WBJEE 2015 Chemistry Question Paper with Answer and Solution

(B) Let $y$ be the depth of a hole below the water surface. The velocity of efflux is $v = \sqrt{2gy}$.
The height of the hole from the ground is $H_{hole} = h - y + h/2 = \frac{3h}{2} - y$.
The time taken for water to reach the plane $PQ$ is $t = \sqrt{\frac{2 H_{hole}}{g}} = \sqrt{\frac{2(3h/2 - y)}{g}} = \sqrt{\frac{3h - 2y}{g}}$.
The horizontal range $x$ is given by $x = v \cdot t = \sqrt{2gy} \cdot \sqrt{\frac{3h - 2y}{g}} = \sqrt{2y(3h - 2y)} = \sqrt{6hy - 4y^2}$.
To maximize $x$,we differentiate with respect to $y$ and set it to zero: $\frac{dx}{dy} = \frac{1}{2\sqrt{6hy - 4y^2}} \cdot (6h - 8y) = 0$.
This gives $6h - 8y = 0$,or $y = \frac{3h}{4}$.
The depth $y$ is measured from the top. The heights of the holes from the base are $0, h/4, h/2, 3h/4$.
Thus,the depths $y$ from the top are $h, 3h/4, h/2, h/4$ for holes $1, 2, 3, 4$ respectively.
The range is maximum when $y = 3h/4$,which corresponds to hole number $2$.

(A) The number of lone pairs on the central atoms are calculated as follows:
$1$. In $H_{2}O$,Oxygen has $6$ valence electrons,$2$ are used in bonding with $H$,leaving $4$ electrons or $2$ lone pairs.
$2$. In $SnCl_{2}$,Tin $(Sn)$ has $4$ valence electrons,$2$ are used in bonding with $Cl$,leaving $2$ electrons or $1$ lone pair.
$3$. In $PCl_{3}$,Phosphorus $(P)$ has $5$ valence electrons,$3$ are used in bonding with $Cl$,leaving $2$ electrons or $1$ lone pair.
$4$. In $XeF_{2}$,Xenon $(Xe)$ has $8$ valence electrons,$2$ are used in bonding with $F$,leaving $6$ electrons or $3$ lone pairs.
Thus,the number of lone pairs are $2, 1, 1$ and $3$ respectively.

(C) To determine the number of $sp$-hybridised carbon atoms,we look for carbons involved in two double bonds (allene type) or one triple bond.
$1$. The central carbon in the $CH_2=C=CH-$ group is $sp$-hybridised.
$2$. The carbon in the $-C\equiv CH$ group is $sp$-hybridised.
$3$. The terminal carbon in the $-C\equiv CH$ group is $sp$-hybridised.
$4$. The carbon in the $-C\equiv N$ group is $sp$-hybridised.
Thus,there are $4$ $sp$-hybridised carbon atoms in the given structure.

(A) For the reaction,$2 SO_{2(g)} + O_{2(g)} \rightleftharpoons 2 SO_{3(g)}$.
Given that,$\Delta G^{\circ} = -690.9 R$ and $T = 300 \ K$.
The relationship between standard Gibbs free energy change and equilibrium constant is $\Delta G^{\circ} = -RT \ln K$.
Substituting the values: $-690.9 R = -R \times 300 \times \ln K$.
Dividing both sides by $-R$: $690.9 = 300 \times \ln K$.
$\ln K = \frac{690.9}{300} = 2.303$.
Since $\ln K = 2.303 \log K$,we have $2.303 \log K = 2.303$.
Therefore,$\log K = 1$,which implies $K = 10^1 = 10$.
The unit of $K$ is $(atm)^{\Delta n}$,where $\Delta n = 2 - (2 + 1) = -1$.
Thus,$K = 10 \ atm^{-1}$.

(A) The electronic configurations of the given elements are:
$Be (Z=4): 1s^2 2s^2$
$B (Z=5): 1s^2 2s^2 2p^1$
$Mg (Z=12): [Ne] 3s^2$
$Al (Z=13): [Ne] 3s^2 3p^1$
To find the second ionisation potential,we look at the configuration of the $+1$ ions:
$Be^+: 1s^2 2s^1$
$B^+: 1s^2 2s^2$
$Mg^+: [Ne] 3s^1$
$Al^+: [Ne] 3s^2$
Removing the second electron from $B^+$ $(2s^2)$ is the most difficult because it involves removing an electron from a completely filled $2s$ subshell,which is highly stable. Therefore,$B$ has the maximum second ionisation potential.

(B) The reaction of methyltrichloroacetate $(Cl_{3}CCO_{2}Me)$ with sodium methoxide $(NaOMe)$ involves the nucleophilic attack of the methoxide ion $(^-OMe)$ on the carbonyl carbon of the ester.
This leads to the formation of a tetrahedral intermediate,which then collapses to eliminate the trichloromethyl carbanion $(^-CCl_{3})$.
The trichloromethyl carbanion $(^-CCl_{3})$ is unstable and subsequently loses a chloride ion $(Cl^-)$ to generate dichlorocarbene $(:CCl_{2})$.
Therefore,the final reactive intermediate generated in this process is a carbene.

(C) To determine the aromaticity,we use the Huckel rule: a cyclic,planar,fully conjugated system with $(4n+2)$ $\pi$-electrons is aromatic,where $n$ is an integer $(0, 1, 2, \dots)$.
$(i)$ Cyclopropenyl chloride reacts with $SbCl_5$ to form the cyclopropenyl cation,which has $2$ $\pi$-electrons $(n=0)$. This is aromatic.
$(ii)$ Cyclopentadiene reacts with sodium metal to form the cyclopentadienyl anion,which has $6$ $\pi$-electrons $(n=1)$. This is aromatic.
$(iii)$ $7$-Bromocyclohepta-$1,3,5$-triene reacts with water to form the tropylium cation,which has $6$ $\pi$-electrons $(n=1)$. This is aromatic.
$(iv)$ Cyclopentadienylamine reacts with $HNO_2$ to form a $4$ $\pi$-electron system (cyclopentadienyl cation),which is anti-aromatic (not aromatic).
Therefore,there are $3$ aromatic species generated.

(C) The enantiomeric excess $(ee)$ is defined as the difference between the percentages of the two enantiomers in a mixture.
$ee = |\% \text{ of enantiomer } 1 - \% \text{ of enantiomer } 2|$
Given the percentages are $85 \%$ and $15 \%$,the calculation is:
$ee = 85 \% - 15 \% = 70 \%$

(C) In the Lassaigne's test,the sodium extract containing $NaCN$ is treated with $FeSO_4$ followed by $FeCl_3$ and acidified with $H_2SO_4$.
The chemical reactions are as follows:
$Na + C + N \rightarrow NaCN$
$FeSO_4 + 2NaCN \rightarrow Fe(CN)_2 + Na_2SO_4$
$Fe(CN)_2 + 4NaCN \rightarrow Na_4[Fe(CN)_6]$ (Sodium ferrocyanide)
$3Na_4[Fe(CN)_6] + 4FeCl_3 \rightarrow Fe_4[Fe(CN)_6]_3 + 12NaCl$
The compound $Fe_4[Fe(CN)_6]_3$ is known as ferric ferrocyanide,which gives the characteristic Prussian blue colour.

(C) The reaction of $HBr$ with $2-$methyl$-1,3-$butadiene (isoprene) involves electrophilic addition to the conjugated diene system.
$H^{+}$ adds to the terminal carbon to form the most stable carbocation,which is a resonance-stabilized allylic carbocation.
At room temperature,the thermodynamic product,which is the more substituted alkene,is the major product.
This corresponds to the $1,4-$addition product,which is $1-$bromo$-3-$methyl$-2-$butene.

(B) The reaction of $1,4-$dimethylbenzene ($p-$xylene) with anhydrous $AlCl_{3}$ and $HCl$ is an example of an isomerization reaction.
Under these acidic conditions,the methyl groups undergo rearrangement to form the more thermodynamically stable isomer.
$1,3-$dimethylbenzene ($m-$xylene) is more stable than $1,4-$dimethylbenzene ($p-$xylene) due to the reduced steric hindrance and electronic factors in the meta-substituted product.
Therefore,the reaction produces $1,3-$dimethylbenzene.

(D) Nuclear iodination of aromatic compounds is a reversible electrophilic substitution reaction.
$I_{2}$ is a weak electrophile and the reaction produces $HI$ as a byproduct,which is a strong reducing agent and can reduce the iodinated product back to the starting material.
To drive the reaction forward,an oxidizing agent like $HNO_{3}$ is used to oxidize $HI$ to $I_{2}$,thereby preventing the reverse reaction.
$2HI + 2HNO_{3} \rightarrow I_{2} + 2NO_{2} + 2H_{2}O$.
Thus,$I_{2} / HNO_{3}$ is the best reagent for this purpose.

(B) The synthesis of $1$-butyl-$4$-nitrobenzene involves the following steps:
$1$. Friedel-Crafts acylation of benzene with butanoyl chloride $(CH_3CH_2CH_2COCl)$ in the presence of $AlCl_3$ gives $1$-phenylbutan-$1$-one.
$2$. Clemmensen reduction of the ketone using $Zn/Hg$ and $HCl$ with heat reduces the carbonyl group to a methylene group,yielding butylbenzene.
$3$. Nitration of butylbenzene using a mixture of concentrated $HNO_3$ and $H_2SO_4$ introduces the nitro group at the para position because the alkyl group is an ortho/para-directing group.

(D) Given: $pH = 5.74$,$pK_a = 4.74$.
Using the Henderson-Hasselbalch equation: $pH = pK_a + \log \frac{[Salt]}{[Acid]}$.
Substituting the values: $5.74 = 4.74 + \log \frac{[CH_3COONa]}{[CH_3COOH]}$.
$1 = \log \frac{[CH_3COONa]}{[CH_3COOH]}$.
Therefore,$\frac{[CH_3COONa]}{[CH_3COOH]} = 10^1 = 10$.
Since the concentrations of both solutions are equal $(0.1 \ N)$,the ratio of their volumes is equal to the ratio of their concentrations: $\frac{V_{acid}}{V_{salt}} = \frac{[CH_3COOH]}{[CH_3COONa]} = \frac{1}{10}$.

(A) Alkaline earth metals impart characteristic colours to the Bunsen flame due to the excitation of electrons to higher energy levels.
$A$. Calcium $(Ca)$ imparts a brick red colour to the flame $(p)$.
$B$. Strontium $(Sr)$ imparts a crimson red colour to the flame $(r)$.
$C$. Barium $(Ba)$ imparts an apple green colour to the flame $(q)$.
Therefore,the correct matching sequence is $A-p, B-r, C-q$.

(A) The solubility of metal chlorides in water depends on their lattice energy and hydration energy.
$NaCl$ is highly soluble due to high hydration energy.
$HgCl_{2}$ and $CuCl_{2}$ are soluble in water.
$Hg_{2}Cl_{2}$ (mercurous chloride),$CuCl$ (cuprous chloride),and $AgCl$ (silver chloride) are known to be insoluble in water.

(A) In an aqueous alkaline medium,the reduction of the hydroperoxide ion $(HO_2^-)$ by two electrons produces hydroxide ions $(OH^{-})$.
The balanced half-reaction is: $HO_2^- + H_2O + 2e^- \rightarrow 3OH^{-}$.

(A) The basic character of metal oxides depends on the electropositivity of the metal.
As we move down a group,the metallic character increases,and as we move across a period from left to right,the metallic character decreases.
$K$ is an alkali metal (Group $1$),while $Ba$,$Ca$,and $Mg$ are alkaline earth metals (Group $2$).
Within Group $2$,the metallic character increases as we go down the group: $Mg < Ca < Ba$.
Since $K$ is more electropositive than the Group $2$ metals,$K_2O$ is the most basic.
Therefore,the correct decreasing order of basic character is $K_2O > BaO > CaO > MgO$.

(B) The formula for $RMS$ velocity is $v_{rms} = \sqrt{\frac{3RT}{M}}$.
Since $v_{rms} \propto \sqrt{T}$,we have $\frac{v_1}{v_2} = \sqrt{\frac{T_1}{T_2}}$.
Given $T_1 = 27^{\circ} C = 300 \ K$ and $v_1 = 1000 \ m / s$.
Given $T_2 = 600 \ K$.
Substituting the values: $\frac{1000}{v_2} = \sqrt{\frac{300}{600}} = \sqrt{\frac{1}{2}} = \frac{1}{1.414}$.
Therefore,$v_2 = 1000 \times 1.414 = 1414 \ m / s$.

(B) gas can be liquefied at a temperature $T$ only when it is below its critical temperature,$T_{c}$.
At temperatures below $T_{c}$,the gas can be liquefied by applying sufficient pressure,specifically when the pressure $p$ is greater than the critical pressure,$p_{c}$.
Therefore,the condition for liquefaction is $T < T_{c}$ and $p > p_{c}$.

(B) Surface tension $(\gamma)$ is defined as force per unit length: $\gamma = \frac{F}{l} = \frac{N}{m} = \frac{kg \ m \ s^{-2}}{m} = kg \ s^{-2}$.
Coefficient of viscosity $(\eta)$ is defined by Newton's law of viscosity: $F = \eta A \frac{dv}{dx}$,which gives $\eta = \frac{F}{A (dv/dx)}$.
The units are $\frac{N}{m^2 (s^{-1})} = N \ s \ m^{-2} = (kg \ m \ s^{-2}) \ s \ m^{-2} = kg \ m^{-1} \ s^{-1}$.
Thus,the units are $kg \ s^{-2}$ and $kg \ m^{-1} \ s^{-1}$ respectively.

(C) For the reaction $X_2Y_{4(l)} \rightarrow 2XY_{2(g)}$
$\Delta n_g = \text{number of gaseous products} - \text{number of gaseous reactants} = 2 - 0 = 2$
Given,$\Delta U = 2 \ kcal = 2000 \ cal$,$\Delta S = 20 \ cal \ K^{-1}$,$T = 300 \ K$,and $R = 2 \ cal \ K^{-1} \ mol^{-1}$
Using the relation $\Delta H = \Delta U + \Delta n_g RT$
$\Delta H = 2000 + (2 \times 2 \times 300) = 2000 + 1200 = 3200 \ cal$
Now,using the Gibbs free energy equation $\Delta G = \Delta H - T \Delta S$
$\Delta G = 3200 - (300 \times 20) = 3200 - 6000 = -2800 \ cal$

(A) The reaction of a Grignard reagent $(RMgBr)$ with triethyl orthoformate $(HC(OEt)_3)$ proceeds via a nucleophilic substitution mechanism.
$1$. The nucleophilic $R^-$ group from the Grignard reagent attacks the electrophilic carbon of the triethyl orthoformate,displacing one ethoxide group $(-OEt)$ to form an acetal intermediate,$RCH(OEt)_2$.
$2$. Upon acidic hydrolysis $(H_3O^{+})$,the acetal intermediate undergoes hydrolysis to yield an aldehyde $(RCHO)$ as the final product $P$.

(C) The given reaction is an example of an intramolecular Cannizzaro reaction. In this reaction,phthalaldehyde reacts with a base $(OH^-)$ to undergo disproportionation,where one aldehyde group is reduced to a primary alcohol $(-CH_2OH)$ and the other is oxidized to a carboxylate ion $(-COO^-)$. The final product is the ortho-hydroxymethylbenzoate ion.

(C) The rate law can be expressed as $Rate = k[A]^x[B]^y$.
When the concentration of $A$ is doubled,the rate doubles,which implies $2^x = 2$,so $x = 1$.
When both concentrations are increased by four times,the rate increases by four times: $4^x \times 4^y = 4$.
Substituting $x = 1$,we get $4^1 \times 4^y = 4$,which means $4^y = 1$,so $y = 0$.
The total order of the reaction is $x + y = 1 + 0 = 1$.

(D) According to the Arrhenius equation,$k = A e^{-E_a / RT}$,the rate constant $k$ increases with temperature $T$.
This is primarily because the fraction of molecules possessing energy equal to or greater than the activation energy $(E_a)$ increases significantly with temperature.
Additionally,the collision frequency of reactant molecules also increases with temperature,which contributes to the increase in the rate of reaction.

(C) Optical isomerism is exhibited by coordination compounds that lack a plane of symmetry and a center of symmetry.
$trans-[CrCl_2(ox)_2]^{3-}$ has a plane of symmetry,making it superimposable on its mirror image,and thus it is optically inactive.
$cis-[CrCl_2(ox)_2]^{3-}$,$[Co(en)_3]^{3+}$,and $[Co(ox)(en)_2]^{+}$ lack a plane of symmetry and exhibit optical isomerism.

(A) The brown ring test involves the reaction of cold $FeSO_{4}$ solution with $NO$ gas to form the brown-colored complex $[Fe(H_{2}O)_{5}(NO)]SO_{4}$.
$FeSO_{4} + NO + 5H_{2}O \rightarrow [Fe(H_{2}O)_{5}(NO)]SO_{4}$
In this complex,iron is in the $+1$ oxidation state ($d^7$ configuration). The $NO$ ligand acts as $NO^+$,making the iron center $Fe^+$. The complex exhibits a magnetic moment of $3.89 \ BM$,which corresponds to $3$ unpaired electrons,confirming it is paramagnetic.

(A) The cell reaction is: $2Ag_{(s)} + Zn^{2+}(0.1 \ M) \longrightarrow 2Ag^{+}(0.1 \ M) + Zn_{(s)}$
Using the Nernst equation at $298 \ K$:
$E_{cell} = E^{\circ}_{cell} - \frac{0.0591}{n} \log \frac{[Ag^{+}]^{2}}{[Zn^{2+}]}$
Here,$n = 2$,$[Ag^{+}] = 0.1 \ M$,and $[Zn^{2+}] = 0.1 \ M$.
$E_{cell} = -1.562 - \frac{0.0591}{2} \log \frac{(0.1)^{2}}{0.1}$
$E_{cell} = -1.562 - 0.02955 \log(0.1)$
$E_{cell} = -1.562 - 0.02955 \times (-1)$
$E_{cell} = -1.562 + 0.02955 = -1.53245 \ V \approx -1.532 \ V$

(A) The relationship between equivalent conductance $(\lambda_{eq})$,specific conductance $(K)$,and concentration $(C)$ is given by:
$\lambda_{eq} = \frac{K \times 1000}{C}$
Rearranging the formula to find the ratio of equivalent conductance to specific conductance:
$\frac{\lambda_{eq}}{K} = \frac{1000}{C}$
Given that the concentration $C = 0.01 \ N$:
$\frac{\lambda_{eq}}{K} = \frac{1000}{0.01} = 10^5 \ cm^3 \ eq^{-1}$
Therefore,the correct option is $A$.

(A) In the extraction of gold,the metal is leached with a dilute solution of $NaCN$ in the presence of air (oxygen) to form the complex $X = [Au(CN)_2]^-$.
Then,gold is recovered by displacement with zinc,which forms the complex $Y = [Zn(CN)_4]^{2-}$.
The chemical equations are:
$4Au + 8CN^- + 2H_2O + O_2 \rightarrow 4[Au(CN)_2]^- + 4OH^-$
$2[Au(CN)_2]^- + Zn \rightarrow [Zn(CN)_4]^{2-} + 2Au$
Thus,$X$ is $[Au(CN)_2]^-$ and $Y$ is $[Zn(CN)_4]^{2-}$.

(A) Roasted copper pyrite on smelting with sand produces $FeSiO_{3}$ as fusible slag and $Cu_{2}S$ as matte.
For removing the gangue,silica $(SiO_{2})$ is added as a flux to react with iron oxide $(FeO)$ formed during the roasting process to form iron silicate $(FeSiO_{3})$ slag.
The chemical reactions are:
$2FeS + 3O_{2} \longrightarrow 2FeO + 2SO_{2}$
$FeO + SiO_{2} \longrightarrow FeSiO_{3}$
Copper matte mainly consists of a mixture of $Cu_{2}S$ and $FeS$.

(A) The reaction involves the treatment of $2$-bromobenzyl chloride with ammonia $(NH_3)$ in ethanol $(EtOH)$. This is a nucleophilic substitution reaction $(S_N2)$ where the nucleophile $NH_3$ attacks the electrophilic carbon of the $-CH_2Cl$ group,displacing the chloride ion $(Cl^-)$. The bromine atom attached to the benzene ring is relatively inert towards nucleophilic substitution under these conditions. Therefore,the $-CH_2Cl$ group is converted into a $-CH_2NH_2$ group,resulting in the formation of $2$-bromobenzylamine.

(A) Alanine is a neutral amino acid with an isoelectric point $(pI)$ of approximately $6.0$.
In an acidic medium ($pH$ $2-4$),which is below the $pI$,the amino group becomes protonated,and the molecule exists as a cation: $H_{3}N^{+}CH(CH_{3})COOH$ (Structure $I$).
In a basic medium ($pH$ $9-11$),which is above the $pI$,the carboxyl group becomes deprotonated,and the molecule exists as an anion: $H_{2}NCH(CH_{3})COO^{-}$ (Structure $II$).
Therefore,the correct pair is $I$ and $II$.

(A) When sulphuryl chloride $(SO_{2}Cl_{2})$ reacts with white phosphorus $(P_{4})$,it produces phosphorus pentachloride $(PCl_{5})$ and sulphur dioxide $(SO_{2})$.
The balanced chemical equation for this reaction is:
$P_{4} + 10 SO_{2}Cl_{2} \longrightarrow 4 PCl_{5} + 10 SO_{2}$

(A) The ionisation potential of noble gases decreases down the group as the atomic size increases.
Because of the lower ionisation energy of heavier noble gases like $Xe$,they can be oxidised by highly electronegative elements like fluorine.
Therefore,only the heavier noble gases form such compounds,and this occurs because they form compounds with electronegative ligands like $F$.
While the octet rule provides stability,it is not the reason why $Xe$ forms fluorides; rather,it is the ease of ionisation of heavier noble gases.

(A) The sequence of reactions is as follows:
$1$. $2AgNO_{3} + Na_{2}S_{2}O_{3} \rightarrow Ag_{2}S_{2}O_{3} \downarrow (X) + 2NaNO_{3}$
$2$. $Ag_{2}S_{2}O_{3} + 3Na_{2}S_{2}O_{3} \rightarrow 2Na_{3}[Ag(S_{2}O_{3})_{2}] (Y)$
$3$. $2Na_{3}[Ag(S_{2}O_{3})_{2}] + H_{2}O \xrightarrow{\Delta} Ag_{2}S (Z) + Na_{2}SO_{4} + 2Na_{2}S_{2}O_{3} + H_{2}SO_{4}$
Thus,$X = Ag_{2}S_{2}O_{3}$,$Y = Na_{3}[Ag(S_{2}O_{3})_{2}]$,and $Z = Ag_{2}S$.

(A) Given that,
Mass of a single $Ag$-atom $= m$.
$Ag$ metal crystallises in an $fcc$ lattice.
The number of atoms per unit cell in an $fcc$ lattice is $Z = 4$.
Therefore,the total mass of the unit cell $= Z \times m = 4m$.
The volume of the unit cell with edge length $a$ is $V = a^{3}$.
Density $(\rho) = \frac{\text{Mass of unit cell}}{\text{Volume of unit cell}} = \frac{4m}{a^{3}}$.

(B) The osmotic pressure $(\pi)$ is given by the equation: $\pi = CRT.$
Comparing this with the equation of a straight line $y = mx,$ where $y = \pi,$ $x = C,$ and the slope $m = RT.$
Given that the slope is $291 \ R,$
$RT = 291 \ R$
$T = 291 \ K$
To convert the temperature to Celsius: $T(^{\circ} C) = 291 - 273 = 18^{\circ} C.$

(A) The atomic number of cerium $(Ce)$ is $58$.
The ground state electronic configuration of $Ce$ is $[Xe] 4f^{1} 5d^{1} 6s^{2}$.
To form the $Ce^{3+}$ ion,three electrons are removed: two from the $6s$ orbital and one from the $5d$ orbital.
Thus,the electronic configuration of $Ce^{3+}$ is $[Xe] 4f^{1}$.

(C) Fog is a type of colloid where a liquid is dispersed in a gas.
Therefore,the dispersed phase is $liquid$ and the dispersion medium is $gas$.