MHT CET 2012 Chemistry Question Paper with Answer and Solution

(B) When a sudden jerk is given to string $C$,an impulsive force is applied to it.
Due to the inertia of the $1\, kg$ mass,it does not immediately move downward.
This impulsive force creates a very high tension in the lower string $C$ (the portion $BC$).
Since the tension in $C$ exceeds its breaking strength before the impulse can be transmitted through the mass to the upper string $A$,the string $BC$ will break first.

(C) The reaction of sodium acetate with sodalime $(NaOH + CaO)$ is a decarboxylation reaction.
$CH_{3}COONa + NaOH \xrightarrow{\Delta} CH_{4} + Na_{2}CO_{3}$
Here,sodium acetate reacts with sodalime to produce methane gas and sodium carbonate.

(B) The central iodine atom in $I_{3}^{-}$ undergoes $sp^{3}d$ hybridization.
It has $3$ lone pairs of electrons in the equatorial positions and $2$ bonding pairs in the axial positions.
According to $VSEPR$ theory,the presence of $3$ lone pairs results in a linear geometry with a bond angle of $180^{\circ}$.

(A) The structure of but$-2-$enedioic acid is $HOOC-CH=CH-COOH$.
In this molecule,each carbon atom is bonded to three other atoms (or groups) and has one double bond.
Specifically,the two carbonyl carbons are bonded to one oxygen (double bond),one hydroxyl group,and one carbon atom,making them $sp^{2}$ hybridized.
The two alkene carbons are each bonded to one hydrogen atom,one carbon atom,and one other carbon atom (double bond),also making them $sp^{2}$ hybridized.
Therefore,all $C$ atoms in but$-2-$enedioic acid are $sp^{2}$ hybridized.

(A) The $C-H$ bond length depends on the hybridization of the carbon atom.
As the $s$-character in the hybrid orbital increases,the bond length decreases.
In acetylene $(HC \equiv CH)$,the carbon is $sp$ hybridized ($50\% \ s$-character).
In ethylene $(CH_2=CH_2)$,the carbon is $sp^2$ hybridized ($33.3\% \ s$-character).
In methane $(CH_4)$ and ethane $(CH_3-CH_3)$,the carbon is $sp^3$ hybridized ($25\% \ s$-character).
Since $sp$ hybridization has the highest $s$-character,the $C-H$ bond length is the shortest in acetylene.

(B) The bond energy is directly proportional to the bond order. The bond order can be calculated using the formula: $\text{Bond Order} = \frac{N_b - N_a}{2}$,where $N_b$ is the number of bonding electrons and $N_a$ is the number of antibonding electrons.
$(A)$ $O_{2}^{2-}$: Total electrons = $18$. Configuration: $KK(\sigma 2s)^2, (\sigma^* 2s)^2, (\sigma 2p_z)^2, (\pi 2p_x)^2, (\pi 2p_y)^2, (\pi^* 2p_x)^2, (\pi^* 2p_y)^2$. Bond order = $\frac{10-8}{2} = 1$.
$(B)$ $O_{2}^{+}$: Total electrons = $15$. Configuration: $KK(\sigma 2s)^2, (\sigma^* 2s)^2, (\sigma 2p_z)^2, (\pi 2p_x)^2, (\pi 2p_y)^2, (\pi^* 2p_x)^1$. Bond order = $\frac{10-5}{2} = 2.5$.
$(C)$ $O_{2}^{-}$: Total electrons = $17$. Configuration: $KK(\sigma 2s)^2, (\sigma^* 2s)^2, (\sigma 2p_z)^2, (\pi 2p_x)^2, (\pi 2p_y)^2, (\pi^* 2p_x)^2, (\pi^* 2p_y)^1$. Bond order = $\frac{10-7}{2} = 1.5$.
$(D)$ $O_{2}$: Total electrons = $16$. Configuration: $KK(\sigma 2s)^2, (\sigma^* 2s)^2, (\sigma 2p_z)^2, (\pi 2p_x)^2, (\pi 2p_y)^2, (\pi^* 2p_x)^1, (\pi^* 2p_y)^1$. Bond order = $\frac{10-6}{2} = 2$.
Comparing the bond orders: $O_{2}^{+} (2.5) > O_{2} (2) > O_{2}^{-} (1.5) > O_{2}^{2-} (1)$.
Since $O_{2}^{+}$ has the highest bond order,it has the highest bond energy.

(A) The bond energy is directly proportional to the bond order. Let us calculate the bond order for each species using Molecular Orbital Theory:
$(A)$ $N_{2}^{2-}$: Total electrons = $14 + 2 = 16$. Configuration: $KK(\sigma 2s)^2 (\sigma^* 2s)^2 (\sigma 2p_z)^2 (\pi 2p_x)^2 (\pi 2p_y)^2 (\pi^* 2p_x)^1 (\pi^* 2p_y)^1$. Bond order = $(8 - 6) / 2 = 1$.
$(B)$ $N_{2}^{-}$: Total electrons = $14 + 1 = 15$. Configuration: $KK(\sigma 2s)^2 (\sigma^* 2s)^2 (\sigma 2p_z)^2 (\pi 2p_x)^2 (\pi 2p_y)^2 (\pi^* 2p_x)^1$. Bond order = $(8 - 5) / 2 = 1.5$.
$(C)$ $N_{2}^{+}$: Total electrons = $14 - 1 = 13$. Configuration: $KK(\sigma 2s)^2 (\sigma^* 2s)^2 (\sigma 2p_z)^1 (\pi 2p_x)^2 (\pi 2p_y)^2$. Bond order = $(7 - 2) / 2 = 2.5$.
$(D)$ $N_{2}$: Total electrons = $14$. Configuration: $KK(\sigma 2s)^2 (\sigma^* 2s)^2 (\pi 2p_x)^2 (\pi 2p_y)^2 (\sigma 2p_z)^2$. Bond order = $(8 - 2) / 2 = 3$.
Since $N_{2}^{2-}$ has the lowest bond order $(1)$,it has the least bond energy.

(C) For $K_{2}Cr_{2}O_{7}$,let the oxidation state of $Cr$ be $x$.
The sum of oxidation states of all atoms in a neutral molecule is $0$.
$2(+1) + 2x + 7(-2) = 0$
$2 + 2x - 14 = 0$
$2x - 12 = 0$
$2x = 12$
$x = +6$
Therefore,the oxidation state of $Cr$ in $K_{2}Cr_{2}O_{7}$ is $+6$.

(B) Tetraethyl lead $(TEL)$ is added to petrol as an anti-knocking agent.
It functions by increasing the octane number of the fuel,which prevents premature ignition or 'knocking' in internal combustion engines.

(D) $LPG$ (Liquefied Petroleum Gas) is a mixture of hydrocarbon gases used as a fuel.
It mainly consists of $butane$ $(C_4H_{10})$ and $propane$ $(C_3H_8)$.
Among the given options,$butane$ is the primary component.

(B) Ionic solvents (like water) are polar in nature. According to the principle of 'like dissolves like',polar solutes dissolve in polar solvents.
$CH_3OH$ (methanol) is a polar molecule capable of forming hydrogen bonds with water,making it soluble in ionic/polar solvents.
$C_6H_6$ (benzene),$CCl_4$ (carbon tetrachloride),and $C_5H_{12}$ (pentane) are non-polar organic compounds and are insoluble in ionic solvents.

(B) Among the given options,$CH_3COOH$ (acetic acid) is a weak acid because it partially dissociates in water to release $H^+$ ions.
$C_6H_6$ (benzene),$CH_2=CH_2$ (ethene),and $CH_3COCH_3$ (acetone) are not considered acids in the context of aqueous solutions as they do not readily donate $H^+$ ions.

(A) Aromatic compounds satisfy the $(4n+2) \pi$ $H$ückel rule.
$(A)$ Cyclobutadiene: Number of $\pi$ electrons $= 4$. Since it is not equal to $4n+2$,it is anti-aromatic,not aromatic.
$(B)$ Pyridine: Number of $\pi$ electrons $= 6$ ($4n+2$,where $n=1$). The lone pair on $N$ is not involved in delocalization.
$(C)$ Furan: Number of $\pi$ electrons $= 4 + 2 = 6$ ($4n+2$,where $n=1$).
$(D)$ Thiophene: Number of $\pi$ electrons $= 4 + 2 = 6$ ($4n+2$,where $n=1$).
Thus,except cyclobutadiene,all other species are aromatic.

(B) The alkenyl groups are derived by removing one hydrogen atom from an alkene. For $C_{4}H_{8}$ (butenes),the possible isomers are $1$-butene,$2$-butene,and $2$-methylpropene.
Removing a hydrogen atom from these structures gives the following $C_{4}H_{7}$ groups:
$(i)$ $CH_{3}CH_{2}CH=CH-$ ($1$-butenyl)
$(ii)$ $CH_{3}CH=CHCH_{2}-$ ($2$-butenyl)
$(iii)$ $CH_{2}=CHCH_{2}CH_{2}-$ ($3$-butenyl)
$(iv)$ $CH_{3}C(CH_{3})=CH-$ ($2$-methyl$-1-$propenyl)
$(v)$ $CH_{2}=C(CH_{3})CH_{2}-$ ($2$-methyl$-2-$propenyl)
Thus,there are $5$ possible alkenyl groups.

(A) The molecular formula $C_{4}H_{10}O$ corresponds to saturated acyclic compounds (alcohols and ethers). The structural isomers are as follows:
Alcohols:
$1$. $CH_{3}CH_{2}CH_{2}CH_{2}OH$ ($n$-butanol)
$2$. $CH_{3}CH_{2}CH(OH)CH_{3}$ (butan-$2$-ol)
$3$. $(CH_{3})_{2}CHCH_{2}OH$ (isobutanol)
$4$. $(CH_{3})_{3}COH$ (tert-butanol)
Ethers:
$5$. $CH_{3}OCH_{2}CH_{2}CH_{3}$ (methyl propyl ether)
$6$. $CH_{3}OCH(CH_{3})_{2}$ (methyl isopropyl ether)
$7$. $CH_{3}CH_{2}OCH_{2}CH_{3}$ (diethyl ether)
There are a total of $7$ structural isomers.

(C) Pentane has the molecular formula $C_5H_{12}$. Isomers must have the same molecular formula but different structural arrangements.
$1.$ $n$-pentane: $CH_3-CH_2-CH_2-CH_2-CH_3$ $(C_5H_{12})$
$2.$ $2,2$-dimethylpropane: $CH_3-C(CH_3)_2-CH_3$ $(C_5H_{12})$
$3.$ $2$-methylbutane: $CH_3-CH(CH_3)-CH_2-CH_3$ $(C_5H_{12})$
$4.$ $2,3$-dimethylbutane: $CH_3-CH(CH_3)-CH(CH_3)-CH_3$ $(C_6H_{14})$
Since $2,3$-dimethylbutane has $6$ carbon atoms and the formula $C_6H_{14}$,it is an isomer of hexane,not pentane. Thus,option $C$ is correct.

(B) Hydrocarbons are separated from petroleum based on their different boiling points using the $Fractional \ distillation$ method.

(B) Alkanes are saturated hydrocarbons and generally do not undergo addition reactions,which is why they do not decolourise bromine water or dilute $KMnO_{4}$ solution. They also do not typically undergo polymerization. However,alkanes do undergo elimination reactions,specifically dehydrogenation or cracking,when heated to high temperatures $(500-700^{\circ} C)$ in the absence of air. For example,propane undergoes cracking to form propene,ethene,methane,and hydrogen. Therefore,the statement that alkanes do not undergo elimination reactions is false.

(C) The rate of electrophilic aromatic substitution,such as sulphonation,depends on the electron density of the benzene ring.
$CH_3$ group in Toluene is an electron-donating group ($+I$ effect and hyperconjugation),which increases the electron density of the benzene ring,making it more reactive towards electrophiles compared to Benzene,Chlorobenzene (deactivating),and Nitrobenzene (strongly deactivating).
Therefore,Toluene undergoes sulphonation most easily.

(B) conjugate acid is formed by adding a proton $(H^{+})$ to a base.
For the species $HS^{-}$,adding a proton $(H^{+})$ results in $H_2S$.
The reaction is: $HS^{-} + H^{+} \rightleftharpoons H_2S$.
Therefore,the conjugate acid of $HS^{-}$ is $H_2S$.

(A) For a weak acid like acetic acid,the dissociation constant $K_{a}$ is related to the degree of ionization $\alpha$ and concentration $C$ by the formula: $K_{a} = C \alpha^{2}$.
Given $K_{a} = 1.8 \times 10^{-5}$ and $C = 0.4 \ M$.
Substituting the values: $1.8 \times 10^{-5} = 0.4 \times \alpha^{2}$.
$\alpha^{2} = \frac{1.8 \times 10^{-5}}{0.4} = 4.5 \times 10^{-5} = 45 \times 10^{-6}$.
$\alpha = \sqrt{45 \times 10^{-6}} \approx 6.71 \times 10^{-3}$.

(C) $pH = -\log [H^{+}]$
$pH = -\log (3.98 \times 10^{-6})$
$pH = -(\log 3.98 + \log 10^{-6})$
$pH = -(0.60 - 6)$
$pH = 5.40$

(A) Ethanol cannot be dried using anhydrous $CaCl_{2}$ because it reacts with $CaCl_{2}$ to form an addition compound (solvate). The reaction is as follows:
$CaCl_{2} + 4C_{2}H_{5}OH \rightarrow CaCl_{2} \cdot 4C_{2}H_{5}OH$ (solid)

(C) Let the oxidation number of $C$ be $x$.
For $CH_{2}Cl_{2}$:
$x + 2(+1) + 2(-1) = 0$
$x + 2 - 2 = 0$
$x = 0$
For $CCl_{4}$:
$x + 4(-1) = 0$
$x - 4 = 0$
$x = 4$
Thus,the oxidation numbers are $0$ and $4$ respectively.

(D) Hydrolysis of metal chlorides depends on the ionic character of the $M-Cl$ bond.
As we move down the group from $Be$ to $Ba$,the ionic character of the metal chloride increases due to the decrease in polarizing power of the cation.
$BeCl_{2}$ is covalent and undergoes hydrolysis,whereas $BaCl_{2}$ is highly ionic and is the least susceptible to hydrolysis among the given options.
Therefore,$BaCl_{2}$ is the least hydrolysed.

(B) Gypsum is a naturally occurring mineral with the chemical formula $CaSO_{4} \cdot 2H_{2}O$.

(C) The relation between volume strength and percentage strength of $H_2O_2$ is given by:
$\text{Volume strength} = 5.6 \times \text{Molarity}$
$\text{Molarity} = \frac{\% \text{ strength} \times 10}{\text{Molar mass of } H_2O_2} = \frac{6.8 \times 10}{34} = 2 \ M$
$\text{Volume strength} = 5.6 \times 2 = 11.2 \ V$
Alternatively,using the direct formula:
$\text{Volume strength} = \frac{112}{34} \times \% \text{ concentration} = \frac{112}{34} \times 6.8 = 22.4 \ V$

(D) The general equation for the combustion of alkanes is $C_n H_{2n+2} + (\frac{3n+1}{2}) O_2 \rightarrow n CO_2 + (n+1) H_2O + \text{Heat}$.
Greater the value of $n$ (number of carbon atoms),greater will be the amount of heat produced upon combustion.
Comparing the given alkanes: Methane $(n=1)$,Ethane $(n=2)$,Propane $(n=3)$,and Butane $(n=4)$.
Since Butane has the highest number of carbon atoms $(n=4)$,it releases the maximum energy upon combustion.

(B) Phenol reacts with bromine water $(Br_2/H_2O)$ to undergo electrophilic aromatic substitution,yielding a white precipitate of $2,4,6-$tribromophenol.
The reaction is:
$C_6H_5OH + 3Br_2(aq) \rightarrow C_6H_2Br_3OH(s) + 3HBr(aq)$

(C) $2,4,6-$trinitrophenol is commonly known as picric acid. The presence of three electron-withdrawing $-NO_2$ groups at the ortho and para positions significantly increases the acidity of the phenolic $-OH$ group by stabilizing the phenoxide ion through resonance and inductive effects. Therefore,it is strongly acidic in nature.

(B) The basic nature of an amine is due to the presence of a lone pair of electrons on the nitrogen atom,which is available for bond formation with a Lewis acid.
In $CH_3NH_2$,the $+I$ effect of the $-CH_3$ group increases the electron density on the nitrogen atom,making it more available for donation compared to $NH_3$.
In $C_6H_5NH_2$,the lone pair is involved in resonance with the benzene ring,reducing its availability.
In $CH_3CONH_2$,the lone pair is involved in resonance with the carbonyl group $(C=O)$,significantly reducing its basicity.
Therefore,$CH_3NH_2$ is the strongest base among the given options.

(B) The structures of the given acids are:
$(A)$ Carbamic acid: $NH_2COOH$ (contains $-COOH$ group)
$(B)$ Barbituric acid: $C_4H_4N_2O_3$ (a cyclic amide,does not contain $-COOH$ group)
$(C)$ Lactic acid: $CH_3CH(OH)COOH$ (contains $-COOH$ group)
$(D)$ Succinic acid: $HOOCCH_2CH_2COOH$ (contains $-COOH$ group)
Therefore,barbituric acid is the correct answer as it is a pyrimidine derivative and not a carboxylic acid.

(B) $HCHO$ (formaldehyde) is the most reactive aldehyde in its homologous series due to the absence of electron-donating alkyl groups and minimal steric hindrance. Therefore,the statement that $HCHO$ is the least reactive is false.
$40 \%$ solution of $HCHO$ is indeed called formalin.
Branching in iso-valeraldehyde decreases the surface area compared to $n$-valeraldehyde,leading to a lower boiling point.
Ketones generally have higher boiling points than isomeric aldehydes due to the higher dipole moment of the carbonyl group.

(D) The alkaline hydrolysis of an ester is known as saponification.
When an ester is heated with aqueous $NaOH$,a sodium salt of the carboxylic acid and an alcohol are formed.
The reaction is represented as:
$R-COOR' + NaOH \rightarrow R-COO^-Na^+ + R'OH$

(A) The atomic number of $Cu$ is $29$. The electronic configuration is $[Ar] 3d^{10} 4s^1$.
In $[Cu(NH_3)_4]^{2+}$,the oxidation state of $Cu$ is $+2$. The electronic configuration of $Cu^{2+}$ is $[Ar] 3d^9$.
In the presence of $4$ $NH_3$ ligands,one electron from the $3d$ orbital is promoted to the $4p$ orbital to facilitate $dsp^2$ hybridization.
Thus,the hybridization is $dsp^2$.

(B) Chlorofluorocarbons,specifically dichlorodifluoromethane $(CCl_2F_2)$,are commonly used as refrigerants in refrigerators and air conditioners.

(B) The acidity of carboxylic acids is influenced by the presence of electron-withdrawing groups (EWGs) such as the chloro group $(-Cl)$.
These groups exert a $-I$ (negative inductive) effect,which stabilizes the carboxylate anion by dispersing the negative charge.
As the number of electron-withdrawing chlorine atoms increases,the $-I$ effect becomes stronger,leading to greater stabilization of the conjugate base and thus higher acidity.
The order of acidity is: $CCl_3COOH > CHCl_2COOH > CH_2ClCOOH > CH_3COOH$.
Therefore,trichloroacetic acid is the strongest acid among the given options.

(B) All the given groups are $o-$ and $p-$ directing.
However,in the $-OH$ group,the lone pair of electrons on the $O$-atom participates in resonance,which significantly increases the electron density at the $o-$ and $p-$ positions of the benzene ring.
This makes the $-OH$ group a strongly activating and strongly $o-$ and $p-$ directing group compared to the others.

(D) Chain isomerism in ketones requires a minimum of $5$ carbon atoms in the main chain to allow for branching.
For example,pentan-$2$-one $(CH_3COCH_2CH_2CH_3)$ can form a chain isomer,$3$-methylbutan-$2$-one $(CH_3COCH(CH_3)_2)$,by branching the carbon chain.
Thus,the minimum number of carbon atoms required is $5$.

(B) Iron pyrite,$FeS_{2}$,is commonly known as fool's gold due to its metallic luster and pale brass-yellow hue,which resembles gold.

(A) The ores of iron are:
$1$. Haematite: $Fe_2O_3$
$2$. Siderite: $FeCO_3$
$3$. Limonite: $Fe_2O_3 \cdot 3H_2O$
Malachite is an ore of copper with the formula $CuCO_3 \cdot Cu(OH)_2$.
Therefore,malachite is not an iron ore.

(C) The given reaction is $C_{6}H_{6} + CH_{3}Cl \xrightarrow{\text{Anhy. } AlCl_{3}} C_{6}H_{5}CH_{3} + HCl$.
This is an electrophilic aromatic substitution reaction where a methyl group is introduced into the benzene ring using an alkyl halide in the presence of a Lewis acid catalyst $(AlCl_{3})$.
This reaction is known as the Friedel-Crafts alkylation reaction.

(A) $Pb^{2+}$ ion is capable of precipitating $Cl^{-}$,$I^{-}$,and $SO_{4}^{2-}$ ions as insoluble salts.
$Pb^{2+} + 2Cl^{-} \longrightarrow PbCl_{2} \downarrow$ (White precipitate)
$Pb^{2+} + 2I^{-} \longrightarrow PbI_{2} \downarrow$ (Yellow precipitate)
$Pb^{2+} + SO_{4}^{2-} \longrightarrow PbSO_{4} \downarrow$ (White precipitate)
While $Ba^{2+}$ precipitates $SO_{4}^{2-}$,it does not form insoluble precipitates with $Cl^{-}$ and $I^{-}$ in aqueous solutions.

(A) The Haber process is an industrial method used for the large-scale production of ammonia $(NH_3)$.
In this process,nitrogen gas reacts with hydrogen gas in the presence of an iron catalyst at high temperature and pressure.
The chemical equation is: $N_{2(g)} + 3H_{2(g)} \rightleftharpoons 2NH_{3(g)}$

(C) Nitrous oxide,$N_{2}O$,is commonly known as laughing gas.

(A) When $BiCl_{3}$ is diluted with water,it undergoes hydrolysis to form a white precipitate of bismuth oxychloride $(BiOCl)$.
The chemical reaction is: $BiCl_{3} + H_{2}O \rightleftharpoons BiOCl \downarrow + 2HCl$
Here,$BiOCl$ is the white precipitate.

(C) Xenon forms the following stable fluorides: $XeF_{2}$,$XeF_{4}$,and $XeF_{6}$.
$XeF_{5}$ is not a known stable compound of xenon.

(B) $Nylon-6,6$ is a condensation polymer formed by the reaction between hexamethylene diamine $(nH_2N-(CH_2)_6-NH_2)$ and adipic acid $(nHOOC-(CH_2)_4-COOH)$.
Since the repeating units are linked by amide bonds $(-CONH-)$,it is classified as a polyamide.

(C) Natural rubber is a linear polymer of isoprene ($2-$methyl-$1,3-$butadiene).
It is formed by the polymerization of isoprene units.
The reaction is as follows:
$nCH_2=C(CH_3)-CH=CH_2 \xrightarrow{\text{Polymerization}} [CH_2-C(CH_3)=CH-CH_2]_n$

(C) For $NaCl$: $n(Cl^-) = M \times V = 3.0 \ M \times 0.300 \ L = 0.9 \ mol$.
For $BaCl_{2}$: $BaCl_{2} \rightarrow Ba^{2+} + 2Cl^-$.
$n(Cl^-) = 2 \times M \times V = 2 \times 4.0 \ M \times 0.200 \ L = 1.6 \ mol$.
Total moles of $Cl^- = 0.9 \ mol + 1.6 \ mol = 2.5 \ mol$.
Total volume of solution = $300 \ mL + 200 \ mL = 500 \ mL = 0.5 \ L$.
Molar concentration of $Cl^- = \frac{\text{Total moles}}{\text{Total volume}} = \frac{2.5 \ mol}{0.5 \ L} = 5.0 \ M$.

(C) The magnetic moment $(\mu)$ is calculated using the formula $\mu = \sqrt{n(n+2)} \text{ BM}$,where $n$ is the number of unpaired electrons.
$1$. For $Mn^{6+}$ $([Ar]3d^1)$: $n = 1$,$\mu = \sqrt{1(1+2)} = 1.73 \text{ BM}$.
$2$. For $Ni^{2+}$ $([Ar]3d^8)$: $n = 2$,$\mu = \sqrt{2(2+2)} = 2.83 \text{ BM}$.
$3$. For $Fe^{3+}$ $([Ar]3d^5)$: $n = 5$,$\mu = \sqrt{5(5+2)} = 5.91 \text{ BM}$.
$4$. For $Ag^{+}$ $([Kr]4d^{10})$: $n = 0$,$\mu = 0 \text{ BM}$.
Thus,$Fe^{3+}$ has the maximum number of unpaired electrons $(n=5)$ and consequently shows the maximum magnetic moment.