JEE Main 2017 Chemistry Question Paper with Answer and Solution

(B) Given $f(x) = ax^2 + bx + c$.
Since $f(x+y) = f(x) + f(y) + xy$,putting $x=0, y=0$ gives $f(0) = f(0) + f(0) + 0$,so $f(0) = 0$. Thus,$c = 0$.
Given $a+b+c = 3$,and since $c=0$,we have $a+b = 3$.
Also,$f(x+y) = a(x+y)^2 + b(x+y) = a(x^2 + 2xy + y^2) + bx + by = (ax^2 + bx) + (ay^2 + by) + 2axy$.
Comparing this with $f(x+y) = f(x) + f(y) + xy$,we get $2axy = xy$,which implies $2a = 1$,so $a = 1/2$.
Since $a+b = 3$,$b = 3 - 1/2 = 5/2$.
Thus,$f(n) = \frac{1}{2}n^2 + \frac{5}{2}n = \frac{n^2 + 5n}{2}$.
Now,$\sum_{n=1}^{10} f(n) = \sum_{n=1}^{10} \frac{n^2 + 5n}{2} = \frac{1}{2} \left[ \sum_{n=1}^{10} n^2 + 5 \sum_{n=1}^{10} n \right]$.
Using the formulas $\sum n^2 = \frac{n(n+1)(2n+1)}{6}$ and $\sum n = \frac{n(n+1)}{2}$:
$\sum_{n=1}^{10} n^2 = \frac{10 \times 11 \times 21}{6} = 385$.
$\sum_{n=1}^{10} n = \frac{10 \times 11}{2} = 55$.
Therefore,$\sum_{n=1}^{10} f(n) = \frac{1}{2} [385 + 5(55)] = \frac{1}{2} [385 + 275] = \frac{660}{2} = 330$.

(B) Let $F(x) = \tan^{-1} \left( \frac{6x\sqrt{x}}{1 - 9x^3} \right)$.
We can rewrite the argument as $\frac{2(3x^{3/2})}{1 - (3x^{3/2})^2}$.
Using the identity $2\tan^{-1}(\theta) = \tan^{-1} \left( \frac{2\theta}{1 - \theta^2} \right)$,we get $F(x) = 2\tan^{-1}(3x^{3/2})$.
Now,differentiate $F(x)$ with respect to $x$:
$F'(x) = 2 \cdot \frac{1}{1 + (3x^{3/2})^2} \cdot \frac{d}{dx}(3x^{3/2})$.
$F'(x) = 2 \cdot \frac{1}{1 + 9x^3} \cdot (3 \cdot \frac{3}{2} x^{1/2})$.
$F'(x) = 2 \cdot \frac{1}{1 + 9x^3} \cdot \frac{9}{2} \sqrt{x} = \frac{9\sqrt{x}}{1 + 9x^3}$.
Given $F'(x) = \sqrt{x} \cdot g(x)$,we have $\sqrt{x} \cdot g(x) = \frac{9\sqrt{x}}{1 + 9x^3}$.
Therefore,$g(x) = \frac{9}{1 + 9x^3}$.

(C) Let $v_f$ be the final speed of the body after $t = 10 \; s$.
Given that the final kinetic energy is $\frac{1}{8} mv_0^2$,we have:
$\frac{1}{2} m v_f^2 = \frac{1}{8} m v_0^2 \Rightarrow v_f^2 = \frac{1}{4} v_0^2 \Rightarrow v_f = \frac{v_0}{2} = 5 \; m/s$.
The equation of motion is $F = m \frac{dv}{dt} = -kv^2$.
Rearranging the terms,we get $\frac{dv}{v^2} = -\frac{k}{m} dt$.
Integrating both sides from $t = 0$ to $t = 10 \; s$ and $v = 10 \; m/s$ to $v = 5 \; m/s$:
$\int_{10}^{5} v^{-2} dv = -\frac{k}{10^{-2}} \int_{0}^{10} dt$.
$[-v^{-1}]_{10}^{5} = -100k [t]_{0}^{10}$.
$-(\frac{1}{5} - \frac{1}{10}) = -100k(10)$.
$-(\frac{1}{10}) = -1000k$.
$k = \frac{1}{10000} = 10^{-4} \; kg \cdot m^{-1}$.

(B) The formula for surface tension is $T = \frac{rhg}{2} \times 10^3$. Here,$r = D/2$,so $\frac{\Delta r}{r} = \frac{\Delta D}{D}$.
The relative error in $T$ is given by $\frac{\Delta T}{T} = \frac{\Delta r}{r} + \frac{\Delta h}{h}$.
Assuming the least count for both $D$ and $h$ is $0.01 \times 10^{-2} \ m$ (based on the precision of the given values):
$\frac{\Delta r}{r} = \frac{0.01 \times 10^{-2}}{1.25 \times 10^{-2}} = \frac{0.01}{1.25} = 0.008$.
$\frac{\Delta h}{h} = \frac{0.01 \times 10^{-2}}{1.45 \times 10^{-2}} = \frac{0.01}{1.45} \approx 0.00689$.
Percentage error $= \left( \frac{\Delta T}{T} \right) \times 100 = (0.008 + 0.00689) \times 100 = 0.01489 \times 100 = 1.489\%$.
Rounding to the nearest value,we get $1.5\%$.

(C) According to the first law of thermodynamics,$\Delta U = Q + W$.
In an adiabatic process,there is no exchange of heat between the system and the surroundings,so $Q = 0$.
Therefore,$\Delta U = W_{\text{adiabatic}}$.

(D) According to Bohr's theory,the radius of the $n^{th}$ orbit is given by the formula:
$r_n = 0.529 \times \frac{n^2}{Z} \, \mathring{A}$
For the second Bohr orbit of the hydrogen atom,$n = 2$ and $Z = 1$.
Substituting these values:
$r_2 = 0.529 \times \frac{2^2}{1} \, \mathring{A}$
$r_2 = 0.529 \times 4 \, \mathring{A} = 2.116 \, \mathring{A}$
Rounding to two decimal places,we get $2.12 \, \mathring{A}$.

(B) For the salt of a weak acid and a weak base,the $pH$ is calculated using the formula:
$pH = 7 + \frac{1}{2} pK_a - \frac{1}{2} pK_b$
Given values are $pK_a = 3.2$ and $pK_b = 3.4$.
Substituting these values into the formula:
$pH = 7 + \frac{1}{2}(3.2) - \frac{1}{2}(3.4)$
$pH = 7 + 1.6 - 1.7$
$pH = 6.9$

(C) The mass of hydrogen in a $75 \ kg$ human body is calculated as: $\text{Mass of } H = \frac{10}{100} \times 75 \ kg = 7.5 \ kg$.
When all $^1H$ atoms (atomic mass $\approx 1$) are replaced by $^2H$ atoms (atomic mass $\approx 2$),the mass of the hydrogen component doubles.
New mass of hydrogen $= 7.5 \ kg \times 2 = 15 \ kg$.
Therefore,the weight gain is $15 \ kg - 7.5 \ kg = 7.5 \ kg$.

(D) Zinc oxide $(ZnO)$ is an amphoteric oxide,meaning it can react with both acids and bases.
In reaction $(A)$,$ZnO + Na_2O \rightarrow Na_2ZnO_2$,$ZnO$ acts as an acid because it reacts with the base $Na_2O$.
In reaction $(B)$,$ZnO + CO_2 \rightarrow ZnCO_3$,$ZnO$ acts as a base because it reacts with the acidic oxide $CO_2$.
Therefore,$ZnO$ acts as an acid in $(A)$ and a base in $(B)$.

(A) Lithium $(Li)$ and Magnesium $(Mg)$ exhibit a diagonal relationship due to similar ionic sizes and charge-to-size ratios.
$1$. Both form nitrides ($Li_3N$ and $Mg_3N_2$) by direct combination with nitrogen.
$2$. Both form soluble bicarbonates in solution.
$3$. Both $LiNO_3$ and $Mg(NO_3)_2$ decompose on heating to give their respective oxides,$NO_2$,and $O_2$.
$4$. Magnesium forms a basic carbonate $(4MgCO_3 \cdot Mg(OH)_2 \cdot 5H_2O)$,whereas lithium does not form a basic carbonate; it only forms a normal carbonate $(Li_2CO_3)$.
Therefore,the statement that both form basic carbonates is incorrect.

(D) $3-$Methylpent$-2-$ene $(CH_3-CH=C(CH_3)-CH_2-CH_3)$ reacts with $HBr$ in the presence of peroxide via anti-Markovnikov addition to form $2-$bromo$-3-$methylpentane $(CH_3-CH(Br)-CH(CH_3)-CH_2-CH_3)$.
The product $CH_3-CH(Br)-CH(CH_3)-CH_2-CH_3$ contains two chiral centers at $C2$ and $C3$.
Since the molecule is unsymmetrical,the number of stereoisomers is given by the formula $2^n$,where $n$ is the number of chiral centers.
Here,$n = 2$,so the number of stereoisomers $= 2^2 = 4$.

(C) We are given the following equations:
$(1) C_{(graphite)} + O_{2(g)} \rightarrow CO_{2(g)}; \Delta_rH^o = -393.5 \, kJ \, mol^{-1}$
$(2) H_{2(g)} + \frac{1}{2} O_{2(g)} \rightarrow H_2O_{(l)}; \Delta_rH^o = -285.8 \, kJ \, mol^{-1}$
$(3) CO_{2(g)} + 2H_2O_{(l)} \rightarrow CH_{4(g)} + 2O_{2(g)}; \Delta_rH^o = + 890.3 \, kJ \, mol^{-1}$
We need to find $\Delta_rH^o$ for the reaction:
$C_{(graphite)} + 2H_{2(g)} \rightarrow CH_{4(g)}$
To obtain this,we perform the operation: $(1) + 2 \times (2) + (3)$
$\Delta_rH^o = (-393.5) + 2 \times (-285.8) + 890.3$
$\Delta_rH^o = -393.5 - 571.6 + 890.3$
$\Delta_rH^o = -965.1 + 890.3 = -74.8 \, kJ \, mol^{-1}$

(D) The sodium salt of oxalic acid,$Na_2C_2O_4$ $(X)$,reacts with conc. $H_2SO_4$ to produce $CO$ and $CO_2$ gases,causing effervescence.
$Na_2C_2O_4 + H_2SO_4 \to Na_2SO_4 + H_2C_2O_4$
$H_2C_2O_4 \xrightarrow{Conc. H_2SO_4} H_2O + CO \uparrow + CO_2 \uparrow$
$X$ reacts with $CaCl_2$ to form a white precipitate of calcium oxalate $(CaC_2O_4)$:
$Na_2C_2O_4 + CaCl_2 \to CaC_2O_4 \downarrow + 2NaCl$
Oxalate ions $(C_2O_4^{2-})$ decolourise acidic $KMnO_4$ solution due to the redox reaction:
$5C_2O_4^{2-} + 2MnO_4^{-} + 16H^{+} \to 10CO_2 \uparrow + 2Mn^{2+} + 8H_2O$
Thus,$'X'$ is $Na_2C_2O_4$.

(B) Paramagnetism is due to the presence of unpaired electrons.
$1.$ $NO$: Total electrons = $15$. It has an odd number of electrons,so it must have at least one unpaired electron,making it paramagnetic.
$2.$ $CO$: Total electrons = $14$. All electrons are paired in its molecular orbitals,so it is diamagnetic (not paramagnetic).
$3.$ $O_2$: Total electrons = $16$. According to Molecular Orbital $(MO)$ theory,it has two unpaired electrons in its antibonding $\pi^* 2p_x$ and $\pi^* 2p_y$ orbitals,so it is paramagnetic.
$4.$ $B_2$: Total electrons = $10$. It has two unpaired electrons in its $\pi 2p_x$ and $\pi 2p_y$ orbitals,so it is paramagnetic.

(D) Aromatic compounds are highly stable due to the delocalization of $\pi$ electrons over the entire ring. For a molecule to be aromatic,it must follow $H$ückel's rule,having $(4n+2) \pi$ electrons in a planar,fully conjugated system.
$A)$ Benzene is aromatic ($6 \pi$ electrons).
$B)$ Furan is aromatic ($6 \pi$ electrons,including the lone pair on oxygen).
$C)$ Pyridine is aromatic ($6 \pi$ electrons).
$D)$ $4,4-$dimethylcyclohexa$-2,5-$dien$-1-$one has an $sp^3$ hybridized carbon at the $4$-position,which breaks the continuous conjugation of the ring. Therefore,it is non-aromatic and shows the least resonance stabilization compared to the other aromatic options.

(C) According to the World Health Organization $(WHO)$ and Indian standards for drinking water,the maximum permissible concentration of fluoride $(F^-)$ is $1.0 \; ppm$. Concentrations above $2 \; ppm$ cause fluorosis (brown mottling of teeth).
The maximum permissible limit for nitrate $(NO_3^-)$ is $50 \; ppm$. Since the given concentration is $50 \; ppm$,it is at the threshold limit and generally considered acceptable.
The maximum permissible limit for sulphate $(SO_4^{2-})$ is $500 \; ppm$. Since the given concentration is $100 \; ppm$,it is well within the safe limit.
Therefore,only fluoride $(F^-)$ at $10 \; ppm$ makes the water unsuitable for drinking.

(B) The balanced chemical equation for the reaction is:
$M_{2}CO_{3} + 2HCl \rightarrow 2MCl + H_{2}O + CO_{2}$
According to the stoichiometry of the reaction,$1\, mol$ of $M_{2}CO_{3}$ produces $1\, mol$ of $CO_{2}$.
Therefore,the moles of $M_{2}CO_{3}$ reacted is equal to the moles of $CO_{2}$ produced,which is $0.01186\, mol$.
Using the formula: $\text{Molar mass} = \frac{\text{Mass}}{\text{Moles}}$
$\text{Molar mass} = \frac{1\, g}{0.01186\, mol} \approx 84.3\, g\, mol^{-1}$.

(A) Isoelectronic species are atoms or ions that have the same number of electrons.
$O^{2-}: 8 + 2 = 10$ electrons
$F^{-}: 9 + 1 = 10$ electrons
$Na^{+}: 11 - 1 = 10$ electrons
$Mg^{2+}: 12 - 2 = 10$ electrons
Since all species in this group have $10$ electrons,they are isoelectronic.

(A) The formation of Nylon-$6$ involves the hydrolysis of caprolactam.
Caprolactam is heated with water at high temperature,which leads to the hydrolysis of the amide bond to form $\epsilon$-amino caproic acid.
This amino acid then undergoes condensation polymerisation to form Nylon-$6$.

(C) Let $N_0$ be the initial number of nuclei of $A$ at $t = 0$.
At time $t$,the number of nuclei of $A$ remaining is $N_A = N_0 e^{-\lambda t}$.
The number of nuclei of $B$ formed is $N_B = N_0 - N_A = N_0(1 - e^{-\lambda t})$.
The ratio is given by $\frac{N_B}{N_A} = \frac{N_0(1 - e^{-\lambda t})}{N_0 e^{-\lambda t}} = e^{\lambda t} - 1$.
Given $\frac{N_B}{N_A} = 0.3$,we have $0.3 = e^{\lambda t} - 1$,which implies $e^{\lambda t} = 1.3$.
Taking the natural logarithm on both sides: $\lambda t = \ln(1.3)$.
Since the decay constant $\lambda = \frac{\ln 2}{T}$,we substitute this into the equation:
$\left(\frac{\ln 2}{T}\right) t = \ln(1.3)$.
Solving for $t$: $t = T \frac{\ln(1.3)}{\ln 2} = T \frac{\log(1.3)}{\log 2}$.

(C) Let $N_0$ be the initial number of nuclei of $A$ at $t = 0$.
At time $t$,the number of nuclei of $A$ remaining is $N_A = N_0 e^{-\lambda t}$.
The number of nuclei of $B$ formed is $N_B = N_0 - N_A = N_0(1 - e^{-\lambda t})$.
The ratio is given as $\frac{N_B}{N_A} = 0.3$.
Substituting the expressions: $\frac{N_0(1 - e^{-\lambda t})}{N_0 e^{-\lambda t}} = 0.3$.
This simplifies to $\frac{1}{e^{-\lambda t}} - 1 = 0.3$,which means $e^{\lambda t} - 1 = 0.3$.
So,$e^{\lambda t} = 1.3$.
Taking the natural logarithm on both sides: $\lambda t = \ln(1.3)$.
Since the decay constant $\lambda = \frac{\ln 2}{T}$,we have $\frac{\ln 2}{T} \cdot t = \ln(1.3)$.
Solving for $t$: $t = T \frac{\ln(1.3)}{\ln 2} = T \frac{\log(1.3)}{\log 2}$.

(B) The given differential equation is $y dx - (x + 3y^2) dy = 0$.
Rearranging the terms,we get $y dx = (x + 3y^2) dy$.
Dividing by $y dy$,we get $\frac{dx}{dy} = \frac{x}{y} + 3y$.
This is a linear differential equation in $x$ of the form $\frac{dx}{dy} + P(y)x = Q(y)$,where $P(y) = -\frac{1}{y}$ and $Q(y) = 3y$.
The integrating factor is $IF = e^{\int P(y) dy} = e^{-\int \frac{1}{y} dy} = e^{-\ln y} = \frac{1}{y}$.
The solution is given by $x \cdot IF = \int Q(y) \cdot IF dy + C$.
Substituting the values,$x \cdot \frac{1}{y} = \int 3y \cdot \frac{1}{y} dy + C$.
$\frac{x}{y} = \int 3 dy + C = 3y + C$.
Thus,$x = 3y^2 + Cy$.
Since the curve passes through $(1, 1)$,we have $1 = 3(1)^2 + C(1)$,which gives $C = -2$.
The equation of the curve is $x = 3y^2 - 2y$.
Checking the options,for $y = \frac{1}{3}$,$x = 3(\frac{1}{9}) - 2(\frac{1}{3}) = \frac{1}{3} - \frac{2}{3} = -\frac{1}{3}$.
Thus,the curve passes through $\left( -\frac{1}{3}, \frac{1}{3} \right)$.

(B) Using the distributive law: $(\sim p) \vee (p \wedge \sim q) \equiv ((\sim p) \vee p) \wedge ((\sim p) \vee (\sim q))$.
Since $((\sim p) \vee p) \equiv T$ (a tautology),the expression simplifies to $T \wedge ((\sim p) \vee (\sim q))$.
This is equivalent to $(\sim p) \vee (\sim q)$,which is $\sim (p \wedge q)$.
Alternatively,checking the truth table:

$p$	$q$	$(\sim p) \vee (p \wedge \sim q)$	$p \to \sim q$
$T$	$T$	$F$	$F$
$T$	$F$	$T$	$T$
$F$	$T$	$T$	$T$
$F$	$F$	$T$	$T$

Comparing the columns,the proposition is equivalent to $p \to \sim q$.

(A) The balanced chemical equation is: $FeCl_3 (aq) + 3NaOH (aq) \to Fe(OH)_3 (s) + 3NaCl (aq)$.
Molar mass of $Fe(OH)_3 = 56 + 3 \times (16 + 1) = 56 + 51 = 107\, g\, mol^{-1}$.
Moles of $Fe(OH)_3 = \frac{\text{mass}}{\text{molar mass}} = \frac{2.14\, g}{107\, g\, mol^{-1}} = 0.02\, mol$.
From the stoichiometry of the reaction,$1\, mole$ of $FeCl_3$ produces $1\, mole$ of $Fe(OH)_3$.
Therefore,moles of $FeCl_3$ present = $0.02\, mol$.
Volume of $FeCl_3$ solution = $100\, mL = 0.1\, L$.
Molarity of $FeCl_3 = \frac{\text{moles of solute}}{\text{volume of solution in } L} = \frac{0.02\, mol}{0.1\, L} = 0.2\, M$.

(B) Real gases behave ideally under conditions of low pressure and high volume,as the intermolecular forces become negligible and the volume of gas molecules becomes insignificant compared to the total volume of the container.
At $Boyle's$ temperature,the effects of intermolecular forces and molecular volume cancel out,leading to ideal behavior.
However,at very low temperatures,the kinetic energy of gas molecules decreases,making intermolecular forces significant,which causes real gases to deviate from ideal behavior.
Therefore,the statement that real gases show ideal behavior at very low temperature is incorrect.

(D) The reaction is $A_{(g)} \to A_{(l)}$.
The relationship between enthalpy change and internal energy change is $\Delta H = \Delta U + \Delta n_g RT$.
Given $\Delta H = -3RT$.
Here,$\Delta n_g = n_{g, \text{products}} - n_{g, \text{reactants}} = 0 - 1 = -1$.
Substituting the values: $\Delta H = \Delta U - RT$.
$-3RT = \Delta U - RT$.
$\Delta U = -3RT + RT = -2RT$.
Comparing the magnitudes: $|\Delta H| = |-3RT| = 3RT$ and $|\Delta U| = |-2RT| = 2RT$.
Therefore,$|\Delta H| > |\Delta U|$.

(D) For the Lyman series,the shortest wavelength occurs at $n_1 = 1$ and $n_2 = \infty$.
Using the Rydberg formula: $\frac{1}{\lambda} = R Z^2 \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right)$.
For hydrogen $(Z=1)$: $\frac{1}{A} = R(1)^2 \left( \frac{1}{1^2} - \frac{1}{\infty^2} \right) = R$,so $R = \frac{1}{A}$.
For the Paschen series,the longest wavelength occurs at the first line,where $n_1 = 3$ and $n_2 = 4$.
For $He^{+}$ $(Z=2)$: $\frac{1}{\lambda} = R(2)^2 \left( \frac{1}{3^2} - \frac{1}{4^2} \right)$.
$\frac{1}{\lambda} = 4R \left( \frac{1}{9} - \frac{1}{16} \right) = 4R \left( \frac{16-9}{144} \right) = 4R \left( \frac{7}{144} \right) = \frac{7R}{36}$.
Substituting $R = \frac{1}{A}$: $\frac{1}{\lambda} = \frac{7}{36A}$.
Therefore,$\lambda = \frac{36A}{7}$.

(C) For an acidic buffer,the Henderson-Hasselbalch equation is given by:
$pH = pK_a + \log \frac{[Salt]}{[Acid]}$
Given $K_a = 10^{-5}$,so $pK_a = -\log(10^{-5}) = 5$.
Given $pH = 6$.
Substituting these values into the equation:
$6 = 5 + \log \frac{[Salt]}{[Acid]}$
$6 - 5 = \log \frac{[Salt]}{[Acid]}$
$1 = \log \frac{[Salt]}{[Acid]}$
Taking the antilog on both sides:
$\frac{[Salt]}{[Acid]} = 10^1 = 10$
Therefore,the ratio of salt to acid concentration is $10: 1$.

(C) The process involves three steps:
$1$. Cooling $1 \ mol$ of liquid water from $5 \ ^\circ C$ to $0 \ ^\circ C$: $\Delta H_1 = C_p(l) \times \Delta T = 75.3 \ J \ mol^{-1} \ K^{-1} \times (0 - 5) \ K = -376.5 \ J \ mol^{-1} = -0.3765 \ kJ \ mol^{-1}$.
$2$. Freezing $1 \ mol$ of liquid water at $0 \ ^\circ C$ to ice at $0 \ ^\circ C$: $\Delta H_2 = -\Delta _{fus}H = -6 \ kJ \ mol^{-1}$.
$3$. Cooling $1 \ mol$ of ice from $0 \ ^\circ C$ to $-5 \ ^\circ C$: $\Delta H_3 = C_p(s) \times \Delta T = 36.8 \ J \ mol^{-1} \ K^{-1} \times (-5 - 0) \ K = -184 \ J \ mol^{-1} = -0.184 \ kJ \ mol^{-1}$.
Total enthalpy change $\Delta H = \Delta H_1 + \Delta H_2 + \Delta H_3 = -0.3765 - 6 - 0.184 = -6.5605 \ kJ \ mol^{-1}$.
The magnitude of the enthalpy change is $6.56 \ kJ \ mol^{-1}$.

(D) To determine the magnetic nature,we calculate the total number of electrons and fill the molecular orbitals:
$NO^{+} : 14 \text{ electrons} \Rightarrow KK, \sigma(2s)^2, \sigma^*(2s)^2, (\pi 2p_x)^2 = (\pi 2p_y)^2, (\sigma 2p_z)^2$ (All paired,diamagnetic)
$CO : 14 \text{ electrons} \Rightarrow KK, \sigma(2s)^2, \sigma^*(2s)^2, (\pi 2p_x)^2 = (\pi 2p_y)^2, (\sigma 2p_z)^2$ (All paired,diamagnetic)
$O_2^{2-} : 18 \text{ electrons}$ $\Rightarrow KK, \sigma(2s)^2, \sigma^*(2s)^2, \sigma(2p_z)^2, (\pi 2p_x)^2 = (\pi 2p_y)^2, \pi^*(2p_x)^2 = \pi^*(2p_y)^2$ (All paired,diamagnetic)
$B_2 : 10 \text{ electrons} \Rightarrow KK, \sigma(2s)^2, \sigma^*(2s)^2, \pi(2p_x)^1 = \pi(2p_y)^1$ (Contains unpaired electrons,paramagnetic)
Therefore,$B_2$ is paramagnetic.

(D) To determine the hybridization,we calculate the steric number $(SN)$ for the central atom: $SN = \text{Number of sigma bonds} + \text{Number of lone pairs}$.
$(a)$ In $BrF_5$,$Br$ has $7$ valence electrons. It forms $5$ bonds with $F$ atoms and has $1$ lone pair. $SN = 5 + 1 = 6$,which corresponds to $sp^3d^2$ hybridization.
$(b)$ In $SF_6$,$S$ has $6$ valence electrons. It forms $6$ bonds with $F$ atoms and has $0$ lone pairs. $SN = 6 + 0 = 6$,which corresponds to $sp^3d^2$ hybridization.
$(c)$ In $[CrF_6]^{3-}$,$Cr^{3+}$ has $d^3$ configuration. It forms $6$ coordinate bonds with $F^-$ ligands. The hybridization is $d^2sp^3$ (inner orbital complex),which is equivalent to $sp^3d^2$ in terms of geometry (octahedral).
$(d)$ In $PF_5$,$P$ has $5$ valence electrons. It forms $5$ bonds with $F$ atoms and has $0$ lone pairs. $SN = 5 + 0 = 5$,which corresponds to $sp^3d$ hybridization.
Therefore,$PF_5$ does not display $sp^3d^2$ hybridization.

(C) The marble of the Taj Mahal is primarily composed of calcium carbonate $(CaCO_3)$.
Pollutant gases like $SO_2$ and $NO_2$ react with moisture in the atmosphere to form sulfuric acid $(H_2SO_4)$ and nitric acid $(HNO_3)$,respectively.
These acids cause acid rain,which reacts with the marble to form soluble salts,leading to the discolouration and loss of lustre of the monument.

(D) An oxidizing agent is a substance that gains electrons and gets reduced in a chemical reaction.
In the reaction $PbS + 4H_2O_2 \to PbSO_4 + 4H_2O$:
$1$. The oxidation state of sulfur $(S)$ in $PbS$ changes from $-2$ to $+6$ in $PbSO_4$. This is an oxidation process.
$2$. The oxidation state of oxygen in $H_2O_2$ changes from $-1$ to $-2$ in $H_2O$. This is a reduction process.
Since $H_2O_2$ is reduced,it acts as an oxidizing agent.
In options $A$,$B$,and $C$,$H_2O_2$ acts as a reducing agent because the oxygen in $H_2O_2$ is oxidized from $-1$ to $0$ (in $O_2$).

(A) The large jump between the $2^{nd}$ and $3^{rd}$ ionization enthalpies for both elements ($A: 1757$ to $14847$; $B: 1450$ to $7731$) indicates that both elements have $2$ valence electrons,placing them in group $2$ (alkaline earth metals).
Ionization enthalpy decreases down a group as the atomic size increases.
Since the ionization enthalpy values for element $B$ are lower than those for element $A$,element $B$ must be located below element $A$ in group $2$.

(C) $(i)$ $\mathop {Al^{3+}}\limits_{\text{Most stable}}$ $\xrightarrow{E^o = -1.66} Al \xleftarrow{E^o = +0.55} \mathop {Al^{+}}\limits_{\text{Less stable}}$
$(ii)$ $\mathop {Tl^{3+}}\limits_{\text{Less stable}}$ $\xrightarrow{E^o = +1.26} Tl \xleftarrow{E^o = -0.34} \mathop {Tl^{+}}\limits_{\text{Most stable}}$
For $Al$,the reduction potential of $Al^{3+}/Al$ is $-1.66 \ V$,indicating $Al^{3+}$ is highly stable. For $Tl$,the reduction potential of $Tl^{+}/Tl$ is $-0.34 \ V$,while $Tl^{3+}/Tl$ is $+1.26 \ V$. $A$ more negative reduction potential for the $M^{+}/M$ couple indicates higher stability of the $M^{+}$ ion relative to the metal. Comparing the stability of $Tl^{+}$ and $Al^{+}$,$Tl^{+}$ is more stable due to the inert pair effect and its lower reduction potential compared to $Al^{+}$. Thus,$Tl^{+}$ is more stable than $Al^{+}$.

(A) $6Li_{(s)} + N_{2(g)} \to 2Li_3N_{(s)}$
$Li_3N_{(s)} + 3H_2O_{(l)} \to 3LiOH_{(aq)} + NH_{3(g)}$
$CuSO_{4(aq)} + 4NH_{3(g)} \to [Cu(NH_3)_4]SO_{4(aq)}$ (deep blue complex)
Thus,$M$ is $Li$ and $B$ is $NH_3$.

(B) Partition chromatography is based on the continuous differential partitioning of components of a mixture between a stationary phase and a mobile phase.
In partition chromatography,the stationary phase is typically a liquid held on a solid support,whereas in adsorption chromatography,the stationary phase is a finely divided solid adsorbent.
Therefore,the statement that the stationary phase is a finely divided solid adsorbent is incorrect for partition chromatography.

(B) $1$. Identify the parent chain: The parent chain is a cyclohexane ring.
$2$. Numbering: Number the ring to give the lowest possible locants to the substituents.
$3$. If we start numbering from the carbon with two methyl groups as $1$,the ethyl group gets position $2$. This gives the locant set $(1, 1, 2)$.
$4$. Alphabetical order: Ethyl comes before methyl.
$5$. Therefore,the name is $2-$ethyl$-1,1-$dimethylcyclohexane.

(C) For the function to be continuous at $x = \frac{\pi}{2}$,we must have $\lim_{x \to \frac{\pi}{2}} f(x) = f(\frac{\pi}{2})$.
First,calculate the limit: $\lim_{x \to \frac{\pi}{2}} (\frac{4}{5})^{\frac{\tan 4x}{\tan 5x}}$.
Let $x = \frac{\pi}{2} + h$,where $h \to 0$.
Then $\tan 4x = \tan(4(\frac{\pi}{2} + h)) = \tan(2\pi + 4h) = \tan 4h \approx 4h$.
And $\tan 5x = \tan(5(\frac{\pi}{2} + h)) = \tan(\frac{5\pi}{2} + 5h) = \cot 5h = \frac{1}{\tan 5h} \approx \frac{1}{5h}$.
Thus,$\frac{\tan 4x}{\tan 5x} = \frac{\tan 4h}{\cot 5h} = \tan 4h \cdot \tan 5h \to 0 \cdot \infty$ (indeterminate form).
Actually,$\frac{\tan 4x}{\tan 5x} = \frac{\tan(2\pi + 4h)}{\tan(\frac{5\pi}{2} + 5h)} = \frac{\tan 4h}{\cot 5h} = \tan 4h \cdot \tan 5h$. As $h \to 0$,this limit is $0 \cdot 0 = 0$.
So,$\lim_{x \to \frac{\pi}{2}} f(x) = (\frac{4}{5})^0 = 1$.
Equating to $f(\frac{\pi}{2}) = k + \frac{2}{5}$,we get $k + \frac{2}{5} = 1$.
Therefore,$k = 1 - \frac{2}{5} = \frac{3}{5}$.

(A) For a conditional statement of the form $p \to q$,the contrapositive is defined as $\sim q \to \sim p$.
Here,$p$ is 'two numbers are not equal' and $q$ is 'their squares are not equal'.
Therefore,$\sim q$ is 'the squares of two numbers are equal' and $\sim p$ is 'the numbers are equal'.
Thus,the contrapositive is: 'If the squares of two numbers are equal,then the numbers are equal'.

(B) For an ideal gas,enthalpy $H$ is a function of temperature only,i.e.,$H = f(T)$.
Since the process is isothermal,the change in temperature $\Delta T = 0$,therefore $\Delta H = n C_p \Delta T = 0$,which means enthalpy remains constant.
For an expansion process,the final volume $V_f$ is greater than the initial volume $V_i$ $(V_f > V_i)$.
The change in entropy for an ideal gas is given by $\Delta S = n R \ln(V_f / V_i)$.
Since $V_f / V_i > 1$,$\ln(V_f / V_i) > 0$,therefore $\Delta S > 0$,which means entropy increases.

(D) The reaction is: $NH_3 + HCl \to NH_4Cl$
Initial moles of $HCl = 0.2 \, M \times 0.025 \, L = 0.005 \, mol$
Initial moles of $NH_3 = 0.2 \, M \times 0.050 \, L = 0.010 \, mol$
Since $HCl$ is the limiting reagent,it reacts completely with $0.005 \, mol$ of $NH_3$ to form $0.005 \, mol$ of $NH_4Cl$.
Remaining moles of $NH_3 = 0.010 - 0.005 = 0.005 \, mol$
This forms a basic buffer solution containing $NH_3$ (base) and $NH_4Cl$ (salt).
Using the Henderson-Hasselbalch equation for a basic buffer:
$pOH = pK_b + \log \frac{[Salt]}{[Base]}$
Since the volume is the same for both,the ratio of concentrations is equal to the ratio of moles:
$pOH = 4.75 + \log \frac{0.005}{0.005} = 4.75 + \log(1) = 4.75$
$pH = 14 - pOH = 14 - 4.75 = 9.25$

$(B)$ The radius of an orbit in a hydrogen-like atom is given by $r = 0.529 \times \frac{n^2}{Z} \mathring{A}$.
Given $r = 211.6 \text{ pm} = 2.116 \mathring{A}$ and for hydrogen $Z = 1$.
Substituting the values: $2.116 = 0.529 \times n^2$.
$n^2 = \frac{2.116}{0.529} = 4$.
$n = 2$.
Since the electron transitions from higher orbitals to the $n = 2$ orbit, this corresponds to the Balmer series.

(C) The density $(\rho)$ of a gas is given by the formula $\rho = \frac{PM}{RT}$,where $P$ is pressure,$M$ is molar mass,$R$ is the gas constant,and $T$ is temperature.
Given that the temperature $T$ is constant for both gases.
For $N_2$ gas: $\rho_{N_2} = \frac{P_{N_2} \times M_{N_2}}{RT} = \frac{4 \times 28}{RT}$.
For the unknown gas: $\rho_{gas} = \frac{P_{gas} \times M_{gas}}{RT} = \frac{2 \times M_{gas}}{RT}$.
According to the problem,$\rho_{gas} = 2 \times \rho_{N_2}$.
Substituting the expressions: $\frac{2 \times M_{gas}}{RT} = 2 \times \left( \frac{4 \times 28}{RT} \right)$.
Canceling $RT$ from both sides: $2 \times M_{gas} = 8 \times 28$.
$M_{gas} = 4 \times 28 = 112\, g\, mol^{-1}$.

(C) Let the volume of the $45\%$ acid solution be $V \ mL$. Then the volume of the $20\%$ acid solution is $(800 - V) \ mL$.
Using the principle of conservation of mass for the acid:
$\frac{V \times 45}{100} + \frac{(800 - V) \times 20}{100} = \frac{800 \times 29.875}{100}$
$0.45V + 160 - 0.2V = 239$
$0.25V = 79$
$V = \frac{79}{0.25} = 316 \ mL$

(D) For the process $A \rightarrow B$:
Heat absorbed,$q_{AB} = +5\, J$
Work done by the gas,$w_{AB} = -8\, J$ (since work is done by the gas,it is negative in the system's perspective for $\Delta U = q + w$)
Change in internal energy,$\Delta U_{AB} = q_{AB} + w_{AB} = 5 + (-8) = -3\, J$
For the reverse process $B \rightarrow A$:
Since internal energy is a state function,$\Delta U_{BA} = -\Delta U_{AB} = -(-3) = +3\, J$
Heat evolved,$q_{BA} = -3\, J$
Using the first law of thermodynamics: $\Delta U_{BA} = q_{BA} + w_{BA}$
$3 = -3 + w_{BA}$
$w_{BA} = +6\, J$
Since the work done $w_{BA}$ is positive,it means work is done by the surrounding on the gas.

(C) Ionization enthalpy $(IE)$ depends on atomic size and electronic stability.
Smaller atomic size leads to higher $IE$ because the valence electrons are more strongly attracted by the nucleus.
Additionally,half-filled and fully-filled orbitals possess extra stability,requiring more energy to remove an electron.
Comparing the given configurations: $[Ne]\, 3s^2\, 3p^1$,$[Ne]\, 3s^2\, 3p^2$,and $[Ne]\, 3s^2\, 3p^3$ belong to the $3^{rd}$ period,while $[Ar]\, 3d^{10}\, 4s^2\, 4p^3$ belongs to the $4^{th}$ period.
Elements in the $3^{rd}$ period have smaller atomic radii than those in the $4^{th}$ period,resulting in higher $IE$ for $3^{rd}$ period elements.
Among the $3^{rd}$ period elements,$[Ne]\, 3s^2\, 3p^3$ has a half-filled $p$-orbital,which provides extra stability,making its ionization enthalpy the highest.

(D) According to Le Chatelier's principle,the equilibrium position is affected by changes in the concentration of gaseous or aqueous species. Solids and pure liquids have constant activity and their concentration does not change the equilibrium position.

Perturbation	Effect on Equilibrium
$A$. Removal of $CO$	Shifts to the left
$B$. Removal of $CO_2$	Shifts to the right
$C$. Addition of $CO_2$	Shifts to the left
$D$. Addition of $Fe_2O_3$	No change

Since $Fe_2O_3$ is a solid,its addition or removal does not affect the equilibrium position.

(C)

Compound	Nature
$KO_2$	Superoxide
$BaO_2$	Peroxide
$SiO_2$	Oxide
$CsO_2$	Superoxide

An oxide is a chemical compound that contains at least one oxygen atom and one other element in its chemical formula,where the oxygen is in the $-2$ oxidation state. $SiO_2$ (silicon dioxide) is a typical oxide,whereas $KO_2$ and $CsO_2$ are superoxides (containing the $O_2^-$ ion) and $BaO_2$ is a peroxide (containing the $O_2^{2-}$ ion).

(D) Greenhouse gases are those gases that absorb and emit radiation within the thermal $IR$ range.
This process is the fundamental cause of the greenhouse effect.
The primary greenhouse gases include $CO_2, CH_4, N_2O,$ and $O_3$.

(C) . Nitration reactions are carried out in the presence of concentrated $HNO_3$ and concentrated $H_2SO_4$.
$(b)$. Aniline acts as a base. In the presence of $H_2SO_4$,it undergoes protonation to form the anilinium ion $(-NH_3^+)$.
$(c)$. The anilinium ion is a strongly deactivating group and is meta-directing in nature due to the positive charge on the nitrogen atom,which exerts a strong $-I$ effect.
$(d)$. Consequently,during the nitration of aniline in an acidic medium,a significant amount of the meta-nitro product is formed.

(B) The reactivity of halides in $S_N1$ reaction depends on the stability of the carbocation formed as the rate-determining step.
The carbocations formed are:
$(I)$ $CH_3-CH^+-CH_2-CH_3$ ($2^\circ$ carbocation)
$(II)$ $CH_3-CH_2-CH_2^+$ ($1^\circ$ carbocation)
$(III)$ $p-CH_3O-C_6H_4-CH_2^+$ (Resonance stabilized benzylic carbocation with strong $+M$ effect of $-OCH_3$ group).
The stability order of these carbocations is $(II) < (I) < (III)$.
Therefore,the increasing order of reactivity for the $S_N1$ reaction is $(II) < (I) < (III)$.

(A) Nylon-$6$ is prepared by heating $\varepsilon$-caprolactam with water at a high temperature $(533-543 \ K)$.
In this process,the cyclic amide ring of $\varepsilon$-caprolactam undergoes a hydrolysis reaction to form an amino acid,which then undergoes polymerization to form Nylon-$6$.

(A) Bromine water is decolourized by compounds containing carbon-carbon double or triple bonds (unsaturation) due to electrophilic addition.
Treatment with $tert-BuONa$ (a strong base) typically induces elimination reactions $(E2)$ in alkyl halides to form alkenes.
Option $A$ is cyclohexyl bromomethyl ether. Treatment with $tert-BuONa$ leads to substitution (Williamson ether synthesis) rather than elimination because the $\beta$-carbon lacks hydrogens,resulting in a saturated ether that does not react with bromine water.
Options $B$,$C$,and $D$ contain $\beta$-hydrogens and undergo elimination to form alkenes,which will decolourize bromine water.

(D) In a face-centred cubic $(fcc)$ unit cell,the atoms touch each other along the face diagonal.
Let the radius of the atoms be $r$ and the edge length of the unit cell be $a$.
The face diagonal of the cube is given by $\sqrt{a^2 + a^2} = \sqrt{2} a$.
Since the atoms touch along the face diagonal,the length of the face diagonal is equal to $4r$.
Therefore,$4r = \sqrt{2} a$,which gives $r = \frac{\sqrt{2} a}{4} = \frac{a}{2\sqrt{2}}$.
The closest approach between two atoms is the distance between their centers,which is $2r$.
Thus,$2r = 2 \times \frac{a}{2\sqrt{2}} = \frac{a}{\sqrt{2}}$.

(D) According to the Arrhenius equation,$k = A e^{-E_a / RT}$.
For reactions $R_1$ and $R_2$ with identical pre-exponential factors $A$:
$k_1 = A e^{-E_{a1} / RT}$
$k_2 = A e^{-E_{a2} / RT}$
Dividing $k_2$ by $k_1$:
$\frac{k_2}{k_1} = \frac{A e^{-E_{a2} / RT}}{A e^{-E_{a1} / RT}} = e^{(E_{a1} - E_{a2}) / RT}$
Taking the natural logarithm on both sides:
$\ln(k_2/k_1) = \frac{E_{a1} - E_{a2}}{RT}$
Given $E_{a1} - E_{a2} = 10 \, kJ \, mol^{-1} = 10,000 \, J \, mol^{-1}$,$R = 8.314 \, J \, mol^{-1} \, K^{-1}$,and $T = 300 \, K$:
$\ln(k_2/k_1) = \frac{10,000}{8.314 \times 300} \approx \frac{10,000}{2494.2} \approx 4.009 \approx 4$.

(A) The conversion involves the selective transformation of an aldehyde group to a tertiary alcohol while preserving the ketone group or transforming it appropriately.
$1$. First,the aldehyde group is oxidized to a carboxylic acid using Tollens' reagent,$[Ag(NH_3)_2]^+ OH^-$.
$2$. Next,the carboxylic acid is esterified to a methyl ester using $H^+/CH_3OH$.
$3$. Finally,the reaction with excess $CH_3MgBr$ (Grignard reagent) attacks both the ketone and the ester groups to form the final diol product.

(B) The Tyndall effect is an optical phenomenon observed in colloidal systems.
The two essential conditions for the observation of the Tyndall effect are:
$1$. The diameter of the dispersed particles is not much smaller than the wavelength of the light used.
$2$. The refractive indices of the dispersed phase and the dispersion medium differ greatly in magnitude.
Thus,conditions $2$ and $4$ are correct.

(A) reducing sugar is a carbohydrate that possesses a free aldehyde or ketone group,or a hemiacetal group that can open to form such a group in solution.
In an aqueous $KOH$ solution,esters of sugars (like the one shown in option $A$) undergo base-catalyzed hydrolysis.
The hydrolysis of the ester group at the anomeric carbon releases a free hydroxyl group,regenerating the hemiacetal structure.
This hemiacetal structure exists in equilibrium with its open-chain form containing a free aldehyde or ketone group,thus exhibiting reducing properties.
Therefore,the compound in option $A$ acts as a reducing sugar after hydrolysis in aqueous $KOH$.

(A) In the reaction $XeF_4 + O_2F_2 \rightarrow XeF_6 + O_2$,the oxidation state of $Xe$ changes from $+4$ to $+6$ (oxidation) and the oxidation state of $O$ changes from $+1$ to $0$ (reduction).
Since both oxidation and reduction occur,this is a redox reaction.
$\stackrel{+4}{Xe}F_4 + \stackrel{+1}{O_2}F_2 \rightarrow \stackrel{+6}{Xe}F_6 + \stackrel{0}{O_2}$

(C) When chlorine gas reacts with cold and dilute aqueous $NaOH$,it undergoes a disproportionation reaction.
The chemical equation is:
$Cl_{2} + 2NaOH \text{ (cold/dilute)} \rightarrow NaCl + NaOCl + H_{2}O$
In this reaction,$Cl_{2}$ is reduced to $Cl^{-}$ (in $NaCl$) and oxidized to $ClO^{-}$ (in $NaOCl$).
Therefore,the products are $Cl^{-}$ and $ClO^{-}$.

(B) The reaction of $1,2-diphenyl-1-bromoethane$ with a strong base like potassium tert-butoxide $(t-BuOK)$ in the presence of heat $(\Delta)$ undergoes a dehydrohalogenation reaction.
This is an $E_2$ elimination reaction where the base abstracts a proton from the $\beta$-carbon,leading to the formation of a double bond between the $\alpha$ and $\beta$ carbons.
The product formed is $1,2-diphenylethene$ $(C_6H_5CH=CHC_6H_5)$,which is a stable conjugated alkene.
The reaction is:
$C_6H_5CH(Br)CH_2C_6H_5 + t-BuOK \xrightarrow{\Delta} C_6H_5CH=CHC_6H_5 + t-BuOH + KBr$

(D) The association reaction is: $2 \ CH_{3}COOH \rightleftharpoons (CH_{3}COOH)_{2}$
Initial moles: $1 \quad 0$
At equilibrium: $1 - \alpha \quad \alpha / 2$
Total moles at equilibrium: $1 - \alpha + \alpha / 2 = 1 - \alpha / 2$
Van't Hoff factor $i = 1 - \alpha / 2$
Using the formula: $\Delta T_{f} = i \times K_{f} \times m$
Molality $m = \frac{0.2 / 60}{20 / 1000} = \frac{0.2 \times 1000}{60 \times 20} = \frac{200}{1200} = 0.1667 \ m$
$0.45 = (1 - \alpha / 2) \times 5.12 \times 0.1667$
$0.45 = (1 - \alpha / 2) \times 0.8535$
$1 - \alpha / 2 = 0.5272$
$\alpha / 2 = 0.4728$
$\alpha = 0.9456$
Percentage association $= 94.56 \% \approx 94.6 \%$

(D) The number of moles of $CoCl_3 \cdot 6H_2O$ is $0.1 \ M \times 0.1 \ L = 0.01 \ mol$.
The number of moles of $AgCl$ precipitated is $\frac{1.2 \times 10^{22}}{6.022 \times 10^{23}} \approx 0.02 \ mol$.
Since $0.01 \ mol$ of complex produces $0.02 \ mol$ of $AgCl$,$1 \ mol$ of complex produces $2 \ mol$ of $AgCl$.
This indicates that there are $2$ chloride ions outside the coordination sphere.
Therefore,the complex is $[Co(H_2O)_5 Cl]Cl_2 \cdot H_2O$.

(A) $DIBAL-H$ (Diisobutylaluminium hydride) is a selective reducing agent. It is commonly used to reduce esters or lactones (cyclic esters) to aldehydes. In the given reaction,the lactone ring is opened and reduced to an aldehyde group,while the hydroxyl group is formed at the position where the oxygen was attached to the ring. The carboxylic acid group $(-COOH)$ remains unaffected under these specific reaction conditions.

(A) The reducing strength of a species is inversely proportional to its standard reduction potential. $A$ lower (more negative) standard reduction potential indicates a higher tendency to undergo oxidation,making it a stronger reducing agent.
Comparing the given values:
$E^o_{Cl_2/Cl^-} = 1.36 \ V$
$E^o_{Cr^{3+}/Cr} = -0.74 \ V$
$E^o_{Cr_2O_7^{2-}/Cr^{3+}} = 1.33 \ V$
$E^o_{MnO_4^-/Mn^{2+}} = 1.51 \ V$
Among the species listed,$Cr$ has the lowest reduction potential $(-0.74 \ V)$.
Therefore,$Cr$ is the strongest reducing agent.

(A) Brownian movement is the random motion of particles suspended in a fluid (a liquid or gas) resulting from their collision with the fast-moving atoms or molecules in the fluid.
Since smaller particles have less mass,they are more easily affected by these collisions,making the Brownian movement more pronounced for smaller particles compared to bigger particles.
Therefore,option $A$ is the correct statement.
Metal sulphides are lyophobic (not lyophilic).
Hardy-Schulze law states that the coagulating power of an ion depends on its valency,not its size.
Charcoal adsorbs gases that are more easily liquefiable; $H_2S$ is more easily liquefiable than $Cl_2$ (due to higher boiling point),so $H_2S$ is adsorbed more.

(A) The standard Gibbs free energy change is given by $\Delta G^o = -nFE^o$.
For reaction $(i)$: $Fe^{2+} + 2e^- \to Fe$,$E^o = -0.47 \ V$,so $\Delta G^o_1 = -2 \times F \times (-0.47) = 0.94 \ F$.
For reaction $(ii)$: $Fe^{3+} + e^- \to Fe^{2+}$,$E^o = +0.77 \ V$,so $\Delta G^o_2 = -1 \times F \times (+0.77) = -0.77 \ F$.
For the overall reaction $(iii)$: $Fe^{3+} + 3e^- \to Fe$,the sum of the reactions is $(i) + (ii)$.
Therefore,$\Delta G^o_3 = \Delta G^o_1 + \Delta G^o_2 = 0.94 \ F - 0.77 \ F = 0.17 \ F$.
Using $\Delta G^o_3 = -nFE^o_{Fe^{3+}/Fe}$,where $n = 3$:
$0.17 \ F = -3 \times F \times E^o_{Fe^{3+}/Fe}$
$E^o_{Fe^{3+}/Fe} = \frac{0.17 \ F}{-3 \ F} = -0.057 \ V$.

(C) Molar mass of $Na_2SO_4 = (2 \times 23) + 32 + (4 \times 16) = 142 \ g \ mol^{-1}$.
Van't Hoff factor $(i)$ for $Na_2SO_4 \to 2Na^+ + SO_4^{2-}$ is $i = 1 + (n-1)\alpha$,where $n=3$ and $\alpha = 0.815$.
$i = 1 + (3-1) \times 0.815 = 1 + 2 \times 0.815 = 2.63$.
Freezing point depression formula: $\Delta T_f = i \times K_f \times m$.
$3.82 = 2.63 \times 1.86 \times \left( \frac{5 / 142}{x / 1000} \right)$.
$3.82 = 2.63 \times 1.86 \times \frac{5000}{142 \times x}$.
$x = \frac{2.63 \times 1.86 \times 5000}{142 \times 3.82} \approx 45.07 \ g$.

(B) Using the Arrhenius equation: $\ln(\frac{k_2}{k_1}) = \frac{E_a}{R} [\frac{T_2 - T_1}{T_1 T_2}]$.
For reaction $A$: $\ln(2) = \frac{E_{a,A}}{R} [\frac{310 - 300}{300 \times 310}] = \frac{E_{a,A}}{R} [\frac{10}{93000}]$.
For reaction $B$: $\ln(2) = \frac{E_{a,B}}{R} [\frac{\Delta T}{300(300 + \Delta T)}]$.
Given $E_{a,B} = 2 E_{a,A}$,we substitute $\ln(2)$ from reaction $A$ into reaction $B$:
$\frac{E_{a,A}}{R} [\frac{10}{93000}] = \frac{2 E_{a,A}}{R} [\frac{\Delta T}{300(300 + \Delta T)}]$.
$\frac{10}{93000} = \frac{2 \Delta T}{300(300 + \Delta T)}$.
$\frac{1}{9300} = \frac{\Delta T}{150(300 + \Delta T)}$.
$150(300 + \Delta T) = 9300 \Delta T$.
$45000 + 150 \Delta T = 9300 \Delta T$.
$9150 \Delta T = 45000$.
$\Delta T = \frac{45000}{9150} \approx 4.92 \, K$.

(A) To determine the highest oxidation state,we calculate the oxidation number for each metal:
$1$. $MnO_4^-$: Let $x$ be the oxidation state of $Mn$. $x + 4(-2) = -1 \implies x = +7$. $CrO_2Cl_2$: Let $y$ be the oxidation state of $Cr$. $y + 2(-2) + 2(-1) = 0 \implies y = +6$. Both $Mn$ $(+7)$ and $Cr$ $(+6)$ are in their highest possible oxidation states.
$2$. $[NiCl_4]^{2-}$: $Ni$ is $+2$; $[CoCl_4]^{2-}$: $Co$ is $+2$.
$3$. $[Fe(CN)_6]^{3-}$: $Fe$ is $+3$; $[Cu(CN)_4]^{2-}$: $Cu$ is $+2$.
$4$. $[FeCl_4]^-$: $Fe$ is $+3$; $Co_2O_3$: $Co$ is $+3$.
Thus,the pair $MnO_4^-$ and $CrO_2Cl_2$ contains metals in their highest oxidation states.

(D) For peroxodisulphuric acid $(H_2S_2O_8)$: The structure contains $4$ $S=O$ bonds and $2$ $S-OH$ bonds.
For pyrosulphuric acid $(H_2S_2O_7)$: The structure contains $4$ $S=O$ bonds and $2$ $S-OH$ bonds.
Therefore,the number of $S=O$ and $S-OH$ bonds in $H_2S_2O_8$ are $4$ and $2$ respectively,and in $H_2S_2O_7$ are $4$ and $2$ respectively.

(D) Group $IV$ cations include $Co^{2+}, Ni^{2+}, Mn^{2+},$ and $Zn^{2+}$.
When $H_2S$ is passed through a basic solution of $Zn^{2+}$,a white precipitate of $ZnS$ is formed.
This precipitate dissolves in dilute $HCl$ to form $ZnCl_2$: $ZnS + 2HCl \rightarrow ZnCl_2 + H_2S$.
$ZnCl_2$ reacts with $NaOH$ to form a white precipitate of $Zn(OH)_2$: $Zn^{2+} + 2OH^{-} \rightarrow Zn(OH)_2 \downarrow$ (White).
With potassium ferrocyanide $(K_4[Fe(CN)_6])$,$Zn^{2+}$ ions form a bluish-white precipitate of $K_2Zn_3[Fe(CN)_6]_2$: $3Zn^{2+} + 2K^{+} + 2[Fe(CN)_6]^{4-} \rightarrow K_2Zn_3[Fe(CN)_6]_2 \downarrow$ (Bluish-white).

(B) When the given mixture is shaken with $1 \, M \, HCl$,the amine gets protonated to form a salt,which is a cation $\left( RNH_2CH_3^{\oplus} \right)$.
This salt is ionic and therefore dissolves in the aqueous layer $(H_2O)$ rather than the organic solvent.
Among the given compounds,$(II)$ is an amine ($N$-methylcyclohexylamine),which reacts with $HCl$ to form a water-soluble salt.
The other compounds (sulfide,ether,and ketone) do not react with $1 \, M \, HCl$ to form ionic species and remain in the organic layer.
Therefore,the correct option is $(II)$.

(C) Enzyme inhibition can be either reversible or irreversible.
In case of irreversible inhibition,the inhibitor dissociates very slowly from its target enzyme because it has become tightly bound to the enzyme,either covalently or non-covalently.
Some irreversible inhibitors are important drugs like $Penicillin$ and $Aspirin$.
Thus,"drug induced poisoning" occurs when a drug binds irreversibly to the active site of the enzyme,permanently blocking its function.

(B) Friedel-Crafts reaction requires the formation of a stable carbocation or an acylium ion intermediate.
In the case of vinyl chloride $(CH_2=CH-Cl)$,the chlorine atom is directly attached to an $sp^2$ hybridized carbon atom.
The lone pair of electrons on the chlorine atom participates in resonance with the double bond,giving the $C-Cl$ bond partial double bond character.
This makes the cleavage of the $C-Cl$ bond very difficult,and the resulting vinyl carbocation $(CH_2=CH^+)$ is highly unstable due to the $sp$ hybridization of the positively charged carbon.
Therefore,vinyl chloride does not undergo Friedel-Crafts reaction.

(B) Essential amino acids are those that cannot be synthesized by the human body and must be obtained through the diet. $Valine$ is an essential amino acid. Other examples include $Histidine$,$Isoleucine$,$Leucine$,$Lysine$,$Methionine$,$Phenylalanine$,$Threonine$,and $Tryptophan$.

(D) The reaction involves an intramolecular esterification (lactonization) between the carboxylic acid group $(-COOH)$ and the hydroxymethyl group $(-CH_2OH)$ in the presence of an acid catalyst $(HCl_{(g)})$.
$1$. The carboxylic acid group is protonated by $HCl$,making the carbonyl carbon more electrophilic.
$2$. The oxygen atom of the $-CH_2OH$ group acts as a nucleophile and attacks the activated carbonyl carbon.
$3$. This leads to the formation of a cyclic ester (lactone).
$4$. Water is eliminated in the process to form the final lactone product.
The structure shown in option $D$ represents the correct cyclic lactone formed by the reaction of the carboxylic acid and the hydroxymethyl group.

(D) The reaction is a dehydrohalogenation reaction involving the elimination of two molecules of $HBr$ from $2,4$-dibromohexane using alcoholic $KOH$ and heat.
According to $Saytzeff's$ rule,the most stable alkene is formed as the major product.
In this case,the formation of $2,4$-hexadiene $(CH_3-CH=CH-CH=CH-CH_3)$ is favored because it is a conjugated diene,which provides extra stability due to resonance compared to isolated or cumulated dienes.

(B) The reaction is an $E2$ elimination reaction.
The ethoxide ion $(C_2H_5O^-)$ acts as a strong base and abstracts a $\beta$-hydrogen.
The hydrogen on the carbon adjacent to the benzene ring is more acidic due to the electron-withdrawing effect of the phenyl group.
Removal of this hydrogen leads to the formation of $C_6H_5-CH=C(CH_3)-CH_2-CH_3$.
This product is the major product because the double bond is in conjugation with the benzene ring,which provides significant resonance stabilization.

(B) The starting material is $2-(2-hydroxyethyl)phenol$,which contains both a phenolic $-OH$ group and an aliphatic $-OH$ group.
$K_2CO_3$ is a mild base that selectively deprotonates the more acidic phenolic $-OH$ group $(pK_a \approx 10)$ compared to the aliphatic $-OH$ group $(pK_a \approx 16)$.
The resulting phenoxide ion is resonance-stabilized by the benzene ring,making it the preferred site for nucleophilic substitution.
Upon addition of $CH_3I$ $(1. \, eq.)$,the phenoxide ion acts as a nucleophile and attacks the methyl iodide via an $S_N2$ mechanism to form the methyl ether at the phenolic position.
Therefore,the major product is $1-(2-hydroxyethyl)-2-methoxybenzene$.

(B) The cell reaction is: $M(s) + 3Ag^{+}(aq) \to M^{3+}(aq) + 3Ag(s)$.
The number of electrons transferred is $n = 3$.
Using the Nernst equation: $E_{cell} = E^{o}_{cell} - \frac{0.0591}{n} \log \frac{[M^{3+}]}{[Ag^{+}]^{3}}$.
Substituting the given values: $0.421 = E^{o}_{cell} - \frac{0.0591}{3} \log \frac{0.001}{(0.01)^{3}}$.
$0.421 = E^{o}_{cell} - 0.0197 \log \frac{10^{-3}}{10^{-6}} = E^{o}_{cell} - 0.0197 \log(10^{3})$.
$0.421 = E^{o}_{cell} - 0.0197 \times 3 = E^{o}_{cell} - 0.0591$.
$E^{o}_{cell} = 0.421 + 0.0591 = 0.4801 \ V \approx 0.48 \ V$.
Since $E^{o}_{cell} = E^{o}_{cathode} - E^{o}_{anode} = E^{o}_{Ag^{+}/Ag} - E^{o}_{M^{3+}/M}$.
$0.48 = 0.80 - E^{o}_{M^{3+}/M}$.
$E^{o}_{M^{3+}/M} = 0.80 - 0.48 = 0.32 \ V$.

(C) According to the Freundlich adsorption isotherm:
$\log \frac{x}{m} = \log k + \frac{1}{n} \log P$
Comparing this with the equation of a straight line $y = mx + c$,the slope is $\frac{1}{n}$.
Given that the slope is $0.5$,we have $\frac{1}{n} = 0.5 = \frac{1}{2}$.
Substituting this into the adsorption equation $\frac{x}{m} = k \cdot P^{1/n}$,we get:
$\frac{x}{m} = k \cdot P^{1/2}$
This implies that the adsorption is proportional to the square root of pressure $(P^{1/2})$.

(A) Using the Arrhenius equation: $\ln \frac{k_2}{k_1} = \frac{E_a}{R} \left( \frac{1}{T_1} - \frac{1}{T_2} \right)$
Given: $\frac{k_2}{k_1} = 4$,$T_1 = 300 \, K$,$T_2 = 310 \, K$,$R = 8.314 \, J \, mol^{-1} \, K^{-1}$,$\ln 2 = 0.693$.
$\ln 4 = \frac{E_a}{8.314} \left( \frac{310 - 300}{310 \times 300} \right)$
$2 \ln 2 = \frac{E_a}{8.314} \left( \frac{10}{93000} \right)$
$2 \times 0.693 = \frac{E_a}{8.314} \times \frac{1}{9300}$
$E_a = 1.386 \times 8.314 \times 9300 \, J \, mol^{-1} \approx 107200 \, J \, mol^{-1} = 107.2 \, kJ \, mol^{-1}$.

(C) Molar mass of $CHCl_3 = 12 + 1 + 3(35.5) = 119.5 \ g \ mol^{-1}$.
Molar mass of $CH_2Cl_2 = 12 + 2 + 2(35.5) = 85 \ g \ mol^{-1}$.
Moles of $CHCl_3 = \frac{11.95 \ g}{119.5 \ g \ mol^{-1}} = 0.1 \ mol$.
Moles of $CH_2Cl_2 = \frac{8.5 \ g}{85 \ g \ mol^{-1}} = 0.1 \ mol$.
Total moles $= 0.1 + 0.1 = 0.2 \ mol$.
Mole fraction of $CHCl_3$ $(x_{CHCl_3}) = \frac{0.1}{0.2} = 0.5$.
Mole fraction of $CH_2Cl_2$ $(x_{CH_2Cl_2}) = \frac{0.1}{0.2} = 0.5$.
Partial pressure of $CHCl_3$ $(P_{CHCl_3}) = x_{CHCl_3} \times P^{\circ}_{CHCl_3} = 0.5 \times 200 = 100 \ mm \ Hg$.
Partial pressure of $CH_2Cl_2$ $(P_{CH_2Cl_2}) = x_{CH_2Cl_2} \times P^{\circ}_{CH_2Cl_2} = 0.5 \times 415 = 207.5 \ mm \ Hg$.
Total pressure $(P_{total}) = P_{CHCl_3} + P_{CH_2Cl_2} = 100 + 207.5 = 307.5 \ mm \ Hg$.
Mole fraction of $CHCl_3$ in vapour phase $(y_{CHCl_3}) = \frac{P_{CHCl_3}}{P_{total}} = \frac{100}{307.5} \approx 0.325$.

(C) The partial hydrolysis of $XeF_6$ is given by the reaction: $XeF_6 + H_2O \to XeOF_4 + 2HF$.
Similarly,$XeF_6$ reacts with silica $(SiO_2)$ to form $XeOF_4$ and $SiF_4$: $2XeF_6 + SiO_2 \to 2XeOF_4 + SiF_4$.
Thus,the compound '$X$' is $XeOF_4$ (Xenon oxytetrafluoride).

(D) The structure of pyrophosphoric acid $(H_4P_2O_7)$ consists of two $PO_4$ tetrahedra linked by an oxygen atom $(P-O-P)$.
Each phosphorus atom is bonded to two $OH$ groups,one terminal oxygen atom via a double bond,and one bridging oxygen atom.
Therefore,the total number of $P-OH$ bonds is $4$ (two per phosphorus atom).
To calculate the oxidation state of $P$ in $H_4P_2O_7$:
Let the oxidation state of $P$ be $x$.
$4(+1) + 2(x) + 7(-2) = 0$
$4 + 2x - 14 = 0$
$2x = 10$
$x = +5$
Thus,the number of $P-OH$ bonds is $4$ and the oxidation state of phosphorus is $+5$.

(D)

Ion	$E^o \ (M^{3+}/M^{2+}) \ (V)$
$Ti^{2+}$	$-0.37$
$V^{2+}$	$-0.26$
$Cr^{2+}$	$-0.41$
$Mn^{2+}$	$+1.57$

The liberation of hydrogen gas from dilute acids by metal ions depends on the reduction potential of the $M^{3+}/M^{2+}$ couple.
If the $E^o$ value is negative,the $M^{2+}$ ion can be oxidized to $M^{3+}$ while reducing $H^+$ to $H_2$ gas.
Among the given ions,$Mn^{2+}$ has a highly positive $E^o$ value $(+1.57 \ V)$,which means it is very stable and does not easily lose an electron to reduce $H^+$ ions.
Therefore,$Mn^{2+}$ does not liberate hydrogen gas.

(C) The number of $\pi$-bonds in the structures are as follows:
$1.$ $H_2SO_3$ (Sulfurous acid): It has one $S=O$ double bond,so it contains $1$ $\pi$-bond.
$2.$ $H_2SO_4$ (Sulfuric acid): It has two $S=O$ double bonds,so it contains $2$ $\pi$-bonds.
$3.$ $H_2S_2O_7$ (Pyrosulfuric acid): It has four $S=O$ double bonds (two on each sulfur atom),so it contains $4$ $\pi$-bonds.
Therefore,the correct decreasing sequence of the number of $\pi$-bonds is $H_2S_2O_7 (4) > H_2SO_4 (2) > H_2SO_3 (1)$.

(A) The structure of the metal carbonyl complex $[Co_2(CO)_8]$ in the solid state consists of two $Co(CO)_4$ units linked by a metal-metal bond.
It contains one $Co-Co$ bond.
There are six terminal $CO$ ligands (three on each $Co$ atom) and two bridging $CO$ ligands.
Therefore,the correct option is $A$.

(A) The molecular formula $C_8H_8O_2$ corresponds to $p$-methoxybenzaldehyde $(CH_3OC_6H_4CHO)$.
$1$. It contains an aldehyde group without $\alpha$-hydrogens,which allows it to undergo the Cannizzaro reaction with conc. $NaOH$ to form $p$-methoxybenzyl alcohol and sodium $p$-methoxybenzoate.
$2$. It reacts with acetophenone $(C_6H_5COCH_3)$ in the presence of a base to undergo a cross-aldol condensation,forming a single product because the aldehyde lacks $\alpha$-hydrogens,ensuring it acts only as the electrophile.

(B) $1$. $\alpha-D$-glucose and $\beta-D$-glucose differ only in the configuration at the $C-1$ carbon (anomeric carbon),so they are anomers. Statement $A$ is correct.
$2$. Enantiomers are non-superimposable mirror images of each other. $\alpha-D$-glucose and $\beta-D$-glucose are diastereomers,not enantiomers. Statement $B$ is incorrect.
$3$. Cellulose is a linear polymer of $\beta-D$-glucose units joined by $\beta-1,4$-glycosidic linkages. Statement $C$ is correct.
$4$. The pentaacetate of glucose does not contain a free $-OH$ group at the $C-1$ position,meaning it cannot form an open-chain aldehyde structure to react with hydroxylamine $(NH_2OH)$. Statement $D$ is correct.
Therefore,the incorrect statement is $B$.

(A) biodegradable polymer is one that can be broken down by microorganisms.
$Nylon-2-nylon-6$ is a biodegradable polyamide copolymer of glycine $(H_2N-CH_2-COOH)$ and aminocaproic acid $(H_2N-(CH_2)_5-COOH)$.
The structure is:
$[-NH-CH_2-CO-NH-(CH_2)_5-CO-]_n$
This corresponds to the structure given in option $A$.

(A) $I$. $C_2H_5OH$: Exhibits strong intermolecular hydrogen bonding,which leads to the highest boiling point.
$II$. $C_2H_5Cl$: Exhibits dipole-dipole interactions. It is more polar than the ether $(IV)$,resulting in a higher boiling point.
$III$. $C_2H_5CH_3$ (Propane): Only weak London dispersion forces exist,leading to the lowest boiling point.
$IV$. $C_2H_5OCH_3$ (Methoxyethane): Exhibits weaker dipole-dipole interactions compared to $C_2H_5Cl$.
Therefore,the increasing order of boiling points is: $(III) < (IV) < (II) < (I)$.

(D) $III$ gives immediate turbidity with Lucas reagent $(Anhy. ZnCl_2 + Conc. HCl)$,which indicates that $III$ is a tertiary $(3^\circ)$ alcohol.
$A$ $3^\circ$ alcohol is formed when a ketone reacts with a Grignard reagent $(CH_3MgBr)$. Therefore,$II$ must be a ketone.
Since $II$ is formed from $I (C_3H_6Cl_2)$ by reaction with aqueous $KOH$,$I$ must be a geminal dihalide. Thus,$I$ is $2,2$-dichloropropane $(CH_3-C(Cl)_2-CH_3)$.
The reaction sequence is:
$CH_3-C(Cl)_2-CH_3 (I)$ $\xrightarrow{KOH_{(aq)}} [CH_3-C(OH)_2-CH_3]$ $\xrightarrow{-H_2O} CH_3-CO-CH_3 (II)$
$CH_3-CO-CH_3 (II) \xrightarrow{(i) CH_3MgBr, (ii) H_2O/H^{+}} (CH_3)_3C-OH (III)$
$(CH_3)_3C-OH (III) \xrightarrow{Lucas\,Reagent} \text{Immediate turbidity}$.

(A) The basic strength of amines depends on the availability of the lone pair of electrons on the nitrogen atom for protonation.
$(II)$ Pyrrole: The lone pair on nitrogen is involved in the aromatic sextet,making it non-basic.
$(I)$ Aniline: The lone pair on nitrogen is delocalized into the benzene ring due to resonance,reducing its availability for protonation.
$(IV)$ Cyclohexylamine: This is a primary aliphatic amine. The lone pair is localized on the nitrogen atom,making it more basic than aniline.
$(III)$ $N$-Methylpiperidine: This is a tertiary aliphatic amine. It is more basic than primary aliphatic amines due to the electron-donating inductive effect $(+I)$ of the alkyl groups,which increases the electron density on the nitrogen atom.
Therefore,the increasing order of basic strength is: $II < I < IV < III$.

(D) The stability of an oxidation state can be inferred from the standard electrode potential $(E^0)$. $A$ more negative $E^0$ value for the reduction half-reaction $(M^n+ + ne^- \rightarrow M)$ indicates that the metal is more easily oxidized,meaning the higher oxidation state is more stable relative to the metal.
For $Al$: $E^0(Al^{3+}/Al) = -1.66 \ V$ and $E^0(Al^{+}/Al) = +0.55 \ V$. Since $E^0(Al^{3+}/Al) < E^0(Al^{+}/Al)$,$Al^{3+}$ is much more stable than $Al^{+}$.
For $Tl$: $E^0(Tl^{3+}/Tl) = +1.26 \ V$ and $E^0(Tl^{+}/Tl) = -0.34 \ V$. Since $E^0(Tl^{+}/Tl) < E^0(Tl^{3+}/Tl)$,$Tl^{+}$ is more stable than $Tl^{3+}$.
Comparing $Tl^{+}$ and $Al^{+}$: $Tl^{+}$ has a negative reduction potential $(-0.34 \ V)$,making it relatively stable,whereas $Al^{+}$ has a positive reduction potential $(+0.55 \ V)$,making it highly unstable and a strong oxidizing agent.
Therefore,$Tl^{+}$ is more stable than $Al^{+}$.