MHT CET 2022 Chemistry Question Paper with Answer and Solution

(C) According to Molecular Orbital Theory $(MOT)$,the electronic configuration of the $O_2$ molecule ($16$ electrons) is:
$\sigma 1s^2, \sigma^* 1s^2, \sigma 2s^2, \sigma^* 2s^2, \sigma 2p_z^2, \pi 2p_x^2 = \pi 2p_y^2, \pi^* 2p_x^1 = \pi^* 2p_y^1$.
Bond order is calculated as:
$\text{Bond Order} = \frac{N_b - N_a}{2}$
Where $N_b$ is the number of bonding electrons and $N_a$ is the number of antibonding electrons.
$N_b = 10$
$N_a = 6$
$\text{Bond Order} = \frac{10 - 6}{2} = \frac{4}{2} = 2$.
Thus,the bond order of $O_2$ is $2$.

(D) The $CO$ molecule has a triple bond between the carbon and oxygen atoms,represented as $:C \equiv O:$.
Using the Molecular Orbital Theory,the bond order is calculated as:
$B.O. = \frac{1}{2} (N_b - N_a) = \frac{1}{2} (10 - 4) = 3$.
Therefore,the bond order of $CO$ is $3$.

(A) The ionization enthalpy $(I.E.)$ is the energy required to remove an electron from an isolated gaseous atom.
In the periodic table,as we move down a group,the atomic size increases and the valence electrons are further from the nucleus,which leads to a decrease in $I.E.$
All the given elements $(Li, K, Rb, Cs)$ belong to Group $1$ (Alkali metals).
Since $Li$ is at the top of the group,it has the smallest atomic size and therefore the highest $I.E.$ among the given options.

(A) According to $Fajan's$ rule,the covalent character of an ionic bond increases with an increase in the polarising power of the cation.
Polarising power is directly proportional to the charge density of the cation.
The charge on the cations is $Al^{+3} > Mg^{+2} > Na^{+}$.
Therefore,the polarising power follows the order $Al^{+3} > Mg^{+2} > Na^{+}$.
Consequently,the decreasing order of covalent character is $AlCl_3 > MgCl_2 > NaCl$.

(C) non-benzenoid aromatic compound is a cyclic,planar,conjugated system that follows $H$ückel's rule ($4n+2$ $\pi$ electrons) but does not contain a benzene ring.
$1$. Benzene,Phenol,and Naphthalene are all benzenoid aromatic compounds as they contain at least one benzene ring.
$2$. Tropone $(C_7H_6O)$ contains a seven-membered ring. Its resonance structure involves a dipolar form where the seven-membered ring becomes a tropylium cation $(C_7H_7^+)$,which is aromatic ($6$ $\pi$ electrons) and non-benzenoid.
Therefore,the correct option is $C$.

(C) molecule is non-polar if its net dipole moment is zero.
$HCl$ is a polar molecule due to the electronegativity difference between $H$ and $Cl$.
$NH_3$ has a pyramidal geometry with a lone pair on nitrogen,resulting in a net dipole moment.
$ICl$ is a polar molecule due to the electronegativity difference between $I$ and $Cl$.
$C_6H_6$ (Benzene) is a planar,symmetric molecule where the individual bond dipoles cancel each other out,making it non-polar.

(D) The dipole moment $(\mu)$ depends on the electronegativity difference between atoms and the geometry of the molecule.
$1$. $CH_3CH_2CH_3$ is a non-polar hydrocarbon with a very low dipole moment.
$2$. $CH_3OCH_3$ (dimethyl ether) has a dipole moment of approximately $1.30 \ D$.
$3$. $CH_3Cl$ (chloromethane) has a dipole moment of approximately $1.86 \ D$.
$4$. $CH_3CN$ (acetonitrile) has a very high dipole moment of approximately $3.92 \ D$ due to the strong electron-withdrawing effect of the cyano group $(-CN)$ and the linear geometry of the $C-C \equiv N$ bond,which creates a large charge separation.
Therefore,$CH_3CN$ has the maximum dipole moment.

(A) The electronegativity of $F$ $(4.0)$ is greater than that of $N$ $(3.0)$,while the electronegativity of $N$ $(3.0)$ is greater than that of $H$ $(2.1)$.
In $NH_3$,the orbital dipole (due to the lone pair) and the resultant dipole moment of the three $N-H$ bonds are in the same direction,leading to a higher net dipole moment $(1.46 \ D)$.
In $NF_3$,the orbital dipole and the resultant dipole moment of the three $N-F$ bonds are in opposite directions,which partially cancels out,leading to a lower net dipole moment $(0.24 \ D)$.
Therefore,the statement that 'Fluorine is less electronegative than nitrogen' is incorrect.

(D) Intramolecular hydrogen bonding occurs within the same molecule.
In $Ethylene \ glycol$ $(HO-CH_2-CH_2-OH)$,the hydrogen atom of one hydroxyl group forms a hydrogen bond with the oxygen atom of the other hydroxyl group within the same molecule,as shown in the structure:
$CH_2(OH)-CH_2(OH)$ forming a cyclic structure via hydrogen bonding.
Ammonia $(NH_3)$,Hydrogen fluoride $(HF)$,and Water $(H_2O)$ exhibit intermolecular hydrogen bonding (between different molecules).

(A) The dipole-dipole interaction strength depends on the polarity of the molecule,which is determined by the electronegativity difference between the bonded atoms.
$F$ is the most electronegative element among the halogens,making the $H-F$ bond the most polar.
Therefore,$HF$ exhibits the highest dipole-dipole interaction among the given molecules.

(B) For a weak monobasic acid,the dissociation constant $K_a$ is given by the formula $K_a = C \alpha^2$,where $C$ is the concentration and $\alpha$ is the degree of dissociation.
Given,$C = 0.05 \ M$ and percentage dissociation $= 0.001 \%$.
Therefore,$\alpha = \frac{0.001}{100} = 10^{-5}$.
Substituting the values into the formula:
$K_a = 0.05 \times (10^{-5})^2$
$K_a = 0.05 \times 10^{-10}$
$K_a = 5 \times 10^{-12}$.

(B) For a weak acid,the dissociation constant $K_a$ is given by the formula $K_a = C \alpha^2$,where $C$ is the concentration and $\alpha$ is the degree of dissociation.
Given: $C = 0.1 \ M$ and $\alpha = 1.5 \% = \frac{1.5}{100} = 0.015$.
Substituting the values:
$K_a = 0.1 \times (0.015)^2$
$K_a = 0.1 \times 0.000225$
$K_a = 2.25 \times 10^{-5}$

(A) For a weak monobasic acid,the dissociation constant $K_a$ is given by the formula $K_a = C \alpha^2$,where $C$ is the molar concentration and $\alpha$ is the degree of dissociation.
Given: $C = 0.1 \ M$ and $\alpha = 2 \% = 0.02$.
Substituting the values: $K_a = 0.1 \times (0.02)^2$.
$K_a = 0.1 \times 0.0004 = 4.0 \times 10^{-5}$.

(D) For a weak base,the dissociation constant $K_b$ is given by the formula $K_b = \alpha^2 C$,where $\alpha$ is the degree of dissociation and $C$ is the molar concentration.
Given $\alpha = 1.2 \% = 0.012$ and $C = 0.2 \ M$.
Substituting the values: $K_b = (0.012)^2 \times 0.2$.
$K_b = 0.000144 \times 0.2 = 2.88 \times 10^{-5}$.

(D) For a weak monobasic acid,the dissociation constant $K_a$ is given by the formula $K_a = C \alpha^2$,where $C$ is the concentration and $\alpha$ is the degree of dissociation.
Given: $C = 0.02 \ M$ and percent dissociation = $3 \%$.
Therefore,$\alpha = 3 / 100 = 0.03$.
Substituting the values into the formula:
$K_a = 0.02 \times (0.03)^2$
$K_a = 0.02 \times 0.0009$
$K_a = 1.8 \times 10^{-5}$

(D) For a weak acid,the dissociation constant $K_a$ is given by the formula $K_a = C \alpha^2$,where $C$ is the concentration and $\alpha$ is the degree of dissociation.
Given,$C = 0.01 \ M = 10^{-2} \ M$.
The degree of dissociation $\alpha = \frac{0.001}{100} = 10^{-5}$.
Substituting the values into the formula:
$K_a = (10^{-2}) \times (10^{-5})^2$
$K_a = 10^{-2} \times 10^{-10}$
$K_a = 10^{-12}$.

(B) The atomic number of copper $(Cu)$ is $29$.
Its electronic configuration is $[Ar] 3d^{10} 4s^1$.
Since the valence electrons enter the $4s$ orbital,it belongs to Period $4$.
It is a transition metal located in Group $11$ of the $d$-block.

(C) The elements of group $1$ (alkali metals) have $1$ electron in their valence shell.
Among the given options,$Rb$ (Rubidium) belongs to group $1$,while $Ca$,$Ra$,and $Ba$ belong to group $2$ (alkaline earth metals).

(D) In a period,as we move from left to right,the atomic radius generally decreases due to an increase in effective nuclear charge.
For the elements $Si$,$P$,$Cl$,and $Ar$ belonging to the $3^{rd}$ period,the atomic radius decreases from $Si$ to $Cl$.
However,$Ar$ is a noble gas and its atomic radius is measured as the van der Waals radius,which is significantly larger than the covalent radii of the other elements in the same period.
Therefore,among the given options,$Cl$ has the smallest atomic radius.

(B) Amphoteric oxides are those that react with both acids and bases to form salt and water.
Among the given options,$BeO$ (Beryllium oxide) is amphoteric in nature.
$MgO$,$BaO$,and $Li_2O$ are basic oxides.

(D) The electronic configuration of $Cr (Z=24)$ is $[Ar] 3d^5 4s^1$.
This configuration is adopted because half-filled $d$-orbitals $(d^5)$ provide extra stability.
In the $3d$ subshell,there are $5$ electrons,each occupying one of the $5$ $d$-orbitals singly.
Therefore,the number of unpaired electrons in the $d$-orbitals is $5$.

(C) good reducing agent is a substance that readily loses electrons and has a low standard reduction potential. Among the given options,$Li$ (Lithium) has the most negative standard reduction potential $(E^\circ = -3.04 \ V)$,making it the strongest reducing agent in aqueous solution.

(C) The atomic number of Chromium $(Cr)$ is $24$.
Its expected electronic configuration is $[Ar] 3d^4 4s^2$.
However,due to the extra stability of half-filled $d$-orbitals,the actual electronic configuration is $[Ar] 3d^5 4s^1$.
In this configuration,there are $5$ unpaired electrons in the $3d$ subshell and $1$ unpaired electron in the $4s$ subshell.
Therefore,the total number of unpaired electrons is $5 + 1 = 6$.

(B) The standard reduction potential $(E^\circ)$ is a measure of the tendency of a species to gain electrons.
Lithium $(Li)$ has the most negative standard reduction potential of approximately $-3.04 \ V$ among the alkali metals.
This is due to the very high hydration energy of the small $Li^+$ ion,which compensates for its high sublimation and ionization energy,making it the strongest reducing agent in aqueous solution.

(A) The principle of green chemistry states that unnecessary derivatization (such as the use of blocking groups,protection/deprotection,or temporary modification of physical/chemical processes) should be avoided whenever possible because these steps require additional reagents and generate waste. Therefore,the statement that protecting and deprotecting functional groups reduces the number of steps is incorrect and against the principles of green chemistry.

(D) Green chemistry is a strategy to design products and processes that minimize the use and generation of hazardous substances.
Key principles include:
$1$. Prevention of waste generation.
$2$. Atom economy.
$3$. Designing safer chemicals.
$4$. Use of renewable feedstocks.
$5$. Catalysis.
Option $D$ states that it aims to dump waste products,which is the opposite of the green chemistry principle of waste valorization and circular economy,where waste from one process is used as a raw material for another.

(C) According to the principles of green chemistry,reactions should be designed to maximize energy efficiency. Carrying out reactions at high temperature and high pressure consumes a significant amount of energy,which contradicts the principle of energy efficiency. Therefore,statement $C$ is $NOT$ true.

(C) The given structure is $(CH_3)_3N$,which consists of a nitrogen atom bonded to three methyl groups.
According to $IUPAC$ nomenclature for tertiary amines,the longest carbon chain attached to the nitrogen is selected as the parent alkane,and the other alkyl groups are named as $N-$substituents.
Here,the parent chain is methane $(CH_4)$,and there are two methyl groups attached to the nitrogen atom.
Therefore,the $IUPAC$ name is $N, N-$dimethylmethanamine.

(C) $1$. Identify the longest carbon chain: The longest chain contains $4$ carbon atoms,so the parent alkane is butane.
$2$. Number the chain: Numbering should be done such that the substituents get the lowest possible locants. Numbering from left to right gives the bromine at position $2$ and the methyl group at position $3$.
$3$. Assign names to substituents: There is a bromo group at position $2$ and a methyl group at position $3$.
$4$. Combine: The $IUPAC$ name is $2-$bromo$-3-$methylbutane.

(D) The $IUPAC$ name prop$-2-$en$-1-$amine indicates a three-carbon chain (prop) with a double bond at the second position ($-2-$en) and an amine group at the first position ($-1-$amine).
The structure is $CH_2=CH-CH_2-NH_2$.

(A) $1$. The principal functional group is the ketone,which gets the lowest possible locant. Thus,the carbonyl carbon is assigned position $1$.
$2$. Numbering the ring to give the substituents the lowest possible locants,we move towards the chlorine atom to give it position $2$.
$3$. The methyl group is then at position $4$.
$4$. Combining these,the $IUPAC$ name is $2-$chloro$-4-$methylcyclohexanone.

(C) The structure $CH_3CH(CH_3)CH_3$ is $iso-$butane (or $2-$methylpropane),not $n-$butane.
The structure of $n-$butane is $CH_3CH_2CH_2CH_3$.
Therefore,option $(C)$ is the incorrect match.

(A) $1$. Identify the principal functional group: The $-COOH$ group has the highest priority,so the parent compound is benzoic acid.
$2$. Number the ring: Start numbering from the carbon attached to the $-COOH$ group as $C-1$. Number the ring to give the lowest possible locants to the substituents.
$3$. Assign positions: The $-OH$ group is at position $3$ and the $-CH_3$ group is at position $4$.
$4$. Name the compound: Combining these,the $IUPAC$ name is $3-$hydroxy$-4-$methylbenzoic acid.

(B) Aliphatic compounds generally burn with a non-sooty (clean) flame because they have a lower carbon-to-hydrogen ratio compared to aromatic compounds. Aromatic compounds,due to their high carbon content,typically burn with a sooty flame. Therefore,the statement that aliphatic compounds burn with a sooty flame is incorrect.

(D) Chlorine is an electronegative atom and exerts an $-I$ (inductive) effect,not a $+I$ effect,due to its ability to withdraw electron density from the carbon chain.
Therefore,the statement in option $D$ is incorrect.

(A) The $-COOH$ group contains a carbonyl group $(C=O)$ attached to the carbon chain,which acts as an electron-withdrawing group via resonance ($-R$ effect).
In contrast,$-OH$,$-OR$,and $-NHR$ groups contain atoms with lone pairs of electrons that donate electron density into the ring or system via resonance ($+R$ effect).

(A) The stability of a carbon free radical $(CFR)$ is directly proportional to the number of $\alpha$-hydrogen atoms present due to hyperconjugation.
$1. \dot{C}H_3$ (methyl radical): $0 \ \alpha-H$
$2. CH_3-\dot{C}H_2$ (ethyl radical): $3 \ \alpha-H$
$3. (CH_3)_2\dot{C}H$ (isopropyl radical): $6 \ \alpha-H$
$4. (CH_3)_3\dot{C}$ (tert-butyl radical): $9 \ \alpha-H$
Since the stability order is $(CH_3)_3\dot{C} > (CH_3)_2\dot{C}H > CH_3-\dot{C}H_2 > \dot{C}H_3$,the methyl radical $\dot{C}H_3$ is the least stable.

(D) Naphthalene consists of two fused benzene rings.
It contains $5$ double bonds in its structure.
Each double bond contributes $2$ $\pi$ electrons.
Therefore,the total number of $\pi$ electrons = $5 \times 2 = 10$ $\pi$ electrons.
This satisfies $H$ückel's rule ($4n + 2 = 10$,where $n = 2$),confirming its aromatic nature.

(D) Pyridine is a heterocyclic aromatic compound with the formula $C_5H_5N$.
It contains three double bonds in the ring,each contributing $2$ $\pi$ electrons.
The lone pair on the nitrogen atom is in an $sp^2$ hybridized orbital and is not part of the aromatic $\pi$ system.
Therefore,the total number of $\pi$ electrons is $3 \times 2 = 6$.

(D) Hyperconjugation requires the presence of at least one $\alpha-H$ atom on a carbon atom adjacent to a double bond or a carbocation.
$A) \ CH_3-CH=CH_2$: Has $3 \ \alpha-H$ atoms.
$B) \ CH_3-C(CH_3)=C(CH_3)-CH_3$: Has $12 \ \alpha-H$ atoms.
$C) \ (CH_3)_3C^+$: Has $9 \ \alpha-H$ atoms.
$D) \ CH_2=CH_2$: Has $0 \ \alpha-H$ atoms.
Since $CH_2=CH_2$ has no $\alpha-H$ atoms,hyperconjugation is not observed.

(D) Position isomerism occurs when the position of a functional group or a double bond changes while the carbon skeleton remains the same.
In $CH_3-CH_2-CH=CH_2$ (but$-1-$ene),the double bond is at the $C-1$ position.
In $CH_3-CH=CH-CH_3$ (but$-2-$ene),the double bond is at the $C-2$ position.
Since the parent carbon chain is the same and only the position of the double bond differs,they are position isomers.

(A) substance is optically active if it contains at least one chiral center (an asymmetric carbon atom bonded to four different groups).
$A$. $CH_3-CH(Br)-CH_2-CH_3$: The central carbon is bonded to $-H$,$-CH_3$,$-Br$,and $-CH_2CH_3$. Since all four groups are different,it is a chiral center. Thus,it is optically active.
$B$. $(CH_3)_2-C(Br)-CH_3$: The central carbon is bonded to two identical $-CH_3$ groups. It is achiral.
$C$. $CH_3-(CH_2)_3-CH_2-Br$: The carbon attached to $-Br$ is bonded to two identical $-H$ atoms. It is achiral.
$D$. $(CH_3)_2-CH-CH(Br)-CH(CH_3)_2$: While this molecule has chiral centers,option $A$ is the simplest and most direct example of a chiral molecule often used in textbooks to demonstrate optical activity. However,based on the provided solution image,option $A$ is the intended answer.

(B) molecule is chiral if it contains at least one chiral center (a carbon atom bonded to four different groups).
In $CH_3-CH(Cl)-CH_2-CH_3$ ($2$-chlorobutane),the second carbon atom is bonded to four different groups: $-H$,$-Cl$,$-CH_3$,and $-CH_2CH_3$.
Since it has one chiral center,it is a chiral molecule.
Other options do not have a chiral center.

(A) chiral carbon atom is a carbon atom bonded to four different groups.
In $CH_3-CH(Cl)-CH_2-CH_3$,the second carbon atom is bonded to $-H$,$-Cl$,$-CH_3$,and $-CH_2CH_3$.
Since all four groups are different,this carbon is chiral.

(D) Metamerism arises due to the presence of different alkyl groups attached to the same polyvalent functional group (like $-O-$,$-S-$,$-NH-$,etc.).
In $Ethoxyethane$ $(CH_3CH_2-O-CH_2CH_3)$,the oxygen atom is attached to two ethyl groups.
In $1-Methoxypropane$ $(CH_3-O-CH_2CH_2CH_3)$,the oxygen atom is attached to a methyl group and a propyl group.
Since the alkyl groups attached to the oxygen atom are different,these compounds are metamers.

(A) For a compound to exhibit cis-trans isomerism,each carbon atom involved in the double bond must be attached to two different groups.
In the compound $H_2C=C(R)_2$,the first carbon atom is attached to two identical hydrogen atoms.
Since at least one carbon atom of the double bond does not have two different groups attached to it,this compound cannot exhibit cis-trans isomerism.

(B) The mineral barite is chemically known as barium sulfate,which has the chemical formula $BaSO_4$.
Therefore,the elements present in barite are Barium $(Ba)$,Sulfur $(S)$,and Oxygen $(O)$.

(B) The cleavage of the $C-X$ bond in these haloarenes occurs via Nucleophilic Aromatic Substitution $(Ar-S_N)$ reactions.
These reactions are facilitated by the presence of electron-withdrawing groups $(-NO_2)$ on the benzene ring,which stabilize the carbanion intermediate formed during the reaction.
The rate of $Ar-S_N$ reaction is directly proportional to the number of electron-withdrawing groups ($-M$ and $-I$ effect) present on the ring.
Compound $(I)$ has one $-NO_2$ group.
Compound $(II)$ has two $-NO_2$ groups.
Compound $(III)$ has three $-NO_2$ groups.
Therefore,the reactivity order is $III > II > I$.

(B) The given compound is bicyclo[$2.2$.$1$]heptane.
$A$ primary $(1^{\circ})$ carbon atom is a carbon atom bonded to only one other carbon atom.
In this cyclic structure,every carbon atom is bonded to at least two other carbon atoms.
Specifically,the bridgehead carbons are tertiary $(3^{\circ})$ and the other carbons are secondary $(2^{\circ})$.
Therefore,there are zero primary carbon atoms present in the compound.

(B) Alkanes with high molecular weight,specifically $C_{36}H_{74}$ and above,are known as paraffin wax or asphaltic residues,which are commonly used for road surfacing and waterproofing.

(A) The given reaction is a Hofmann elimination reaction.
In this reaction,a quaternary ammonium hydroxide is heated,leading to the formation of an alkene and a tertiary amine.
The hydroxide ion $(OH^-)$ acts as a base and abstracts a proton from the $\beta$-carbon atom of the alkyl group attached to the nitrogen.
According to the Hofmann rule,the less substituted alkene is formed as the major product.
In the given reactant,the alkyl groups attached to the nitrogen are one propyl group $(CH_3CH_2CH_2-)$ and three ethyl groups $(-CH_2CH_3)$.
When the base abstracts a proton from the $\beta$-carbon of an ethyl group,ethene $(H_2C=CH_2)$ is formed as the alkene product $A$.

(C) $KCN$ is an ionic compound,so it provides $CN^-$ ions which act as an ambident nucleophile. The carbon atom is more nucleophilic,leading to the formation of alkyl cyanide $(C_2H_5CN)$.
Therefore,the reaction $C_2H_5Br + KCN \rightarrow C_2H_5NC + KBr$ does not occur as written; it should produce $C_2H_5CN$ (ethyl cyanide).
$AgCN$ is covalent,so the nitrogen atom is the nucleophilic site,leading to the formation of alkyl isocyanide $(C_2H_5NC)$.
$AgNO_2$ is covalent,leading to nitroalkane $(C_2H_5NO_2)$.
$KNO_2$ is ionic,leading to alkyl nitrite $(C_2H_5ONO)$.

(A) The Carbylamine reaction is a chemical test for the detection of primary amines. In this reaction,a primary amine is heated with chloroform $(CHCl_3)$ and alcoholic potassium hydroxide $(KOH)$ to form an isocyanide (carbylamine),which has a foul smell.
The reaction is: $R-NH_2 + CHCl_3 + 3KOH \xrightarrow{\Delta, \text{alc.}} R-NC + 3KCl + 3H_2O$.

(A) Fructose is a ketohexose with the molecular formula $C_6H_{12}O_6$.
In its cyclic form,known as fructofuranose,the ring is formed by the reaction between the ketone group at $C-2$ and the hydroxyl group at $C-5$.
This creates a five-membered ring containing four carbon atoms and one oxygen atom.
Therefore,the $C-2$ and $C-5$ atoms are linked through an oxygen atom to form the furanose ring structure.

(D) Ribose is a monosaccharide,therefore it cannot be further hydrolysed into smaller carbohydrate units.

(D) Fructose is a ketohexose that is levorotatory in nature,which is why it is commonly known as laevulose.

(D) The structure of glucose is $CHO-(CHOH)_4-CH_2OH$.
In this structure,the $-CH_2OH$ group is a primary alcoholic group because the carbon atom is attached to only one other carbon atom.
There is $1$ such primary alcoholic group.
The four $-CHOH$ groups are secondary alcoholic groups because each carbon atom is attached to two other carbon atoms.
Thus,there are $4$ secondary alcoholic groups.
Therefore,the number of primary and secondary alcoholic groups is $1$ and $4$ respectively.

(D) The open-chain structure of glucose is $CHO-(CHOH)_4-CH_2OH$.
In this structure,the four $-CHOH-$ groups are attached to two other carbon atoms,making the hydroxyl groups attached to these carbons secondary $(2^{\circ})$ hydroxyl groups.
The $-CH_2OH$ group at the end is a primary $(1^{\circ})$ hydroxyl group.
Therefore,there are $4$ secondary hydroxyl groups in glucose.

(A) Glucose contains an aldehyde group $(-CHO)$ at the $C-1$ position and a primary alcohol group $(-CH_2OH)$ at the $C-6$ position.
Bromine water $(Br_2-H_2O)$ is a mild oxidizing agent.
It selectively oxidizes the aldehyde group $(-CHO)$ of glucose to a carboxylic acid group $(-COOH)$,forming gluconic acid.
Therefore,the carbon atom at the $C-1$ position is converted to a $-COOH$ group.

(B) Glucose contains an aldehyde group $(-CHO)$ at $C-1$ and a primary alcoholic group $(-CH_2OH)$ at $C-6$.
When glucose is treated with dilute nitric acid $(dil. HNO_3)$,it undergoes oxidation to form a dicarboxylic acid known as saccharic acid (or glucaric acid).
In this reaction,both the terminal aldehyde group and the primary alcoholic group are oxidized to carboxylic acid $(-COOH)$ groups.
The reaction is represented as:
$CHO-(CHOH)_4-CH_2OH + [O] \xrightarrow{dil. HNO_3} COOH-(CHOH)_4-COOH$
Thus,the correct reagent is dilute nitric acid.

(D) The enzyme $Glucose \ isomerase$ is responsible for the conversion of $glucose$ into $fructose$.

(D) Sucrose is a disaccharide with the formula $C_{12}H_{22}O_{11}$.
When treated with concentrated $H_2SO_4$,it undergoes dehydration.
The concentrated $H_2SO_4$ acts as a strong dehydrating agent and removes water molecules from the sucrose.
The reaction is: $C_{12}H_{22}O_{11} \xrightarrow{\text{conc. } H_2SO_4} 12C + 11H_2O$.
Thus,the products obtained are sugar charcoal $(12C)$ and water $(11H_2O)$.

(A) $ \text{Amylase}$ is the enzyme responsible for the hydrolysis of $ \text{starch}$ into simpler sugars like $ \text{maltose}$ and $ \text{glucose}$.
$ \text{Proteases}$ hydrolyze proteins.
$ \text{Glucose isomerase}$ converts $ \text{glucose}$ to $ \text{fructose}$.
$ \text{Insulin}$ is a hormone.

(B) The chemical equation for the acid hydrolysis of sucrose is:
$C_{12}H_{22}O_{11} + H_2O \xrightarrow{H^{+}} C_6H_{12}O_6 (\text{glucose}) + C_6H_{12}O_6 (\text{fructose})$
From the stoichiometry of the reaction,$1 \ mol$ of sucrose reacts with $1 \ mol$ of $H_2O$ to produce $1 \ mol$ of glucose and $1 \ mol$ of fructose.
Therefore,to obtain $1 \ mol$ of glucose,we require $1 \ mol$ of $H_2O$.
The molar mass of water $(H_2O)$ is $18 \ g/mol$.
Mass of water = $\text{moles} \times \text{molar mass} = 1 \ mol \times 18 \ g/mol = 18 \ g$.

(B) The hydrolysis products of the given carbohydrates are as follows:
$1$. Sucrose: $1$ mole of glucose + $1$ mole of fructose.
$2$. Stachyose: $2$ moles of galactose + $1$ mole of glucose + $1$ mole of fructose.
$3$. Raffinose: $1$ mole of galactose + $1$ mole of glucose + $1$ mole of fructose.
$4$. Lactose: $1$ mole of galactose + $1$ mole of glucose.
Therefore,$Stachyose$ yields the highest number ($2$ moles) of galactose molecules upon hydrolysis.

(A) Essential amino acids are those that cannot be synthesized by the human body and must be obtained through the diet.
Among the given options,$Valine$ is an essential amino acid.
The list of essential amino acids includes: $1$. $Leucine$,$2$. $Isoleucine$,$3$. $Lysine$,$4$. $Methionine$,$5$. $Phenylalanine$,$6$. $Threonine$,$7$. $Tryptophan$,$8$. $Valine$,and $9$. $Histidine$.
$Tyrosine$,$Proline$,and $Alanine$ are non-essential amino acids,meaning they can be synthesized by the body.

(D) Among the given options,$Cysteine$ is a sulfur-containing amino acid. Its chemical structure is $HS-CH_2-CH(NH_2)-COOH$.

(D) The amino acids are organic compounds containing an amino group $(-NH_2)$ and a carboxyl group $(-COOH)$ attached to the same carbon atom.
Glycine is the simplest amino acid with the structure $NH_2-CH_2-COOH$.
Therefore,it contains both amino and carboxyl groups.

(D) Essential amino acids are those that cannot be synthesized by the human body and must be obtained through the diet.
There are $10$ essential amino acids in total.

(C) Amino acids that are not synthesized in the human body are called essential amino acids and must be obtained through the diet. $Phenylalanine$ is an essential amino acid,whereas $Cysteine$,$Asparagine$,and $Serine$ are non-essential amino acids that can be synthesized by the body.

(A) Nitrogenous bases in nucleic acids are classified into two types: purines and pyrimidines.
$1$. Pyrimidines are heterocyclic compounds containing two nitrogen atoms in a six-membered ring. Examples include $Cytosine$,$Uracil$,and $Thymine$.
$2$. Purines are heterocyclic compounds consisting of a six-membered ring fused to a five-membered ring,containing four nitrogen atoms. Examples include $Adenine$ and $Guanine$.
Therefore,$Guanine$ is a purine derivative,not a pyrimidine derivative.

(B) Nucleotides are composed of a nitrogenous base, a phosphate group, and a pentose sugar.
In $RNA$ (Ribonucleic Acid), the pentose sugar present is $\beta-D$-ribose.
In $DNA$ (Deoxyribonucleic Acid), the pentose sugar present is $\beta-D-2$-deoxyribose.
Therefore, the sugar present in the nucleotide of $RNA$ is $D$-Ribose.

(D) Polynucleotides,such as $DNA$ and $RNA$,are formed by the polymerization of nucleotides.
Each nucleotide consists of a nitrogenous base,a phosphate group,and a sugar molecule.
The sugar present in these nucleotides is a $5$-carbon sugar,known as a pentose sugar.
Specifically,$RNA$ contains $\beta-D-ribose$ and $DNA$ contains $\beta-D-2-deoxyribose$,both of which are pentose sugars.

(D) Nitrogenous bases in nucleic acids are classified into two types: purines and pyrimidines.
Purines have a double-ring structure (a six-membered ring fused to a five-membered ring),while pyrimidines have a single-ring structure.
Among the given options,$Thymine$,$Uracil$,and $Cytosine$ are pyrimidines (single-ring).
$Guanine$ is a purine and possesses a double-ring structure.
Therefore,the correct option is $D$.

(C) In nucleosides,the nitrogen atom at position $1$ $(N-1)$ of the pyrimidine base is covalently bonded to the $C-1'$ carbon atom of the furanose sugar.

(A) To determine the number of $N$ atoms in each compound,we examine their chemical structures:
$1$. Guanine $(C_5H_5N_5O)$: It contains $5$ nitrogen atoms.
$2$. Thymine $(C_5H_6N_2O_2)$: It contains $2$ nitrogen atoms.
$3$. Cytosine $(C_4H_5N_3O)$: It contains $3$ nitrogen atoms.
$4$. Uracil $(C_4H_4N_2O_2)$: It contains $2$ nitrogen atoms.
Comparing these,Guanine has the highest number of nitrogen atoms $(5)$.
Therefore,the correct option is $A$.

(C) In a nucleoside,the nitrogenous base is linked to the $C1'$ position of the sugar (ribose or deoxyribose) via a $\beta$-glycosidic linkage.
For purine bases (like adenine or guanine),this linkage occurs at the $N9$ nitrogen atom.
For pyrimidine bases (like cytosine,thymine,or uracil),this linkage occurs at the $N1$ nitrogen atom.
As shown in the provided structure,the furanose ring is attached to the $N9$ position of the purine base.

(C) The reaction of benzoyl chloride $(C_6H_5COCl)$ with water $(H_2O)$ is a hydrolysis reaction.
In this reaction,the chlorine atom $(-Cl)$ of the acid chloride is replaced by a hydroxyl group $(-OH)$ from water.
The chemical equation is: $C_6H_5COCl + H_2O \rightarrow C_6H_5COOH + HCl$.
Here,$C_6H_5COOH$ is benzoic acid,which is the product $B$.

(A) Citric acid is a tricarboxylic acid found in citrus fruits like lemon and orange.

(A) Aromatic carboxylic acids can be prepared from alkyl benzenes by vigorous oxidation of the alkyl side chain using strong oxidizing agents.
Alkaline potassium permanganate $(KMnO_4/OH^-)$ and chromic acid $(H_2CrO_4)$ are well-known strong oxidizing agents used for this purpose.
Dilute nitric acid $(HNO_3)$ is also capable of oxidizing the alkyl side chain of alkyl benzenes to a carboxylic acid group.
Diborane $(B_2H_6)$ is a reducing agent,typically used for hydroboration-oxidation reactions to convert alkenes into alcohols,and it is not used for the oxidation of alkyl benzenes to carboxylic acids.
Therefore,the correct answer is $A$.

(A) The chemical structure of citric acid is $HOOC-CH_2-C(OH)(COOH)-CH_2-COOH$.
By observing the structure,we can identify the functional groups present:
$1$. There is one alcoholic $-OH$ group attached to the central carbon atom.
$2$. There are three carboxylic acid $-COOH$ groups present in the molecule.
Therefore,the number of alcoholic $-OH$ and $-COOH$ groups in citric acid are $1$ and $3$ respectively.

(A) $L$-Lactic acid is primarily found in curd,produced by the fermentation of lactose by lactic acid bacteria.

(C) The reaction of a Grignard reagent $(R-Mg-X)$ with solid carbon dioxide ($CO_2$,dry ice) followed by acidic hydrolysis is a standard method for the preparation of carboxylic acids.
Step $1$: $R-Mg-X + CO_2 \xrightarrow{\text{dry ether}} R-COOMgX$ (Carboxylato magnesium halide).
Step $2$: $R-COOMgX + H_2O \xrightarrow{\text{dil } HCl} R-COOH + Mg(X)OH$ (Carboxylic acid).
Therefore,compound $A$ is $CO_2$ (Solid).

(C) Phthalic acid is benzene-$1,2$-dicarboxylic acid.
Its chemical structure consists of a benzene ring with two carboxylic acid $(-COOH)$ groups attached at the ortho positions.
Therefore,the number of $-COOH$ groups present in phthalic acid is $2$.

(C) The reaction of two molecules of acetic acid $(CH_3COOH)$ in the presence of a dehydrating agent like phosphorus pentoxide $(P_2O_5)$ and heat $(\Delta)$ results in the removal of one water molecule $(H_2O)$.
This process is known as dehydration.
The reaction is as follows:
$2 CH_3COOH \xrightarrow{\Delta, P_2O_5} (CH_3CO)_2O H_2O$
Here,$(CH_3CO)_2O$ is acetic anhydride.
Therefore,product $A$ is acetic anhydride.

(C) When two molecules of carboxylic acid $(R-COOH)$ are heated in the presence of a dehydrating agent like phosphorus pentoxide $(P_2O_5)$,they undergo a condensation reaction to form an acid anhydride by the elimination of a water molecule $(H_2O)$.
The reaction is represented as:
$2R-COOH \xrightarrow{P_2O_5, \Delta} (RCO)_2O + H_2O$
Thus,option $C$ is the correct reaction for the formation of acid anhydride.

(D) The bond dissociation enthalpy depends on the bond length and the repulsion between lone pairs of electrons on the bonded atoms.
In $F-F$,the atoms are very small,which leads to significant inter-electronic repulsion between the lone pairs of the two fluorine atoms.
This repulsion weakens the $F-F$ bond,making its bond dissociation enthalpy lower than that of $Cl-Cl$.
Therefore,the correct decreasing order of bond dissociation enthalpy is $Cl-Cl > Br-Br > F-F > I-I$.

(D) The bond dissociation enthalpy of halogens generally decreases down the group due to an increase in atomic size.
However,$F_2$ is an exception because of the small size of the fluorine atom.
In $F_2$,the lone pairs on the fluorine atoms experience significant inter-electronic repulsion,which weakens the $F-F$ bond.
Therefore,$F_2$ has a lower bond dissociation enthalpy than $Cl_2$ and $Br_2$,but $I_2$ has the lowest bond dissociation enthalpy due to the largest atomic size and longest bond length.

(A) The bond dissociation enthalpy of halogens generally decreases down the group due to an increase in atomic size.
However,$F_2$ has an exceptionally low bond dissociation enthalpy due to the high inter-electronic repulsion between the lone pairs of the small $F$ atoms.
The correct order of bond dissociation enthalpy is: $I_2 < F_2 < Br_2 < Cl_2$.
Therefore,$Cl_2$ has the highest bond dissociation enthalpy.

(A) The $C-Br$ bond length in methyl bromide $(CH_3Br)$ is experimentally determined to be approximately $193 \ pm$. This value is consistent with the covalent radii of carbon $(77 \ pm)$ and bromine $(114 \ pm)$.

(A) For the reaction $4 NH_{3(g)} + 5 O_{2(g)} \rightarrow 4 NO_{(g)} + 6 H_2O_{(g)}$,the rate expression is given by:
$-\frac{1}{4} \frac{d[NH_3]}{dt} = -\frac{1}{5} \frac{d[O_2]}{dt} = +\frac{1}{4} \frac{d[NO]}{dt} = +\frac{1}{6} \frac{d[H_2O]}{dt}$
Given that the rate of formation of $NO$ is $\frac{d[NO]}{dt} = 3.6 \times 10^{-3} \ mol \ L^{-1} \ sec^{-1}$.
Equating the terms for $NO$ and $H_2O$:
$\frac{1}{4} \frac{d[NO]}{dt} = \frac{1}{6} \frac{d[H_2O]}{dt}$
$\frac{d[H_2O]}{dt} = \frac{6}{4} \times \frac{d[NO]}{dt} = 1.5 \times 3.6 \times 10^{-3} \ mol \ L^{-1} \ sec^{-1}$
$= 5.4 \times 10^{-3} \ mol \ L^{-1} \ sec^{-1}$

(C) The rate of a reaction is defined as the change in concentration of a reactant or product per unit time.
Concentration is typically expressed in $mol \ dm^{-3}$ (or $M$) and time $(t)$ is expressed in seconds $(s)$,minutes $(min)$,or hours $(h)$.
Therefore,the unit for the rate of reaction is $\frac{\text{concentration}}{\text{time}} = \frac{mol \ dm^{-3}}{t} = mol \ dm^{-3} \ t^{-1}$.

(D) The rate expression for the given reaction $2 \ NO_{(g)} + O_{2_{(g)}} \rightarrow 2 \ NO_{2_{(g)}}$ is given by:
$-\frac{1}{2} \frac{d[NO]}{dt} = \frac{1}{2} \frac{d[NO_2]}{dt}$.
The rate of consumption of $NO_{(g)}$ is defined as $-\frac{d[NO]}{dt}$.
From the expression,$-\frac{d[NO]}{dt} = \frac{d[NO_2]}{dt}$.
Given that $\frac{d[NO_2]}{dt} = 0.052 \ mol \ dm^{-3} \ s^{-1}$,
Therefore,$-\frac{d[NO]}{dt} = 0.052 \ mol \ dm^{-3} \ s^{-1}$.

(A) For a general reaction $aA + bB \rightarrow cC + dD$,the rate of reaction is given by:
$r = -\frac{1}{a} \frac{d[A]}{dt} = -\frac{1}{b} \frac{d[B]}{dt} = \frac{1}{c} \frac{d[C]}{dt} = \frac{1}{d} \frac{d[D]}{dt}$
For the reaction $2 N_2 O_{5(g)} \rightarrow 4 NO_{2(g)} + O_{2(g)}$,the rate expression is:
$r = -\frac{1}{2} \frac{d[N_2 O_5]}{dt} = \frac{1}{4} \frac{d[NO_2]}{dt} = \frac{d[O_2]}{dt}$
Comparing this with the given options,option $A$ is the correct expression for the rate of reaction.

(C) The rate of reaction is given by the expression: $Rate = -\frac{1}{2} \frac{d[A]}{dt} = -\frac{d[B]}{dt} = \frac{1}{2} \frac{d[C]}{dt}$.
Given that the rate of disappearance of $A$ is $-\frac{d[A]}{dt} = 0.076 \ mol \ dm^{-3} \ s^{-1}$.
Substituting this into the relation $-\frac{d[B]}{dt} = \frac{1}{2} \frac{d[A]}{dt}$:
$-\frac{d[B]}{dt} = \frac{1}{2} \times 0.076 \ mol \ dm^{-3} \ s^{-1} = 0.038 \ mol \ dm^{-3} \ s^{-1}$.

(A) The rate expression for the reaction $2 NO_{(g)} + O_{2(g)} \rightarrow 2 NO_{2(g)}$ is given by:
$-\frac{1}{2} \frac{d[NO]}{dt} = -\frac{d[O_2]}{dt} = \frac{1}{2} \frac{d[NO_2]}{dt}$
Given that the rate of formation of $NO_2$ is $\frac{d[NO_2]}{dt} = 0.052 \ mol \ dm^{-3} \ s^{-1}$.
Equating the rate of disappearance of $O_2$ to the rate of formation of $NO_2$:
$-\frac{d[O_2]}{dt} = \frac{1}{2} \times \frac{d[NO_2]}{dt}$
$-\frac{d[O_2]}{dt} = \frac{1}{2} \times 0.052 \ mol \ dm^{-3} \ s^{-1} = 0.026 \ mol \ dm^{-3} \ s^{-1}$.

(B) The rate of a chemical reaction is defined as the change in concentration of any of the reactants or products per unit time.
Mathematically,for a reaction $A \rightarrow B$,the rate can be expressed as:
Rate $= -\frac{d[A]}{dt} = \frac{d[B]}{dt}$.
Thus,it is expressed in terms of both the rate of consumption of reactants and the rate of formation of products.

(D) For a general reaction $aA + bB \rightarrow cC + dD$,the rate of reaction is expressed as $-\frac{1}{a} \frac{d[A]}{dt} = -\frac{1}{b} \frac{d[B]}{dt} = \frac{1}{c} \frac{d[C]}{dt} = \frac{1}{d} \frac{d[D]}{dt}$.
Given: $\frac{1}{3} \frac{d[X]}{dt} = -\frac{1}{2} \frac{d[Y]}{dt} = -\frac{d[Z]}{dt}$.
This implies that $X$ is a product (positive sign) with a stoichiometric coefficient of $3$,and $Y$ and $Z$ are reactants (negative sign) with stoichiometric coefficients of $2$ and $1$ respectively.
Thus,the reaction is $2Y + Z \rightarrow 3X$.

(A) The rate of reaction is given by the expression: $-\frac{d[N_2]}{dt} = -\frac{1}{3} \frac{d[H_2]}{dt} = \frac{1}{2} \frac{d[NH_3]}{dt}$.
Given that the rate of disappearance of $N_2$ is $-\frac{d[N_2]}{dt} = 2.22 \times 10^{-3} \ mol \ dm^{-3} \ s^{-1}$.
From the stoichiometric relationship,we have: $\frac{d[NH_3]}{dt} = 2 \times (-\frac{d[N_2]}{dt})$.
Substituting the given value: $\frac{d[NH_3]}{dt} = 2 \times (2.22 \times 10^{-3} \ mol \ dm^{-3} \ s^{-1}) = 4.44 \times 10^{-3} \ mol \ dm^{-3} \ s^{-1}$.

(A) For a general reaction $aA + bB \rightarrow cC + dD$,the rate of reaction is given by $r = -\frac{1}{a} \frac{d[A]}{dt} = -\frac{1}{b} \frac{d[B]}{dt} = \frac{1}{c} \frac{d[C]}{dt} = \frac{1}{d} \frac{d[D]}{dt}$.
Applying this to the reaction $2 A + B \rightarrow C + 3 D$,we get:
$r = -\frac{1}{2} \frac{d[A]}{dt} = -\frac{d[B]}{dt} = \frac{d[C]}{dt} = \frac{1}{3} \frac{d[D]}{dt}$.
Comparing this with the given options,both $A$ and $B$ are mathematically equivalent expressions for the rate of reaction.