JEE Main 2021 Chemistry Question Paper with Answer and Solution

(B) The reaction of an alkene with $KMnO_4 / H^{+}$ (oxidative cleavage) or ozonolysis ($O_3$ followed by $Zn/H_2O$) breaks the double bond.
Compound $'A'$ $(C_4H_8)$ yields $C_3H_6O$ (acetone) as one of the products.
$2-$Methylpropene $(CH_3-C(CH_3)=CH_2)$ undergoes oxidative cleavage with $KMnO_4 / H^{+}$ to form acetone $(CH_3COCH_3)$ and $CO_2$.
Similarly,ozonolysis of $2-$Methylpropene gives acetone $(CH_3COCH_3)$ and formaldehyde $(HCHO)$.
Since the question specifies that $'A'$ yields $'B'$ $(C_3H_6O)$,$2-$Methylpropene is the correct structure for $'A'$.

(B) Compounds that are more acidic than $H_{2}CO_{3}$ (carbonic acid) will react with $NaHCO_{3}$ to liberate $CO_{2}$.
$(CH_{3})_{3}NH^{+}Cl^{-}$ is a salt of a weak base and a strong acid,making the cation $(CH_{3})_{3}NH^{+}$ acidic enough to react with $HCO_{3}^{-}$:
$(CH_{3})_{3}NH^{+}Cl^{-} + NaHCO_{3} \longrightarrow (CH_{3})_{3}N + NaCl + H_{2}CO_{3}$
$H_{2}CO_{3} \longrightarrow H_{2}O + CO_{2} \uparrow$
Therefore,option $B$ is the correct answer.

(D) The enthalpy of bond dissociation at $298.2 \ K$ for hydrogen $(H_2)$ is approximately $435.88 \ kJ \ mol^{-1}$.
The enthalpy of bond dissociation at $298.2 \ K$ for deuterium $(D_2)$ is approximately $443.35 \ kJ \ mol^{-1}$.
Comparing these values,we find that $E_{H} \approx 435.88$ and $E_{D} \approx 443.35$.
Calculating the difference: $E_{D} - E_{H} = 443.35 - 435.88 = 7.47 \ kJ \ mol^{-1} \approx 7.5 \ kJ \ mol^{-1}$.
Therefore,the relationship is $E_{H} \simeq E_{D} - 7.5$.

(D) Resonating structures must maintain the same number of paired and unpaired electrons and the same atomic positions.
In the given structures,we are looking for the one that violates the rules of resonance,such as exceeding the octet rule for nitrogen or having an invalid charge distribution.
Option $D$ shows a structure where the nitrogen atom is bonded to two oxygen atoms and a carbon atom,and it carries a positive charge. In this structure,the central carbon atom also carries a positive charge. This arrangement is unstable and does not represent a valid resonance contributor for the nitroalkene system because it creates an adjacent positive-positive charge repulsion and violates the stability criteria for resonance structures.

(A) Dry cleaning of clothes typically involves the use of solvents that can dissolve grease and oil stains without damaging the fabric.
$Cl_{2}C=CCl_{2}$ (tetrachloroethene),$CCl_{4}$ (carbon tetrachloride),and liquid $CO_{2}$ are commonly used as dry-cleaning agents.
$H_{2}O_{2}$ (hydrogen peroxide) is primarily used as a bleaching agent in laundry to whiten fabrics,not as a solvent for dry cleaning.

(C) Statement-$I$ is incorrect because $Be(OH)_2$ is amphoteric in nature and dissolves in alkali to form beryllates.
Statement-$II$ is correct because the solubility of alkaline earth metal hydroxides in water increases down the group as the lattice energy decreases more rapidly than the hydration energy.
Therefore,Statement-$I$ is incorrect and Statement-$II$ is correct.

(A) Total energy provided by the source per second $= \frac{1000 \, J}{10 \, s} = 100 \, J/s$.
Energy of one photon $E = \frac{hc}{\lambda} = \frac{6.626 \times 10^{-34} \, Js \times 3 \times 10^8 \, m/s}{400 \times 10^{-9} \, m} = 4.9695 \times 10^{-19} \, J$.
Number of photons incident per second $= \frac{\text{Total energy per second}}{\text{Energy of one photon}} = \frac{100}{4.9695 \times 10^{-19}} \approx 2.012 \times 10^{20}$.
Since one photon ejects one electron,the number of electrons ejected per second is $2.012 \times 10^{20}$.
Comparing this with $x \times 10^{20}$,we get $x \approx 2$.

(C) According to Hess's Law,the enthalpy of sublimation is the sum of the enthalpy of fusion and the enthalpy of vaporization:
$\Delta H_{\text{sub}} = \Delta H_{\text{fus}} + \Delta H_{\text{vap}}$
Given:
$\Delta H_{\text{fus}} = 2.8 \ kJ \ mol^{-1}$
$\Delta H_{\text{vap}} = 98.2 \ kJ \ mol^{-1}$
Therefore:
$\Delta H_{\text{sub}} = 2.8 + 98.2 = 101 \ kJ \ mol^{-1}$

(D) The balanced chemical equation for the combustion of butane is:
$C_{4}H_{10} + \frac{13}{2} O_{2} \longrightarrow 4 CO_{2} + 5 H_{2}O$
Calculate the moles of water produced:
$\text{Moles of } H_{2}O = \frac{72.0 \ g}{18.0 \ g/mol} = 4.0 \ mol$
From the stoichiometry of the reaction,$1 \ mol$ of $C_{4}H_{10}$ produces $5 \ mol$ of $H_{2}O$.
Therefore,the moles of $C_{4}H_{10}$ required to produce $4.0 \ mol$ of $H_{2}O$ is:
$\text{Moles of } C_{4}H_{10} = \frac{1}{5} \times 4.0 = 0.8 \ mol$
The molar mass of butane $(C_{4}H_{10})$ is $(4 \times 12) + (10 \times 1) = 58 \ g/mol$.
Calculate the mass of butane:
$\text{Mass of } C_{4}H_{10} = 0.8 \ mol \times 58 \ g/mol = 46.4 \ g$
Expressing $46.4 \ g$ in the form $.... \times 10^{-1} \ g$:
$46.4 \ g = 464 \times 10^{-1} \ g$
Thus,the nearest integer value is $464$.

(A) The reaction is $A + B \rightleftharpoons 2C$.
Initial concentrations: $[A] = 1 \ M, [B] = 1 \ M, [C] = 1 \ M$.
Let the change in concentration be $x$.
At equilibrium: $[A] = 1 - x, [B] = 1 - x, [C] = 1 + 2x$.
$K_c = \frac{[C]^2}{[A][B]} = \frac{(1+2x)^2}{(1-x)(1-x)} = 100$.
Taking the square root on both sides: $\frac{1+2x}{1-x} = 10$.
$1 + 2x = 10 - 10x$.
$12x = 9$,so $x = \frac{9}{12} = 0.75$.
Equilibrium concentration of $C = 1 + 2x = 1 + 2(0.75) = 1 + 1.5 = 2.5 \ M$.
$2.5 \ M = 25 \times 10^{-1} \ M$.
Thus,$X = 25$.

(C) Given:
Volume $V = 4.00 \times 10^{3} \ m^{3} = 4.00 \times 10^{3} \times 10^{3} \ L = 4.00 \times 10^{6} \ L$
Pressure $P = 1.0 \ atm$
Temperature $T = 300 \ K$
Gas constant $R = 0.083 \ L \ atm \ K^{-1} \ mol^{-1}$
Using the ideal gas equation $PV = nRT$,where $n = \frac{mass (m)}{Molar \ mass (M)}$:
$n = \frac{PV}{RT} = \frac{1.0 \times 4.00 \times 10^{6}}{0.083 \times 300} \ mol$
$n = \frac{4.00 \times 10^{6}}{24.9} \ mol \approx 1.6064 \times 10^{5} \ mol$
Molar mass of $CH_{4} = 12 + 4 \times 1 = 16 \ g \ mol^{-1}$
Mass $m = n \times M = 1.6064 \times 10^{5} \times 16 \ g$
$m = 25.7024 \times 10^{5} \ g$
Comparing with $X \times 10^{5} \ g$,we get $X = 25.7024$.
Rounding to the nearest integer,$X = 26$.

(C) The structure of the molecule is $CH_3-CH_2-CH_2-C \equiv CH$.
To find the number of $\sigma$ bonds,we count all single bonds and one bond from each multiple bond.
- $C-H$ bonds: $3$ (in $CH_3$) + $2$ (in $CH_2$) + $2$ (in $CH_2$) + $1$ (in $CH$) = $8$ $\sigma$ bonds.
- $C-C$ bonds: $1$ $(C-C)$ + $1$ $(C-C)$ + $1$ $(C-C)$ = $3$ $\sigma$ bonds.
- $C \equiv C$ bond: $1$ $\sigma$ bond.
Total $\sigma$ bonds = $8 + 3 + 1 = 12$.

(C) $F^{-}$,$O^{2-}$,and $N^{3-}$ are all isoelectronic species with $10$ electrons each.
In isoelectronic species,the ionic radius increases as the nuclear charge (number of protons) decreases.
The number of protons in $N^{3-}$,$O^{2-}$,and $F^{-}$ are $7$,$8$,and $9$ respectively.
Since $N^{3-}$ has the lowest nuclear charge,it experiences the least nuclear attraction on its valence electrons,resulting in the largest ionic radius.
Therefore,the order of ionic radii is $N^{3-} > O^{2-} > F^{-}$.
Thus,the ionic radius of $N^{3-}$ is bigger than both $O^{2-}$ and $F^{-}$.

(A) $1$. Analyze the number of $\pi$-bonds in each species:
- $CO_{3}^{2-}$: Has one $\pi$-bond (delocalized over three oxygen atoms).
- $O_{2}$: Has one $\pi$-bond (in its ground state,$O=O$).
- $SO_{2}$: Has two $\pi$-bonds (one $\pi$-bond per $S=O$ bond).
- $SO_{3}$: Has three $\pi$-bonds.
$2$. Analyze the number of canonical forms:
- $CO_{3}^{2-}$: Has $3$ equivalent canonical forms.
- $O_{2}$: Does not exhibit resonance in the standard sense of multiple canonical forms for a single Lewis structure in this context.
$3$. Conclusion:
$CO_{3}^{2-}$ is the species that contains one $\pi$-bond and exhibits $3$ canonical forms,which is the maximum among the given options.

(A) According to Molecular Orbital Theory $(MOT)$,the bond order is calculated as: $\text{Bond Order} = \frac{N_b - N_a}{2}$,where $N_b$ is the number of bonding electrons and $N_a$ is the number of antibonding electrons.
For $O_{2}^{+} (15 \ e^-)$: $\text{Bond Order} = \frac{10 - 5}{2} = 2.5$
For $O_{2} (16 \ e^-)$: $\text{Bond Order} = \frac{10 - 6}{2} = 2.0$
For $O_{2}^{-} (17 \ e^-)$: $\text{Bond Order} = \frac{10 - 7}{2} = 1.5$
For $O_{2}^{2-} (18 \ e^-)$: $\text{Bond Order} = \frac{10 - 8}{2} = 1.0$
Thus,the correct sequence is $O_{2}^{+} > O_{2} > O_{2}^{-} > O_{2}^{2-}$.

(A) The acidity of a compound depends on the stability of its conjugate base.
When cyclopentadiene loses a proton $(H^+)$,it forms a cyclopentadienyl anion.
This anion has $6 \pi$ electrons and is cyclic and planar,making it aromatic according to $H$ückel's rule.
Since the conjugate base is aromatic,it is exceptionally stable,which makes cyclopentadiene the strongest acid among the given options.

(C) Transition metals of group $7, 8,$ and $9$ do not form hydrides. This is known as the hydride gap.
Among the given options,$Fe$ (group $8$),$Mn$ (group $7$),and $Co$ (group $9$) belong to the hydride gap.
$Cr$ (group $6$) does not fall into this gap and can form interstitial hydrides.

(A) $Li_2CO_3$ decomposes easily on heating due to the high polarizing power of the small $Li^{+}$ cation,which imparts significant covalent character to the bond.
$(b)$ $NaHCO_3$ (sodium bicarbonate) is used in fire extinguishers because it releases $CO_2$ gas upon heating.
$(c)$ $K^{+}$ ions are the most abundant cations found within the intracellular fluid of cells.
$(d)$ $CsI$ (cesium iodide) has poor water solubility because both $Cs^{+}$ and $I^{-}$ are large ions,resulting in a low lattice energy and low hydration energy,making the dissolution process energetically unfavorable.
Therefore,the correct matching is: $(a)-(iv), (b)-(iii), (c)-(ii), (d)-(i)$.

(D) To find the change in oxidation state,we calculate the oxidation number of the central atom in both the reactant and the product.
$A$: In $C_{2}O_{4}^{2-}$,$2x + 4(-2) = -2$ $\Rightarrow 2x = 6$ $\Rightarrow x = +3$. In $CO_{2}$,$x + 2(-2) = 0 \Rightarrow x = +4$. Change = $|4 - 3| = 1$.
$B$: In $CrO_{4}^{2-}$,$x + 4(-2) = -2 \Rightarrow x = +6$. In $Cr^{3+}$,$x = +3$. Change = $|3 - 6| = 3$.
$C$: In $Cr_{2}O_{7}^{2-}$,$2x + 7(-2) = -2$ $\Rightarrow 2x = 12$ $\Rightarrow x = +6$. In $Cr^{3+}$,$x = +3$. Change per $Cr$ atom = $|3 - 6| = 3$.
$D$: In $MnO_{4}^{-}$,$x + 4(-2) = -1 \Rightarrow x = +7$. In $Mn^{2+}$,$x = +2$. Change = $|2 - 7| = 5$.
Thus,the process with a change of $5$ is $MnO_{4}^{-} \rightarrow Mn^{2+}$.

(B) Statement $I$ is false because $CFCs$ are broken down by powerful $UV$ radiation,not visible radiation,and they release chlorine free radicals,not chlorine gas.
$CF_{2}Cl_{2(g)} \xrightarrow{UV} \dot{Cl}_{(g)} + \dot{C}F_{2}Cl_{(g)}$
$\dot{Cl}_{(g)} + O_{3(g)} \rightarrow \dot{Cl}O_{(g)} + O_{2(g)}$
$\dot{Cl}O_{(g)} + O_{(g)} \rightarrow \dot{Cl}_{(g)} + O_{2(g)}$
Statement $II$ is false because atmospheric ozone reacts with nitric oxide to produce nitrogen dioxide and oxygen,not nitrogen and oxygen gases.
$NO_{(g)} + O_{3(g)} \rightarrow NO_{2(g)} + O_{2(g)}$

(B) $Ba(OH)_{2}$ is a strong base and dissociates completely as: $Ba(OH)_{2} \rightarrow Ba^{2+} + 2OH^{-}$.
Given concentration of $Ba(OH)_{2} = 0.005 \, M$.
Therefore,$[OH^{-}] = 2 \times 0.005 \, M = 0.01 \, M = 10^{-2} \, M$.
At $298 \, K$,the ionic product of water is $K_{w} = [H_{3}O^{+}][OH^{-}] = 10^{-14}$.
Substituting the value of $[OH^{-}]$:
$[H_{3}O^{+}] = \frac{10^{-14}}{10^{-2}} = 10^{-12} \, M$.
Thus,the concentration is $1 \times 10^{-12} \, mol \, L^{-1}$.

(C) The uncertainty in velocity $\Delta v$ is calculated as:
$\Delta v = \frac{0.02}{100} \times 5 \times 10^{6} \ m \ s^{-1} = 10^{3} \ m \ s^{-1}$
According to Heisenberg's uncertainty principle:
$\Delta x \cdot m \cdot \Delta v \geq \frac{h}{4 \pi}$
Substituting the given values:
$\Delta x \times (9.1 \times 10^{-31} \ kg) \times (10^{3} \ m \ s^{-1}) = \frac{6.63 \times 10^{-34} \ J \ s}{4 \times 3.14 \times 9.1 \times 10^{-31} \ kg}$
$\Delta x \times 9.1 \times 10^{-28} = \frac{6.63 \times 10^{-34}}{114.296 \times 10^{-31}}$
$\Delta x = \frac{6.63 \times 10^{-34}}{114.296 \times 10^{-31} \times 9.1 \times 10^{-31}} \approx 5.79 \times 10^{-9} \ m$
Given $\Delta x = x \times 10^{-9} \ m$,we get $x \approx 5.79$. Rounding to the nearest integer,$x = 6$ (Note: Re-evaluating the calculation: $\Delta x = \frac{6.63 \times 10^{-34}}{4 \times 3.14 \times 9.1 \times 10^{-31} \times 10^{3}} = \frac{6.63 \times 10^{-34}}{114.296 \times 10^{-28}} \approx 0.058 \times 10^{-6} = 58 \times 10^{-9} \ m$. Thus,$x = 58$).

(B) The number $0.00340$ contains leading zeros which are not significant.
According to the rules for significant figures,the trailing zeros in a decimal number are significant.
Therefore,the significant figures are $3, 4,$ and $0$.
The total number of significant figures is $3$.

(C) According to Gay-Lussac's Law,for a fixed volume of gas,pressure is directly proportional to temperature: $\frac{P_{1}}{T_{1}} = \frac{P_{2}}{T_{2}}$.
Given: $P_{1} = 300 \ kPa = 300 \times 10^{3} \ Pa$,$T_{1} = 27 + 273 = 300 \ K$,$P_{2} = 1.2 \times 10^{6} \ Pa$.
Substituting the values: $\frac{300 \times 10^{3}}{300} = \frac{1.2 \times 10^{6}}{T_{2}}$.
$1000 = \frac{1.2 \times 10^{6}}{T_{2}} \Rightarrow T_{2} = \frac{1.2 \times 10^{6}}{1000} = 1200 \ K$.
Converting to Celsius: $T(^{\circ} C) = 1200 - 273 = 927^{\circ} C$.

(D) Mass of organic compound = $0.8 \ g$.
Percentage of nitrogen = $42 \ \%$.
Mass of nitrogen = $\frac{42}{100} \times 0.8 = 0.336 \ g$.
Moles of nitrogen = $\frac{0.336}{14} = 0.024 \ mol$.
Since $1 \ mol$ of $N$ produces $1 \ mol$ of $NH_3$,moles of $NH_3 = 0.024 \ mol$.
The reaction is $2NH_3 + H_2SO_4 \rightarrow (NH_4)_2SO_4$.
From the stoichiometry,$2 \ mol$ of $NH_3$ react with $1 \ mol$ of $H_2SO_4$.
Therefore,moles of $H_2SO_4$ required = $\frac{0.024}{2} = 0.012 \ mol$.
Using $M = \frac{n}{V(L)}$,we have $1 = \frac{0.012}{V(L)}$.
$V(L) = 0.012 \ L = 12 \ mL$.

(A) According to the first law of thermodynamics,$\Delta U = q + w$.
Here,the system does work,so $w = -200 \, J$.
The system absorbs heat,so $q = +150 \, J$.
Substituting these values into the equation: $\Delta U = 150 \, J + (-200 \, J) = -50 \, J$.
The magnitude of the change in internal energy is $|\Delta U| = |-50 \, J| = 50 \, J$.

(D) The reaction of $cis-but-2-ene$ with $Br_2$ in $CCl_4$ proceeds via an anti-addition mechanism.
This results in the formation of a racemic mixture of $(2R, 3R)-2,3-dibromobutane$ and $(2S, 3S)-2,3-dibromobutane$.
Since these two enantiomers are stereoisomers,the total number of stereoisomers formed as product $'P'$ is $2$.

(C) Eutrophication is the process where nutrient enrichment leads to excessive growth of algae,which consumes dissolved oxygen during decomposition.
This process leads to a decrease in the oxygen level of water.
Therefore,the statement that eutrophication leads to an increase in the oxygen level is incorrect.

(A) The stability of carbocations depends on factors like resonance,hyperconjugation,and the electronegativity of the carbon atom bearing the positive charge.
$A$ (Benzyl carbocation) is the most stable due to resonance stabilization by the benzene ring.
$C$ $(CH_3-CH_2^+)$ is a primary alkyl carbocation,which is stabilized by the inductive effect $(+I)$ of the methyl group.
$B$ $(CH_2=CH^+)$ is a vinyl carbocation where the positive charge is on an $sp^2$ hybridized carbon,which is more electronegative than an $sp^3$ carbon.
$D$ $(HC \equiv C^+)$ is an acetylenic carbocation where the positive charge is on an $sp$ hybridized carbon,which is the most electronegative among the three $(sp > sp^2 > sp^3)$.
Since a positive charge is less stable on a more electronegative atom,the order of stability for $B, C,$ and $D$ is $C > B > D$.
Combining these,the overall order of stability is $A > C > B > D$.

(D) According to Fajan's rule,the covalent character in an ionic bond increases with the polarizing power of the cation.
Lithium $(Li^+)$ has a very small ionic radius,which results in a high charge density and high polarizing power.
This high polarizing power allows the $Li^+$ ion to distort the electron cloud of the halide anion,thereby introducing significant covalent character into the lithium halides.
Therefore,both Assertion $A$ and Reason $R$ are true,and $R$ is the correct explanation of $A$.

(D) The iodination of alkanes is a reversible reaction because the byproduct $HI$ is a strong reducing agent and reduces the alkyl iodide back to the alkane.
To make the reaction irreversible,the $HI$ must be removed or oxidized as it is formed.
This can be achieved by adding a strong oxidizing agent such as concentrated $HNO_3$ or concentrated $HIO_3$,which oxidizes $HI$ to $I_2$.

(B) $NaOH$ is a strong base,so it is $(ii)$ Basic.
$Be(OH)_{2}$ is amphoteric,so it is $(iii)$ Amphoteric.
$Ca(OH)_{2}$ is a strong base,so it is $(ii)$ Basic.
$B(OH)_{3}$ (or $H_{3}BO_{3}$) is a weak Lewis acid,so it is $(i)$ Acidic.
$Al(OH)_{3}$ is amphoteric,so it is $(iii)$ Amphoteric.
Therefore,the correct matching is: $(a)-(ii), (b)-(iii), (c)-(ii), (d)-(i), (e)-(iii)$.

(B) Staggered and eclipsed conformers of ethane are generated by rotation around the $C-C$ single bond.
These different spatial arrangements are known as conformational isomers or rotamers.
Since they are interconvertible by rotation about a single bond,they are classified as rotamers.

(D) Statement $I$ is true: Rutherford's model could not explain the stability of the atom or the origin of the line spectrum of hydrogen.
Statement $II$ is true: Bohr's model assumes that electrons move in well-defined circular orbits with fixed radii and velocities,which directly contradicts Heisenberg's uncertainty principle,which states that it is impossible to determine both the position and momentum of a subatomic particle simultaneously with absolute precision.
Therefore,both statements are true.

(A) To find the oxidation state of $P$ (let it be $x$):
For $H_{4}P_{2}O_{7}$: $4(+1) + 2x + 7(-2) = 0 \implies 4 + 2x - 14 = 0 \implies 2x = 10 \implies x = +5$.
For $H_{4}P_{2}O_{5}$: $4(+1) + 2x + 5(-2) = 0 \implies 4 + 2x - 10 = 0 \implies 2x = 6 \implies x = +3$.
For $H_{4}P_{2}O_{6}$: $4(+1) + 2x + 6(-2) = 0 \implies 4 + 2x - 12 = 0 \implies 2x = 8 \implies x = +4$.
Thus,the oxidation states are $5, 3$ and $4$ respectively.

(A) In gaseous triethylamine,the nitrogen atom is $sp^3$ hybridized with one lone pair. Due to the repulsion between the bulky ethyl groups,the $C-N-C$ bond angle is increased from the ideal tetrahedral angle of $109.5^\circ$ to approximately $108^\circ$ to $110^\circ$ depending on the specific experimental conditions,but $108^\circ$ is the standard accepted value in this context.

(C) $PCl_{5(g)} \rightleftharpoons PCl_{3(g)} + Cl_{2(g)} \quad K_{c} = 1.844$
$\text{Initial moles at } t = 0: \quad 3.0 \ \text{moles of } PCl_{5}$
$\text{Moles at equilibrium } (t = eq): \quad (3-x) \ \text{for } PCl_{5}, \ x \ \text{for } PCl_{3}, \ x \ \text{for } Cl_{2}$
$\text{Equilibrium constant expression: } K_{c} = \frac{[PCl_{3}][Cl_{2}]}{[PCl_{5}]} = \frac{x^{2}}{3-x} = 1.844$
$x^{2} + 1.844x - 5.532 = 0$
$\text{Using quadratic formula: } x = \frac{-b + \sqrt{b^{2} - 4ac}}{2a} = \frac{-1.844 + \sqrt{(1.844)^{2} + 4(5.532)}}{2}$
$x = \frac{-1.844 + \sqrt{3.400 + 22.128}}{2} = \frac{-1.844 + 5.052}{2} \approx 1.604$
$\text{Moles of } PCl_{5} \text{ at equilibrium} = 3 - 1.604 = 1.396 \ \text{moles}$
$1.396 \ \text{moles} = 1396 \times 10^{-3} \ \text{moles} \approx 1400 \times 10^{-3} \ \text{moles}$

(D) The vaporization process is represented as: $H_{2}O_{(\ell)} \rightleftharpoons H_{2}O_{(g)}$
From the first law of thermodynamics,the relation between enthalpy change and internal energy change is given by: $\Delta H = \Delta U + \Delta n_{g} RT$
Rearranging the equation: $\Delta H - \Delta U = \Delta n_{g} RT$
For $1 \, mol$ of water vaporization,the change in the number of moles of gas is: $\Delta n_{g} = n_{g(product)} - n_{g(reactant)} = 1 - 0 = 1$
Given $T = 100^{\circ} C = 373 \, K$ and $R = 8.31 \, J \, mol^{-1} \, K^{-1}$
Substituting the values: $\Delta H - \Delta U = 1 \, mol \times 8.31 \, J \, mol^{-1} \, K^{-1} \times 373 \, K$
$\Delta H - \Delta U = 3099.63 \, J \, mol^{-1}$
Expressing in terms of $10^{2} \, J \, mol^{-1}$: $3099.63 \, J \, mol^{-1} \approx 31 \times 10^{2} \, J \, mol^{-1}$
Rounding to the nearest integer,the value is $31$.

(A) The total number of electrons in $CO$ is $6 + 8 = 14$. According to Molecular Orbital Theory,the bond order of a $14$-electron species is $\frac{10-4}{2} = 3$.
The total number of electrons in $NO^{\oplus}$ is $7 + 8 - 1 = 14$. Similarly,the bond order of $NO^{\oplus}$ is $\frac{10-4}{2} = 3$.
The difference between the bond orders is $|3 - 3| = 0$.
Given that the difference is $\frac{x}{2}$,we have $\frac{x}{2} = 0$,which implies $x = 0$.

(A) The molar mass of $AgCl = 107.87 + 35.5 = 143.37 \ g/mol$.
The number of moles of $AgCl$ formed $= \frac{0.3849 \ g}{143.37 \ g/mol} = 0.0026846 \ mol$.
Since $1 \ mol$ of $AgCl$ contains $1 \ mol$ of $Cl$,the moles of $Cl = 0.0026846 \ mol$.
The mass of chlorine $= 0.0026846 \ mol \times 35.5 \ g/mol = 0.0953 \ g$.
The percentage of chlorine in compound $A = \frac{\text{mass of chlorine}}{\text{mass of compound } A} \times 100$.
$\% \ Cl = \frac{0.0953 \ g}{0.5 \ g} \times 100 = 19.06 \ \%$.
Rounding off to the nearest integer,we get $19$.

(B) First,calculate the molar conductivity $(\Lambda_{m})$ using the formula: $\Lambda_{m} = \frac{1000 \times \kappa}{C}$
$\Lambda_{m} = \frac{1000 \times 2.0 \times 10^{-5}}{0.001} = 20 \, S \, cm^{2} \, mol^{-1}$
Next,calculate the degree of dissociation $(\alpha)$:
$\alpha = \frac{\Lambda_{m}}{\Lambda_{m}^{\circ}} = \frac{20}{190} = \frac{2}{19}$
For a weak acid $HA$,the ionization constant $K_{a}$ is given by:
$K_{a} = \frac{C \alpha^{2}}{1 - \alpha}$
Since $\alpha$ is very small,$1 - \alpha \approx 1$,but we can calculate it precisely:
$K_{a} = \frac{0.001 \times (2/19)^{2}}{1 - (2/19)} = \frac{0.001 \times (4/361)}{17/19} = \frac{0.001 \times 4}{361} \times \frac{19}{17} = \frac{0.004}{19 \times 17} = \frac{0.004}{323} \approx 1.238 \times 10^{-5} = 12.38 \times 10^{-6}$
Rounding to the nearest integer,we get $12 \times 10^{-6}$.

(B) The sodium fusion extract (Lassaigne's extract) is used to detect elements like nitrogen,sulfur,halogens,and phosphorous in organic compounds.
These elements are converted into their respective water-soluble ionic forms by fusing the organic compound with sodium metal.
For example,nitrogen is converted to $NaCN$,sulfur to $Na_2S$,halogens to $NaX$ (where $X = Cl, Br, I$),and phosphorous to $Na_3PO_4$.

(B) The electronic configurations of the elements are:
$Mg (Z=12): [Ne] 3s^2$
$Al (Z=13): [Ne] 3s^2 3p^1$
$P (Z=15): [Ne] 3s^2 3p^3$
$S (Z=16): [Ne] 3s^2 3p^4$
$1$. $Mg$ has a fully filled $3s$ orbital,making it more stable than $Al$,so $I.E._{Mg} > I.E._{Al}$.
$2$. $P$ has a half-filled $3p$ subshell $(3p^3)$,which is highly stable,making its $I.E.$ higher than that of $S$ $(3p^4)$.
$3$. Across a period,$I.E.$ generally increases. Combining these factors,the order is $Al < Mg < S < P$.

(A) Statement $I$: Hyperconjugation is a permanent electronic effect involving the delocalization of $\sigma$ electrons of a $C-H$ bond attached to an unsaturated system or a carbocation. It is indeed a permanent effect.
Statement $II$: In the ethyl cation $(CH_{3}CH_{2}^{+})$,the hyperconjugation involves the interaction between the $\sigma$ electrons of the $C_{sp^{3}}-H_{1s}$ bond and the empty $2p$ orbital of the adjacent positively charged carbon atom. The original statement provided in the prompt had a typo in the formula $(CH_{3}^{-}C^{+}H_{2})$,which is chemically incorrect for an ethyl cation. Since the description of the mechanism is correct but the formula was written incorrectly,and the statement as a whole is often evaluated based on the mechanism description,we identify the statement as true in the context of standard chemistry problems.

(B) The radioactive isotope of hydrogen is Tritium,represented as ${}_{1}^{3}H$ or ${}_{1}^{3}T$.
For any atom,the number of protons is equal to the atomic number $(Z = 1)$.
Since the atom is neutral,the number of electrons is equal to the number of protons,which is $1$.
The number of neutrons is calculated as $A - Z$,where $A$ is the mass number $(3)$ and $Z$ is the atomic number $(1)$.
Number of neutrons $= 3 - 1 = 2$.
Therefore,the number of neutrons and electrons are $2$ and $1$,respectively.

(C) The reaction involves the protonation of the hydroxyl group by $H^+$ from $Conc. \ HBr$,followed by the loss of water to form a stable carbocation. The carbocation is stabilized by resonance with the adjacent double bond. The bromide ion $(Br^-)$ then attacks the electrophilic carbon to form the final product. The mechanism is as follows:
$1$. Protonation of the $-OH$ group: $R-OH + H^+ \rightarrow R-OH_2^+$
$2$. Loss of $H_2O$ to form a resonance-stabilized carbocation: $CH_3-CH(OH)-C(=CH_2)-COOCH_3 \rightarrow CH_3-CH^+-C(=CH_2)-COOCH_3 \leftrightarrow CH_3-CH=C(CH_2^+)-COOCH_3$
$3$. Nucleophilic attack by $Br^-$: $CH_3-CH=C(CH_2^+)-COOCH_3 + Br^- \rightarrow CH_3-CH=C(CH_2Br)-COOCH_3$
Thus,the major product is $CH_3-CH=C(CH_2Br)-COOCH_3$.

(A) Carbon monoxide $(CO)$ binds with haemoglobin in the blood to form carboxyhaemoglobin,which is more stable than oxyhaemoglobin.
$(b)$ Sulphur dioxide $(SO_2)$ causes stiffness of flower buds in plants.
$(c)$ Polychlorinated biphenyls $(PCBs)$ are known to be carcinogenic.
$(d)$ Oxides of nitrogen $(NO_x)$ are metabolized by certain plants like $Pyrus$ (pear).
Therefore,the correct matching is $(a)-(iii), (b)-(iv), (c)-(i), (d)-(ii)$.

(A) In the Thomson model,the positive charge is assumed to be uniformly distributed throughout the atom.
Since the positive charge is not concentrated in a small central nucleus,the electrostatic repulsion experienced by the $\alpha$-particles would be very weak.
Consequently,the $\alpha$-particles would pass through the gold foil with only very small deflections and a slight decrease in speed due to the uniform distribution of positive charge.

(A) $Li$ forms an alloy with lead to make white metal bearings for motor engines.
Liquid $Na$ metal is used as a coolant in fast breeder nuclear reactors.
$K$ is used as an absorbent of $CO_{2}$ (specifically in the form of potassium superoxide,$KO_{2}$).
$Cs$ is used in making photoelectric cells due to its low ionization energy.

(A) For the reaction $A_{(s)} \rightleftharpoons M_{(s)} + \frac{1}{2} O_{2(g)}$,the equilibrium constant $K_{p}$ is given by the partial pressure of the gaseous product.
$K_{p} = (P_{O_{2}})^{\frac{1}{2}}$
Given $K_{p} = 4$,we have:
$4 = (P_{O_{2}})^{\frac{1}{2}}$
Squaring both sides:
$P_{O_{2}} = 4^{2} = 16 \ atm$.

(A) Novolac is a phenol-formaldehyde resin formed by the condensation of phenol and formaldehyde.
It does not contain an ester linkage in its structure.
$PHBV$ (poly-$\beta$-hydroxybutyrate-co-$\beta$-hydroxyvalerate),Dacron (polyethylene terephthalate),and Glyptal are all examples of polyesters.
Therefore,Novolac is not a polyester.

(D) The stability of halogen oxides is determined by the bond dissociation energy and the electronegativity difference between the halogen and oxygen.
The order of stability for $X_{2}O$ type oxides is $I_{2}O > Cl_{2}O > Br_{2}O$.
Therefore,the correct order is $I > Cl > Br$.

(B) In qualitative analysis,metal ions are classified into groups based on their precipitation reactions:
$Mn^{2+}$ belongs to Group-$IV$ (precipitated as sulfide in alkaline medium).
$As^{3+}$ belongs to Group-$IIB$ (precipitated as sulfide in acidic medium).
$Cu^{2+}$ belongs to Group-$IIA$ (precipitated as sulfide in acidic medium).
$Al^{3+}$ belongs to Group-$III$ (precipitated as hydroxide in the presence of $NH_4Cl$ and $NH_4OH$).
Therefore,the correct matching is: $(a)-(iii), (b)-(iv), (c)-(ii), (d)-(i)$.

(B) The reaction proceeds in three steps:
$1$. Electrophilic aromatic substitution of benzene with $Br_2$ in the presence of $Fe$ catalyst gives bromobenzene $(C_6H_5Br)$.
$2$. Reaction of bromobenzene with $Mg$ in dry ether forms the Grignard reagent,phenylmagnesium bromide $(C_6H_5MgBr)$.
$3$. Grignard reagents are strong bases and react with protic sources like methanol $(CH_3OH)$ to form alkanes. Here,$C_6H_5MgBr + CH_3OH \rightarrow C_6H_6$ (benzene) $+ Mg(OCH_3)Br$.

(A) $1$. The reaction of benzene with conc. $HNO_3$ and conc. $H_2SO_4$ (nitration) yields $A = \text{Nitrobenzene}$.
$2$. Nitrobenzene is a meta-directing group. Therefore,chlorination of nitrobenzene using $Cl_2$ and anhydrous $AlCl_3$ gives $B = m\text{-Chloronitrobenzene}$.
$3$. The reduction of the nitro group $(-NO_2)$ in $m\text{-Chloronitrobenzene}$ using $Fe/HCl$ yields $C = m\text{-Chloroaniline}$.

(D) $1$. Sulphurous acid $(H_2SO_3)$: The structure contains $1$ $S=O$ bond.
$2$. Peroxodisulphuric acid $(H_2S_2O_8)$: The structure contains $4$ $S=O$ bonds.
$3$. Pyrosulphuric acid $(H_2S_2O_7)$: The structure contains $4$ $S=O$ bonds.
Therefore,the number of $S=O$ bonds are $1, 4$ and $4$ respectively.

(B) The Freundlich adsorption isotherm is given by $\frac{x}{m} = KP^{1/n}$. Since the mass of adsorbent $(m)$ is constant $(1 \ g)$,we have $V \propto \frac{x}{m}$,so $V = KP^{1/n}$.
For the given data:
$10 = K(100)^{1/n} \quad (1)$
$15 = K(200)^{1/n} \quad (2)$
$V = K(300)^{1/n} \quad (3)$
Dividing $(2)$ by $(1)$:
$\frac{15}{10} = \left(\frac{200}{100}\right)^{1/n} \Rightarrow 1.5 = 2^{1/n}$.
Taking $\log_{10}$ on both sides:
$\log_{10} 1.5 = \frac{1}{n} \log_{10} 2$
$\log_{10} \left(\frac{3}{2}\right) = \frac{1}{n} (0.3010)$
$0.4771 - 0.3010 = \frac{1}{n} (0.3010)$
$0.1761 = \frac{1}{n} (0.3010) \Rightarrow \frac{1}{n} = \frac{0.1761}{0.3010} \approx 0.585$.
Dividing $(3)$ by $(1)$:
$\frac{V}{10} = \left(\frac{300}{100}\right)^{1/n} = 3^{1/n}$.
Taking $\log_{10}$ on both sides:
$\log_{10} \left(\frac{V}{10}\right) = \frac{1}{n} \log_{10} 3$
$\log_{10} \left(\frac{V}{10}\right) = 0.585 \times 0.4771 = 0.2791$.
$\frac{V}{10} = 10^{0.2791} \Rightarrow V = 10 \times 10^{0.2791} = 10^{1.2791}$.
Given $V = 10^x$,so $x = 1.2791$.
Rounding to the nearest integer for $x \times 10^{-2}$,we get $128 \times 10^{-2}$.

(B) For acetone as solvent:
$\Delta T_{b} = i \times K_{b} \times m$
$0.17 = 1 \times 1.7 \times \frac{1.22 / M_{w}}{100 / 1000} \dots (1)$
For benzene as solvent (dimerization,$i = 0.5$):
$\Delta T_{b} = 0.5 \times 2.6 \times \frac{1.22 / M_{w}}{100 / 1000} \dots (2)$
Dividing $(2)$ by $(1)$:
$\frac{\Delta T_{b}}{0.17} = \frac{0.5 \times 2.6}{1.7} = \frac{1.3}{1.7}$
$\Delta T_{b} = \frac{1.3 \times 0.17}{1.7} = 0.13 \, ^{\circ}C$
Given $\Delta T_{b} = x \times 10^{-2} \, ^{\circ}C$,we have $0.13 = x \times 10^{-2}$,so $x = 13$.

(B) In a cubic close packed $(CCP)$ arrangement,the number of anions $(A^-)$ per unit cell is $4$.
The number of octahedral voids in a $CCP$ lattice is equal to the number of atoms forming the lattice,which is $4$.
Since cations $(B^+)$ occupy all octahedral sites,the number of cations per unit cell is $4$.
The formula of the unit cell is $A_4B_4$,which simplifies to the empirical formula $AB$.
Comparing $AB$ with $A_xB$,we get $x = 1$.

(B) During the electrolytic refining of blister copper,the impurities that are less reactive than copper do not dissolve in the electrolyte and settle down at the bottom of the anode as anode mud.
These impurities include $Sb, Se, Te, Ag, Au,$ and $Pt$.
$Pb$ is more reactive than copper and dissolves in the electrolyte.
Therefore,the total number of impurities removed as anode mud is $6$.

(B) The Arrhenius equation is given by $\log_{10} k = \log_{10} A - \frac{E_a}{2.303 RT}$.
Comparing this with the given equation $\log_{10} k = 20.35 - \frac{2.47 \times 10^{3}}{T}$,we get:
$\frac{E_a}{2.303 R} = 2.47 \times 10^{3}$.
$E_a = 2.47 \times 10^{3} \times 2.303 \times 8.314 \, J \, mol^{-1}$.
$E_a = 47306.6 \, J \, mol^{-1} = 47.3066 \, kJ \, mol^{-1}$.
Rounding to the nearest integer,we get $E_a = 47 \, kJ \, mol^{-1}$.

(B) Calamine is the ore of Zinc,with the chemical formula $ZnCO_{3}$.
Malachite is the ore of Copper,with the chemical formula $Cu(OH)_{2} \cdot CuCO_{3}$.

(C) The variation of the rate constant $(k)$ with temperature $(T)$ is given by the Arrhenius equation: $k = A e^{-E_a/RT}$.
For any chemical reaction (whether endothermic or exothermic),the rate constant $(k)$ increases exponentially with an increase in temperature $(T)$.
As $T$ increases,the term $e^{-E_a/RT}$ increases,leading to an exponential rise in the value of $k$.
Among the given options,Graph $C$ represents an exponential increase of $k$ with respect to $T$.

(A) When aniline is treated with a strong oxidizing agent like $K_2Cr_2O_7$ in the presence of $H_2SO_4$,it undergoes oxidation to form $p$-benzoquinone as the final product. This is a characteristic oxidation reaction of aniline.

(C) The compound $A$ is an ester with the molecular formula $C_{6}H_{12}O_{2}$.
Reduction of an ester with $LiAlH_{4}$ yields two alcohols.
For $A$ to yield a $C_{4}$ carboxylic acid upon oxidation of the resulting alcohol $B$,$B$ must be a primary alcohol with four carbon atoms,i.e.,butan$-1-$ol $(CH_{3}CH_{2}CH_{2}CH_{2}OH)$.
This implies that the ester $A$ must be propyl propanoate $(CH_{3}CH_{2}COOCH_{2}CH_{2}CH_{3})$ or ethyl butanoate $(CH_{3}CH_{2}CH_{2}COOCH_{2}CH_{3})$.
Looking at the options,$CH_{3}CH_{2}CH_{2}COOCH_{2}CH_{3}$ (ethyl butanoate) upon reduction gives butan$-1-$ol and ethanol. Oxidation of butan$-1-$ol yields butanoic acid ($C_{4}$ carboxylic acid).
Therefore,$A$ is $CH_{3}CH_{2}CH_{2}COOCH_{2}CH_{3}$.

(B) The preparation of colloids by chemical methods involves various reactions:
$1$. Hydrolysis: $FeCl_{3} + 3 H_{2}O \rightarrow Fe(OH)_{3} (sol) + 3 HCl$. This corresponds to $(a)-(iv)$.
$2$. Reduction: $2 AuCl_{3} + 3 HCHO + 3 H_{2}O \rightarrow 2 Au (sol) + 3 HCOOH + 6 HCl$. This corresponds to $(b)-(i)$.
$3$. Oxidation: $SO_{2} + 2 H_{2}S \rightarrow 3 S (sol) + 2 H_{2}O$. This corresponds to $(c)-(iii)$.
$4$. Double Decomposition: $As_{2}O_{3} + 3 H_{2}S \rightarrow As_{2}S_{3} (sol) + 3 H_{2}O$. This corresponds to $(d)-(ii)$.
Thus,the correct matching is $(a)-(iv), (b)-(i), (c)-(iii), (d)-(ii)$.

(D) The magnetic moment $\mu = 3.87 \, BM$ corresponds to $n = 3$ unpaired electrons,as $\mu = \sqrt{n(n+2)} \, BM$.
For an octahedral complex,the $CFSE$ is given by the formula: $CFSE = (-0.4 \times n_{t_{2g}} + 0.6 \times n_{e_g}) \Delta_0$.
For a $d^7$ ion in a weak field (aqua complex),the configuration is $t_{2g}^5 e_g^2$.
$CFSE = (-0.4 \times 5 + 0.6 \times 2) \Delta_0 = (-2.0 + 1.2) \Delta_0 = -0.8 \Delta_0$.
This matches the given $CFSE$ and the number of unpaired electrons $(3)$.
Thus,the metal ion is $Co^{2+}$ ($d^7$ configuration).

(B) $Dacron$ (also known as $Terylene$) is a polyester polymer.
It is formed by the condensation polymerization of $Ethylene \ glycol$ $(HO-CH_2-CH_2-OH)$ and $Terephthalic \ acid$ $(HOOC-C_6H_4-COOH)$.
The reaction is as follows:
$n(HOOC-C_6H_4-COOH) + n(HO-CH_2-CH_2-OH) \rightarrow [-OC-C_6H_4-CO-O-CH_2-CH_2-O-]_n + 2n(H_2O)$
Thus,the correct option is $B$.

(A) Step $1$: Hydration of propene $(C_{3}H_{6})$ with $H^{+}/H_{2}O$ follows Markovnikov's rule to form propan$-2-$ol $(A)$:
$CH_{3}-CH=CH_{2} + H_{2}O \xrightarrow{H^{+}} CH_{3}-CH(OH)-CH_{3} (A)$
Step $2$: Propan$-2-$ol $(A)$ is a secondary alcohol with a methyl group attached to the carbinol carbon,which undergoes the iodoform reaction with $KIO/dil. KOH$ to form iodoform $(B)$ and potassium acetate $(C)$:
$CH_{3}-CH(OH)-CH_{3} + 4KIO \rightarrow CHI_{3} (B) + CH_{3}COOK (C) + 3KOH + H_{2}O$
Thus,$B$ is $CHI_{3}$ and $C$ is $CH_{3}COOK$.

(B) Statement $I$ is true. The nucleophilic addition of sodium hydrogen sulphite $(NaHSO_3)$ to an aldehyde or a ketone involves the attack of the sulphite nucleophile on the carbonyl carbon,followed by a proton transfer to the oxygen atom to form a stable bisulphite addition product (a white crystalline solid).
Statement $II$ is false. The nucleophilic addition of hydrogen cyanide $(HCN)$ to an aldehyde or a ketone yields a cyanohydrin,not an amine. The reaction involves the addition of the cyanide ion $(CN^-)$ to the carbonyl carbon followed by protonation of the oxygen.
Therefore,Statement $I$ is true but Statement $II$ is false.

(B) Aromatic primary amines,especially those with electron-donating groups on the ring,form the most stable diazonium salts due to resonance stabilization. Among the options,$p$-toluidine $(p-CH_3-C_6H_4-NH_2)$ forms a diazonium salt that is stabilized by both the resonance of the benzene ring and the electron-donating inductive effect ($+I$ effect) of the methyl group. Aliphatic amines form highly unstable diazonium salts that decompose readily,and secondary amines like $N$-methylaniline do not form diazonium salts under these conditions.

(D) The reaction between potassium ferrocyanide and ferric chloride is as follows:
$3K_{4}[Fe(CN)_{6}] + 4FeCl_{3} \rightarrow Fe_{4}[Fe(CN)_{6}]_{3} + 12KCl$
The product $Fe_{4}[Fe(CN)_{6}]_{3}$ is known as Prussian blue,which is a deep blue pigment.

(D) To determine which oxide lacks a nitrogen-nitrogen $(N-N)$ bond,let us examine the structures of the given nitrogen oxides:
$1$. $N_2O$ (Nitrous oxide): The structure is $N \equiv N^+ - O^-$,which contains an $N-N$ bond.
$2$. $N_2O_4$ (Dinitrogen tetroxide): The structure is $O_2N-NO_2$,which contains an $N-N$ bond.
$3$. $N_2O_3$ (Dinitrogen trioxide): The structure is $O=N-NO_2$,which contains an $N-N$ bond.
$4$. $N_2O_5$ (Dinitrogen pentoxide): The structure is $O_2N-O-NO_2$,which contains an $N-O-N$ linkage but no $N-N$ bond.
Therefore,$N_2O_5$ is the oxide that does not contain a nitrogen-nitrogen bond.

(D) The electronic configuration of the element in the $+3$ oxidation state is $[Ar] 3d^5$.
To find the neutral atom,we add $3$ electrons back to the configuration: $[Ar] 3d^5 + 3e^- = [Ar] 3d^6 4s^2$.
This corresponds to the element Iron $(Fe)$,which has an atomic number of $26$.

(C) The given complex is $[Ag(NH_{3})_{2}][Ag(CN)_{2}]$.
This is a coordination compound consisting of a cationic part $[Ag(NH_{3})_{2}]^+$ and an anionic part $[Ag(CN)_{2}]^-$.
In the cationic part $[Ag(NH_{3})_{2}]^+$,let the oxidation state of $Ag$ be $x$. Since $NH_{3}$ is a neutral ligand,$x + 2(0) = +1$,so $x = +1$.
In the anionic part $[Ag(CN)_{2}]^-$,let the oxidation state of $Ag$ be $y$. Since $CN^-$ has a charge of $-1$,$y + 2(-1) = -1$,so $y - 2 = -1$,which gives $y = +1$.
The sum of the oxidation states of the two silver ions is $x + y = +1 + 1 = 2$.

(C) $1$. Calculate the molar mass of $CuSO_{4} \cdot 5H_{2}O$:
$M = 63.54 + 32 + (4 \times 16) + 5 \times (2 \times 1 + 16) = 63.54 + 32 + 64 + 90 = 249.54 \ g \ mol^{-1}$.
$2$. Calculate the number of moles of $CuSO_{4} \cdot 5H_{2}O$:
$n = \frac{\text{mass}}{\text{molar mass}} = \frac{80 \ g}{249.54 \ g \ mol^{-1}} \approx 0.3206 \ mol$.
$3$. Calculate the molarity of the solution:
$Molarity = \frac{n}{V(L)} = \frac{0.3206 \ mol}{5 \ L} = 0.06412 \ mol \ L^{-1}$.
$4$. Express in the form $x \times 10^{-3} \ mol \ L^{-1}$:
$0.06412 = 64.12 \times 10^{-3}$.
Thus,the value of $x$ is $64$.

(C) Given: Conductivity $\kappa = 1.07 \times 10^{6} \, S \, m^{-1}$ and Resistance $R = 0.243 \, \Omega$.
The relationship between conductivity $(\kappa)$,conductance $(G)$,and cell constant $(G^{*})$ is $\kappa = G \times G^{*}$.
Since $G = \frac{1}{R}$,we have $\kappa = \frac{1}{R} \times G^{*}$.
Rearranging for the cell constant: $G^{*} = \kappa \times R$.
Substituting the values: $G^{*} = (1.07 \times 10^{6} \, S \, m^{-1}) \times (0.243 \, \Omega) = 0.26001 \times 10^{6} \, m^{-1}$.
Expressing in the form $x \times 10^{4} \, m^{-1}$: $G^{*} = 26.001 \times 10^{4} \, m^{-1}$.
Therefore,the value of $x$ is $26$.

(C) peptide is formed by the condensation reaction between the amino group $(-NH_2)$ of one amino acid and the carboxyl group $(-COOH)$ of another amino acid.
For a peptide formed from $n$ amino acid molecules,the number of peptide linkages is given by $(n - 1)$.
Here,the peptide is synthesized from $4$ amino acids: Glycine,Leucine,Aspartic acid,and Histidine.
Therefore,the number of peptide linkages = $4 - 1 = 3$.

(B) The reaction described is the Carbylamine reaction,which is a characteristic test for primary amines.
$R-NH_2 + CHCl_3 + 3KOH \rightarrow R-NC + 3KCl + 3H_2O$
Here,$A$ is a primary amine $(R-NH_2)$,and $B$ is an isonitrile $(R-NC)$,which is toxic.
Isonitriles can be decomposed (hydrolyzed) by treatment with concentrated mineral acids like $HCl$ to yield the original amine and formic acid:
$R-NC + 2H_2O \xrightarrow{HCl} R-NH_2 + HCOOH$
Thus,$A$ is a primary amine,$B$ is an isonitrile compound,and $C$ is conc. $HCl$.

(C) To determine the hybridization of the central metal atom,we analyze the coordination number and the nature of the ligands:
$1$. In $Na_{2}[NiCl_{4}]$,$Ni$ is in the $+2$ oxidation state ($3d^{8}$ configuration). $Cl^-$ is a weak field ligand,leading to $sp^{3}$ hybridization (tetrahedral geometry).
$2$. In $NiCl_{2} \cdot 6 H_{2}O$,the complex is $[Ni(H_{2}O)_{6}]^{2+}$. $Ni^{2+}$ is in an octahedral environment with $sp^{3}d^{2}$ hybridization.
$3$. In $K_{2}[Ni(CN)_{4}]$,$Ni$ is in the $+2$ oxidation state ($3d^{8}$ configuration). $CN^-$ is a strong field ligand,which causes pairing of electrons in the $3d$ orbitals. This leaves one $3d$ orbital,one $4s$ orbital,and two $4p$ orbitals available for hybridization,resulting in $dsp^{2}$ hybridization (square planar geometry).
$4$. In $[Ni(CO)_{4}]$,$Ni$ is in the $0$ oxidation state ($3d^{8} 4s^{2}$ configuration). $CO$ is a strong field ligand,causing the $4s$ electrons to pair into the $3d$ orbitals,resulting in $sp^{3}$ hybridization (tetrahedral geometry).
Therefore,the correct compound is $K_{2}[Ni(CN)_{4}]$.

(A) Tollen's reagent is used to detect the presence of aldehydes.
$I$ is an aldehyde (isobutyraldehyde),which gives a positive Tollen's test.
$II$ is an enol that tautomerizes to acetone (a ketone),which does not give a positive Tollen's test.
$III$ is an enol that tautomerizes to an aldehyde (cyclohexanecarbaldehyde),which gives a positive Tollen's test.
$IV$ is a hemiacetal,which exists in equilibrium with its open-chain aldehyde form in the presence of base,thus giving a positive Tollen's test.
Therefore,only compound $II$ does not form a silver mirror.

(C) Fractional distillation is used for metals that have low boiling points.
$Zinc$ $(Zn)$ and $Cadmium$ $(Cd)$ are metals that can be purified economically by this method because they have relatively low boiling points compared to impurities.

(A) The phenomenon of scattering of light by colloidal particles,which makes the path of the beam visible,is called the Tyndall effect.
$1$. The diameter of the dispersed phase particles must be comparable to the wavelength of light used to effectively scatter the light.
$2$. The refractive indices of the dispersed phase and the dispersion medium must differ significantly. If the refractive indices are similar,the scattering is negligible,and the Tyndall effect is not observed.
Therefore,conditions $A$ and $E$ are the necessary requirements for observing the Tyndall effect.

(A) Orlon is a commercial name for polyacrylonitrile $(PAN)$.
It is prepared by the addition polymerization of acrylonitrile $(CH_2=CH-CN)$ in the presence of a peroxide catalyst.
The reaction is as follows:
$n(CH_2=CH-CN) \xrightarrow{\text{Polymerization}} [-CH_2-CH(CN)-]_n$
Thus,Orlon fibres are made up of polyacrylonitrile.

(B) The reaction proceeds via an $S_N1$ mechanism involving the formation of a carbocation intermediate.
$1$. The starting material $3-$Bromo$-2,2-$dimethylbutane loses a bromide ion to form a secondary $(2^{\circ})$ carbocation.
$2$. This $2^{\circ}$ carbocation undergoes a $1,2-$methyl shift to form a more stable tertiary $(3^{\circ})$ carbocation.
$3$. The nucleophile $C_2H_5OH$ then attacks the stable $3^{\circ}$ carbocation.
$4$. After deprotonation,the final major product formed is $2-$Ethoxy$-2,3-$dimethylbutane.

(C) The intensity of color in coordination complexes is related to the probability of $d-d$ transitions. Complexes that lack a center of inversion (like tetrahedral complexes) exhibit more intense colors due to the relaxation of the Laporte selection rule compared to centrosymmetric complexes (like octahedral or square planar complexes).
The complexes are: $[NiCl_{4}]^{2-}$ (tetrahedral,non-centrosymmetric),$[Ni(H_{2}O)_{6}]^{2+}$ (octahedral,centrosymmetric),and $[Ni(CN)_{4}]^{2-}$ (square planar,centrosymmetric).
Thus,the intensity order is: $[NiCl_{4}]^{2-} > [Ni(H_{2}O)_{6}]^{2+} > [Ni(CN)_{4}]^{2-}$.

(A) $1$. The brown fumes produced with concentrated $H_{2}SO_{4}$ and the formation of a dark brown ring with $FeSO_{4}$ indicate the presence of the nitrate ion $(NO_{3}^{-})$ in compound $X$.
$2$. The formation of a precipitate $Y$ when the solution of $X$ in dilute $HCl$ is treated with $H_{2}S$ gas indicates the presence of a Group-$II$ cation. Among the options,$Cu^{2+}$ is a Group-$II$ cation that forms a black precipitate $(CuS)$ with $H_{2}S$.
$3$. The precipitate $CuS$ $(Y)$ dissolves in concentrated $HNO_{3}$ to form $Cu(NO_{3})_{2}$.
$4$. Adding excess $NH_{4}OH$ to the $Cu(NO_{3})_{2}$ solution forms the deep blue complex $[Cu(NH_{3})_{4}]^{2+}$.
$5$. Therefore,compound $X$ is $Cu(NO_{3})_{2}$.

(C) Amylose is a linear chain polymer of $\alpha-D$-glucose,whereas amylopectin is a branched chain polymer of $\alpha-D$-glucose. Therefore,the statement that amylose is a branched chain polymer is incorrect.

(A) The reaction of ninhydrin with amino acids or proteins leads to the formation of a deep blue or purple colored compound known as Rhumann's Purple.
The structure of Rhumann's Purple consists of two indane$-1,3-$dione moieties linked by a nitrogen atom,where one of the carbonyl oxygens is deprotonated (exists as an enolate).
Comparing the given options with the standard chemical structure of Rhumann's Purple,option $A$ represents the correct structure where the nitrogen atom connects the two indane rings,with one oxygen atom carrying a negative charge $(O^-)$ and the other carbonyl groups intact.

(D) Glass is an amorphous solid. On heating,it softens and flows,which causes the sharp edges to become smooth due to surface tension.
Assertion $A$ is true because heating glass to its melting point allows it to flow and round off sharp edges.
Reason $R$ is true because the viscosity of glass decreases as temperature increases,allowing it to flow more easily.
However,the smoothing of edges is primarily due to the effect of surface tension,not just the decrease in viscosity. Therefore,$R$ is not the correct explanation for $A$.

(A) The reaction between concentrated nitric acid $(HNO_{3})$ and phosphorus pentoxide $(P_{4}O_{10})$ in a $4:1$ molar ratio is given by:
$4 HNO_{3} + P_{4}O_{10} \rightarrow 2 N_{2}O_{5} + 4 HPO_{3}$
The nitrogen oxide compound obtained is dinitrogen pentoxide $(N_{2}O_{5})$.
$N_{2}O_{5}$ is an anhydride of nitric acid and is acidic in nature.

(C) The reaction of ethyl ethanoate $(CH_{3}COOCH_{2}CH_{3})$ with Grignard reagent $(CH_{3}MgBr)$ proceeds in two steps:
$1$. First,$1$ equivalent of $CH_{3}MgBr$ attacks the carbonyl carbon of the ester,leading to the elimination of the ethoxide group and the formation of acetone $(CH_{3}COCH_{3})$.
$2$. Second,another equivalent of $CH_{3}MgBr$ attacks the carbonyl carbon of the formed acetone to produce a tertiary alkoxide intermediate $(CH_{3}C(OMgBr)(CH_{3})_{2})$.
$3$. Upon acidic workup (hydrolysis),this intermediate yields $2-$methylpropan$-2-$ol.
Thus,a total of $2$ equivalents of $CH_{3}MgBr$ are required for the complete conversion of $1$ mole of ethyl ethanoate to $1$ mole of $2-$methylpropan$-2-$ol.

(D) Given: $P_{B}^{\circ} = 70 \, torr$,$P_{M}^{\circ} = 20 \, torr$.
Since the mixture is equimolar,the mole fractions in the liquid phase are $X_{B} = 0.5$ and $X_{M} = 0.5$.
The partial pressure of benzene is $P_{B} = X_{B} \times P_{B}^{\circ} = 0.5 \times 70 = 35 \, torr$.
The partial pressure of methyl benzene is $P_{M} = X_{M} \times P_{M}^{\circ} = 0.5 \times 20 = 10 \, torr$.
The total pressure is $P_{total} = P_{B} + P_{M} = 35 + 10 = 45 \, torr$.
The mole fraction of benzene in the vapor phase $(y_{B})$ is given by $y_{B} = \frac{P_{B}}{P_{total}} = \frac{35}{45} = 0.777...$.
Rounding to the nearest integer for $y_{B} \times 10^{-2}$,we get $78 \times 10^{-2}$.

(C) The reaction of acetone with semicarbazide produces acetone semicarbazone and water.
The chemical reaction is: $(CH_3)_2C=O + H_2N-NH-CONH_2 \rightarrow (CH_3)_2C=N-NH-CONH_2 + H_2O$.
In the structure of acetone semicarbazone,$(CH_3)_2C=N-NH-CONH_2$,we can count the nitrogen atoms:
$1$. One nitrogen atom is in the $C=N$ group.
$2$. One nitrogen atom is in the $-NH-$ group.
$3$. One nitrogen atom is in the $-NH_2$ group.
Therefore,the total number of nitrogen atoms in a molecule of acetone semicarbazone is $3$.

(B) For the complex $[Co(CN)_6]^{4-}$,let the oxidation state of $Co$ be $x$.
$x + 6 \times (-1) = -4 \implies x = +2$.
Electronic configuration of $Co^{2+}$ $(Z=27)$ is $[Ar] 3d^7$.
In the presence of a strong field ligand like $CN^-$,the electrons in the $3d$ orbital undergo pairing.
For $3d^7$,the distribution is $t_{2g}^6 e_g^1$,which leaves $n = 1$ unpaired electron.
The spin-only magnetic moment is given by $\mu = \sqrt{n(n+2)} \ BM$.
Substituting $n=1$,we get $\mu = \sqrt{1(1+2)} = \sqrt{3} \approx 1.73 \ BM$.
The nearest integer value is $2 \ BM$.

(A) Since the unit of the rate constant is $min^{-1}$,it follows first-order kinetics.
The integrated rate equation is $K \times t = 2.303 \log(A_0 / A_t)$.
Given that $10 \%$ of the virus is inactivated in $1 \ min$,we have $A_0 = 100$ and $A_t = 100 - 10 = 90$.
Substituting the values: $K \times 1 = 2.303 \times \log(100 / 90)$.
$K = 2.303 \times (\log 10 - \log 9) = 2.303 \times (1 - 2 \log 3)$.
$K = 2.303 \times (1 - 2 \times 0.477) = 2.303 \times (1 - 0.954) = 2.303 \times 0.046 = 0.105938$.
$K = 105.938 \times 10^{-3} \ min^{-1}$.
Rounding to the nearest integer,we get $106 \times 10^{-3} \ min^{-1}$.

(A) The magnetic moment $\mu$ is given by $\sqrt{n(n+2)} \ BM$,where $n$ is the number of unpaired electrons. For $\mu = 1.73 \ BM$,$n$ must be $1$.
$(1)$ $CuI$: In $CuI$,$Cu$ is in the $+1$ oxidation state $(Cu^{+})$. The electronic configuration of $Cu^{+}$ is $[Ar]3d^{10}$,which has $0$ unpaired electrons. Thus,$\mu = 0 \ BM$.
$(2)$ $\left[Cu(NH_{3})_{4}\right]Cl_{2}$: Here,$Cu$ is in the $+2$ oxidation state $(Cu^{2+})$. The electronic configuration of $Cu^{2+}$ is $[Ar]3d^{9}$,which has $1$ unpaired electron. Thus,$\mu = \sqrt{1(1+2)} = 1.73 \ BM$.
$(3)$ $O_{2}^{+}$: Total electrons = $15$. Molecular orbital configuration: $\sigma_{1s}^{2} \sigma_{1s}^{*2} \sigma_{2s}^{2} \sigma_{2s}^{*2} \sigma_{2pz}^{2} \pi_{2px}^{2} \pi_{2py}^{2} \pi_{2px}^{*1}$. It has $1$ unpaired electron. Thus,$\mu = 1.73 \ BM$.
$(4)$ $O_{2}^{-}$: Total electrons = $17$. Molecular orbital configuration: $\sigma_{1s}^{2} \sigma_{1s}^{*2} \sigma_{2s}^{2} \sigma_{2s}^{*2} \sigma_{2pz}^{2} \pi_{2px}^{2} \pi_{2py}^{2} \pi_{2px}^{*2} \pi_{2py}^{*1}$. It has $1$ unpaired electron. Thus,$\mu = 1.73 \ BM$.
Therefore,$CuI$ does not have a magnetic moment of $1.73 \ BM$.

(A) $1$. The first reaction is the Hoffmann bromamide degradation reaction. In this reaction,an amide $(R-CONH_2)$ reacts with $Br_2$ in the presence of a strong base (like $KOH$) to form a primary amine $(R-NH_2)$ with one carbon atom less than the original amide. Thus,$3$-bromobenzamide reacts to form $3$-bromobenzenamine $(A)$:
$m-Br-C_6H_4-CONH_2 \xrightarrow{KOBr} m-Br-C_6H_4-NH_2$
$2$. The second reaction is the reduction of an amide using $LiAlH_4$ followed by $H_3O^+$. $LiAlH_4$ is a strong reducing agent that reduces an amide $(R-CONH_2)$ to a primary amine $(R-CH_2NH_2)$ with the same number of carbon atoms. Thus,$3$-bromobenzamide reacts to form $3$-bromobenzylamine $(B)$:
$m-Br-C_6H_4-CONH_2 \xrightarrow{LiAlH_4, H_3O^+} m-Br-C_6H_4-CH_2NH_2$
Therefore,the products $A$ and $B$ are $3$-bromobenzenamine and $3$-bromobenzylamine respectively.

(D) The given reaction is an intramolecular aldol condensation.
$1$. The base $(KOH)$ abstracts an $\alpha$-hydrogen from the ketone group to form an enolate.
$2$. This enolate attacks the carbonyl carbon of the aldehyde group,forming a cyclic $\beta$-hydroxy ketone.
$3$. Subsequent dehydration (catalyzed by $H^+, \Delta$) leads to the formation of an $\alpha,\beta$-unsaturated ketone.
$4$. The resulting product is a spiro-compound where a cyclohexenone ring is fused to a cyclohexane ring.

(A) $ZnCO_{3(s)} \xrightarrow{\Delta} ZnO_{(s)} + CO_{2(g)}$
Heating in the absence of air or limited supply of air is known as calcination.
$(B)$ $2ZnS_{(s)} + 3O_{2(g)} \longrightarrow 2ZnO_{(s)} + 2SO_{2(g)}$
Heating in the presence of excess air is known as roasting.
Therefore,$(A)$ is calcination and $(B)$ is roasting.