MHT CET 2025 Chemistry Question Paper with Answer and Solution

(C) In an acetylene molecule $(HC \equiv CH)$,each carbon atom undergoes $sp$ hybridization.
Each carbon atom forms one $sp-sp$ sigma bond with the other carbon atom.
Each carbon atom also forms one $sp-s$ sigma bond with a hydrogen atom.
Therefore,the $C-H$ bond is formed by the overlap of an $sp$ hybridized orbital of carbon and an $s$ orbital of hydrogen.

(C) To determine the shapes of the molecules, we use the $VSEPR$ theory:
$1$. $H_2O$: The central oxygen atom has $2$ bond pairs and $2$ lone pairs, resulting in a bent ($V$-shaped) geometry.
$2$. $SCl_2$: The central sulfur atom has $2$ bond pairs and $2$ lone pairs, resulting in a bent ($V$-shaped) geometry.
Both $H_2O$ and $SCl_2$ have a bent shape.
$3$. $NH_3$ is trigonal pyramidal, while $SO_2$ is bent.
$4$. $XeF_4$ is square planar, while $SF_4$ is see-saw.
$5$. $PCl_5$ is trigonal bipyramidal, while $BrF_5$ is square pyramidal.
Therefore, the correct pair is $H_2O$ and $SCl_2$.

(C) To determine the geometry,we calculate the steric number using the formula: $Steric \ Number = \frac{1}{2} (V + M - C + A)$,where $V$ is the number of valence electrons of the central atom,$M$ is the number of monovalent atoms,$C$ is the cationic charge,and $A$ is the anionic charge.
For $XeOF_4$:
$V = 8$ (for $Xe$),$M = 4$ (for $F$),$O$ is divalent (not counted in $M$).
$Steric \ Number = \frac{1}{2} (8 + 4) = 6$.
$A$ steric number of $6$ corresponds to $sp^3d^2$ hybridization,which has an octahedral electron geometry.
With $5$ bonding pairs ($4$ $Xe-F$ and $1$ $Xe=O$) and $1$ lone pair,the molecular geometry is square pyramidal.

(D) The electronic configuration of $N_2$ ($14$ electrons) is $\sigma 1s^2, \sigma^* 1s^2, \sigma 2s^2, \sigma^* 2s^2, \pi 2p_x^2 = \pi 2p_y^2, \sigma 2p_z^2$.
Bond order = $\frac{1}{2}(N_b - N_a) = \frac{1}{2}(10 - 4) = 3$.
Since all electrons are paired,$N_2$ is diamagnetic.
$O_2$ ($16$ electrons) has a bond order of $2$ and is paramagnetic due to two unpaired electrons in $\pi^* 2p$ orbitals.
$N_2$ has $10$ bonding electrons,while $O_2$ has $8$ bonding electrons.
Therefore,the statement '$N_2$ is diamagnetic' is correct,and the statement '$N_2$ is paramagnetic' would be incorrect.
Given the options,the statement '$N_2$ is diamagnetic' is correct,but if the option was written as 'It is dimagnetic' (a typo for diamagnetic),it is still the intended correct property.
However,all statements $A$,$B$,$C$,and $D$ are actually correct descriptions of $N_2$.
If we assume the question asks for the incorrect statement,there might be a typo in the provided options.
Based on standard chemistry,$N_2$ is diamagnetic,so option $D$ is a correct statement.

(A) Isoelectronic species are those that have the same number of electrons.
$1$. For $Ne$ $(Z=10)$: Electrons = $10$. For $O^{2-}$ $(Z=8)$: Electrons = $8 + 2 = 10$. Both have $10$ electrons,so they are isoelectronic.
$2$. For $Cl^{-}$ $(Z=17)$: Electrons = $17 + 1 = 18$. For $Ca$ $(Z=20)$: Electrons = $20$. Not isoelectronic.
$3$. For $Ar$ $(Z=18)$: Electrons = $18$. For $F^{-}$ $(Z=9)$: Electrons = $9 + 1 = 10$. Not isoelectronic.
$4$. For $K^{+}$ $(Z=19)$: Electrons = $19 - 1 = 18$. For $Al^{3+}$ $(Z=13)$: Electrons = $13 - 3 = 10$. Not isoelectronic.
Therefore,the correct pair is $Ne$ and $O^{2-}$.

(D) To determine the number of electrons in antibonding molecular orbitals,we write the molecular orbital configuration for each molecule:
$1$. $Li_2$ ($6$ electrons): $\sigma 1s^2, \sigma^* 1s^2, \sigma 2s^2$. Antibonding electrons = $2$ (in $\sigma^* 1s$).
$2$. $N_2$ ($14$ electrons): $\sigma 1s^2, \sigma^* 1s^2, \sigma 2s^2, \sigma^* 2s^2, \pi 2p_x^2 = \pi 2p_y^2, \sigma 2p_z^2$. Antibonding electrons = $2+2 = 4$.
$3$. $O_2$ ($16$ electrons): $\sigma 1s^2, \sigma^* 1s^2, \sigma 2s^2, \sigma^* 2s^2, \sigma 2p_z^2, \pi 2p_x^2 = \pi 2p_y^2, \pi^* 2p_x^1 = \pi^* 2p_y^1$. Antibonding electrons = $2+2+2 = 6$.
$4$. $F_2$ ($18$ electrons): $\sigma 1s^2, \sigma^* 1s^2, \sigma 2s^2, \sigma^* 2s^2, \sigma 2p_z^2, \pi 2p_x^2 = \pi 2p_y^2, \pi^* 2p_x^2 = \pi^* 2p_y^2$. Antibonding electrons = $2+2+4 = 8$.
Thus,$F_2$ contains the maximum number of electrons in antibonding molecular orbitals.

(B) The total number of electrons in an $F_2$ molecule is $18$ $(9 + 9)$.
The molecular orbital configuration for $F_2$ is: $\sigma 1s^2, \sigma^* 1s^2, \sigma 2s^2, \sigma^* 2s^2, \sigma 2p_z^2, \pi 2p_x^2 = \pi 2p_y^2, \pi^* 2p_x^2 = \pi^* 2p_y^2$.
Counting the electrons in bonding molecular orbitals (those without an asterisk): $2 (\sigma 1s) + 2 (\sigma 2s) + 2 (\sigma 2p_z) + 4 (\pi 2p_x, \pi 2p_y) = 10$.
Counting the electrons in antibonding molecular orbitals (those with an asterisk): $2 (\sigma^* 1s) + 2 (\sigma^* 2s) + 4 (\pi^* 2p_x, \pi^* 2p_y) = 8$.
Therefore,the number of electrons in bonding and antibonding molecular orbitals are $10$ and $8$ respectively.

$(A)$ To determine the bond length, we look at the bond order and the atomic radii of the atoms involved.
$1$. The bond orders are: $N_2$ $(3)$, $O_2$ $(2)$, and $Cl_2$ $(1)$.
$2$. Bond length is inversely proportional to bond order for similar atoms, but here we must also consider atomic size.
$3$. The atomic radius of $Cl$ $(99 \text{ pm})$ is significantly larger than that of $N$ $(75 \text{ pm})$ and $O$ $(73 \text{ pm})$.
$4$. Due to the large atomic size of $Cl$, the $Cl-Cl$ bond length $(199 \text{ pm})$ is much greater than the $O=O$ bond length $(121 \text{ pm})$ and the $N \equiv N$ bond length $(110 \text{ pm})$.
$5$. Comparing $O_2$ and $N_2$, $N_2$ has a higher bond order $(3)$ than $O_2$ $(2)$, so $N_2$ has a shorter bond length.
$6$. Therefore, the decreasing order of bond length is $Cl_2 > O_2 > N_2$.

(C) The total number of electrons in $NO^{+}$ is $7 (N) + 8 (O) - 1 = 14$ electrons.
According to Molecular Orbital Theory,the electronic configuration of $NO^{+}$ is: $\sigma 1s^2, \sigma^* 1s^2, \sigma 2s^2, \sigma^* 2s^2, \sigma 2p_z^2, \pi 2p_x^2 = \pi 2p_y^2$.
The bond order is calculated as: $\text{Bond Order} = \frac{N_b - N_a}{2} = \frac{10 - 4}{2} = \frac{6}{2} = 3$.

(C) Lattice enthalpy is directly proportional to the product of the charges of the ions and inversely proportional to the inter-ionic distance (sum of ionic radii).
$Lattice \ Enthalpy \propto \frac{|q_+ q_-|}{r_+ + r_-}$.
Comparing the given compounds:
$LiCl$: $Li^+$ $(+1)$,$Cl^-$ $(-1)$
$NaCl$: $Na^+$ $(+1)$,$Cl^-$ $(-1)$
$BeF_2$: $Be^{2+}$ $(+2)$,$F^-$ $(-1)$
$CaCl_2$: $Ca^{2+}$ $(+2)$,$Cl^-$ $(-1)$
$BeF_2$ has the highest charge product $(|(+2) \times (-1)| = 2)$ and the smallest ionic radii sum ($Be^{2+}$ and $F^-$ are smaller than $Ca^{2+}$ and $Cl^-$). Therefore,$BeF_2$ has the highest lattice enthalpy.

(B) According to Fajan's rule,covalent character increases with:
$1$. Smaller size of the cation.
$2$. Larger size of the anion.
Comparing the given compounds:
- The cations are $Li^+$ and $Na^+$. Since $Li^+$ is smaller than $Na^+$,$Li^+$ compounds have more covalent character.
- The anions are $Cl^-$ and $I^-$. Since $I^-$ is larger than $Cl^-$,$I^-$ compounds have more covalent character.
Therefore,$LiI$ has the smallest cation and the largest anion,resulting in the maximum covalent character.

(B) The polarity of a molecule is determined by its net dipole moment $(\mu)$.
$1$. $H_2S$ has a bent geometry with a dipole moment of approximately $0.95 \ D$.
$2$. $NH_3$ has a trigonal pyramidal geometry with a dipole moment of approximately $1.47 \ D$.
$3$. $NF_3$ has a trigonal pyramidal geometry,but the electronegativity difference between $N$ and $F$ results in bond dipoles that partially cancel out,leading to a lower dipole moment of approximately $0.24 \ D$.
$4$. $CHCl_3$ has a tetrahedral geometry with a dipole moment of approximately $1.01 \ D$.
Comparing these values,$NH_3$ has the highest net dipole moment among the given options,making it the most polar molecule.

(D) In a metallic solid,the metal atoms lose their valence electrons to form a sea of mobile electrons. The electrostatic force of attraction between these positively charged metal ions (cations) and the surrounding sea of delocalized electrons is known as a metallic bond.

(C) Intermolecular hydrogen bonding occurs in molecules where a hydrogen atom is covalently bonded to a highly electronegative atom like $F$,$O$,or $N$.
In $H_2O$,$NH_3$,and $HF$,the hydrogen atom is bonded to $O$,$N$,and $F$ respectively,allowing for hydrogen bonding.
In $Br_2$,the molecule consists of two bromine atoms bonded together. Since there is no hydrogen atom bonded to a highly electronegative atom,$Br_2$ cannot form intermolecular hydrogen bonding.
Therefore,the correct option is $C$.

(C) Dinitrogen $(N_2)$ is a non-polar homonuclear diatomic molecule.
In non-polar molecules,the only intermolecular forces present are London dispersion forces (also known as induced dipole $-$ induced dipole interactions).
Therefore,the correct option is $C$.

(B) The unit of the rate constant $(k)$ is $s^{-1}$.
For a reaction of order $n$,the unit of the rate constant is $(mol \ L^{-1})^{1-n} \ s^{-1}$.
Given the unit is $s^{-1}$,we have $1-n = 0$,which implies $n = 1$.
Therefore,the reaction is a first-order reaction.

(A) Atomic radius decreases across a period (from left to right) and increases down a group.
$Na$ $(Z=11)$ and $Mg$ $(Z=12)$ belong to the $3^{rd}$ period,where $Mg$ is to the right of $Na$,so $Mg < Na$.
$Na$,$K$,and $Rb$ belong to Group $1$. The order of atomic radii down the group is $Na < K < Rb$.
Combining these,we compare the periods: $Mg$ ($3^{rd}$ period) is smaller than $Na$ ($3^{rd}$ period),and $Na$ is smaller than $K$ ($4^{th}$ period),which is smaller than $Rb$ ($5^{th}$ period).
Thus,the correct increasing order is $Mg < Na < K < Rb$.

(D) Electronegativity generally increases as we move up a group in the periodic table due to a decrease in atomic size and an increase in effective nuclear charge.
All the given elements ($Be$,$Mg$,$Ca$,$Sr$) belong to Group $2$ (Alkaline Earth Metals).
The order of elements from top to bottom in Group $2$ is $Be$,$Mg$,$Ca$,$Sr$.
Since $Be$ is at the top of this group,it has the smallest atomic size and the highest electronegativity among the given options.

(A) Electronegativity is the tendency of an atom to attract a shared pair of electrons towards itself.
In the periodic table,electronegativity generally decreases down a group as the atomic size increases and the effective nuclear charge decreases.
All the given elements ($Li$,$Na$,$K$,$Rb$) belong to Group $1$ (alkali metals).
Since $Li$ is at the top of the group,it has the smallest atomic size and the highest effective nuclear charge among the given options.
Therefore,$Li$ has the highest electronegativity.

(D) In $KMnO_4$,the oxidation state of $Mn$ is $+7$.
In $MnSO_4$,the oxidation state of $Mn$ is $+2$.
The reduction half-reaction is: $MnO_4^- + 8H^+ + 5e^- \rightarrow Mn^{2+} + 4H_2O$.
This shows that $1 \ mol$ of $KMnO_4$ requires $5 \ mol$ of electrons ($5 \ F$ of charge) to be reduced to $Mn^{2+}$.
Therefore,for $2 \ mol$ of $KMnO_4$,the total charge required is $2 \times 5 \ F = 10 \ F$.

(D) The electrolysis of $50 \%$ sulfuric acid $(H_{2}SO_{4})$ at high current density yields peroxydisulfuric acid $(H_{2}S_{2}O_{8})$,also known as Marshall's acid,at the anode.
$2H_{2}SO_{4} \rightarrow H_{2}S_{2}O_{8} + H_{2(g)}$
Upon further hydrolysis of peroxydisulfuric acid,it reacts with water to produce hydrogen peroxide $(H_{2}O_{2})$ and sulfuric acid.
$H_{2}S_{2}O_{8} + 2H_{2}O \rightarrow 2H_{2}SO_{4} + H_{2}O_{2}$
Thus,the final product obtained after hydrolysis is hydrogen peroxide $(H_{2}O_{2})$.

(D) According to Kohlrausch's law,the molar conductivity at infinite dilution for $NH_4OH$ is given by: $\Lambda^0_{m}(NH_4OH) = \Lambda^0_{m}(NH_4Cl) + \Lambda^0_{m}(NaOH) - \Lambda^0_{m}(NaCl)$.
Substituting the given values: $\Lambda^0_{m}(NH_4OH) = 130 + 213 - 109 = 234 \ S \ cm^2 \ mol^{-1}$.
The degree of dissociation $(\alpha)$ is calculated as: $\alpha = \frac{\Lambda^m_c}{\Lambda^0_m} = \frac{9.0}{234}$.
The percent dissociation is: $\alpha \times 100 = \frac{9.0 \times 100}{234} = \frac{900}{234} = \frac{100}{26} \%$.

(D) The depletion of the ozone layer is primarily caused by chlorofluorocarbons $(CFCs)$.
In the stratosphere,$CFCs$ absorb ultraviolet $(UV)$ radiation and undergo photolysis to release chlorine radicals $(Cl^{\bullet})$.
These chlorine radicals then react with ozone $(O_3)$ to break it down into oxygen $(O_2)$ and chlorine monoxide $(ClO^{\bullet})$.
Therefore,the reactive species produced during the process of ozone depletion involving $CFCs$ are chlorine radicals.

(B) Nitric oxide $(NO)$ reacts with ozone $(O_3)$ in the stratosphere to form nitrogen dioxide $(NO_2)$ and oxygen $(O_2)$.
This reaction is represented as: $NO(g) + O_3(g) \rightarrow NO_2(g) + O_2(g)$.
This process contributes to the depletion of the ozone layer.

(A) Nitrogen oxides,particularly nitric oxide $(NO)$,are released from the exhaust systems of supersonic jet aeroplanes flying in the lower stratosphere.
These gases react with ozone $(O_3)$ to form nitrogen dioxide $(NO_2)$ and oxygen $(O_2)$,leading to the depletion of the ozone layer.
The reaction is: $NO(g) + O_3(g) \rightarrow NO_2(g) + O_2(g)$.

(D) In the context of Green Chemistry,$H_2O$ is considered an environmentally friendly solvent. It is non-toxic,non-flammable,and does not contribute to air pollution or the generation of hazardous waste compared to organic solvents like $CH_2Cl_2$,$CHCl_3$,or $CCl_4$.

(C) The principle of green chemistry that advocates for the use of chemicals derived from plant-based or biological sources is known as the $Use \ of \ renewable \ feedstocks$. This principle aims to reduce reliance on depleting fossil fuel resources by utilizing sustainable materials.

(A) green solvent is a solvent that is environmentally friendly and minimizes the impact on human health and the environment.
Supercritical $CO_2$ is widely used as a green solvent in various industrial processes,such as decaffeination and dry cleaning,because it is non-toxic,non-flammable,and easily recyclable.
$CHCl_3$,$CH_2Cl_2$,and $CCl_4$ are chlorinated solvents that are toxic and harmful to the environment.

(B) The principle of 'Reduce derivatives' in green chemistry states that unnecessary derivatization (such as the use of blocking groups,protection/deprotection,or temporary modification of physical/chemical processes) should be minimized or avoided if possible. This is because such steps require additional reagents and generate waste,making the process less efficient and less environmentally friendly.

(C) Neopentane is $2,2$-dimethylpropane,which has the molecular formula $C_5H_{12}$.
Its structural formula is $CH_3-C(CH_3)_2-CH_3$.
In a bond line formula,the central carbon atom is bonded to four methyl groups,forming a cross-like shape where the central point represents the quaternary carbon atom and the four lines represent the methyl groups.
This corresponds to the structure shown in image $232310-$c.

(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.

(A) $1$. Identify the longest carbon chain containing the double bond: The chain has $5$ carbons,so the parent alkane is pentene.
$2$. Number the chain to give the double bond the lowest possible locant: Numbering from the right gives the double bond position $2$ (between $C2$ and $C3$).
$3$. Identify substituents: There is a bromo group $(-Br)$ at position $4$ and a methyl group $(-CH_3)$ at position $4$.
$4$. Combine the parts: $4-$Bromo$-4-$methylpent$-2-$ene.

(B) According to the $IUPAC$ priority order for functional groups,the priority is as follows:
$1$. Carboxylic acids $(-COOH)$
$2$. Esters $(-COOR)$
$3$. Acid halides $(-COX)$
$4$. Amides $(-CONH_2)$
$5$. Nitriles $(-CN)$
$6$. Aldehydes $(-CHO)$
$7$. Ketones $(-CO-)$
$8$. Alcohols $(-OH)$
$9$. Amines $(-NH_2)$
$10$. Alkenes/Alkynes $(-C=C-, -C \equiv C-)$
Comparing the given options:
$-CONH_2$ (Amide) has a higher priority than $-CN$ (Nitrile),$-NH_2$ (Amine),and $-C \equiv C-$ (Alkyne).
Therefore,the correct option is $B$.

(C) $1$. Identify the principal functional group: The $-OH$ group is the principal functional group,so the parent chain is a cyclopentane ring,making it a cyclopentanol.
$2$. Number the ring: The carbon attached to the $-OH$ group is assigned position $1$.
$3$. Number the substituents: To give the lowest possible locants to the methyl groups,we number the ring in a way that the methyl groups are at positions $2$ and $5$. Thus,the name is $2,5-$dimethylcyclopentan$-1-$ol (or simply $2,5-$dimethylcyclopentanol).

(C) $1$. Identify the principal functional group: The compound contains a benzene ring with $-OH$,$-CHO$,and $-CH_3$ groups. According to $IUPAC$ priority rules,the aldehyde $(-CHO)$ group has higher priority than the hydroxyl $(-OH)$ group. Therefore,the parent compound is benzaldehyde.
$2$. Numbering the ring: The carbon atom attached to the $-CHO$ group is assigned position $1$. We then number the ring to give the lowest possible locants to the substituents. If we start from the $-CHO$ group as $1$,the $-OH$ group is at position $3$ and the $-CH_3$ group is at position $5$. Thus,the name is $3-$hydroxy$-5-$methylbenzaldehyde.

(A) $1$. Identify the principal functional group: The $-OH$ group is the principal functional group,so the parent chain is a cyclobutanol. The carbon attached to the $-OH$ group is assigned position $1$.
$2$. Number the ring: To give the substituents the lowest possible locants,we number the ring starting from the carbon with the $-OH$ group $(C-1)$. Moving towards the ethyl group gives the substituents positions $2$ and $4$ (ethyl at $2$,methyl at $4$). Moving the other way would give positions $2$ and $4$ as well (methyl at $2$,ethyl at $4$).
$3$. Apply alphabetical order: According to $IUPAC$ rules,substituents are listed alphabetically. Ethyl $(E)$ comes before methyl $(M)$. Therefore,we choose the numbering that assigns $2$ to the ethyl group and $4$ to the methyl group.
$4$. Final Name: The compound is $2-$ethyl$-4-$methylcyclobutanol.

(B) According to the $IUPAC$ priority order for functional groups,the priority sequence is: $-COOH > -SO_3H > -COOR > -COCl > -CONH_2 > -CN > -CHO > -CO > -OH > -NH_2 > -OR > -R > -X$.
Comparing the given options:
$1$. $-SO_3H$ (Sulfonic acid)
$2$. $-COCl$ (Acid chloride)
$3$. $-CHO$ (Aldehyde)
$4$. $-OH$ (Alcohol)
Among the given choices,$-SO_3H$ appears earliest in the priority list,thus it has the highest priority.

(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.

(D) $1$. Identify the principal functional group: The $-COOH$ group (carboxylic acid) has higher priority than the $-CHO$ (aldehyde) and $-CH_3$ (methyl) groups. Therefore,the parent chain is a benzoic acid derivative.
$2$. Number the ring: Assign position $1$ to the carbon attached to the $-COOH$ group. Number the ring to give the lowest possible locants to the substituents. Moving towards the $-CH_3$ group gives it position $2$,and the $-CHO$ group gets position $5$.
$3$. Name the substituents: The $-CHO$ group at position $5$ is named as 'formyl',and the $-CH_3$ group at position $2$ is named as 'methyl'.
$4$. Combine the parts: The name is $5-$formyl$-2-$methylbenzoic acid.

(B) Functional isomers are compounds that have the same molecular formula but different functional groups.
In option $B$,Butan-$1$-ol $(C_4H_{10}O)$ is an alcohol,while $1$-Methoxypropane $(C_4H_{10}O)$ is an ether.
Since they have the same molecular formula but different functional groups ($-OH$ vs $-O-$),they are functional isomers.

(D) heteroatom is any atom other than carbon or hydrogen in an organic molecule.
$Furan$ $(C_4H_4O)$ contains an oxygen atom in its ring.
$THF$ (Tetrahydrofuran,$C_4H_8O$) contains an oxygen atom in its ring.
$4H$-Pyran $(C_5H_6O)$ contains an oxygen atom in its ring.
$Pyrrole$ $(C_4H_5N)$ contains a nitrogen atom as the heteroatom,not oxygen.
Therefore,the correct answer is $D$.

(C) The given structure is a cyclobutane ring with a methoxy group $(-OCH_3)$ at position $1$ and two methyl groups $(-CH_3)$ at position $2$.
According to $IUPAC$ nomenclature rules for cyclic ethers:
$1$. The ring is numbered starting from the carbon attached to the methoxy group as $1$.
$2$. The next carbon is numbered $2$ to give the lowest possible locants to the substituents.
$3$. Thus,the methoxy group is at position $1$ and two methyl groups are at position $2$.
$4$. The name is $1-$Methoxy-$2,2-$dimethylcyclobutane.

(A) The chemical formulas for the given compounds are as follows:
$1$. Thiophene: $C_4H_4S$ (contains sulfur,no nitrogen).
$2$. Pyridine: $C_5H_5N$ (contains nitrogen).
$3$. Pyrrole: $C_4H_5N$ (contains nitrogen).
$4$. Piperidine: $C_5H_{11}N$ (contains nitrogen).
Therefore,Thiophene is the compound that does not contain nitrogen.

(D) Undecane is an alkane with the molecular formula $C_{11}H_{24}$.
Dodecane is an alkane with the molecular formula $C_{12}H_{26}$.
The difference between these two consecutive members of the homologous series is a $CH_2$ group.
The molar mass of a $CH_2$ group is $(1 \times 12.01) + (2 \times 1.008) \approx 14.02 \ g \ mol^{-1}$,which is approximately $14 \ g \ mol^{-1}$.

(C) Pyrogallol is a common name for $1,2,3$-trihydroxybenzene.
According to $IUPAC$ nomenclature rules,when three hydroxyl $(-OH)$ groups are attached to a benzene ring,the compound is named as a benzene derivative with the suffix $-triol$.
The positions of the $-OH$ groups are $1, 2,$ and $3$.
Therefore,the correct $IUPAC$ name is Benzene-$1,2,3$-triol.

(D) In the carbinol system,methyl alcohol $(CH_3OH)$ is considered the parent compound and is called $carbinol$.
Other alcohols are named as derivatives of $carbinol$ by replacing the hydrogen atoms of the $CH_3$ group with alkyl groups.
$tert-butyl$ alcohol has the structure $(CH_3)_3C-OH$.
Here,all three hydrogen atoms of the $CH_3$ group in $carbinol$ are replaced by three methyl groups.
Therefore,it is named as $trimethyl$ $carbinol$.

(B) homologous series is a group of organic compounds having the same functional group and similar chemical properties,where each successive member differs by a $CH_2$ group.
The molar mass of a $CH_2$ group is calculated as:
$M(CH_2) = M(C) + 2 \times M(H) = 12 \ g/mol + 2 \times 1 \ g/mol = 14 \ g/mol$.
Therefore,the difference in molar masses of two neighbouring members of a homologous series is $14 \ g$.

(C) The given structure is $2,3$-dimethylbutane.
Its molecular formula is $C_6H_{14}$.
In one molecule,there are $6$ carbon atoms and $14$ hydrogen atoms.
Therefore,in $n$ moles of the molecule,the number of moles of $C$ atoms is $6 n$ and the number of moles of $H$ atoms is $14 n$.

(A) The given structure is styrene $(C_6H_5-CH=CH_2)$.
Counting the atoms in one molecule:
Carbon $(C)$ atoms: $6$ (in benzene ring) $+ 2$ (in vinyl group) $= 8$ atoms.
Hydrogen $(H)$ atoms: $5$ (in benzene ring) $+ 3$ (in vinyl group) $= 8$ atoms.
Therefore,the molecular formula is $C_8H_8$.
For $n$ moles of the molecule,the number of moles of $C$ atoms $= 8 \times n = 8n$.
The number of moles of $H$ atoms $= 8 \times n = 8n$.
Thus,the correct option is $A$.

(B) Aromatic compounds are classified into two types: benzenoid and non-benzenoid.
Benzenoid compounds contain at least one benzene ring, such as $Aniline$, $Naphthalene$, and $Phenol$.
Non-benzenoid aromatic compounds possess aromaticity but do not contain a benzene ring.
$Tropone$ (cycloheptatrienone) is a classic example of a non-benzenoid aromatic compound because it follows Hückel's rule ($4n+2$ $\pi$ electrons) and has a cyclic, planar, conjugated structure without a benzene ring.

(C) The given reaction is a $Cross-Cannizzaro$ reaction between formaldehyde $(HCHO)$ and benzaldehyde $(C_6H_5CHO)$ in the presence of concentrated $NaOH$.
In a $Cross-Cannizzaro$ reaction,the more reactive aldehyde (formaldehyde) undergoes oxidation to form a salt of the carboxylic acid,while the less reactive aldehyde (benzaldehyde) undergoes reduction to form an alcohol.
$1$. Formaldehyde $(HCHO)$ is more reactive towards nucleophilic attack,so it gets oxidized to sodium formate $(HCOONa)$,which upon acidification $(H_3O^+)$ gives methanoic acid $(HCOOH)$.
$2$. Benzaldehyde $(C_6H_5CHO)$ gets reduced to phenylmethanol $(C_6H_5CH_2OH)$.
Therefore,the products are methanoic acid and phenylmethanol.

(C) The reduction of nitriles with diisobutylaluminium hydride $(DIBAL-H)$ followed by acid hydrolysis is a standard method to prepare aldehydes.
$DIBAL-H$ selectively reduces the nitrile group $(-CN)$ to an imine intermediate,which upon hydrolysis yields the corresponding aldehyde.
In the case of $Hex-3-enenitrile$ $(CH_3CH_2CH=CHCH_2CN)$,the nitrile group is reduced to an aldehyde group $(-CHO)$ while the carbon-carbon double bond $(C=C)$ remains unaffected.
Therefore,the product formed is $Hex-3-enal$ $(CH_3CH_2CH=CHCH_2CHO)$.

(D) The boiling point of aldehydes and ketones increases with an increase in the molecular mass due to the increase in the magnitude of intermolecular van der Waals forces of attraction.
Among the given options,$Hexanal$ $(C_6H_{12}O)$ has the highest molecular mass compared to $Ethanal$ $(C_2H_4O)$,$Propanal$ $(C_3H_6O)$,and $Pentanal$ $(C_5H_{10}O)$.
Therefore,$Hexanal$ has the highest boiling point.

(C) The reduction of the carbonyl group $(>C=O)$ of aldehydes and ketones to a methylene group $(-CH_2-)$ using zinc-amalgam $(Zn-Hg)$ and concentrated hydrochloric acid $(HCl)$ is known as the $Clemmensen \ reduction$.
The general reaction is: $R-CO-R' + 4[H] \xrightarrow{Zn-Hg/HCl} R-CH_2-R' + H_2O$.

(C) Schiff's reagent is a specific test used to detect the presence of aldehydes.
Aldehydes react with Schiff's reagent to produce a characteristic pink or magenta colouration.
Among the given options,$Propanal$ $(CH_3CH_2CHO)$ is an aldehyde.
Propanol is an alcohol,Propanone is a ketone,and Propanoic acid is a carboxylic acid; none of these react with Schiff's reagent to give the pink colouration.
Therefore,the correct option is $C$.

(B) The haloform reaction is given by compounds containing the $CH_3CO-$ group or compounds that can be oxidized to this group (like $CH_3CH(OH)-$ group).
$1$. Ethanal $(CH_3CHO)$ contains the $CH_3CO-$ group and gives a positive haloform test.
$2$. Propanal $(CH_3CH_2CHO)$ does not contain the $CH_3CO-$ group and cannot be oxidized to it,so it does not give a positive haloform test.
$3$. Propanone $(CH_3COCH_3)$ contains the $CH_3CO-$ group and gives a positive haloform test.
$4$. Butanone $(CH_3COCH_2CH_3)$ contains the $CH_3CO-$ group and gives a positive haloform test.
Therefore,Propanal does not exhibit the haloform reaction.

(B) The Gatterman-Koch reaction is a specific formylation reaction used to introduce a formyl group $(-CHO)$ onto an aromatic ring.
In this reaction,the arene is treated with carbon monoxide $(CO)$ and hydrogen chloride $(HCl)$ in the presence of anhydrous aluminum chloride $(AlCl_3)$ and cuprous chloride $(CuCl)$ as a catalyst.
This combination generates the reactive electrophile $HCO^+$,which performs electrophilic aromatic substitution on the arene to yield an aromatic aldehyde.
Therefore,the correct reagent system is $CO, HCl$ (anhyd. $AlCl_3$).

(A) The solubility of organic compounds in water depends on their ability to form hydrogen bonds with water molecules.
$1$. Alcohols $(R-OH)$ can form strong hydrogen bonds with water due to the highly electronegative oxygen atom.
$2$. Amines $(R-NH_2)$ can also form hydrogen bonds with water, but the $N-H \cdots O$ hydrogen bond is weaker than the $O-H \cdots O$ hydrogen bond because nitrogen is less electronegative than oxygen.
$3$. Alkanes are non-polar hydrocarbons and cannot form hydrogen bonds with water, making them essentially insoluble.
Therefore, the correct decreasing order of solubility is $Alcohols > Amines > Alkanes$.

(A) The $pK_{b}$ value is inversely proportional to the basic strength of the compound. $A$ lower $pK_{b}$ value indicates a stronger base.
$N$-Ethylethanamine is a secondary aliphatic amine,$(C_2H_5)_2NH$.
Propan-$2$-amine is a primary aliphatic amine,$CH_3CH(NH_2)CH_3$.
$NH_3$ is ammonia.
Benzenamine (aniline) is an aromatic amine where the lone pair on nitrogen is involved in resonance with the benzene ring,making it the least basic.
Among aliphatic amines,secondary amines are generally more basic than primary amines due to the inductive effect of alkyl groups and solvation factors.
Therefore,$N$-Ethylethanamine is the strongest base among the given options and has the lowest $pK_{b}$ value.

(C) secondary amine is an amine where the nitrogen atom is bonded to two carbon atoms (alkyl or aryl groups) and one hydrogen atom,represented by the general formula $R_2NH$.
$1$. Ethane-$1,2$-diamine $(NH_2-CH_2-CH_2-NH_2)$ is a primary diamine.
$2$. Propan-$2$-amine $(CH_3-CH(NH_2)-CH_3)$ is a primary amine.
$3$. $N$-Methylmethanamine $(CH_3-NH-CH_3)$ has two methyl groups attached to the nitrogen atom,making it a secondary amine.
$4$. $N,N$-Dimethylmethanamine $(CH_3-N(CH_3)_2)$ is a tertiary amine.
Therefore,the correct option is $C$.

(D) The stability of substituted ammonium ions in the gas phase is determined by the inductive effect of the alkyl groups.
Alkyl groups are electron-donating ($+I$ effect).
As the number of alkyl groups attached to the nitrogen atom increases,the positive charge on the nitrogen is dispersed more effectively due to the electron-donating nature of the alkyl groups.
Therefore,the stability order is $R_3NH^{+} > R_2NH_2^{+} > R-NH_3^{+} > NH_4^{+}$.
Thus,$R_3NH^{+}$ is the most stable species.

(A) The reaction of benzene diazonium chloride $(C_6H_5N_2^+Cl^-)$ with ethanol $(C_2H_5OH)$ is a reduction reaction.
In this reaction,the diazonium group is replaced by a hydrogen atom,and ethanol is oxidized to acetaldehyde $(CH_3CHO)$.
The reaction is: $C_6H_5N_2^+Cl^- + C_2H_5OH \rightarrow C_6H_6 + N_2 + CH_3CHO + HCl$.
Thus,product $A$ is benzene $(C_6H_6)$.

(D) The reaction of an amide with bromine and an aqueous or alcoholic solution of potassium hydroxide is known as the $Hofmann$ bromamide degradation reaction.
In this reaction,acetamide $(CH_3CONH_2)$ reacts with $Br_2$ and excess $KOH$ to form methylamine $(CH_3NH_2)$.
The chemical equation is:
$CH_3CONH_2 + Br_2 + 4KOH \rightarrow CH_3NH_2 + K_2CO_3 + 2KBr + 2H_2O$.
Thus,the product obtained is methylamine $(CH_3NH_2)$.

(B) Azo coupling is an electrophilic aromatic substitution reaction where a diazonium salt reacts with an electron-rich aromatic compound (like phenol or aniline) to form an azo compound $(R-N=N-R')$.
In option $B$,$C_6H_5N_2Cl$ (benzenediazonium chloride) reacts with $C_6H_5OH$ (phenol) in a basic medium $(OH^-)$ to produce $p$-hydroxyazobenzene,which is a classic example of an azo coupling reaction.
Option $A$ represents the diazotization of aniline.
Option $C$ represents the Balz-Schiemann reaction.
Option $D$ represents the Gattermann reaction.

(C) The basicity of amines in an aqueous solution depends on the combined effect of the inductive effect $(+I)$,solvation effect,and steric hindrance.
For methyl-substituted amines in an aqueous medium,the order of basicity is $(CH_3)_2NH > CH_3NH_2 > (CH_3)_3N > NH_3$.
$(CH_3)_2NH$ (dimethylamine) is the strongest base because it provides the optimal balance between the electron-donating inductive effect of two methyl groups and the stability provided by solvation in water,while minimizing steric hindrance compared to trimethylamine.

(C) The $pK_{b}$ value is inversely proportional to the basic strength of the amine. $A$ higher $pK_{b}$ value indicates a weaker base.
$1$. $(CH_3)_2NH$ is a secondary aliphatic amine,which is a strong base due to the electron-donating inductive effect of two methyl groups.
$2$. $(CH_3)_3N$ is a tertiary aliphatic amine,which is also a strong base.
$3$. $C_6H_5CH_2NH_2$ is a primary amine where the lone pair is not in conjugation with the benzene ring,making it more basic than aniline.
$4$. $C_6H_5NH_2$ (aniline) is the weakest base among the given options because the lone pair on the nitrogen atom is involved in resonance with the benzene ring,reducing its availability for protonation.
Since $C_6H_5NH_2$ is the weakest base,it has the highest $pK_{b}$ value.

(D) Step $1$: Aniline reacts with $NaNO_2 + HCl$ at $273-278 \ K$ to form $A$,which is Benzenediazonium chloride $(C_6H_5N_2^+Cl^-)$.
Step $2$: When Benzenediazonium chloride is heated with water $(\Delta, H_2O)$,it undergoes hydrolysis to form $B$,which is Phenol $(C_6H_5OH)$,along with the evolution of Nitrogen gas $(N_2 \uparrow)$ and Hydrochloric acid $(HCl)$.
Therefore,product $B$ is Phenol.

(A) In aqueous solution,the basic strength of methyl-substituted amines depends on the combined effect of inductive effect,solvation effect,and steric hindrance.
For methyl amines,the order of basicity is $(CH_3)_2NH > CH_3NH_2 > (CH_3)_3N > NH_3$.
However,looking at the provided options,option $A$ represents the correct increasing order: $NH_3 < (CH_3)_3N < CH_3NH_2 < (CH_3)_2NH$.

(D) The reduction of nitroethane $(CH_3CH_2NO_2)$ to ethylamine $(CH_3CH_2NH_2)$ using $Sn$ and $HCl$ involves the replacement of two oxygen atoms with two hydrogen atoms and the addition of two more hydrogen atoms to the nitrogen atom.
The balanced chemical equation is:
$CH_3CH_2NO_2 + 6[H] \xrightarrow{Sn/HCl} CH_3CH_2NH_2 + 2H_2O$.
From the stoichiometry of the reaction,$1 \ mole$ of nitroethane requires $6 \ moles$ of nascent hydrogen atoms for complete reduction to ethylamine.

(B) primary amine has the general formula $R-NH_2$.
$A$ secondary amine has the general formula $R_2NH$.
$A$ tertiary amine has the general formula $R_3N$.
$1$. Phenylmethanamine $(C_6H_5CH_2NH_2)$ is a primary amine.
$2$. $N$-Methylethanamine $(CH_3CH_2NHCH_3)$ is a secondary amine.
$3$. $N,N$-Dimethylethanamine $(CH_3CH_2N(CH_3)_2)$ is a tertiary amine.
$4$. Prop-$2$-en-$1$-amine $(CH_2=CHCH_2NH_2)$ is a primary amine.
Therefore,the correct option is $B$.

(D) The given reaction sequence is the Stephen reduction followed by hydrolysis.
$1$. Ethanenitrile $(CH_3CN)$ reacts with $SnCl_2$ and $HCl$ to form an imine intermediate $(CH_3CH=NH)$,which is compound '$A$'.
$2$. The imine intermediate '$A$' upon acid hydrolysis $(H_3O^{+})$ yields Ethanal $(CH_3CHO)$ as the final product '$B$' along with ammonium chloride $(NH_4Cl)$.
$3$. The reaction is: $CH_3CN$ $\xrightarrow{SnCl_2, HCl} CH_3CH=NH$ $\xrightarrow{H_3O^{+}} CH_3CHO + NH_4Cl$.
Therefore,the product '$B$' is Ethanal.

(B) In an aqueous solution,the basic strength of amines is determined by a combination of inductive effect,solvation effect (hydrogen bonding),and steric hindrance.
For methyl-substituted amines,the order of basicity is $(CH_3)_2NH > CH_3NH_2 > (CH_3)_3N$.
$1.$ $(CH_3)_2NH$ (secondary amine) is the most basic due to the combined effect of the electron-donating $+I$ effect of two methyl groups and sufficient solvation of the conjugate acid.
$2.$ $CH_3NH_2$ (primary amine) is less basic than the secondary amine but more basic than the tertiary amine in aqueous media.
$3.$ $(CH_3)_3N$ (tertiary amine) is the least basic among these because the steric hindrance of the three methyl groups prevents effective solvation of the protonated amine by water molecules,which outweighs the inductive effect.

(A) The $pK_{b}$ value is inversely proportional to the basic strength of the amine. $A$ lower $pK_{b}$ value indicates a stronger base.
$1$. $C_6H_5NH_2$ (Aniline) is a weak base due to the resonance of the lone pair on nitrogen with the benzene ring.
$2$. $(CH_3)_3N$ is a tertiary amine,which is less basic than secondary amines due to steric hindrance.
$3$. $C_6H_5CH_2NH_2$ (Benzylamine) is more basic than aniline because the lone pair is not in conjugation with the benzene ring.
$4$. $C_2H_5NH_2$ (Ethylamine) is a primary aliphatic amine. In the gas phase,basicity increases with the number of alkyl groups,but in aqueous solution,$C_2H_5NH_2$ is a strong base due to the inductive effect and solvation.
Comparing these,$C_2H_5NH_2$ is the strongest base among the given options,hence it has the lowest $pK_{b}$ value.

(B) The reaction between benzene diazonium chloride and phenol is an electrophilic substitution reaction known as coupling reaction.
This reaction occurs in a mild alkaline medium (pH $9-10$).
In this medium,phenol is converted into phenoxide ion,which is more reactive towards the electrophilic attack by the diazonium cation,leading to the formation of $p$-hydroxyazobenzene.

(D) The reaction of primary amines with chloroform $(CHCl_3)$ and ethanolic potassium hydroxide $(KOH)$ is known as the Carbylamine reaction.
This reaction produces isocyanides (carbylamines),which have a characteristic foul smell.
Secondary and tertiary amines do not undergo the Carbylamine reaction.
Among the given options,$CH_3CH_2NH_2$ is a primary amine,while $(CH_3)_3N$ is a tertiary amine and $(CH_3)_2NH$ and $(CH_3CH_2)_2NH$ are secondary amines.
Therefore,$CH_3CH_2NH_2$ will produce a foul smell.

(B) primary amine is an amine where the nitrogen atom is attached to only one carbon atom $(R-NH_2)$.
$A$: $N$-Methylmethanamine is $(CH_3)_2NH$,which is a secondary amine.
$B$: $4$-Bromobenzenamine is $Br-C_6H_4-NH_2$,where the nitrogen is attached to one aromatic ring carbon,making it a primary amine.
$C$: $N$-Phenylbenzenamine is $(C_6H_5)_2NH$,which is a secondary amine.
$D$: $N$-Ethyl-$N$-methylpropan-$2$-amine is a tertiary amine because the nitrogen is bonded to three different alkyl groups.
Therefore,the correct option is $B$.

(C) In the aqueous phase,the basic strength of amines depends on three factors: inductive effect,solvation effect,and steric hindrance.
For alkyl amines (where $R = CH_3$ or $C_2H_5$),the combined effect of these factors results in the order:
$R_2NH > RNH_2 > R_3N > NH_3$ (for $R = CH_3$) or $R_2NH > R_3N > RNH_2 > NH_3$ (for $R = C_2H_5$).
However,considering the general trend often taught for alkyl amines in aqueous solution,the secondary amine is the most basic.
Given the options provided,the correct order reflecting the higher basicity of secondary amines is $NH_3 < RNH_2 < R_3N < R_2NH$.

(A) The stability of substituted ammonium ions depends on the number of alkyl groups attached to the nitrogen atom.
Alkyl groups are electron-donating ($+I$ effect),which increases the electron density on the nitrogen atom and stabilizes the positive charge.
As the number of alkyl groups increases,the positive charge on the nitrogen is better dispersed,leading to greater stability.
$NH_4^{+}$ has no alkyl groups,$R-NH_3^{+}$ has one,$R_2NH_2^{+}$ has two,and $R_3NH^{+}$ has three.
Therefore,$NH_4^{+}$ is the least stable species among the given options because it lacks the stabilizing effect of electron-donating alkyl groups.

(D) tertiary amine is an amine where the nitrogen atom is bonded to three carbon atoms $(R_3N)$.
$1$. Cyclohexanamine is a primary amine $(R-NH_2)$.
$2$. Ethane-$1,2$-diamine is a primary diamine $(H_2N-CH_2-CH_2-NH_2)$.
$3$. $N$-Phenylbenzeneamine (Diphenylamine) is a secondary amine $(Ph-NH-Ph)$.
$4$. $N$-Ethyl-$N$-methylpropan-$2$-amine has the nitrogen atom bonded to an ethyl group,a methyl group,and a propan-$2$-yl group. Since the nitrogen is bonded to three carbon atoms,it is a tertiary amine.

(B) Hofmann's exhaustive alkylation is a process where a primary amine $(RNH_2)$ reacts with an excess of an alkyl halide $(RX)$ to form a quaternary ammonium salt $(R_4N^{+} X^{-})$.
Option $B$ correctly represents this reaction: $RNH_2 + 3RX \rightarrow R_4N^{+} X^{-} + 3HX$.

(A) The reduction of nitroalkanes with $Sn/HCl$ (or $Fe/HCl$) is a standard method for the preparation of primary amines.
The reaction for nitroethane is:
$CH_3CH_2NO_2 + 6[H] \xrightarrow{Sn/HCl} CH_3CH_2NH_2 + 2H_2O$.
Thus,the product obtained is ethanamine.

(C) The Hofmann bromamide degradation reaction is represented as: $R-CONH_2 + Br_2 + 4KOH \rightarrow R-NH_2 + K_2CO_3 + 2KBr + 2H_2O$.
In this reaction,the amide group $(-CONH_2)$ is converted into an amine group $(-NH_2)$.
The carbon atom of the carbonyl group is lost as potassium carbonate $(K_2CO_3)$.
The loss in molar mass corresponds to the mass of the carbonyl group $(CO)$ that is removed from the amide to form the amine.
The molar mass of $CO$ is $12 + 16 = 28 \ g \ mol^{-1}$.
Therefore,the loss in molar mass is $28 \ g \ mol^{-1}$.

(D) In the cyclic structure of glucopyranose,the carbon atom that is part of the hemiacetal group is known as the anomeric carbon. This carbon is the one that was the carbonyl carbon (aldehyde group) in the open-chain form of glucose. In glucopyranose,this is the $C-1$ carbon,which is bonded to both an oxygen atom in the ring and a hydroxyl group $(-OH)$.

(C) ribonucleoside consists of a ribose sugar molecule attached to a nitrogenous base.
In a ribose sugar $(C_5H_{10}O_5)$,there are $-OH$ groups at the $C_2$,$C_3$,and $C_5$ positions.
When the ribose sugar forms a nucleoside,the $-OH$ group at the $C_1$ position is replaced by the nitrogenous base.
Therefore,the remaining $-OH$ groups are at the $C_2$,$C_3$,and $C_5$ positions.
Thus,the total number of $-OH$ groups in a ribonucleoside is $3$.

(B) The sugar molecule present in an $RNA$ nucleotide is $\beta-D-ribose$.
This is a pentose sugar,which means it contains $5$ carbon atoms in its structure.
Therefore,the total number of carbon atoms in the sugar molecule of an $RNA$ nucleotide is $5$.

(B) Raffinose is a trisaccharide with the formula $C_{18}H_{32}O_{16}$.
On complete hydrolysis,it yields one molecule each of glucose,galactose,and fructose.
The reaction is: $C_{18}H_{32}O_{16} + 2H_2O$ $\rightarrow C_6H_{12}O_6 (\text{glucose}) + C_6H_{12}O_6 (\text{galactose}) + C_6H_{12}O_6 (\text{fructose})$.

(B) The reaction of glucose with $Br_2$ water (bromine water) is a mild oxidation reaction.
$Br_2$ water acts as a mild oxidizing agent that specifically oxidizes the aldehydic group $(-CHO)$ of glucose to a carboxylic acid group $(-COOH)$,resulting in the formation of gluconic acid.
This specific oxidation confirms the presence of an aldehydic group in the glucose molecule.

(C) The constituents of the given carbohydrates are as follows:
$1$. Sucrose: Glucose + Fructose
$2$. Maltose: Glucose + Glucose
$3$. Lactose: Glucose + Galactose
$4$. Raffinose: Glucose + Fructose + Galactose
$5$. Stachyose: Glucose + Fructose + $2$ Galactose
Therefore,both $Raffinose$ and $Stachyose$ contain $Galactose$ as one of their constituent units.

(D) Sucrose $(C_{12}H_{22}O_{11})$ on hydrolysis produces $1$ mole of glucose and $1$ mole of fructose.
$C_{12}H_{22}O_{11} + H_2O \rightarrow \text{Glucose} + \text{Fructose}$.
Maltose $(C_{12}H_{22}O_{11})$ produces $2$ moles of glucose on hydrolysis.
$C_{12}H_{22}O_{11} + H_2O \rightarrow 2 \text{Glucose}$.
Therefore,maltose produces double the quantity of glucose compared to sucrose.

(A) $1$. Hydrolysis of $1 \ mole$ of $Sucrose$ $(C_{12}H_{22}O_{11})$ yields $1 \ mole$ of $Glucose$ and $1 \ mole$ of $Fructose$.
$2$. Hydrolysis of $1 \ mole$ of $Lactose$ $(C_{12}H_{22}O_{11})$ yields $1 \ mole$ of $Glucose$ and $1 \ mole$ of $Galactose$.
$3$. Hydrolysis of $1 \ mole$ of $Maltose$ $(C_{12}H_{22}O_{11})$ yields $2 \ moles$ of $Glucose$.
$4$. Comparing the glucose yield: $Sucrose$ gives $1 \ mole$,$Lactose$ gives $1 \ mole$,and $Maltose$ gives $2 \ moles$.
$5$. Therefore,$Sucrose$ and $Lactose$ both produce $1 \ mole$ of glucose per mole of carbohydrate.

(A) Hydrolysis of sucrose $(C_{12}H_{22}O_{11})$ in the presence of an acid or the enzyme invertase yields an equimolar mixture of glucose and fructose.
This mixture is known as invert sugar because the sign of optical rotation changes from dextrorotatory $(+)$ to levorotatory $(-)$ during the process.
Therefore,sucrose is the correct answer.

(A) 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 $\alpha-D$-glucose and the $C-2$ of $\beta-D$-fructose.
This is a non-reducing sugar because both the anomeric carbons are involved in the linkage.

(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.

(A) Amylose is a linear polymer of $\alpha-D$-glucose units.
These glucose units are linked together by $\alpha-1,4$-glycosidic linkages.
Amylopectin,on the other hand,contains both $\alpha-1,4$ and $\alpha-1,6$-glycosidic linkages due to its branched structure.

(C) Cellulose is a polysaccharide consisting of a linear chain of several hundred to many thousands of $\beta(1 \rightarrow 4)$ linked $D$-glucose units.
Humans lack the necessary enzymes (cellulases) to hydrolyze the $\beta$-glycosidic linkages in cellulose.
Therefore,it cannot be digested and acts as dietary fiber (roughage),which adds bulk to the food and facilitates bowel movement in the human digestive system.

(D) The general structure of an $\alpha$-amino acid is $R-CH(NH_2)-COOH$.
For serine,the side chain $(R)$ is a hydroxymethyl group,which is represented as $HO-CH_2-$.
Therefore,the correct option is $D$.

(C) Essential amino acids are those that cannot be synthesized by the human body and must be obtained through the diet.
Among the given options,$Histidine$ is an essential amino acid.
$Tyrosine$,$Serine$,and $Glycine$ are non-essential amino acids as they can be synthesized by the body.

(A) Amino acids are classified as acidic,basic,or neutral based on the number of amino $(-NH_2)$ and carboxyl $(-COOH)$ groups present in their structure.
$1$. $Glu$ $(Glutamic \ acid)$ contains two $-COOH$ groups and one $-NH_2$ group,making it an acidic amino acid.
$2$. $Arg$ $(Arginine)$ and $Lys$ $(Lysine)$ contain more amino groups than carboxyl groups,making them basic amino acids.
$3$. $Trp$ $(Tryptophan)$ contains one $-NH_2$ group and one $-COOH$ group in its structure,making it a neutral amino acid.
Therefore,the correct option is $A$.

(B) Amino acids that cannot be synthesized by the human body and must be obtained through the diet are called essential amino acids.
Among the given options,$Tryptophan$ is an essential amino acid.
Tyrosine,Glutamine,and Glycine are non-essential amino acids,meaning they can be synthesized by the human body.