TS EAMCET 2019 Chemistry Question Paper with Answer and Solution

(A) The given pair of lines is $x^2+2 \sqrt{2} x y+k y^2=0$. Comparing with $ax^2+2hxy+by^2=0$,we have $a=1, h=\sqrt{2}, b=k$.
The angle $\theta$ between the lines is given by $\tan \theta = \left| \frac{2\sqrt{h^2-ab}}{a+b} \right|$.
Given $\theta = 45^{\circ}$,so $\tan 45^{\circ} = 1 = \left| \frac{2\sqrt{2-k}}{1+k} \right|$.
Squaring both sides: $1 = \frac{4(2-k)}{(1+k)^2}$ $\Rightarrow (1+k)^2 = 8-4k$ $\Rightarrow 1+k^2+2k = 8-4k$ $\Rightarrow k^2+6k-7=0$.
Solving for $k$: $(k+7)(k-1)=0$. Since $k>0$,we get $k=1$.
The pair of lines is $x^2+2\sqrt{2}xy+y^2=0$. The equation of the angle bisectors is $\frac{x^2-y^2}{a-b} = \frac{xy}{h} \Rightarrow \frac{x^2-y^2}{1-1} = \frac{xy}{\sqrt{2}}$,which is not the standard form. Using $\frac{x^2-y^2}{a-b} = \frac{xy}{h}$ is invalid when $a=b$. The correct bisector equation is $h(x^2-y^2) = (a-b)xy$. Since $a=b=1$,we have $0 = \sqrt{2}(x^2-y^2)$,so $x^2-y^2=0$,which gives $x-y=0$ and $x+y=0$.
The triangle is formed by $x-y=0, x+y=0$ and $x+2y+1=0$. The vertices are $(0,0)$,$(-1,0)$,and $(-1/3, -1/3)$.
The area is $\frac{1}{2} |x_1(y_2-y_3) + x_2(y_3-y_1) + x_3(y_1-y_2)| = \frac{1}{2} |0 + (-1)(-1/3-0) + (-1/3)(0-0)| = \frac{1}{2} |1/3| = \frac{1}{6}$.

(C) Let $I = \int_0^{\frac{\pi}{4}} \frac{\cos^2 x}{\cos^2 x + 4\sin^2 x} dx$.
Divide the numerator and denominator by $\cos^2 x$:
$I = \int_0^{\frac{\pi}{4}} \frac{1}{1 + 4\tan^2 x} dx$.
Let $u = \tan x$,then $du = \sec^2 x dx$. Since $\sec^2 x = 1 + \tan^2 x = 1 + u^2$,we have $dx = \frac{du}{1+u^2}$.
When $x = 0, u = 0$. When $x = \frac{\pi}{4}, u = 1$.
$I = \int_0^1 \frac{1}{(1+4u^2)(1+u^2)} du$.
Using partial fractions: $\frac{1}{(1+4u^2)(1+u^2)} = \frac{A}{1+4u^2} + \frac{B}{1+u^2}$.
$1 = A(1+u^2) + B(1+4u^2)$.
For $u^2 = -1, 1 = B(1-4) \Rightarrow B = -\frac{1}{3}$.
For $u^2 = -\frac{1}{4}, 1 = A(1-\frac{1}{4}) \Rightarrow A = \frac{4}{3}$.
$I = \int_0^1 (\frac{4/3}{1+4u^2} - \frac{1/3}{1+u^2}) du = [\frac{4}{3} \cdot \frac{1}{2} \tan^{-1}(2u) - \frac{1}{3} \tan^{-1}(u)]_0^1$.
$I = [\frac{2}{3} \tan^{-1}(2u) - \frac{1}{3} \tan^{-1}(u)]_0^1 = \frac{2}{3} \tan^{-1}(2) - \frac{1}{3} \tan^{-1}(1) = \frac{2}{3} \tan^{-1}(2) - \frac{1}{3} \cdot \frac{\pi}{4} = \frac{2}{3} \tan^{-1}(2) - \frac{\pi}{12}$.

(B) For a given principal quantum number $n=4$,the possible subshells are $4s$,$4p$,$4d$,and $4f$.
Each subshell contains at least one orbital with $m_l=0$ (specifically the $ns$,$np_z$,$nd_{z^2}$,and $nf_{z^3}$ orbitals).
There are $4$ such orbitals in the $n=4$ shell that have $m_l=0$.
Since each orbital can hold only one electron with a specific spin $m_s=\frac{1}{2}$,the total number of electrons is $4 \times 1 = 4$.

(A) The compound '$A$' must contain an alkene group to decolourise $Br_2 / CCl_4$ and a primary amine group $(-NH_2)$ to release $N_2$ gas upon reaction with nitrous acid $(HNO_2)$.
$1$. Reaction with $HNO_2$: Primary aliphatic amines react with $HNO_2$ to form unstable diazonium salts,which decompose to release $N_2$ gas and form alcohols.
$2$. Reaction with $Br_2 / CCl_4$: The presence of a carbon-carbon double bond $(C=C)$ leads to the addition of bromine,which causes the reddish-brown colour of $Br_2$ to disappear.
Among the given options,Cyclopent$-3-$en$-1-$amine contains both a primary amine group and a double bond,satisfying both conditions.

(A)

Ionic species	Steric Number/Hybridisation/Geometry
$BH_4^{-}$	$4, sp^3$,Tetrahedral
$NH_2^{-}$	$4, sp^3$,$V$-shaped
$CO_3^{2-}$	$3, sp^2$,Trigonal Planar
$H_3O^{+}$	$4, sp^3$,Pyramidal

The central atom in $BH_4^{-}$ is $B$ (Boron). The steric number is $4$ ($4$ bond pairs,$0$ lone pairs),which corresponds to $sp^3$ hybridisation and a Tetrahedral geometry. Thus,$BH_4^{-}$ is the correct answer.

(D) $XeF_4$ has $4$ bond pairs and $2$ lone pairs of electrons on $Xe$,resulting in $sp^3d^2$ hybridisation and a square planar geometry.
$XeF_6$ has $6$ bond pairs and $1$ lone pair of electrons on $Xe$,resulting in $sp^3d^3$ hybridisation and a distorted octahedral geometry due to the presence of the lone pair.
Thus,option $D$ is correct.

(A) . Trigonal planar geometry is observed in $BF_3$ $(III)$.
$B$. Tetrahedral geometry is observed in $CCl_4$ $(IV)$.
$C$. Trigonal bipyramidal geometry is observed in $PCl_5$ $(I)$.
$D$. Octahedral geometry is observed in $SF_6$ $(II)$.
Therefore,the correct match is $A-III, B-IV, C-I, D-II$.

(B) To determine the linear structure,we analyze the hybridization and lone pairs on the central atom:
$I. SnCl_2$: $Sn$ has $4$ valence electrons. It forms $2$ sigma bonds and has $1$ lone pair. Due to the lone pair,it has a bent ($V$-shaped) structure.
$II. BeF_2$: $Be$ has $2$ valence electrons,both involved in bonding with $F$ atoms. It has no lone pair and is $sp$ hybridized,resulting in a linear structure.
$III. SO_2$: $S$ has $6$ valence electrons. It forms $2$ sigma bonds and $2$ pi bonds with $O$ atoms,leaving $1$ lone pair on $S$. This results in a bent structure.
$IV. NO_2^+$: The nitronium ion $(NO_2^+)$ has $N$ as the central atom with $4$ valence electrons (after losing one). It forms $2$ double bonds with $O$ atoms. It has no lone pair and is $sp$ hybridized,resulting in a linear structure.
$V. C_2H_2$: Each $C$ atom is $sp$ hybridized,forming $1$ sigma bond with $H$ and $1$ sigma bond with the other $C$ atom,plus $2$ pi bonds. The molecule is linear.
Thus,$II, IV,$ and $V$ have linear structures. The correct option is $(b)$.

(B) According to molecular orbital theory,species with all paired electrons in their electronic configuration are diamagnetic in nature.
$He_2^{+} \rightarrow$ Total $3$ electrons,configuration $\sigma 1s^2, \sigma^* 1s^1$. It contains one unpaired electron and is paramagnetic.
$H_2 \rightarrow$ Total $2$ electrons,configuration $\sigma 1s^2$. All electrons are paired,hence it is diamagnetic.
$H_2^{+} \rightarrow$ Total $1$ electron,configuration $\sigma 1s^1$. It contains one unpaired electron and is paramagnetic.
$H_2^{-} \rightarrow$ Total $3$ electrons,configuration $\sigma 1s^2, \sigma^* 1s^1$. It contains one unpaired electron and is paramagnetic.
$He \rightarrow$ Total $2$ electrons,configuration $\sigma 1s^2$. All electrons are paired,hence it is diamagnetic.
Therefore,$H_2$ and $He$ are diamagnetic. The total count is $2$.

(B) Bond order is calculated using the formula: $\text{Bond Order} = \frac{N_b - N_a}{2}$,where $N_b$ is the number of bonding electrons and $N_a$ is the number of antibonding electrons.

Species	Bond Order
$N_2^{-}$	$2.5$
$NO^{-}$	$2$
$C_2^{-}$	$2.5$
$N_2^{+}$	$2.5$
$C_2^{2-}$	$3$
$CN^{+}$	$2$

From the table,the ions with a bond order of $2.5$ are $N_2^{-}$,$C_2^{-}$,and $N_2^{+}$.
Therefore,there are $3$ such ions.

(A) The bond order is calculated using the formula: $\text{Bond order} = \frac{1}{2} (N_b - N_a)$,where $N_b$ is the number of bonding electrons and $N_a$ is the number of antibonding electrons.
$A. N_2^+$ ($13$ electrons): $\sigma 1s^2 < \sigma^* 1s^2 < \sigma 2s^2 < \sigma^* 2s^2 < \pi 2p_x^2 = \pi 2p_y^2 < \sigma 2p_z^1$. Bond order = $\frac{9-4}{2} = 2.5$ $(IV)$.
$B. CO$ ($14$ electrons): $\sigma 1s^2 < \sigma^* 1s^2 < \sigma 2s^2 < \sigma^* 2s^2 < \sigma 2p_z^2 < \pi 2p_x^2 = \pi 2p_y^2$. Bond order = $\frac{10-4}{2} = 3.0$ $(V)$.
$C. O_2$ ($16$ electrons): $\sigma 1s^2 < \sigma^* 1s^2 < \sigma 2s^2 < \sigma^* 2s^2 < \sigma 2p_z^2 < \pi 2p_x^2 = \pi 2p_y^2 < \pi^* 2p_x^1 = \pi^* 2p_y^1$. Bond order = $\frac{10-6}{2} = 2.0$ $(III)$.
$D. O_2^-$ ($17$ electrons): $\sigma 1s^2 < \sigma^* 1s^2 < \sigma 2s^2 < \sigma^* 2s^2 < \sigma 2p_z^2 < \pi 2p_x^2 = \pi 2p_y^2 < \pi^* 2p_x^2 = \pi^* 2p_y^1$. Bond order = $\frac{10-7}{2} = 1.5$ $(II)$.
Therefore,the correct match is $A-IV, B-V, C-III, D-II$.

(D) $LiCl$ exhibits both ionic and covalent character.
$LiCl$ is ionic because $Li$ is a metal and $Cl$ is a non-metal,resulting in an electrostatic force of attraction between the positive $(Li^{+})$ and negative $(Cl^{-})$ ions.
$LiCl$ also exhibits covalent character because the small size of the $Li^{+}$ cation causes significant polarization of the larger $Cl^{-}$ anion,as described by Fajan's rule.
Therefore,the bonding in $LiCl$ is both ionic and covalent.

(B) The dipole moments of the given compounds are as follows:
$(a)$ For $HBr$,$\mu = 0.82 \ D$.
$(b)$ For acetone $(CH_3COCH_3)$,the electron-releasing methyl groups increase the electron density towards the oxygen atom,resulting in a net dipole moment of $\mu_{net} = 2.69 \ D$.
$(c)$ For $H_2S$,$\mu_{net} = 0.97 \ D$.
$(d)$ For $COCl_2$,$\mu_{net} = 1.17 \ D$.
Comparing these values,acetone $(CH_3COCH_3)$ has the highest dipole moment.

(C) $BF_3$ has a zero dipole moment due to its symmetrical trigonal planar geometry.
In $NH_3$,$N$ is more electronegative than $H$. The bond dipoles of the three $N-H$ bonds and the lone pair on the nitrogen atom are in the same direction,which reinforces each other,resulting in a high net dipole moment.
In $NF_3$,$F$ is more electronegative than $N$. The bond dipoles of the three $N-F$ bonds are in the opposite direction to the dipole moment of the lone pair on the nitrogen atom,which partially cancels out the net dipole moment.
Therefore,the dipole moment of $NH_3$ is greater than that of $NF_3$,and $BF_3$ has the lowest (zero) dipole moment.
The correct order is $NH_3 > NF_3 > BF_3$.

(D) Key Idea: The strength of hydrogen bonding is directly proportional to the electronegativity difference between the $H$-atom and the atom to which it is covalently bonded,as well as the electronegativity of the atom forming the hydrogen bond.
Since the electronegativity of $F$ is the highest among $N$,$O$,and $F$,the $H-F$ bond is the most polar,and the interaction in $F-H \cdots F$ is the strongest.
Therefore,option $D$ is the correct answer.

(A) The reaction is $0.5 C_{(s)} + 0.5 CO_{2(g)} \rightleftharpoons CO_{(g)}$.
Let the initial moles of $CO_{2(g)}$ be $1 \ mol$.
At equilibrium,$50 \%$ of $CO_{2(g)}$ is converted,so $0.5 \ mol$ of $CO_{2(g)}$ remains.
The amount of $CO_{(g)}$ formed is $1 \ mol$ (based on stoichiometry: $0.5 \ mol$ of $CO_2$ produces $1 \ mol$ of $CO$).
Total moles at equilibrium = $0.5 \ (CO_2) + 1 \ (CO) = 1.5 \ mol$.
Mole fraction of $CO_2$ $(x_{CO_2})$ = $0.5 / 1.5 = 1/3$.
Mole fraction of $CO$ $(x_{CO})$ = $1 / 1.5 = 2/3$.
Partial pressure $P_{CO_2} = (1/3) \times 12 \ atm = 4 \ atm$.
Partial pressure $P_{CO} = (2/3) \times 12 \ atm = 8 \ atm$.
$K_p = \frac{P_{CO}}{(P_{CO_2})^{0.5}} = \frac{8}{(4)^{0.5}} = \frac{8}{2} = 4 \ atm^{0.5}$.
Thus,the value of $K_p$ is $4$.

(B) The reaction is: $CH_3NH_2 + HCl \rightarrow CH_3NH_3^+ + Cl^-$
Initial moles: $0.1 \ mol$ of $CH_3NH_2$ and $0.08 \ mol$ of $HCl$.
After reaction: $0.02 \ mol$ of $CH_3NH_2$ remains and $0.08 \ mol$ of $CH_3NH_3Cl$ is formed.
This forms a basic buffer solution.
The $pOH$ is calculated using the Henderson-Hasselbalch equation: $pOH = pK_b + \log \left( \frac{[Salt]}{[Base]} \right)$.
$pK_b = -\log(5 \times 10^{-4}) = 4 - \log 5 = 4 - 0.7 = 3.3$.
$pOH = 3.3 + \log \left( \frac{0.08}{0.02} \right) = 3.3 + \log(4) = 3.3 + 0.6 = 3.9$.

(B) The given reaction is $H_2 + I_2 \rightleftharpoons 2 HI$.
The reverse reaction is $2 HI \rightleftharpoons H_2 + I_2$.
The equilibrium constant for the reverse reaction is given by $K = \frac{[H_2][I_2]}{[HI]^2}$.
Given concentrations are $[H_2] = 1.14 \times 10^{-2} \ mol \ L^{-1}$,$[I_2] = 0.12 \times 10^{-2} \ mol \ L^{-1}$,and $[HI] = 2.52 \times 10^{-2} \ mol \ L^{-1}$.
Substituting these values into the expression:
$K = \frac{(1.14 \times 10^{-2}) \times (0.12 \times 10^{-2})}{(2.52 \times 10^{-2})^2}$
$K = \frac{0.1368 \times 10^{-4}}{6.3504 \times 10^{-4}}$
$K \approx 0.02154$.
Rounding to two significant figures,$K = 0.021$.

(D) The relationship between the equilibrium constant $K_p$ and temperature $T$ is given by the van 't Hoff equation:
$\ln K_p = -\frac{\Delta H^\circ}{R} \cdot \frac{1}{T} + \frac{\Delta S^\circ}{R}$
where $\Delta H^\circ$ is the standard enthalpy change of the reaction,$R$ is the gas constant,and $\Delta S^\circ$ is the standard entropy change.
For an endothermic reaction,$\Delta H^\circ > 0$.
Comparing this with the linear equation $y = mx + c$,where $y = \ln K_p$ and $x = \frac{1}{T}$,the slope $m = -\frac{\Delta H^\circ}{R}$.
Since $\Delta H^\circ > 0$ and $R > 0$,the slope $m = -\frac{\Delta H^\circ}{R}$ must be negative.
Therefore,the plot of $\ln K_p$ versus $\frac{1}{T}$ for an endothermic reaction is a straight line with a negative slope.
Thus,option $(d)$ is correct.

(A) The element with $Z=120$ is a hypothetical element with the $IUPAC$ name $\text{unbinilium}$.
Its electronic configuration is $[Og] 8s^2$,where $[Og]$ represents the configuration of Oganesson $(Z=118)$.
Since it has two electrons in its outermost $s$-orbital $(8s^2)$,it belongs to group $2$ of the $s$-block in the periodic table.
Therefore,option $(A)$ is the correct answer.

(A) For isoelectronic species,the ionic radius decreases as the nuclear charge (number of protons) increases because the electrons are more strongly attracted to the nucleus.

$Species$	$N^{3-}, O^{2-}, Na^{+}, Mg^{2+}$
$Number$ of electrons	$10, 10, 10, 10$
$Number$ of protons	$7, 8, 11, 12$

As the number of protons increases,the effective nuclear charge increases,leading to a smaller ionic radius.
The order of ionic radius is: $Mg^{2+} < Na^{+} < O^{2-} < N^{3-}$.
Thus,$Mg^{2+}$ has the least ionic radius and $N^{3-}$ has the highest ionic radius.

(A) All the given species $(S^{2-}, P^{3-}, Cl^{-}, Ca^{2+}, Ar, K^{+})$ are isoelectronic,meaning they all contain $18$ electrons.
For isoelectronic species,the ionic radius decreases as the nuclear charge (atomic number,$Z$) increases.
The atomic numbers are: $P (Z=15), S (Z=16), Cl (Z=17), Ar (Z=18), K (Z=19), Ca (Z=20)$.
Since the number of electrons is constant $(18)$,the species with the lowest atomic number will have the weakest nuclear attraction for the electrons,resulting in the largest radius.
Therefore,the decreasing order of radius is: $P^{3-} > S^{2-} > Cl^{-} > Ar > K^{+} > Ca^{2+}$.

(B) When an atom loses one or more electrons to form a cation,the number of protons remains the same,but the number of electrons decreases. This results in an increased effective nuclear charge,which leads to a stronger electrostatic attraction between the nucleus and the remaining electrons,pulling them closer and causing the ionic radius to be smaller than the atomic radius.

(C) The ionic radii of isoelectronic species decrease as the nuclear charge increases.
For the given ions,$K^{+}$ ($18$ electrons),$Li^{+}$ ($2$ electrons),$Mg^{2+}$ ($10$ electrons),and $Al^{3+}$ ($10$ electrons) have different electronic configurations.
Comparing them based on their positions in the periodic table:
$K^{+}$ is in the $4^{th}$ period,so it has the largest radius.
$Li^{+}$ is in the $2^{nd}$ period.
$Mg^{2+}$ and $Al^{3+}$ are isoelectronic ($10$ electrons each) and belong to the $3^{rd}$ period.
For isoelectronic species,the ionic radius decreases as the atomic number $(Z)$ increases: $Mg^{2+}$ $(Z=12)$ > $Al^{3+}$ $(Z=13)$.
Thus,the correct order is $K^{+} > Li^{+} > Mg^{2+} > Al^{3+}$.

(D) The explanation of the given statements is as follows:
$I$. Both beryllium $(Be)$ and barium $(Ba)$ belong to the same group (group-$2$). As we move down the group,the size of the atom/ion increases,and the covalent character decreases. Thus,$Ba$ compounds are generally less covalent than $Be$ compounds. Therefore,statement $I$ is not correct.
$II$. The electron gain enthalpy of $He$ is positive due to its very small size and fully-filled electronic configuration $(1s^2)$. Hence,statement $II$ is correct.
$III$. The oxidation state of $O$ in $OF_2$ is $+2$,whereas the oxidation state of $O$ in $Na_2O$ is $-2$. Hence,statement $III$ is not correct.
$IV$. Both $Na^{+}$ and $F^{-}$ are isoelectronic species (both have $10$ electrons). However,$Na^{+}$ has $11$ protons while $F^{-}$ has $9$ protons. Due to the higher nuclear charge in $Na^{+}$,it exerts a greater attractive force on the electrons,making its radius smaller than that of $F^{-}$. Hence,statement $IV$ is correct.
Since statements $I$ and $III$ are not correct,the correct option is $(D)$.

(B) In the structure of $CuSO_4 \cdot 5 H_2 O$,there are $5$ water molecules present as water of crystallization.
Out of these $5 H_2 O$ molecules,$4 H_2 O$ molecules are directly coordinated to the $Cu^{2+}$ ion through coordinate covalent bonds.
The fifth $H_2 O$ molecule is held in the crystal lattice by hydrogen bonding between the $SO_4^{2-}$ ion and the coordinated water molecules.
Therefore,only $1$ water molecule is hydrogen bonded.
Thus,option $(b)$ is the correct answer.

(A) $1$. Identify the principal functional group: The carboxylic acid group $(-COOH)$ has the highest priority,so the parent chain is a heptanoic acid derivative,and the carbon of $-COOH$ is $C-1$.
$2$. Number the chain: Starting from the carboxylic acid carbon as $C-1$,the chain is numbered as follows: $C-1$ $(-COOH)$,$C-2$ $(-CHBr)$,$C-3$ $(-CH_2)$,$C-4$ $(-CH_2)$,$C-5$ $(C=O)$,$C-6$ $(-CH_2)$,$C-7$ $(-CH_2OH)$.
$3$. Identify substituents and their positions: There is a bromo group at $C-2$,an oxo group at $C-5$,and a hydroxy group at $C-7$.
$4$. Assemble the name: Alphabetical order for substituents is Bromo,Hydroxy,Oxo. Thus,the name is $2-$Bromo$-7-$hydroxy$-5-$oxoheptanoic acid.

(D) The irritant red haze in traffic and congested places is due to the oxides of nitrogen $(NO_x)$.
Nitrogen dioxide $(NO_2)$ is a poisonous gas with a reddish-brown color.
When it is mixed with aerosols in the atmosphere,it causes a reddish-brown haze in congested areas.
Thus,option $(D)$ is correct.

(C) The reaction sequence shows that $NO$ and $NO_2$ participate in the catalytic cycle that maintains the concentration of $O_3$ in the atmosphere.
The formation of formaldehyde $(H_2C=O)$ from methane $(CH_4)$ is directly dependent on the concentration of ozone $(O_3)$.
Since the concentration of $O_3$ is regulated by the $NO_x$ cycle ($NO$ and $NO_2$),the overall formation of formaldehyde is controlled by the presence of $O_3, NO$ and $NO_2$.
Therefore,option $(C)$ is correct.

(D) The enamel of teeth is composed of hydroxyapatite,which has the formula $Ca_{10}(PO_4)_6(OH)_2$.
When it reacts with $F^{-}$ ions,the $OH^{-}$ ions are replaced by $F^{-}$ ions to form fluorapatite:
$Ca_{10}(PO_4)_6(OH)_2 + 2 F^{-} \longrightarrow Ca_{10}(PO_4)_6 F_2 + 2 OH^{-}$
This product,$Ca_{10}(PO_4)_6 F_2$,can be represented as $3[Ca_3(PO_4)_2] \cdot CaF_2$.
Therefore,option $(D)$ is the correct answer.

(D) The correct matches are as follows:
$(A)$ Bleaching of paper $\rightarrow$ $(II)$ $H_2O_2$ is used as a bleaching agent for paper.
$(B)$ Eye irritant $\rightarrow$ $(IV)$ $PAN$ (Peroxyacetyl nitrate) is a well-known eye irritant.
$(C)$ Freons $\rightarrow$ $(I)$ $CF_2Cl_2$ (Dichlorodifluoromethane) is a common type of freon.
$(D)$ Herbicide $\rightarrow$ $(III)$ $Na_2AsO_3$ (Sodium arsenite) is used as an herbicide.
Therefore,the correct sequence is $A-II, B-IV, C-I, D-III$.
Hence,option $(D)$ is the correct answer.

(B) In the $cis-isomer$,the methyl groups are on the same side of the double bond,resulting in a net dipole moment.
In the $trans-isomer$,the methyl groups are on opposite sides,causing the individual bond dipoles to cancel each other out,resulting in a net dipole moment of $0.00 \ D$.
Due to the higher polarity,the $cis-isomer$ has stronger intermolecular dipole-dipole interactions compared to the $trans-isomer$,leading to a higher boiling point.
The boiling point of $cis-but-2-ene$ is $277 \ K$ and the dipole moment of $trans-but-2-ene$ is $0.00 \ D$.

(A) Non-charged molecular species are the most stable. Therefore,structure $(I)$ has the maximum stability.
In the charged resonating structures $(II)$ and $(III)$,the stability depends on the distance between the positive and negative charges.
Structure $(II)$ has the negative charge at the ortho position relative to the positive charge,while structure $(III)$ has the negative charge at the para position.
Since the distance between the opposite charges is smaller in $(II)$ than in $(III)$,the electrostatic attraction is stronger in $(II)$.
Thus,the stability order is $I > II > III$.

(A) Groups that exhibit $-R$ effect (electron-withdrawing groups) decrease the electron density on the benzene ring. These include: $-CN, -NO_2, -CHO$.
Groups that exhibit $+R$ effect (electron-donating groups) increase the electron density on the benzene ring. These include: $-OR, -NHCOR, -X, -NH_2$ (Note: Halogens like $-X$ exhibit $-I$ effect but $+R$ effect).

(A) The acid dissociation constant $(K_a)$ values for the given acids are as follows:
$A$. Phenol $(C_6H_5OH)$: $K_a \approx 1.0 \times 10^{-10}$ (Match $III$)
$B$. Benzoic acid $(C_6H_5COOH)$: $K_a \approx 6.5 \times 10^{-5}$ (Match $IV$)
$C$. Hypochlorous acid $(HClO)$: $K_a \approx 3.0 \times 10^{-8}$ (Match $II$)
$D$. Acetic acid $(CH_3COOH)$: $K_a \approx 1.75 \times 10^{-5}$ (Match $V$)
Thus,the correct matching is $A-III, B-IV, C-II, D-V$.

(A) The stability of polycyclic aromatic hydrocarbons is determined by the resonance energy per ring.
Benzene $(A)$: $36 \ kcal/mol$ per ring.
Naphthalene $(B)$: $61 \ kcal/mol$ for $2$ rings,so $61/2 = 30.5 \ kcal/mol$ per ring.
Anthracene $(C)$: $84 \ kcal/mol$ for $3$ rings,so $84/3 = 28 \ kcal/mol$ per ring.
Phenanthrene $(D)$: $92 \ kcal/mol$ for $3$ rings,so $92/3 = 30.67 \ kcal/mol$ per ring.
Comparing the values: $28 (C) < 30.5 (B) < 30.67 (D) < 36 (A)$.
Therefore,the order of stability is $C < B < D < A$.

(C) No bond resonance (hyperconjugation) depends on the number of $\alpha-H$ atoms.
For but$-1-$ene $(CH_3-CH_2-CH=CH_2)$,the $\alpha$-carbon is the $CH_2$ group attached to the double bond. It has $2$ $\alpha-H$ atoms. Therefore,it gives $2$ no bond resonating structures.
For a $3^{\circ}$ carbocation with methyl $(-CH_3)$,ethyl $(-CH_2CH_3)$,and isobutyl $(-CH_2CH(CH_3)_2)$ groups attached to the cationic carbon,the total number of $\alpha-H$ atoms is calculated as follows:
- From methyl group: $3$ $\alpha-H$
- From ethyl group: $2$ $\alpha-H$
- From isobutyl group: $2$ $\alpha-H$
Total $\alpha-H = 3 + 2 + 2 = 7$.
Thus,it forms $7$ no bond resonating structures.
Hence,option $(C)$ is correct.

(B) Key Idea: Conditions for a species to be aromatic are:
$(I)$ The structure must follow $H$ückel's rule,i.e.,it must have $(4n+2) \pi$-electrons,where $n$ is an integer $(0, 1, 2, 3, \dots)$.
$(II)$ The structure must be a planar cyclic ring.
$(A)$ Cyclopentadienyl anion: It has $6 \pi$-electrons $(n=1)$ and is planar. Thus,it is aromatic.
$(B)$ $1,2-$dihydronaphthalene: It is not fully conjugated and has $8 \pi$-electrons. Thus,it is not aromatic.
$(C)$ Tropylium cation: It has $6 \pi$-electrons $(n=1)$ and is planar. Thus,it is aromatic.
$(D)$ Phenanthrene: It has $14 \pi$-electrons $(n=3)$ and is planar. Thus,it is aromatic.
$(E)$ Cyclopentadienyl cation: It has $4 \pi$-electrons ($n=1$ for $4n$ rule). Thus,it is anti-aromatic (not aromatic).
$(F)$ Cyclopropenyl anion: It has $4 \pi$-electrons ($n=0$ for $4n$ rule). Thus,it is anti-aromatic (not aromatic).
Therefore,species $(B)$,$(E)$,and $(F)$ are not aromatic. Hence,option $(b)$ is the correct answer.

(C) The degree of unsaturation for $C_4H_6$ is $4 - (6/2) + 1 = 2$. The possible isomers are:
$(i)$ $CH_3-CH_2-C \equiv CH$ (but-$1$-yne)
$(ii)$ $CH_3-C \equiv C-CH_3$ (but-$2$-yne)
$(iii)$ $CH_2=CH-CH=CH_2$ (buta-$1,3$-diene)
$(iv)$ $CH_3-CH=C=CH_2$ (buta-$1,2$-diene)
$(v)$ cyclobutene
$(vi)$ $1$-methylcyclopropene
$(vii)$ $3$-methylcyclopropene
$(viii)$ methylenecyclopropane
$(ix)$ bicyclo[$1.1.0$]butane
Thus,there are $9$ possible isomers. Hence,option $(C)$ is correct.

(D) Gasoline is a mixture of hydrocarbons,primarily $C_8H_{18}$,present in crude oil along with many other components.
Since these components have different boiling points,they are separated by the process of fractional distillation.
Hence,option $(D)$ is the correct answer.

(D) The $Carius$ method is the standard analytical technique used for the quantitative estimation of halogens in organic compounds.
In this method,a known mass of the organic compound is heated with fuming $HNO_3$ in the presence of $AgNO_3$ in a hard glass tube known as a $Carius$ tube.
The halogen present in the compound reacts with $AgNO_3$ to form a precipitate of silver halide $(AgX)$,which is then filtered,washed,dried,and weighed to calculate the percentage of the halogen.

(C) Step $1$: Dehydrohalogenation of $2$-bromopropane with alcoholic $KOH$ gives propene $(A)$ as the major product.
$CH_3-CH(Br)-CH_3 \xrightarrow{alc. KOH} CH_3-CH=CH_2 (A) + HBr$
Step $2$: Addition of $HBr$ to propene in the presence of peroxide $( (C_6H_5CO)_2O_2 )$ follows the Anti-Markownikoff rule (Kharasch effect).
$CH_3-CH=CH_2 + HBr \xrightarrow{peroxide} CH_3-CH_2-CH_2-Br (B)$
Thus,the final product $B$ is $1$-bromopropane.
Hence,option $C$ is correct.

(A-IV, B-II, C-III, D-I) . $1,6$-dibromohexane reacts with $Zn$ to undergo intramolecular cyclization to form cyclohexane. Thus,$A-iv$.
$B$. Ethanol $(C_2H_5OH)$ reacts with concentrated $H_2SO_4$ at $443 \ K$ to undergo dehydration to form ethene $(H_2C=CH_2)$. Thus,$B-ii$.
$C$. Propene reacts with $HBr$ in the presence of peroxide (anti-Markovnikov addition) to form $1$-bromopropane $(CH_3-CH_2-CH_2-Br)$. Thus,$C-iii$.
$D$. $1,1$-dibromopropane reacts with $NaNH_2$ (a strong base) to undergo dehydrohalogenation to form propyne $(H_3C-C \equiv CH)$. Thus,$D-i$.
Therefore,the correct matching is $A-iv, B-ii, C-iii, D-i$.

(C) $(I)$ The mixture of $NaOH + CaO$ is called sodalime. When it reacts with the sodium salt of a carboxylic acid,it gives the corresponding alkane. This reaction is called decarboxylation of salts of carboxylic acids: $R-COONa + NaOH \xrightarrow[\Delta]{CaO} R-H + Na_2CO_3$ (Alkane).
$(II)$ This reaction is called Clemmensen reduction. It converts aldehydes and ketones to the corresponding alkane: $CH_3COCH_3 \xrightarrow[conc. HCl]{Zn-Hg} CH_3CH_2CH_3$ (Propane,an alkane).
$(III)$ The reaction of $CH_3C \equiv CCH_3$ with $LiAlH_4$ does not typically reduce alkynes to alkanes; it is generally unreactive toward isolated alkynes,or may lead to complex mixtures depending on conditions,but it is not a standard method for alkane synthesis.
$(IV)$ The reaction of a tertiary alkyl halide like $(CH_3)_3CCl$ with $NaBH_4$ is generally slow or does not occur under standard conditions,as $NaBH_4$ is a mild reducing agent typically used for aldehydes and ketones,not alkyl halides.
Therefore,only reactions $(I)$ and $(II)$ produce alkanes. However,reviewing the options provided,if we re-evaluate the reactivity,$(IV)$ can sometimes undergo reduction to alkanes with specific reagents,but based on standard textbook reactions,$(I)$ and $(II)$ are the primary alkane-producing reactions. Given the options,$(C)$ is the most plausible choice if $(IV)$ is considered to proceed under specific conditions.

(A) In Kolbe's electrolysis,the aqueous solution of sodium or potassium salt of a carboxylic acid is electrolyzed. The reaction is: $2RCOO^{-}Na^{+} + 2H_2O \rightarrow R-R + 2CO_2 + H_2 + 2NaOH$.
At the anode,the carboxylate ion loses an electron to form a radical,which then decarboxylates and dimerizes to form an alkane: $2RCOO^{-}$ $\rightarrow 2RCOO^{\bullet} + 2e^{-}$ $\rightarrow 2R^{\bullet} + 2CO_2$ $\rightarrow R-R + 2CO_2$.
Since the product $R-R$ is formed by the coupling of two identical alkyl groups $(R)$,the resulting hydrocarbon always contains an even number of carbon atoms.
At the cathode,water is reduced to produce hydrogen gas: $2H_2O + 2e^{-} \rightarrow H_2 + 2OH^{-}$.
Therefore,hydrocarbons with an even number of carbon atoms are produced at the anode.
Thus,option $(A)$ is the correct answer.

(A) The hydrocarbon has the molecular formula $C_4H_6$.
Cyclobutene $(C_4H_6)$ reacts with $Br_2$ to decolourise it,reacts with $HBr$ to form bromocyclobutane,and undergoes ozonolysis $(O_3 / Zn + H_2O)$ to produce butane$-1,4-$dial $(C_4H_6O_2)$.
Therefore,the correct structure is cyclobutene.

(A) The given reaction is the oxidative cleavage of cyclopentene to form glutaric acid $(HOOC-(CH_2)_3-COOH)$.
Strong oxidizing agents like hot acidic potassium permanganate $(KMnO_4-H_2SO_4 / \Delta)$ cause the oxidative cleavage of the carbon-carbon double bond in alkenes to produce carboxylic acids.
Thus,option $(A)$ is correct.

(C) The reaction is an oxymercuration-demercuration of an alkene.
$1$. In the first step,$Hg(OAc)_2$ in aqueous $THF$ reacts with the alkene to form an organomercurial intermediate via an electrophilic addition of $Hg(OAc)^+$ followed by the attack of water.
$2$. In the second step,$NaBH_4$ reduces the $C-Hg$ bond to a $C-H$ bond,resulting in the Markovnikov addition of water across the double bond.
$3$. For methylenecyclohexane,the addition of water follows Markovnikov's rule,where the $OH$ group attaches to the more substituted carbon (the tertiary carbon of the ring).
$4$. Thus,the product formed is $1$-methylcyclohexanol.
Hence,option $(C)$ is the correct answer.

(A) Synthesis gas is a mixture of $CO$ and $H_2$.
To produce methanol $(CH_3OH)$,the chemical reaction is: $CO(g) + 2H_2(g) \longrightarrow CH_3OH(g)$.
According to the stoichiometry of the reaction,$1$ mole of $CO$ reacts with $2$ moles of $H_2$.
Therefore,the ratio of $CO:H_2$ in the synthesis gas should be $1: 2$.
Hence,option $(A)$ is correct.

(A) The balanced chemical equation for the reaction between $KMnO_4$ and $H_2O_2$ in an acidic medium is:
$2 \ KMnO_4 + 3 \ H_2SO_4 + 5 \ H_2O_2 \longrightarrow K_2SO_4 + 2 \ MnSO_4 + 8 \ H_2O + 5 \ O_2$
From the stoichiometry,$2 \ \text{moles of } KMnO_4$ react with $5 \ \text{moles of } H_2O_2$.
Given: $n(KMnO_4) = \text{Molarity} \times \text{Volume} = 0.02 \ M \times 1 \ L = 0.02 \ \text{moles}$.
Therefore,$n(H_2O_2) = \frac{5}{2} \times 0.02 = 0.05 \ \text{moles}$.
Mass of $H_2O_2 = 0.05 \ \text{mol} \times 34 \ g/mol = 1.7 \ g$.
$10 \ \text{vol } H_2O_2$ means $1 \ L$ of $H_2O_2$ solution gives $10 \ L$ of $O_2$ at $STP$.
Since $1 \ L$ of $H_2O_2$ contains $30.3 \ g$ of $H_2O_2$ (as $68 \ g$ of $H_2O_2$ gives $22.4 \ L$ of $O_2$,so $10 \ L$ of $O_2$ corresponds to $\frac{68 \times 10}{22.4} \approx 30.35 \ g$),the concentration is $\frac{30.35 \ g}{1000 \ mL}$.
Volume of $H_2O_2$ solution required $= \frac{1.7 \ g}{30.35 \ g/L} \approx 0.056 \ L = 56 \ mL$.

(B) In the given reaction,carboxylic acid $(R-COOH)$ is treated with diborane $(B_2H_6)$ followed by acidic hydrolysis $(H_2O/H^{\oplus})$.
$B_2H_6$ is a selective reducing agent that reduces carboxylic acids to primary alcohols $(R-CH_2OH)$ without affecting other functional groups like esters or halides.
Therefore,the product is $R-CH_2OH$.

(C) The reaction of benzyl phenyl ether with $HI$ proceeds via the protonation of the ether oxygen atom.
Following protonation,the bond between the oxygen and the benzylic carbon breaks to form a stable benzylic carbocation $(C_6H_5CH_2^+)$ and phenol $(C_6H_5OH)$.
The benzylic carbocation is resonance-stabilized.
Finally,the iodide ion $(I^-)$ attacks the benzylic carbocation to form benzyl iodide $(C_6H_5CH_2I)$.
Thus,the major products are benzyl iodide and phenol.
This corresponds to option $C$.

(A) The reaction sequence is as follows:
$1$. $C_2H_2$ (acetylene) when passed through a red hot iron tube undergoes cyclic polymerization to form benzene. Benzene reacts with $CO$ and $HCl$ in the presence of anhydrous $AlCl_3/CuCl$ (Gattermann-Koch reaction) to form $X$ (Benzaldehyde).
$2$. Benzaldehyde $(X)$ reacts with aniline $(Y)$ to form a Schiff's base.
$3$. Benzaldehyde $(X)$ undergoes nitration with $HNO_3/H_2SO_4$ at $273 \ K$ to form $Z$ ($m$-nitrobenzaldehyde) because the $-CHO$ group is meta-directing.
Therefore,$X$ is benzaldehyde,$Y$ is aniline,and $Z$ is $m$-nitrobenzaldehyde. The correct option is $(a)$.

(C) The reaction is an intramolecular benzoin condensation followed by rearrangement.
$CN^-$ acts as a nucleophile and attacks the aldehyde carbonyl group.
This is followed by an intramolecular attack on the ketone carbonyl group.
After hydrolysis and rearrangement,the final product $E$ is formed,which is a cyclic compound as shown in the reaction mechanism.

(C) When benzaldehyde reacts with $NH_2OH$ (hydroxylamine),an oxime is formed.
Further,the oxime undergoes dehydration upon heating with acetic anhydride to yield benzonitrile.
$C_6H_5CHO + NH_2OH \rightarrow C_6H_5CH=NOH + H_2O$
$C_6H_5CH=NOH \xrightarrow{Ac_2O, \Delta} C_6H_5CN + H_2O$
Thus,option $(C)$ is correct.

(C) The reaction of benzaldehyde with concentrated $NaOH$ is a Cannizzaro reaction.
In this reaction,two molecules of a non-enolizable aldehyde undergo base-induced disproportionation to form a primary alcohol and a carboxylic acid salt.
Benzaldehyde,being a non-enolizable aldehyde,reacts with concentrated $NaOH$ to produce benzyl alcohol and sodium benzoate.
Therefore,option $(C)$ is correct.

(B) Step $1$: Aldol condensation of $CH_3CH_2CHO$ (propanal) in the presence of $NaOH$ gives $\beta$-hydroxy aldehyde,which upon heating $( \Delta, H^{\oplus} )$ undergoes dehydration to form an $\alpha,\beta$-unsaturated aldehyde: $CH_3CH_2CH=C(CH_3)CHO$.
Step $2$: The final step involves hydrogenation using $H_2/Ni$ at $573 \ K$. This reagent reduces both the carbon-carbon double bond and the aldehyde group to a primary alcohol.
Thus,the final product $(P)$ is $CH_3CH_2CH_2CH(CH_3)CH_2OH$.

(B) $MnO_2$ is a mild oxidizing agent and does not oxidize ketones like $CH_3COCH_3$ to carboxylic acids.
$(b)$ $Ph-CCl_3$ undergoes hydrolysis with aqueous $NaOH$ followed by acidification to form benzoic acid $(PhCOOH)$. This is a correct reaction.
$(c)$ Alkyl halides react with $Mg$ to form Grignard reagents $(RMgX)$,which react with $CO_2$ followed by hydrolysis to yield carboxylic acids. This is a correct reaction.
$(d)$ Vigorous oxidation of internal alkenes like $CH_3CH=CHCH_3$ with acidified $K_2Cr_2O_7$ cleaves the double bond to form two molecules of acetic acid $(CH_3COOH)$. This is a correct reaction.
Therefore,reactions $(b)$,$(c)$,and $(d)$ produce carboxylic acids.

(B) In a standard $NaCl$ crystal structure,$Na^{+}$ ions occupy the octahedral voids (edge centres and body centre) and $Cl^{-}$ ions occupy the $FCC$ lattice points (corners and face centres).
However,based on the specific distribution provided in the question:
$Na^{+}$ ions at corners $(8 \times \frac{1}{8} = 1)$ and face centres $(6 \times \frac{1}{2} = 3)$,total $Na^{+} = 1 + 3 = 4$.
$Cl^{-}$ ions at edge centres $(12 \times \frac{1}{4} = 3)$ and body centre $(1 \times 1 = 1)$,total $Cl^{-} = 3 + 1 = 4$.
Since there are $4$ $Na^{+}$ ions and $4$ $Cl^{-}$ ions per unit cell,the number of $NaCl$ formula units is $4$.

(C) When $C_6H_5CN$ reacts with $Na / C_2H_5OH$,$C_6H_5CN$ undergoes reduction and the $-C \equiv N$ group changes to $-CH_2-NH_2$. Thus,the product $(P)$ is $C_6H_5CH_2NH_2$ (Benzylamine).
The reaction occurs as follows:
$C_6H_5CN + 4[H] \rightarrow C_6H_5CH_2NH_2$
When $C_6H_5CN$ reacts with $(i)$ $HCl$ (conc. cold) and (ii) $Br_2 / KOH$,it follows the partial hydrolysis of nitrile to amide,followed by the Hoffmann-Bromamide degradation reaction. The final product $(Q)$ is aniline $(C_6H_5NH_2)$.
The whole reaction is as follows:
$C_6H_5CN$ $\xrightarrow{H_2O, OH^{-}} C_6H_5CONH_2$ $\xrightarrow{Br_2 / KOH} C_6H_5NH_2$
Hence,option $(C)$ is the correct answer.

(D) The reaction sequence is as follows:
$1$. $Phthalic \ acid + 2NH_3 \rightarrow Ammonium \ phthalate (X)$
$2$. $Ammonium \ phthalate (X) \xrightarrow{\Delta} Phthalamide (Y) + 2H_2O$
$3$. $Phthalamide (Y) \xrightarrow{\text{heating}} Phthalimide (Z) + NH_3$
Thus,$Z$ is Phthalimide.

(D) The reaction between benzene diazonium chloride and $o$-toluidine ($2$-methylaniline) at $273 \ K$ and $pH \ 4-5$ is an electrophilic aromatic substitution reaction known as coupling reaction.
In this reaction,the diazonium cation acts as an electrophile.
The $-NH_2$ group is a strong activating group and is ortho/para directing.
Due to the steric hindrance caused by the methyl group at the ortho position,the electrophile attacks the para position with respect to the $-NH_2$ group.
Therefore,the major product is $4$-amino-$3$-methylazobenzene.

(D) Reducing sugars are those carbohydrates that contain a free hemiacetal or hemiketal group,which can act as a reducing agent.
$(A)$ Sucrose is a non-reducing sugar because both anomeric carbons are involved in the glycosidic linkage.
$(B)$ Maltose is a reducing sugar as it contains a free hemiacetal group.
$(C)$ Lactose is a reducing sugar as it contains a free hemiacetal group.
$(D)$ Fructose is a reducing sugar because it can tautomerize to glucose in alkaline solution due to the presence of an $\alpha$-hydroxy ketone group.
Therefore,$(B)$,$(C)$,and $(D)$ are reducing sugars.

(A) When glucose is oxidised by a strong oxidising agent like $HNO_3$,both the terminal aldehyde group $(-CHO)$ and the primary alcoholic group $(-CH_2OH)$ are oxidised to carboxylic acid groups $(-COOH)$.
This results in the formation of a dicarboxylic acid known as saccharic acid or glucaric acid.
The reaction is represented as:
$CHO(CHOH)_4 CH_2OH + [O] \xrightarrow{Conc. HNO_3} COOH(CHOH)_4 COOH + H_2O$.

(C) Glucose $(CHO(CHOH)_4CH_2OH)$ on oxidation with concentrated nitric acid $(HNO_3)$ gets oxidized to a dicarboxylic acid,which is known as saccharic acid (glucaric acid).
The reaction is as follows:
$CHO(CHOH)_4CH_2OH \xrightarrow{HNO_3} COOH(CHOH)_4COOH$
Thus,$HNO_3$ is the reagent that oxidizes glucose into saccharic acid.
Hence,option $(C)$ is the correct answer.

(B) Analgesic: Analgesics are medications designed to relieve pain. $Codeine$ is an example of an analgesic.
Tranquilizer: Tranquilizers are chemical compounds used for the treatment of stress and mental diseases. $Phenelzine$ is an example of a tranquilizer.
Antibiotic: Antibiotics are drugs used to treat infections. $Prontosil$ is an example of an antibiotic.
Antihistamine: Antihistamines block the action of histamine during allergic reactions. $Terfenadine$ is an example of an antihistamine.
Therefore,the correct matching is $A-III, B-I, C-IV, D-II$.

(A) Convulsions are the result of abnormal electrical activity in the brain.
Deficiency of vitamin $B_6$ (pyridoxine) is known to cause convulsions,especially in infants.
Therefore,the correct option is $A$.

(A) Gabriel phthalimide synthesis is used for the preparation of aliphatic primary amines without the loss of carbons. It involves nucleophilic substitution $(S_{N}2)$ of alkyl halides by the anion formed by the phthalimide.
Phthalimide on treatment with ethanolic $KOH$ forms potassium salt of phthalimide which on heating with $RX$ followed by either alkaline hydrolysis or hydrazinolysis with hydrazine produces the corresponding $1^{\circ}$ amine.

(D) The reaction of $R-C \equiv N$ with $H_3O^+$ (acidic hydrolysis) typically requires heating to proceed to the carboxylic acid stage. Under mild conditions,it often stops at the amide stage $(R-CONH_2)$. Thus,statement $(A)$ is generally considered incorrect as it implies a simple,mild reaction.
When alkyl cyanide $(R-C \equiv N)$ is hydrolysed in an alkaline aqueous medium,it undergoes hydrolysis to form a carboxylate salt $(R-COO^-Na^+)$ and ammonia $(NH_3)$. Thus,statement $(B)$ is correct.

(A) The reaction sequence is as follows:
$1$. Acetic acid $(CH_3COOH)$ reacts with $NH_3$ followed by heating to form acetamide $(CH_3CONH_2)$.
$2$. Acetamide reacts with benzenesulfonyl chloride $(C_6H_5SO_2Cl)$ in the presence of pyridine to form $N$-acetylbenzenesulfonamide $(CH_3CONHSO_2C_6H_5)$ along with water $(H_2O)$ and hydrogen chloride $(HCl)$.
$3$. The reaction is: $CH_3CONH_2 + C_6H_5SO_2Cl \xrightarrow{Pyridine} CH_3CONHSO_2C_6H_5 + H_2O + HCl$.
$4$. Thus,$X = CH_3CONHSO_2C_6H_5$,$Y = H_2O$,and $Z = HCl$.

(A-IV, B-II, C-I, D-V) To find the correct match,we determine the hybridisation of each compound/ion:
$A. sp^3d$: $PCl_5$ $(IV)$ has $5$ bond pairs and $0$ lone pairs,total $5$ hybrid orbitals,corresponding to $sp^3d$ hybridisation.
$B. sp^3d^2$: $SF_6$ $(II)$ has $6$ bond pairs and $0$ lone pairs,total $6$ hybrid orbitals,corresponding to $sp^3d^2$ hybridisation.
$C. dsp^2$: $[PtCl_4]^{2-}$ $(I)$ has $4$ bond pairs and $0$ lone pairs,total $4$ hybrid orbitals,corresponding to $dsp^2$ hybridisation.
$D. dsp^3$: $ClF_3$ $(V)$ has $3$ bond pairs and $2$ lone pairs,total $5$ hybrid orbitals,which can be $sp^3d$ or $dsp^3$ (depending on the model,but $ClF_3$ is commonly associated with $sp^3d$ geometry). However,based on standard matching options provided in such questions,$D$ matches $V$ $(ClF_3)$.
Thus,the correct matching is: $A-IV, B-II, C-I, D-V$.

(C) The reaction sequence is as follows:
$1$. $N$-acylation of aniline with $(CH_3CO)_2O$ in the presence of pyridine yields acetanilide $(C_6H_5NHCOCH_3)$.
$2$. Nitration of acetanilide using concentrated $HNO_3$ and concentrated $H_2SO_4$ occurs primarily at the para-position due to the steric hindrance of the bulky $-NHCOCH_3$ group,forming $p$-nitroacetanilide.
$3$. Finally,acid-catalyzed hydrolysis $(H_2O/H^+)$ of $p$-nitroacetanilide removes the acetyl group to yield $p$-nitroaniline ($4$-nitroaniline) as the major product.

(D) Gold number is defined as the minimum mass of a protective colloid in $mg$ that prevents the coagulation of $10 \ mL$ of a gold sol when $1 \ mL$ of $10 \% NaCl$ solution is added to it.
Given,the gold number of Haemoglobin $= 0.03$.
This means $0.03 \ mg$ of Haemoglobin is required to protect $10 \ mL$ of gold sol against $1 \ mL$ of $10 \% NaCl$.
For $50 \ mL$ of gold sol,the amount of protective colloid required is $5 \times 0.03 \ mg = 0.15 \ mg$.
Since the amount of $NaCl$ added is $5 \ mL$ (which is $5 \times$ the standard amount),the protection required is proportional to the volume of the gold sol.
Thus,the mass of Haemoglobin required is $0.15 \ mg$.

(C) For a zero-order reaction,the integrated rate equation is given as:
$k = \frac{[R_0] - [R]}{t}$
where $k$ is the rate constant,$[R_0]$ is the initial concentration,and $[R]$ is the concentration at time $t$.
At half-life time,$t = t_{1/2}$,the concentration of the reactant is $[R] = \frac{[R_0]}{2}$.
Substituting these values into the rate equation:
$k = \frac{[R_0] - \frac{[R_0]}{2}}{t_{1/2}}$
Thus,the correct expression is $k = \frac{[R_0] - \frac{1}{2}[R_0]}{t_{1/2}}$.
Hence,option $(C)$ is the correct answer.

(C) The decomposition reaction is $O_{3(g)} \longrightarrow O_{2(g)} + O_{(g)}$.
Initial pressure: $100 \ atm, \ 0, \ 0$.
Pressure after time $t$: $(100-x), \ x, \ x$.
Given: $k = 1.0 \times 10^{-3} \ s^{-1}$,$t = 38.38 \ min = 38.38 \times 60 \ s = 2302.8 \ s$.
Using the first-order rate equation: $k = \frac{2.303}{t} \log \frac{p_i}{p_f}$.
$1.0 \times 10^{-3} = \frac{2.303}{2302.8} \log \frac{100}{100-x}$.
$\log \frac{100}{100-x} = \frac{1.0 \times 10^{-3} \times 2302.8}{2.303} \approx 1$.
$\frac{100}{100-x} = 10^1 = 10$.
$100 = 1000 - 10x \implies 10x = 900 \implies x = 90 \ atm$.
Partial pressure of $O_3 = 100 - 90 = 10 \ atm$.
Partial pressure of $O_2 = 90 \ atm$.
Partial pressure of $O = 90 \ atm$.
Thus,the correct option is $C$.

(B) Given,rate constant $(k)$ for first order $= 0.2303 \ min^{-1}$.
For a first order reaction,the time $(t)$ is given by:
$t = \frac{2.303}{k} \log \left( \frac{a}{a-x} \right)$
where $a$ is the initial concentration and $x$ is the amount reacted.
Let the initial concentration $(a) = 1$.
Given that the fraction completed is $9/10$,so $x = 0.9$.
The remaining concentration $(a-x) = 1 - 0.9 = 0.1$.
Substituting the values into the equation:
$t = \frac{2.303}{0.2303} \log \left( \frac{1}{0.1} \right)$
$t = 10 \log(10)$
Since $\log(10) = 1$,we get $t = 10 \times 1 = 10 \ min$.
Therefore,the correct option is $(b)$.

(C) For most chemical reactions,the rate constant approximately doubles when the temperature is increased by $10^{\circ}C$.
This is known as the temperature coefficient,which is defined as the ratio of rate constants at temperatures differing by $10^{\circ}C$ (usually at $T+10$ and $T$).
Mathematically,$\text{Temperature Coefficient} = \frac{k_{T+10}}{k_T} \approx 2$ to $3$.
In standard textbook problems,this value is typically taken as $2$.

(A) Codeine,$(D)$ morphine,and $(F)$ heroin are classified as opiates or narcotics. These substances are primarily used for pain relief.
$(B)$ Thymine is a nitrogenous base found in $DNA$.
$(C)$ Epinephrine is a hormone and neurotransmitter released during extreme stress.
$(E)$ Thiamine is a vitamin $(Vitamin \ B_1)$.
Therefore,the correct combination of opiates is $(A, D, F)$,which corresponds to option $(A)$.

(D) To obtain the maximum percentage of the terminal alkene (Hoffman product) in a dehydrohalogenation reaction,a bulky base is required.
Steric hindrance prevents the base from abstracting a proton from the more substituted $\beta$-carbon,favoring the abstraction of a proton from the less substituted $\beta$-carbon.
Among the given options,the potassium alkoxide derived from $3$-ethyl-$3$-pentanol,which is $(C_2H_5)_3CO^-K^+$,is the bulkiest base.
Therefore,it will yield the highest percentage of the terminal alkene.

(A) Methacetin ($4$-methoxy-acetanilide) is a derivative of acetanilide and belongs to the class of analgesics and antipyretics.
It acts by reducing fever and relieving pain.
Therefore,it is classified as an antipyretic drug.
Thus,option $(A)$ is correct.

(B) Antioxidants are substances that retard the oxidative degradation of food. They act as sacrificial materials by being more readily oxidized than the food material they protect. They are often used as food additives to prevent rancidity.
Among the given options,$BHT$ (butylated hydroxy toluene) is a well-known antioxidant used in food preservation.
Sulfanilamide is a sulfa drug (antibacterial),Veronal is a barbiturate (tranquilizer),and Saccharin is an artificial sweetener.
Therefore,option $(b)$ is the correct answer.

(B) Saccharin is an artificial sweetener that is about $550$ times sweeter than cane sugar.
Its sodium salt is highly soluble in water.
It is of great value for diabetic patients as it provides no food energy.
It is excreted as such in the urine without being metabolized by the body,and it is considered safe for consumption.
Therefore,statements $(B)$ and $(C)$ are true.

(C) In group $16$,the electron gain enthalpy becomes less negative as we move down the group due to an increase in atomic size.
However,oxygen $(O)$ has an exceptionally low electron gain enthalpy due to its small size and strong inter-electronic repulsions.
Therefore,sulfur $(S)$ has the highest (most negative) electron gain enthalpy,and polonium $(Po)$ or tellurium $(Te)$ depending on the context,but generally,oxygen is the lowest in the group.
Comparing the options,sulfur $(S)$ has the highest and oxygen $(O)$ has the lowest electron gain enthalpy in group $16$.
Thus,option $(C)$ is correct.

(A) In the periodic table,as we move from left to right,the atomic size decreases due to an increase in effective nuclear charge.
For the $4f$-series (lanthanoids),the decrease in atomic size is less pronounced due to the poor shielding effect of $4f$-electrons,which leads to lanthanoid contraction.
Conversely,the $3d$-series elements show a more rapid decrease in atomic size compared to the $4f$-series because $3d$-electrons provide better shielding than $4f$-electrons,but the overall effective nuclear charge increase is more dominant in the $3d$ transition series.
Therefore,both Assertion $(A)$ and Reason $(R)$ are correct,and $(R)$ is the correct explanation of $(A)$.

(A-V, B-II, C-VI, D-IV) Oxidation state of $V$ in $VOCl_2$:
Let oxidation state of $V = x$.
$x + (-2) + 2(-1) = 0$ $\Rightarrow x - 4 = 0$ $\Rightarrow x = +4$.
Thus,$(A)$ matches with $(V)$.
$(B)$ Oxidation state of $Mn$ in $MnO_4^{2-}$:
Let oxidation state of $Mn = x$.
$x + 4(-2) = -2$ $\Rightarrow x - 8 = -2$ $\Rightarrow x = +6$.
Electronic configuration of $Mn^{6+}$ is $[Ar] 3d^1 4s^0$. It has $1$ unpaired electron.
Thus,$(B)$ matches with $(II)$.
$(C)$ Oxidation state of $Ni$ in $[NiCl_4]^{2-}$:
Let oxidation state of $Ni = x$.
$x + 4(-1) = -2 \Rightarrow x = +2$.
Electronic configuration of $Ni^{2+}$ is $[Ar] 3d^8 4s^0$. It has $2$ unpaired electrons.
Thus,$(C)$ matches with $(VI)$.
$(D)$ All lanthanide ions exhibit an oxidation state of $+3$.
Thus,$(D)$ matches with $(IV)$.

(C) Biodegradable polymers are polymers that can be decomposed by microorganisms or enzymes under aerobic or anaerobic conditions.
$A$. Nylon-$6,6$ is a synthetic polyamide and is non-biodegradable.
$B$. Nylon-$6$ is a synthetic polyamide and is non-biodegradable.
$C$. Nylon-$2$-nylon-$6$ is an alternating copolymer of glycine and amino caproic acid,which is a well-known biodegradable polymer.
$D$. Bakelite is a thermosetting phenol-formaldehyde resin and is non-biodegradable.
Therefore,the correct option is $C$.

(C) The given compound is $[Ag(NH_3)_2][Ag(CN)_2]$.
In this coordination entity,the cation is $[Ag(NH_3)_2]^+$ and the anion is $[Ag(CN)_2]^-$.
For the cation $[Ag(NH_3)_2]^+$,the name is diammine silver $(I)$.
For the anion $[Ag(CN)_2]^-$,the name is dicyanoargentate $(I)$.
Combining these,the $IUPAC$ name is diammine silver $(I)$ dicyanoargentate $(I)$.

(D) The name tris(ethane-$1,2$-diamine)cobalt$(III)$ sulphate indicates a coordination complex.
$1$. The central metal is Cobalt $(Co)$ with an oxidation state of $+3$.
$2$. The ligand is ethane-$1,2$-diamine,abbreviated as $en$,which is a neutral bidentate ligand. Since there are three such ligands,the coordination sphere is $[Co(en)_3]^{3+}$.
$3$. The anion is sulphate,$SO_4^{2-}$.
$4$. To balance the charges,we need two $[Co(en)_3]^{3+}$ cations and three $SO_4^{2-}$ anions.
$5$. Thus,the chemical formula is $[Co(en)_3]_2(SO_4)_3$.
Hence,option $D$ is the correct answer.

(B) The crystal field splitting energy,$\Delta_0$,depends on the nature of the ligand and the oxidation state of the metal ion.
Ligands are arranged in the spectrochemical series based on their field strength: $I^{\ominus} < Br^{\ominus} < SCN^{\ominus} < Cl^{\ominus} < F^{\ominus} < OH^{\ominus} < C_2O_4^{2-} < H_2O < NCS^{\ominus} < EDTA^{4-} < NH_3 < en < CN^{\ominus} < CO$.
In the given complexes,the metal ion is $Co^{3+}$ in all cases.
Comparing the ligands: $F^{\ominus}$ (weak field),$C_2O_4^{2-}$ (moderate field),$H_2O$ (moderate field),and $NH_3$ (strong field).
Since $NH_3$ is the strongest ligand among the options provided,the complex $[Co(NH_3)_6]^{3+}$ will have the highest magnitude of $\Delta_0$.
Therefore,the correct option is $B$.

(A) . $Co^{2+} \longrightarrow (V)$ Pink colour
$B$. $Fe^{2+} \longrightarrow (IV)$ Pale green colour
$C$. $Ni^{2+} \longrightarrow (II)$ Dark green colour
$D$. $Cu^{2+} \longrightarrow (III)$ Blue colour
Hence,the correct matching is $A-V, B-IV, C-II, D-III$.
Therefore,option $(A)$ is the correct answer.

(C) The rules for naming a coordination compound are:
$1.$ Name the counterion first: $Ammonium$.
$2.$ Name the ligands in alphabetical order: $diaqua$ $(H_2O)$ and $oxalato$ $(C_2O_4^{2-})$. Since $oxalato$ is a polydentate ligand,we use $bis$ for the prefix.
$3.$ Name the central metal atom: Since the complex is anionic,we add the suffix $-ate$ to $Nickel$,resulting in $nickelate$.
$4.$ Determine the oxidation state of $Ni$: Let $x$ be the oxidation state of $Ni$. $2(+1) + x + 2(-2) + 2(0) = 0 \implies 2 + x - 4 = 0 \implies x = +2$.
$5.$ Combine the parts: $Ammonium$ $diaquabis(oxalato)nickelate(II)$.

(D) $A)$ The reaction of hydrochloric acid with sodium sulphide produces hydrogen sulphide gas: $2HCl(aq) + Na_2S(s) \rightarrow 2NaCl(aq) + H_2S(g) \uparrow$.
$B)$ The reaction of concentrated sulphuric acid with a mixture of sodium chloride and manganese dioxide produces chlorine gas: $MnO_2(s) + 2NaCl(s) + 3H_2SO_4(conc.) \rightarrow 2NaHSO_4(aq) + MnSO_4(aq) + 2H_2O(l) + Cl_2(g) \uparrow$.

(A) To determine the magnetic moment,we calculate the number of unpaired electrons $(n)$ using the formula $\mu = \sqrt{n(n+2)} \ BM$.
$1$. $Ni(CN)_4^{2-}$: $Ni^{2+}$ is $d^8$. $CN^-$ is a strong field ligand,causing pairing. $n = 0$,$\mu = 0 \ BM$.
$2$. $Cu(H_2O)_6^{2+}$: $Cu^{2+}$ is $d^9$. $n = 1$,$\mu = \sqrt{1(1+2)} = \sqrt{3} \approx 1.73 \ BM$.
$3$. $NiCl_4^{2-}$: $Ni^{2+}$ is $d^8$. $Cl^-$ is a weak field ligand,no pairing. $n = 2$,$\mu = \sqrt{2(2+2)} = \sqrt{8} \approx 2.83 \ BM$.
$4$. $Fe(H_2O)_6^{2+}$: $Fe^{2+}$ is $d^6$. $H_2O$ is a weak field ligand. $n = 4$,$\mu = \sqrt{4(4+2)} = \sqrt{24} \approx 4.90 \ BM$.
The increasing order of magnetic moments is $Ni(CN)_4^{2-} < Cu(H_2O)_6^{2+} < NiCl_4^{2-} < Fe(H_2O)_6^{2+}$.

(A) $V^{4+}$ $(3d^1)$,$Ti^{3+}$ $(3d^1)$,and $Ni^{2+}$ $(3d^8)$ have unpaired electrons in their $d$-subshell. Therefore,these ions show $d-d$ transition and exhibit characteristic colours.
$Ti^{4+}$ $(3d^0)$ does not have any unpaired electrons,therefore it is colourless.
Matching the ions with their colours:
$A. V^{4+}$: $IV. Blue$
$B. Ti^{3+}$: $II. Purple$
$C. Ti^{4+}$: $I. Colourless$
$D. Ni^{2+}$: $III. Green$
Thus,the correct matching is $A-IV, B-II, C-I, D-III$.

(A) The common oxidation state of $f$-block elements is $+3$.
$Eu$ $(Z=63)$ has the electronic configuration $[Xe] 4f^7 6s^2$. It shows a stable $+2$ oxidation state due to the stable half-filled $4f^7$ configuration.
$Tb$ $(Z=65)$ has the electronic configuration $[Xe] 4f^9 6s^2$. It shows a stable $+4$ oxidation state due to the stable half-filled $4f^7$ configuration after losing four electrons.
Therefore,the other stable oxidation states for $Eu$ and $Tb$ are $+2$ and $+4$ respectively.
Hence,option $A$ is the correct answer.

(C) Due to lanthanide contraction,the ionic radius of lanthanide ions $(Ln^{3+})$ decreases as the atomic number increases from $Ce$ $(Z=58)$ to $Gd$ $(Z=64)$.
As the size of the $Ln^{3+}$ ion decreases,the covalent character of the $Ln-OH$ bond increases according to Fajan's rule.
Consequently,the basic strength of the hydroxides decreases as we move from $Ce(OH)_3$ to $Gd(OH)_3$.
Therefore,the correct increasing order of basic character is $Gd(OH)_3 < Nd(OH)_3 < Ce(OH)_3$.

(B) The reaction of $AgCl$ with $NH_4OH$ is a complexation reaction.
$AgCl + 2NH_4OH \rightarrow [Ag(NH_3)_2]^{+} + Cl^{-} + 2H_2O$
$AgCl$ is a white-colored,water-insoluble compound.
In the presence of $NH_3$ (from $NH_4OH$),its solubility increases due to the formation of the soluble complex $[Ag(NH_3)_2]Cl$.

(A) reaction is spontaneous if the standard cell potential $E^{\circ}_{cell}$ is positive,which corresponds to a negative Gibbs free energy change $\Delta G^{\circ} = -nFE^{\circ}_{cell}$.
For system $(A)$: $Cu(s) + 2Ag^{+}(aq) \rightarrow Cu^{2+}(aq) + 2Ag(s)$.
$E^{\circ}_{cell} = E^{\circ}_{cathode} - E^{\circ}_{anode} = E^{\circ}_{Ag^{+}/Ag} - E^{\circ}_{Cu^{2+}/Cu} = 0.80 \ V - 0.34 \ V = 0.46 \ V$.
Since $E^{\circ}_{cell} > 0$,the reaction is spontaneous.
For system $(B)$: $Ag(s) + Cu^{2+}(aq) \rightarrow Ag^{+}(aq) + Cu(s)$.
$E^{\circ}_{cell} = E^{\circ}_{Cu^{2+}/Cu} - E^{\circ}_{Ag^{+}/Ag} = 0.34 \ V - 0.80 \ V = -0.46 \ V$.
Since $E^{\circ}_{cell} < 0$,the reaction is non-spontaneous.
Therefore,the copper-silver nitrate interface shows a spontaneous reaction.

(D) The half-cell reaction is: $O_{2(g)} + 4H^+_{(aq)} + 4e^- \rightarrow 2H_2O_{(l)}$
For $HA \rightleftharpoons H^+ + A^-$,$[H^+] = \sqrt{K_a \times C} = \sqrt{10^{-4} \times 1} = 10^{-2} \ M$.
Using the Nernst equation: $E = E^{\circ} - \frac{0.0592}{n} \log \frac{1}{P_{O_2} [H^+]^4}$.
Here,$n = 4$,$P_{O_2} = 1 \ atm$,and $[H^+] = 10^{-2} \ M$.
$E = 1.23 - \frac{0.0592}{4} \log \frac{1}{1 \times (10^{-2})^4}$.
$E = 1.23 - \frac{0.0592}{4} \log (10^8)$.
$E = 1.23 - \frac{0.0592}{4} \times 8$.
$E = 1.23 - 0.0592 \times 2 = 1.23 - 0.1184 = 1.1116 \ V \approx 1.112 \ V$.

(B) The Nernst equation is given by:
$E = E^{\circ} - \frac{R T}{n F} \ln \frac{[P]}{[A]}$
Rearranging the equation:
$E - E^{\circ} = -\frac{R T}{n F} \ln \frac{[P]}{[A]}$
$E - E^{\circ} = \frac{R T}{n F} \ln \frac{[A]}{[P]}$
Multiplying by $\frac{n F}{R T}$:
$\frac{n F}{R T} (E - E^{\circ}) = \ln \frac{[A]}{[P]}$
Taking the exponential of both sides:
$\frac{[A]}{[P]} = \exp \left(\frac{n F}{R T} (E - E^{\circ})\right)$
Thus,option $B$ is the correct answer.