JEE Main 2020 Chemistry Question Paper with Answer and Solution

(A) The molar mass of ferrous sulphate heptahydrate $(FeSO_4 \cdot 7H_2O)$ is calculated as: $55.85 + 32.0 + 4 \times 16.0 + 7 \times 18.0 = 277.85 \; g/mol$.
Given,$10 \; ppm$ of iron in $100 \; kg$ of wheat means:
$10 = \frac{\text{mass of Fe (in mg)}}{\text{mass of wheat (in kg)}}$
Mass of $Fe$ required $= 10 \times 100 = 1000 \; mg = 1 \; g$.
In $FeSO_4 \cdot 7H_2O$,the mass fraction of $Fe$ is $\frac{55.85}{277.85}$.
Let $w$ be the mass of the salt in grams:
$w \times \frac{55.85}{277.85} = 1 \; g$
$w = \frac{277.85}{55.85} \approx 4.97 \; g$.

(A) Let the original observations be $x_1, x_2, \dots, x_{10}$. Given mean $\bar{x} = 20$ and s.d. $\sigma = 2$.
When each observation is transformed to $y_i = p x_i - q$,the new mean $\bar{y} = p \bar{x} - q$ and the new s.d. $\sigma_y = |p| \sigma$.
Given $\bar{y} = \frac{20}{2} = 10$ and $\sigma_y = \frac{2}{2} = 1$.
Substituting the values: $10 = 20p - q \dots (i)$ and $1 = |p| \times 2 \dots (ii)$.
From $(ii)$,$|p| = \frac{1}{2}$,so $p = \frac{1}{2}$ or $p = -\frac{1}{2}$.
Case $1$: If $p = \frac{1}{2}$,then $10 = 20(\frac{1}{2}) - q$ $\Rightarrow 10 = 10 - q$ $\Rightarrow q = 0$. But given $q \neq 0$,this is rejected.
Case $2$: If $p = -\frac{1}{2}$,then $10 = 20(-\frac{1}{2}) - q$ $\Rightarrow 10 = -10 - q$ $\Rightarrow q = -20$.

(A) The $IE$ values indicate that the metal belongs to the $I^{st}$ group because the second $IE$ is very high,suggesting the presence of only one valence electron.
Therefore,the metal hydroxide will be of the type $MOH$.
The reaction with $HCl$ is:
$MOH + HCl \rightarrow MCl + H_{2}O$
From the stoichiometry,$1 \; mol$ of $MOH$ reacts with $1 \; mol$ of $HCl$.
The reaction with $H_{2}SO_{4}$ is:
$2MOH + H_{2}SO_{4} \rightarrow M_{2}SO_{4} + 2H_{2}O$
Dividing by $2$ to match $1 \; mol$ of $MOH$:
$MOH + \frac{1}{2} H_{2}SO_{4} \rightarrow \frac{1}{2} M_{2}SO_{4} + H_{2}O$
From the stoichiometry,$1 \; mol$ of $MOH$ reacts with $0.5 \; mol$ of $H_{2}SO_{4}$.
Thus,$1$ mole of $HCl$ and $0.5$ mole of $H_{2}SO_{4}$ are required.

(C) racemic mixture is produced when a chiral center is generated in a reaction,and both enantiomers are formed in equal amounts.
$A$. $CH_3-CH(CH_3)-CH=CH_2 + HCl \rightarrow$ The carbocation formed undergoes rearrangement to form a stable tertiary carbocation,which upon attack by $Cl^-$ produces a chiral center,resulting in a racemic mixture.
$B$. $CH_3-CO-CH_2-CH_3 + HCN \rightarrow$ The addition of $HCN$ to the carbonyl group creates a chiral center at the carbon atom,resulting in a racemic mixture of cyanohydrins.
$C$. The reaction of $3$-methylcyclohex-$1$-ene with $HCl$ involves the formation of a carbocation at the $C-1$ position. The carbocation is planar,and the attack of $Cl^-$ can occur from either side,but due to the existing chiral center at $C-3$,the products formed are diastereomers,not a racemic mixture.
$D$. $CH_3-CH_2-CH=CH_2 + HBr \rightarrow$ The addition of $HBr$ creates a chiral center at the $C-2$ position,resulting in a racemic mixture.

(A) The reaction sequence is as follows:
$1$. $A$ reacts with $Br_2, h\nu$ to form a brominated product.
$2$. Treatment with $KOH$ (alc.) causes dehydrohalogenation to form an alkene.
$3$. Ozonolysis $(O_3, (CH_3)_2S)$ of the alkene cleaves the ring to form a dialdehyde.
$4$. Intramolecular aldol condensation followed by dehydration with $NaOH (aq) + \Delta$ yields the final product,which is cyclopent$-1-$ene$-1-$carbaldehyde.
$5$. Working backwards from the final product,the precursor dialdehyde is hexane$-1,6-$dial. The alkene formed after step (ii) is cyclohexene. The brominated product is bromocyclohexane. Therefore,$A$ must be cyclohexane.

(C) Biochemical oxygen demand $(BOD)$ is defined as the amount of dissolved oxygen required by aerobic bacteria to break down the organic matter present in a certain volume of a water sample at a specific temperature over a period of $5$ days.

(A) Lithium has the highest hydration enthalpy among alkali metals due to its smallest ionic size.
$(b)$ Lithium chloride $(LiCl)$ is soluble in pyridine because it possesses significant covalent character due to polarization (Fajans' rule).
$(c)$ Lithium reacts with ethyne $(C_2H_2)$ to form lithium ethynide $(Li_2C_2)$. Thus,statement $(c)$ is incorrect.
$(d)$ Both lithium and magnesium react slowly with $H_2O$ due to their high ionization energy and protective oxide layers. Thus,statements $(a), (c)$ and $(d)$ are correct.

(A) Ionic conductance is directly proportional to the concentration of dissolved ions in water.
Distilled water is purified through distillation,which removes dissolved salts and minerals,resulting in the lowest concentration of ions.
Therefore,distilled water exhibits the lowest ionic conductance compared to well water,saline water,or sea water,which contain significant amounts of dissolved electrolytes.

(A) From the figure,count the number of squares $(A)$ and circles $(B)$:
Number of squares $(A)$ = $4$
Number of circles $(B)$ = $8$
Assuming the reaction $A \rightleftharpoons B$,the equilibrium constant $K$ is given by:
$K = \frac{[B]}{[A]} = \frac{8}{4} = 2$
Therefore,the correct option is $A$.

(A) The dissociation of $Cr(OH)_3$ is given by: $Cr(OH)_{3(s)} \rightleftharpoons Cr^{3+}_{(aq)} + 3OH^-_{(aq)}$.
Let the solubility of $Cr(OH)_3$ be $s \ mol/L$.
Then,$[Cr^{3+}] = s$ and $[OH^-] = 3s$.
The solubility product expression is $K_{sp} = [Cr^{3+}][OH^-]^3$.
Substituting the values: $K_{sp} = (s)(3s)^3 = 27s^4$.
Given $K_{sp} = 6.0 \times 10^{-31}$,we have $27s^4 = 6.0 \times 10^{-31}$.
Thus,$s^4 = \frac{6.0 \times 10^{-31}}{27} = \frac{60 \times 10^{-32}}{27} \approx 2.22 \times 10^{-32}$.
The concentration of hydroxide ions is $[OH^-] = 3s$.
Since $s = (2.22 \times 10^{-32})^{1/4}$,then $[OH^-] = 3 \times (2.22 \times 10^{-32})^{1/4} = (3^4 \times 2.22 \times 10^{-32})^{1/4} = (81 \times 2.22 \times 10^{-32})^{1/4} = (179.82 \times 10^{-32})^{1/4} = (17.982 \times 10^{-31})^{1/4} \approx (18 \times 10^{-31})^{1/4}$.

(D) The chemical reactions are as follows:
$(a)$ $Zn(s) + 2HCl(aq) \longrightarrow ZnCl_{2}(aq) + H_{2}(g)$
$(b)$ $Zn(s) + 2NaOH(aq) + 2H_{2}O(l) \longrightarrow Na_{2}[Zn(OH)_{4}](aq) + H_{2}(g)$
From the stoichiometry of both balanced equations,$1 \ mole$ of $Zn$ produces $1 \ mole$ of $H_{2}$ gas in both cases.
Since the amount of $Zn$ $(5 \ g)$ is the same in both reactions,the number of moles of $H_{2}$ produced will be identical.
Therefore,the ratio of the volumes of $H_{2}$ evolved is $1 : 1$.

(B) The reaction of $B$-trichloroborazine $(A)$ with $LiBH_{4}$ in tetrahydrofuran $(THF)$ is a reduction reaction that yields borazine $(B)$,which is known as inorganic benzene:
$H_{3}N_{3}B_{3}Cl_{3} + 3LiBH_{4} \rightarrow H_{3}N_{3}B_{3}H_{3} (B) + 3LiCl + 3BH_{3}THF$.
The reaction of $(A)$ with a Grignard reagent like $MeMgBr$ $(C)$ results in the methylation of the boron atoms:
$H_{3}N_{3}B_{3}Cl_{3} + 3MeMgBr (C) \rightarrow H_{3}N_{3}B_{3}(Me)_{3} + 3MgBrCl$.
Thus,$(B)$ is borazine and $(C)$ is $MeMgBr$.

(D) Benzene $(C_{6}H_{6})$ consists of $6$ carbon atoms,each of which is $sp^{2}$ hybridized.
Each $sp^{2}$ hybridized carbon atom forms $3$ $sp^{2}$ hybrid orbitals.
Therefore,the total number of $sp^{2}$ hybrid orbitals in a molecule of benzene is $6 \times 3 = 18$.

(A) Entropy $(S)$ is a state function and its value depends on the state of the system,which is defined by variables like temperature $(T)$,pressure $(P)$,and volume $(V)$.
Since $S = f(T, P, V)$,entropy is a function of temperature.
Similarly,the change in entropy $(\Delta S = S_2 - S_1)$ also depends on the initial and final states of the system,which are defined by temperature.
Therefore,both $S$ and $\Delta S$ are functions of temperature.

(C) The heat of combustion $(HOC)$ is inversely proportional to the stability of the isomer.
For the given alkadienes:
$(a)$ is the (trans,trans) isomer,which is the most stable due to minimum steric hindrance.
$(b)$ is the (trans,cis) isomer,which has intermediate stability.
$(c)$ is the (cis,cis) isomer,which is the least stable due to maximum steric hindrance.
Stability order: $a > b > c$.
Since $HOC$ $\propto \frac{1}{\text{Stability}}$,the order of heat of combustion is: $c > b > a$.

(A) $CCl_4$ is a molecular solid held together by weak van der Waals forces,which explains its low melting point.
Since it consists of neutral molecules and lacks free ions or delocalized electrons,it is a poor conductor of electricity in both solid and liquid states.

(B) First,calculate the final concentrations of the ions in the mixture (total volume = $300 \ mL + 100 \ mL = 400 \ mL$):
$[Pb^{2+}] = \frac{300 \ mL \times 0.134 \ M}{400 \ mL} = 0.1005 \ M$
$[Cl^-] = \frac{100 \ mL \times 0.4 \ M}{400 \ mL} = 0.1 \ M$
Next,calculate the reaction quotient $Q$ for the dissociation $PbCl_{2(s)} \rightleftharpoons Pb^{2+}_{(aq)} + 2Cl^-_{(aq)}$:
$Q = [Pb^{2+}][Cl^-]^2$
$Q = (0.1005) \times (0.1)^2$
$Q = 0.1005 \times 0.01 = 1.005 \times 10^{-3}$
Comparing $Q$ with $K_{sp}$ $(1.6 \times 10^{-5})$:
$1.005 \times 10^{-3} > 1.6 \times 10^{-5}$
Therefore,$Q > K_{sp}$.

(A) The magnetic moment $(\mu)$ is given by the formula $\mu = \sqrt{n(n+2)} \; B.M$,where $n$ is the number of unpaired electrons. For $\mu = 1.73 \; B.M$,we have $n = 1$.

Species	Unpaired electrons $(n)$
$O_{2}^{+}$	$1$
$O_{2}^{-}$	$1$
$O_{2}$	$2$

Since both $O_{2}^{+}$ and $O_{2}^{-}$ have $1$ unpaired electron,they both exhibit a magnetic moment of $1.73 \; B.M$.

(D) The enthalpy of atomisation $(\Delta H_{atom})$ is the energy required to convert one mole of a substance into its gaseous atoms.
For $Br_{2(l)}$,the process is:
$Br_{2(l)}$ $\xrightarrow{\Delta H_{vap}} Br_{2(g)}$ $\xrightarrow{\Delta H_{BE}} 2Br_{(g)}$
Therefore,$\Delta H_{atom} = \Delta H_{vap} + \Delta H_{BE}$.
Given $\Delta H_{atom} = x$ and $\Delta H_{BE} = y$,we have $x = \Delta H_{vap} + y$.
Since the enthalpy of vaporisation $(\Delta H_{vap})$ is always positive,it follows that $x > y$.

(C) Basicity $\propto \frac{1}{\text{Stability of conjugate base}}$.
Stability order of the given carbanions:
$(v)$ $CN^{-}$: The negative charge is on the nitrogen atom (more electronegative) and it is also stabilized by the $-I$ effect of the triple bond. It is the most stable.
$(iii)$ $HC \equiv C^{-}$: The negative charge is on an $sp$ hybridized carbon atom. It is more stable than $sp^2$ or $sp^3$ hybridized carbons.
$(ii)$ $H_2C=CH-CH_2^{-}$: The negative charge is stabilized by resonance (delocalization).
$(iv)$ $CH_3^{-}$: The negative charge is on an $sp^3$ hybridized carbon atom.
$(i)$ $(CH_3)_3C^{-}$: The negative charge is on an $sp^3$ hybridized carbon atom,which is further destabilized by the $+I$ effect of three methyl groups. It is the least stable.
Stability order: $(v) > (iii) > (ii) > (iv) > (i)$.
Since Basicity $\propto \frac{1}{\text{Stability}}$,the increasing order of basicity is: $(v) < (iii) < (ii) < (iv) < (i)$.

(A) $Be$ has the electronic configuration $1s^{2} 2s^{2}$.
$B$ has the electronic configuration $1s^{2} 2s^{2} 2p^{1}$.
Statement $(I)$ is correct: It is easier to remove a $2p$ electron than a $2s$ electron because the $2p$ orbital is higher in energy and less stable.
Statement $(II)$ is correct: The $2p$ electron in $B$ experiences greater shielding from the $1s^{2} 2s^{2}$ core compared to the $2s$ electrons in $Be$,which are less shielded.
Statement $(III)$ is correct: The $2s$ orbital is closer to the nucleus and has higher penetration power than the $2p$ orbital.
Statement $(IV)$ is incorrect: $B$ has a smaller atomic radius than $Be$ due to the increase in effective nuclear charge across the period.
Therefore,statements $(I), (II),$ and $(III)$ are correct.

(D) To identify the nature of the oxides,we consider their periodic trends:
$1$. Non-metallic oxides are generally acidic (e.g.,$Cl_2O, N_2O_3, SO_3, P_4O_{10}$).
$2$. Metallic oxides are generally basic (e.g.,$MgO, CaO, Na_2O, Li_2O$).
$3$. Certain oxides like $Al_2O_3$ and $ZnO$ are amphoteric,meaning they react with both acids and bases.
Evaluating the options:
- Option $A$: $MgO$ (Basic),$Cl_2O$ (Acidic),$Al_2O_3$ (Amphoteric). This matches the required order (Acidic,Basic,Amphoteric) incorrectly.
- Option $B$: $Cl_2O$ (Acidic),$CaO$ (Basic),$P_4O_{10}$ (Acidic). This does not contain an amphoteric oxide.
- Option $C$: $Na_2O$ (Basic),$SO_3$ (Acidic),$Al_2O_3$ (Amphoteric). This does not match the required order.
- Option $D$: $N_2O_3$ (Acidic),$Li_2O$ (Basic),$Al_2O_3$ (Amphoteric). This matches the required order (Acidic,Basic,Amphoteric).

(C) The Friedel-Crafts reaction is an electrophilic aromatic substitution. It requires an activated or moderately deactivated aromatic ring.
$1$. $Benzamide$ has a strongly electron-withdrawing $-CONH_2$ group,which deactivates the ring.
$2$. $Aniline$ $(-NH_2)$ and $Phenol$ $(-OH)$ contain lone pairs that react with the Lewis acid catalyst $(AlCl_3)$ to form a complex,which strongly deactivates the ring and prevents the reaction.
$3$. $Chlorobenzene$ is a deactivated ring due to the $-I$ effect of the chlorine atom,but it is still capable of undergoing Friedel-Crafts reactions,unlike the other options which are either strongly deactivated or form complexes with the catalyst. Therefore,among the given choices,$Chlorobenzene$ will produce the highest yield.

(D) substance can act as both an oxidising and a reducing agent if the central atom is in an intermediate oxidation state.
$1$. In $H_2O_2$,the oxidation state of $O$ is $-1$,which can increase to $0$ (reducing) or decrease to $-2$ (oxidising).
$2$. In $H_2SO_3$,the oxidation state of $S$ is $+4$,which can increase to $+6$ (reducing) or decrease to lower values (oxidising).
$3$. In $HNO_2$,the oxidation state of $N$ is $+3$,which can increase to $+5$ (reducing) or decrease to lower values (oxidising).
$4$. In $H_3PO_4$,the oxidation state of $P$ is $+5$,which is its maximum oxidation state. Therefore,it can only act as an oxidising agent and cannot be further oxidised to act as a reducing agent.

(A) According to the Bohr quantization condition, the circumference of the orbit is an integral multiple of the de Broglie wavelength: $2 \pi r = n \lambda$.
For the $n^{th}$ Bohr orbit, the radius is given by $r = n^{2} a_{0}$, where $a_{0}$ is the Bohr radius.
For the $4^{th}$ orbit $(n = 4)$, the radius is $r = 4^{2} a_{0} = 16 a_{0}$.
Substituting these values into the quantization equation: $2 \pi (16 a_{0}) = 4 \lambda$.
Solving for $\lambda$: $\lambda = \frac{32 \pi a_{0}}{4} = 8 \pi a_{0}$.

(D) Hardness in terms of $CaCO_{3}$ equivalents is calculated by converting the concentration of the salt to the equivalent concentration of $CaCO_{3}$.
Given concentration of $MgSO_{4} = 10^{-3} \; M = 10^{-3} \; mol/L$.
Since $1 \; mol$ of $MgSO_{4}$ is equivalent to $1 \; mol$ of $CaCO_{3}$ (both have a valency factor of $2$),the concentration of $CaCO_{3}$ equivalents is $10^{-3} \; mol/L$.
Mass of $CaCO_{3} = 10^{-3} \; mol \times 100 \; g/mol = 0.1 \; g$ in $1 \; L$ of water.
$ppm$ is defined as parts per million,which is $mg$ of solute per $L$ of solution.
$0.1 \; g = 100 \; mg$.
Therefore,the hardness is $100 \; ppm$.

(D) Given:
Density $(d)$ = $1.4 \; g/mL$
Mass percentage = $63\%$
Molar mass of $HNO_{3}$ $(M_{w})$ = $63 \; g/mol$
Formula for Molarity $(M)$:
$M = \frac{\text{mass percentage} \times d \times 10}{\text{Molar mass}}$
Calculation:
$M = \frac{63 \times 1.4 \times 10}{63}$
$M = 1.4 \times 10 = 14 \; M$
Therefore,the molarity is $14 \; M$.

(A) The molecular formula of histamine is $C_5H_9N_3$.
The molar mass of histamine is $(5 \times 12) + (9 \times 1) + (3 \times 14) = 60 + 9 + 42 = 111 \ g/mol$.
The mass of nitrogen in one mole of histamine is $3 \times 14 = 42 \ g$.
The mass percentage of nitrogen is calculated as:
$\text{Mass } \% = (\frac{\text{Mass of nitrogen}}{\text{Molar mass of histamine}}) \times 100$
$\text{Mass } \% = (\frac{42}{111}) \times 100 = 37.8378 \% \approx 37.84 \%$.

(A) Let $I = \int \frac{d \theta}{\cos ^{2} \theta(\tan 2 \theta+\sec 2 \theta)}$.
Using $\tan 2\theta = \frac{2\tan \theta}{1-\tan^2 \theta}$ and $\sec 2\theta = \frac{1+\tan^2 \theta}{1-\tan^2 \theta}$,we get:
$I = \int \frac{\sec^2 \theta d \theta}{\frac{2\tan \theta}{1-\tan^2 \theta} + \frac{1+\tan^2 \theta}{1-\tan^2 \theta}} = \int \frac{(1-\tan^2 \theta) \sec^2 \theta d \theta}{(1+\tan \theta)^2}$.
Let $\tan \theta = t$,then $\sec^2 \theta d \theta = dt$.
$I = \int \frac{1-t^2}{(1+t)^2} dt = \int \frac{(1-t)(1+t)}{(1+t)^2} dt = \int \frac{1-t}{1+t} dt$.
$I = \int \frac{2-(1+t)}{1+t} dt = \int \left( \frac{2}{1+t} - 1 \right) dt$.
$I = 2 \log_e |1+t| - t + C$.
Substituting $t = \tan \theta$,we get $I = -\tan \theta + 2 \log_e |1+\tan \theta| + C$.
Comparing this with $\lambda \tan \theta + 2 \log_e |f(\theta)| + C$,we find $\lambda = -1$ and $f(\theta) = 1+\tan \theta$.

(C) Let the thickness of the ice be $h \; cm$.
The radius of the iron ball is $r = 10 \; cm$.
The total radius of the sphere (iron ball + ice) is $R = 10 + h$.
The volume of the ice layer $V$ is given by the difference between the volume of the total sphere and the volume of the iron ball:
$V = \frac{4}{3} \pi (10 + h)^{3} - \frac{4}{3} \pi (10)^{3}$.
Differentiating both sides with respect to time $t$:
$\frac{dV}{dt} = \frac{4}{3} \pi \cdot 3(10 + h)^{2} \cdot \frac{dh}{dt} = 4 \pi (10 + h)^{2} \frac{dh}{dt}$.
Given that the ice melts at a rate of $\frac{dV}{dt} = 50 \; cm^{3}/min$,we have:
$50 = 4 \pi (10 + h)^{2} \frac{dh}{dt}$.
When the thickness $h = 5 \; cm$:
$50 = 4 \pi (10 + 5)^{2} \frac{dh}{dt} = 4 \pi (15)^{2} \frac{dh}{dt} = 4 \pi (225) \frac{dh}{dt} = 900 \pi \frac{dh}{dt}$.
Therefore,$\frac{dh}{dt} = \frac{50}{900 \pi} = \frac{1}{18 \pi} \; cm/min$.

(B) Let $z = \frac{1+\sin \frac{2 \pi}{9}+i \cos \frac{2 \pi}{9}}{1+\sin \frac{2 \pi}{9}-i \cos \frac{2 \pi}{9}}$.
Using $\sin \theta = \cos(\frac{\pi}{2}-\theta)$ and $\cos \theta = \sin(\frac{\pi}{2}-\theta)$,we have $\sin \frac{2 \pi}{9} = \cos \frac{5 \pi}{18}$ and $\cos \frac{2 \pi}{9} = \sin \frac{5 \pi}{18}$.
So,$z = \frac{1+\cos \frac{5 \pi}{18}+i \sin \frac{5 \pi}{18}}{1+\cos \frac{5 \pi}{18}-i \sin \frac{5 \pi}{18}}$.
Using $1+\cos \theta = 2 \cos^2 \frac{\theta}{2}$ and $\sin \theta = 2 \sin \frac{\theta}{2} \cos \frac{\theta}{2}$,we get:
$z = \frac{2 \cos^2 \frac{5 \pi}{36} + 2i \sin \frac{5 \pi}{36} \cos \frac{5 \pi}{36}}{2 \cos^2 \frac{5 \pi}{36} - 2i \sin \frac{5 \pi}{36} \cos \frac{5 \pi}{36}} = \frac{\cos \frac{5 \pi}{36} + i \sin \frac{5 \pi}{36}}{\cos \frac{5 \pi}{36} - i \sin \frac{5 \pi}{36}}$.
$z = \frac{e^{i 5 \pi / 36}}{e^{-i 5 \pi / 36}} = e^{i 10 \pi / 36} = e^{i 5 \pi / 18}$.
Therefore,$z^3 = (e^{i 5 \pi / 18})^3 = e^{i 15 \pi / 18} = e^{i 5 \pi / 6}$.
$z^3 = \cos \frac{5 \pi}{6} + i \sin \frac{5 \pi}{6} = -\frac{\sqrt{3}}{2} + i \frac{1}{2} = -\frac{1}{2}(\sqrt{3}-i)$.

(D) Given the equation $\cos^{4} \theta + \sin^{4} \theta + \lambda = 0$,we can write $\lambda = -(\sin^{4} \theta + \cos^{4} \theta)$.
Using the identity $\sin^{4} \theta + \cos^{4} \theta = (\sin^{2} \theta + \cos^{2} \theta)^{2} - 2 \sin^{2} \theta \cos^{2} \theta$,we get $\sin^{4} \theta + \cos^{4} \theta = 1 - \frac{1}{2} \sin^{2} 2\theta$.
Thus,$\lambda = -(1 - \frac{1}{2} \sin^{2} 2\theta) = \frac{1}{2} \sin^{2} 2\theta - 1$.
Since $0 \le \sin^{2} 2\theta \le 1$,we have $0 \le \frac{1}{2} \sin^{2} 2\theta \le \frac{1}{2}$.
Subtracting $1$ from all parts,we get $-1 \le \frac{1}{2} \sin^{2} 2\theta - 1 \le -\frac{1}{2}$.
Therefore,$\lambda \in [-1, -\frac{1}{2}]$.

(C) The truth value of the implication $(p \wedge q) \rightarrow (\sim q \vee r)$ is $F$ (False).
An implication $A \rightarrow B$ is false only when $A$ is $T$ (True) and $B$ is $F$ (False).
Therefore,$(p \wedge q) = T$ and $(\sim q \vee r) = F$.
For $(p \wedge q) = T$,both $p$ and $q$ must be $T$.
Now,substitute $q = T$ into the second condition: $(\sim T \vee r) = F$.
This simplifies to $(F \vee r) = F$.
For this disjunction to be $F$,$r$ must be $F$.
Thus,the truth values are $p = T, q = T, r = F$.

(C) Let $x = \sin^{-1}(\frac{4}{5})$,$y = \sin^{-1}(\frac{5}{13})$,and $z = \sin^{-1}(\frac{16}{65})$.
Converting to $\tan^{-1}$ form:
$x = \tan^{-1}(\frac{4}{3})$,$y = \tan^{-1}(\frac{5}{12})$,$z = \tan^{-1}(\frac{16}{63})$.
Now,calculate $x + y = \tan^{-1}(\frac{4}{3}) + \tan^{-1}(\frac{5}{12}) = \tan^{-1}(\frac{4/3 + 5/12}{1 - (4/3)(5/12)}) = \tan^{-1}(\frac{16/12 + 5/12}{1 - 20/36}) = \tan^{-1}(\frac{21/12}{16/36}) = \tan^{-1}(\frac{7/4}{4/9}) = \tan^{-1}(\frac{63}{16})$.
Then,$(x + y) + z = \tan^{-1}(\frac{63}{16}) + \tan^{-1}(\frac{16}{63})$.
Since $\tan^{-1}(a) + \tan^{-1}(1/a) = \frac{\pi}{2}$ for $a > 0$,we have $(x + y) + z = \frac{\pi}{2}$.
Finally,$2 \pi - (x + y + z) = 2 \pi - \frac{\pi}{2} = \frac{3 \pi}{2}$.

(B) Given equation: $(a+\sqrt{2} b \cos x)(a-\sqrt{2} b \cos y)=a^{2}-b^{2}$
Expanding the left side:
$a^{2} - \sqrt{2}ab \cos y + \sqrt{2}ab \cos x - 2b^{2} \cos x \cos y = a^{2} - b^{2}$
Subtracting $a^{2}$ from both sides:
$-\sqrt{2}ab \cos y + \sqrt{2}ab \cos x - 2b^{2} \cos x \cos y = -b^{2}$
Divide by $-1$:
$\sqrt{2}ab \cos y - \sqrt{2}ab \cos x + 2b^{2} \cos x \cos y = b^{2}$
Differentiating both sides with respect to $y$:
$\sqrt{2}ab(-\sin y) - \sqrt{2}ab(-\sin x \frac{dx}{dy}) + 2b^{2}[\cos x(-\sin y) + \cos y(-\sin x \frac{dx}{dy})] = 0$
At point $\left(\frac{\pi}{4}, \frac{\pi}{4}\right)$,$\sin x = \sin y = \frac{1}{\sqrt{2}}$ and $\cos x = \cos y = \frac{1}{\sqrt{2}}$:
$-\sqrt{2}ab(\frac{1}{\sqrt{2}}) + \sqrt{2}ab(\frac{1}{\sqrt{2}} \frac{dx}{dy}) + 2b^{2}[-\frac{1}{2} - \frac{1}{2} \frac{dx}{dy}] = 0$
$-ab + ab \frac{dx}{dy} - b^{2} - b^{2} \frac{dx}{dy} = 0$
$\frac{dx}{dy}(ab - b^{2}) = ab + b^{2}$
$\frac{dx}{dy} = \frac{ab + b^{2}}{ab - b^{2}} = \frac{b(a+b)}{b(a-b)} = \frac{a+b}{a-b}$

(D) The lines $L_{1}$ and $L_{2}$ are coplanar if the determinant of the matrix formed by the vector connecting a point on each line and the direction vectors of the lines is zero.
Point on $L_{1}$ is $P_{1}(-1, 2, 1)$ and direction vector is $\vec{v}_{1} = (2, -1, 1)$.
Point on $L_{2}$ is $P_{2}(-2, -1, -1)$ and direction vector is $\vec{v}_{2} = (\alpha, 5-\alpha, 1)$.
The vector connecting $P_{1}$ and $P_{2}$ is $\vec{P_{1}P_{2}} = (-2 - (-1), -1 - 2, -1 - 1) = (-1, -3, -2)$.
For coplanarity,$\left|\begin{array}{ccc} -1 & -3 & -2 \\ 2 & -1 & 1 \\ \alpha & 5-\alpha & 1 \end{array}\right| = 0$.
Expanding the determinant:
$-1(-1 - (5-\alpha)) + 3(2 - \alpha) - 2(2(5-\alpha) - \alpha(-1)) = 0$
$-1(\alpha - 6) + 6 - 3\alpha - 2(10 - 2\alpha + \alpha) = 0$
$-\alpha + 6 + 6 - 3\alpha - 20 + 2\alpha = 0$
$-2\alpha - 8 = 0 \Rightarrow \alpha = -4$.
Substituting $\alpha = -4$ into $L_{2}$,we get $\frac{x+2}{-4} = \frac{y+1}{5-(-4)} = \frac{z+1}{1} \Rightarrow \frac{x+2}{-4} = \frac{y+1}{9} = \frac{z+1}{1}$.
Checking the options,for point $(2, -10, -2)$:
$\frac{2+2}{-4} = \frac{4}{-4} = -1$,$\frac{-10+1}{9} = \frac{-9}{9} = -1$,$\frac{-2+1}{1} = -1$.
Since all ratios are equal,the line $L_{2}$ passes through $(2, -10, -2)$.

(B) Given the differential equation: $\frac{5+e^x}{2+y} \frac{dy}{dx} = -e^x$.
Separating the variables,we get: $\int \frac{dy}{2+y} = \int \frac{-e^x}{e^x+5} dx$.
Integrating both sides,we obtain: $\ln(y+2) = -\ln(e^x+5) + C_1$.
This simplifies to: $\ln(y+2) + \ln(e^x+5) = C_1$,which implies $(y+2)(e^x+5) = C$.
Given the condition $y(0)=1$,we substitute $x=0$ and $y=1$: $(1+2)(e^0+5) = C \Rightarrow 3(1+5) = C \Rightarrow C = 18$.
Thus,the equation is $(y+2)(e^x+5) = 18$,or $y+2 = \frac{18}{e^x+5}$.
To find $y(\ln 13)$,substitute $x = \ln 13$: $y+2 = \frac{18}{e^{\ln 13}+5} = \frac{18}{13+5} = \frac{18}{18} = 1$.
Therefore,$y = 1 - 2 = -1$.

(A) Let the two remaining observations be $x$ and $y$.
Given the mean $\bar{x} = 8$ for $n = 7$ observations:
$\frac{2 + 4 + 10 + 12 + 14 + x + y}{7} = 8$
$42 + x + y = 56 \implies x + y = 14$ $(i)$
Given the variance $\sigma^2 = 16$:
$\sigma^2 = \frac{\sum x_i^2}{n} - (\bar{x})^2$
$16 = \frac{2^2 + 4^2 + 10^2 + 12^2 + 14^2 + x^2 + y^2}{7} - 8^2$
$16 + 64 = \frac{4 + 16 + 100 + 144 + 196 + x^2 + y^2}{7}$
$80 \times 7 = 460 + x^2 + y^2$
$560 = 460 + x^2 + y^2 \implies x^2 + y^2 = 100$ $(ii)$
We know that $(x - y)^2 = 2(x^2 + y^2) - (x + y)^2$
$(x - y)^2 = 2(100) - (14)^2 = 200 - 196 = 4$
$|x - y| = \sqrt{4} = 2$

(B) Given $f(x) = \frac{a-x}{a+x}$.
We are given $(f \circ f)(x) = x$.
$f(f(x)) = \frac{a - f(x)}{a + f(x)} = \frac{a - \frac{a-x}{a+x}}{a + \frac{a-x}{a+x}}$
$= \frac{a(a+x) - (a-x)}{a(a+x) + (a-x)} = \frac{a^2 + ax - a + x}{a^2 + ax + a - x} = x$
$a^2 + ax - a + x = x(a^2 + ax + a - x)$
$a^2 + ax - a + x = a^2x + ax^2 + ax - x^2$
Rearranging the terms: $x^2(a-1) + x(a^2-1) - a(a-1) = 0$.
For this to hold for all $x$,the coefficients must be zero.
$a-1 = 0 \Rightarrow a = 1$.
Thus,$f(x) = \frac{1-x}{1+x}$.
Now,$f(-\frac{1}{2}) = \frac{1 - (-1/2)}{1 + (-1/2)} = \frac{3/2}{1/2} = 3$.

(C) Toilet cleaning liquid typically contains $HCl$ (hydrochloric acid).
To neutralize the acid on the skin,a mild base like aqueous $NaHCO_{3}$ (sodium bicarbonate) is used.
$NaOH$ is avoided because it is highly corrosive and can cause severe chemical burns.

(C) In the $3^{rd}$ period,as we move from left to right,the metallic character decreases and non-metallic character increases.
Consequently,the nature of oxides changes from basic to amphoteric to acidic.
$X$ (basic oxide) is on the left,$Y$ (amphoteric oxide) is in the middle,and $Z$ (acidic oxide) is on the right.
Since atomic number increases from left to right in a period,the order of atomic numbers is $X < Y < Z$.

(D) For a given principal quantum number $n=4$,the possible values of azimuthal quantum number $\ell$ are $0, 1, 2, 3$.
These correspond to the subshells $4s, 4p, 4d, 4f$ respectively.
The magnetic quantum number $m$ ranges from $-\ell$ to $+\ell$.
For $4s$ $(\ell=0)$: $m=0$.
For $4p$ $(\ell=1)$: $m=-1, 0, +1$.
For $4d$ $(\ell=2)$: $m=-2, -1, 0, +1, +2$.
For $4f$ $(\ell=3)$: $m=-3, -2, -1, 0, +1, +2, +3$.
The value $m=-2$ is present in the $4d$ and $4f$ subshells.
Therefore,the number of subshells associated with $n=4$ and $m=-2$ is $2$.

(B) Both $Li$ and $Mg$ exhibit diagonal relationship and show similar chemical properties.
$Li$ and $Mg$ do not form solid hydrogencarbonates ($LiHCO_3$ and $Mg(HCO_3)_2$ are unstable in solid state).
Both $Li$ and $Mg$ react directly with nitrogen to form their respective nitrides,$Li_3N$ and $Mg_3N_2$.

(B) The interaction energy $(E)$ depends on the distance $(r)$ between particles as follows:
$1$. Ion-ion interaction: $E \propto \frac{1}{r}$
$2$. Dipole-dipole interaction: $E \propto \frac{1}{r^{3}}$
$3$. London dispersion forces: $E \propto \frac{1}{r^{6}}$
Therefore,the correct matching is $I-a, II-c, III-d$.

(A) For $SF_4$,the central sulfur atom has $6$ valence electrons.
It forms $4$ $\sigma$ bonds with fluorine atoms and has $1$ lone pair of electrons.
Total electron pairs = $4 + 1 = 5$.
According to $VSEPR$ theory,a steric number of $5$ corresponds to a trigonal bipyramidal electron geometry.

(A) The combustion reaction of ethanol is: $C_2H_5OH_{(\ell)} + 3O_{2_{(g)}} \longrightarrow 2CO_{2_{(g)}} + 3H_2O_{(\ell)}$
The change in the number of moles of gaseous species is $\Delta n_g = n_{p(g)} - n_{r(g)} = 2 - 3 = -1$.
The relationship between enthalpy change and internal energy change is given by $\Delta H = \Delta U + (\Delta n_g) RT$.
Given $\Delta H = -327 \ kcal = -327000 \ cal$,$T = 27 + 273 = 300 \ K$,and $R = 2 \ cal \ mol^{-1} \ K^{-1}$.
Substituting the values: $-327000 = \Delta U + (-1) \times 2 \times 300$.
$-327000 = \Delta U - 600$.
$\Delta U = -327000 + 600 = -326400 \ cal$.
The heat evolved at constant volume is equal to the magnitude of the change in internal energy,which is $326400 \ cal$.

(D) Given mass ratios: $C : H = 4 : 1$ and $C : O = 3 : 4$.
To find the combined mass ratio $C : H : O$,we equate the $C$ part: $C : H = 12 : 3$ and $C : O = 12 : 16$.
Thus,$C : H : O = 12 : 3 : 16$.
Dividing by atomic masses $(C=12, H=1, O=16)$,the mole ratio is $C : H : O = (12/12) : (3/1) : (16/16) = 1 : 3 : 1$.
The empirical formula is $CH_3O$. Since the compound is a saturated acyclic organic compound,the molecular formula is $C_2H_6O_2$ (e.g.,ethylene glycol).
The combustion reaction is: $C_2H_6O_2 + 2.5 O_2 \longrightarrow 2 CO_2 + 3 H_2O$.
For $1$ mole of $X$,$2.5$ moles of $O_2$ are required.
For $2$ moles of $X$,the moles of $O_2$ required $= 2 \times 2.5 = 5$ moles.

(A) According to Einstein's photoelectric equation: $E = W + K.E._{max}$
$K.E._{max} = E - W$
Energy of incident photon $E = \frac{hc}{\lambda} = \frac{6.63 \times 10^{-34} \times 3 \times 10^{8}}{300 \times 10^{-9}} = 6.63 \times 10^{-19} \ J$
Given work function $W = 4.41 \times 10^{-19} \ J$
$K.E._{max} = 6.63 \times 10^{-19} - 4.41 \times 10^{-19} = 2.22 \times 10^{-19} \ J$
Converting to $10^{-21} \ J$: $2.22 \times 10^{-19} \ J = 222 \times 10^{-21} \ J$
Thus,the value is $222$.

(A) $1$. Identify the principal functional group: The compound contains a carboxylic acid $(-COOH)$ group and an aldehyde $(-CHO)$ group. According to $IUPAC$ priority rules,the carboxylic acid group has higher priority than the aldehyde group. Therefore,the parent chain must include the carbon of the $-COOH$ group,and the suffix will be $-oic \ acid$.
$2$. Number the carbon chain: Start numbering from the carboxylic acid carbon as $C-1$. The longest chain containing the principal functional group and the double bond is a $6-$carbon chain.
$3$. Identify substituents: The aldehyde group at $C-6$ is treated as a substituent and named as $oxo-$. There is a methyl group at $C-2$ and another methyl group at $C-5$.
$4$. Assemble the name: The substituents are $2,5-$dimethyl and $6-oxo$. The double bond is at $C-3$. Combining these,the name is $2,5-$dimethyl$-6-$oxo$-hex-3-$enoic acid.

(A) The quantum nature of atoms is demonstrated by phenomena where energy is quantized,such as the photoelectric effect,atomic spectra,and black-body radiation.
$1$. The photoelectric effect (option $B$) shows that light interacts as discrete packets of energy (photons).
$2$. Atomic spectra (option $C$) show that electrons in atoms occupy discrete energy levels.
$3$. Black-body radiation (option $D$) shows that energy emission is quantized.
However,the internal energy of a gas like $Ar$ (option $A$) increases continuously with temperature according to classical kinetic theory,which is a classical,not a quantum,manifestation. Therefore,the figure representing the internal energy of $Ar$ versus temperature is not a direct manifestation of the quantum nature of atoms.

(C) Penicillin has a core structure consisting of a $\beta$-lactam ring fused to a thiazolidine ring. By examining the structure,we can identify the chiral centers (carbon atoms bonded to four different groups). As shown in the structure,there are $3$ chiral centers marked with an asterisk $(*)$. Therefore,the total number of chiral centers in penicillin is $3$.

(C) The chemical formula of the complex is $[Co(NH_{3})_{6}]Cl_{3}$.
Each mole of $[Co(NH_{3})_{6}]Cl_{3}$ contains $3$ moles of ionizable $Cl^{-}$ ions.
The reaction with $AgNO_{3}$ is: $[Co(NH_{3})_{6}]Cl_{3} + 3AgNO_{3} \rightarrow [Co(NH_{3})_{6}](NO_{3})_{3} + 3AgCl$.
Number of moles of $[Co(NH_{3})_{6}]Cl_{3} = \frac{0.3 \; g}{267.46 \; g/mol} \approx 0.0011216 \; mol$.
Since $1$ mole of complex reacts with $3$ moles of $AgNO_{3}$,moles of $AgNO_{3}$ required $= 3 \times 0.0011216 = 0.0033648 \; mol$.
Using the molarity formula $M = \frac{n}{V(L)}$,we have $V(L) = \frac{n}{M} = \frac{0.0033648}{0.125} = 0.0269184 \; L$.
Converting to $mL$,$V = 0.0269184 \times 1000 = 26.92 \; mL$.

(A) The reduction half-reaction is: $O_{2(g)} + 4 H^+ + 4 e^- \rightarrow 2 H_2O_{(l)} ; E_{red}^{0} = 1.23 \ V$
Using the Nernst equation:
$E = E^0 - \frac{0.0591}{n} \log \frac{1}{[H^+]^4}$
Given $pH = 5$,so $[H^+] = 10^{-5} \ M$.
$E = 1.23 - \frac{0.0591}{4} \log \frac{1}{(10^{-5})^4}$
$E = 1.23 - \frac{0.0591}{4} \log (10^{20})$
$E = 1.23 - \frac{0.0591 \times 20}{4}$
$E = 1.23 - 0.0591 \times 5$
$E = 1.23 - 0.2955 = 0.9345 \ V$
Note: The question asks for the potential of the oxidation reaction $2 H_2O \rightarrow O_2 + 4 H^+ + 4 e^-$. The reduction potential calculated is $0.9345 \ V$. The oxidation potential is $-0.9345 \ V$. However,based on standard textbook problems of this type,the question often implies the reduction potential of the $O_2/H_2O$ couple at the given $pH$. Re-evaluating the provided options,$1.52 \ V$ is the standard answer for $pH = 0$ to $pH = 5$ shift calculations in specific contexts,but mathematically $0.9345 \ V$ is correct for the reduction potential.

(B) $(I) [Cr(H_2O)_6]^{2+}$: $Cr^{2+} \Rightarrow [Ar] 3d^4$. $H_2O$ is a weak field ligand. Unpaired $e^- = 4$. Magnetic moment $\mu = \sqrt{4(4+2)} = \sqrt{24} \ BM \approx 4.89 \ BM$.
$(II) [Fe(CN)_6]^{4-}$: $Fe^{2+} \Rightarrow [Ar] 3d^6$. $CN^-$ is a strong field ligand. Unpaired $e^- = 0$. Magnetic moment $\mu = 0 \ BM$.
$(III) [Fe(C_2O_4)_3]^{3-}$: $Fe^{3+} \Rightarrow [Ar] 3d^5$. Given $\Delta_0 > P$,it is a low spin complex. Unpaired $e^- = 1$. Magnetic moment $\mu = \sqrt{1(1+2)} = \sqrt{3} \ BM \approx 1.73 \ BM$.
$(IV) [CoCl_4]^{2-}$: $Co^{2+} \Rightarrow [Ar] 3d^7$. Tetrahedral complex,$Cl^-$ is a weak field ligand. Unpaired $e^- = 3$. Magnetic moment $\mu = \sqrt{3(3+2)} = \sqrt{15} \ BM \approx 3.87 \ BM$.
Comparing the values: $4.89 (I) > 3.87 (IV) > 1.73 (III) > 0 (II)$.
Therefore,the correct order is $(I) > (IV) > (III) > (II)$.

(D) When a mixture of gases is introduced into a closed vessel containing charcoal,the charcoal acts as an adsorbent.
As time passes,the gas molecules get adsorbed onto the surface of the charcoal.
Since the number of gas molecules in the gaseous phase decreases due to adsorption,the pressure exerted by the gas mixture in the vessel decreases with time until it reaches an equilibrium state.
Therefore,the graph showing a decrease in pressure over time is the correct representation.

(D) $PHBV$ stands for Poly $\beta$-hydroxybutyrate-co-$\beta$-hydroxyvalerate.
The monomers involved are $3$-hydroxybutanoic acid and $3$-hydroxypentanoic acid.
Both of these monomers contain a chiral carbon atom (the carbon atom attached to the $-OH$ group),making them chiral.

(D) The bond length of the $C-Cl$ bond depends on the extent of resonance.
In $Cl-CH=CH-NO_2$,the $-NO_2$ group is a strong electron-withdrawing group ($-M$ effect),while the $Cl$ atom acts as an electron donor ($+M$ effect).
This creates a strong conjugation effect where the lone pair on $Cl$ is delocalized towards the $NO_2$ group,increasing the double bond character of the $C-Cl$ bond.
Greater double bond character leads to a shorter bond length.
Therefore,$Cl-CH=CH-NO_2$ has the shortest $C-Cl$ bond.

(A) To determine the basicity,we look at the availability of the lone pair on the nitrogen atom:
$1$. $(III)$ Cyclohexylamine: The nitrogen is attached to an $sp^3$ hybridized carbon. There is no resonance,and the alkyl group is electron-donating,making it the most basic.
$2$. $(II)$ Pyridine: The lone pair is in an $sp^2$ orbital,which is more electronegative than $sp^3$,making it less basic than cyclohexylamine but more basic than aniline.
$3$. $(I)$ Aniline: The lone pair on the nitrogen is involved in resonance with the benzene ring,significantly reducing its availability for protonation.
$4$. $(IV)$ Pyrrole: The lone pair on the nitrogen is part of the aromatic sextet ($6\pi$ electrons),making it unavailable for protonation. Thus,it is the least basic.
Therefore,the decreasing order of basicity is $(III) > (II) > (I) > (IV)$.

(B) The reaction of $[P]$ with $NaNO_2/HCl$ at $0-5^oC$ followed by coupling with $\beta$-naphthol to form a colored solid indicates that $[P]$ is a primary aromatic amine.
The reaction of $[P]$ with $Br_2/H_2O$ to form $C_7H_6NBr_3$ (a tribromo derivative) confirms that the amino group is activating the ring towards electrophilic substitution.
Comparing the options:
$m$-toluidine ($3$-methylaniline) reacts with $NaNO_2/HCl$ to form a diazonium salt,which then couples with $\beta$-naphthol to form an azo dye (colored solid).
It also reacts with $Br_2/H_2O$ to form $2,4,6$-tribromo-$3$-methylaniline $(C_7H_6NBr_3)$.
Therefore,$[P]$ is $3$-methylaniline.

(D) $\text{Molisch's test}$ is a general test for carbohydrates. $A$ and $B$ are carbohydrates as they give a positive $\text{Molisch's test}$.
$\text{Barfoed test}$ is used to distinguish monosaccharides from disaccharides; monosaccharides give a positive test.
Since $A$ gives a negative $\text{Barfoed test}$,it is a disaccharide (e.g.,$\text{Lactose}$),and since $B$ gives a positive test,it is a monosaccharide (e.g.,$\text{Glucose}$).
$\text{Biuret test}$ is positive for proteins. Since $C$ gives a positive $\text{Biuret test}$,it is a protein (e.g.,$\text{Albumin}$).
Therefore,$A = \text{Lactose}, B = \text{Glucose}, C = \text{Albumin}$.

(D) The complex $[Co(NH_{3})_{4}Cl_{2}]^{+}$ exhibits geometrical isomerism,existing as $cis$ and $trans$ isomers.
In the $trans$ isomer,the two $Cl$ ligands are at opposite positions ($180^{\circ}$ angle).
In the $cis$ isomer,the two $Cl$ ligands are adjacent to each other,resulting in a $Cl-Co-Cl$ bond angle of $90^{\circ}$.

(C) $1$. Calculate the freezing point of the solution:
$\Delta T_{f} = K_{f} \times m = 2.0 \; K \; kg \; mol^{-1} \times 0.5 \; mol \; kg^{-1} = 1.0 \; K$.
Freezing point of solution $T = 273 - 1 = 272 \; K$.
$2$. Calculate the initial pressure of the gas:
Using $PV = nRT$,$P = \frac{nRT}{V} = \frac{0.1 \; mol \times 0.08 \; dm^{3} \; atm \; K^{-1} \; mol^{-1} \times 272 \; K}{1.0 \; dm^{3}} = 2.176 \; atm$.
$3$. After withdrawing the stoppers,the piston is frictionless and exposed to atmospheric pressure $(P_{ext} = 1 \; atm)$. The gas will expand until its internal pressure equals the external pressure.
Using Boyle's Law $(P_{1}V_{1} = P_{2}V_{2})$:
$2.176 \; atm \times 1.0 \; L = 1.0 \; atm \times V_{2}$.
$V_{2} = 2.176 \; L \approx 2.18 \; L$.

(B) The mass of $O_2$ is $10.30 \; mg = 10.30 \times 10^{-3} \; g$.
The volume of seawater is $1 \; L = 1000 \; mL$.
The density of seawater is $1.03 \; g/mL$.
The mass of seawater = $\text{density} \times \text{volume} = 1.03 \; g/mL \times 1000 \; mL = 1030 \; g$.
The concentration in $ppm$ is calculated as:
$ppm = \frac{\text{mass of solute}}{\text{mass of solution}} \times 10^{6}$
$ppm = \frac{10.30 \times 10^{-3} \; g}{1030 \; g} \times 10^{6} = \frac{10.30}{1030} \times 10^{3} = 0.01 \times 1000 = 10$.

(D) The rate constant $k$ is inversely proportional to the time $t$ required for a specific change (doubling of population),so $k \propto 1/t$.
Using the Arrhenius equation in the form $\ln(k_2 / k_1) = \frac{E_a}{R} [\frac{1}{T_1} - \frac{1}{T_2}]$,we substitute $k_2/k_1 = t_1/t_2$:
$\ln(t_1 / t_2) = \frac{E_a}{R} [\frac{1}{T_1} - \frac{1}{T_2}]$
$\ln(60 / 40) = \frac{E_a}{8.3} [\frac{1}{300} - \frac{1}{400}]$
$\ln(1.5) = \frac{E_a}{8.3} [\frac{400 - 300}{120000}]$
$0.405 = \frac{E_a}{8.3} \times \frac{100}{120000}$
$0.405 = \frac{E_a}{8.3} \times \frac{1}{1200}$
$E_a = 0.405 \times 8.3 \times 1200 = 4033.8 \; J / mol \approx 4.03 \; kJ / mol$.
Comparing with the given options,the closest value is $3.98 \; kJ / mol$.

(C) In a chromate ion $(CrO_4^{2-})$,the structure consists of one $Cr$ atom bonded to four oxygen atoms. There are two double bonds and two single bonds,totaling $6$ $Cr-O$ bonds $(4\sigma + 2\pi)$.
In a dichromate ion $(Cr_2O_7^{2-})$,there are two $Cr$ atoms linked by one bridging oxygen atom $(Cr-O-Cr)$. Each $Cr$ atom is also bonded to three terminal oxygen atoms. This results in a total of $12$ $Cr-O$ bonds $(8\sigma + 4\pi)$.
The sum of the total number of $Cr-O$ bonds is $6 + 12 = 18$.

(D) is formed by the reaction of $A$ with $CH_3MgBr$.
Since $B$ gives $2\text{-methyl-2-butene}$ $(CH_3-C(CH_3)=CH-CH_3)$ upon heating with $Cu$ at $573 \text{ K}$,$B$ must be a tertiary alcohol,specifically $2\text{-methyl-2-butanol}$ $(CH_3-C(OH)(CH_3)-CH_2-CH_3)$.
Therefore,$A$ must be butanone $(CH_3-CO-CH_2-CH_3)$.
The molecular formula of $A$ is $C_4H_8O$.
The molar mass of $A = (4 \times 12) + (8 \times 1) + 16 = 72 \text{ g/mol}$.
The mass of carbon in $A = 4 \times 12 = 48 \text{ g}$.
The mass percentage of carbon $= \frac{48}{72} \times 100 = 66.67\%$.

(A) The reaction sequence starts with $(A)$,which gives a positive iodoform test,indicating the presence of a $CH_3CO-$ group.
$1$. $(A)$ reacts with $CH_3MgBr$ followed by $H_3O^+$ to form a tertiary alcohol.
$2$. Dehydration with conc. $H_2SO_4/\Delta$ yields a cyclic compound $(B)$ (an indene derivative).
$3$. Ozonolysis $(O_3/Zn, H_2O)$ of $(B)$ breaks the double bond to form a dicarbonyl compound.
Based on the provided solution image,$(A)$ is $3$-methyl-$4$-phenylbut-$3$-en-$2$-one. This structure contains the $CH_3CO-$ group (positive iodoform test) and matches the reaction pathway shown.

(D) The rate constant $k$ is given by the Arrhenius equation: $k = A e^{-\frac{E_a}{RT}}$.
For the uncatalysed reaction at $700 \ K$: $k_1 = A e^{-\frac{E_a}{R \times 700}}$.
For the catalysed reaction at $500 \ K$: $k_2 = A e^{-\frac{E_a - 30}{R \times 500}}$.
Since the rate remains unchanged,$k_1 = k_2$,which implies:
$-\frac{E_a}{700R} = -\frac{E_a - 30}{500R}$.
Simplifying the equation:
$\frac{E_a}{700} = \frac{E_a - 30}{500}$.
$5E_a = 7(E_a - 30)$.
$5E_a = 7E_a - 210$.
$2E_a = 210 \implies E_a = 105 \ kJ/mol$.
The activation energy for the catalysed reaction is $E_a - 30 = 105 - 30 = 75 \ kJ/mol$.

(A) $(i)$ Blue-violet color with ninhydrin $\rightarrow$ indicates the presence of a free amino group (amino acid derivative). Aspartame and Alitame contain amino groups,while Saccharin and Sucralose do not.
$(ii)$ Lassaigne extract of $C$ gives a positive $AgNO_{3}$ test,indicating the presence of chlorine $(Cl)$. The negative $Fe_{4}[Fe(CN)_{6}]_{3}$ test indicates the absence of nitrogen $(N)$. Sucralose is the only sweetener among the options that contains $Cl$ and lacks $N$.
$(iii)$ Lassaigne extract of $B$ and $D$ gives a positive sodium nitroprusside test,which indicates the presence of sulfur $(S)$. Saccharin and Alitame both contain sulfur atoms.
Combining these: $C$ is Sucralose. $A$ and $D$ are amino acid derivatives (Aspartame and Alitame). Since $D$ contains $S$ (positive nitroprusside test),$D$ is Alitame. Therefore,$A$ is Aspartame. Consequently,$B$ must be Saccharin. The correct mapping is $A$: Aspartame,$B$: Saccharin,$C$: Sucralose,$D$: Alitame.

(B) In an Ellingham diagram,a metal $A$ can reduce the oxide of another metal $B$ $(BO_{2})$ if the line for the formation of $AO_{2}$ lies below the line for the formation of $BO_{2}$.
From the given diagram,the line for $A + O_{2} \rightarrow AO_{2}$ intersects the line for $B + O_{2} \rightarrow BO_{2}$ at $1400^{\circ} C$.
Above $1400^{\circ} C$,the line for $AO_{2}$ formation is below the line for $BO_{2}$ formation,meaning the $\Delta G^{\circ}$ for the reaction $A + BO_{2} \rightarrow AO_{2} + B$ becomes negative.
Therefore,$A$ reduces $BO_{2}$ at temperatures $> 1400^{\circ} C$.

(B) The complex $[Pd(F)(Cl)(Br)(I)]^{2-}$ is a square planar complex of the type $[M(abcd)]$. It has $3$ geometrical isomers. Thus,$n = 3$.
The formula for the second complex becomes $[Fe(CN)_6]^{3- - 6} = [Fe(CN)_6]^{3-}$.
In $[Fe(CN)_6]^{3-}$,the oxidation state of $Fe$ is $+3$. The electronic configuration of $Fe^{3+}$ is $3d^5$.
Since $CN^-$ is a strong field ligand,it causes pairing of electrons in the $3d$ orbitals. The configuration becomes $t_{2g}^5 e_g^0$.
Number of unpaired electrons $(n_{u}) = 1$.
Spin-only magnetic moment $= \sqrt{n_{u}(n_{u}+2)} \ BM = \sqrt{1(1+2)} \ BM = \sqrt{3} \ BM \approx 1.73 \ BM$.
$CFSE = [(-0.4 \times n_{t2g}) + (0.6 \times n_{eg})] \ \Delta_{0} = [(-0.4 \times 5) + (0.6 \times 0)] \ \Delta_{0} = -2.0 \ \Delta_{0}$.

(B) $1$. The starting material is $m$-bromoaniline. Treatment with $NaNO_2/HCl$ at $273-278 \ K$ performs diazotization to form $m$-bromobenzenediazonium chloride $(X)$.
$2$. Reaction of $X$ with $Cu_2Br_2$ (Sandmeyer reaction) replaces the diazonium group with a bromine atom,yielding $1,3$-dibromobenzene $(Y)$.
$3$. Finally,nitration of $1,3$-dibromobenzene using $HNO_3/H_2SO_4$ occurs. Since the bromine atom is ortho/para-directing,the nitro group enters the position ortho to one bromine and para to the other,which is the $4$-position relative to one of the bromine atoms,leading to $1,3$-dibromo$-4-$nitrobenzene as the major product.

(C) Step $1$: Hydration of alkyne $(X)$
$CH_3-CH(CH_3)-C\equiv CH + H_2O \xrightarrow{HgSO_4, H_2SO_4} CH_3-CH(CH_3)-CO-CH_3$ ($X$ is $3-methylbutan-2-one$)
Step $2$: Grignard reaction
$CH_3-CH(CH_3)-CO-CH_3 + C_2H_5MgBr \xrightarrow{H_2O} CH_3-CH(CH_3)-C(OH)(C_2H_5)-CH_3$
Step $3$: Dehydration
$CH_3-CH(CH_3)-C(OH)(C_2H_5)-CH_3 \xrightarrow{conc. H_2SO_4, \Delta} CH_3-C(CH_3)=C(C_2H_5)-CH_3$ (Major product $Y$ follows Saytzeff rule)

(A) The spin-only magnetic moment of $3.83 \ BM$ corresponds to $n = 3$ unpaired electrons,indicating $Cr$ is in the $+3$ oxidation state. Thus,the formula is $Cr(H_{2}O)_{6}Cl_{3}$.
Possible isomers are:
$1$. $[Cr(H_{2}O)_{6}]Cl_{3}$: Reacts with $AgNO_{3}$ but shows no geometrical isomerism.
$2$. $[Cr(H_{2}O)_{5}Cl]Cl_{2} \cdot H_{2}O$: Reacts with $AgNO_{3}$ but shows no geometrical isomerism.
$3$. $[Cr(H_{2}O)_{4}Cl_{2}]Cl \cdot 2H_{2}O$: Reacts with $AgNO_{3}$ and shows geometrical isomerism (cis and trans forms).
$4$. $[Cr(H_{2}O)_{3}Cl_{3}] \cdot 3H_{2}O$: Does not react with $AgNO_{3}$ (as no $Cl^-$ is outside the coordination sphere) and shows geometrical isomerism.
The complex $[Cr(H_{2}O)_{4}Cl_{2}]Cl \cdot 2H_{2}O$ satisfies both conditions. Its $IUPAC$ name is Tetraaquadichloridochromium $(III)$ chloride dihydrate.

(B) The atomic number of $Eu$ is $63$. The electronic configuration of $Eu$ is $[Xe] 4f^{7} 6s^{2}$.
For $Eu^{2+}$,two electrons are removed from the $6s$ orbital,resulting in $[Xe] 4f^{7}$.
The atomic number of $Ce$ is $58$. The electronic configuration of $Ce$ is $[Xe] 4f^{1} 5d^{1} 6s^{2}$.
For $Ce^{3+}$,three electrons are removed (two from $6s$ and one from $5d$),resulting in $[Xe] 4f^{1}$.

(B) According to Faraday's law of electrolysis,the number of gram equivalents of substances deposited or liberated at electrodes is equal for the same quantity of electricity passed.
$1.$ Number of gram equivalents of $Ag = \frac{\text{mass}}{\text{equivalent mass}} = \frac{108 \ g}{108 \ g \ mol^{-1}} = 1 \ eq$.
$2.$ Therefore,gram equivalents of $O_2(g)$ produced = $1 \ eq$.
$3.$ The reaction for the evolution of $O_2$ from water is: $2H_2O \rightarrow O_2 + 4H^+ + 4e^-$. The n-factor for $O_2$ is $4$.
$4.$ Moles of $O_2 = \frac{\text{gram equivalents}}{\text{n-factor}} = \frac{1}{4} = 0.25 \ mol$.
$5.$ At $273 \ K$ and $1 \ bar$ pressure $(STP)$,the molar volume of an ideal gas is $22.7 \ L \ mol^{-1}$.
$6.$ Volume of $O_2 = 0.25 \ mol \times 22.7 \ L \ mol^{-1} = 5.675 \ L$.

(C) The formula for freezing point depression is $\Delta T_{f} = i \times K_{f} \times m$.
Here,$\Delta T_{f} = 0.2 \ K$,$K_{f} = 2 \ K \ kg \ mol^{-1}$,and for $NaCl$,the van't Hoff factor $i = 2$.
The molality $m = \frac{w / 58.5}{0.6 \ kg}$,where $w$ is the mass of $NaCl$ in grams.
Substituting the values: $0.2 = 2 \times 2 \times \frac{w}{58.5 \times 0.6}$.
$0.2 = 4 \times \frac{w}{35.1}$.
$w = \frac{0.2 \times 35.1}{4} = 1.755 \ g$.

(A) The reaction involves the nitration of $2$-methyl$-4-$nitrophenol using concentrated $HNO_3$ and concentrated $H_2SO_4$.
The $-OH$ group is a strong activating group and is ortho/para directing.
The $-CH_3$ group is also ortho/para directing.
The $-NO_2$ group is meta directing.
In $2$-methyl$-4-$nitrophenol,the positions ortho to the $-OH$ group are position $6$ (which is vacant) and position $3$ (which is also vacant).
However,position $6$ is less sterically hindered compared to position $3$,which is adjacent to the $-CH_3$ group.
Therefore,the incoming $-NO_2^+$ electrophile attacks the $6$-position to form $2$-methyl$-4,6-$dinitrophenol as the major product.

(A) The acidity of the protons depends on the stability of the conjugate base formed after the removal of the proton.
$1$. Proton $b$ is part of a carboxylic acid group $(-COOH)$. The conjugate base (carboxylate ion,$-COO^\theta$) is stabilized by equivalent resonance.
$2$. Protons $c$ and $d$ are phenolic protons ($-OH$ attached to an aromatic ring). The conjugate base is a phenoxide ion,which is stabilized by resonance,but less so than the carboxylate ion.
$3$. Proton $a$ is an acetylenic proton $(-C \equiv CH)$. The conjugate base is an acetylide ion $(-C \equiv C^\theta)$,which is the least stable among these due to the lack of resonance stabilization.
$4$. Comparing $c$ and $d$: Proton $c$ is ortho to an electron-withdrawing $-NO_2$ group (via the aromatic ring system),which increases its acidity compared to $d$ through the inductive effect.
Thus,the order of acidity is $b > c > d > a$.

(B) Cast iron is the most impure form of iron,containing about $3-4.5 \%$ carbon. It is used for the manufacture of wrought iron and steel by removing impurities through oxidation.

(C) is propanal $(CH_{3}CH_{2}CHO)$. Reaction with $CH_{3}MgBr$ gives butan$-2-$ol $(CH_{3}CH_{2}CH(OH)CH_{3})$ as $C$.
Butan$-2-$ol is a $2^{\circ}$ alcohol,gives positive Iodoform test,and Lucas test turbidity in $5$ minutes.
$B$ is acetone $(CH_{3}COCH_{3})$. Reaction with $CH_{3}MgBr$ gives $2$-methylpropan$-2-$ol $(CH_{3}C(CH_{3})(OH)CH_{3})$ as $D$.
$2$-methylpropan$-2-$ol is a $3^{\circ}$ alcohol,gives immediate turbidity with Lucas reagent,and negative Iodoform test.
Thus,$C$ is $CH_{3}CH_{2}CH(OH)CH_{3}$ and $D$ is $CH_{3}C(CH_{3})(OH)CH_{3}$.

(A) For $[XeF_5]^-$,the central atom $Xe$ has $8$ valence electrons. Adding $5$ electrons from $F$ atoms and $1$ electron for the negative charge gives $14$ electrons,which corresponds to $7$ electron pairs. The hybridization is $sp^3d^3$. Due to the presence of $2$ lone pairs,the shape is pentagonal planar.
For $XeO_3F_2$,the central atom $Xe$ has $8$ valence electrons. $3$ oxygen atoms form double bonds (using $6$ electrons) and $2$ fluorine atoms form single bonds (using $2$ electrons). This results in $5$ bonding pairs and $0$ lone pairs. The hybridization is $sp^3d$,and the shape is trigonal bipyramidal.

(D) In $E_{2}$-elimination,a base removes a proton from the $\beta$-carbon while the leaving group departs from the $\alpha$-carbon.
For $3$-bromo-$2$-fluoropentane $(CH_3-CH_2-CH(Br)-CH(F)-CH_3)$,there are two potential $\beta$-carbons: $C_{2}$ (bearing $F$) and $C_{4}$ (bearing $H$).
$Br^-$ is a much better leaving group than $F^-$. Therefore,the elimination occurs primarily at the $C_{3}$ position (where $Br$ is attached) and the $C_{2}$ position (where $H$ is attached).
The base abstracts the more acidic proton from $C_{2}$ (which is adjacent to the electron-withdrawing $F$ atom).
This leads to the formation of $CH_3-CH_2-CH=C(F)-CH_3$ as the major product,which is a stable alkene with $5$ $\alpha$-hydrogens.

(A) The crystal field splitting energy $(\Delta_{oh})$ is directly proportional to the strength of the ligand.
The spectrochemical series for the given ligands is: $F^{-} < NCS^{-} < NH_3$.
Therefore,the order of $\Delta_{oh}$ is: $[MF_6]^{(-6+n)} < [M(NCS)_6]^{(-6+n)} < [M(NH_3)_6]^{n+}$.
Since $\Delta_{oh} = \frac{hc}{\lambda_{max}}$,the wavelength of maximum absorption $(\lambda_{max})$ is inversely proportional to $\Delta_{oh}$.
Thus,the order of $\lambda_{max}$ is: $[MF_6]^{(-6+n)} > [M(NCS)_6]^{(-6+n)} > [M(NH_3)_6]^{n+}$.
From the given graph,the $\lambda_{max}$ values are in the order $A < B < C$.
Matching the values: $A$ corresponds to $[MF_6]^{(-6+n)}$ (complex $ii$),$B$ corresponds to $[M(NCS)_6]^{(-6+n)}$ (complex $i$),and $C$ corresponds to $[M(NH_3)_6]^{n+}$ (complex $iii$).
Therefore,the correct match is $A-(ii), B-(i), C-(iii)$.

(A) Reaction $(1)$ is an $SN_{1}$ reaction. The rate law is given as $\text{rate} = k[t-BuBr]$. Since the rate is independent of the concentration of the nucleophile/base,changing the concentration of the base will have no effect on reaction $(1)$.
Reaction $(2)$ is an $E_{2}$ reaction. The rate law is given as $\text{rate} = k[t-BuBr][OH^{\ominus}]$. Since the rate depends on the concentration of the base,changing the concentration of the base will affect the rate of reaction $(2)$.
Therefore,the statement that changing the concentration of base will have no effect on reaction $(1)$ is true.

(A) From the rate law expression: $Rate = k [A]^x [B]^y$
Using the data from experiments $I, II,$ and $III$:
$6.00 \times 10^{-3} = k (0.1)^x (0.1)^y \dots(1)$
$2.40 \times 10^{-2} = k (0.1)^x (0.2)^y \dots(2)$
$1.20 \times 10^{-2} = k (0.2)^x (0.1)^y \dots(3)$
Dividing $(3)$ by $(1)$: $2^x = 2 \implies x = 1$
Dividing $(2)$ by $(1)$: $2^y = 4 \implies y = 2$
Thus,the rate law is $Rate = k [A]^1 [B]^2$.
For experiment $IV$:
$7.20 \times 10^{-2} = k (X)^1 (0.2)^2$
Using $k$ from experiment $I$: $k = \frac{6.00 \times 10^{-3}}{(0.1)(0.1)^2} = 6 \ L^2 \ mol^{-2} \ min^{-1}$
$7.20 \times 10^{-2} = 6 \times X \times 0.04 \implies X = \frac{7.20 \times 10^{-2}}{0.24} = 0.3 \ M$
For experiment $V$:
$2.88 \times 10^{-1} = 6 \times (0.3) \times Y^2$
$Y^2 = \frac{2.88 \times 10^{-1}}{1.8} = 0.16 \implies Y = 0.4 \ M$
Therefore,$X = 0.3$ and $Y = 0.4$.

(B) The reaction of the ether $'A'$ $(C_9H_{10}O)$ with conc. $HI$ leads to cleavage.
Compound $'A'$ is benzyl vinyl ether $(C_6H_5CH_2-O-CH=CH_2)$.
Upon treatment with $HI$,the ether bond cleaves to form benzyl iodide ($C_6H_5CH_2I$,compound $'B'$) and vinyl alcohol ($CH_2=CHOH$,compound $'C'$).
Compound $'B'$ $(C_6H_5CH_2I)$ reacts with $AgNO_3$ to give a yellow precipitate of $AgI$.
Compound $'C'$ $(CH_2=CHOH)$ is unstable and tautomerizes to acetaldehyde ($CH_3CHO$,compound $'D'$).
Acetaldehyde $(CH_3CHO)$ gives a positive iodoform test because it contains the $CH_3CO-$ group.
Therefore,the structure of $'A'$ is benzyl vinyl ether.

(C) Raw mango shrinks in a concentrated salt solution due to the net movement of water molecules from the mango (lower solute concentration) to the salt solution (higher solute concentration) through the semi-permeable cell membranes of the mango. This phenomenon is known as $Osmosis$.

(B) $[Ni(NH_3)_2Cl_2]$ is a tetrahedral complex,which does not exhibit geometrical or optical isomerism.
It also does not exhibit structural isomerism.
$[Ni(NH_3)_4(H_2O)_2]^{2+}$ (octahedral) shows geometrical isomerism.
$[Pt(NH_3)_2Cl_2]$ (square planar) shows geometrical isomerism.
$[Ni(en)_3]^{2+}$ (octahedral) shows optical isomerism.

(B) Adsorption is an exothermic process. As adsorption proceeds,the number of available active sites on the adsorbent decreases,resulting in less heat being evolved. Thus,$\Delta H$ becomes less negative.
$(b)$ $NH_3$ is a polar molecule with a higher van der Waals constant $'a'$ compared to $N_2$,leading to stronger forces of attraction and greater adsorption.
$(c)$ Adsorption occurs to satisfy the residual forces on the surface of the adsorbent. Therefore,as adsorption proceeds,the residual forces on the surface decrease,not increase.
$(d)$ According to Le Chatelier's principle,for an exothermic process,an increase in temperature shifts the equilibrium in the backward direction,decreasing the concentration of the adsorbed phase.

(C) Hydrolysis of $\text{Sucrose}$ yields $D\text{-glucose}$ and $D\text{-fructose}$.
$\text{Seliwanoff's test}$ is a chemical test used to distinguish between aldose and ketose sugars.
Ketoses (like $\text{fructose}$) react with $\text{Seliwanoff's reagent}$ (resorcinol in $HCl$) to produce a cherry-red colored complex.

(B) The disproportionation reaction is:
$2 Cu ^{+}( aq ) \longrightarrow Cu ( s ) + Cu ^{2+}( aq )$
This can be split into two half-reactions:
Oxidation: $Cu ^{+}( aq ) \longrightarrow Cu ^{2+}( aq ) + e ^{-} \quad (E^{0} = -0.16 \ V)$
Reduction: $Cu ^{+}( aq ) + e ^{-} \longrightarrow Cu ( s ) \quad (E^{0} = 0.52 \ V)$
$E_{cell}^{0} = E_{cathode}^{0} - E_{anode}^{0} = 0.52 \ V - 0.16 \ V = 0.36 \ V$
At equilibrium,the relationship between $E_{cell}^{0}$ and the equilibrium constant $K$ is given by:
$E_{cell}^{0} = \frac{RT}{nF} \ln K$
Here,$n = 1$ (number of electrons transferred).
$\ln K = \frac{E_{cell}^{0} \times n}{RT/F} = \frac{0.36 \times 1}{0.025} = 14.4$
Expressing $14.4$ as $\times 10^{-1}$:
$14.4 = 144 \times 10^{-1}$
Thus,the value is $144$.

(C) For $K_{2}Cr_{2}O_{7}$: $2(+1) + 2x + 7(-2) = 0 \implies 2 + 2x - 14 = 0 \implies 2x = 12 \implies x = +6$.
For $KMnO_{4}$: $(+1) + y + 4(-2) = 0 \implies 1 + y - 8 = 0 \implies y = +7$.
For $K_{2}FeO_{4}$: $2(+1) + z + 4(-2) = 0 \implies 2 + z - 8 = 0 \implies z = +6$.
The sum is $x + y + z = 6 + 7 + 6 = 19$.

(B) The basic gas $(C)$ is ammonia $(NH_{3})$.
According to the Haber process,$N_{2} + 3H_{2} \rightleftharpoons 2NH_{3}$. Thus,gas $(B)$ must be $N_{2}$.
Let us analyze the thermal decomposition of the given compounds:
$(1)$ $(NH_{4})_{2}Cr_{2}O_{7} \xrightarrow{\Delta} N_{2} \uparrow + Cr_{2}O_{3} + 4H_{2}O \uparrow$ (Produces $N_{2}$)
$(2)$ $Pb(NO_{3})_{2} \xrightarrow{\Delta} PbO + 2NO_{2} \uparrow + \frac{1}{2}O_{2} \uparrow$ (Does not produce $N_{2}$)
$(3)$ $2NaN_{3} \xrightarrow{\Delta} 2Na + 3N_{2} \uparrow$ (Produces $N_{2}$)
$(4)$ $NH_{4}NO_{2} \xrightarrow{\Delta} N_{2} \uparrow + 2H_{2}O \uparrow$ (Produces $N_{2}$)
Since $(A)$ must produce $N_{2}$ gas,$(A)$ cannot be $Pb(NO_{3})_{2}$.

(C) $(I)$ Under weak field ligands,octahedral $Mn(II)$ and tetrahedral $Ni(II)$ complexes are both high spin.
$(II)$ Tetrahedral $Ni(II)$ complexes are rarely low spin because $Ni(II)$ prefers square planar geometry with strong field ligands,which are low spin.
$(III)$ With strong field ligands,$Mn(II)$ $(d^5)$ complexes can be low spin (unpaired electrons $= 1$).
$(IV)$ Aqueous solution of $Mn(II)$ ions is pink in color,not yellow. Therefore,statement $(IV)$ is incorrect.

(B) The spin-only magnetic moment is given by $\mu = \sqrt{n(n+2)} \ BM.$ For $\mu = 4.90 \ BM,$ $n = 4$ unpaired electrons.
For a $d^{6}$ ion,$4$ unpaired electrons are possible in a high-spin tetrahedral complex $(e^{3}t_{2}^{3})$ or a high-spin octahedral complex $(t_{2g}^{4}e_{g}^{2})$.
In a tetrahedral complex $[M(H_{2}O)_{4}]^{2+}$,the configuration is $e^{3}t_{2}^{3}$.
$CFSE = (3 \times -0.6 \Delta_{t}) + (3 \times 0.4 \Delta_{t}) = -1.8 \Delta_{t} + 1.2 \Delta_{t} = -0.6 \Delta_{t}.$
In an octahedral complex $[M(H_{2}O)_{6}]^{2+}$,the configuration is $t_{2g}^{4}e_{g}^{2}$.
$CFSE = (4 \times -0.4 \Delta_{0}) + (2 \times 0.6 \Delta_{0}) = -1.6 \Delta_{0} + 1.2 \Delta_{0} = -0.4 \Delta_{0}.$
Comparing with the given options,option $B$ matches the tetrahedral geometry and calculated $CFSE$.

(B) $1$. The reaction of $2$-methyl-$2H$-chromene with excess $HBr$ under heating leads to the cleavage of the ether ring. The $C-O$ bond breaks,and $HBr$ adds across the double bond and the ring-opened position,resulting in a phenolic compound with a dibromoalkyl side chain: $2-(3,5-dibromopentyl)phenol$ (or similar depending on regioselectivity,but the provided solution image shows the specific transformation).
$2$. Treatment with alcoholic $KOH$ followed by $H^+$ causes dehydrohalogenation,forming a conjugated diene side chain attached to the phenol ring: $2-(penta-1,3-dienyl)phenol$.
$3$. Finally,reductive ozonolysis $(O_3, Zn/H_2O)$ cleaves the double bonds in the side chain. The terminal double bond yields formaldehyde $(HCHO)$,the internal double bond yields glyoxal $(CHO-CHO)$,and the side chain attached to the benzene ring yields salicylaldehyde $(2-hydroxybenzaldehyde)$.
$4$. Among the options,the major aromatic product $C$ is salicylaldehyde.

(C) When a non-volatile solute like glucose is added to a solvent,the number of solvent molecules at the surface decreases.
This results in a decrease in the rate of evaporation,meaning fewer water molecules leave the solution.
Consequently,the vapour pressure of the solution decreases.

(A) In the compound $CH_3-CH(C_2H_5)-CH_2-Br$,the chiral center is located at the second carbon atom $(C2)$.
The nucleophilic substitution by the $OH^{-}$ ion occurs at the first carbon atom $(C1)$,where the $Br$ atom is attached.
Since the bonds connected to the chiral center are not broken or modified during the reaction,the spatial arrangement (configuration) around the chiral center remains unchanged.
This results in the retention of configuration.

(A) The reactivity of carbonyl compounds towards nucleophilic addition of $HCN$ depends on the electrophilicity of the carbonyl carbon and steric hindrance. Electron-withdrawing groups ($-I$ or $-R$ effect) increase the electrophilicity of the carbonyl carbon,thereby increasing reactivity. Electron-donating groups ($+R$ or $+I$ effect) decrease the electrophilicity,thereby decreasing reactivity.
$(i)$ $m$-Methoxybenzaldehyde: $-OCH_3$ group shows $-I$ effect (electron-withdrawing) from the meta position.
(ii) $o$-Nitrobenzaldehyde: $-NO_2$ group shows both $-I$ and $-R$ effects (strong electron-withdrawing).
(iii) $o$-Methoxybenzaldehyde: $-OCH_3$ group shows $+R$ effect (strong electron-donating) from the ortho position.
(iv) $m$-Nitrobenzaldehyde: $-NO_2$ group shows only $-I$ effect from the meta position.
Comparing the effects:
- (iii) has a strong electron-donating group $(+R)$,making it the least reactive.
- $(i)$ has a weak electron-withdrawing group $(-I)$.
- (iv) has a stronger electron-withdrawing group ($-I$ only).
- (ii) has the strongest electron-withdrawing effect ($-I$ and $-R$),making it the most reactive.
Thus,the increasing order of reactivity is: $(iii) < (i) < (iv) < (ii)$.