GUJCET 2014 Chemistry Question Paper with Answer and Solution

(D) Reduction of $Propanoic \ acid$ $(CH_3CH_2COOH)$ gives $Propan-1-ol$ (primary alcohol).
Reduction of $Propanal$ $(CH_3CH_2CHO)$ gives $Propan-1-ol$ (primary alcohol).
Reduction of $Methyl \ propanoate$ $(CH_3CH_2COOCH_3)$ gives $Propan-1-ol$ and $Methanol$ (both primary alcohols).
Reduction of $Propan-2-one$ $(CH_3COCH_3)$ gives $Propan-2-ol$,which is a secondary alcohol.
Therefore,$Propan-2-one$ does not give a primary alcohol on reduction.

(A) In the oxidation reaction of alcohols,the $O-H$ bond is cleaved,and a $C-H$ bond is broken to form a carbonyl group $(C=O)$. The $C-O$ bond remains intact.
In dehydration,$PBr_3$ reaction,and reaction with $HCl$ (Lucas test),the $C-O$ bond is cleaved to form alkenes or alkyl halides.
Therefore,the correct answer is $A$.

(D) The solubility of alcohols in water depends on the ability to form hydrogen bonds with water molecules.
As the number of hydroxyl $(-OH)$ groups increases,the number of hydrogen bonds formed with water increases,leading to higher solubility.
$A$: Secondary butyl alcohol $(C_4H_9OH)$ has $1$ $-OH$ group.
$B$: Tertiary butyl alcohol $(C_4H_9OH)$ has $1$ $-OH$ group.
$C$: Ethylene glycol $(HO-CH_2-CH_2-OH)$ has $2$ $-OH$ groups.
$D$: Glycerol $(HO-CH_2-CH(OH)-CH_2-OH)$ has $3$ $-OH$ groups.
Since glycerol has the maximum number of $-OH$ groups,it forms the most extensive hydrogen bonding with water,making it the most soluble among the given options.
Therefore,the correct option is $D$.

(C) The reaction described is the $Cannizzaro$ reaction.
Aldehydes that do not contain an $\alpha$-hydrogen atom undergo self-oxidation and reduction (disproportionation) when treated with concentrated alkali.
$Benzaldehyde$ $(C_6H_5CHO)$ has no $\alpha$-hydrogen.
$Trimethylacetaldehyde$ $((CH_3)_3CCHO)$ has no $\alpha$-hydrogen.
$Formaldehyde$ $(HCHO)$ has no $\alpha$-hydrogen.
$Dimethyl$ $acetaldehyde$ (isobutyraldehyde,$(CH_3)_2CHCHO$) contains one $\alpha$-hydrogen atom attached to the $\alpha$-carbon.
Therefore,$Dimethyl$ $acetaldehyde$ does not undergo the $Cannizzaro$ reaction.
The correct option is $C$.

(C) Fehling's solution is prepared by mixing two solutions,Fehling $A$ and Fehling $B$,in equal amounts just before use.
Fehling $A$ is an aqueous copper$(II)$ sulfate solution.
Fehling $B$ is an alkaline solution of sodium potassium tartrate,also known as Rochelle salt.
Therefore,the correct option is $C$.

(B) The acidic strength of carboxylic acids depends on the stability of the conjugate base (carboxylate ion).
Electron-withdrawing groups (like $Cl$) increase the acidity by stabilizing the negative charge on the carboxylate ion through the inductive effect ($-I$ effect).
As the number of chlorine atoms increases,the $-I$ effect increases,leading to greater stabilization of the carboxylate ion.
Therefore,the order of acidic strength is $Cl_3CCOOH > Cl_2CHCOOH > ClCH_2COOH > CH_3COOH$.

(B) For a first-order reaction, the integrated rate equation is $t = \frac{2.303}{k} \log\left(\frac{[A]_0}{[A]}\right)$.
Given $k = 2.303 \times 10^{-2} \text{ s}^{-1}$ and $[A] = \frac{[A]_0}{10}$.
Substituting these values, we get $t = \frac{2.303}{2.303 \times 10^{-2}} \log\left(\frac{[A]_0}{[A]_0 / 10}\right)$.
$t = \frac{1}{10^{-2}} \log(10) = 10^2 \times 1 = 100 \text{ s}$.

(B) The Arrhenius equation is given by $k = A e^{-E_a / RT}$.
Taking the logarithm on both sides to the base $10$,we get $\log k = \log A - \frac{E_a}{2.303 RT}$.
Comparing this with the equation of a straight line $y = mx + c$,where $y = \log k$ and $x = \frac{1}{T}$,the slope $m$ is equal to $\frac{- E_a}{2.303 R}$.

(B) For a first-order reaction,the rate constant $k$ is given by $k = \frac{2.303}{t} \log \frac{[A]_0}{[A]}$.
At half-life $(t = t_{1/2})$,the concentration $[A] = \frac{[A]_0}{2}$.
Substituting these values: $k = \frac{2.303}{t_{1/2}} \log \frac{[A]_0}{[A]_0/2} = \frac{2.303}{t_{1/2}} \log 2$.
Thus,$t_{1/2} = \frac{2.303 \times 0.3010}{k} = \frac{0.693}{k}$.
Since $t_{1/2}$ depends only on the rate constant $k$ and not on the initial concentration $[A]_0$,it is independent of the concentration.

(A) ligand with only one coordination site is called a monodentate ligand.
$O^{2-}$ is a monodentate ligand as it has only one donor atom.
$CO_3^{2-}$ (carbonate) is a bidentate ligand.
$SO_4^{2-}$ (sulfate) is a bidentate ligand.
$C_2O_4^{2-}$ (oxalate) is a bidentate ligand.
Therefore,the correct option is $A$.

(B) The stability of a coordination complex depends on the oxidation state of the central metal ion and the nature of the ligands.
$1$. Higher oxidation state of the central metal ion generally leads to greater stability due to stronger electrostatic attraction.
$2$. $[Co(NH_3)_6]^{2+}$ contains $Co^{2+}$,whereas $[Co(CN)_6]^{3-}$ and $[Co(NH_3)_6]^{3+}$ contain $Co^{3+}$.
$3$. Among the given options,$[Co(NH_3)_6]^{2+}$ has the lowest oxidation state $(+2)$,making it the least stable complex compared to those with $Co^{3+}$ ions.
Therefore,the correct option is $B$.

(D) Optical isomerism is shown by complexes that lack a plane of symmetry or center of inversion (i.e.,they are chiral).
$1$. $[Cr(C_2O_4)_3]^{3-}$ is an octahedral complex of the type $[M(AA)_3]$,which is chiral and shows optical isomerism.
$2$. $Cis-[Pt(Br)_2(en)_2]^{2+}$ is an octahedral complex of the type $Cis-[M(AA)_2X_2]$,which lacks a plane of symmetry and shows optical isomerism.
$3$. $[CrCl_2(NH_3)_2en]^+$ exists in various geometric isomers,and the specific isomer with the correct arrangement (lacking symmetry) shows optical isomerism.
$4$. $[Cr(NH_3)_4SO_4]^+$ is an octahedral complex of the type $[M(NH_3)_4(SO_4)]^+$. This complex possesses a plane of symmetry (the plane containing the $SO_4$ ligand and the two trans $NH_3$ ligands),making it achiral. Therefore,it does not show optical isomerism.

(A) transition element is defined as an element which has an incompletely filled $d$-orbital in its ground state or in any of its oxidation states.
$Au$ (Gold) has the ground state electronic configuration $[Xe] 4f^{14} 5d^{10} 6s^1$. However,in its $+3$ oxidation state,it has the configuration $[Xe] 4f^{14} 5d^8$,which has an incompletely filled $d$-orbital.
$Zn$,$Cd$,and $Hg$ belong to group $12$ and have the general electronic configuration $(n-1)d^{10} ns^2$. They do not have incompletely filled $d$-orbitals in their ground state or common oxidation states $(+2)$,and thus are not considered transition elements.
Therefore,$Au$ is the correct answer.

(A) The basic strength of metallic hydroxides depends on the ionic character of the $M-OH$ bond.
As the size of the metal cation increases,the ionic character of the $M-OH$ bond increases,leading to higher basicity.
$Al(OH)_3$ is amphoteric and is the least basic among the given compounds.
Among the lanthanoids,the basic strength decreases as the ionic radius decreases due to lanthanoid contraction $(Ce^{3+} > Lu^{3+})$.
$Ca(OH)_2$ is a strong base compared to the lanthanoid hydroxides.
Therefore,the correct order of increasing basic strength is $Al(OH)_3 < Lu(OH)_3 < Ce(OH)_3 < Ca(OH)_2$.

(B) The correct answer is $B$.
Pyrophoric Misch metal is an alloy of lanthanoid metals (about $95\%$) and iron (about $5\%$) with traces of $S, C, Ca,$ and $Al$.
It is highly pyrophoric,meaning it produces sparks when struck or scratched,which makes it ideal for use in gas lighters and bullet tips.

(C) The correct answer is $C$.
Metallic conductivity (or electronic conductivity) is due to the movement of electrons through the metal lattice.
As the temperature of a metal increases,the positive ions (kernels) in the lattice begin to vibrate more vigorously.
This increased vibration causes more collisions between the moving electrons and the vibrating ions,which increases the resistance of the metal.
Therefore,the electrical conductivity of a metal decreases,not increases,with an increase in temperature.

(C) In the electrolysis of a dilute aqueous solution of $NaCl$,water undergoes oxidation and reduction at the electrodes.
At the cathode,the reduction of water occurs: $2H_2O(l) + 2e^- \rightarrow H_{2(g)} + 2OH^-(aq)$.
At the anode,the oxidation of water occurs: $2H_2O(l) \rightarrow O_{2(g)} + 4H^+(aq) + 4e^-$.
Thus,$H_{2(g)}$ is produced at the cathode and $O_{2(g)}$ is produced at the anode.

(C) The $S_{N}2$ reaction is a second-order reaction where the rate depends on the concentration of both the substrate and the nucleophile.
Statement $C$ is incorrect because the rate law for an $S_{N}2$ reaction is given by $\text{Rate} = k[\text{Substrate}][\text{Nucleophile}]$,meaning it is dependent on both concentrations.

(C) The Swartz reaction is a method used to prepare alkyl fluorides by heating alkyl chlorides or alkyl bromides in the presence of metallic fluorides like $AgF$,$Hg_2F_2$,$CoF_2$,or $SbF_3$.
In the given options,the reaction $CH_3Br + AgF \longrightarrow CH_3F + AgBr$ represents the Swartz reaction.
Note: The reactions in options $A$ and $B$ are Finkelstein reactions,and the reaction in option $D$ is the Wurtz reaction.