WBJEE 2012 Chemistry Question Paper with Answer and Solution

(C) The structure of $but-2-ene$ is $CH_3-CH=CH-CH_3$.
In this molecule:
$C_1$ is $sp^3$ hybridized (attached to $3$ $H$ atoms and $C_2$).
$C_2$ is $sp^2$ hybridized (attached to $H$,$C_1$,and double-bonded to $C_3$).
$C_3$ is $sp^2$ hybridized.
$C_4$ is $sp^3$ hybridized.
Therefore,the $C_1-C_2$ bond is formed by the overlap of $sp^3$ orbital of $C_1$ and $sp^2$ orbital of $C_2$,making it a $sp^3-sp^2$ $\sigma$-bond.

(C) In $POCl_{3}$,the central phosphorus atom is bonded to one oxygen atom via a double bond and three chlorine atoms via single bonds.
Total number of electron pairs around the central atom = $4$ (three $P-Cl$ $\sigma$-bonds and one $P=O$ bond,where the double bond counts as one region of electron density for hybridization).
Since there are $4$ bonding pairs and no lone pairs,the hybridization is $sp^{3}$.
Therefore,the hybridization is $sp^{3}$ and the number of lone pairs is $0$.

(B) The total number of electrons in $B_2$ is $5 + 5 = 10 \ e^-$.
According to Molecular Orbital Theory,the electronic configuration is: $\sigma 1s^2, \sigma^* 1s^2, \sigma 2s^2, \sigma^* 2s^2, \pi 2p_x^1, \pi 2p_y^1$.
Due to the presence of $2$ unpaired electrons in the $\pi 2p$ bonding molecular orbitals,$B_2$ exhibits paramagnetic behaviour.

(A) In the $CO$ molecule,the electronegativity difference suggests a dipole moment,but the actual dipole moment is very small $(0.11 \ D)$.
This is because the $\sigma$-electron drift from $C$ to $O$ (due to the lone pair on $C$) is almost completely nullified by the back-donation of $\pi$-electrons from $O$ to $C$ through $\pi$-bonding.
Thus,the net dipole moment is nearly zero,making it practically non-polar.

(C) The equilibrium constant ($K_c$ or $K_p$) is a characteristic value for a given reaction at a specific temperature. It depends only on the temperature and is independent of the initial concentrations of the reactants or products,pressure,or the presence of a catalyst. Therefore,if the concentration of any reactant is doubled,the equilibrium constant remains unchanged.

(A) The given species $Ca^{2+}$,$Cl^{-}$,and $S^{2-}$ are isoelectronic,meaning they all contain $18$ electrons.

Species	Atomic number $(Z)$	Number of electrons
$Ca^{2+}$	$20$	$18$
$Cl^{-}$	$17$	$18$
$S^{2-}$	$16$	$18$

For isoelectronic species,the ionic radius decreases as the nuclear charge (atomic number) increases. Since the atomic numbers are $Z(S^{2-}) = 16$,$Z(Cl^{-}) = 17$,and $Z(Ca^{2+}) = 20$,the order of increasing radius is $Ca^{2+} < Cl^{-} < S^{2-}$.

(C) In acidic medium,the reduction reaction is:
$Cr_{2}O_{7}^{2-} + 14H^{+} + 6e^{-} \longrightarrow 2Cr^{3+} + 7H_{2}O$
Here,the oxidation state of $Cr$ changes from $+6$ to $+3$.
Since there are two $Cr$ atoms in $K_{2}Cr_{2}O_{7}$,the total change in oxidation state is $2 \times (6 - 3) = 6$.
Therefore,the n-factor is $6$.
Equivalent weight = $\frac{\text{Molecular weight}}{\text{n-factor}} = \frac{M}{6}$.

(D) When $H_{2}O_{2}$ reacts with an acidified solution of $K_{2}Cr_{2}O_{7}$,it forms chromium pentoxide $(CrO_{5})$.
The reaction is: $K_{2}Cr_{2}O_{7} + H_{2}SO_{4} + 4H_{2}O_{2} \rightarrow K_{2}SO_{4} + 5H_{2}O + 2CrO_{5}$.
$CrO_{5}$ is unstable in aqueous solution but is stabilized in ether,which gives a characteristic blue color to the ethereal layer.

\((m+1)\) vernier division \(=m\) main scale division One division on vernier scale \(=\left(\frac{m}{m+1}\right)\) division on main scale Vernier constant \(=\left(1-\frac{m}{m+1}\right) d=\frac{d}{m+1}\)

(B) The stability of carbocations is determined by resonance and inductive effects.
$(I)$ $Ph_{2}\stackrel{+}{C} CH_{2} Me$ is a $3^{\circ}$ benzylic carbocation stabilized by two phenyl rings.
$(II)$ $Ph CH_{2} CH_{2} \stackrel{+}{C} HPh$ is a $2^{\circ}$ benzylic carbocation stabilized by one phenyl ring.
$(III)$ $Ph_{2} CH \stackrel{+}{C} HMe$ is a $2^{\circ}$ carbocation with two phenyl groups attached to the adjacent carbon,providing some inductive stabilization.
$(IV)$ $Ph_{2} C(Me) \stackrel{+}{C} H_{2}$ is a $1^{\circ}$ carbocation.
Comparing these,the resonance stabilization from phenyl rings is the dominant factor. The order of stability is $(I)$ > $(II)$ > $(III)$ > $(IV)$.

(C) The stability of alkenes is determined by the number of hyperconjugative structures,which depends on the number of $\alpha$-hydrogen atoms attached to the $sp^2$ hybridized carbon atoms.
In $Me_2C=CH_2$ ($2$-methylprop$-1-$ene),there are two $Me$ groups attached to the double-bonded carbon,providing $3 + 3 = 6$ $\alpha$-hydrogen atoms.
In $MeCH_2CH=CH_2$ (but$-1-$ene),there is only one $MeCH_2$ group attached to the double-bonded carbon,providing $2$ $\alpha$-hydrogen atoms.
Since $Me_2C=CH_2$ has more $\alpha$-hydrogen atoms $(6)$ compared to $MeCH_2CH=CH_2$ $(2)$,it exhibits a greater hyperconjugative effect,making it more stable.

(A) Electrophiles are electron-deficient species (neutral or cationic).
They attack electron-rich $C$-atoms.

(D) The Friedel-Crafts reaction is an electrophilic aromatic substitution reaction.
It requires an electron-rich aromatic ring to react efficiently with the electrophile $(CH_3^+)$.
Nitrobenzene and acetophenone contain electron-withdrawing groups ($-NO_2$ and $-COCH_3$ respectively),which deactivate the benzene ring towards electrophilic substitution.
Benzene is moderately reactive.
Toluene contains a methyl group $(-CH_3)$,which is an electron-donating group due to the $+I$ effect and hyperconjugation.
This increases the electron density of the benzene ring,making it the most reactive among the given options towards electrophilic substitution.

(D) $(+)$-lactic acid and $(-)$-lactic acid are non-superimposable mirror images of each other.
They possess a chiral carbon atom and rotate the plane of plane-polarized light in opposite directions.
Therefore,they are enantiomers,which is a type of optical isomerism.

(B) The compound $MeCOCH_{2}CO_{2}Et$ is ethyl acetoacetate,a $\beta$-keto ester.
It can form an enol by the migration of an $\alpha$-hydrogen to the carbonyl oxygen.
The most stable enol form is the one that is stabilized by intramolecular hydrogen bonding and conjugation.
In $MeC(OH)=CHCO_{2}Et$,the enol double bond is conjugated with the ester carbonyl group,and it forms a stable six-membered ring through intramolecular hydrogen bonding between the hydroxyl group and the ester carbonyl oxygen.
Therefore,$MeC(OH)=CHCO_{2}Et$ is the most contributing tautomeric enol form.

(C) The structure of phosphorous acid $(H_{3}PO_{3})$ contains one $P=O$ bond,two $P-OH$ bonds,and one $P-H$ bond.
Protons attached directly to oxygen atoms in $P-OH$ groups are acidic because the oxygen atom is highly electronegative,which facilitates the release of the $H^+$ ion.
The hydrogen atom directly bonded to the phosphorus atom $(P-H)$ is not acidic because the electronegativity difference between $P$ and $H$ is very small.
Therefore,$H_{3}PO_{3}$ has $2$ acidic protons.

(C) Ammonium acetate $(CH_3COONH_4)$ is a salt formed from a weak acid $(CH_3COOH)$ and a weak base $(NH_4OH)$.
For a salt of a weak acid and a weak base,the pH of the solution is given by the formula:
$pH = 7 + \frac{1}{2} pK_a - \frac{1}{2} pK_b$
Since the pH of such a salt solution is independent of the concentration (dilution),the pH remains constant upon dilution.
Given that $pK_a \approx pK_b$,the equation simplifies to:
$pH = 7 + \frac{1}{2} (pK_a - pK_b) = 7 + 0 = 7$
Therefore,the pH of the resulting solution is $7$.

(B) The reaction is: $CH_3COOH + NaOH \rightarrow CH_3COONa + H_2O$
Initial millimoles of $CH_3COOH = 20 \ mL \times 0.1 \ N = 2 \ mmol$.
Initial millimoles of $NaOH = 10 \ mL \times 0.1 \ N = 1 \ mmol$.
After the reaction,$1 \ mmol$ of $CH_3COOH$ remains and $1 \ mmol$ of $CH_3COONa$ is formed.
This forms an acidic buffer solution.
Using Henderson-Hasselbalch equation:
$pH = pK_a + \log \frac{[Salt]}{[Acid]}$
$pH = 4.74 + \log \frac{1 \ mmol}{1 \ mmol}$
$pH = 4.74 + \log(1) = 4.74 + 0 = 4.74$.

(C) Let the weight of each gas be $x \ g$.
Number of moles of $CH_{4} = \frac{x}{16} \ mol$.
Number of moles of $H_{2} = \frac{x}{2} \ mol$.
Total moles = $\frac{x}{16} + \frac{x}{2} = \frac{x + 8x}{16} = \frac{9x}{16} \ mol$.
The mole fraction of $H_{2}$ is $X_{H_{2}} = \frac{n_{H_{2}}}{n_{total}} = \frac{x/2}{9x/16} = \frac{x}{2} \times \frac{16}{9x} = \frac{8}{9}$.
According to Dalton's Law,the partial pressure is proportional to the mole fraction.
Therefore,the fraction of total pressure exerted by $H_{2} = \frac{8}{9}$.

(B) Given lines are:
$3x + 4y = 9$ $(i)$
$y = mx + 1$ (ii)
Substituting (ii) into $(i)$:
$3x + 4(mx + 1) = 9$
$3x + 4mx + 4 = 9$
$x(3 + 4m) = 5$
$x = \frac{5}{3 + 4m}$
For $x$ to be an integer,$(3 + 4m)$ must be a divisor of $5$.
The divisors of $5$ are $\pm 1, \pm 5$.
Case $1$: $3 + 4m = 1 \implies 4m = -2 \implies m = -0.5$ (not an integer).
Case $2$: $3 + 4m = -1 \implies 4m = -4 \implies m = -1$ (integer).
Case $3$: $3 + 4m = 5 \implies 4m = 2 \implies m = 0.5$ (not an integer).
Case $4$: $3 + 4m = -5 \implies 4m = -8 \implies m = -2$ (integer).
Thus,the possible integer values for $m$ are $-1$ and $-2$.
The number of such integer values is $2$.

(C) The ionization energy $(IE)$ for a hydrogen-like species is given by the formula: $IE_n = 13.6 \times Z^2 / n^2 \ eV$.
For the second ionization of $He$ $(He^+ \rightarrow He^{2+} + e^-)$,the electron is removed from the $n=1$ state of a $He^+$ ion,which has an atomic number $Z=2$.
Substituting these values: $IE = 13.6 \times (2^2 / 1^2) \ eV = 13.6 \times 4 \ eV = 54.4 \ eV$.

(D) According to the $\text{HEISENBERG}$ uncertainty principle,the product of uncertainty in position and momentum is given by $\Delta x \cdot \Delta p \geq \frac{h}{4 \pi}$.
Since $\Delta p = m \Delta v$,substituting this into the equation gives $\Delta x \cdot m \Delta v \geq \frac{h}{4 \pi}$,which simplifies to $\Delta x \cdot \Delta v \geq \frac{h}{4 \pi m}$.
Additionally,the principle can be expressed in terms of energy and time as $\Delta E \cdot \Delta t \geq \frac{h}{4 \pi}$.
Therefore,the expression $\Delta E \cdot \Delta x \geq \frac{h}{4 \pi}$ is incorrect.

(A) When water evaporates,it changes from a liquid state to a gaseous state.
In the gaseous state,the molecules have more freedom of movement compared to the liquid state.
Therefore,the disorder or entropy of the system increases during evaporation.

(B) The spontaneity of a chemical reaction is determined by the Gibbs free energy equation: $\Delta G = \Delta H - T \Delta S$.
If $\Delta G < 0$,the reaction is spontaneous.
If $\Delta G > 0$,the reaction is non-spontaneous.
Therefore,the change in entropy $(\Delta S)$ along with the change in enthalpy $(\Delta H)$ determines the spontaneity of the reaction.

(B) Phenol is acidic enough to react with strong bases like $NaOH$ to form sodium phenoxide $(C_{6}H_{5}ONa)$,which is water-soluble.
Phenol is a weaker acid than carbonic acid $(H_{2}CO_{3})$,therefore it cannot displace $CO_{2}$ from $NaHCO_{3}$.

(B) $Me_{3}CONa + EtCl \rightarrow Me_{3}COEt + NaCl$
Williamson synthesis involves the reaction of an alkoxide with a primary alkyl halide to form an ether.
Since $EtCl$ is a primary $(1^{\circ})$ alkyl halide and $Me_{3}CONa$ is a bulky alkoxide,this reaction proceeds via an $S_{N}2$ mechanism to yield the desired ether.
Option $A$ would result in elimination $(E2)$ because $Me_{3}CCl$ is a tertiary alkyl halide.
Therefore,the correct option is $B$.

(D) The iodoform test is given by compounds containing the $CH_{3}CO-$ group or compounds that can be oxidized to this group,such as secondary alcohols with the $CH_{3}CH(OH)-$ group.
Option $A$: $CH_{3} CH_{2} CH_{2} CH_{2} CHO$ is a pentanal,which does not contain the $CH_{3}CO-$ group.
Option $B$: $CH_{3} CH_{2} CO CH_{2} CH_{3}$ is pentan-$3$-one,which lacks the $CH_{3}CO-$ group.
Option $C$: $CH_{3} CH_{2} CH_{2} CH_{2} CH_{2} OH$ is pentan-$1$-ol,a primary alcohol that does not form the required methyl ketone upon oxidation.
Option $D$: $CH_{3} CH_{2} CH_{2} CH(OH) CH_{3}$ is pentan-$2$-ol. It contains the $CH_{3}CH(OH)-$ group,which is oxidized by $I_{2}/NaOH$ to pentan-$2$-one $(CH_{3} CH_{2} CH_{2} CO CH_{3})$,a methyl ketone that subsequently undergoes the iodoform reaction to produce $CHI_{3}$ (iodoform).

(A) The reaction of $CH_3CHO$ (acetaldehyde) with $Al(OEt)_3$ (aluminum ethoxide) is known as the Tishchenko reaction.
This reaction involves the disproportionation of two molecules of an aldehyde to form an ester.
Specifically,$2CH_3CHO \xrightarrow{Al(OEt)_3} CH_3COOCH_2CH_3$ (ethyl acetate).
Therefore,the product formed is ethyl acetate $(CH_3COOCH_2CH_3)$.

(A) Molecularity of a reaction is defined as the number of reacting species (atoms,ions,or molecules) taking part in an elementary reaction,which must collide in order to bring about a chemical reaction.
Molecularity is always a whole number $(1, 2, 3, ...)$ and can never be zero,fractional,or negative.
Therefore,the statement that it may be a fractional number is incorrect.

(D) The radioactive decay of $_{11}Na^{24}$ occurs via $\beta^-$-decay.
In $\beta^-$-decay,a neutron is converted into a proton,increasing the atomic number by $1$ while the mass number remains constant.
The nuclear reaction is: $_{11}Na^{24} \longrightarrow _{12}Mg^{24} + _{-1}\beta^{0}$.

(B) $_{82}Pb = [Xe] 4f^{14} 5d^{10} 6s^{2} 6p^{2}$
$_{83}Bi = [Xe] 4f^{14} 5d^{10} 6s^{2} 6p^{3}$
Due to the inert pair effect,the $6s^{2}$ electrons are held tightly by the nucleus due to relativistic contraction and poor shielding by $4f$ and $5d$ electrons.
Consequently,$Pb$ shows a stable bivalency ($+2$ oxidation state) and $Bi$ shows a stable trivalency ($+3$ oxidation state) instead of their group oxidation states of $+4$ and $+5$ respectively.

(A) In the brown ring complex $[Fe(H_2O)_5(NO)]SO_4$,the iron is in the $+1$ oxidation state $(Fe^{+})$.
To balance the charge of the complex,the nitric oxide ligand $(NO)$ acts as a nitrosonium ion $(NO^{+})$.
Thus,the oxidation state of $Fe$ is $+1$ and $NO$ is $+1$.

(A) In the electrochemical series,metals are arranged in the increasing order of their standard reduction potential.
The standard reduction potential $(E^{\circ})$ of $Li^{+} / Li$ is $-3.05 \ V$.
The standard reduction potential $(E^{\circ})$ of $Cu^{2+} / Cu$ is $+0.34 \ V$.
Since $-3.05 \ V < +0.34 \ V$,$Li$ has a lower standard reduction potential than $Cu$.
Therefore,$Li$ is placed higher in the electrochemical series than $Cu$.

(D) According to Faraday's law of electrolysis,the mass $m$ deposited is given by $m = \frac{E \times I \times t}{F}$.
Here,$I = C$,$t = t$,and $F = 96500 \ C/mol$.
The equivalent weight $E$ of $Cu$ in $CuSO_4$ is $\frac{\text{Atomic weight}}{\text{Valency factor}} = \frac{63.5}{2} = 31.75$.
Substituting these values,we get $m = \frac{31.75 \times C \times t}{96500}$.

(A) The rate of $S_{N}1$ reaction depends on the stability of the carbocation intermediate formed during the rate-determining step.
$S_{N}1$ reactions proceed via the formation of a carbocation,and the stability order of carbocations is $3^{\circ} > 2^{\circ} > 1^{\circ}$.
Therefore,the reactivity order for $S_{N}1$ reactions is $3^{\circ} > 2^{\circ} > 1^{\circ}$ alkyl halides.
Among the given options:
$A$. tert-butyl chloride is a $3^{\circ}$ alkyl halide.
$B$. $1-$chlorobutane is a $1^{\circ}$ alkyl halide.
$C$. $2-$methyl$-1-$chloropropane is a $1^{\circ}$ alkyl halide.
$D$. $2-$chlorobutane is a $2^{\circ}$ alkyl halide.
Since tert-butyl chloride forms the most stable $3^{\circ}$ carbocation,it reacts most efficiently via the $S_{N}1$ mechanism.

(B) In aniline $(C_6H_5NH_2)$,the lone pair of electrons on the nitrogen atom is delocalized into the benzene ring through resonance.
This reduces the availability of the lone pair for protonation,making aniline a weaker base.
In contrast,in methylamine $(CH_3NH_2)$,the electron-donating methyl group increases the electron density on the nitrogen atom,making it more basic than aniline.

(D) In the presence of heat (boiling point conditions) or $UV$ light,chlorine reacts with the side chain of toluene via a free radical substitution mechanism.
When excess $Cl_{2}$ is used,it leads to the progressive substitution of all three hydrogen atoms on the methyl group.
The reaction proceeds as follows:
$C_6H_5CH_3$ $\xrightarrow{Cl_2, \Delta} C_6H_5CH_2Cl$ $\xrightarrow{Cl_2, \Delta} C_6H_5CHCl_2$ $\xrightarrow{Cl_2, \Delta} C_6H_5CCl_3$.
Thus,the final product is $C_6H_5CCl_3$,which is known as trichloromethylbenzene or benzotrichloride.

(A) Toluene $(C_{6}H_{5}CH_{3})$ and chlorobenzene $(C_{6}H_{5}Cl)$ both undergo electrophilic aromatic substitution (nitration) when treated with a mixture of conc. $H_{2}SO_{4}$ and conc. $HNO_{3}$.
The methyl group $(-CH_{3})$ in toluene is an activating group,which increases the electron density of the benzene ring,making it more reactive towards electrophilic substitution compared to the benzene ring in chlorobenzene.
The chlorine atom $(-Cl)$ in chlorobenzene is a deactivating group due to its strong $-I$ effect,which decreases the electron density of the benzene ring,making it less reactive towards electrophilic substitution.
Since toluene is more reactive than chlorobenzene,it will undergo nitration at a faster rate,resulting in the formation of $p$-nitrotoluene in excess compared to $p$-nitrochlorobenzene.

(D) When $Acetone$ $(CH_3COCH_3)$ and $Chloroform$ $(CHCl_3)$ are mixed,they form intermolecular hydrogen bonds between the oxygen atom of acetone and the hydrogen atom of chloroform.
This interaction is stronger than the original $Acetone-Acetone$ and $Chloroform-Chloroform$ interactions.
As a result,the escaping tendency of the molecules decreases,leading to a negative deviation from Raoult's law.

(A) Elevation in boiling point is given by $\Delta T_{b} = i \times K_{b} \times m$.
For $NaCl$ $(58.4 \ g \approx 1 \text{ mole})$,$i = 2$.
For glucose $(180 \ g = 1 \text{ mole})$,$i = 1$.
Since both solutes are dissolved in the same amount of solvent ($1000 \ mL \approx 1 \text{ kg}$ of water),their molality $(m)$ is the same.
Therefore,$\Delta T_{b} \propto i$.
Since $i_{NaCl} > i_{glucose}$,the $NaCl$ solution will show a higher elevation of boiling point.

(D) The preparation of $Me_{3}CCOOH$ (pivalic acid) from a Grignard reagent involves the reaction of a Grignard reagent with dry ice $(CO_{2})$.
The reaction is as follows:
$O=C=O + (CH_{3})_{3}CMgI \rightarrow (CH_{3})_{3}C-COOMgI$
Followed by hydrolysis:
$(CH_{3})_{3}C-COOMgI + H_{2}O \rightarrow (CH_{3})_{3}C-COOH + Mg(OH)I$
Thus,treating $1 \ mol$ of dry ice with $1 \ mol$ of $Me_{3}CMgI$ yields $Me_{3}CCOOH$.

(C) The formula for normality is $N = \frac{w \times 1000}{E \times V(mL)}$.
Here,$N = 1/20$,$V = 1000 \ mL$,and the equivalent weight $E$ of oxalic acid $(H_2C_2O_4 \cdot 2H_2O)$ is $63 \ g/eq$.
Substituting the values: $w = \frac{N \times E \times V}{1000} = \frac{1}{20} \times 63 \times \frac{1000}{1000} = \frac{63}{20} \ g$.