GUJCET 2023 Chemistry Question Paper with Answer and Solution

(C) The chemical reaction between ethanol and sodium metal is:
$2C_2H_5OH + 2Na \rightarrow 2C_2H_5ONa + H_2$
From the stoichiometry,$2 \ mol$ of ethanol produces $1 \ mol$ of $H_2$ gas.
At $S.T.P.$,$1 \ mol$ of any gas occupies $22400 \ mL$.
Therefore,$22400 \ mL$ of $H_2$ is produced by $2 \times 46 \ g$ of ethanol.
$280 \ mL$ of $H_2$ is produced by:
$\frac{2 \times 46 \times 280}{22400} \ g$
$= \frac{92 \times 280}{22400} \ g$
$= \frac{92}{80} \ g = 1.15 \ g$.
Thus,the correct option is $C$.

(B) In a faintly alkaline or neutral medium,the permanganate ion $(MnO_4^-)$ acts as an oxidizing agent and oxidizes the iodide ion $(I^-)$ to the iodate ion $(IO_3^-)$.
The balanced chemical equation for this reaction is:
$2MnO_4^- + I^- + H_2O \rightarrow 2MnO_2 + IO_3^- + 2OH^-$
Therefore,the product obtained is $IO_3^-$.

(D) The reaction of alcohols with $Cu$ at $573 \ K$ is a dehydrogenation reaction,not dehydration.
$1$. Primary alcohols $(RCH_2OH)$ form aldehydes.
$2$. Secondary alcohols $(R_2CHOH)$ form ketones.
$3$. Tertiary alcohols $(R_3COH)$ undergo dehydration to form alkenes because they cannot undergo dehydrogenation due to the absence of $\alpha$-hydrogen.
Since the question asks for the reaction with $Cu$ at $573 \ K$,tertiary alcohols specifically undergo dehydration under these conditions. Therefore,the correct option is $D$.

(A) The reaction between a strong base/nucleophile like $CH_3ONa$ (sodium methoxide) and a tertiary alkyl halide like $(CH_3)_3CBr$ (tert-butyl bromide) proceeds primarily via an $E2$ elimination mechanism.
Because the substrate is sterically hindered,the nucleophilic substitution $(S_N2)$ is suppressed.
Therefore,the major product formed is the alkene,which is $2$-methylpropene (isobutylene).
Thus,the correct option is $A$.

(D) The Cannizzaro reaction is given by aldehydes that do not possess an $\alpha$-hydrogen atom.
$HCHO$ (formaldehyde) has no $\alpha$-hydrogen.
Benzaldehyde $(C_6H_5CHO)$ has no $\alpha$-hydrogen.
$1$-methylcyclohexanecarbaldehyde has no $\alpha$-hydrogen at the carbonyl carbon position.
$CH_3CHO$ (acetaldehyde) possesses three $\alpha$-hydrogen atoms,so it undergoes aldol condensation instead of the Cannizzaro reaction.
Therefore,the correct option is $D$.

(B) The acidity of benzoic acid derivatives depends on the electronic effects of the substituents attached to the benzene ring.
$1$. The $-NO_2$ group in $(I)$ is a strong electron-withdrawing group ($-I$ and $-M$ effect),which stabilizes the carboxylate anion,increasing acidity.
$2$. The $-OCH_3$ group in $(II)$ is an electron-donating group ($+M$ effect),which destabilizes the carboxylate anion,decreasing acidity.
$3$. Benzoic acid $(III)$ serves as the reference point.
Therefore,the order of acidity is $4$-Nitrobenzoic acid $(I)$ > Benzoic acid $(III)$ > $4$-Methoxybenzoic acid $(II)$.
The correct option is $B$.

(C) The compound $CH_2=CH-CHO$ is an unsaturated aldehyde.
Its $IUPAC$ name is $Prop-2-enal$.
Its common name is $Acrolein$.

(B) The nitrogenous bases present in $DNA$ are Adenine $(A)$,Guanine $(G)$,Cytosine $(C)$,and Thymine $(T)$.
Uracil $(U)$ is a nitrogenous base found in $RNA$ instead of Thymine.
Therefore,Uracil is not present in the $DNA$ structure.
Thus,the correct option is $B$.

(D) Glucose is an aldohexose $(C_6H_{12}O_6)$.
When heated with $HI$,it undergoes reduction to form $n-hexane$,confirming the presence of a straight chain of $6$ carbon atoms.
It reacts with hydroxylamine $(NH_2OH)$ to form an oxime,confirming the presence of a carbonyl group.
However,glucose exists in a pyranose structure (a $6$-membered ring) in its cyclic form,not a furanose structure (a $5$-membered ring). Fructose exists in a furanose structure.
Therefore,the statement that glucose contains a furanose structure is incorrect.

(A) The rate law for the reaction is given by: $Rate = k[A]^1[B]^2$.
If the concentration of $B$ is increased two times,the new concentration becomes $[B'] = 2[B]$.
The new rate $Rate'$ will be: $Rate' = k[A]^1[2B]^2 = 4 \times k[A]^1[B]^2$.
Therefore,the rate of reaction increases by $4-$Times.
Hence,the correct option is $A$.

(C) The overall order of the reaction is $n = \frac{1}{2} + \frac{3}{2} = 2$.
The unit of the rate constant for a reaction of order $n$ is given by the formula $(Mol \cdot L^{-1})^{1-n} \cdot Sec^{-1}$.
Substituting $n = 2$ into the formula,we get $(Mol \cdot L^{-1})^{1-2} \cdot Sec^{-1} = (Mol \cdot L^{-1})^{-1} \cdot Sec^{-1} = Mol^{-1} \cdot L \cdot Sec^{-1}$.

(A) The integrated rate equation for a first-order reaction is given by: $\ln \frac{[R]_0}{[R]} = Kt$.
Converting to base $10$ logarithm: $\log \frac{[R]_0}{[R]} = \frac{Kt}{2.303}$.
Comparing this with the equation of a straight line $y = mx + c$,where $y = \log \frac{[R]_0}{[R]}$ and $x = t$,the slope $m$ is equal to $\frac{K}{2.303}$.
Therefore,the graph of $\log \frac{[R]_0}{[R]}$ versus $t$ gives a slope of $\frac{K}{2.303}$.

(B) ligand is an ion or molecule that binds to a central metal atom or ion to form a coordination complex. The essential requirement for a species to act as a ligand is the presence of at least one lone pair of electrons to donate to the metal center.
$1$. $NO$ (Nitric oxide) has a lone pair on the nitrogen atom and can act as a ligand.
$2$. $NH_4^+$ (Ammonium ion) has all its valence electrons involved in covalent bonding with hydrogen atoms. It has no lone pair of electrons to donate,so it cannot act as a ligand.
$3$. $H_2N-CH_2-CH_2-NH_2$ (Ethylenediamine) has lone pairs on both nitrogen atoms and acts as a bidentate ligand.
$4$. $CO$ (Carbon monoxide) has a lone pair on the carbon atom and acts as a ligand.
Therefore,$NH_4^+$ is not expected to be a ligand.

(D) In an octahedral complex,the $d$-orbitals split into two sets: $t_{2g}$ (lower energy) and $e_g$ (higher energy).
For a $d^4$ ion,if the crystal field splitting energy $(\Delta_0)$ is greater than the pairing energy $(P)$,the electrons will prefer to pair up in the lower energy $t_{2g}$ orbitals rather than occupying the higher energy $e_g$ orbitals.
This results in a low-spin configuration of $t_{2g}^4 e_g^0$.
Therefore,the condition for this configuration is $\Delta_0 > P$.

(B) The given complexes are $[Cr(H_2O)_6]Cl_3$ and $[Cr(H_2O)_5Cl]Cl_2 \cdot H_2O$.
In these complexes,the water molecule acts as a ligand in the first case,while in the second case,one water molecule is present as a solvent of crystallization (lattice water) outside the coordination sphere.
This type of isomerism,where the solvent molecule (like $H_2O$) can be either inside or outside the coordination sphere,is known as solvate isomerism (also called hydrate isomerism).

(A) The electronic configuration of Gadolinium ($Gd$,$Z=64$) is $[Xe] 4f^7 5d^1 6s^2$.
In this configuration,one electron enters the $5d$ orbital due to the stability provided by the half-filled $4f^7$ subshell.
For other lanthanoids like $Eu$ $(Z=63)$,$Tb$ $(Z=65)$,and $Dy$ $(Z=66)$,the $5d$ orbital remains empty in their ground state configurations as electrons fill the $4f$ subshell.

(A) During the electrolysis of a highly concentrated $H_2SO_4$ solution,the concentration of $SO_4^{2-}$ ions is very high.
Due to this high concentration,the oxidation of $SO_4^{2-}$ ions becomes more favorable than the oxidation of water at the anode.
The reaction occurring at the anode is: $2SO_4^{2-}{_{\text{(aq)}}} \rightarrow S_2O_8^{2-}{_{\text{(aq)}}} + 2e^{-}$.
Therefore,the correct option is $A$.

(A) The formula for conductivity $(\kappa)$ is given by: $\kappa = G^* \times G$,where $G^*$ is the cell constant and $G$ is the conductance.
Conductance $(G)$ is the reciprocal of resistance $(R)$: $G = \frac{1}{R} = \frac{1}{100 \ \Omega} = 0.01 \ S$.
Given conductivity $\kappa = 1.29 \ S/m$.
Substituting the values into the formula: $1.29 \ S/m = G^* \times 0.01 \ S$.
Therefore,$G^* = \frac{1.29}{0.01} \ m^{-1} = 129 \ m^{-1}$.

(D) The cell reaction is as follows:
Anode (Oxidation): $Mg_{(s)} \rightarrow Mg_{(aq)}^{2+} + 2e^-$
Cathode (Reduction): $Cl_{2_{(g)}} + 2e^- \rightarrow 2Cl_{(aq)}^{-}$
Overall reaction: $Mg_{(s)} + Cl_{2_{(g)}} \rightarrow Mg_{(aq)}^{2+} + 2Cl_{(aq)}^{-}$
The Nernst equation is given by: $E_{cell} = E_{cell}^0 - \frac{0.059}{n} \log Q$
Here,$n = 2$ and the reaction quotient $Q = \frac{[Mg^{2+}][Cl^{-}]^2}{P_{Cl_2}}$.
Since $P_{Cl_2} = 1 \ bar$,$Q = [Mg^{2+}][Cl^{-}]^2$.
Thus,the correct equation is $E_{cell} = E_{cell}^0 - \frac{0.059}{2} \log [Mg^{2+}][Cl^{-}]^2$.

(C) To determine the reaction mechanism,we analyze the rate law based on the experimental data.
Let the rate law be $\text{Rate} = k[S]^x[Nu]^y$.
From experiment $1$ and $2$,when $[S]$ is doubled and $[Nu]$ is constant,the rate doubles ($2.2 \times 10^{-3}$ to $4.4 \times 10^{-3}$),so $x = 1$.
From experiment $1$ and $3$,when $[Nu]$ is doubled and $[S]$ is constant,the rate doubles ($2.2 \times 10^{-3}$ to $4.4 \times 10^{-3}$),so $y = 1$.
The overall rate law is $\text{Rate} = k[S]^1[Nu]^1$.
Since the reaction is first order with respect to both the substrate and the nucleophile,the total order is $2$.
This indicates a bimolecular nucleophilic substitution reaction,which is the $S_{N}2$ mechanism.

(C) The preparation of Freon-$12$ $(CCl_2F_2)$ from carbon tetrachloride $(CCl_4)$ is achieved by the Swarts reaction.
In this reaction,$CCl_4$ is treated with antimony trifluoride $(SbF_3)$ in the presence of antimony pentachloride $(SbCl_5)$ as a catalyst.
The chemical equation is: $3CCl_4 + 2SbF_3 \xrightarrow{SbCl_5} 3CCl_2F_2 + 2SbCl_3$.

(C) The formula for osmotic pressure is $\pi = i \times C \times R \times T$.
For $NaCl$,the van't Hoff factor $(i)$ is $2$ because it dissociates into $Na^+$ and $Cl^-$ ions.
Given the concentration $C = 0.02 \ M$.
Substituting these values: $\pi = 2 \times 0.02 \times RT = 0.04 \ RT$.

(C) The boiling point elevation is a colligative property,which depends on the van't Hoff factor $(i)$ of the solute. The formula is $\Delta T_b = i \times K_b \times m$.
Since the molality $(m)$ is the same for all solutions,the solution with the highest van't Hoff factor $(i)$ will have the highest boiling point.
$1.$ $KNO_3 \rightarrow K^+ + NO_3^-$,$i = 2$.
$2.$ $\text{Urea}$ is a non-electrolyte,$i = 1$.
$3.$ $K_4[Fe(CN)_6] \rightarrow 4K^+ + [Fe(CN)_6]^{4-}$,$i = 5$.
$4.$ $NH_4NO_3 \rightarrow NH_4^+ + NO_3^-$,$i = 2$.
Comparing the values,$K_4[Fe(CN)_6]$ has the highest $i$ value $(i = 5)$,therefore it has the highest boiling point.

(C) The Van't Hoff factor $(i)$ for a strong electrolyte like $NaCl$ is given by the formula $i = 1 + (n-1)\alpha$,where $n$ is the number of ions produced per formula unit and $\alpha$ is the degree of dissociation.
For $NaCl$,$n = 2$. As the concentration of the solution decreases,the degree of dissociation $(\alpha)$ increases due to the decrease in inter-ionic attractions.
Therefore,as concentration decreases from $0.1 \ M$ to $0.001 \ M$,$\alpha$ increases,which leads to an increase in the value of $i$.
Thus,the order of the Van't Hoff factor is $i_{A} < i_{B} < i_{C}$.