MHT CET 2023 Chemistry Question Paper with Answer and Solution

(B) The total number of electrons in an $F_2$ molecule is $18$.
The molecular orbital configuration is: $(\sigma 1s)^2 (\sigma^* 1s)^2 (\sigma 2s)^2 (\sigma^* 2s)^2 (\sigma 2p_z)^2 (\pi 2p_x)^2 (\pi 2p_y)^2 (\pi^* 2p_x)^2 (\pi^* 2p_y)^2$.
Number of bonding electrons $(N_b)$ $= 10$.
Number of antibonding electrons $(N_a)$ $= 8$.
Bond order $= \frac{N_b - N_a}{2} = \frac{10 - 8}{2} = 1$.

(D) The total number of electrons in $N_2^{+}$ is $13$.
The molecular orbital configuration is: $(\sigma 1s)^2 (\sigma^* 1s)^2 (\sigma 2s)^2 (\sigma^* 2s)^2 (\pi 2p_x)^2 (\pi 2p_y)^2 (\sigma 2p_z)^1$.
Number of bonding electrons $(N_b)$ $= 9$.
Number of antibonding electrons $(N_a)$ $= 4$.
Bond order $= \frac{N_b - N_a}{2} = \frac{9 - 4}{2} = \frac{5}{2} = 2.5$.

(A) The atomic number of copper $(Cu)$ is $29$.
The ground state electronic configuration of $Cu$ is $[Ar] 3d^{10} 4s^1$.
In the $3d$ subshell,all $10$ electrons are paired.
In the $4s$ subshell,there is $1$ electron,which is unpaired.
Therefore,the number of unpaired electrons in $Cu$ is $1$.

(C) The boiling point of haloalkanes depends on the magnitude of van der Waals forces,which increase with an increase in the size and mass of the halogen atom.
For the given alkyl group $(-CH_3)$,the order of boiling points is $CH_3I > CH_3Br > CH_3Cl > CH_3F$.
Therefore,$CH_3I$ (Iodomethane) has the highest boiling point.

(B) Intramolecular hydrogen bonding occurs when a stable five-membered or six-membered ring is formed within the molecule.
Benzoic acid,benzaldehyde,and phenol do not possess the necessary structural arrangement to form such rings,and thus they cannot form intramolecular hydrogen bonds.
In the case of salicylaldehyde,the hydroxyl group $(-OH)$ is ortho to the aldehyde group $(-CHO)$. This proximity allows the hydrogen atom of the hydroxyl group to form a hydrogen bond with the oxygen atom of the carbonyl group,resulting in a stable six-membered ring as shown below:
(Image: $231309-$s)

(C) Intermolecular hydrogen bonding requires a hydrogen atom covalently bonded to a highly electronegative atom like $N$,$O$,or $F$.
Trimethylamine,$N(CH_3)_3$,contains only $C-N$ bonds and no $N-H$ bonds.
Therefore,it cannot act as a hydrogen bond donor,and it does not develop intermolecular hydrogen bonding.

(D) Given: Percent dissociation $= 1.42 \%$,Concentration $C = 0.05 \ M$.
Degree of dissociation $\alpha = \frac{1.42}{100} = 0.0142$.
For a weak base,the dissociation constant $K_b$ is given by the formula $K_b = \alpha^2 C$ (assuming $\alpha << 1$).
$K_b = (0.0142)^2 \times 0.05$.
$K_b = 0.00020164 \times 0.05$.
$K_b = 1.0082 \times 10^{-5} \approx 1.0 \times 10^{-5}$.

(A) For a reaction to be spontaneous,the Gibbs free energy change $\Delta G$ must be negative,i.e.,$\Delta G < 0$.
Given the equation $\Delta G = \Delta H - T\Delta S < 0$.
Convert $\Delta H$ to Joules: $\Delta H = -84.2 \ kJ = -84200 \ J$.
Substituting the values: $-84200 \ J - T(-200 \ J \ K^{-1}) < 0$.
$-84200 + 200T < 0$.
$200T < 84200$.
$T < \frac{84200}{200} \ K$.
$T < 421 \ K$.
Thus,the highest temperature at which the reaction proceeds in the forward direction is $421 \ K$.

(B) reaction intermediate is a species that is produced in one step of a reaction mechanism and consumed in a subsequent step.
In step $(i)$,$ClO_2^{-}$ is formed.
In step $(ii)$,$ClO_2^{-}$ is consumed.
Therefore,$ClO_2^{-}$ is the reaction intermediate.

(C) $Ra$ $(Radium)$ is a radioactive element that has been historically used in radiotherapy for the treatment of cancer.
Its isotopes emit radiation that can destroy cancerous cells.

(B) The transition elements from $Sc$ $(Z=21)$ to $Zn$ $(Z=30)$ belong to the $3d$ series.
These elements are located in the $4^{th}$ period and span from Group $3$ to Group $12$ in the long form of the periodic table.

(B) The atomic number of copper $(Cu)$ is $29$.
Its electronic configuration is $[Ar] 3d^{10} 4s^1$.
Since the highest principal quantum number is $n = 4$,it belongs to period-$4$.
As a $d$-block element with $11$ valence electrons ($10$ in $d$-subshell + $1$ in $s$-subshell),it belongs to group-$11$.

(B) The ionisation enthalpy decreases down a group due to an increase in atomic size.
Therefore,the ionisation enthalpy of $Ne$ is greater than that of $Ar$.
Across a period,ionisation enthalpy generally increases with an increase in atomic number.
Comparing the elements in the same period ($3^{rd}$ period),the order is $S < Cl < Ar$.
Combining these trends,the correct decreasing order of ionisation enthalpy is:
$Ne > Ar > Cl > S$.

(B) The electronic configuration of $Mn$ $(Z=25)$ is $[Ar] \ 3d^5 \ 4s^2$.
In this configuration,the $d$-orbital is half-filled $(d^5)$.
Therefore,$Mn$ possesses half-filled $d$-orbitals.

(B) Given: $K_{a} = 1.8 \times 10^{-5}$ and $\alpha = 0.02$.
For a weak monobasic acid,the relationship between dissociation constant $(K_{a})$,degree of dissociation $(\alpha)$,and molar concentration $(c)$ is given by the formula:
$K_{a} = \alpha^2 c$
Rearranging the formula to solve for concentration $(c)$:
$c = \frac{K_{a}}{\alpha^2}$
Substituting the given values:
$c = \frac{1.8 \times 10^{-5}}{(0.02)^2} = \frac{1.8 \times 10^{-5}}{4 \times 10^{-4}} = 0.45 \times 10^{-1} \ M = 4.5 \times 10^{-2} \ M$.

(A) For a weak base,the dissociation constant $K_b$ is given by the formula $K_b = c \alpha^2$,where $c$ is the concentration and $\alpha$ is the degree of dissociation.
Given: $c = 0.001 \ M = 10^{-3} \ M$ and $\alpha = 2 \% = 0.02 = 2 \times 10^{-2}$.
Substituting the values: $K_b = (10^{-3}) \times (2 \times 10^{-2})^2$.
$K_b = 10^{-3} \times 4 \times 10^{-4} = 4 \times 10^{-7}$.

(A) Given: $K_{a} = 2.25 \times 10^{-6}$ and concentration $c = 0.01 \ M$.
For a weak monobasic acid,the degree of dissociation $\alpha$ is given by $\alpha = \sqrt{\frac{K_{a}}{c}}$.
Substituting the values: $\alpha = \sqrt{\frac{2.25 \times 10^{-6}}{0.01}} = \sqrt{2.25 \times 10^{-4}} = 0.015$.
Percent dissociation = $\alpha \times 100 = 0.015 \times 100 = 1.5 \%$.

(B) According to the principles of green chemistry,good atom economy means that most of the atoms of the reactants are incorporated into the desired product,minimizing waste. Therefore,the statement that 'good atom economy means very few atoms of reactants are incorporated in the product' is false.

(B) $\text{Percent atom economy} = \frac{\text{Formula weight of the desired product}}{\text{Sum of formula weight of all the reactants used in the reaction}} \times 100$
$= \frac{175 \ u}{225 \ u} \times 100$
$= 77.7 \%$

(B) $1$. Identify the principal functional group: The compound is an ether,where the methoxy group $(-OCH_3)$ is attached to a cyclobutane ring.
$2$. Number the ring: The ring is a cyclobutane. We must assign the lowest possible locants to the substituents.
$3$. If we assign position $1$ to the carbon attached to the methoxy group,the methyl groups are at position $3$. This gives the locants $1, 3, 3$.
$4$. If we assign position $1$ to the carbon attached to the methyl groups,the methoxy group is at position $3$. This gives the locants $1, 1, 3$.
$5$. Comparing the sets $(1, 3, 3)$ and $(1, 1, 3)$,the set $(1, 1, 3)$ is lower.
$6$. Therefore,the methoxy group is at position $3$ and the two methyl groups are at position $1$.
$7$. The $IUPAC$ name is $3-$methoxy$-1,1-$dimethylcyclobutane.

(C) $1$. Identify the longest carbon chain containing the double bond. The longest chain has $6$ carbons,so the parent alkane is hexane,and the alkene is hex$-3-$ene.
$2$. Number the chain from the end that gives the double bond the lowest locant. Numbering from left to right gives the double bond at position $3$.
$3$. Identify the substituents: $A$ bromine atom $(-Br)$ is at position $4$ and a methyl group $(-CH_3)$ is at position $3$.
$4$. Combine the parts alphabetically: $4-$Bromo$-3-$methylhex$-3-$ene.

(B) $1$. Identify the parent chain: The parent ring is a cyclobutane ring.
$2$. Identify substituents: There is a methoxy group $(-OCH_3)$ and two methyl groups $(-CH_3)$.
$3$. Numbering: Number the ring to give the lowest possible locants to the substituents. Starting from the carbon with the two methyl groups as $1$,the methoxy group is at position $3$.
$4$. Name: The compound is $3-$methoxy$-1,1-$dimethylcyclobutane.

(B) $1$. Identify the principal functional group: The $-OH$ group is the principal functional group,so the parent chain is a cyclopentanol. The carbon attached to the $-OH$ group is assigned position $1$.
$2$. Number the ring: To give the lowest possible locants to the substituents (ethyl and methyl groups),we number the ring starting from the carbon with the $-OH$ group as $1$. Moving clockwise,the ethyl group is at position $2$ and the methyl group is at position $5$.
$3$. Alphabetical order: Substituents are listed alphabetically. Since 'ethyl' comes before 'methyl',the name is $2-$ethyl$-5-$methylcyclopentanol.

(B) $1$. Identify the longest carbon chain containing the functional group $(-NH_2)$ and the double bond. The chain has $5$ carbon atoms,so the parent alkane is pentane.
$2$. Number the chain starting from the end closer to the $-NH_2$ group to give it the lowest possible locant. Thus,the $-NH_2$ group is at position $2$.
$3$. The double bond starts at position $3$. Therefore,the suffix for the alkene is $-3-$ene and for the amine is $-2-$amine.
$4$. Combining these,the $IUPAC$ name is Pent$-3-$en$-2-$amine.

(B) The structure of allylamine is $CH_2=CH-CH_2-NH_2$.
In this molecule,the principal functional group is the amine $(-NH_2)$,which is assigned the lowest possible number.
The carbon chain contains $3$ carbon atoms with a double bond starting at position $2$.
Therefore,the $IUPAC$ name is Prop-$2$-en-$1$-amine.

(C) The given molecule is $CH_3-CH(NH_2)-CH_2-COOH$.
$1$. The principal functional group is the carboxylic acid group $(-COOH)$,so the parent chain is a butanoic acid derivative.
$2$. Numbering starts from the $-COOH$ carbon as $C-1$. Thus,the amino group $(-NH_2)$ is at $C-3$.
$3$. The correct $IUPAC$ name is $3-$aminobutanoic acid.
$4$. Since the amino group is on the $\beta-$carbon relative to the carboxyl group,it is a $\beta-$amino acid.
Therefore,statement $C$ is true.

(C) $1$. Identify the longest carbon chain containing the principal functional group $(-OH)$ and the double bond. The chain has $8$ carbons,so the parent alkane is octane.
$2$. Number the chain from the end that gives the lowest locant to the principal functional group $(-OH)$. Numbering from right to left gives the $-OH$ group at position $3$.
$3$. The double bond starts at carbon $4$,so it is named as oct$-4-$en.
$4$. There is a methyl group at position $6$.
$5$. Combining these,the $IUPAC$ name is $6-$methyloct$-4-$en$-3-$ol.

(A) $1$. Identify the longest carbon chain containing the double bond. The longest chain has $7$ carbon atoms,so the parent alkane is heptane,and with the double bond,it is hept$-3-$ene.
$2$. Number the chain from the end that gives the lowest locant to the double bond. Numbering from the right gives the double bond position $3$.
$3$. Identify the substituents: there is a bromo group at position $2$ and methyl groups at positions $4$ and $5$.
$4$. Combining these,the $IUPAC$ name is $2-$Bromo$-4,5-$dimethylhept$-3-$ene.

(A) $1$. Identify the principal functional group: The compound contains a ketone group,so the suffix is $-one$ and the parent chain is a cyclohexane ring,making it a cyclohexanone.
$2$. Number the ring: The carbonyl carbon of the ketone is assigned position $1$. We number the ring to give the substituents the lowest possible locants. Moving clockwise,the chlorine atom is at position $2$ and the methyl group is at position $4$.
$3$. Assemble the name: The substituents are $2-$chloro and $4-$methyl. Alphabetical order is followed for naming substituents ($C$ before $M$).
$4$. Thus,the $IUPAC$ name is $2-$chloro$-4-$methylcyclohexanone.

(C) The priority order for functional groups in $IUPAC$ nomenclature is as follows:
$-COOH > -SO_3H > -COOR > -COCl > -CONH_2 > -CN > -CHO > >C=O > -OH > -NH_2 > -C=C- > -C\equiv C-$
Comparing the given options:
$1$. $-COOR$ (Ester)
$2$. $-COCl$ (Acyl chloride)
$3$. $>C=O$ (Ketone)
$4$. $-CONH_2$ (Amide)
According to the priority order,the ketone group $(>C=O)$ has the lowest priority among the given choices.

(B) The structure of benzene$-1,3-$diol consists of a benzene ring with two hydroxyl $(-OH)$ groups attached at the $1$ and $3$ positions.
This specific isomer is commonly known as Resorcinol.
Catechol is benzene$-1,2-$diol,Quinol (or hydroquinone) is benzene$-1,4-$diol,and Pyrogallol is benzene$-1,2,3-$triol.

(B) The chemical structure of crotonyl alcohol is $CH_3CH=CHCH_2OH$.
In the $IUPAC$ nomenclature,the longest carbon chain containing the functional group $(-OH)$ and the double bond is selected.
The numbering starts from the carbon attached to the $-OH$ group to give it the lowest possible locant.
Thus,the chain has $4$ carbons,the double bond starts at position $2$,and the alcohol group is at position $1$.
The correct $IUPAC$ name is $but-2-en-1-ol$.

(D) $1$. The structure is $CH_3-CH(OH)-C(=O)-CH_2-CH_3$. The ketone group $(>C=O)$ has higher priority than the hydroxyl group $(-OH)$ according to $IUPAC$ nomenclature rules. Thus,the ketone is the principal functional group and $-OH$ is a substituent. Option $(B)$ is false.
$2$. The $-OH$ group is attached to the $C-2$ carbon,which is bonded to four single bonds,making it $sp^3$ hybridized. Option $(A)$ is false.
$3$. Counting the atoms: $5$ carbons,$10$ hydrogens,and $2$ oxygens. The molecular formula is $C_5H_{10}O_2$. Option $(C)$ is false.
$4$. The longest carbon chain containing the ketone group has $5$ carbons (pentane). Numbering from the left gives the hydroxyl group at position $2$ and the ketone at position $3$. The $IUPAC$ name is $2-$hydroxypentan$-3-$one. Option $(D)$ is true.

(D) $1$. Identify the principal functional group: The $-OH$ group is the principal functional group,so the parent chain is a cyclopentanol.
$2$. Numbering the ring: The carbon atom attached to the $-OH$ group is assigned position $1$. The ring is then numbered to give the lowest possible locants to the substituents.
$3$. If we number clockwise,the substituents are at positions $2$ (ethyl) and $5$ (methyl). If we number counter-clockwise,the substituents are at positions $2$ (methyl) and $5$ (ethyl).
$4$. Applying the lowest locant rule: Comparing the sets $(2, 5)$ and $(2, 5)$,we look at alphabetical order. Ethyl comes before methyl.
$5$. Therefore,we assign the lower number to the substituent that comes first alphabetically. Thus,the ethyl group is at position $2$ and the methyl group is at position $5$.
$6$. The $IUPAC$ name is $2-$Ethyl$-5-$methylcyclopentanol.

(B) $1$. Identify the longest carbon chain containing the double bond. The longest chain has $7$ carbon atoms,so the parent alkane is heptane,and with the double bond,it is hept$-3-$ene.
$2$. Number the chain from the end that gives the double bond the lowest possible locant. Numbering from the right gives the double bond position $3$.
$3$. Identify the substituents: there is an iodo group at position $2$ and methyl groups at positions $4$ and $5$.
$4$. Combine these to get the $IUPAC$ name: $2-$Iodo$-4,5-$dimethylhept$-3-$ene.

(C) In the carbinol system,alcohols are considered as derivatives of methyl alcohol $(CH_3OH)$,which is called carbinol.
Isopropyl alcohol has the structure $(CH_3)_2CHOH$.
In this structure,two hydrogen atoms of the carbinol group $(-CH_2OH)$ are replaced by two methyl groups $(-CH_3)$.
Therefore,it is named as dimethyl carbinol.

(A) $1$. Identify the longest carbon chain containing the double bond. The longest chain has $6$ carbon atoms,so the parent alkane is hexane. Since there is a double bond at position $3$,it is hex$-3-$ene.
$2$. Number the chain from the end that gives the lowest locant to the double bond. Numbering from right to left gives the double bond at position $3$.
$3$. Identify the substituents: there is a bromo group $(-Br)$ at position $3$ and a methyl group $(-CH_3)$ at position $4$.
$4$. Combine these to get the $IUPAC$ name: $3-$Bromo$-4-$methylhex$-3-$ene.

(B) $1$. Identify the principal functional group: The compound is an ether,where the methoxy group $(-OCH_3)$ is attached to a cyclobutane ring.
$2$. Number the ring: To give the lowest possible locants to the substituents,we start numbering from the carbon attached to the methoxy group or the dimethyl group. Following the $IUPAC$ rule of lowest locant set,we number the carbon with the two methyl groups as $1$ and the carbon with the methoxy group as $3$.
$3$. Name the substituents: There are two methyl groups at position $1$ and a methoxy group at position $3$.
$4$. Combine the parts: The name is $3-$methoxy$-1,1-$dimethylcyclobutane.

(D) The given compound is $HO-CH_2-CH_2-CH_2-CH(CH_3)-CH(CH_3)_2$.
This is a chain of $6$ carbon atoms in the main backbone with a hydroxyl group at the first carbon and methyl substituents at the $4^{th}$ and $5^{th}$ positions.
Counting from the $OH$ group:
$C_1$ is $CH_2OH$
$C_2$ is $CH_2$
$C_3$ is $CH_2$
$C_4$ is $CH(CH_3)$
$C_5$ is $CH(CH_3)_2$
Looking at the options,image $232466-$d represents a $6$-carbon chain with an $OH$ group at the end and two methyl groups at the $4^{th}$ and $5^{th}$ positions relative to the $OH$ group.

(C) The correct answer is $C$.
In the Newman projection formula,the molecule is viewed along the $C-C$ bond axis.
The carbon atom nearer to the observer is represented by a point,and the rear carbon atom (further from the observer) is represented by a circle around that point.

(A) The structure of isobutane is $CH_3-CH(CH_3)-CH_3$.
$A$ tertiary carbon atom is a carbon atom bonded to three other carbon atoms.
In isobutane,the central carbon atom is bonded to three methyl groups $(-CH_3)$.
Therefore,there is $1$ tertiary carbon atom in a molecule of isobutane.

(B) Aniline,naphthalene,and phenol are benzenoid aromatic compounds because they contain at least one benzene ring.
In contrast,tropone is a nonbenzenoid aromatic compound because it is aromatic but does not contain a benzene ring.

(C) Enantiomers are non-superimposable mirror images of each other. Therefore,the statement that they are superimposable is incorrect.

(C) molecule is chiral if it contains at least one chiral carbon atom,which is a carbon atom bonded to four different groups.
$A$. $2-$Bromopropane: $CH_3-CH(Br)-CH_3$. The central carbon is bonded to two identical methyl groups. It is achiral.
$B$. $2-$Bromo$-2-$methylbutane: $CH_3-C(Br)(CH_3)-CH_2-CH_3$. The $C-2$ carbon is bonded to two identical methyl groups. It is achiral.
$C$. $2-$Bromo$-3-$methylbutane: $CH_3-CH(Br)-CH(CH_3)-CH_3$. The $C-2$ carbon is bonded to $-H$,$-Br$,$-CH_3$,and $-CH(CH_3)_2$ groups. Since all four groups are different,it is a chiral molecule.
$D$. $3-$Bromopentane: $CH_3-CH_2-CH(Br)-CH_2-CH_3$. The $C-3$ carbon is bonded to two identical ethyl groups. It is achiral.
Therefore,the correct option is $C$.

(C) An alkane with the molecular formula $C_4H_{10}$ (butane) exhibits exactly two structural isomers: $n$-butane (butane) and $2$-methylpropane (isobutane).

$2$ Carbon atoms	Ethane: $0$ structural isomers
$3$ Carbon atoms	Propane: $0$ structural isomers
$4$ Carbon atoms	Butane: $2$ structural isomers ($n$-butane and $2$-methylpropane)
$6$ Carbon atoms	Hexane: $5$ structural isomers

(C) molecule is chiral if it contains at least one carbon atom bonded to four different groups (a chiral center).
$A$. $2-$Iodopropane: $CH_3-CH(I)-CH_3$. The central carbon is bonded to two identical $-CH_3$ groups. Not chiral.
$B$. $2-$Iodo$-2-$methylbutane: $CH_3-C(I)(CH_3)-CH_2-CH_3$. The central carbon is bonded to two identical $-CH_3$ groups. Not chiral.
$C$. $2-$Iodo$-3-$methylbutane: $CH_3-CH(I)-CH(CH_3)_2$. The carbon at position $2$ is bonded to $-H$,$-I$,$-CH_3$,and $-CH(CH_3)_2$. Since all four groups are different,it is a chiral molecule.
$D$. $3-$Iodopentane: $CH_3-CH_2-CH(I)-CH_2-CH_3$. The central carbon is bonded to two identical $-CH_2CH_3$ groups. Not chiral.
Therefore,the correct option is $C$.

(C) An alkane with three structural isomers is pentane $(C_5H_{12})$.
The three isomers are:
$1$. $n$-Pentane: $CH_3-CH_2-CH_2-CH_2-CH_3$
$2$. Isopentane ($2$-methylbutane): $CH_3-CH(CH_3)-CH_2-CH_3$
$3$. Neopentane ($2$,$2$-dimethylpropane): $CH_3-C(CH_3)_2-CH_3$
In one molecule of pentane,there are $5$ carbon atoms.
Therefore,in $n$ moles of pentane,the number of moles of '$C$' atoms is $5 \times n = 5n$.

(C) The number of structural isomers for alkanes increases with the number of carbon atoms. Based on the data:
$1$. $C_3H_8$ (Propane): $0$ structural isomers.
$2$. $C_4H_{10}$ (Butane): $2$ structural isomers ($n$-butane and isobutane).
$3$. $C_5H_{12}$ (Pentane): $3$ structural isomers ($n$-pentane,isopentane,and neopentane).
$4$. $C_6H_{14}$ (Hexane): $5$ structural isomers.
Therefore,the alkane with three structural isomers is $C_5H_{12}$.

(C) Enantiomers are defined as non-superimposable mirror images of each other. Therefore,the statement that they are superimposable is false.

(C) The Wurtz reaction involves the coupling of two alkyl halide molecules in the presence of sodium metal and dry ether.
For isopropyl bromide $(CH_3-CH(Br)-CH_3)$,two moles react to form $2,3-$dimethylbutane:
$2CH_3-CH(Br)-CH_3 + 2Na \rightarrow CH_3-CH(CH_3)-CH(CH_3)-CH_3 + 2NaBr$.

(D) The ring structure of fructose is a hemiketal,not a hemiacetal.
Fructose $(C_6H_{12}O_6)$ is a laevorotatory ketohexose.
It acts as a reducing sugar due to the presence of an $\alpha$-hydroxy ketone group which can tautomerize to an aldehyde in basic solution.

(D) Glucose exists in two cyclic forms,$\alpha$ and $\beta$,which are formed by intramolecular hemiacetal formation between the aldehyde group and one of the alcoholic groups at $C-5$ within the molecule.
This cyclic structure is responsible for the mutarotation and the existence of two anomers.
Thus,the open chain structure of glucose does not explain the existence of $\alpha$ and $\beta$ forms of glucose.

(B) The specific rotation of $D$-fructose is $-92.4^{\circ}$.
Since the value is negative,it indicates that fructose is levorotatory,which is why it is commonly referred to as levulose.

(A) In an acetylation reaction,the $H$ atom of an $(-OH)$ group is replaced by an acetyl group $(-COCH_3)$.
This results in an increase in molecular mass by $[(12+16+12+3 \times 1)-1] = 42 \ u$ per $-OH$ group.
Given the total increase in molecular mass is $84 \ u$.
Therefore,the number of $-OH$ groups $= \frac{84 \ u}{42 \ u} = 2$.
Among the given options,an aldotriose (e.g.,glyceraldehyde) contains two alcoholic $-OH$ groups,which upon acetylation will increase the molecular mass by $2 \times 42 \ u = 84 \ u$.

(A) non-reducing sugar is a carbohydrate that does not have a free aldehyde or ketone group to act as a reducing agent.
$Sucrose$ is a disaccharide composed of $Glucose$ and $Fructose$ units linked by a glycosidic bond between their respective anomeric carbons ($C1$ of $Glucose$ and $C2$ of $Fructose$).
Since both anomeric carbons are involved in the glycosidic linkage,$Sucrose$ cannot open to form an aldehyde or ketone group,making it a non-reducing sugar.
$Maltose$,$Lactose$,and $Glucose$ all contain at least one free anomeric carbon,making them reducing sugars.

(B) When glucose $(C_6H_{12}O_6)$ is oxidized with dilute nitric acid $(HNO_3)$,both the terminal aldehyde group $(-CHO)$ and the primary alcoholic group $(-CH_2OH)$ are oxidized to carboxylic acid groups $(-COOH)$.
This results in the formation of saccharic acid (also known as glucaric acid),which is a dicarboxylic acid with the structure $COOH-(CHOH)_4-COOH$.

(B) Cellulose is a linear polysaccharide consisting of a long chain of $D$-glucose units joined by $\beta-1,4$-glycosidic linkages.
These linkages are formed between the $C1$ carbon of one glucose unit and the $C4$ carbon of the adjacent glucose unit in the $\beta$-configuration.

(A) Maltose is a disaccharide composed of two $ \alpha-D-glucose $ units.
These two glucose units are linked together by an $ \alpha-1,4-glycosidic $ linkage.
In this linkage,the $ C1 $ of one glucose molecule is connected to the $ C4 $ of the other glucose molecule.

(C) $Maltose$ and $Lactose$ are disaccharides.
$Raffinose$ is a trisaccharide.
$Stachyose$ is a tetrasaccharide.

(B) Sucrose is a disaccharide formed by the condensation of one molecule of $\alpha$-$D$-glucose and one molecule of $\beta$-$D$-fructose.
The glycosidic linkage is formed between the $C-1$ of $\alpha$-glucose and the $C-2$ of $\beta$-fructose.
This bond involves the anomeric carbon atoms of both monosaccharides,which is why sucrose is a non-reducing sugar.

(C) Lactose is a disaccharide composed of $\beta-D-galactose$ and $\beta-D-glucose$.
The glycosidic linkage is formed between the $C-1$ of $\beta-D-galactose$ and the $C-4$ of $\beta-D-glucose$.
Therefore,the linkage is a $\beta-1,4-glycosidic$ linkage.

(D) $Stachyose$ is a tetrasaccharide.
Upon hydrolysis,it yields four monosaccharide units: two units of $D-galactose$,one unit of $D-glucose$,and one unit of $D-fructose$.
$Glycogen$ and $Cellulose$ are polysaccharides,while $Ribose$ is a monosaccharide.

(A) Lactose is a disaccharide composed of $\beta$-$D$-galactose and $\beta$-$D$-glucose units.
In lactose,the glycosidic linkage is formed between the $C-1$ carbon of $\beta$-$D$-galactose and the $C-4$ carbon of $\beta$-$D$-glucose.
Therefore,the linkage is a $\beta-1,4$-glycosidic linkage.

(D) Sucrose,maltose,and lactose are disaccharides,which consist of two monosaccharide units.
Raffinose is a trisaccharide,which consists of three monosaccharide units (galactose,glucose,and fructose).

(A) Basic amino acids contain more than one amino group and fewer carboxyl groups.
$Lysine$,$Arginine$,and $Histidine$ are basic amino acids.
$Proline$ is an imino acid and is neutral in nature.
Therefore,$Proline$ is not a basic amino acid.

(A) Proteins are classified into two types based on their molecular shape: fibrous proteins and globular proteins.
Fibrous proteins have a thread-like structure and are generally insoluble in water,such as keratin and myosin.
Globular proteins have a spherical shape and are generally soluble in water,such as insulin,legumelin,and serum albumin.
Therefore,$Myosin$ is a fibrous protein,not a globular protein.

(D) $Insulin$,$egg \ albumin$ and $serum \ albumin$ are globular proteins,whereas $keratin$ is a fibrous protein.

(A) Glycylalanine is formed by the condensation of two amino acids,glycine and alanine,through a peptide bond.
Since it consists of two amino acid residues,it is classified as a dipeptide.

(C) The open-chain structure of ribose is given by the formula $CHO-(CHOH)_3-CH_2OH$.
In this structure,there are $3$ hydroxyl groups attached to the middle carbons in the $(CHOH)_3$ unit and $1$ hydroxyl group attached to the terminal $CH_2OH$ group.
Therefore,the total number of $-OH$ groups in one molecule of ribose is $3 + 1 = 4$.

(D) The boiling point of carboxylic acids increases with an increase in molar mass due to stronger intermolecular forces (hydrogen bonding and van der Waals forces).
Among the given compounds,the molar masses are:
$1$. Formic acid $(HCOOH)$: $46 \ g/mol$
$2$. Acetic acid $(CH_3COOH)$: $60 \ g/mol$
$3$. Butyric acid $(C_3H_7COOH)$: $88 \ g/mol$
$4$. Valeric acid $(C_4H_9COOH)$: $102 \ g/mol$
Since Formic acid has the lowest molar mass,it has the lowest boiling point.

(C) The reaction of an alkyl halide with a silver salt of a carboxylic acid is a standard method for the preparation of esters. This is a nucleophilic substitution reaction ($S_N2$ mechanism).
The reaction is as follows:
$CH_3CH_2Br + CH_3CH_2COOAg \xrightarrow{\Delta} CH_3CH_2COOCH_2CH_3 + AgBr \downarrow$
Here,$CH_3CH_2COOAg$ is silver propanoate. Thus,the reagent '$A$' is silver propanoate.

(A) The boiling point depends on the strength of intermolecular forces,primarily hydrogen bonding.
$N-H$ bonds in amines are less polar than $O-H$ bonds in alcohols,leading to weaker hydrogen bonding in amines.
Carboxylic acids contain a $-COOH$ group,which allows for the formation of stable dimeric structures through strong intermolecular hydrogen bonding.
Therefore,the strength of hydrogen bonding follows the order: $Amines < Alcohols < Carboxylic \ acids$.
Consequently,the increasing order of boiling points is $Amines < Alcohols < Carboxylic \ acids$.

(B) The boiling point of a compound depends on the strength of intermolecular forces.
Carboxylic acids,such as $CH_3COOH$,form stable intermolecular hydrogen-bonded dimers,which significantly increase their boiling points compared to alcohols,aldehydes,and ketones of comparable molecular mass.
Therefore,$Ethanoic \ acid$ has the highest boiling point among the given options.

(C) dicarboxylic acid contains two carboxylic acid $(-COOH)$ functional groups.
Valeric acid $(CH_3(CH_2)_3COOH)$,Caproic acid $(CH_3(CH_2)_4COOH)$,and Butyric acid $(CH_3(CH_2)_2COOH)$ are monocarboxylic acids.
Glutaric acid has the structure $HOOC-(CH_2)_3-COOH$,which contains two $-COOH$ groups,making it a dicarboxylic acid.

(B) dicarboxylic acid contains two carboxyl $(-COOH)$ groups.
Malonic acid is $HOOC-CH_2-COOH$.
Glutaric acid is $HOOC-(CH_2)_3-COOH$.
Succinic acid is $HOOC-(CH_2)_2-COOH$.
Caproic acid is a monocarboxylic acid with the formula $CH_3(CH_2)_4COOH$,containing only one $-COOH$ group.
Therefore,the correct answer is $B$.

(C) The reaction of benzoic acid with phosphorus pentachloride $(PCl_5)$ is a standard method for the preparation of acid chlorides. The chemical equation is:
$C_6H_5COOH + PCl_5 \rightarrow C_6H_5COCl + POCl_3 + HCl$
Here,$C_6H_5COCl$ is benzoyl chloride,$POCl_3$ is phosphorus oxychloride,and $HCl$ is hydrogen chloride. Thus,the reagent is $PCl_5$.

(D) Monocarboxylic acids contain only one $-COOH$ group in their structure.
$A$. Malonic acid $(HOOC-CH_2-COOH)$ is a dicarboxylic acid,while propionic acid $(CH_3-CH_2-COOH)$ is a monocarboxylic acid.
$B$. Valeric acid $(CH_3(CH_2)_3COOH)$ is a monocarboxylic acid,while succinic acid $(HOOC-(CH_2)_2-COOH)$ is a dicarboxylic acid.
$C$. Acetic acid $(CH_3COOH)$ is a monocarboxylic acid,while adipic acid $(HOOC-(CH_2)_4-COOH)$ is a dicarboxylic acid.
$D$. Butyric acid $(CH_3(CH_2)_2COOH)$ and caproic acid $(CH_3(CH_2)_4COOH)$ both contain only one $-COOH$ group,so they are both monocarboxylic acids.

(C) The reaction of $2-\text{hydroxybenzoic acid}$ (Salicylic acid) with acetic anhydride in the presence of an acid catalyst $(H^{+})$ is an acetylation reaction.
In this reaction,the phenolic $-OH$ group of salicylic acid is acetylated to form $2-\text{acetoxybenzoic acid}$,which is commonly known as Aspirin,along with the byproduct acetic acid.
Therefore,$A$ is Salicylic acid.

(D) Anhydrides undergo hydrolysis with water to yield carboxylic acids.
The reaction is as follows:
$(CH_3 CO)_2 O + H_2 O \longrightarrow 2 CH_3 COOH$
Thus,the product obtained is acetic acid $(CH_3 COOH)$.

(A) The hydrolysis of $Methyl$ propanoate $(CH_3CH_2COOCH_3)$ with dil. $NaOH$ yields sodium propanoate $(CH_3CH_2COONa)$ and methanol $(CH_3OH)$.
Subsequent acidification of sodium propanoate with conc. $HCl$ yields propanoic acid $(CH_3CH_2COOH)$.

(C) Ice is the solid form of water $(H_2O)$.
In ice,water molecules are held together by hydrogen bonding,which is a type of intermolecular force.
Therefore,ice is classified as a molecular solid.

(C) The boiling point of haloalkanes depends on the magnitude of van der Waals forces,which increase with an increase in the molecular mass of the halogen atom.
Since the molecular mass of iodine is the highest among the given halogens,$CH_3I$ has the strongest intermolecular forces.
Therefore,the order of boiling points is $CH_3I > CH_3Br > CH_3Cl > CH_3F$.
Thus,$CH_3I$ (Iodomethane) has the highest boiling point.

(B) For a given alkyl group,the boiling point increases with the increasing atomic mass of the halogen atom.
As the size of the halogen atom increases,the magnitude of van der Waals forces increases,leading to a higher boiling point.
Therefore,the boiling point of the given alkyl halides decreases in the order: $CH_3I > CH_3Br > CH_3Cl > CH_3F$.
Thus,$CH_3F$ (Fluoromethane) has the lowest boiling point.

(D) The degree of dissociation $(\alpha)$ is defined as the ratio of molar conductivity at a specific concentration $(\Lambda_c)$ to the limiting molar conductivity $(\Lambda_0)$.
$\alpha = \frac{\Lambda_c}{\Lambda_0}$
Given:
$\Lambda_c = 15.0 \ \Omega^{-1} \ cm^2 \ mol^{-1}$
$\Lambda_0 = 300 \ \Omega^{-1} \ cm^2 \ mol^{-1}$
Substituting the values:
$\alpha = \frac{15.0}{300} = 0.05$

(B) The given reaction is $3 I^{-} + S_2 O_8^{2-} \rightarrow I_3^{-} + 2 SO_4^{2-}$.
According to the rate law expression,the rate of reaction is given by: $\text{Rate} = -\frac{1}{3} \frac{d[I^{-}]}{dt} = \frac{1}{2} \frac{d[SO_4^{2-}]}{dt}$.
We are given $\frac{d[SO_4^{2-}]}{dt} = 2.2 \times 10^{-2} \ mol \ dm^{-3} \ s^{-1}$.
Rearranging the expression to find $-\frac{d[I^{-}]}{dt}$: $-\frac{d[I^{-}]}{dt} = \frac{3}{2} \times \frac{d[SO_4^{2-}]}{dt}$.
Substituting the given value: $-\frac{d[I^{-}]}{dt} = \frac{3}{2} \times 2.2 \times 10^{-2} = 3.3 \times 10^{-2} \ mol \ dm^{-3} \ s^{-1}$.

(C) The rate of reaction is expressed as:
$-\frac{1}{2} \frac{\Delta [N_2O_5]}{\Delta t} = \frac{1}{4} \frac{\Delta [NO_2]}{\Delta t} = \frac{\Delta [O_2]}{\Delta t}$
Given that $-\frac{\Delta [N_2O_5]}{\Delta t} = x \ mol \ dm^{-3} \ s^{-1}$.
Substituting this into the rate expression:
$\frac{1}{2} (x) = \frac{1}{4} \frac{\Delta [NO_2]}{\Delta t}$
$\frac{\Delta [NO_2]}{\Delta t} = 4 \times \frac{x}{2} = 2 x \ mol \ dm^{-3} \ s^{-1}$
Therefore,the average rate of formation of $NO_{2_{(g)}}$ is $2 x \ mol \ dm^{-3} \ s^{-1}$.

(D) The rate of reaction for $2A + B \longrightarrow 3C$ is given by: $\text{Rate} = -\frac{1}{2} \frac{d[A]}{dt} = -\frac{d[B]}{dt} = \frac{1}{3} \frac{d[C]}{dt}$.
Given,the rate of appearance of $C$ is $\frac{d[C]}{dt} = 1.3 \times 10^{-4} \ mol \ L^{-1} \ s^{-1}$.
From the rate expression,$-\frac{1}{2} \frac{d[A]}{dt} = \frac{1}{3} \frac{d[C]}{dt}$.
Therefore,the rate of disappearance of $A$ is $-\frac{d[A]}{dt} = \frac{2}{3} \frac{d[C]}{dt}$.
$-\frac{d[A]}{dt} = \frac{2}{3} \times (1.3 \times 10^{-4} \ mol \ L^{-1} \ s^{-1}) = 0.866 \times 10^{-4} \ mol \ L^{-1} \ s^{-1} = 8.66 \times 10^{-5} \ mol \ L^{-1} \ s^{-1}$.

(A) For a general reaction $aA + bB \rightarrow cC + dD$,the average rate is given by: $\text{Rate} = -\frac{1}{a} \frac{\Delta[A]}{\Delta t} = -\frac{1}{b} \frac{\Delta[B]}{\Delta t} = \frac{1}{c} \frac{\Delta[C]}{\Delta t} = \frac{1}{d} \frac{\Delta[D]}{\Delta t}$.
For the reaction $N_{2(g)} + 3H_{2(g)} \rightarrow 2NH_{3(g)}$,the stoichiometric coefficients are $1$,$3$,and $2$ respectively.
Thus,the rate expression is: $-\frac{\Delta[N_2]}{\Delta t} = -\frac{1}{3} \frac{\Delta[H_2]}{\Delta t} = \frac{1}{2} \frac{\Delta[NH_3]}{\Delta t}$.

(A) The rate law is given as $\text{rate} = k[A][B]^2$.
Given values are $k = 6.25 \ mol^{-2} \ dm^6 \ s^{-1}$,$[A] = 1 \ M$,and $[B] = 0.2 \ M$.
Substituting these values into the rate equation:
$\text{Rate} = 6.25 \times (1) \times (0.2)^2$
$\text{Rate} = 6.25 \times 1 \times 0.04$
$\text{Rate} = 0.25 \ mol \ dm^{-3} \ s^{-1}$.

(D) The rate of reaction is given by the expression:
$\text{Rate} = -\frac{1}{2} \frac{d[N_2 O_5]}{dt} = +\frac{1}{4} \frac{d[NO_2]}{dt} = \frac{d[O_2]}{dt}$
We are given that $-\frac{d[N_2 O_5]}{dt} = 0.02 \ mol \ dm^{-3} \ s^{-1}$.
To find the rate of formation of $NO_2$,we use the relation:
$\frac{1}{4} \frac{d[NO_2]}{dt} = -\frac{1}{2} \frac{d[N_2 O_5]}{dt}$
$\frac{d[NO_2]}{dt} = 2 \times \left( -\frac{d[N_2 O_5]}{dt} \right)$
$\frac{d[NO_2]}{dt} = 2 \times 0.02 \ mol \ dm^{-3} \ s^{-1} = 0.04 \ mol \ dm^{-3} \ s^{-1}$

(B) The given rate law is: $\text{Rate} = k[CH_3Br][OH^{-}]$
If the concentration of both reactants is doubled,the new concentrations become $2[CH_3Br]$ and $2[OH^{-}]$.
New rate $(\text{Rate})_{1} = k \times (2[CH_3Br]) \times (2[OH^{-}])$
$(\text{Rate})_{1} = 4 \times k[CH_3Br][OH^{-}]$
$(\text{Rate})_{1} = 4 \times \text{Rate}$
Therefore,the rate of reaction increases by a factor of $4$.

(C) The rate law for the reaction is given by: $\text{Rate} = k[A]^2[B]$
Rearranging to solve for the rate constant $k$: $k = \frac{\text{Rate}}{[A]^2[B]}$
Substituting the given values: $k = \frac{1.8 \times 10^{-2} \ mol \ dm^{-3} \ s^{-1}}{(0.2 \ mol \ dm^{-3})^2 \times (0.1 \ mol \ dm^{-3})}$
$k = \frac{1.8 \times 10^{-2}}{0.04 \times 0.1} = \frac{1.8 \times 10^{-2}}{4 \times 10^{-3}} = 4.5 \ mol^{-2} \ dm^6 \ s^{-1}$

(A) For a first-order reaction,the rate constant $k$ is given by the formula: $k = \frac{2.303}{t} \log_{10} \frac{[A]_0}{[A]_t}$.
Given that the concentration decreases by $90 \%$,the remaining concentration $[A]_t$ is $10 \%$ of the initial concentration $[A]_0$.
Therefore,$[A]_t = 0.1 [A]_0$ or $\frac{[A]_0}{[A]_t} = 10$.
Substituting the values $t = 30 \ minutes$ and $\frac{[A]_0}{[A]_t} = 10$ into the formula:
$k = \frac{2.303}{30} \log_{10} (10)$.
Since $\log_{10} (10) = 1$,we get $k = \frac{2.303}{30} \approx 0.07676 \ minute^{-1}$.
Rounding to two significant figures,$k = 7.7 \times 10^{-2} \ minute^{-1}$.

(A) For the first order reaction,$A \rightarrow B$,the rate law is given by: $\text{Rate} = k[A]$.
$k = \frac{\text{Rate}}{[A]}$.
Given,$\text{Rate} = 5.4 \times 10^{-6} \ mol \ dm^{-3} \ s^{-1}$ and $[A] = 0.3 \ M$ (or $0.3 \ mol \ dm^{-3}$).
$k = \frac{5.4 \times 10^{-6}}{0.3} \ s^{-1} = 1.8 \times 10^{-5} \ s^{-1}$.

(A) For a first order reaction,the rate constant $k$ is given by the formula: $k = \frac{2.303}{t} \log_{10} \frac{[A]_0}{[A]_t}$.
Given that $80 \%$ of the reactant has reacted,if the initial concentration $[A]_0 = 100$,then the remaining concentration $[A]_t = 100 - 80 = 20$.
The time taken $t = 15 \ minute$.
Substituting these values into the formula:
$k = \frac{2.303}{15} \log_{10} \frac{100}{20}$
$k = \frac{2.303}{15} \log_{10}(5)$
Using $\log_{10}(5) \approx 0.699$:
$k = \frac{2.303 \times 0.699}{15}$
$k \approx 0.1073 \ minute^{-1}$
Rounding to two decimal places,$k \approx 0.11 \ minute^{-1}$.

(A) The rate law expression is determined by the order of the reaction with respect to each reactant.
Given that the reaction is second order in $NO$ and first order in $Cl_2$,the rate law is expressed as:
Rate $= k[NO]^2 [Cl_2]^1 = k[NO]^2 [Cl_2]$

(A) For a first order reaction,the rate constant $k$ is related to the half-life $t_{1/2}$ by the formula:
$k = \frac{0.693}{t_{1/2}}$
Given that the half-life $t_{1/2} = 40 \ minute$:
$k = \frac{0.693}{40} = 1.733 \times 10^{-2} \ minute^{-1}$

(B) For a first-order reaction,the integrated rate equation is $t = \frac{2.303}{k} \log_{10} \frac{[A]_0}{[A]_t}$.
Given that $99.9 \%$ of the reaction is complete,if $[A]_0 = 100$,then $[A]_t = 100 - 99.9 = 0.1$.
Substituting the values: $t = \frac{2.303}{0.576} \log_{10} \frac{100}{0.1}$.
$t = \frac{2.303}{0.576} \log_{10} (1000) = \frac{2.303}{0.576} \times 3$.
$t = 11.99 \approx 12 \ minutes$.

(D) For a first order reaction,the rate constant $k$ is given by the formula: $k = \frac{2.303}{t} \log_{10} \frac{[A]_0}{[A]_t}$.
Given,$[A]_0 = 100 \%$,$[A]_t = 100 - 20 = 80 \%$,and $t = 23.03 \ minutes$.
Substituting these values into the equation:
$k = \frac{2.303}{23.03} \log_{10} \left( \frac{100}{80} \right)$.
$k = 0.1 \times \log_{10}(1.25)$.
Since $\log_{10}(1.25) \approx 0.0969$,we get:
$k = 0.1 \times 0.0969 = 0.00969 \ minute^{-1}$.
Therefore,$k = 9.69 \times 10^{-3} \ minute^{-1}$.

(A) For a first order reaction,$20 \%$ of the reactant has decomposed.
If the initial concentration $[A]_0 = 100$,then the concentration at time $t$,$[A]_t = 100 - 20 = 80$.
The rate constant $k$ is given by the formula: $k = \frac{2.303}{t} \log_{10} \frac{[A]_0}{[A]_t}$.
Substituting the values: $k = \frac{2.303}{15} \log_{10} \frac{100}{80} = \frac{2.303}{15} \log_{10} (1.25)$.
Using $\log_{10} (1.25) \approx 0.0969$:
$k = \frac{2.303}{15} \times 0.0969 \approx 0.01488 \ minute^{-1}$.
Thus,$k = 1.488 \times 10^{-2} \ minute^{-1}$.