WBJEE 2009 Chemistry Question Paper with Answer and Solution

(C) The function $f(x) = e^{x - [x]}$ is a periodic function with period $T = 1$,where $[x]$ denotes the greatest integer function.
Using the property of definite integrals for periodic functions,$\int_{0}^{nT} f(x) \, dx = n \int_{0}^{T} f(x) \, dx$,we have:
$\int_{0}^{1000} e^{x - [x]} \, dx = 1000 \int_{0}^{1} e^{x - [x]} \, dx$.
For $0 < x < 1$,the value of $[x] = 0$. Therefore,$e^{x - [x]} = e^x$.
Substituting this into the integral:
$1000 \int_{0}^{1} e^x \, dx = 1000 [e^x]_{0}^{1}$.
Evaluating the definite integral:
$1000 (e^1 - e^0) = 1000 (e - 1)$.

(A) The given differential equation is $(x \log x) \frac{dy}{dx} + y = 2 \log x$.
Dividing both sides by $(x \log x)$,we get:
$\frac{dy}{dx} + \frac{1}{x \log x} y = \frac{2 \log x}{x \log x} = \frac{2}{x}$.
This is a linear differential equation of the form $\frac{dy}{dx} + Py = Q$,where $P = \frac{1}{x \log x}$ and $Q = \frac{2}{x}$.
The integrating factor $(I.F.)$ is given by $e^{\int P dx}$.
$I.F. = e^{\int \frac{1}{x \log x} dx}$.
Let $u = \log x$,then $du = \frac{1}{x} dx$.
So,$\int \frac{1}{x \log x} dx = \int \frac{1}{u} du = \log u = \log(\log x)$.
Therefore,$I.F. = e^{\log(\log x)} = \log x$.

(B) The energy stored in a capacitor is given by $U = \frac{Q^2}{2C}$,where $Q$ is the charge and $C$ is the capacitance.
Let the initial charge be $Q$. The initial energy is $U_1 = \frac{Q^2}{2C}$.
When the charge is increased by $2 \, C$,the new charge is $Q' = Q + 2$.
The new energy is $U_2 = \frac{(Q+2)^2}{2C}$.
Given that the energy increases by $21 \%$,the new energy is $U_2 = U_1 + 0.21 U_1 = 1.21 U_1$.
Substituting the expressions for $U_1$ and $U_2$: $\frac{(Q+2)^2}{2C} = 1.21 \left( \frac{Q^2}{2C} \right)$.
Canceling $\frac{1}{2C}$ from both sides,we get $(Q+2)^2 = 1.21 Q^2$.
Taking the square root of both sides: $Q + 2 = \sqrt{1.21} Q$.
$Q + 2 = 1.1 Q$.
$0.1 Q = 2$.
$Q = \frac{2}{0.1} = 20 \, C$.

(D) Given: Poisson's ratio $\sigma = 0.5$. The percentage decrease in cross-sectional area is $\frac{\Delta A}{A} \times 100 = 4 \%$,so $\frac{\Delta A}{A} = 0.04$.
Since the area $A = \pi r^2$,the fractional change in area is related to the fractional change in radius by $\frac{\Delta A}{A} = 2 \frac{\Delta r}{r}$.
Thus,$2 \frac{\Delta r}{r} = 0.04$,which gives the lateral strain $\frac{\Delta r}{r} = 0.02$.
Poisson's ratio is defined as $\sigma = - \frac{\text{Lateral Strain}}{\text{Longitudinal Strain}} = - \frac{\Delta r / r}{\Delta l / l}$.
Taking magnitudes,$0.5 = \frac{0.02}{\Delta l / l}$.
Therefore,the longitudinal strain $\frac{\Delta l}{l} = \frac{0.02}{0.5} = 0.04$.
The percentage increase in length is $\frac{\Delta l}{l} \times 100 = 0.04 \times 100 = 4 \%$.

(C) Given that the horizontal component $B_H$ is $\frac{1}{\sqrt{3}}$ times the vertical component $B_V$.
So,$B_H = \frac{1}{\sqrt{3}} B_V$,which implies $\frac{B_V}{B_H} = \sqrt{3}$.
The angle of dip $\delta$ is defined by the relation $\tan \delta = \frac{B_V}{B_H}$.
Substituting the value,we get $\tan \delta = \sqrt{3}$.
Since $\tan 60^{\circ} = \sqrt{3}$,the angle of dip $\delta = 60^{\circ}$.

(A) In $NF_3$,the central nitrogen atom is bonded to three fluorine atoms by three $\sigma$-bonds and has one lone pair of electrons.
To calculate the hybridisation,we use the formula:
$\text{Steric Number} = \text{Number of } \sigma \text{-bonds} + \text{Number of lone pairs}$.
$\text{Steric Number} = 3 + 1 = 4$.
$A$ steric number of $4$ corresponds to $sp^3$ hybridisation.

(C) The structure of the compound is $H-C(4) \equiv C(3)-C(2)H=C(1)H_2$.
For $C-2$: It is bonded to one $H$ atom,one $C-3$ atom (via a single bond),and one $C-1$ atom (via a double bond). The number of sigma bonds is $3$ and there are no lone pairs,so the hybridization is $sp^2$.
For $C-3$: It is bonded to one $C-4$ atom (via a triple bond) and one $C-2$ atom (via a single bond). The number of sigma bonds is $2$ and there are no lone pairs,so the hybridization is $sp$.
Therefore,the hybridization of $C-2$ and $C-3$ are $sp^2$ and $sp$ respectively.

(B) The relationship between $K_p$ and $K_c$ is given by $K_p = K_c(RT)^{\Delta n_g}$.
Here,$\Delta n_g$ is the change in the number of moles of gaseous products and reactants.
$\Delta n_g = (\sum n_g)_{\text{products}} - (\sum n_g)_{\text{reactants}}$.
For the reaction $SO_2(g) + \frac{1}{2} O_2(g) \rightleftharpoons SO_3(g)$:
$\Delta n_g = 1 - (1 + \frac{1}{2}) = 1 - \frac{3}{2} = -\frac{1}{2}$.
Comparing this with $K_p = K_c(RT)^x$,we get $x = -\frac{1}{2}$.

(A) Electron is a magnesium-based alloy.
It typically consists of $Mg (95 \%)$,$Zn (4.5 \%)$,and $Cu (0.5 \%)$.
Therefore,the correct option is $A$.

(B) $KMnO_4$ (Potassium permanganate) acts as a strong oxidizing agent. It is widely used as an antiseptic and disinfectant in dilute solutions for treating skin infections and wounds.

(D) The degree of unsaturation (or double bond equivalent) is calculated using the formula: $U = C + 1 - \frac{H + X - N}{2}$.
For a stable organic compound,the degree of unsaturation must be an integer ($0$,$1$,$2$,...).
For $C_4H_{10}O_4$: $U = 4 + 1 - \frac{10}{2} = 0$.
For $C_4H_8O_4$: $U = 4 + 1 - \frac{8}{2} = 1$.
For $C_4H_7ClO_4$: $U = 4 + 1 - \frac{7+1}{2} = 1$.
For $C_4H_9O_4$: $U = 4 + 1 - \frac{9}{2} = 0.5$.
Since the degree of unsaturation cannot be $0.5$ for a stable organic molecule,$C_4H_9O_4$ does not represent a valid organic compound.

(A) The stability of carbocations is determined by factors such as resonance,hyperconjugation,and inductive effect.
$Ph_3C^{+}$ (triphenylmethyl carbocation) is the most stable among the given options because the positive charge on the central carbon atom is delocalized over three phenyl rings through resonance.
This extensive delocalization provides significant stabilization compared to the other carbocations,which are stabilized only by hyperconjugation or simple resonance.

(C) The reaction is $C_3H_8 \xrightarrow{-2H, +2Cl} C_3H_6Cl_2$.
The structural isomers of $C_3H_6Cl_2$ are:
$1,1$-dichloropropane
$2,2$-dichloropropane
$1,2$-dichloropropane
$1,3$-dichloropropane
Among these,$1,2$-dichloropropane contains a chiral carbon atom (the $C-2$ atom),which means it exists as a pair of enantiomers (optical isomers).
Therefore,the total number of isomers is $4$ structural isomers + $1$ additional enantiomer = $5$ isomers.

(C) To determine the relationship between the two structures,we assign the $R/S$ configuration to the chiral centers.
For the first structure $(I)$:
The top chiral center has $-OH$ $(1)$,$-CH_2OH$ $(2)$,$-CH_3$ $(3)$,and $-H$ $(4)$. With $-H$ on the horizontal,the configuration is $R$.
The bottom chiral center has $-OH$ $(1)$,$-CH(OH)CH_3$ $(2)$,$-CH_2OH$ $(3)$,and $-H$ $(4)$. The configuration is $R$.
For the second structure $(II)$:
Rotating the Fischer projection by $180^{\circ}$ in the plane of the paper does not change the configuration of the chiral centers.
After a $180^{\circ}$ rotation of structure $(II)$,it becomes identical to structure $(I)$.
Therefore,both structures represent the same molecule.

(C) The reaction of $HBr$ with $2-$pentene $(CH_3-CH=CH-CH_2-CH_3)$ is an electrophilic addition reaction.
According to Markovnikov's rule,the proton $(H^+)$ adds to the carbon atom of the double bond that already has more hydrogen atoms,or in this case,to form the more stable carbocation intermediate.
$2-$pentene is a symmetrical alkene with respect to the double bond position relative to the chain ends,but the two carbons of the double bond are not equivalent.
Protonation at $C_2$ gives a secondary carbocation at $C_3$ $(CH_3-CH^+-CH_2-CH_2-CH_3)$,and protonation at $C_3$ gives a secondary carbocation at $C_2$ $(CH_3-CH_2-CH^+-CH_2-CH_3)$.
Both carbocations are secondary and have similar stability.
Therefore,the bromide ion $(Br^-)$ can attack either carbocation,resulting in a mixture of $2-$bromopentane and $3-$bromopentane.

(C) Acetylene $(HC \equiv CH)$ is a terminal alkyne with acidic hydrogen atoms.
When passed through ammoniacal cuprous chloride $(Cu_2Cl_2)$,it forms a red precipitate of copper$(I)$ acetylide:
$HC \equiv CH + Cu_2Cl_2 \rightarrow CuC \equiv CCu \downarrow (\text{red ppt}) + 2HCl$.
Ethene $(CH_2=CH_2)$ does not react with ammoniacal $Cu_2Cl_2$ and passes through unchanged.
Therefore,the mixture can be separated using ammoniacal $Cu_2Cl_2$.

(C) The blackening of oil paintings is caused by the formation of $PbS$ (lead sulfide) due to atmospheric $H_2S$.
This can be restored by the action of $H_2 O_2$ (hydrogen peroxide),which oxidizes the black $PbS$ into white $PbSO_4$ (lead sulfate).
The chemical reaction is:
$PbS + 4H_2 O_2 \rightarrow PbSO_4 + 4H_2 O$

(A) Let $H$ be the horizontal component and $V$ be the vertical component of the earth's magnetic field.
Given that $H = \sqrt{3} V$.
The angle of dip $\theta$ is defined by the relation $\tan \theta = \frac{V}{H}$.
Substituting the given value of $H$ into the formula:
$\tan \theta = \frac{V}{\sqrt{3} V} = \frac{1}{\sqrt{3}}$.
Since $\tan 30^{\circ} = \frac{1}{\sqrt{3}}$,we have $\theta = 30^{\circ}$.

(C) Silicon is the second most abundant element in the Earth's crust,but it does not occur in the free state in nature.
It is always found in the combined state,primarily as silica $(SiO_2)$ and silicates.
Therefore,the statement that silicon occurs in the free state in nature is incorrect.

(C) In a $P_4$ molecule of white phosphorus,each phosphorus atom is $sp^3$ hybridized.
These four phosphorus atoms are linked to each other by single covalent bonds.
They are arranged at the four corners of a regular tetrahedron,with bond angles of $60^{\circ}$.

(D) When $Cl_2$ gas is passed through a hot and concentrated $NaOH$ solution,the reaction is:
$3Cl_2 + 6NaOH \rightarrow 5NaCl + NaClO_3 + 3H_2O$
In this reaction,the oxidation state of chlorine changes from $0$ in $Cl_2$ to $-1$ in $NaCl$ (reduction) and $+5$ in $NaClO_3$ (oxidation).
Since the same element is simultaneously oxidized and reduced,this is a disproportionation reaction.

(B) The correct answer is $HNO_2$.
In $HNO_2$ (nitrous acid),the oxidation state of nitrogen is $+3$.
Since the oxidation state of nitrogen ranges from $-3$ to $+5$,nitrogen in $HNO_2$ can be oxidized to $+5$ (acting as a reducing agent) or reduced to lower oxidation states (acting as an oxidizing agent).
Additionally,$NO_2^-$ (nitrite ion) acts as a ligand and is capable of forming complex compounds with transition metals.

(A) The atomic weight of nitrogen is $14 \ g/mol$. In an organic molecule,there must be at least one nitrogen atom present.
Given that the compound contains $20 \%$ nitrogen by mass.
Let the molecular weight be $M$.
Then,$\frac{14}{M} \times 100 = 20$.
Solving for $M$: $M = \frac{14 \times 100}{20} = 70 \ g/mol$.
Thus,the molecular weight is $70$.

(A) The initial $pH$ of $0.01 \ M \ HCl$ is $-\log(0.01) = 2$.
To decrease the $pH$,the concentration of $H^+$ ions must increase.
Option $A$: Adding $5 \ mL$ of $1 \ M \ HCl$ adds $5 \ mmol$ of $H^+$ to the existing $0.5 \ mmol$. The new concentration becomes $\frac{5.5 \ mmol}{55 \ mL} = 0.1 \ M$. The new $pH$ is $-\log(0.1) = 1$. Since $1 < 2$,the $pH$ decreases.
Option $B$: Adding $50 \ mL$ of $0.01 \ M \ HCl$ results in the same concentration $(0.01 \ M)$,so the $pH$ remains $2$.
Option $C$: Adding $50 \ mL$ of $0.002 \ M \ HCl$ dilutes the solution,decreasing the $H^+$ concentration and increasing the $pH$.
Option $D$: Adding $Mg$ reacts with $HCl$ to consume $H^+$ ions,increasing the $pH$.

(B) We have,$\cos 15^{\circ} \cdot \cos 7.5^{\circ} \cdot \sin 7.5^{\circ}$
$= \frac{1}{2} \cos 15^{\circ} (2 \sin 7.5^{\circ} \cos 7.5^{\circ})$
$= \frac{1}{2} \cos 15^{\circ} \sin(2 \times 7.5^{\circ})$
$= \frac{1}{2} \cos 15^{\circ} \sin 15^{\circ}$
$= \frac{1}{4} (2 \sin 15^{\circ} \cos 15^{\circ})$
$= \frac{1}{4} \sin(2 \times 15^{\circ})$
$= \frac{1}{4} \sin 30^{\circ}$
$= \frac{1}{4} \times \frac{1}{2} = \frac{1}{8}$

(A) Given parametric equations are $x = 5t^2 + 2$ and $y = 10t + 4$.
From the second equation,$10t = y - 4$,so $t = \frac{y - 4}{10}$.
Substituting this into the first equation: $x = 5\left(\frac{y - 4}{10}\right)^2 + 2$.
$x - 2 = 5 \cdot \frac{(y - 4)^2}{100} = \frac{(y - 4)^2}{20}$.
Thus,$(y - 4)^2 = 20(x - 2)$.
This is of the form $(y - k)^2 = 4a(x - h)$,where $(h, k) = (2, 4)$ is the vertex and $4a = 20$,so $a = 5$.
The focus of this parabola is $(h + a, k) = (2 + 5, 4) = (7, 4)$.

(C) The Rydberg formula for the hydrogen spectrum is given by $\Delta E = R h c \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right)$.
For the Paschen series,the electron transitions occur from higher energy levels to the third energy level.
Therefore,the value of $n_1$ is $3$ and $n_2$ can be any integer greater than $3$,i.e.,$n_2 = 4, 5, 6, \dots$.

(A) The total energy $E$ for $N$ moles of photons is given by $E = N_A \times h \times \nu$.
Here,$N_A = 6.022 \times 10^{23} \ mol^{-1}$,$h = 6.626 \times 10^{-34} \ J \cdot s$,and $\nu = 2500 \ s^{-1}$.
$E = (6.022 \times 10^{23}) \times (6.626 \times 10^{-34}) \times 2500 \ J$.
$E \approx 9.97 \times 10^{-7} \ J$.
Since $1 \ J = 10^7 \ erg$,$E \approx 9.97 \times 10^{-7} \times 10^7 \ erg = 9.97 \ erg \approx 10 \ erg$.

(A) The enthalpy change is defined as $\Delta H = \Delta E + \Delta(PV)$.
For a system where pressure $(P)$ is constant,the expression becomes $\Delta H = \Delta E + P \Delta V$.
Therefore,the relation is valid under the condition of constant pressure.

(B) The enthalpy of neutralization for $1 \ g$ equivalent of a strong acid with a strong base is constant at $13.7 \ kcal$.
Since the volumes and molarities are equal,let the volume be $V \ L$ and molarity be $M \ M$.
For $HCl$,the number of equivalents is $M \times V \times 1 = MV$.
For $H_2SO_4$,the number of equivalents is $M \times V \times 2 = 2MV$.
Since $H_2SO_4$ provides twice the number of equivalents of $H^+$ ions compared to $HCl$,the heat liberated by $H_2SO_4$ $(y)$ will be twice the heat liberated by $HCl$ $(x)$.
Therefore,$y = 2x$ or $x = \frac{y}{2}$.

(D) The second law of thermodynamics,specifically the Kelvin-Planck statement,asserts that it is impossible for any device that operates on a thermodynamic cycle to receive heat from a single thermal reservoir and produce a net amount of work.
In other words,heat cannot be completely converted into work in a cyclic process without leaving some effect on the surroundings.

(B) The relationship between the standard Gibbs free energy change $(\Delta G^0)$ and the equilibrium constant $(K)$ is given by the equation: $\Delta G^0 = -RT \ln K$.
Rearranging this equation to solve for $K$:
$\ln K = -\frac{\Delta G^0}{RT}$.
Taking the exponential of both sides:
$K = e^{-\Delta G^0 / RT}$.

(B) The given differential equation is $x \log x \frac{dy}{dx} + y = 2 \log x$.
Dividing both sides by $x \log x$,we get:
$\frac{dy}{dx} + \frac{1}{x \log x} y = \frac{2}{x}$.
This is a linear differential equation of the form $\frac{dy}{dx} + P(x)y = Q(x)$,where $P(x) = \frac{1}{x \log x}$.
The integrating factor $(IF)$ is given by $IF = e^{\int P(x) dx}$.
$IF = e^{\int \frac{1}{x \log x} dx}$.
Let $u = \log x$,then $du = \frac{1}{x} dx$.
$IF = e^{\int \frac{1}{u} du} = e^{\log u} = u = \log x$.
Thus,the integrating factor is $\log x$.

(C) Total number of ways to choose $3$ numbers from $20$ is given by $^{20}C_3 = \frac{20 \times 19 \times 18}{3 \times 2 \times 1} = 1140$.
Favorable cases are sets of consecutive numbers: $(1, 2, 3), (2, 3, 4), \dots, (18, 19, 20)$.
There are $18$ such sets.
Therefore,the required probability is $\frac{18}{1140} = \frac{3}{190}$.

(B) The reaction between an alcohol $(R-OH)$ and a Grignard reagent $(R'-MgX)$ is an acid-base reaction.
Alcohols contain a weakly acidic hydrogen atom attached to an oxygen atom.
Grignard reagents $(R'-MgX)$ contain a highly nucleophilic/basic alkyl group $(R'^- )$.
The $R'^- $ group abstracts the acidic proton from the alcohol to form an alkane $(R'-H)$ and an alkoxide salt $(R-OMgX)$.
The reaction is: $R-OH + R'-MgX \rightarrow R-OMgX + R'-H$ (Alkane).

(C) Phenol is a weaker acid than carbonic acid $(H_2CO_3)$.
Therefore,it cannot displace carbonic acid from its salts like $Na_2CO_3$.
Consequently,phenol does not liberate $CO_2$ gas from $Na_2CO_3$ solution.
Thus,the statement in option $C$ is incorrect.

(A) In the Cannizzaro reaction,an aldehyde lacking an $\alpha$-hydrogen undergoes disproportionation in the presence of a concentrated base to form an alcohol and a carboxylic acid salt. No new $C-C$ bond is formed in this process.
For example: $2HCHO + 50\% NaOH \rightarrow CH_3OH + HCOONa$.

(D) $2$-pentanone $(CH_3COCH_2CH_2CH_3)$ contains a $CH_3CO-$ group,which makes it capable of undergoing the iodoform test with $I_2 / NaOH$.
$3$-pentanone $(CH_3CH_2COCH_2CH_3)$ does not contain a $CH_3CO-$ group and therefore does not give the iodoform test.
Since the iodoform test is not listed among the options $(A)$,$(B)$,or $(C)$,the correct choice is $(D)$.

(D) The term $carbylamine$ is another name for an isocyanide or isonitrile,which has the general chemical formula $RNC$.
In the carbylamine reaction,primary amines react with chloroform $(CHCl_3)$ and an alcoholic base $(KOH)$ to form isocyanides $(RNC)$,which are characterized by their foul smell.

(A) The double helical structure of $DNA$ is primarily stabilized by $hydrogen$ bonds between the complementary nitrogenous bases ($Adenine$ with $Thymine$ and $Guanine$ with $Cytosine$).
These $hydrogen$ bonds hold the two strands of the $DNA$ double helix together.

(C) If $n_t$ is the number of radioactive atoms present at time $t$,radioactive decay follows the law:
$n_t = n_0 e^{-\lambda t}$
Taking the natural logarithm on both sides:
$\ln n_t = \ln n_0 - \lambda t$
Differentiating with respect to time $t$:
$\frac{d}{d t} (\ln n_t) = -\lambda$
Since $\lambda$ (decay constant) is a constant,the expression $\frac{d \ln n_t}{d t}$ is constant.

(C) For a reaction of order $n$,the half-life $T_{1/2}$ is related to the initial concentration $a_0$ by the expression: $T_{1/2} \propto \frac{1}{a_0^{n-1}}$.
From the given graph,$T_{1/2}$ is directly proportional to $\frac{1}{a_0}$.
This means $T_{1/2} \propto (a_0)^{-1}$.
Comparing the exponents of $a_0$,we get $n - 1 = 1$,which implies $n = 2$.
Therefore,the order of the reaction is $2$.

(A) For a first order reaction,the rate equation is $t = \frac{2.303}{k} \log \frac{100}{100 - x}$.
For $10 \%$ completion $(x = 10)$: $20 = \frac{2.303}{k} \log \frac{100}{90} = \frac{2.303}{k} \log (10/9)$ $(i)$.
For $19 \%$ completion $(x = 19)$: $t = \frac{2.303}{k} \log \frac{100}{100 - 19} = \frac{2.303}{k} \log \frac{100}{81} = \frac{2.303}{k} \log (10/9)^2 = 2 \times \frac{2.303}{k} \log (10/9)$ (ii).
Dividing equation (ii) by equation $(i)$:
$\frac{t}{20} = \frac{2 \times \frac{2.303}{k} \log (10/9)}{\frac{2.303}{k} \log (10/9)} = 2$.
Therefore,$t = 20 \times 2 = 40 \ min$.

(B) Radioactive decay follows first-order kinetics. The decay constant $k = \frac{0.693}{t_{1/2}} = \frac{0.693}{10} \ h^{-1}$.
Using the first-order integrated rate equation: $k = \frac{2.303}{t} \log \frac{N_0}{N_t}$.
Given $N_0 = 1 \ g$-atom,$t = 4 \ h$,and $k = 0.0693 \ h^{-1}$.
$0.0693 = \frac{2.303}{4} \log \frac{1}{N_t} \implies \log \frac{1}{N_t} = \frac{0.0693 \times 4}{2.303} \approx 0.12036$.
$\log N_t = -0.12036 = \bar{1}.87964 \implies N_t = 10^{-0.12036} \approx 0.7579 \ g$-atoms.
Number of atoms $= N_t \times N_A = 0.7579 \times 6.022 \times 10^{23} \approx 4.56 \times 10^{23} \ \text{atoms}$.

(D) In the complex $\left[Cu(NH_3)_4\right]^{2+}$,the central metal ion is $Cu^{2+}$.
$Cu^{2+}$ has the electronic configuration $[Ar] 3d^9$.
During the formation of the complex,one electron from the $3d$ orbital is promoted to the $4p$ orbital to allow for $dsp^2$ hybridization.
This results in a square planar geometry for the complex.

(C) In the industrial Bayer process,the sodium aluminate solution is cooled and seeded with freshly precipitated aluminium hydroxide $(Al(OH)_3)$.
This seeding induces the precipitation of $Al(OH)_3$ from the supersaturated solution.
Additionally,carbon dioxide $(CO_2)$ is often bubbled through the solution to neutralize the excess sodium hydroxide,which further promotes the precipitation of aluminium hydroxide.
Therefore,both methods are commonly practised in industrial settings.

(B) In a group,as we move downwards,the metallic character increases and the non-metallic character decreases.
Consequently,the acidic nature of oxides decreases down the group.
Among the given oxides of Group $15$ elements,$P$ is at the top of the group,followed by $As$,$Sb$,and $Bi$.
Therefore,$P_2O_5$ is the most acidic oxide.

(A) The catalyst used for the polymerization of olefins (such as ethylene) is the Ziegler-Natta catalyst.
It is typically a mixture of titanium tetrachloride $(TiCl_4)$ or titanium trichloride $(TiCl_3)$ and an organoaluminum compound like triethylaluminum $((C_2H_5)_3Al)$.

(D) Orlon is also known as polyacrylonitrile $(PAN)$.
The monomer used for the preparation of orlon is acrylonitrile,which has the chemical formula $CH_2=CH-CN$.

(B) The emission of one $\alpha$ particle $({}_{2}^{4}He)$ decreases the mass number by $4$ and the atomic number by $2$.
The emission of one $\beta$ particle $({}_{-1}^{0}e)$ does not change the mass number but increases the atomic number by $1$.
For the product ${}_{89}^{227}Ac$ from ${}_{92}^{235}U$:
Change in mass number $= 235 - 227 = 8$,which corresponds to $2 \alpha$ particles $(2 \times 4 = 8)$.
Change in atomic number $= 92 - 89 = 3$.
With $2 \alpha$ particles,the atomic number decreases by $2 \times 2 = 4$. To reach $89$,we need to increase the atomic number by $1$ using one $\beta$ particle $(92 - 4 + 1 = 89)$.
Thus,the decay involves $2 \alpha$ and $1 \beta$ particle,resulting in ${}_{89}^{227}Ac$.