WBJEE 2010 Chemistry Question Paper with Answer and Solution

(A) We know that $\tan(90^\circ - \theta) = \cot \theta$ and $\cot(90^\circ - \theta) = \tan \theta$.
Given expression: $\frac{\cot 54^\circ}{\tan 36^\circ} + \frac{\tan 20^\circ}{\cot 70^\circ}$
Using the identity $\tan 36^\circ = \tan(90^\circ - 54^\circ) = \cot 54^\circ$ and $\cot 70^\circ = \cot(90^\circ - 20^\circ) = \tan 20^\circ$:
$= \frac{\cot 54^\circ}{\cot 54^\circ} + \frac{\tan 20^\circ}{\tan 20^\circ}$
$= 1 + 1 = 2$.

(B) According to Graham's Law of Diffusion,$\frac{r_1}{r_2} = \sqrt{\frac{M_2}{M_1}}$.
Given $\frac{r_{H_2}}{r_{hydrocarbon}} = 3\sqrt{3}$,where $M_{H_2} = 2 \ g/mol$.
So,$3\sqrt{3} = \sqrt{\frac{M_{hydrocarbon}}{2}}$.
Squaring both sides: $(3\sqrt{3})^2 = \frac{M_{hydrocarbon}}{2} \Rightarrow 27 = \frac{M_{hydrocarbon}}{2}$.
$M_{hydrocarbon} = 54 \ g/mol$.
The molecular formula is $C_nH_{2n-2}$,so $12n + (2n-2) = 54$.
$14n - 2 = 54$ $\Rightarrow 14n = 56$ $\Rightarrow n = 4$.

(A) For a compound to exhibit keto-enol tautomerism,it must possess at least one $\alpha$-hydrogen atom attached to an $\alpha$-carbon (the carbon atom adjacent to the carbonyl group).
$1$. $C_6H_5COC_6H_5$ (Benzophenone): The carbonyl carbon is attached to two phenyl groups. There are no $\alpha$-hydrogen atoms present on the carbons adjacent to the carbonyl group. Therefore,it cannot show keto-enol tautomerism.
$2$. $C_6H_5COCH_3$ (Acetophenone): It has three $\alpha$-hydrogen atoms on the methyl group attached to the carbonyl carbon,so it exhibits tautomerism.
$3$. $C_6H_5COCH_2COCH_3$ and $CH_3COCH_2COCH_3$: Both contain active methylene groups $(-CH_2-)$ between two carbonyl groups,which have acidic $\alpha$-hydrogens,allowing for keto-enol tautomerism.
Thus,the correct option is $A$.

(B) According to Graham's Law of diffusion,$\frac{r_1}{r_2} = \sqrt{\frac{M_2}{M_1}}$.
Given $\frac{r_{H_2}}{r_{gas}} = 3\sqrt{3}$,where $M_{H_2} = 2 \ g/mol$.
So,$3\sqrt{3} = \sqrt{\frac{M_{gas}}{2}}$.
Squaring both sides,$(3\sqrt{3})^2 = \frac{M_{gas}}{2} \implies 9 \times 3 = \frac{M_{gas}}{2} \implies 27 = \frac{M_{gas}}{2}$.
Thus,$M_{gas} = 54 \ g/mol$.
The molecular formula is $C_n H_{2n-2}$,so $12n + (2n - 2) = 54$.
$14n - 2 = 54 \implies 14n = 56 \implies n = 4$.

(A) The initial kinetic energy at the point of projection $O$ is given by:
$K = \frac{1}{2} mu^2$
where $m$ is the mass of the body and $u$ is the initial velocity of projection.
At the highest point (i.e.,at maximum height $H$),the vertical component of velocity becomes zero,and the velocity is only the horizontal component:
$v = u \cos \beta$
Therefore,the kinetic energy at the highest point is:
$K^{\prime} = \frac{1}{2} mv^2 = \frac{1}{2} m(u \cos \beta)^2 = \frac{1}{2} mu^2 \cos^2 \beta$
According to the problem,the kinetic energy at the highest point is $\frac{3}{4}$ of the initial kinetic energy:
$K^{\prime} = \frac{3}{4} K$
Substituting the expressions:
$\frac{1}{2} mu^2 \cos^2 \beta = \frac{3}{4} \left( \frac{1}{2} mu^2 \right)$
$\cos^2 \beta = \frac{3}{4}$
Taking the square root on both sides:
$\cos \beta = \frac{\sqrt{3}}{2}$
$\beta = \cos^{-1} \left( \frac{\sqrt{3}}{2} \right) = 30^{\circ}$

(A) In $PCl_5$,the central phosphorus atom $(P)$ has $5$ valence electrons. It forms $5$ covalent bonds with $5$ chlorine atoms.
Using the formula for steric number: $\text{Steric Number} = \frac{1}{2} (V + M - C + A)$,where $V = 5$ (valence electrons of $P$),$M = 5$ (monovalent atoms),$C = 0$ (cationic charge),and $A = 0$ (anionic charge).
$\text{Steric Number} = \frac{1}{2} (5 + 5) = 5$.
$A$ steric number of $5$ corresponds to $sp^3d$ hybridisation,resulting in a trigonal bipyramidal geometry.

(C) The molecular orbital configuration of $CO$ ($14$ electrons) is: $\sigma(1s)^2, \sigma^*(1s)^2, \sigma(2s)^2, \sigma^*(2s)^2, \pi(2p_x)^2 = \pi(2p_y)^2, \sigma(2p_z)^2$.
Number of bonding electrons $(N_b)$ = $10$.
Number of antibonding electrons $(N_a)$ = $4$.
Bond Order = $\frac{N_b - N_a}{2} = \frac{10 - 4}{2} = 3$.

(C) The dipole moment of a molecule is the vector sum of the individual bond dipoles. For chlorobenzene,the dipole moment is $1.5 \ D$. Let the dipole moment of a $C-Cl$ bond be $\mu$.
For $o-$dichlorobenzene,the two $C-Cl$ bonds are at an angle of $60^{\circ}$ to each other.
The resultant dipole moment is given by $\mu_{net} = \sqrt{\mu^2 + \mu^2 + 2\mu^2 \cos(60^{\circ})}$.
Since $\cos(60^{\circ}) = 0.5$,we have $\mu_{net} = \sqrt{2\mu^2 + 2\mu^2(0.5)} = \sqrt{3\mu^2} = \mu \sqrt{3}$.
Given $\mu = 1.5 \ D$,the dipole moment of $o-$dichlorobenzene is $1.5 \times 1.732 \approx 2.54 \ D$.

(D) The strength of hydrogen bonding in a compound depends on the number of $-OH$ groups available for intermolecular interaction.
Propan$-1,2,3-$triol (glycerol) contains three $-OH$ groups,which allows for a greater extent of intermolecular hydrogen bonding compared to the other options.
Therefore,it exhibits the strongest hydrogen bonding.

(C) The relationship between $K_P$ and $K_C$ is given by the equation: $K_P = K_C(RT)^{\Delta n}$.
For the reaction $CaCO_{3(s)} \rightleftharpoons CaO_{(s)} + CO_{2(g)}$,the change in the number of moles of gaseous species is $\Delta n = n_{p(g)} - n_{r(g)} = 1 - 0 = 1$.
Given $K_P = 167$,$R = 0.0821 \ L \ atm \ K^{-1} \ mol^{-1}$,and $T = 800 + 273 = 1073 \ K$.
Substituting these values into the formula: $K_C = \frac{K_P}{(RT)^{\Delta n}} = \frac{167}{(0.0821 \times 1073)^1}$.
$K_C = \frac{167}{88.0933} \approx 1.896$.
Rounding to two decimal places,we get $K_C = 1.89$.

(C) The basic character of oxides depends on the metallic character of the element.
Metallic character decreases from left to right across a period,so the basic character of oxides also decreases from left to right.
Metallic character increases down a group,so the basic character of oxides increases from top to bottom.
Therefore,the basic character of oxides decreases from left to right and increases from top to bottom.

(C) The first ionization energy $(IE_1)$ generally increases across a period due to increasing effective nuclear charge. However,nitrogen $(N)$ has a stable half-filled $2p^3$ electronic configuration,which makes it harder to remove an electron compared to oxygen $(O)$.
Thus,$IE_1$ of $N > O$.
Since fluorine $(F)$ has a smaller atomic size and higher effective nuclear charge than oxygen,$IE_1$ of $F > O$.
Combining these,the correct order is $N > O < F$.

(C) The fusion of a manganous salt $(Mn^{2+})$ with $KNO_3$ and solid $NaOH$ leads to the formation of potassium manganate $(K_2MnO_4)$.
The chemical reaction is: $Mn^{2+} + 2NO_3^{-} + 4OH^{-} \rightarrow MnO_4^{2-} + 2NO_2^{-} + 2H_2O$.
In $MnO_4^{2-}$,the oxidation state of $Mn$ is calculated as: $x + 4(-2) = -2$,which gives $x = +6$.
Therefore,the oxidation number of $Mn$ changes from $+2$ to $+6$.

No Solution Available

(A) Keto-enol tautomerism requires the presence of at least one $\alpha$-hydrogen atom adjacent to the carbonyl group $(C=O)$.
In $C_6H_5COC_6H_5$ (benzophenone),the carbonyl carbon is attached to two phenyl groups $(C_6H_5)$.
There is no $\alpha$-carbon with an $\alpha$-hydrogen atom available in this molecule.
Therefore,it cannot exhibit keto-enol tautomerism.

(D) The reaction of an alkane with chlorine in the presence of ultra-violet light is a free radical substitution reaction.
For an alkane to form only one mono-chloro-alkane,all the hydrogen atoms in the alkane must be equivalent.
In $Neopentane$ $(2,2-dimethylpropane)$,all $12$ hydrogen atoms are attached to equivalent primary carbon atoms.
Therefore,replacing any one of these hydrogen atoms with a chlorine atom results in the same product,$1-chloro-2,2-dimethylpropane$.

(A) Ortho-hydrogen and para-hydrogen are nuclear spin isomers of the hydrogen molecule $(H_2)$.
In ortho-hydrogen,the nuclear spins of the two protons are parallel,whereas in para-hydrogen,the nuclear spins are antiparallel.
Because they have the same electronic structure,their chemical properties are identical.
However,due to the difference in their nuclear spin states,they exhibit different physical properties,such as different thermal conductivities and specific heats.

(C) The molecule $B_2H_6$ (Diborane) contains two bridging hydrogen atoms.
In $B_2H_6$,each boron atom is $sp^3$ hybridized.
Two hydrogen atoms act as bridges between the two boron atoms,forming two $3c-2e$ (three-center-two-electron) bonds,which are also known as hydrogen bridge bonds or banana bonds.

(B) Step $1$: Calculate the total millimoles of $H^{+}$ ions.
Millimoles of $H^{+}$ from $HCl = 0.1 \ (M) \times 2 \ mL \times 1 = 0.2 \ mmol$.
Millimoles of $H^{+}$ from $H_2SO_4 = 0.1 \ (M) \times 2 \ mL \times 2 = 0.4 \ mmol$.
Total millimoles of $H^{+} = 0.2 + 0.4 = 0.6 \ mmol$.
Step $2$: Calculate the concentration of $H^{+}$ in the final mixture.
Total volume $= 2 \ mL + 2 \ mL + 2 \ mL = 6 \ mL$.
$[H^{+}] = \frac{\text{Total millimoles}}{\text{Total volume in } mL} = \frac{0.6}{6} = 0.1 \ (M)$.
Step $3$: Calculate the $pH$.
$pH = -\log_{10} [H^{+}] = -\log_{10} (0.1) = 1$.

(B) Given $K_a = 10^{-5}$,so $pK_a = -\log(10^{-5}) = 5$.
At one-third neutralization,the amount of salt formed is $1/3$ and the amount of acid remaining is $2/3$ of the initial concentration.
According to the Henderson-Hasselbalch equation:
$pH = pK_a + \log \frac{[\text{Salt}]}{[\text{Acid}]}$
$pH = 5 + \log \frac{1/3}{2/3}$
$pH = 5 + \log \frac{1}{2}$
$pH = 5 - \log 2$.

(A) The balanced chemical equation is: $2 Na_2S_2O_3 + I_2 \longrightarrow Na_2S_4O_6 + 2 NaI$.
In this reaction,the oxidation state of sulfur in $Na_2S_2O_3$ changes from $+2$ to $+2.5$ in $Na_2S_4O_6$.
The change in oxidation state per sulfur atom is $0.5$. Since there are $2$ sulfur atoms in $Na_2S_2O_3$,the total change in oxidation state (n-factor) is $2 \times 0.5 = 1$.
The equivalent weight $E$ is given by the formula $E = \frac{M}{n\text{-factor}}$,where $M$ is the molar mass.
Thus,$E = \frac{M}{1} = M$.
Therefore,the equivalent weight is equal to the molar mass of sodium thiosulphate.

(A) $1$. Calculate the molar mass of $NaOH$: $Na(23) + O(16) + H(1) = 40 \ g/mol$.
$2$. Calculate the number of moles of $NaOH$: $n = \frac{\text{mass}}{\text{molar mass}} = \frac{4 \ g}{40 \ g/mol} = 0.1 \ mol$.
$3$. Convert the volume of the solution from $mL$ to $L$: $V = \frac{250 \ mL}{1000} = 0.25 \ L$.
$4$. Calculate the molarity $(M)$: $M = \frac{n}{V} = \frac{0.1 \ mol}{0.25 \ L} = 0.4 \ M$.

(B) According to Graham's Law of diffusion,$\frac{r_1}{r_2} = \sqrt{\frac{M_2}{M_1}}$.
Given,$\frac{r_{H_2}}{r_{C_n H_{2n-2}}} = 3\sqrt{3} = \sqrt{27}$.
Therefore,$\sqrt{\frac{M_{C_n H_{2n-2}}}{M_{H_2}}} = \sqrt{27}$.
Since $M_{H_2} = 2 \ g/mol$,we have $\frac{M_{C_n H_{2n-2}}}{2} = 27$.
$M_{C_n H_{2n-2}} = 54 \ g/mol$.
The molar mass of $C_n H_{2n-2}$ is $12n + (2n - 2) = 14n - 2$.
Equating the two,$14n - 2 = 54$,which gives $14n = 56$,so $n = 4$.

(A) The $n$-th term of the series is $T_n = (2n-1)^3$.
We need to find the sum $S_n = \sum_{k=1}^{n} (2k-1)^3$.
Expanding the term: $(2k-1)^3 = 8k^3 - 12k^2 + 6k - 1$.
Now,$S_n = \sum_{k=1}^{n} (8k^3 - 12k^2 + 6k - 1) = 8 \sum k^3 - 12 \sum k^2 + 6 \sum k - \sum 1$.
Using the standard summation formulas:
$\sum k^3 = \frac{n^2(n+1)^2}{4}$,$\sum k^2 = \frac{n(n+1)(2n+1)}{6}$,$\sum k = \frac{n(n+1)}{2}$,$\sum 1 = n$.
Substituting these values:
$S_n = 8 \left[ \frac{n^2(n+1)^2}{4} \right] - 12 \left[ \frac{n(n+1)(2n+1)}{6} \right] + 6 \left[ \frac{n(n+1)}{2} \right] - n$.
$S_n = 2n^2(n^2+2n+1) - 2n(2n^2+3n+1) + 3n^2+3n - n$.
$S_n = 2n^4 + 4n^3 + 2n^2 - 4n^3 - 6n^2 - 2n + 3n^2 + 2n$.
$S_n = 2n^4 - n^2 = n^2(2n^2-1)$.

(A) Given the equation: $\sin 6 \theta + \sin 4 \theta + \sin 2 \theta = 0$
Group the terms: $(\sin 6 \theta + \sin 2 \theta) + \sin 4 \theta = 0$
Using the sum-to-product formula $\sin A + \sin B = 2 \sin \frac{A+B}{2} \cos \frac{A-B}{2}$:
$2 \sin 4 \theta \cos 2 \theta + \sin 4 \theta = 0$
Factor out $\sin 4 \theta$:
$\sin 4 \theta (2 \cos 2 \theta + 1) = 0$
This gives two cases:
Case $1$: $\sin 4 \theta = 0 \implies 4 \theta = n \pi \implies \theta = \frac{n \pi}{4}$
Case $2$: $2 \cos 2 \theta + 1 = 0 \implies \cos 2 \theta = -\frac{1}{2}$
Since $\cos \frac{2 \pi}{3} = -\frac{1}{2}$,we have $\cos 2 \theta = \cos \frac{2 \pi}{3}$
The general solution for $\cos x = \cos \alpha$ is $x = 2 n \pi \pm \alpha$
So,$2 \theta = 2 n \pi \pm \frac{2 \pi}{3}$
Dividing by $2$,we get $\theta = n \pi \pm \frac{\pi}{3}$
Thus,the general values are $\theta = \frac{n \pi}{4}$ and $\theta = n \pi \pm \frac{\pi}{3}$.

(D) The number of protons $(p)$ determines the atomic number $(Z)$.
Here,$Z = 16$,which corresponds to the element Sulfur $(S)$.
The number of electrons $(e^-)$ is $18$.
The net charge is calculated as: $\text{Charge} = \text{Number of protons} - \text{Number of electrons} = 16 - 18 = -2$.
Therefore,the species is $S^{2-}$.

(B) Sommerfeld introduced elliptical orbits to explain the fine structure of spectral lines in the hydrogen atom. According to his model,the electron moves in an elliptical orbit with the nucleus at one of the foci. The trajectory is a closed ellipse-like curve,which is narrower at the perihelion (closest point to the nucleus) and flatter at the aphelion (farthest point from the nucleus).

(A) The $yz$ plane is defined by the equation $x = 0$.
An orbital will have zero probability of finding an electron in the $yz$ plane if it has a nodal plane coinciding with the $yz$ plane.
The $p_x$ orbital has its electron density distributed along the $x$-axis,and its nodal plane is the $yz$ plane.
Therefore,the probability of finding an electron in the $yz$ plane for the $p_x$ orbital is zero.

(A) The Gibbs free energy is defined as $G = H - TS$.
Taking the differential,we get $dG = dH - TdS - SdT$.
Since $H = U + PV$,the differential is $dH = dU + PdV + VdP$.
From the first law of thermodynamics,$dU = TdS - PdV$.
Substituting $dU$ into the expression for $dH$,we get $dH = (TdS - PdV) + PdV + VdP = TdS + VdP$.
Now,substituting $dH$ back into the expression for $dG$:
$dG = (TdS + VdP) - TdS - SdT$.
Simplifying this,we obtain $dG = VdP - SdT$.

(C) Let the equation of the ellipse be $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$,where $a > b$.
The foci are $S(-ae, 0)$ and $T(ae, 0)$,and the end point of the minor axis is $B(0, b)$.
The distance $ST = 2ae$ and the distance $SB = \sqrt{(ae)^2 + b^2} = a$.
Since $\triangle STB$ is an equilateral triangle,all sides are equal,so $ST = SB = TB$.
Thus,$2ae = a$,which implies $e = \frac{1}{2}$.
Alternatively,in $\triangle OSB$ (where $O$ is the origin),$\tan(\angle OSB) = \frac{OB}{OS} = \frac{b}{ae}$.
Since $\triangle STB$ is equilateral,$\angle OSB = 30^\circ$,so $\frac{b}{ae} = \tan(30^\circ) = \frac{1}{\sqrt{3}}$.
Thus,$b^2 = \frac{a^2 e^2}{3}$.
Using $b^2 = a^2(1 - e^2)$,we get $a^2(1 - e^2) = \frac{a^2 e^2}{3}$.
$1 - e^2 = \frac{e^2}{3}$ $\Rightarrow 1 = \frac{4e^2}{3}$ $\Rightarrow e^2 = \frac{3}{4}$ $\Rightarrow e = \frac{\sqrt{3}}{2}$.
Re-evaluating the condition: If $ST = SB$,then $2ae = a \Rightarrow e = 1/2$. If $ST = TB$ and $SB = TB$,then $2ae = \sqrt{a^2 e^2 + b^2}$.
Given the standard interpretation of such problems,$e = 1/2$ is the correct result.

(C) Given $y = (1+x)(1+x^2)(1+x^4) \dots (1+x^{2^n})$.
Taking the natural logarithm on both sides:
$\ln y = \ln(1+x) + \ln(1+x^2) + \ln(1+x^4) + \dots + \ln(1+x^{2^n})$.
Differentiating both sides with respect to $x$:
$\frac{1}{y} \frac{dy}{dx} = \frac{1}{1+x} + \frac{2x}{1+x^2} + \frac{4x^3}{1+x^4} + \dots + \frac{2^n x^{2^n-1}}{1+x^{2^n}}$.
Thus,$\frac{dy}{dx} = y \left[ \frac{1}{1+x} + \frac{2x}{1+x^2} + \dots + \frac{2^n x^{2^n-1}}{1+x^{2^n}} \right]$.
At $x=0$,$y = (1+0)(1+0) \dots (1+0) = 1$.
Substituting $x=0$ into the derivative expression:
$\left(\frac{dy}{dx}\right)_{x=0} = 1 \cdot \left[ \frac{1}{1+0} + 0 + 0 + \dots + 0 \right] = 1 \cdot 1 = 1$.

(A) The slope of the normal to the curve $y=f(x)$ at a point $(x_0, y_0)$ is given by $m_n = \tan(\theta)$,where $\theta$ is the angle the normal makes with the positive $x$-axis.
Given $\theta = 3\pi/4$,the slope of the normal is $m_n = \tan(3\pi/4) = -1$.
We know that the slope of the normal is also related to the derivative of the function by the formula $m_n = -1/f'(x_0)$.
Substituting the given point $(3,4)$,we have $m_n = -1/f'(3)$.
Equating the two expressions for the slope of the normal: $-1/f'(3) = -1$.
Solving for $f'(3)$,we get $f'(3) = 1$.

(A) Let $I = \int \frac{\log \sqrt{x}}{3 x} dx$.
Using the property $\log \sqrt{x} = \log(x^{1/2}) = \frac{1}{2} \log x$,the integral becomes:
$I = \int \frac{\frac{1}{2} \log x}{3 x} dx = \frac{1}{6} \int \frac{\log x}{x} dx$.
Let $u = \log x$,then $du = \frac{1}{x} dx$.
Substituting these into the integral:
$I = \frac{1}{6} \int u du = \frac{1}{6} \cdot \frac{u^2}{2} + C = \frac{u^2}{12} + C$.
Substituting $u = \log x$ back:
$I = \frac{(\log x)^2}{12} + C$.
Alternatively,using the substitution $t = \log \sqrt{x} = \frac{1}{2} \log x$,then $dt = \frac{1}{2x} dx$,which implies $\frac{1}{x} dx = 2 dt$.
$I = \int \frac{t}{3} (2 dt) = \frac{2}{3} \int t dt = \frac{2}{3} \cdot \frac{t^2}{2} + C = \frac{t^2}{3} + C = \frac{1}{3} (\log \sqrt{x})^2 + C$.
Thus,the correct option is $A$.

(C) The reaction of $PCl_5$ with alcohols $(R-OH)$ yields alkyl chlorides $(R-Cl)$:
$C_2 H_5 OH + PCl_5 \rightarrow C_2 H_5 Cl + POCl_3 + HCl$
Thus,$X$ is $C_2 H_5 OH$.
The reaction of $PCl_5$ with carboxylic acids $(R-COOH)$ yields acid chlorides $(R-COCl)$:
$CH_3 CO_2 H + PCl_5 \rightarrow CH_3 COCl + POCl_3 + HCl$
Thus,$Y$ is $CH_3 CO_2 H$.
Therefore,the correct option is $C$.

(C) The reaction of benzenediazonium chloride $(C_6H_5N_2^+Cl^-)$ with boiling water is a standard method for the preparation of phenol.
The reaction proceeds as follows:
$C_6H_5N_2^+Cl^- + H_2O \xrightarrow{\Delta} C_6H_5OH + N_2 \uparrow + HCl$
Here,the diazonium group is replaced by a hydroxyl group $(-OH)$,resulting in the formation of phenol.

(C) When acetone $(CH_3COCH_3)$ is saturated with dry $HCl$ gas,it undergoes self-aldol condensation followed by dehydration to form mesityl oxide $(CH_3COCH=C(CH_3)_2)$.
The reaction is as follows:
$2CH_3COCH_3 \xrightarrow{HCl(g)} CH_3COCH=C(CH_3)_2 + H_2O$
Note: Phorone is formed as a minor product,but the major product is mesityl oxide.

(D) Vitamin $C$ is chemically known as $Ascorbic \ acid$. It is a water-soluble vitamin found in various foods and is essential for the synthesis of collagen and the functioning of the immune system.

(C) The radioactive decay follows first-order kinetics. The decay constant $\lambda$ is given by: $\lambda = \frac{2.303}{t} \log \frac{N_0}{N_t}$.
Given that radioactivity decreases by $90 \%$,the remaining amount $N_t = N_0 - 0.90 N_0 = 0.10 N_0$.
Substituting the values: $\lambda = \frac{2.303}{10} \log \frac{N_0}{0.10 N_0} = \frac{2.303}{10} \log(10) = \frac{2.303}{10} \times 1 = 0.2303 \ year^{-1}$.
The half-life $t_{1/2}$ is given by: $t_{1/2} = \frac{0.693}{\lambda}$.
$t_{1/2} = \frac{0.693}{0.2303} \approx 3.01 \ years$.
Therefore,the half-life is approximately $3 \ years$.

(B) The reaction is $RCl + H_2O \rightarrow ROH + HCl$.
Since water is present in large excess,its concentration remains practically constant during the reaction.
This makes the reaction a pseudo-first-order reaction,where the order of reaction is $1$.
However,the molecularity is determined by the number of reacting species colliding simultaneously in the elementary step,which is $2$ ($RCl$ and $H_2O$).
Therefore,the molecularity is $2$ and the order of reaction is $1$.

(A) For an exothermic reaction,the energy of the products is lower than the energy of the reactants.
As shown in the energy profile diagram,the activation energy for the forward reaction $(E_{a,f})$ is the energy difference between the transition state and the reactants.
The activation energy for the backward reaction $(E_{a,b})$ is the energy difference between the transition state and the products.
Since the products are at a lower energy level than the reactants,the energy barrier to reach the transition state from the products $(E_{a,b})$ is greater than the energy barrier from the reactants $(E_{a,f})$.
Therefore,$(E_{a,b}) > (E_{a,f})$.

(A) Aspirin is chemically known as $2$-acetoxybenzoic acid or acetylsalicylic acid.
It is prepared by the acetylation of salicylic acid with acetic anhydride.
The structure consists of a benzene ring with a carboxylic acid group $(-COOH)$ and an acetoxy group $(-OCOCH_3)$ at the ortho position.

(B) The inert pair effect is the tendency of the two electrons in the outermost $s$-orbital to remain unshared in compounds of post-transition elements,particularly in the $p$-block elements of the $6^{th}$ period and some $5^{th}$ period elements.
$Sn$ ($5^{th}$ period),$Pb$ ($6^{th}$ period),and $In$ ($5^{th}$ period) are $p$-block elements that exhibit the inert pair effect.
$Fe$ is a $d$-block element (transition metal) and does not exhibit the inert pair effect.

(A) Hemoglobin is a complex protein found in red blood cells that is responsible for transporting oxygen throughout the body.
It contains a prosthetic group called heme,which consists of a porphyrin ring coordinated to a central iron ion.
In functional hemoglobin,this iron ion is in the ferrous state,represented as $Fe^{2+}$.

(B) When $AgCl$ reacts with $KCN$,it dissolves to form a soluble complex compound.
The chemical reaction is as follows:
$AgCl(s) + 2KCN(aq) \rightarrow K[Ag(CN)_2](aq) + KCl(aq)$
In this reaction,the silver chloride reacts with potassium cyanide to form the dicyanoargentate$(I)$ complex ion,$[Ag(CN)_2]^-$,which is soluble in water.

(C) The half-cell reaction for a hydrogen electrode is: $2H^+ (aq) + 2e^- \rightarrow H_2 (g)$.
Using the Nernst equation: $E = E^0 - \frac{0.0591}{n} \log \frac{P_{H_2}}{[H^+]^2}$.
Given $E^0 = 0 \ V$,$n = 2$,$P_{H_2} = 1 \ atm$,and $[H^+] = 10^{-pH} = 10^{-10} \ M$.
Substituting the values: $E = 0 - \frac{0.0591}{2} \log \frac{1}{(10^{-10})^2}$.
$E = -\frac{0.0591}{2} \log (10^{20})$.
$E = -\frac{0.0591}{2} \times 20$.
$E = -0.0591 \times 10 = -0.591 \ V$.
Thus,the potential is approximately $-0.59 \ V$.

(A) The abundance of metals in the earth's crust is as follows: $Aluminium$ $(Al)$ is the most abundant metal,accounting for approximately $8.3\%$ by weight. $Iron$ $(Fe)$ is the second most abundant metal. Therefore,the correct option is $A$.

(C) The reaction proceeds in two steps:
$1$. Nitrobenzene $(C_6H_5NO_2)$ is reduced by $NH_4Cl / Zn$ dust to form phenylhydroxylamine $(C_6H_5NHOH)$.
$2$. Phenylhydroxylamine is then oxidized by $H_2SO_4 / Na_2Cr_2O_7$ to form nitrosobenzene $(C_6H_5NO)$.

(C) The structure of Pyrophosphoric acid $(H_4P_2O_7)$ consists of two phosphate units linked by an oxygen atom,forming a $P-O-P$ bond.
Its structure is $(HO)_2P(O)-O-P(O)(OH)_2$.
Other acids like Hypophosphorus acid $(H_3PO_2)$,Phosphorus acid $(H_3PO_3)$,and Orthophosphoric acid $(H_3PO_4)$ do not contain a $P-O-P$ linkage.

(C) An anhydride is a compound formed by the removal of water from an acid.
For phosphoric acid $(H_3 PO_4)$,the dehydration reaction is:
$4 H_3 PO_4 \longrightarrow P_4 O_{10} + 6 H_2 O$
Thus,$P_4 O_{10}$ is the anhydride of orthophosphoric acid $(H_3 PO_4)$.

(C) The ozone $(O_3)$ molecule is angular in shape and is a resonance hybrid of two contributing structures. Due to resonance,the bond order in ozone is $1.5$,which is intermediate between a single bond $(1)$ and a double bond $(2)$. In molecular oxygen $(O_2)$,the bond order is $2$. Since bond length is inversely proportional to bond order,the $O-O$ bond length in $O_3$ is greater than the $O=O$ bond length in $O_2$. Therefore,statement $C$ is incorrect.

(A) co-polymer is a polymer formed from two or more different types of monomer units.
$Buna-S$ is a co-polymer formed by the polymerisation of $1,3-butadiene$ and styrene.
The structure of $Buna-S$ is: $(-CH_2-CH=CH-CH_2-CH(C_6H_5)-CH_2-)_n$.
Teflon,$PVC$,and Polypropylene are homopolymers.

(D) The radioactive decay process follows the conservation of mass number and atomic number.
$1$. For $\beta$-decay: ${ }_{89}^{227} Ac \xrightarrow{-\beta} { }_{90}^{227} Th + { }_{-1}^{0} e$. Thus,$[A] = { }_{90}^{227} Th$.
$2$. For $\alpha$-decay: ${ }_{90}^{227} Th \xrightarrow{-\alpha} { }_{88}^{223} Ra + { }_{2}^{4} He$. Thus,$[B] = { }_{88}^{223} Ra$.
$3$. Further $\alpha$-decay: ${ }_{88}^{223} Ra \xrightarrow{-\alpha} { }_{86}^{219} Rn + { }_{2}^{4} He$. This confirms the sequence.
Therefore,$[A]$ is ${ }_{90}^{227} Th$ and $[B]$ is ${ }_{88}^{223} Ra$.