GUJCET 2015 Chemistry Question Paper with Answer and Solution

(A) Isophthalic acid is the common name for benzene-$1,3$-dicarboxylic acid.
In this structure,two carboxylic acid groups $(-COOH)$ are attached to the benzene ring at the $1$ and $3$ positions.
Therefore,the correct $IUPAC$ name is benzene-$1,3$-dicarboxylic acid.
The correct option is $A$.

(B) Phenol is a very weak acid $(K_a \approx 10^{-10})$ and it does not react with sodium carbonate $(Na_2CO_3)$ to form salt and evolve $CO_2$ gas.
Therefore,the statement that phenol is neutralised by sodium carbonate is incorrect.
$o$-Nitrophenol exhibits intramolecular hydrogen bonding,whereas $p$-nitrophenol exhibits intermolecular hydrogen bonding,making the boiling point of $o$-nitrophenol lower than that of $p$-nitrophenol.
Phenol is used in the synthesis of drugs like aspirin (an analgesic).
Phenol is more soluble in water than chlorobenzene due to its ability to form hydrogen bonds with water molecules.

(B) Methyl phenyl ether (anisole) is an electron-rich aromatic compound due to the $+M$ effect of the $-OCH_3$ group.
Electrophilic aromatic substitution occurs readily in anisole.
For bromination,$Br_2$ in ethanoic acid $(CH_3COOH)$ is used as the reagent.
The polar solvent $CH_3COOH$ polarizes the $Br-Br$ bond,allowing the reaction to proceed without the need for a strong Lewis acid catalyst like $FeBr_3$,which would otherwise lead to poly-substitution or other side reactions.

(B) The chemical formula of $Picric \ Acid$ is $2,4,6-trinitrophenol$.
It contains a phenolic $-OH$ group and three nitro $(-NO_2)$ groups attached to the benzene ring,but it does not contain a carboxylic acid $(-COOH)$ group.
Ethanoic acid $(CH_3COOH)$,Benzoic acid $(C_6H_5COOH)$,and Salicylic acid $(C_6H_4(OH)COOH)$ all contain the $-COOH$ group.
Therefore,the correct option is $B$.

(C) The acidic strength of carboxylic acids depends on the stability of the conjugate base (carboxylate ion). Electron-withdrawing groups $(EWG)$ increase acidity by stabilizing the negative charge,while electron-donating groups $(EDG)$ decrease acidity.
$A$. $Cl_3C-COOH > Cl_2CH-COOH > ClCH_2-COOH$: Correct,as the number of $EWG$ $(Cl)$ increases,acidity increases.
$B$. $CH_3CH_2CH(Cl)COOH > CH_3CH(Cl)CH_2COOH > CH_2(Cl)CH_2CH_2COOH$: Correct,as the distance of the $EWG$ from the $-COOH$ group increases,the inductive effect decreases,thus acidity decreases.
$C$. $HCOOH > CH_3COOH > C_6H_5COOH$: This order is incorrect. The correct order is $HCOOH > C_6H_5COOH > CH_3COOH$. $C_6H_5COOH$ is more acidic than $CH_3COOH$ because the phenyl group exerts an electron-withdrawing effect through resonance,whereas the methyl group is electron-donating ($+I$ effect).
$D$. $CH_3COOH > CH_3CH_2COOH > (CH_3)_2CHCOOH$: Correct,as the number of alkyl groups $(EDG)$ increases,the $+I$ effect increases,which destabilizes the carboxylate ion and decreases acidity.

(D) In $Maltose$,two $\alpha-D-glucose$ units are joined by a $C_1-C_4$ glycosidic linkage.
In $Lactose$,$\beta-D-galactose$ and $\beta-D-glucose$ are joined by a $C_1-C_4$ glycosidic linkage.
In $Cellulose$,$\beta-D-glucose$ units are joined by a $C_1-C_4$ glycosidic linkage.
In $Amylopectin$,it is a branched-chain polymer of $\alpha-D-glucose$ units. The linear chain is formed by $C_1-C_4$ glycosidic linkages,but the branching occurs due to $C_1-C_6$ glycosidic linkages. Therefore,not all units are joined by $C_1-C_4$ linkages.

(A) Liver is a rich source of many vitamins,including $B_2$,$B_{12}$,and $H$ (Biotin).
However,$B_1$ (Thiamine) is primarily found in whole grains,legumes,and nuts,and is not considered a significant dietary source in liver compared to the others.
Thus,the correct option is $A$.

(A) In a multi-step reaction,the rate of the overall reaction is determined by the slowest step,which is known as the rate-determining step.
Given the mechanism:
$(i)$ $ClO^{-} + ClO^{-} \xrightarrow{K_{1}} ClO_{2}^{-} + Cl^{-}$ (Slow step)
$(ii)$ $ClO_{2}^{-} + ClO^{-} \xrightarrow{K_{2}} ClO_{3}^{-} + Cl^{-}$ (Fast step)
The rate of the reaction is determined by the slow step $(i)$.
Therefore,the rate law expression is given by: $\text{Rate} = K_{1}[ClO^{-}][ClO^{-}] = K_{1}[ClO^{-}]^{2}$.

(C) The rate law for the reaction is given by $Rate = K[X]^m[Y]^n$.
Given that the total order of the reaction is $3$,so $m + n = 3$.
The order of reaction with respect to $X$ is $m = 2$.
Substituting $m = 2$ into the equation $2 + n = 3$,we get $n = 1$.
Therefore,the rate equation is $Rate = K[X]^2[Y]^1$.
Since the rate of reaction can be expressed as $-\frac{d[X]}{dt}$,the differential rate equation is $-\frac{d[X]}{dt} = K[X]^2[Y]$.

(A) In $[Ni(CO)_4]$,the oxidation state of $Ni$ is $0$. The electronic configuration of $Ni$ $(Z=28)$ is $[Ar] 3d^8 4s^2$. Since $CO$ is a strong field ligand,it causes pairing of electrons,resulting in $3d^{10}$ configuration. The hybridisation is $sp^3$ (tetrahedral geometry).
In $[Ni(CN)_4]^{2-}$,the oxidation state of $Ni$ is $+2$. The electronic configuration of $Ni^{2+}$ is $[Ar] 3d^8$. $CN^-$ is a strong field ligand,which forces the two unpaired electrons in the $3d$ orbitals to pair up. This creates one empty $3d$ orbital,which is used for $dsp^2$ hybridisation (square planar geometry).
Therefore,the hybridisation types are $sp^3$ and $dsp^2$ respectively.

(D) To determine the magnetic nature,we calculate the number of unpaired electrons in each complex:
$1$. In $[Co(NH_3)_6]^{3+}$,$Co^{3+}$ is $d^6$. $NH_3$ is a strong field ligand,causing pairing. All electrons are paired,so it is diamagnetic.
$2$. In $[Ni(CO)_4]$,$Ni$ is $d^8 s^2$. $CO$ is a strong field ligand,causing $4s$ electrons to pair into $3d$. All electrons are paired,so it is diamagnetic.
$3$. In $[Ni(CN)_4]^{2-}$,$Ni^{2+}$ is $d^8$. $CN^-$ is a strong field ligand,causing pairing in $3d$ orbitals. All electrons are paired,so it is diamagnetic.
$4$. In $[NiCl_4]^{2-}$,$Ni^{2+}$ is $d^8$. $Cl^-$ is a weak field ligand,so no pairing occurs. There are $2$ unpaired electrons in the $3d$ orbitals,making it paramagnetic.

(A) The spectrochemical series is an arrangement of ligands in order of increasing crystal field splitting energy $(\Delta_o)$.
The correct order of increasing field strength for the given ligands is: $SCN^{-} < F^{-} < NH_3 < en < CO$.
Thus,the correct option is $A$.

(C) The magnetic moment is calculated using the formula $\mu = \sqrt{n(n+2)} \text{ BM}$,where $n$ is the number of unpaired electrons.
$1$. For $V^{3+}$ $([Ar] 3d^2)$: $n = 2$,$\mu = \sqrt{2(4)} = \sqrt{8} \approx 2.83 \text{ BM}$.
$2$. For $Mn^{3+}$ $([Ar] 3d^4)$: $n = 4$,$\mu = \sqrt{4(6)} = \sqrt{24} \approx 4.90 \text{ BM}$.
$3$. For $Fe^{3+}$ $([Ar] 3d^5)$: $n = 5$,$\mu = \sqrt{5(7)} = \sqrt{35} \approx 5.92 \text{ BM}$.
$4$. For $Cu^{3+}$ $([Ar] 3d^8)$: $n = 2$,$\mu = \sqrt{2(4)} = \sqrt{8} \approx 2.83 \text{ BM}$.
Thus,$Fe^{3+}$ has the maximum number of unpaired electrons $(n=5)$,resulting in the maximum magnetic moment.

(A) The basicity of lanthanoid oxides $(Ln_2O_3)$ decreases as the ionic radius of the $Ln^{3+}$ ion decreases across the lanthanoid series due to lanthanoid contraction.
As we move from $Pr$ $(Z=59)$ to $Lu$ $(Z=71)$,the ionic radius decreases.
Therefore,the basic character follows the order: $Pr_2O_3 > Sm_2O_3 > Gd_2O_3 > Lu_2O_3$.
Thus,$Pr_2O_3$ has the maximum basicity.

(B) The magnetic moment $(\mu)$ is calculated using the formula $\mu = \sqrt{n(n+2)} \ BM$,where $n$ is the number of unpaired electrons.
$1$. For $Cr^{3+}$ $([Ar] 3d^3)$: $n = 3$,$\mu = \sqrt{3(3+2)} = \sqrt{15} \approx 3.87 \ BM$.
$2$. For $Fe^{3+}$ $([Ar] 3d^5)$: $n = 5$,$\mu = \sqrt{5(5+2)} = \sqrt{35} \approx 5.92 \ BM$.
$3$. For $Ti^{3+}$ $([Ar] 3d^1)$: $n = 1$,$\mu = \sqrt{1(1+2)} = \sqrt{3} \approx 1.73 \ BM$.
$4$. For $Co^{3+}$ $([Ar] 3d^6)$: In high spin,$n = 4$,$\mu = \sqrt{4(4+2)} = \sqrt{24} \approx 4.90 \ BM$.
Comparing these values,$Fe^{3+}$ has the maximum number of unpaired electrons $(n=5)$,hence it has the maximum theoretical magnetic moment.

(D) $KMnO_4$ is a crystalline substance,not an amorphous one. It forms dark purple,almost black,orthorhombic crystals. Therefore,the statement that it is an amorphous substance is incorrect. The other statements are correct: it acts as an antiseptic,a strong oxidizing agent,and is used for bleaching in textile industries.

(B) According to Faraday's second law of electrolysis,when the same quantity of electricity is passed through different electrolytes connected in series,the masses of the substances deposited are proportional to their equivalent masses.
$\frac{\text{Mass of } Ni}{\text{Mass of } Al} = \frac{\text{Equivalent mass of } Ni}{\text{Equivalent mass of } Al}$
Equivalent mass of $Ni = \frac{\text{Atomic mass of } Ni}{\text{Valency factor}} = \frac{58.5}{2} = 29.25 \ g/eq$
Equivalent mass of $Al = \frac{\text{Atomic mass of } Al}{\text{Valency factor}} = \frac{27}{3} = 9 \ g/eq$
Let the mass of $Ni$ be $w$.
$\frac{w}{18} = \frac{29.25}{9}$
$w = 18 \times \frac{29.25}{9} = 2 \times 29.25 = 58.5 \ g$
Therefore,the weight of nickel obtained is $58.5 \ g$.

(C) The reducing power of a metal is inversely proportional to its standard reduction potential $(E^0)$.
Lower (more negative) $E^0$ values indicate a stronger tendency to lose electrons,making the metal a better reducing agent.
The given values are: $E^0_A = 0.34 \ V$,$E^0_B = -0.80 \ V$,and $E^0_C = -0.46 \ V$.
Comparing these values: $-0.80 \ V < -0.46 \ V < 0.34 \ V$.
Therefore,the order of reducing ability is $B > C > A$.

(B) During the electrolysis of concentrated aqueous $NaCl$ (brine),the following reactions occur:
At the cathode: $2H_2O(l) + 2e^- \rightarrow H_2(g) + 2OH^-(aq)$
At the anode: $2Cl^-(aq) \rightarrow Cl_2(g) + 2e^-$
The net reaction is: $2NaCl(aq) + 2H_2O(l) \rightarrow 2NaOH(aq) + Cl_2(g) + H_2(g)$
Since $NaOH$ is a strong base,the resulting solution is basic in nature.
Therefore,it turns red litmus into blue.

(B) The chemical reaction for the dehydrohalogenation of ethyl chloride is:
$CH_3CH_2Cl + KOH \xrightarrow{\text{Ethanol}, \Delta} CH_2=CH_2 + KCl + H_2O$
Molar mass of $CH_3CH_2Cl = (2 \times 12) + (5 \times 1) + 35.5 = 64.5 \ g \ mol^{-1}$.
Molar mass of the main product,ethene $(CH_2=CH_2)$ $= (2 \times 12) + (4 \times 1) = 28 \ g \ mol^{-1}$.
According to the stoichiometry,$64.5 \ g$ of $CH_3CH_2Cl$ produces $28 \ g$ of ethene.
Therefore,$6.45 \ g$ of $CH_3CH_2Cl$ would theoretically produce:
$\text{Mass} = \frac{6.45 \ g}{64.5 \ g \ mol^{-1}} \times 28 \ g \ mol^{-1} = 2.8 \ g$ of ethene.
Since only $50\%$ of the reagent is used,the actual amount of product obtained is:
$2.8 \ g \times 0.5 = 1.4 \ g$.

(B) The given reaction is $CH_3CH_2Cl + NaI \xrightarrow{\text{acetone}} CH_3CH_2I + NaCl$.
This is a halogen exchange reaction where an alkyl chloride reacts with sodium iodide in the presence of acetone to form an alkyl iodide.
This specific reaction is known as the $Finkelstein$ reaction.

(D) An allylic halide is a compound in which the halogen atom is bonded to an $sp^3$ hybridized carbon atom adjacent to a carbon-carbon double bond $(C=C)$.
$A$. Benzyl chloride $(C_6H_5CH_2Cl)$ is a benzylic halide.
$B$. $1-$Bromoethylbenzene $(C_6H_5CH(Br)CH_3)$ is a benzylic halide.
$C$. Bromobenzene $(C_6H_5Br)$ is an aryl halide.
$D$. $3-$Chlorocyclohex$-1-$ene has the chlorine atom attached to a carbon atom that is adjacent to the $C=C$ double bond,making it an allylic halide.