MHT CET 2024 Chemistry Question Paper with Answer and Solution

(B) Methane $(CH_4)$ involves $sp^3$ hybridization.
Acetylene ($C_2H_2$ or $HC \equiv CH$) involves $sp$ hybridization where each carbon atom is bonded to one hydrogen and one carbon via a triple bond.
Ethylene $(C_2H_4)$ involves $sp^2$ hybridization.
Ammonia $(NH_3)$ involves $sp^3$ hybridization.

(B) According to Molecular Orbital Theory,the bond order is calculated as $\frac{N_b - N_a}{2}$,where $N_b$ is the number of bonding electrons and $N_a$ is the number of antibonding electrons.
For $CO$ molecule,the total number of electrons is $6 + 8 = 14$.
The molecular orbital configuration is $\sigma 1s^2, \sigma^* 1s^2, \sigma 2s^2, \sigma^* 2s^2, \pi 2p_x^2 = \pi 2p_y^2, \sigma 2p_z^2$.
Here,$N_b = 10$ and $N_a = 4$.
Bond order = $\frac{10 - 4}{2} = \frac{6}{2} = 3$.

(A) The electronic configuration of $Li_2$ molecule is $(\sigma 1s)^2 (\sigma^* 1s)^2 (\sigma 2s)^2$.
Bond order of $Li_2$ molecule is calculated as:
$\text{Bond Order} = \frac{N_b - N_a}{2} = \frac{4 - 2}{2} = 1$.
Since all electrons are paired in the $Li_2$ molecule,it is diamagnetic.

(A) The energy of an orbital is determined by the $(n+l)$ rule,where $n$ is the principal quantum number and $l$ is the azimuthal quantum number. The orbital with the lowest $(n+l)$ value has the lowest energy. If $(n+l)$ values are equal,the orbital with the lower $n$ value has lower energy.
Calculating $(n+l)$ for each:
$2p: n=2, l=1 \implies n+l = 3$
$3s: n=3, l=0 \implies n+l = 3$
$3d: n=3, l=2 \implies n+l = 5$
$4p: n=4, l=1 \implies n+l = 5$
Comparing $2p$ and $3s$,both have $(n+l) = 3$. Since $2p$ has a lower $n$ value $(n=2)$ compared to $3s$ $(n=3)$,$2p$ has the lowest energy.

(C) The electronic configuration of the $O_2$ molecule 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 $\frac{N_b - N_a}{2}$,where $N_b$ is the number of bonding electrons and $N_a$ is the number of antibonding electrons.
For $O_2$,$N_b = 10$ and $N_a = 6$.
Bond order $= \frac{10 - 6}{2} = \frac{4}{2} = 2$.

(C) The electronic configuration of $O_2$ molecule is $(\sigma 1s)^2 (\sigma^* 1s)^2 (\sigma 2s)^2 (\sigma^* 2s)^2 (\sigma 2p_z)^2 (\pi 2p_x)^2 (\pi 2p_y)^2 (\pi^* 2p_x)^1 (\pi^* 2p_y)^1$.
Bonding molecular orbitals are those without an asterisk $(*)$.
These are $(\sigma 1s)^2$,$(\sigma 2s)^2$,$(\sigma 2p_z)^2$,$(\pi 2p_x)^2$,and $(\pi 2p_y)^2$.
Total number of electrons in bonding orbitals = $2 + 2 + 2 + 2 + 2 = 10$.

(B) The electronic configuration of $N_2$ molecule ($14$ electrons) is $(\sigma 1s)^2 (\sigma^* 1s)^2 (\sigma 2s)^2 (\sigma^* 2s)^2 (\pi 2p_x)^2 (\pi 2p_y)^2 (\sigma 2p_z)^2$.
Antibonding orbitals are those denoted with an asterisk $(*)$.
These are $(\sigma^* 1s)^2$ and $(\sigma^* 2s)^2$.
Total number of electrons in antibonding orbitals $= 2 + 2 = 4$.

(A) The formula for the radius of the $n^{th}$ orbit of a hydrogen-like species is given by $r_n = \frac{52.9 \times n^2}{Z} \ pm$.
For the $B^{4+}$ ion, the atomic number $Z = 5$ and the orbit number $n = 4$.
Substituting these values into the formula:
$r_4 = \frac{52.9 \times (4)^2}{5} \ pm$
$r_4 = \frac{52.9 \times 16}{5} \ pm$
$r_4 = \frac{846.4}{5} \ pm$
$r_4 = 169.28 \ pm \approx 169.3 \ pm$.

(D) The dipole moment of $CH_3X$ molecules depends on the electronegativity difference between the carbon atom and the halogen atom $(X)$.
As we move down the group from $F$ to $I$,the electronegativity of the halogen decreases,which leads to a decrease in the $C-X$ bond polarity.
Although the bond length increases down the group,the decrease in electronegativity is the dominant factor.
Therefore,the dipole moment decreases in the order: $CH_3F > CH_3Cl > CH_3Br > CH_3I$.
Thus,$CH_3I$ has the lowest dipole moment.

(C) The dipole moment $(\mu)$ of a molecule depends on its geometry and the polarity of its bonds.
$HF$ is a polar molecule with a linear geometry,so $\mu \neq 0$.
$NH_3$ has a trigonal pyramidal geometry with a lone pair on the nitrogen atom,resulting in a net dipole moment $(\mu = 1.48 \ D)$.
$BF_3$ has a trigonal planar geometry. The three $B-F$ bonds are arranged at an angle of $120^{\circ}$ to each other. The vector sum of the dipole moments of the three $B-F$ bonds is zero,making the molecule non-polar $(\mu = 0)$.
$CHCl_3$ is a polar molecule with a tetrahedral geometry where the bond dipoles do not cancel out $(\mu = 1.04 \ D)$.
Therefore,$BF_3$ has a zero dipole moment.

(A) Hydrogen bonding occurs when a hydrogen atom is covalently bonded to a highly electronegative atom (like $N$,$O$,or $F$).
In $NH_3$,the hydrogen atoms are bonded to a highly electronegative nitrogen atom,allowing it to form intermolecular hydrogen bonds with other $NH_3$ molecules.
$C_2H_6$ (ethane) is non-polar and lacks such bonds.
$H_2S$ does not form significant hydrogen bonds because sulfur is not electronegative enough.
$CH_3-O-CH_3$ (dimethyl ether) has an oxygen atom but lacks a hydrogen atom directly bonded to it,so it cannot form hydrogen bonds with itself.

(A) The element located in period-$4$ and group-$12$ is Zinc $(Zn)$.
$Zn$ has an atomic number of $30$.
The electronic configuration of $Zn$ in the ground state is $[Ar] \ 3d^{10} \ 4s^2$.
In this configuration,all $3d$ and $4s$ orbitals are completely filled,meaning there are $0$ unpaired electrons.
In the excited state (e.g.,$Zn^{2+}$ ion),the configuration is $[Ar] \ 3d^{10}$,which also contains $0$ unpaired electrons.
Therefore,the total number of unpaired electrons is $0$.

(A) $1$. Rubidium $(Rb)$: Rubidium is in Group $1$ (alkali metal) and Period $5$. This is a correct match.
$2$. Strontium $(Sr)$: Strontium is in Group $2$ (alkaline earth metals),not Group $1$.
$3$. Caesium $(Cs)$: Caesium is in Group $1$,but it is in Period $6$,not Period $5$.
$4$. Barium $(Ba)$: Barium is in Group $2$ (alkaline earth metals),not Group $1$.

(A) The electronegativity of Group $1$ elements decreases down the group as the atomic size increases and the effective nuclear charge decreases.
Therefore,the decreasing order of electronegativity is $Li > Na > K > Cs$.
Thus,$Li$ has the highest electronegativity among the given elements.

(C) The number of electrons in each species is as follows:
$Ne$ (Atomic number $10$): $10$ electrons.
$O^{2-}$ (Atomic number $8$): $8 + 2 = 10$ electrons.
$Na^{+}$ (Atomic number $11$): $11 - 1 = 10$ electrons.
$Na$ (Atomic number $11$): $11$ electrons.
Since $Ne$,$O^{2-}$,and $Na^{+}$ are isoelectronic (all have $10$ electrons),$Na$ is the species that does not have the same number of electrons.

(C) Cesium $(Cs)$ is the element used in photoelectric cells.
Explanation: Cesium is the preferred metal for photoelectric cells because it can easily convert sunlight into electricity.
It has a very low ionization energy and is the most electropositive element among the alkali metals.
The amount of energy required to eject electrons from a cesium surface is relatively small,only $206.5 \ kJ/mol$.

(B) Electronegativity generally increases across a period from left to right and decreases down a group.
Comparing the given elements: $O$ and $F$ are in the $2^{nd}$ period,while $S$ and $Cl$ are in the $3^{rd}$ period.
Since electronegativity decreases down a group,the elements in the $3^{rd}$ period ($S$ and $Cl$) have lower electronegativity than those in the $2^{nd}$ period ($O$ and $F$).
Between $S$ and $Cl$,$S$ is to the left of $Cl$ in the $3^{rd}$ period,therefore $S$ has the lowest electronegativity among the given options.

(A) The ionization enthalpy $(I.E.)$ decreases down the group due to an increase in atomic size.
Therefore,the $I.E.$ of $Te$ is higher than that of $Po$.
Across a period,$I.E.$ increases with an increase in atomic number.
Comparing the given elements,$Po$ (Polonium) is in Group $16$ and Period $6$,$Te$ (Tellurium) is in Group $16$ and Period $5$,$Br$ (Bromine) is in Group $17$ and Period $4$,and $Kr$ (Krypton) is in Group $18$ and Period $4$.
Since $I.E.$ increases from left to right across a period and decreases down a group,$Po$ has the lowest $I.E.$ among the given elements.

(A) The ionization enthalpy $(IE)$ is highest for noble gases due to their stable electronic configuration. Therefore,the $IE$ of $Ar$ and $He$ is higher than that of the halogens $Cl$ and $I$.
Down a group,$IE$ decreases due to an increase in atomic size.
Comparing $He$ and $Ar$,$He$ has a smaller atomic size,so $IE$ of $He > Ar$.
Across a period,$IE$ increases with an increase in atomic number.
Comparing $Cl$ and $Ar$,$Ar$ has a higher $IE$ than $Cl$.
Thus,the overall order of ionization enthalpy is $I < Cl < Ar < He$.
Therefore,$He$ has the highest ionization enthalpy.

(C) Elements $C$ and $D$ belong to period $4$,while elements $A$ and $B$ belong to period $3$.
Atomic radius increases down a group,so elements in period $4$ are larger than those in period $3$.
Comparing $C$ and $D$ in period $4$,$D$ has a higher effective nuclear charge than $C$ because it is to the right of $C$ in the same period.
Therefore,$D$ is smaller than $C$.
Thus,element $C$ has the largest atomic radius.

(B) Atomic size increases down a group due to the addition of new electron shells and decreases across a period from left to right due to an increase in effective nuclear charge.
In the periodic table,$S$ and $Cl$ belong to period $3$,while $Se$ and $Br$ belong to period $4$.
Since $Se$ and $Br$ are in a higher period,they have larger atomic sizes than $S$ and $Cl$.
Within period $4$,$Se$ is to the left of $Br$,so $Se$ has a larger atomic size than $Br$.
Therefore,the atomic size order is $Cl < S < Br < Se$,making $Se$ the element with the largest atomic size.

(D) $SO_3$ is an acidic oxide.
$NO$ is a neutral oxide.
$Al_2O_3$ is an amphoteric oxide.
$CaO$ is a basic oxide.
Generally,oxides of metals in Group $1$ and Group $2$ are basic,while oxides of non-metals are acidic.

(B) For $PF_6^{-}$: Let the oxidation number of $P$ be $x$.
$x + (6 \times -1) = -1$
$x - 6 = -1$
$x = +5$
For $V_2 O_7^{4-}$: Let the oxidation number of $V$ be $y$.
$2y + (7 \times -2) = -4$
$2y - 14 = -4$
$2y = +10$
$y = +5$
Thus,the oxidation numbers are $+5$ and $+5$ respectively.

(C) The atomic number of $Ti$ is $22$. The electronic configuration of $Ti$ is $[Ar] 3d^2 4s^2$.
In the $+2$ oxidation state,$Ti^{2+}$ loses two electrons from the $4s$ orbital.
Therefore,the electronic configuration of $Ti^{2+}$ is $[Ar] 3d^2$.
Thus,there are $2$ electrons present in the $3d$ orbital.

(D) Let the oxidation number of $Mn$ be $x$.
In $MnO_4^{-}$,the sum of oxidation numbers of all atoms equals the charge on the ion.
$x + (4 \times -2) = -1$
$x - 8 = -1$
$x = +7$
Therefore,the oxidation number of $Mn$ is $+7$.

(C) In $CrO_4^{2-}$,let the oxidation state of $Cr$ be $x$.
$x + 4(-2) = -2$ $\Rightarrow x - 8 = -2$ $\Rightarrow x = +6$.
In $K_2Cr_2O_7$,let the oxidation state of $Cr$ be $x$.
$2(+1) + 2x + 7(-2) = 0$ $\Rightarrow 2 + 2x - 14 = 0$ $\Rightarrow 2x = 12$ $\Rightarrow x = +6$.

(A) The charge on $1$ electron is $e = 1.602 \times 10^{-19} \ C$.
To find the number of electrons $(n)$ required to generate a total charge of $Q = 1 \ C$,we use the formula $Q = n \times e$.
Therefore,$n = \frac{Q}{e} = \frac{1 \ C}{1.602 \times 10^{-19} \ C} \approx 6.242 \times 10^{18}$ electrons.

(B) The degree of dissociation $(\alpha)$ is calculated using the formula: $\alpha = \frac{\Lambda_m^c}{\Lambda_m^0}$
Given: $\Lambda_m^c = 16.5 \ \Omega^{-1} \ cm^2 \ mol^{-1}$ and $\Lambda_m^0 = 390.7 \ \Omega^{-1} \ cm^2 \ mol^{-1}$.
Substituting the values: $\alpha = \frac{16.5}{390.7} = 0.04223$
Therefore,the degree of dissociation is approximately $0.0422$.

(C) The degree of dissociation $(\alpha)$ is given by the ratio of molar conductivity at a specific concentration $(\Lambda_c)$ to the molar conductivity at infinite dilution $(\Lambda_0)$:
$\alpha = \frac{\Lambda_c}{\Lambda_0}$
Given: $\Lambda_c = 3.3 \ \Omega^{-1} \ cm^2 \ mol^{-1}$ and $\Lambda_0 = 132 \ \Omega^{-1} \ cm^2 \ mol^{-1}$.
$\alpha = \frac{3.3}{132} = 0.025$
Percentage dissociation $= \alpha \times 100 = 0.025 \times 100 = 2.5 \%$.

(D) The molar conductivity of weak electrolytes increases as the concentration of the solution decreases.
This is because the degree of dissociation of the weak electrolyte increases with dilution.
Therefore,among the given concentrations,the lowest concentration,$0.001 \ M$,will exhibit the maximum molar conductivity.

(A) Among the given options,$Water$ $(H_2O)$ is the only solvent that is non-toxic,non-flammable,and does not contribute to environmental pollution. It is widely considered a green solvent in chemical processes.

(A) According to the principles of green chemistry,unnecessary derivatization (i.e.,the use of blocking or protecting groups) should be avoided or minimized whenever possible.
These steps require additional reagents and generate additional waste,which contradicts the goal of atom economy and waste reduction.
Therefore,the statement suggesting that protection and deprotection techniques are 'good' is not true in the context of green chemistry.

(B) Atom economy is defined as the ratio of the molecular weight of the desired product to the sum of the molecular weights of all reactants,expressed as a percentage.
Reaction $B$: $C_2H_4 + H_2 \xrightarrow{Ni} C_2H_6$ is an addition reaction where all atoms of the reactants are incorporated into the product.
Atom economy = $\frac{\text{Mol. wt of } C_2H_6}{\text{Mol. wt of } (C_2H_4 + H_2)} \times 100 = \frac{30}{30} \times 100 = 100 \%$.
Since the atom economy is $100 \%$,this reaction exhibits the best atom economy among the given options.

(C) $\text{Percentage atom economy} = \frac{\text{Formula weight of the desired product}}{\text{Sum of formula weight of all the reactants}} \times 100$
$\text{Percentage atom economy} = \frac{70 \ u}{140 \ u} \times 100 = 50 \%$

(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.

(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 $trimethyl$ $carbinol$.

(D) According to $IUPAC$ nomenclature rules,the priority order for functional groups determines the principal functional group. \\ The priority sequence is $-COOH > -CN > -OH > -C \equiv C-$. \\ Therefore,among the given options,$-COOH$ is the principal functional group.

(D) The structure of $(CH_3)_4C$ is a central carbon atom bonded to four methyl groups.
To determine the $IUPAC$ name,we first identify the longest carbon chain,which contains $3$ carbon atoms (propane).
There are two methyl groups attached to the second carbon atom.
Therefore,the $IUPAC$ name is $2,2-$dimethylpropane.

(B) To determine the $IUPAC$ name,follow these steps:
$1$. Identify the longest carbon chain containing the double bond. The longest chain has $6$ carbons,so it is a hexene derivative.
$2$. Number the chain from the end that gives the double bond the lowest possible locant. Numbering from left to right gives the double bond the position $3$.
$3$. Identify substituents: There is a bromo group at position $3$ and a methyl group at position $4$.
$4$. Combine these to get the name: $3-$Bromo$-4-$methylhex$-3-$ene.

(C) $1$. Identify the parent chain: The parent ring is a cyclobutane ring.
$2$. Numbering the ring: Numbering should start such that the substituents get the lowest possible locants. The methoxy group $(-OCH_3)$ and the two methyl groups $(-CH_3)$ are attached to the ring.
$3$. Assigning locants: If we assign the carbon with the two methyl groups as position $1$,the methoxy group is at position $2$. This gives the locants $1, 1, 2$.
$4$. Alphabetical order: 'Methoxy' comes before 'methyl'.
$5$. Combining the parts: The name is $2-$Methoxy$-1,1-$dimethylcyclobutane.

(D) tricarboxylic acid is an organic compound that contains three carboxyl $(-COOH)$ groups.
$1$. Propionic acid $(CH_3CH_2COOH)$ is a monocarboxylic acid.
$2$. Oxalic acid $(HOOC-COOH)$ is a dicarboxylic acid.
$3$. Malonic acid $(HOOC-CH_2-COOH)$ is a dicarboxylic acid.
$4$. Citric acid $(HOOC-CH_2-C(OH)(COOH)-CH_2-COOH)$ contains three carboxyl groups,making it a tricarboxylic acid.

(D) The structural formula for propylene glycerol is $CH_2(OH)-CH(OH)-CH_2(OH)$.
This compound contains three carbon atoms in the main chain and three hydroxyl $(-OH)$ groups attached to each carbon atom.
According to $IUPAC$ nomenclature rules,the parent alkane is propane,and the presence of three $-OH$ groups is indicated by the suffix $-triol$.
The positions of the hydroxyl groups are at carbons $1$,$2$,and $3$.
Therefore,the $IUPAC$ name is Propane-$1,2,3$-triol.

(A) The principal functional group is decided based on the following $IUPAC$ priority order:
$-COOH > -SO_3H > -COOR > -COCl > -CONH_2 > -CN > -CHO > >C=O > -OH > -NH_2 > -C=C- > -C \equiv C-$.
Comparing the given options:
$1$. $-CONH_2$
$2$. $-CN$
$3$. $-OH$
$4$. $-C \equiv C-$
According to the priority order,$-CONH_2$ appears first among the given choices,making it the principal functional group.

(B) The structure of Acrylic acid is $CH_2=CH-COOH$.
It contains a three-carbon chain with a carboxylic acid functional group and a double bond.
The carbon atom of the carboxylic acid group is assigned position $1$.
The double bond starts at position $2$.
Therefore,the $IUPAC$ name is Prop$-2-$enoic acid.

(C) $1$. Identify the longest carbon chain containing the double bond. The chain has $5$ carbon atoms,so the parent alkane is pentane,and the alkene is pent$-2-$ene.
$2$. Number the chain from the end that gives the double bond the lowest possible locant. Numbering from right to left gives the double bond position at $C-2$.
$3$. Identify the substituents: a chloro group at $C-4$ and a methyl group at $C-3$.
$4$. Combine these to get the $IUPAC$ name: $4-$chloro$-3-$methylpent$-2-$ene.

(B) . The given structure is a benzene ring attached to a $-SO_3^-Na^+$ group.
This compound is the sodium salt of benzene sulphonic acid,known as sodium benzene sulphonate.

(D) $1$. Identify the parent carbon chain: The compound is a cyclic structure with $4$ carbon atoms,which is a cyclobutane ring.
$2$. Numbering the ring: Start numbering from the carbon atom with the substituents to give them the lowest possible locants. The carbon with two methyl groups is assigned position $1$,and the carbon with the methoxy group is assigned position $3$.
$3$. Naming the substituents: There are two methyl groups at position $1$ and one methoxy group at position $3$.
$4$. Combining the parts: The name is $3-$methoxy$-1, 1-$dimethylcyclobutane.

(C) The $IUPAC$ name propan$-1-$ol indicates a chain of $3$ carbon atoms with an alcohol $(-OH)$ functional group attached to the first carbon atom.
In a bond line formula,each vertex and endpoint represents a carbon atom.
For propan$-1-$ol $(CH_3-CH_2-CH_2-OH)$,we need a $3$-carbon chain where the $-OH$ group is at the end.
Option $C$ correctly represents this structure,where the terminal carbon is bonded to the $-OH$ group.

(A) According to the $IUPAC$ nomenclature rules for polyfunctional compounds,the priority order for selecting the principal functional group 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 with this standard order,the correct priority sequence is $-COOH > -SO_3H > -COOR$.

(C) $1$. Identify the principal functional group. The $-COOH$ group has higher priority than the $-OH$ group,so the parent chain is benzoic acid.
$2$. Number the benzene ring starting from the carbon attached to the $-COOH$ group as position $1$.
$3$. The $-OH$ group is at position $3$ and the $-C_2H_5$ (ethyl) group is at position $5$.
$4$. Alphabetically,'ethyl' comes before 'hydroxy'.
$5$. Therefore,the $IUPAC$ name is $3-$ethyl$-5-$hydroxybenzoic acid.

(B) The silver mirror test is a characteristic reaction of aldehydes.
Ammoniacal silver nitrate is known as Tollen's reagent.
$CH_3CHO + 2[Ag(NH_3)_2]^+ + 3OH^- \rightarrow CH_3COO^- + 2Ag(s) + 4NH_3 + 2H_2O$.
Ethanal $(CH_3CHO)$ is an aldehyde and thus gives a positive silver mirror test.

(D) An aromatic amine is one where the nitrogen atom is directly attached to an aromatic ring.
$A$ mixed amine (or unsymmetrical amine) has different alkyl or aryl groups attached to the nitrogen atom.
$A$ $3^{\circ}$ (tertiary) amine has three groups (alkyl or aryl) attached to the nitrogen atom.
Let us analyze the options:
$A$: $C_6H_5NHC_2H_5$ is an aromatic,mixed,$2^{\circ}$ amine.
$B$: $(C_2H_5)_3N$ is an aliphatic,simple,$3^{\circ}$ amine.
$C$: $(CH_3)_3N$ is an aliphatic,simple,$3^{\circ}$ amine.
$D$: $C_6H_5N(CH_3)_2$ ($N$,$N$-dimethylaniline) has a phenyl group and two methyl groups attached to the nitrogen. It is aromatic,mixed,and a $3^{\circ}$ amine.
Therefore,the correct option is $D$.

(C) As a consequence of the combined effect of the inductive effect,steric effect,and solvation,the secondary amines are the strongest bases among aliphatic amines in the aqueous phase.
Therefore,the basic strength varies as: $2^{\circ} \text{ amine} > 3^{\circ} \text{ amine} > 1^{\circ} \text{ amine} > \text{ammonia}$.
Hence,among the given options,diethylamine $(2^{\circ})$ has the highest basic strength.

(C) Intermolecular hydrogen bonding requires a hydrogen atom directly bonded to a highly electronegative atom like $N$,$O$,or $F$.
In $Trimethylamine$ $(N(CH_3)_3)$,the nitrogen atom is bonded to three methyl groups and has no hydrogen atom attached to it.
Therefore,it cannot form intermolecular hydrogen bonds.
In contrast,$Cyclohexylamine$ $(C_6H_{11}NH_2)$,$Allylamine$ $(CH_2=CH-CH_2NH_2)$,and $Diphenylamine$ $((C_6H_5)_2NH)$ all contain at least one $N-H$ bond,allowing them to participate in intermolecular hydrogen bonding.

(B) In Hofmann elimination,the quaternary ammonium hydroxide undergoes thermal decomposition to form the least substituted alkene as the major product.
Step $1$: The quaternary ammonium iodide is converted to the corresponding hydroxide using moist $Ag_2O$.
$(CH_3CH_2CH_2)N^{+}(CH_2CH_3)_3 I^{-} + AgOH \rightarrow (CH_3CH_2CH_2)N^{+}(CH_2CH_3)_3 OH^{-} + AgI$
Step $2$: Upon heating,the $OH^{-}$ ion abstracts a $\beta$-hydrogen from the alkyl groups attached to the nitrogen atom.
The available $\beta$-hydrogens are on the propyl group (at the $\beta$-carbon) and the ethyl groups (at the $\beta$-carbon).
Removal of $\beta$-hydrogen from the ethyl group yields ethene $(CH_2=CH_2)$.
Removal of $\beta$-hydrogen from the propyl group yields propene $(CH_3CH=CH_2)$.
According to the Hofmann rule,the less substituted alkene is the major product. Since ethene is less substituted than propene,ethene is the major product.

(A) The reaction is an example of the Hofmann elimination reaction.
$1$ mole of $N, N, N$-triethylpropylammonium iodide reacts with moist $Ag_2O$ to form the corresponding quaternary ammonium hydroxide.
Upon heating,the quaternary ammonium hydroxide undergoes elimination to yield ethene $(CH_2=CH_2)$ and $N, N$-diethylpropylamine.
The stoichiometry of the reaction is $1:1$ with respect to the quaternary ammonium salt and ethene.
Therefore,$n$ moles of $N, N, N$-triethylpropylammonium iodide will yield $n$ moles of ethene.

(C) secondary amine has the general formula $R-NH-R'$. For the molecular formula $C_4H_{11}N$,the possible secondary amine isomers are:
$1$. $CH_3CH_2-NH-CH_2CH_3$ (Diethylamine)
$2$. $CH_3-NH-CH_2CH_2CH_3$ (Methylpropylamine)
$3$. $CH_3-NH-CH(CH_3)_2$ (Methylisopropylamine)
Thus,there are $3$ such isomers.

(C) The reaction sequence is a Stephen reduction followed by hydrolysis.
$1$. Isopropyl cyanide $((CH_3)_2CH-CN)$ reacts with $SnCl_2$ and $HCl$ to form an imine hydrochloride intermediate $(A)$,which is $(CH_3)_2CH-CH=NH \cdot HCl$.
$2$. Subsequent acid hydrolysis $(H_3O^{+})$ of the imine hydrochloride converts the imine group into an aldehyde group,yielding $2-$Methylpropanal as product $B$ along with $NH_4Cl$.
$3$. The reaction is: $(CH_3)_2CH-CN$ $\xrightarrow[HCl]{SnCl_2} (CH_3)_2CH-CH=NH \cdot HCl$ $\xrightarrow{H_3O^{+}} (CH_3)_2CH-CHO + NH_4Cl$.

(C) The correct formula for Phenylhydrazine is $C_6H_5-NH-NH_2$.
Option $C$ is incorrect because $C_6H_5-NH-NH-C_6H_5$ represents $1,2$-diphenylhydrazine.

(D) Aliphatic or aromatic primary amines on heating with chloroform $(CHCl_3)$ and alcoholic potassium hydroxide $(KOH)$ give foul-smelling products called alkyl or aryl isocyanides (carbylamines).
The chemical reaction is:
$R-NH_2 + CHCl_3 + 3KOH \xrightarrow[\text{alc.}]{\Delta} R-NC + 3KCl + 3H_2O$
Here,$R-NC$ is the alkyl isocyanide.

(B) Acylation reaction occurs with primary and secondary amines because they possess at least one hydrogen atom attached to the nitrogen atom,which can be replaced by an acyl group.
$N$-Methylaniline $(C_6H_5NHCH_3)$ is a secondary amine and thus undergoes acylation.
Ethyldimethylamine,$N,N$-Dimethylmethanamine,and $N,N$-Dimethylaniline are all tertiary amines,which lack a hydrogen atom on the nitrogen and therefore do not undergo acylation.

(C) Aliphatic and aromatic primary and secondary amines undergo acylation reactions because they contain replaceable hydrogen atoms attached to the nitrogen atom.
Triethylamine,$(CH_3CH_2)_3N$,is a tertiary amine and does not possess any replaceable hydrogen atom on the nitrogen.
Therefore,it does not react with acetyl chloride.

(A) The basic strength of amines depends on the electron-donating inductive effect of the alkyl group $(R)$.
As the number of alkyl groups increases,the electron density on the nitrogen atom increases,making the amine more basic.
Therefore,the order of basic strength is $NH_3 < RNH_2 < R_2NH$.
Since $pK_b = -\log(K_b)$,a higher basic strength corresponds to a lower $pK_b$ value.
Thus,the decreasing order of $pK_b$ values is $NH_3 > RNH_2 > R_2NH$.

(A) The $pK_{b}$ value is inversely proportional to the basic strength of the amine.
Arylamines are much weaker bases than aliphatic amines due to the delocalization of the lone pair of electrons on the nitrogen atom into the benzene ring.
Among the given options,arylamine is the weakest base,and therefore,it possesses the highest $pK_{b}$ value.

(A) The reaction of benzonitrile $(C_6H_5CN)$ with phenylmagnesium bromide $(C_6H_5MgBr)$ in the presence of dry ether forms an intermediate imine salt $(A)$: $C_6H_5-C(C_6H_5)=NMgBr$.
Upon acid hydrolysis $(H_3O^{+})$,this intermediate is converted into a ketone,specifically benzophenone $(C_6H_5-CO-C_6H_5)$,along with ammonia $(NH_3)$ and magnesium hydroxybromide $(Mg(Br)OH)$.
Therefore,product $B$ is benzophenone.

(C) Gabriel phthalimide synthesis involves the following steps:
$1$. Phthalimide reacts with $alc. KOH$ to form the potassium salt of phthalimide.
$2$. The potassium salt of phthalimide reacts with an alkyl halide $(R-X)$ to form $N$-alkyl phthalimide.
$3$. $N$-alkyl phthalimide undergoes alkaline hydrolysis (using $NaOH_{(aq)}$) to yield a primary amine $(R-NH_2)$ and the sodium salt of phthalic acid (sodium phthalate).
Phthalic acid itself is not formed as a product; instead,its sodium salt is obtained. Therefore,phthalic acid is not obtained at any stage.

(C) The reagent $AlH(i-Bu)_2$ is Diisobutylaluminium hydride ($DIBAL$-$H$).
It is a selective reducing agent that reduces nitriles $(-CN)$ to imines,which upon acidic hydrolysis $(H_3O^+)$ yield aldehydes.
The reaction is: $CH_3-CH=CH-CH_2-CN \xrightarrow[2. H_3O^+]{1. AlH(i-Bu)_2} CH_3-CH=CH-CH_2-CHO$.
Thus,the product formed is $Pent-3-enal$.

(B) primary $(1^{\circ})$ amine is an amine where the nitrogen atom is attached to only one alkyl or aryl group,represented by the general formula $R-NH_2$.
$1$. $N$-methylmethanamine $(CH_3-NH-CH_3)$ is a secondary $(2^{\circ})$ amine.
$2$. Phenylmethanamine $(C_6H_5-CH_2-NH_2)$ has the $-NH_2$ group attached to a carbon atom,making it a primary $(1^{\circ})$ amine.
$3$. $N$-phenylbenzenamine $(C_6H_5-NH-C_6H_5)$ is a secondary $(2^{\circ})$ amine.
$4$. $N$-methylbenzenamine $(C_6H_5-NH-CH_3)$ is a secondary $(2^{\circ})$ amine.
Therefore,Phenylmethanamine is the correct primary amine.

(B) In arylamines,the $-NH_2$ group is attached directly to an aromatic ring. The lone pair of electrons on nitrogen is conjugated to the aromatic ring and is less available for protonation.
In phenylmethanamine (an aliphatic amine),the lone pair of electrons on nitrogen is not involved in resonance with the aromatic ring and is thus more available for protonation.
Therefore,phenylmethanamine is the strongest base among the given amines.

(B) The reaction shown is the conversion of aniline to benzene diazonium chloride.
This process is known as diazotisation.
Aniline reacts with nitrous acid $(HNO_2)$,which is generated in situ by the reaction of $NaNO_2$ and $HCl$ at a low temperature $(273-278 \ K)$,to form benzene diazonium chloride.
Therefore,the correct reagent '$A$' is $NaNO_2+HCl$ at low temperature.

(B) Step $1$: $CH_3Br$ reacts with $KCN$ to form methyl cyanide $(CH_3CN)$ as product $A$.
$CH_3Br + KCN \rightarrow CH_3CN + KBr$
Step $2$: Methyl cyanide $(CH_3CN)$ undergoes reduction with $Na / C_2H_5OH$ (Mendius reduction) to form ethyl amine $(CH_3CH_2NH_2)$ as product $B$.
$CH_3CN + 4[H] \xrightarrow{Na / C_2H_5OH} CH_3CH_2NH_2$
Therefore,the product $B$ is ethyl amine.

(C) The compound $(C_6H_5)_3N$ is known as triphenylamine.
It is classified as an aromatic amine because the nitrogen atom is directly attached to three phenyl groups $(C_6H_5)$,which are aromatic rings.
Since the nitrogen atom is bonded to three identical aryl groups and no hydrogen atoms,it is a tertiary $(3^{\circ})$ simple aromatic amine.

(C) The Hofmann elimination reaction is a process where a quaternary ammonium hydroxide $(R_4N^{+}OH^{-})$ undergoes elimination upon heating to form an alkene and a tertiary amine.
This reaction typically involves the treatment of a quaternary ammonium halide with moist $Ag_2O$ (which generates $AgOH$ to provide the $OH^{-}$ ion),followed by thermal decomposition.

(C) Hinsberg's reagent is benzenesulfonyl chloride.
Its chemical formula is $C_6H_5SO_2Cl$.

(B) The reaction of methylamine $(CH_3NH_2)$ with iodomethane $(CH_3I)$ proceeds via exhaustive alkylation until the quaternary ammonium salt is formed:
$CH_3NH_2 + CH_3I \rightarrow (CH_3)_2NH + HI$
$(CH_3)_2NH + CH_3I \rightarrow (CH_3)_3N + HI$
$(CH_3)_3N + CH_3I \rightarrow (CH_3)_4N^{+}I^{-}$
Summing these steps,$3$ moles of $CH_3I$ are consumed to convert $1$ mole of $CH_3NH_2$ into tetramethylammonium iodide $((CH_3)_4N^{+}I^{-})$.

(A) The reaction shown is the alkaline hydrolysis of $N$-alkyl phthalimide,which is a step in the Gabriel phthalimide synthesis.
The reaction is: $N$-alkyl phthalimide + $2NaOH_{(aq)} \rightarrow \text{Sodium phthalate} + R-NH_2$ (Primary amine).
Therefore,the product '$A$' is a primary amine.

(B) The reaction of benzonitrile $(C_6H_5CN)$ with phenylmagnesium bromide $(C_6H_5MgBr)$ in dry ether proceeds via a nucleophilic addition to the nitrile group.
First,the phenyl group from the Grignard reagent attacks the electrophilic carbon of the nitrile,forming an imine complex intermediate: $C_6H_5CN + C_6H_5MgBr \rightarrow C_6H_5-C(C_6H_5)=N-MgBr$.
Subsequently,acid-catalyzed hydrolysis $(H_3O^+)$ of the imine complex converts the $C=N-MgBr$ group into a carbonyl group $(C=O)$,yielding benzophenone $(C_6H_5-CO-C_6H_5)$ as the final product along with ammonia and magnesium salts.

(D) The correct answer is $D$.
Mendius reduction involves the reduction of nitriles $(R-CN)$ to primary amines $(R-CH_2NH_2)$ using reducing agents like sodium in ethanol or methanol ($Na / C_2H_5OH$ or $Na / CH_3OH$) or lithium aluminum hydride $(LiAlH_4)$.
Explanation of other options:
- $A$: This is Hofmann bromamide degradation,where an amide $(R-CONH_2)$ reacts with bromine and alkali to form a primary amine.
- $B$: This is the $N$-alkylation of amines,where an amine reacts with excess alkyl halide to form a quaternary ammonium salt.
- $C$: This is Hofmann elimination,where a quaternary ammonium hydroxide undergoes thermal decomposition to form an alkene and a tertiary amine.

(D) Mendius reduction is a chemical reaction that involves the reduction of alkyl cyanides (nitriles) to primary amines using sodium in ethanol $(Na/C_2H_5OH)$.
The reaction is represented as: $R-CN + 4[H] \xrightarrow{Na/C_2H_5OH} R-CH_2NH_2$.
This reaction does not involve the loss of a carbon atom as $CO_2$.
Therefore,the statement that one carbon atom is lost as $CO_2$ is incorrect.

(D) Stachyose is a tetrasaccharide with the molecular formula $C_{24}H_{42}O_{21}$.
Upon complete hydrolysis,one mole of stachyose yields two moles of galactose,one mole of glucose,and one mole of fructose.
Therefore,$n$ moles of stachyose will yield $2n$ moles of galactose,$n$ moles of glucose,and $n$ moles of fructose.

(A) Fructose $(C_6H_{12}O_6)$ is a laevorotatory ketohexose.
It exhibits laevorotation with a specific rotation of $[\alpha]_D^{20} = -92.4^{\circ}$.
Due to this property,fructose is commonly referred to as laevulose.

(A) Saccharic acid is formed by the oxidation of both the terminal $-CHO$ and $-CH_2OH$ groups of glucose to $-COOH$ groups.
The structure of saccharic acid is $HOOC-(CHOH)_4-COOH$.
It contains two carboxyl $(-COOH)$ groups at the ends and four hydroxyl $(-OH)$ groups attached to the carbon chain.
Therefore,the correct statement is that it contains two carboxyl and four hydroxyl groups.

(A) Explanation:
- Glucose is an aldohexose,meaning it is a $6$-carbon sugar with an aldehyde group.
- It is a reducing sugar,not nonreducing,because it can reduce Fehling's solution and Tollens' reagent.
- It contains $4$ chiral carbon atoms,not $3$.
- The ring structure of glucose is a hemiacetal,not a hemiketal,because it forms from the reaction of an aldehyde group with a hydroxyl group.

(D) In animals,excess glucose is stored as glycogen,which is a polysaccharide that serves as an energy reserve.
Glycogen has a highly branched structure consisting of glucose units.
Linear chains of glucose units are connected by $\alpha-1,4$-glycosidic linkages.
Branch points in the glycogen molecule are formed by $\alpha-1,6$-glycosidic linkages.

(D) Amylopectin is a branched-chain polymer of $\alpha-D-glucose$.
The linear chain of glucose units is formed by $\alpha-(1 \rightarrow 4)$ glycosidic linkages.
The branching points are formed by $\alpha-(1 \rightarrow 6)$ glycosidic linkages.
Therefore,the correct linkages for the chain and branches are $\alpha-(1 \rightarrow 4)$ and $\alpha-(1 \rightarrow 6)$ respectively.

(A) The false statement is that cellulose is a constituent of cell walls in animal cells.
Cellulose is a polysaccharide that makes up the cell walls of plant cells,but it is not found in animal cells.

(C) When glucose is heated with $HI$ for a long time,it undergoes reduction to form $n$-hexane.
The reaction is:
$CHO(CHOH)_4 CH_2 OH \xrightarrow{\Delta, HI} CH_3-(CH_2)_4-CH_3$
This reaction indicates that all six carbon atoms in glucose are linked in a straight chain.

(B) The formation of cellulose xanthate occurs in the viscose process for manufacturing rayon.
First,cellulose $(Cell-OH)$ reacts with $NaOH$ to form alkali cellulose $(Cell-ONa)$.
Then,alkali cellulose reacts with carbon disulfide $(CS_2)$ to form cellulose xanthate,which has the formula $Cell-O-C(=S)-SNa$.

(B) Maltose is a disaccharide composed of two $\alpha-D-glucose$ units.
The glycosidic linkage is formed between the $C-1$ of one $\alpha-D-glucose$ molecule and the $C-4$ of another $\alpha-D-glucose$ molecule.
Therefore,it is an $\alpha-1,4$ glycosidic linkage.

(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.
Since the reducing groups (aldehyde of glucose and ketone of fructose) are involved in this bond,sucrose is a non-reducing sugar.

(D) Essential amino acids are those that cannot be synthesized by the human body and must be obtained through the diet.
There are nine essential amino acids: $Histidine$,$Isoleucine$,$Leucine$,$Lysine$,$Methionine$,$Phenylalanine$,$Threonine$,$Tryptophan$,and $Valine$.
Among the given options,$Histidine$ is an essential amino acid.

(A) Aspartic acid is an acidic amino acid,which is represented by the one-letter symbol '$D$'.

(D) Amino acids are classified as acidic,basic,or neutral based on the relative number of amino and carboxyl groups in the molecule.
Basic amino acids contain more amino groups than carboxyl groups.
Among the given options,$Histidine$ contains an imidazole ring in its side chain,which has a nitrogen atom capable of accepting a proton. Therefore,it is classified as a basic amino acid.

(B) $Asp$ stands for Aspartic acid,which is an acidic amino acid.
$Arg$ (Arginine) is basic.
$Asn$ (Asparagine) is neutral.
$Ala$ (Alanine) is neutral.

(C) In a polypeptide chain,when $n$ amino acids are linked together,they form $(n-1)$ peptide (amide) bonds.
This is because each peptide bond is formed by the condensation reaction between the carboxyl group of one amino acid and the amino group of the adjacent amino acid.
Therefore,for $n$ amino acids,there are $(n-1)$ such linkages.

(B) Glycine is the simplest amino acid with the formula $NH_2-CH_2-COOH$.
In an aqueous solution,the acidic carboxyl group $(-COOH)$ loses a proton to the basic amino group $(-NH_2)$.
This internal proton transfer results in the formation of a dipolar ion known as a zwitter ion.
The structure of the glycine zwitter ion is $H_3N^{+}-CH_2-COO^{-}$.

(A) Amino acids are classified based on the relative number of amino and carboxyl groups in their molecule.
$Glu$ (Glutamic acid) contains two carboxyl groups and one amino group,making it an acidic amino acid.
$Gly$ (Glycine) is neutral.
$Gln$ (Glutamine) is neutral.
$Arg$ (Arginine) is basic due to the presence of an extra amino group in its side chain.

(A) Myosin is a fibrous protein,not a globular protein. It is primarily involved in muscle contraction and has a long,elongated structure.
The other options are globular proteins:
- $(B)$ Insulin: $A$ hormone involved in regulating blood sugar levels. It is a small globular protein.
- $(C)$ Egg albumin: Also known as albumin,it is a globular protein found in egg whites.
- $(D)$ Legumelin: $A$ globular protein found in legumes,such as peas and beans.
Globular proteins generally have a spherical shape and are water-soluble,whereas fibrous proteins,like myosin,have elongated shapes and are typically insoluble in water.

(B) Fibrous proteins consist of polypeptide chains held together by hydrogen and disulfide bonds to form fiber-like structures. They are generally insoluble in water.
$Myosin$ is a fibrous protein found in muscle tissue.
In contrast,$Legumelin$,$Insulin$,and $Serum \ albumin$ are globular proteins,which are generally soluble in water and have a spherical shape.

(B) Pepsin is a digestive enzyme that catalyzes the hydrolysis of proteins into $\alpha$-amino acids in the stomach.