step is the nitration of benzaldehyde using a nitrating mixture $(HNO_3/H_2SO_4)$.
$3$. The $-CHO$ group is a meta-directing group,so the nitro group $(-NO_2)$ will be introduced at the meta-position relative to the aldehyde group.
$4$. Thus,the final product $X$ is $3$-nitrobenzaldehyde.

(A) In the Etard reaction,chromyl chloride $(CrO_2Cl_2)$ oxidises the methyl group of toluene to a chromium complex in the presence of a solvent like $CS_2$ or $CCl_4$.
This chromium complex,upon subsequent hydrolysis with water $(H_3O^+)$,yields benzaldehyde as the final product.

(B) Nitrogen cannot form pentahalides because it lacks vacant $d$-orbitals in its valence shell to expand its octet. Thus,Assertion $(A)$ is true.
Nitrogen can exhibit a $+5$ oxidation state,for example,in $N_2O_5$ or $HNO_3$. Thus,Reason $(R)$ is true.
However,the inability to form pentahalides is due to the absence of $d$-orbitals,not because it cannot exhibit a $+5$ oxidation state. Therefore,$(R)$ is not the correct explanation of $(A)$.

(D) The order in '$A$' is correct because the bond angle decreases as the electronegativity of the central atom decreases down the group,leading to less repulsion between bond pairs.
The order in '$B$' is correct because the electron density on the central atom decreases as the size of the atom increases,making the lone pair less available for donation.
The order in '$C$' is incorrect because thermal stability decreases down the group due to the increase in bond length and decrease in bond dissociation energy. The correct order is $NH_3 > PH_3 > AsH_3 > SbH_3$.
The order in '$D$' is correct. $NH_3$ has a high boiling point due to intermolecular $H$-bonding,while for the others $(PH_3, AsH_3, SbH_3)$,the boiling point increases down the group due to increasing van der Waals forces (molecular weight).
Thus,the correct orders are $A, B$ and $D$.

(C) Among the allotropes of sulphur,rhombic (yellow) sulphur is more stable than the monoclinic sulphur and is stable at room temperature.
Above $369 \ K$,rhombic sulphur transforms into monoclinic.
$N_2$ is a colourless gas.
$O_3$ is less stable than $O_2$ because its decomposition into $O_2$ is exothermic.
$Cl_2$ is a greenish-yellow gas.

(A) The reaction between sodium nitrite $(NaNO_2)$ and hydrochloric acid $(HCl)$ is as follows:
$2 NaNO_2 + 2 HCl \rightarrow 2 NaCl + NO + NO_2 + H_2O$
In this reaction,the two nitrogen oxides produced are nitric oxide $(NO)$ and nitrogen dioxide $(NO_2)$.

(D) The given reaction is: $P_4 + 3 NaOH + 3 H_2 O \rightarrow PH_3 + 3 NaH_2 PO_2$.
Here,$Q$ is phosphine $(PH_3)$.
Let us analyze the products of the given reactions:
$A$: $Ca_3 P_2 + 6 H_2 O \rightarrow 3 Ca(OH)_2 + 2 PH_3$ (Produces $PH_3$)
$B$: $4 H_3 PO_3 \stackrel{\Delta}{\longrightarrow} 3 H_3 PO_4 + PH_3$ (Produces $PH_3$)
$C$: $PH_4 I + KOH \rightarrow KI + H_2 O + PH_3$ (Produces $PH_3$)
$D$: $PCl_3 + 3 H_2 O \rightarrow H_3 PO_3 + 3 HCl$ (Does not produce $PH_3$)
Therefore,$Q$ is not the product in reaction $D$.

(A) $P_4 + 3 NaOH + 3 H_2 O \xrightarrow{CO_2} 3 NaH_2 PO_2 + PH_3$
$(X = NaH_2 PO_2)$
$NaH_2 PO_2 + HCl_{(aq)} \rightarrow H_3 PO_2 + NaCl$
$(Y = H_3 PO_2)$
$H_3 PO_2$ is hypophosphorous acid or phosphinic acid.
It is a monobasic acid because it has only one $P-OH$ bond.
It acts as a strong reducing agent.
The oxidation state of $P$ in $H_3 PO_2$ is $+1$.
Therefore,the statement that it is called phosphonic acid is incorrect,as phosphonic acid is $H_3 PO_3$.

(D) On moving down the group from $O$ to $Te$,the molecular mass increases,which leads to an increase in the magnitude of van der Waals forces of attraction,causing the boiling point to increase.
However,$H_2O$ exhibits an anomalously high boiling point due to the presence of strong intermolecular hydrogen bonding.
The boiling point order for group $16$ hydrides is: $H_2S < H_2Se < H_2Te < H_2O$.
Therefore,$H_2S$ has the lowest boiling point $(X)$ and $H_2O$ has the highest boiling point $(Y)$.

(C) The $S-O-S$ bond is present in disulphuric acid (also known as pyrosulphuric acid or oleum),which has the chemical formula $H_2S_2O_7$.
In its structure,two $SO_3$ units are linked through an oxygen atom,forming an $S-O-S$ bridge.