JEE Main 2019 Chemistry Question Paper with Answer and Solution

(C) The reaction proceeds in two steps:
$1$. Electrophilic addition of $Cl_2$ across the double bond in the presence of $CCl_4$ yields a vicinal dichloride: $CH_3O-C_6H_4-CH_2-CH_2-CHCl-CH_2Cl$.
$2$. In the presence of anhydrous $AlCl_3$ (a Lewis acid),one of the chlorine atoms is abstracted to form a carbocation. The electron-rich benzene ring (activated by the $-OCH_3$ group) then undergoes intramolecular Friedel-Crafts alkylation.
$3$. The cyclization occurs such that the more stable carbocation intermediate leads to the formation of a six-membered ring,resulting in $6-methoxy-2-chlorotetralin$ (or a related isomer depending on the specific cyclization pathway). Based on the provided options,the product is $C$.

(B) The electronic configuration of the element with $Z = 120$ is $[Og]_{118} \, 8s^2$,where $Og$ is Oganesson.
Since the valence shell configuration is $ns^2$,it belongs to Group $2$ of the periodic table.
Elements in Group $2$ are known as alkaline earth metals.

(A) Let the partial pressure of $B_{(g)}$ be $P_1$ and $E_{(g)}$ be $P_2$.
From the first equilibrium: $A_{(s)} \rightleftharpoons B_{(g)} + C_{(g)}$,the partial pressure of $C_{(g)}$ contributed by this reaction is $P_1$.
From the second equilibrium: $D_{(s)} \rightleftharpoons C_{(g)} + E_{(g)}$,the partial pressure of $C_{(g)}$ contributed by this reaction is $P_2$.
Total partial pressure of $C_{(g)} = P_1 + P_2$.
$K_{p_1} = P_B \cdot P_C = P_1(P_1 + P_2) = x$
$K_{p_2} = P_C \cdot P_E = (P_1 + P_2)P_2 = y$
Adding both equations:
$x + y = P_1(P_1 + P_2) + P_2(P_1 + P_2) = (P_1 + P_2)(P_1 + P_2) = (P_1 + P_2)^2$
So,$(P_1 + P_2) = \sqrt {x + y}$.
Total pressure $P_{total} = P_B + P_C + P_E = P_1 + (P_1 + P_2) + P_2 = 2(P_1 + P_2)$.
Substituting the value: $P_{total} = 2 \sqrt {x + y} \, atm$.

(C) The energy of the incident photon is given by $E = \frac{hc}{\lambda}$.
$E = \frac{6.626 \times 10^{-34} \times 3 \times 10^8}{4000 \times 10^{-10}} = 4.9695 \times 10^{-19} \, J$.
The kinetic energy of the emitted photoelectron is $K.E. = \frac{1}{2}mv^2$.
$K.E. = \frac{1}{2} \times 9 \times 10^{-31} \times (6 \times 10^5)^2 = 1.62 \times 10^{-19} \, J$.
According to Einstein's photoelectric equation,$E = \phi + K.E.$,where $\phi$ is the work function.
$\phi = E - K.E. = 4.9695 \times 10^{-19} - 1.62 \times 10^{-19} = 3.3495 \times 10^{-19} \, J$.
To convert the work function into $eV$,divide by the charge of an electron $(1.6 \times 10^{-19} \, C)$:
$\phi = \frac{3.3495 \times 10^{-19}}{1.6 \times 10^{-19}} \approx 2.093 \, eV \approx 2.1 \, eV$.

(A) The chemical reaction between iodine and concentrated nitric acid is given by:
$I_2 + 10HNO_3 \to 2HIO_3 + 10NO_2 + 4H_2O$.
Here,the product $Y$ is $HIO_3$ (iodic acid).
To find the oxidation state of iodine $(I)$ in $HIO_3$:
Let the oxidation state of $I$ be $x$.
$1 + x + 3(-2) = 0$
$1 + x - 6 = 0$
$x - 5 = 0$
$x = +5$.
Thus,the oxidation state of iodine in $Y$ is $+5$.

(C) The $BOD$ (Biochemical Oxygen Demand) value is a measure of the amount of dissolved oxygen required by bacteria to decompose organic matter in water.
Clean water typically has a $BOD$ value of less than $5 \ ppm$.
Highly polluted water typically has a $BOD$ value of $17 \ ppm$ or more.
Therefore,a water sample with $4 \ ppm$ is considered clean,and a water sample with $18 \ ppm$ is considered highly polluted.

(C) Given the equation $\tan^{-1}(2x) + \tan^{-1}(3x) = \frac{\pi}{4}$.
Using the formula $\tan^{-1}(A) + \tan^{-1}(B) = \tan^{-1}\left(\frac{A+B}{1-AB}\right)$,we have:
$\tan^{-1}\left(\frac{2x+3x}{1-(2x)(3x)}\right) = \frac{\pi}{4}$.
Taking tangent on both sides:
$\frac{5x}{1-6x^2} = \tan\left(\frac{\pi}{4}\right) = 1$.
This simplifies to $5x = 1 - 6x^2$,or $6x^2 + 5x - 1 = 0$.
Factoring the quadratic equation:
$(6x - 1)(x + 1) = 0$.
This gives $x = \frac{1}{6}$ or $x = -1$.
Since the set $A$ is defined for $x \geq 0$,we reject $x = -1$.
Thus,$x = \frac{1}{6}$ is the only solution.
Therefore,the set $A = \{\frac{1}{6}\}$ is a singleton set.

(A) Given $(2x)^{2y} = 4e^{2x - 2y}$.
Taking natural logarithm on both sides:
$2y \ln(2x) = \ln(4) + 2x - 2y$
$2y \ln(2x) + 2y = 2x + 2 \ln 2$
$y(1 + \ln 2x) = x + \ln 2$
$y = \frac{x + \ln 2}{1 + \ln 2x} \quad \dots(1)$
Now,differentiate both sides with respect to $x$:
$\frac{dy}{dx} = \frac{(1 + \ln 2x)(1) - (x + \ln 2)(\frac{1}{x})}{(1 + \ln 2x)^2}$
$\frac{dy}{dx} = \frac{1 + \ln 2x - 1 - \frac{\ln 2}{x}}{(1 + \ln 2x)^2}$
$\frac{dy}{dx} = \frac{\ln 2x - \frac{\ln 2}{x}}{(1 + \ln 2x)^2}$
Multiply both sides by $(1 + \ln 2x)^2$:
$(1 + \ln 2x)^2 \frac{dy}{dx} = \ln 2x - \frac{\ln 2}{x}$
$(1 + \ln 2x)^2 \frac{dy}{dx} = \frac{x \ln 2x - \ln 2}{x}$

(D) Let the $50$ observations be $x_1, x_2, ..., x_{50}$.
Given that the sum of deviations from $30$ is $50$,we have:
$\sum_{i=1}^{50} (x_i - 30) = 50$
Expanding the summation:
$\sum_{i=1}^{50} x_i - \sum_{i=1}^{50} 30 = 50$
$\sum_{i=1}^{50} x_i - (50 \times 30) = 50$
$\sum_{i=1}^{50} x_i - 1500 = 50$
$\sum_{i=1}^{50} x_i = 1550$
The mean $\bar{x}$ is given by $\frac{\sum x_i}{n}$:
$\bar{x} = \frac{1550}{50} = 31$
Therefore,the mean of the observations is $31$.

(C) Let $I = \int_{0}^{a} f(x)g(x) dx$.
Using the property $\int_{0}^{a} h(x) dx = \int_{0}^{a} h(a-x) dx$,we have:
$I = \int_{0}^{a} f(a-x)g(a-x) dx$.
Since $f(x) = f(a-x)$ and $g(a-x) = 4 - g(x)$,we substitute these into the integral:
$I = \int_{0}^{a} f(x)(4 - g(x)) dx$.
$I = 4\int_{0}^{a} f(x) dx - \int_{0}^{a} f(x)g(x) dx$.
$I = 4\int_{0}^{a} f(x) dx - I$.
$2I = 4\int_{0}^{a} f(x) dx$.
$I = 2\int_{0}^{a} f(x) dx$.

(D) Catenation is the ability of an element to form bonds with its own atoms to form long chains or rings.
In the carbon family $(Group \ 14)$,the tendency for catenation decreases down the group due to the decrease in bond dissociation energy of the $M-M$ bond as the atomic size increases.
$Pb$ (lead) does not show catenation because the $Pb-Pb$ bond energy is extremely low,making such bonds unstable.

(A) For an ideal gas undergoing isothermal expansion,the temperature $T$ is constant.
$(a)$ According to Boyle's Law,$P = nRT / V_m$. Thus,$P$ vs $1/V_m$ is a straight line passing through the origin. This represents isothermal expansion.
$(b)$ $P$ vs $V_m$ should be a rectangular hyperbola $(P \propto 1/V_m)$,not a straight line. This plot is incorrect.
$(c)$ For an ideal gas,$PV_m = RT$. Since $T$ is constant,$PV_m$ is constant regardless of $P$. This plot is correct.
$(d)$ For an ideal gas,internal energy $U$ depends only on temperature $(U = f(T))$. Since the process is isothermal,$U$ must be constant. The plot shows $U$ increasing with $V_m$,which is incorrect.
Therefore,plots $(b)$ and $(d)$ do not represent isothermal expansion.

(B) The ability to form $p\pi - p\pi$ multiple bonds depends on the atomic size of the element.
Smaller atoms have effective side-on overlap of their $p$-orbitals.
Among the given elements $(C, Si, Ge, Sn)$,carbon $(C)$ has the smallest atomic size,which allows for the most effective $p\pi - p\pi$ overlap.

(A) According to Hess's Law of constant heat summation,the enthalpy change of a reaction is the same whether it occurs in one step or several steps.
Adding equations $(ii)$ and $(iii)$:
$[C(\text{graphite}) + \frac{1}{2} O_{2(g)}] + [CO_{(g)} + \frac{1}{2} O_{2(g)}] \to CO_{(g)} + CO_{2(g)}$
Canceling $CO_{(g)}$ from both sides and combining $\frac{1}{2} O_{2(g)}$ terms:
$C(\text{graphite}) + O_{2(g)} \to CO_{2(g)}$
This is equation $(i)$.
Therefore,the enthalpy changes follow the same relationship:
$x = y + z$.

(C) The relationship between volume strength and molarity $(M)$ of $H_2O_2$ is given by the formula: $\text{Volume strength} = 11.35 \times M$.
Given $M = 1 \, M$.
Therefore,$\text{Volume strength} = 11.35 \times 1 = 11.35$.

(C) $I.$ Potassium ions $(K^{+})$ are known to activate many enzymes,such as pyruvate kinase.
$II.$ Potassium ions participate in the oxidation of glucose to produce $ATP$ by maintaining the electrochemical gradient necessary for cellular processes.
$III.$ Potassium ions,along with sodium ions $(Na^{+})$,are essential for the transmission of nerve signals by maintaining the resting membrane potential.
Therefore,all three statements are correct.

(D) Photochemical smog is formed by the action of sunlight on unsaturated hydrocarbons and nitrogen oxides.
Its common components include ozone $(O_3)$,nitric oxide $(NO)$,acrolein $(CH_2=CHCHO)$,formaldehyde $(HCHO)$,and peroxyacetyl nitrate ($PAN$,$CH_3-C(=O)OONO_2$).
$CF_2Cl_2$ (Freon-$12$) is a chlorofluorocarbon $(CFC)$ which is responsible for ozone layer depletion in the stratosphere,but it is not a component of photochemical smog.

(C) For an open vessel,the pressure $P$ and volume $V$ remain constant. According to the ideal gas law,$PV = nRT$,which implies $n_1T_1 = n_2T_2$.
Let the initial number of moles be $n_1 = n$ and the initial temperature $T_1 = 27 + 273 = 300\,K$.
Since two-fifths of the air escapes,the remaining moles are $n_2 = n - \frac{2}{5}n = \frac{3}{5}n$.
Substituting these values into the equation: $n \times 300 = \frac{3}{5}n \times T_2$.
Solving for $T_2$: $T_2 = 300 \times \frac{5}{3} = 500\,K$.

(D) The reaction of a vicinal dihalide with alcoholic $KOH$ followed by $NaNH_2$ results in double dehydrohalogenation to form an alkyne.
Step $1$: $CH_3-CH_2-CH(Br)-CH_2-Br \xrightarrow{KOH(alc)} CH_3-CH_2-CH=CHBr$ (Vinyl halide).
Step $2$: $CH_3-CH_2-CH=CHBr \xrightarrow{NaNH_2/liq. NH_3} CH_3-CH_2-C \equiv CH$ (But$-1-$yne).
The major product is but$-1-$yne.

(C) The dissociation of $Ag_2CO_3$ is given by: $Ag_2CO_3(s) \rightleftharpoons 2Ag^+(aq) + CO_3^{2-}(aq)$.
Let the solubility of $Ag_2CO_3$ in $0.1 \ M \ AgNO_3$ be $S' \ M$.
The concentration of $Ag^+$ ions from $AgNO_3$ is $0.1 \ M$,and from $Ag_2CO_3$ is $2S'$.
Total $[Ag^+] = (0.1 + 2S') \approx 0.1 \ M$ (since $S'$ is very small).
$[CO_3^{2-}] = S'$.
The solubility product expression is $K_{sp} = [Ag^+]^2 [CO_3^{2-}]$.
Substituting the values: $8 \times 10^{-12} = (0.1)^2 \times S'$.
$S' = \frac{8 \times 10^{-12}}{0.01} = 8 \times 10^{-10} \ M$.

(A) The ozone layer in the upper stratosphere absorbs harmful ultraviolet $(UV)$ radiation from the sun.
This radiation typically falls in the wavelength range of $200-315\, nm$ (often cited as $200-340\, nm$ depending on the specific $UV$ band classification).
Among the given options,$200-315\, nm$ is the correct range for the radiation filtered by the ozone layer.

(D) According to the Bohr model,the circumference of the $n^{th}$ orbit is given by $2 \pi r_n = n \lambda$,where $\lambda$ is the de Broglie wavelength.
Substituting the expression for the radius of the $n^{th}$ orbit,$r_n = a_0 \frac{n^2}{Z}$,we get:
$2 \pi \left( a_0 \frac{n^2}{Z} \right) = n \lambda$
$\lambda = \frac{2 \pi a_0 n^2}{n Z} = 2 \pi a_0 \frac{n}{Z}$
Given that $\lambda = 1.5 \pi a_0$,we equate the two expressions:
$2 \pi a_0 \frac{n}{Z} = 1.5 \pi a_0$
$\frac{n}{Z} = \frac{1.5 \pi a_0}{2 \pi a_0} = \frac{1.5}{2} = 0.75$

(B) We have the integral $I = \int \frac{3x^{13} + 2x^{11}}{(2x^4 + 3x^2 + 1)^4} dx$.
Divide the numerator and denominator by $x^{16}$ inside the integral:
$I = \int \frac{\frac{3x^{13} + 2x^{11}}{x^{16}}}{(\frac{2x^4 + 3x^2 + 1}{x^4})^4} dx = \int \frac{3x^{-3} + 2x^{-5}}{(2 + 3x^{-2} + x^{-4})^4} dx$.
Let $t = 2 + 3x^{-2} + x^{-4}$.
Then $dt = (-6x^{-3} - 4x^{-5}) dx = -2(3x^{-3} + 2x^{-5}) dx$.
So,$(3x^{-3} + 2x^{-5}) dx = -\frac{1}{2} dt$.
Substituting these into the integral:
$I = \int \frac{-\frac{1}{2} dt}{t^4} = -\frac{1}{2} \int t^{-4} dt = -\frac{1}{2} \left( \frac{t^{-3}}{-3} \right) + C = \frac{1}{6t^3} + C$.
Substituting $t$ back:
$I = \frac{1}{6(2 + 3x^{-2} + x^{-4})^3} + C = \frac{1}{6(\frac{2x^4 + 3x^2 + 1}{x^4})^3} + C = \frac{x^{12}}{6(2x^4 + 3x^2 + 1)^3} + C$.

(D) For isoelectronic species,the number of electrons is the same,so the size is determined by the effective nuclear charge $(Z)$.
As the nuclear charge $(Z)$ increases,the force of attraction between the nucleus and the electrons increases,which pulls the electron cloud closer to the nucleus.
Therefore,the size is inversely proportional to the nuclear charge: $\text{Size} \propto \frac{1}{Z}$.

(B) The reaction of diborane $(B_2H_6)$ with oxygen $(O_2)$ is highly exothermic and produces boron trioxide $(B_2O_3)$ and water:
$B_2H_6 + 3O_2 \longrightarrow B_2O_3 + 3H_2O$
The reaction of diborane $(B_2H_6)$ with water $(H_2O)$ produces boric acid $(H_3BO_3)$ and hydrogen gas $(H_2)$:
$B_2H_6 + 6H_2O \longrightarrow 2H_3BO_3 + 6H_2$
Therefore,the products are $B_2O_3$ and $H_3BO_3$ respectively.

(A) The hardness of water is expressed in terms of $CaCO_3$ equivalents.
$n_{eq}(CaCO_3) = n_{eq}(Ca(HCO_3)_2) + n_{eq}(Mg(HCO_3)_2)$
Given that the n-factor for bicarbonate salts is $2$:
$\frac{w}{100} \times 2 = \frac{0.81}{162} \times 2 + \frac{0.73}{146} \times 2$
$\frac{w}{50} = 0.005 \times 2 + 0.005 \times 2$
$\frac{w}{50} = 0.01 + 0.01 = 0.02$
$w = 0.02 \times 50 = 1.0 \, g$
Hardness in $ppm = \frac{\text{Mass of } CaCO_3 \text{ in } g}{\text{Mass of water in } g} \times 10^6$
Since $100 \, mL$ of water has a mass of $100 \, g$:
$\text{Hardness} = \frac{1.0}{100} \times 10^6 = 10,000 \, ppm$

(B) Using plastic bags is harmful to the environment because plastics are non-biodegradable and cause pollution. Therefore,it is against our responsibility to protect the environment.

(B) The change in enthalpy $(\Delta H)$ for a process at constant pressure is given by the integral of $n \, C_P \, dT$.
Given $n = 3 \, mol$,$C_P = 23 + 0.01 \, T$,$T_1 = 300 \, K$,and $T_2 = 1000 \, K$.
$\Delta H = n \int_{T_1}^{T_2} C_P \, dT = 3 \int_{300}^{1000} (23 + 0.01 \, T) \, dT$.
$\Delta H = 3 [23 \, T + \frac{0.01 \, T^2}{2}]_{300}^{1000}$.
$\Delta H = 3 [23(1000 - 300) + 0.005(1000^2 - 300^2)]$.
$\Delta H = 3 [23(700) + 0.005(1000000 - 90000)]$.
$\Delta H = 3 [16100 + 0.005(910000)]$.
$\Delta H = 3 [16100 + 4550] = 3 [20650] = 61950 \, J$.
Converting to $kJ$,$\Delta H = 61.95 \, kJ \approx 62 \, kJ$.

(D) Hydration enthalpy depends upon the ionic potential (charge/size ratio). As the ionic size increases down the group,the ionic potential decreases,and consequently,the hydration enthalpy decreases. The order of ionic size is $Li^{+} < Na^{+} < K^{+} < Rb^{+} < Cs^{+}$. Therefore,the order of hydration enthalpy is $Li^{+} > Na^{+} > K^{+} > Rb^{+} > Cs^{+}$. The relationship is given by $\Delta_{hyd}H^{\circ} \propto \frac{q}{r}$.

(C) The energy of an orbital is determined by the $(n + l)$ rule.
For $I: n = 4, l = 2, (n + l) = 4 + 2 = 6$ ($4d$ orbital).
For $II: n = 3, l = 2, (n + l) = 3 + 2 = 5$ ($3d$ orbital).
For $III: n = 4, l = 1, (n + l) = 4 + 1 = 5$ ($4p$ orbital).
For $IV: n = 3, l = 1, (n + l) = 3 + 1 = 4$ ($3p$ orbital).
According to the $(n + l)$ rule,lower $(n + l)$ value means lower energy. If $(n + l)$ values are equal,the orbital with lower $n$ has lower energy.
Comparing the values: $IV (4) < II (5, n=3) < III (5, n=4) < I (6)$.
Thus,the increasing order of energy is $IV < II < III < I$.

(A) The dissociation of $Zr_3(PO_4)_4$ is given by: $Zr_3(PO_4)_{4(s)} \rightleftharpoons 3Zr^{4+}{(aq)} + 4PO_4^{3-}{(aq)}$.
Let the molar solubility be $S$. Then the concentrations of the ions are $[Zr^{4+}] = 3S$ and $[PO_4^{3-}] = 4S$.
The solubility product expression is $K_{SP} = [Zr^{4+}]^3 [PO_4^{3-}]^4$.
Substituting the values: $K_{SP} = (3S)^3 \cdot (4S)^4 = (27S^3) \cdot (256S^4) = 6912S^7$.
Solving for $S$: $S^7 = K_{SP} / 6912$,which gives $S = (K_{SP} / 6912)^{1/7}$.

(NONE) The first law of thermodynamics is given by $\Delta U = q + w$.
$1$. For an isochoric process,volume is constant,so $w = 0$,which implies $\Delta U = q$. This is correct.
$2$. For an adiabatic process,$q = 0$,which implies $\Delta U = w$. This is correct.
$3$. For a cyclic process,the internal energy change $\Delta U = 0$,which implies $q = -w$. This is correct.
$4$. For an isothermal process of an ideal gas,$\Delta U = 0$ (since internal energy depends only on temperature),which implies $q = -w$. This is also correct.
Note: All given options are actually correct representations of the first law of thermodynamics for the specified processes. However,if the question implies a standard convention where $w = -P_{ext}\Delta V$,then all statements are valid.

(A) The principal functional group is $-COOH$ (carboxylic acid),which is assigned the lowest locant $(1)$.
Numbering the parent chain starting from the carboxylic acid carbon:
$C^5H_3-C^4H(CH_3)-C^3H(OH)-C^2H_2-C^1OOH$
There is a methyl group at position $4$ and a hydroxy group at position $3$.
According to $IUPAC$ nomenclature rules,substituents are listed in alphabetical order.
'Hydroxy' precedes 'methyl'.
Therefore,the correct name is $3$-hydroxy-$4$-methylpentanoic acid.

(C) The oxidation of the components in an acidic medium involves the following $n$-factors:
$FeSO_4 \rightarrow Fe^{3+} + SO_4^{2-} + e^-$,so $n$-factor $= 1$.
$FeC_2O_4 \rightarrow Fe^{3+} + 2CO_2 + 3e^-$,so $n$-factor $= 3$.
$Fe_2(C_2O_4)_3 \rightarrow 2Fe^{3+} + 6CO_2 + 6e^-$,so $n$-factor $= 6$.
$Fe_2(SO_4)_3$ is already in the highest oxidation state $(Fe^{3+})$,so it does not react.
For $KMnO_4$ in acidic medium,the $n$-factor is $5$.
Using the principle of equivalence: $n_{eq}(KMnO_4) = n_{eq}(FeC_2O_4) + n_{eq}(Fe_2(C_2O_4)_3) + n_{eq}(FeSO_4)$.
Let $x$ be the moles of $KMnO_4$: $x \times 5 = (1 \times 3) + (1 \times 6) + (1 \times 1)$.
$5x = 10$,which gives $x = 2$ moles.

(C) The upper stratosphere contains the ozone layer $(O_3)$,which protects the Earth from harmful ultraviolet $(UV)$ radiation from the Sun. $CFCs$ (chlorofluorocarbons) released into the atmosphere reach the stratosphere,where they undergo photochemical decomposition to release chlorine radicals. These radicals catalyze the destruction of ozone,leading to the formation of ozone holes. Consequently,these ozone holes allow a higher intensity of $UV$ radiation to reach the Earth's surface. Therefore,both statements are correct,and the reason explains the consequence of the phenomenon described in the assertion.

(B) Let $L = \mathop {\lim }\limits_{x \to 0} \frac{{\sin^2 x}}{{\sqrt{2} - \sqrt{1 + \cos x}}}$.
Rationalizing the denominator:
$L = \mathop {\lim }\limits_{x \to 0} \frac{{\sin^2 x (\sqrt{2} + \sqrt{1 + \cos x})}}{{2 - (1 + \cos x)}}$
$L = \mathop {\lim }\limits_{x \to 0} \frac{{\sin^2 x (\sqrt{2} + \sqrt{1 + \cos x})}}{{1 - \cos x}}$
Using the identity $1 - \cos x = 2\sin^2(x/2)$ and $\sin x = 2\sin(x/2)\cos(x/2)$:
$L = \mathop {\lim }\limits_{x \to 0} \frac{{(2\sin(x/2)\cos(x/2))^2 (\sqrt{2} + \sqrt{1 + \cos x})}}{{2\sin^2(x/2)}}$
$L = \mathop {\lim }\limits_{x \to 0} \frac{{4\sin^2(x/2)\cos^2(x/2) (\sqrt{2} + \sqrt{1 + \cos x})}}{{2\sin^2(x/2)}}$
$L = \mathop {\lim }\limits_{x \to 0} 2\cos^2(x/2) (\sqrt{2} + \sqrt{1 + \cos x})$
As $x \to 0$,$\cos(x/2) \to 1$ and $\cos x \to 1$:
$L = 2(1)^2 (\sqrt{2} + \sqrt{1 + 1}) = 2(\sqrt{2} + \sqrt{2}) = 2(2\sqrt{2}) = 4\sqrt{2}$.

(C) The reaction proceeds in two steps:
$1$. Treatment with $t-BuOK$ (a strong base) causes an intramolecular elimination reaction (specifically an $E2$ mechanism) to form an $\alpha,\beta$-unsaturated ketone. The base abstracts an $\alpha$-hydrogen,and the resulting enolate displaces the chloride ion to form a vinyl ketone derivative.
$2$. Treatment with concentrated $H_2SO_4$ and heat induces a Friedel-Crafts type cyclization. The acid protonates the carbonyl oxygen,increasing the electrophilicity of the $\beta$-carbon,which then undergoes an intramolecular electrophilic aromatic substitution (cyclization) onto the benzene ring.
Given the structure,the cyclization occurs at the ortho position relative to the alkyl chain,leading to the formation of a six-membered ring fused to the benzene ring,resulting in $7$-isopropyl-$3,4$-dihydronaphthalen-$2(1H)$-one (or a related isomer depending on the specific substitution pattern). Based on the provided options,the product is represented by option $C$.

(C) The kinetic energy $(K.E.)$ of the photoelectron is given by $K.E. = \frac{p^2}{2m}$.
According to the photoelectric equation,$\frac{hc}{\lambda} = W_o + K.E.$
Given that $K.E. \gg W_o$,we can approximate $\frac{hc}{\lambda} \approx K.E. = \frac{p^2}{2m}$.
For the first case,$\frac{hc}{\lambda} = \frac{p^2}{2m}$.
For the second case,let the new wavelength be $\lambda'$ and momentum be $p' = 1.5 \ p = \frac{3}{2} \ p$.
Then,$\frac{hc}{\lambda'} = \frac{(1.5 \ p)^2}{2m} = \frac{2.25 \ p^2}{2m}$.
Dividing the two equations: $\frac{\lambda}{\lambda'} = \frac{2.25 \ p^2 / 2m}{p^2 / 2m} = 2.25 = \frac{9}{4}$.
Therefore,$\lambda' = \frac{4}{9} \lambda$.

(B) To determine the hybridization,we use the formula: $H = \frac{1}{2}(V + M - C + A)$,where $V$ is the number of valence electrons of the central atom,$M$ is the number of monovalent atoms,$C$ is the cationic charge,and $A$ is the anionic charge.
For $[IF_6]^-$: $H = \frac{1}{2}(7 + 6 + 1) = 7$,which corresponds to $sp^3d^3$ hybridization.
For $[ICl_4]^-$: $H = \frac{1}{2}(7 + 4 + 1) = 6$,which corresponds to $sp^3d^2$ hybridization.
For $[ICl_2]^-$: $H = \frac{1}{2}(7 + 2 + 1) = 5$,which corresponds to $sp^3d$ hybridization.
For $[BrF_2]^-$: $H = \frac{1}{2}(7 + 2 + 1) = 5$,which corresponds to $sp^3d$ hybridization.
Therefore,the ion with $sp^3d^2$ hybridization is $[ICl_4]^-$,which corresponds to option $B$.

(B) The target reaction is: $2SO_{2(g)} + O_{2(g)} \rightleftharpoons 2SO_{3(g)}$
Given reactions:
$(i) \ S_{(s)} + O_{2(g)} \rightleftharpoons SO_{2(g)}; K_1 = 10^{52}$
$(ii) \ 2S_{(s)} + 3O_{2(g)} \rightleftharpoons 2SO_{3(g)}; K_2 = 10^{129}$
To obtain the target reaction,we perform the operation: $(ii) - 2 \times (i)$:
$2SO_{3(g)} - 2SO_{2(g)} = (2S + 3O_2) - 2(S + O_2) = O_{2(g)}$
Thus,the equilibrium constant $K_{eq}$ is given by:
$K_{eq} = \frac{K_2}{(K_1)^2}$
$K_{eq} = \frac{10^{129}}{(10^{52})^2} = \frac{10^{129}}{10^{104}} = 10^{25}$

(A) Beryllium $(Be)$ has a very small atomic size and high ionization enthalpy compared to other alkaline earth metals.
According to Fajans' rules,the small size and high charge density of the $Be^{2+}$ ion result in high polarizing power,which gives its halides $(BeX_2)$ a predominantly covalent character.
Other alkaline earth metal halides are primarily ionic.

(B) According to $IUPAC$ nomenclature for elements with atomic number $Z > 100$,the digits are represented as: $1 = \text{un}$,$1 = \text{un}$,$9 = \text{enn}$.
Therefore,the name is $\text{Ununennium}$.
The symbol is derived from the first letters of the roots: $U$ (for $1$),$u$ (for $1$),$e$ (for $9$).
Thus,the symbol is $Uue$.

(B) In $CH_4$,the number of moles of carbon atoms is $n_C = 1$.
The number of moles of hydrogen atoms is $n_H = 4$.
The total number of moles of atoms in $1 \text{ mole}$ of $CH_4$ is $n_C + n_H = 1 + 4 = 5$.
The percentage composition of carbon by mole is calculated as:
$\text{Percentage of } C = \frac{n_C}{n_C + n_H} \times 100 = \frac{1}{5} \times 100 = 20 \% $.

(B) The reaction of $HCl$ with an alkene typically follows Markovnikov's rule,where the proton adds to the carbon with more hydrogen atoms to form the more stable carbocation.
However,if an electron-withdrawing group $(EWG)$ like $-CF_3$ is attached to the double bond,it destabilizes the carbocation formed at the adjacent carbon.
In the case of $F_3C-CH=CH_2$,the formation of a carbocation at the carbon attached to the $-CF_3$ group $(F_3C-CH^+-CH_3)$ is highly unstable due to the strong inductive effect of the three fluorine atoms.
Therefore,the proton adds to the carbon attached to the $-CF_3$ group,leading to the formation of the more stable carbocation at the terminal carbon $(F_3C-CH_2-CH_2^+)$.
Subsequent attack by the chloride ion $(Cl^-)$ results in the anti-Markovnikov product,$F_3C-CH_2-CH_2Cl$.

(B) For an ideal gas,the change in internal energy is given by $\Delta U = n \times C_v \times \Delta T$.
Given $n = 5 \, mol$,$C_v = 28 \, J \, K^{-1} \, mol^{-1}$,and $\Delta T = 200 \, K - 100 \, K = 100 \, K$.
$\Delta U = 5 \times 28 \times 100 = 14000 \, J = 14 \, kJ$.
For an ideal gas,$\Delta (pV) = nR\Delta T$.
Given $R = 8.0 \, J \, K^{-1} \, mol^{-1}$.
$\Delta (pV) = 5 \times 8 \times 100 = 4000 \, J = 4 \, kJ$.

(C) The reaction involves the alkylation of adenine with $CH_3I$ in the presence of a base.
The base deprotonates the nitrogen atom at the $N-9$ position (marked with $*$ in the mechanism),which is the most acidic site in the adenine molecule.
This deprotonated nitrogen then acts as a nucleophile and attacks the methyl group of $CH_3I$ via an $S_N2$ mechanism,resulting in the formation of $9-methyladenine$ as the major product.

(A) The decomposition reaction of $H_2O_2$ is: $2H_2O_2 \to 2H_2O + O_2 \uparrow$
From the stoichiometry,$2 \times 34 \ g = 68 \ g$ of $H_2O_2$ produces $22.4 \ L$ of $O_2$ at $STP$.
Therefore,$11.2 \ L$ of $O_2$ is produced by $\frac{68 \ g}{22.4 \ L} \times 11.2 \ L = 34 \ g$ of $H_2O_2$.
This means $34 \ g$ of $H_2O_2$ is present in $1 \ L$ $(1000 \ mL)$ of the solution.
Strength in $g/L = 34 \ g/L$.
Percentage strength = $\frac{\text{Strength in } g/L}{10} = \frac{34}{10} = 3.4\%$.

$\textbf{(B) } C_2^{2-}: \sigma_{1s}^2, \sigma^*_{1s}{^2}, \sigma_{2s}^2, \sigma^*_{2s}{^2}, [\pi_{2p_x}^2 = \pi_{2p_y}^2], \sigma_{2p_z}^2. \text{ Bond Order (B.O.)} = \frac{10 - 4}{2} = 3 \text{ (diamagnetic)}.$
$N_2^{2-}: \sigma_{1s}^2, \sigma^*_{1s}{^2}, \sigma_{2s}^2, \sigma^*_{2s}{^2}, \sigma_{2p_z}^2, [\pi_{2p_x}^2 = \pi_{2p_y}^2], [\pi^*_{2p_x}{^1} = \pi^*_{2p_y}{^1}]. \text{ B.O.} = \frac{10 - 6}{2} = 2 \text{ (paramagnetic)}.$
$O_2^{2-}: \sigma_{1s}^2, \sigma^*_{1s}{^2}, \sigma_{2s}^2, \sigma^*_{2s}{^2}, \sigma_{2p_z}^2, [\pi_{2p_x}^2 = \pi_{2p_y}^2], [\pi^*_{2p_x}{^2} = \pi^*_{2p_y}{^2}]. \text{ B.O.} = \frac{10 - 8}{2} = 1 \text{ (diamagnetic)}.$
$O_2: \sigma_{1s}^2, \sigma^*_{1s}{^2}, \sigma_{2s}^2, \sigma^*_{2s}{^2}, \sigma_{2p_z}^2, [\pi_{2p_x}^2 = \pi_{2p_y}^2], [\pi^*_{2p_x}{^1} = \pi^*_{2p_y}{^1}]. \text{ B.O.} = \frac{10 - 6}{2} = 2 \text{ (paramagnetic)}.$
$\text{Since bond length is inversely proportional to bond order, } C_2^{2-} \text{ with B.O. } = 3 \text{ has the shortest bond length and is diamagnetic.}$

(B) The maximum prescribed concentration of copper in drinking water is $3 \ ppm$.
Above this concentration,water becomes toxic.

(B) For $ICl_5$: The central iodine atom has $7$ valence electrons. It forms $5$ bonds with $Cl$ atoms and has $1$ lone pair. The steric number is $5 + 1 = 6$,which corresponds to $sp^3d^2$ hybridization. Due to $5$ bond pairs and $1$ lone pair,the geometry is octahedral,but the shape is square pyramidal.
For $ICl_4^-$: The central iodine atom has $7$ valence electrons,plus $1$ from the negative charge,totaling $8$ electrons. It forms $4$ bonds with $Cl$ atoms and has $2$ lone pairs. The steric number is $4 + 2 = 6$,which corresponds to $sp^3d^2$ hybridization. Due to $4$ bond pairs and $2$ lone pairs,the geometry is octahedral,but the shape is square planar.

(C) For a zero-order reaction,the integrated rate law is given by $C_0 - C_t = Kt$.
First,calculate the rate constant $K$ using the half-life formula for a zero-order reaction: $t_{1/2} = \frac{C_0}{2K}$.
Given $t_{1/2} = 6 \ h$ and $C_0 = 0.2 \ M$,we have $6 = \frac{0.2}{2K}$,which gives $K = \frac{0.2}{12} = \frac{1}{60} \ M \ h^{-1}$.
Now,for the second condition where initial concentration $C_0 = 0.5 \ M$ and final concentration $C_t = 0.2 \ M$:
Using $C_0 - C_t = Kt$,we get $0.5 - 0.2 = Kt$.
$0.3 = Kt$.
Substituting $K = \frac{1}{60}$,we get $0.3 = \frac{1}{60} \times t$.
$t = 0.3 \times 60 = 18 \ h$.

(A) The reaction involves the electrophilic addition of $Br_2$ to the $C=C$ double bond in the presence of $MeOH$ as a nucleophilic solvent.
This is a classic bromomethoxylation reaction.
The $Br_2$ molecule forms a cyclic bromonium ion intermediate with the alkene.
Then,the nucleophilic solvent $MeOH$ attacks the more substituted carbon of the bromonium ion (following Markovnikov-like regioselectivity due to the stability of the developing positive charge),leading to the anti-addition of $Br$ and $OMe$ across the double bond.
Therefore,the product is the one where $Br$ and $OMe$ are added across the double bond,which corresponds to the structure in option $A$.

(C) The identification of amino acids based on functional group tests is as follows:
$1$. Ester test: Serine $(Ser)$ contains a hydroxyl $(-OH)$ group,which can undergo esterification. Thus,$A-R$.
$2$. Carbylamine test: Lysine $(Lys)$ contains a primary amine $(-NH_2)$ group,which gives a positive carbylamine test. Thus,$B-S$.
$3$. Phthalein dye test: Tyrosine $(Tyr)$ contains a phenolic group,which reacts with phthalic anhydride to form a phthalein dye. Thus,$C-P$.
Therefore,the correct matching is $A-R, B-S, C-P$.

(A) The structure of $Co_2(CO)_8$ contains two bridging $CO$ ligands that connect the two cobalt atoms.
Additionally,there is a direct metal-metal bond between the two cobalt atoms ($Co-Co$ bond).
Therefore,the number of bridging $CO$ ligands is $2$ and the number of $Co-Co$ bonds is $1$.

(B) The given reactions represent the chemistry of manganese compounds:
$1$. $2 MnO_2 + 4 KOH + O_2 \rightarrow 2 K_2MnO_4 (Green) + 2 H_2O$. Thus,$A = MnO_2$ and $B = K_2MnO_4$.
$2$. $3 K_2MnO_4 + 4 HCl \rightarrow 2 KMnO_4 (Purple) + MnO_2 + 2 H_2O$. Thus,$C = KMnO_4$.
$3$. $2 KMnO_4 + H_2O + KI \rightarrow 2 MnO_2 + 2 KOH + KIO_3$. Thus,$D = KIO_3$.
Comparing these,$A = MnO_2$ and $D = KIO_3$.

(B) In the $Hall-Heroult$ process,the electrolytic cell consists of a steel vessel lined with carbon,which acts as the cathode.
Therefore,the cathode is made out of carbon.

(A) The extent of adsorption of a gas on a solid surface is directly related to its ease of liquefaction.
Easier liquefaction occurs for gases with higher critical temperatures.
Therefore,the smaller the value of the critical temperature of a gas,the lesser is the extent of its adsorption.
Comparing the given values,the critical temperature of $H_2$ $(33 \ K)$ is the lowest among all the gases listed.
Thus,$H_2$ is the least adsorbed gas.

(A) An organometallic compound is defined as a compound containing at least one direct chemical bond between a metal atom and a carbon atom of an organic group or molecule.
In $Mn_2(CO)_{10}$,the manganese $(Mn)$ atoms are directly bonded to the carbon $(C)$ atoms of the carbonyl $(CO)$ ligands.
Therefore,the presence of the $Mn-C$ bond makes it an organometallic compound.

(A) The spin-only magnetic moment is given by the formula $\mu = \sqrt{n(n+2)} \, BM$,where $n$ is the number of unpaired electrons.
For $\mu = 3.9 \, BM$,we have $\sqrt{n(n+2)} \approx 3.9$,which implies $n = 3$.
Thus,the metal ion $M^{2+}$ must have $3$ unpaired electrons.
Electronic configurations in the presence of the weak field ligand $H_2O$:
$V^{2+} (d^3): t_{2g}^3 e_g^0$ (unpaired electrons $n = 3$)
$Co^{2+} (d^7): t_{2g}^5 e_g^2$ (unpaired electrons $n = 3$)
$Fe^{2+} (d^6): t_{2g}^4 e_g^2$ (unpaired electrons $n = 4$)
$Cr^{2+} (d^4): t_{2g}^3 e_g^1$ (unpaired electrons $n = 4$)
$Mn^{2+} (d^5): t_{2g}^3 e_g^2$ (unpaired electrons $n = 5$)
Therefore,the pair of metal ions with $3$ unpaired electrons is $V^{2+}$ and $Co^{2+}$.

(B) The basicity of an amino acid is determined by the number of amino groups relative to the number of carboxylic acid groups in its side chain.
Lysine contains two amino groups and one carboxylic acid group,making it a basic amino acid.
Histidine is also basic,but Lysine is more basic due to the presence of an aliphatic amino group in its side chain.
Comparing their $pI$ values:

Compound	$pI$ value
$A$. Histidine	$7.6$
$B$. Serine	$5.7$
$C$. Lysine	$9.8$
$D$. Asparagine	$5.4$

$A$ higher $pI$ value indicates a more basic character. Therefore,Lysine is the most basic among the given options.

(A) The reaction of these compounds with alkyl halide is a nucleophilic substitution reaction where the nitrogen atom acts as the nucleophile.
Reactivity depends on the availability of the lone pair on the nitrogen atom.
$(a)$ Benzamide: The lone pair on nitrogen is involved in resonance with the carbonyl group $(C=O)$,making it very weakly nucleophilic.
$(b)$ Phthalimide: The lone pair on nitrogen is involved in resonance with two carbonyl groups,making it the least nucleophilic.
$(c)$ $2$-Cyanoaniline: The $-CN$ group is electron-withdrawing by both $-I$ and $-M$ effects,which reduces the electron density on the nitrogen atom.
$(d)$ Aniline: The lone pair on nitrogen is involved in resonance with the benzene ring,but it is more nucleophilic than the others because it lacks the strong electron-withdrawing carbonyl groups present in $(a)$ and $(b)$ or the strong $-I$ effect of the $-CN$ group in $(c)$.
Thus,the order of nucleophilicity (and reactivity) is $b < a < c < d$.

(B) The freezing point of a substance is related to its ability to form a stable crystal lattice,which is influenced by intermolecular forces and molecular symmetry.
Phthalic acid $(C_8H_6O_4)$ has strong intermolecular hydrogen bonding.
$2$-Naphthol $(C_{10}H_8O)$ also exhibits hydrogen bonding.
Naphthalene $(C_{10}H_8)$ is a highly symmetric planar molecule that packs well in a crystal lattice.
$9$,$10$-Dimethylanthracene is a bulky,non-polar molecule with significant steric hindrance due to the methyl groups at the $9$ and $10$ positions,which disrupts efficient crystal packing,leading to a lower freezing point compared to the others.

(A) The target molecule is $2\text{-phenylbutan-2-ol}$,which is a tertiary $(3^o)$ alcohol.
Option $A$: $HCHO + PhCH(CH_3)CH_2MgX$ gives a primary $(1^o)$ alcohol,$PhCH(CH_3)CH_2CH_2OH$.
Option $B$: $PhCOCH_2CH_3 + CH_3MgX$ gives $CH_3-CH_2-C(Ph)(OH)-CH_3$.
Option $C$: $PhCOCH_3 + CH_3CH_2MgX$ gives $CH_3-CH_2-C(Ph)(OH)-CH_3$.
Option $D$: $CH_3CH_2COCH_3 + PhMgX$ gives $CH_3-CH_2-C(Ph)(OH)-CH_3$.
Thus,option $A$ cannot prepare the target molecule.

(A) The depression in freezing point is given by $\Delta T_f = K_f \times m$,where $m$ is the molality of the solution.
Since the freezing points are equal,$(\Delta T_f)_X = (\Delta T_f)_Y$,which implies $m_X = m_Y$.
For a $4 \%$ aqueous solution of $X$,the mass of $X$ is $4 \ g$ in $96 \ g$ of water. Molality $m_X = \frac{4 \times 1000}{A \times 96}$.
For a $12 \%$ aqueous solution of $Y$,the mass of $Y$ is $12 \ g$ in $88 \ g$ of water. Molality $m_Y = \frac{12 \times 1000}{M_Y \times 88}$.
Equating the two: $\frac{4}{A \times 96} = \frac{12}{M_Y \times 88}$.
Solving for $M_Y$: $M_Y = \frac{12 \times 96 \times A}{4 \times 88} = \frac{3 \times 96 \times A}{88} = \frac{288 \times A}{88} \approx 3.27 \times A$.
Rounding to the nearest integer,the molecular weight of $Y$ is $3A$.

(B) $K_3[Co(CN)_6]$ is an octahedral coordination complex.
In an octahedral geometry,the ligands approach the central metal ion along the $x, y,$ and $z$ axes.
The $d_{x^2 - y^2}$ and $d_{z^2}$ orbitals (collectively known as $e_g$ orbitals) have their electron density lobes directed along the axes.
Therefore,these orbitals directly face the incoming ligands,leading to greater electrostatic repulsion and higher energy compared to the $t_{2g}$ orbitals $(d_{xy}, d_{xz}, d_{yz})$,which have lobes directed between the axes.

(A) The unit of the rate constant is $\mu g/year$,which indicates that the reaction is of zero order.
For a zero order reaction,the integrated rate law is $[A] = [A]_0 - kt$.
Here,$[A]_0 = 5 \ \mu g$,$[A] = 2.5 \ \mu g$,and $k = 0.05 \ \mu g/year$.
Substituting these values: $2.5 \ \mu g = 5 \ \mu g - (0.05 \ \mu g/year) \times t$.
$0.05 \ \mu g/year \times t = 2.5 \ \mu g$.
$t = \frac{2.5 \ \mu g}{0.05 \ \mu g/year} = 50 \ years$.

(A) The number of electrons transferred in the reaction is $n = 2$.
Standard Gibbs free energy change is given by $\Delta_r G^{\ominus} = -nFE^{\ominus} = -2 \times 96500 \times 2 = -386000 \ J = -386 \ kJ$.
Standard entropy change is given by $\Delta_r S^{\ominus} = nF \left( \frac{dE^{\ominus}}{dT} \right) = 2 \times 96500 \times (-5 \times 10^{-4}) = -96.5 \ J \ K^{-1}$.
Using the relation $\Delta_r G^{\ominus} = \Delta_r H^{\ominus} - T\Delta_r S^{\ominus}$,we get $\Delta_r H^{\ominus} = \Delta_r G^{\ominus} + T\Delta_r S^{\ominus}$.
Substituting the values at $T = 300 \ K$:
$\Delta_r H^{\ominus} = -386 \ kJ + 300 \times (-96.5 \times 10^{-3} \ kJ \ K^{-1}) = -386 - 28.95 = -414.95 \ kJ$.
Rounding to the nearest provided option,the value is $-412.8 \ kJ$.

(B) The formation of acetal from an aldehyde and an alcohol is a nucleophilic addition reaction.
This reaction is highly sensitive to steric hindrance.
$Rate \propto \frac{1}{\text{Steric Crowding}}$.
$HCHO$ is the least sterically hindered aldehyde,and $MeOH$ is the least sterically hindered alcohol.
Therefore,the combination of $HCHO$ and $MeOH$ provides the fastest and most efficient reaction to form an acetal.

(B) The given reaction is an intramolecular Aldol condensation followed by dehydration.
$1$. The starting material is $6,6-\text{dimethylheptan-2-one}$ (or a similar keto-aldehyde structure). The base $(OH^-)$ abstracts an $\alpha$-hydrogen from the ketone group to form an enolate.
$2$. This enolate attacks the carbonyl carbon of the aldehyde group,leading to the formation of a cyclic $\beta$-hydroxy ketone,which is product $A$.
$3$. Upon heating with acid $(H_3O^+/\Delta)$,the $\beta$-hydroxy ketone undergoes dehydration (elimination of water) to form an $\alpha,\beta$-unsaturated ketone,which is product $B$.
$4$. Based on the structure,$A$ is $3-\text{hydroxy}-2,2-\text{dimethylcyclohexanone}$ and $B$ is $6,6-\text{dimethylcyclohex}-2-\text{enone}$.

(D) $PHBV$ is a biodegradable aliphatic polyester copolymer.
It is obtained by the copolymerization of $3$-hydroxybutanoic acid and $3$-hydroxypentanoic acid.

(C) Moles of $NaOH = \frac{8 \ g}{40 \ g \ mol^{-1}} = 0.2 \ mol$.
Moles of $H_2O = \frac{18 \ g}{18 \ g \ mol^{-1}} = 1 \ mol$.
Mole fraction of $NaOH = \frac{n_{NaOH}}{n_{NaOH} + n_{H_2O}} = \frac{0.2}{0.2 + 1} = \frac{0.2}{1.2} = 0.167$.
Molality $(m) = \frac{\text{moles of solute}}{\text{mass of solvent in kg}} = \frac{0.2 \ mol}{18 \times 10^{-3} \ kg} = \frac{0.2}{0.018} \approx 11.11 \ mol \ kg^{-1}$.

(C) The electronic configuration of $Mn^{2+}$ is $[Ar] 3d^5$.
For an octahedral complex,the magnetic moment is given by $\mu = \sqrt{n(n+2)} \ BM$,where $n$ is the number of unpaired electrons.
$A$ magnetic moment of $5.9 \ BM$ corresponds to $n = 5$ unpaired electrons.
This indicates that the complex is a high-spin complex,which occurs when the ligand is a weak field ligand.
Among the given options,$NCS^{-}$ is a weak field ligand,while $ethylenediamine$,$CN^{-}$,and $CO$ are strong field ligands that would cause pairing of electrons.
Therefore,the complex is $[Mn(NCS)_6]^{4-}$.

(D) The statement in option $D$ is false because lyophilic sols are inherently stable due to strong interaction between the dispersed phase and the dispersion medium. They cannot be easily coagulated by adding small amounts of an electrolyte. Coagulation of lyophilic sols requires the addition of a large amount of electrolyte or the addition of a suitable solvent (like alcohol or acetone) to dehydrate the particles. In contrast,lyophobic sols are unstable and are easily coagulated by adding small amounts of an electrolyte. Latex is a colloidal solution of rubber particles which are negatively charged,but since the question asks for the false statement,$D$ is the most scientifically inaccurate statement regarding the ease of coagulation.

(A) Histidine is an amino acid with three ionizable groups: the $\alpha$-amino group,the $\alpha$-carboxyl group,and the imidazole side chain.
In a strongly acidic solution $(pH = 2)$,the concentration of $H^{+}$ ions is very high.
At this $pH$,all basic groups will be protonated:
$1$. The carboxyl group exists as $-COOH$ (since $pH < pK_a$ of $-COO^-$).
$2$. The $\alpha$-amino group exists as $-NH_3^+$.
$3$. The imidazole side chain also gets protonated to exist as an imidazolium ion $(-NH^+)$.
Therefore,the structure will have a net charge of $+2$. This corresponds to the structure shown in option $A$.

(D) The reaction proceeds as follows:
$1$. Treatment of the primary amine with $NaNO_2/H^+$ leads to the formation of a diazonium salt,which is unstable and undergoes hydrolysis to form a primary alcohol: $Ar-(CH_2)_3-OH$.
$2$. Oxidation with $CrO_3/H^+$ (Jones reagent) oxidizes the primary alcohol to a carboxylic acid: $Ar-(CH_2)_2-COOH$.
$3$. Treatment with conc. $H_2SO_4$ and heat causes an intramolecular Friedel-Crafts acylation. During this process,the ester group (acetoxy) is hydrolyzed to a phenolic $-OH$ group under acidic conditions. The resulting carboxylic acid cyclizes onto the benzene ring to form a five-membered ketone ring,yielding $5-hydroxy-1-indanone$.

(C) According to Kohlrausch's law,$\wedge _m^o(HA) = \wedge _m^o(HCl) + \wedge _m^o(NaA) - \wedge _m^o(NaCl)$
$\wedge _m^o(HA) = 425.9 + 100.5 - 126.4 = 400 \ S \ cm^2 \ mol^{-1}$
Calculate molar conductivity $\wedge _m$ for $0.001 \ M \ HA$:
$\wedge _m = \frac{\kappa \times 1000}{M} = \frac{5 \times 10^{-5} \times 1000}{0.001} = 50 \ S \ cm^2 \ mol^{-1}$
Degree of dissociation $\alpha$ is given by:
$\alpha = \frac{\wedge _m}{\wedge _m^o} = \frac{50}{400} = 0.125$

(A) The dimerization reaction is: $2C_6H_5COOH \to (C_6H_5COOH)_2$
Degree of association $\alpha = 0.80$.
Van't Hoff factor $i = 1 - \alpha + \frac{\alpha}{n} = 1 - 0.8 + \frac{0.8}{2} = 0.2 + 0.4 = 0.6$.
Depression in freezing point formula: $\Delta T_f = i \times K_f \times m$,where $m$ is molality.
$m = \frac{w \times 1000}{M_{solute} \times W_{solvent(g)}} = \frac{w \times 1000}{122 \times 30}$.
Substituting the values: $2 = 0.6 \times 5 \times \frac{w \times 1000}{122 \times 30}$.
$2 = 3 \times \frac{w \times 100}{122 \times 3} = \frac{w \times 100}{122}$.
$w = \frac{2 \times 122}{100} = 2.44 \ g$.

(A) The reaction of chlorine with hot and concentrated sodium hydroxide is a disproportionation reaction.
$3Cl_2 + 6NaOH \to 5NaCl + NaClO_3 + 3H_2O$
In this reaction,chlorine is both oxidized and reduced.
The products formed are sodium chloride $(NaCl)$ and sodium chlorate $(NaClO_3)$.
Thus,the ionic species present are $Cl^{-}$ and $ClO_3^{-}$.

(B) The reaction is an $E2$ elimination reaction of a secondary alkyl chloride with a strong base $(NaOEt)$.
According to Saytzeff's rule,the major product is the more substituted alkene,as it is more stable.
In the given substrate,$CH_3-CH(Cl)-CH_2-CH_3$ (with a $CO_2Et$ group),the elimination of $HCl$ can occur from either the $CH_3$ group or the $CH_2$ group.
Elimination from the $CH_2$ group leads to a trisubstituted alkene $(CH_3-C(CO_2Et)=CH-CH_3)$,which is more stable than the disubstituted alkene $(CH_3-CH_2-C(CO_2Et)=CH_2)$ formed by elimination from the $CH_3$ group.
Therefore,the major product is $CH_3-C(CO_2CH_2CH_3)=CH-CH_3$.

(B) The reaction involves two steps when $HBr$ is in excess and heat is applied:
$1$. The ether group $(-OCH_3)$ undergoes cleavage by $HBr$ to form a phenol $(-OH)$ and methyl bromide $(CH_3Br)$.
$2$. The alkene group $(-CH=CH-CH_3)$ undergoes electrophilic addition of $HBr$ following Markownikoff's rule.
According to Markownikoff's rule,the electrophile $H^+$ adds to the carbon with more hydrogens,and the nucleophile $Br^-$ adds to the more substituted carbon,resulting in the formation of a stable carbocation intermediate. Thus,the product is $4-(1-bromopropyl)phenol$.

(B) $NaBH_4$ is a selective reducing agent that reduces carbonyl groups (aldehydes and ketones) to alcohols but does not reduce carbon-carbon double bonds $(C=C)$.
In the given reaction,the ketone group of cyclopent$-3-$en$-1-$one is reduced to an alcohol group,while the double bond remains unaffected.
Therefore,the major product is cyclopent$-3-$en$-1-$ol.

(A) The atomic radii of the given lanthanoids are as follows:
$Eu = 199 \text{ pm}$
$Ce = 183 \text{ pm}$
$Nd = 181 \text{ pm}$
$Ho = 176 \text{ pm}$
Therefore, the correct order of atomic radii is $Eu > Ce > Nd > Ho$.

(A) Nylon $6, 6$ is a polyamide formed by the condensation polymerization of two monomers:
$1$. Adipic acid: $HOOC(CH_2)_4COOH$
$2$. Hexamethylene diamine: $H_2N(CH_2)_6NH_2$
Therefore,the correct option is $A$.

(A) Grignard reagents are strong bases as well as nucleophiles. They react with acidic protons (like those in $-COOH$ or $-OH$ groups) before reacting with the carbonyl group.
$(a)$ Benzaldehyde has no acidic protons,so $1$ equivalent of Grignard reagent reacts with the aldehyde group.
$(b)$ $4$-Formylbenzoic acid contains an acidic $-COOH$ group. It consumes $1$ equivalent of Grignard reagent for acid-base reaction and another for the aldehyde,totaling $2$ equivalents.
$(c)$ $4$-Methoxybenzaldehyde has no acidic protons,so $1$ equivalent reacts with the aldehyde group.
$(d)$ $4$-(Hydroxymethyl)benzaldehyde contains an acidic $-OH$ group. It consumes $1$ equivalent for acid-base reaction and another for the aldehyde,totaling $2$ equivalents.
Thus,compounds $(b)$ and $(d)$ will not form a single Grignard product with only $1$ equivalent of reagent.

(C) From the Arrhenius equation,$\ln k = \ln A - \frac{E_a}{R} \left( \frac{1}{T} \right)$.
Comparing this with the equation of a straight line $y = mx + c$,the slope $m = -\frac{E_a}{R}$.
Given slope $= -4606 \ K$,so $\frac{E_a}{R} = 4606 \ K$.
Using the integrated Arrhenius equation: $\ln \left( \frac{k_2}{k_1} \right) = \frac{E_a}{R} \left( \frac{1}{T_1} - \frac{1}{T_2} \right)$.
Here,$T_1 = 400 \ K$,$k_1 = 10^{-5} \ s^{-1}$,$T_2 = 500 \ K$,and $k_2 = ?$.
$\ln \left( \frac{k_2}{10^{-5}} \right) = 4606 \times \left( \frac{1}{400} - \frac{1}{500} \right) = 4606 \times \left( \frac{500 - 400}{200000} \right) = 4606 \times \frac{100}{200000} = 4606 \times \frac{1}{2000} = 2.303$.
Since $\ln 10 \approx 2.303$,we have $\ln \left( \frac{k_2}{10^{-5}} \right) = \ln 10$.
Therefore,$\frac{k_2}{10^{-5}} = 10$,which gives $k_2 = 10^{-4} \ s^{-1}$.

(D) The reaction involves the electrophilic addition of $HCl$ to an alkene.
According to Markovnikov's rule,the electrophile $(H^+)$ adds to the carbon atom of the double bond that has more hydrogen atoms,resulting in the formation of the most stable carbocation intermediate.
In this case,the double bond is between a terminal $CH_2$ group and a tertiary carbon atom.
Addition of $H^+$ to the $CH_2$ group forms a stable tertiary carbocation at the adjacent carbon.
Subsequently,the nucleophile $(Cl^-)$ attacks this tertiary carbocation to form the major product,which is a tertiary alkyl chloride.

(A) The reactivity of carboxylic acid derivatives towards nucleophilic attack by $LiAlH_4$ depends on the leaving group ability and the electrophilicity of the carbonyl carbon.
The order of reactivity is determined by the stability of the leaving group and the resonance stabilization of the carbonyl group:
$1. \text{Acid chloride } (C_2H_5COCl) \text{ is the most reactive due to the excellent leaving group } Cl^- \text{ and strong electron-withdrawing effect.}$
$2. \text{Acid anhydride } ((C_2H_5CO)_2O) \text{ is next, with } RCOO^- \text{ as a good leaving group.}$
$3. \text{Ester } (C_2H_5COOCH_3) \text{ follows, with } CH_3O^- \text{ as a poorer leaving group.}$
$4. \text{Amide } (C_2H_5CONH_2) \text{ is the least reactive due to strong resonance stabilization of the carbonyl group by the nitrogen lone pair.}$
Thus,the increasing order of reactivity is: $(a) < (b) < (d) < (c)$.

(A) Calcination is a process of heating an ore in a limited supply of air to convert carbonates or hydrated oxides into their respective metal oxides.
$ZnO$ and $MgO$ are already present in their oxide forms,hence they do not require calcination.

(B) In a $ccp$ (cubic close-packed) structure,the number of atoms per unit cell is $Z = 4$.
$1$. Number of $B$ atoms $= 4$.
$2$. Number of octahedral voids $= 4$. Since '$A$' occupies half of the octahedral voids,the number of $A$ atoms $= 4 \times \frac{1}{2} = 2$.
$3$. Number of tetrahedral voids $= 2 \times Z = 2 \times 4 = 8$. Since oxygen atoms occupy all the tetrahedral voids,the number of oxygen atoms $= 8$.
The ratio of atoms $A:B:O$ is $2:4:8$,which simplifies to $1:2:4$.
Therefore,the formula of the bimetallic oxide is $AB_2O_4$.

(C) The coupling reaction of benzene diazonium chloride with $1$-naphthol in an alkaline medium is an electrophilic aromatic substitution reaction.
In $1$-naphthol,the $-OH$ group is a strong activating group.
The electrophile,benzene diazonium ion $(PhN_2^+)$,attacks the position ortho or para to the $-OH$ group.
Due to steric hindrance,the attack occurs preferentially at the $4$-position (para position) of the $1$-naphthol ring.
This results in the formation of $4$-phenylazo-$1$-naphthol,which is an orange-red dye.
The correct structure is shown in image $821-c1202$.

(C) The basic strength of amines in the gaseous phase or non-polar solvents is determined by the inductive effect ($+I$ effect) of the alkyl groups attached to the nitrogen atom.
As the number of alkyl groups increases,the electron density on the nitrogen atom increases due to the $+I$ effect,making the lone pair more available for donation.
Therefore,the order of basic strength is $(C_2H_5)_2NH > C_2H_5NH_2 > NH_3$.

(B) The spin-only magnetic moment is calculated using the formula $\mu = \sqrt{n(n+2)} \text{ BM}$,where $n$ is the number of unpaired electrons.

$Complex$	$Configuration \text{ and } n$
$[V(CN)_6]^{4-}$ ($V^{2+}$,$d^3$)	$t_{2g}^3 e_g^0$,$n = 3$
$[Cr(NH_3)_6]^{2+}$ ($Cr^{2+}$,$d^4$)	$t_{2g}^4 e_g^0$,$n = 2$
$[Ru(NH_3)_6]^{3+}$ ($Ru^{3+}$,$d^5$)	$t_{2g}^5 e_g^0$,$n = 1$
$[Fe(CN)_6]^{4-}$ ($Fe^{2+}$,$d^6$)	$t_{2g}^6 e_g^0$,$n = 0$

Since the magnetic moment is directly proportional to the number of unpaired electrons $(n)$,the order is $n=3 > n=2 > n=1 > n=0$.
Therefore,the order is $[V(CN)_6]^{4-} > [Cr(NH_3)_6]^{2+} > [Ru(NH_3)_6]^{3+} > [Fe(CN)_6]^{4-}$.
This corresponds to the order of metal ions: $V^{2+} > Cr^{2+} > Ru^{3+} > Fe^{2+}$.

(C) The solubility profile of the compound '$X$' is as follows:
$1$. Insoluble in water: Indicates it is likely a non-polar or high molecular weight organic compound.
$2$. Insoluble in $5\% \,HCl$: Indicates it is not a basic compound (like an amine).
$3$. Insoluble in $10\% \,NaOH$: Indicates it is not an acidic compound (like a carboxylic acid or a phenol).
$4$. Insoluble in $10\% \,NaHCO_3$: Confirms it is not a strong acid (like a carboxylic acid).
Evaluating the options:
- Oleic acid is a carboxylic acid,so it is soluble in $NaOH$ and $NaHCO_3$.
- $o-$Toluidine is an amine,so it is soluble in $HCl$.
- $m-$Cresol is a phenol,so it is soluble in $NaOH$.
- Benzamide $(C_6H_5CONH_2)$ is a neutral compound. It is insoluble in water,$HCl$,$NaOH$,and $NaHCO_3$. Therefore,it matches the given solubility profile.

(D) The color of lanthanoid ions depends on the presence of unpaired $f$-electrons.
$La^{3+}$ has a configuration of $[Xe] 4f^0$ (no unpaired electrons).
$Gd^{3+}$ has a configuration of $[Xe] 4f^7$ (half-filled,stable,but colorless in many compounds due to high energy transition).
$Lu^{3+}$ has a configuration of $[Xe] 4f^{14}$ (fully filled,no unpaired electrons).
$Sm^{3+}$ has a configuration of $[Xe] 4f^5$,which contains unpaired electrons,allowing for $f-f$ transitions that result in a yellow color.

(D) The reaction involves two steps with excess $HBr$ and heat:
$1$. Cleavage of the ether linkage: The methoxy group $(-OCH_3)$ is protonated by $HBr$ to form an oxonium ion,which then undergoes $S_N2$ attack by $Br^-$ to yield $CH_3Br$ and a phenol derivative,$3$-vinylphenol.
$2$. Electrophilic addition to the alkene: The vinyl group $(-CH=CH_2)$ undergoes electrophilic addition with $HBr$. According to Markovnikov's rule,the proton adds to the terminal carbon to form a stable benzylic carbocation $(Ar-CH^+-CH_3)$. The bromide ion then attacks this carbocation to form the final product,$3-(1-bromoethyl)phenol$.

(C) The compound must satisfy the following conditions:
$1$. Negative test with neutral $FeCl_3$ solution: This indicates the absence of a phenolic $-OH$ group.
$2$. Negative test with Fehling solution: This indicates the absence of an aldehyde $(-CHO)$ group.
$3$. Reacts with Grignard reagent $(RMgX)$: This indicates the presence of an acidic hydrogen (e.g.,$-OH$ group) or a carbonyl group.
$4$. Positive iodoform test: This indicates the presence of a $CH_3CO-$ group or a $CH_3CH(OH)-$ group.
Analyzing the options:
- Option $A$ ($2$-hydroxyacetophenone) contains a phenolic $-OH$ group,so it would give a positive test with neutral $FeCl_3$.
- Option $B$ ($1-(2$-methoxyphenyl)propan-$1$-one) does not contain a $CH_3CO-$ or $CH_3CH(OH)-$ group,so it would not give a positive iodoform test.
- Option $C$ ($1-(2$-hydroxyphenyl)propan-$1$-one) is incorrect as it contains a phenolic group.
- The structure provided in the solution image corresponds to $1-(2$-hydroxyphenyl)ethanol derivative or similar,but based on the standard chemistry of these functional groups,the compound that satisfies all conditions is one that lacks a phenolic group but possesses a $CH_3CH(OH)-$ moiety and a ketone group. Given the options,the structure in option $C$ is often represented in such problems,but if we strictly follow the criteria,the compound is $1-(2$-hydroxyphenyl)propan-$1$-one derivative. However,based on the provided solution image,the correct structure is $1-(2$-hydroxyphenyl)propan-$1$-one.

(B) The rate law is expressed as $r = k[A]^x[B]^y$.
From the data:
$0.045 = k(0.05)^x(0.05)^y$ ...... $(1)$
$0.090 = k(0.10)^x(0.05)^y$ ...... $(2)$
$0.72 = k(0.20)^x(0.10)^y$ ...... $(3)$
Dividing $(2)$ by $(1)$:
$\frac{0.090}{0.045} = \left( \frac{0.10}{0.05} \right)^x$ $\Rightarrow 2 = 2^x$ $\Rightarrow x = 1$.
Dividing $(3)$ by $(2)$:
$\frac{0.72}{0.090} = \left( \frac{0.20}{0.10} \right)^x \left( \frac{0.10}{0.05} \right)^y$
$8 = (2)^1 \times (2)^y$
$8 = 2 \times 2^y$ $\Rightarrow 4 = 2^y$ $\Rightarrow y = 2$.
Thus,the rate law is $r = k[A][B]^2$.

(A) The reaction between phthalic anhydride and chlorobenzene in the presence of $AlCl_3$ and heat is a Friedel-Crafts acylation reaction.
Chlorine $(-Cl)$ is an ortho/para-directing group.
Due to steric hindrance,the para-substitution is favored over ortho-substitution.
Therefore,the major product formed is $2-(4-chlorobenzoyl)benzoic acid$.

(D) The reaction involves the reduction of the carbonyl group of $2$-bromoacetophenone by $NaBH_4$ to form an alkoxide intermediate.
This intermediate then undergoes an intramolecular nucleophilic substitution reaction,where the alkoxide oxygen attacks the carbon bearing the bromine atom,displacing the bromide ion to form an epoxide ring.
Thus,the final product is phenyloxirane.

(A) The Freundlich adsorption isotherm is given by the equation: $\frac{x}{m} = k P^{1/n}$
Taking the logarithm on both sides,we get: $\log \frac{x}{m} = \log k + \frac{1}{n} \log P$
This equation represents a straight line of the form $y = mx + c$,where the slope is $\frac{1}{n}$.
From the given graph,the slope is calculated as: $\text{Slope} = \frac{\text{rise}}{\text{run}} = \frac{2}{3}$.
Therefore,$\frac{1}{n} = \frac{2}{3}$.
Substituting this back into the original equation,we get: $\frac{x}{m} = k P^{2/3}$.
Thus,$\frac{x}{m}$ is proportional to $p^{2/3}$.