AIIMS 2007 Chemistry Question Paper with Answer and Solution

(C) The reaction sequence is as follows:
$1$. $Toluene$ undergoes oxidation to form benzoic acid $(A)$: $C_6H_5CH_3 \xrightarrow{[O]} C_6H_5COOH$.
$2$. Benzoic acid reacts with $SOCl_2$ to form benzoyl chloride $(B)$: $C_6H_5COOH \xrightarrow{SOCl_2} C_6H_5COCl$.
$3$. Benzoyl chloride reacts with $NaN_3$ to form benzoyl azide $(C)$: $C_6H_5COCl \xrightarrow{NaN_3} C_6H_5CON_3$.
$4$. Upon heating,benzoyl azide undergoes Curtius rearrangement to form phenyl isocyanate $(D)$: $C_6H_5CON_3 \xrightarrow{\text{Heat}} C_6H_5NCO + N_2$.
Therefore,$D$ is phenyl isocyanate.

(A) Bond length is inversely proportional to bond order.
Bond order for $O_2$ is $2.0$.
Bond order for $O_3$ is $1.5$ (due to resonance).
Bond order for $O_2^{2-}$ (peroxide ion) is $1.0$.
Since bond length $\propto \frac{1}{\text{bond order}}$,the order of bond length is $O_2 < O_3 < O_2^{2-}$.

(C) Generally,most real gases show the same type of isotherm. The segment $ab$ represents the gaseous state. The line $bc$,which is a horizontal line,shows the equilibrium between liquid and vapour. The pressure corresponding to the line $bc$ is known as the vapour pressure of the liquid. The line $cd$ represents the liquid state.

(A) The critical temperature $(T_C)$ of a gas is defined as the temperature above which a gas cannot be liquefied,regardless of the pressure applied.
It is given by the relation $T_C = \frac{8a}{27Rb}$,which implies $T_C \propto a$,where $a$ is the van der Waals constant representing the magnitude of intermolecular forces of attraction.
$A$ higher critical temperature indicates stronger intermolecular forces,making the gas easier to liquefy.
Comparing the given values: $A = 25\,^{\circ}C$,$B = 10\,^{\circ}C$,$C = -80\,^{\circ}C$,and $D = 15\,^{\circ}C$.
Since gas $A$ has the highest critical temperature $(25\,^{\circ}C)$,it will be liquefied most easily.

(A) The entropy change of the surroundings is given by the formula $\Delta S_{surr} = -\frac{\Delta H_{sys}}{T}$.
For an exothermic reaction,the enthalpy change of the system $\Delta H_{sys}$ is negative.
Therefore,$\Delta S_{surr} = -\frac{(\text{negative value})}{T}$,which results in a positive value.
Thus,$\Delta S_{surr}$ is always positive for an exothermic reaction.

(B) Given: $\Delta H = -92.2\, kJ$,$P = 40\, atm$,$\Delta V = -1\, L$.
Using the relation: $\Delta H = \Delta U + P\Delta V$
Therefore: $\Delta U = \Delta H - P\Delta V$
First,convert $P\Delta V$ from $atm\, L$ to $kJ$ using the conversion factor $1\, atm\, L = 101.325\, J = 0.101325\, kJ$.
$P\Delta V = 40\, atm \times (-1\, L) = -40\, atm\, L$
$P\Delta V = -40 \times 0.101325\, kJ = -4.053\, kJ$
Now,calculate $\Delta U$:
$\Delta U = -92.2\, kJ - (-4.053\, kJ)$
$\Delta U = -92.2\, kJ + 4.053\, kJ = -88.147\, kJ$
Rounding to the nearest integer,we get $\Delta U \approx -88\, kJ$.

(A) According to Hess's Law,the enthalpy change of a reaction is the same whether it occurs in one step or several steps.
Sublimation is the direct conversion of a solid to a vapor.
This process can be represented as a two-step process: Solid $\rightarrow$ Liquid (fusion) followed by Liquid $\rightarrow$ Vapor (vaporization).
Therefore,$\Delta H_{sublimation} = \Delta H_{fusion} + \Delta H_{vap}$.
Given $\Delta H_{fusion} = x$ and $\Delta H_{vap} = y$,we have $\Delta H_{sublimation} = x + y$.

(A) In the titration of a weak acid $(CH_3COOH)$ with a strong base $(NaOH)$,the salt formed $(CH_3COONa)$ undergoes anionic hydrolysis,resulting in a basic solution at the equivalence point $(pH > 7)$.
Due to the excess of free base added beyond the equivalence point,there is a steep rise in $pH$,which is represented by the vertical portion of the titration curve.
Therefore,the vertical line in the graph indicates the alkaline nature of the equivalence point.

(B) The reaction between $NaOH$ (strong base) and $CH_3COOH$ (weak acid) produces sodium acetate $(CH_3COONa)$,which is a salt of a weak acid and a strong base.
$CH_3COOH + NaOH \rightarrow CH_3COONa + H_2O$
Since the salt undergoes anionic hydrolysis,the resulting solution is basic.
For a salt of a weak acid and strong base,the $pH$ is given by $pH = \frac{1}{2} (pK_w + pK_a + \log C)$.
Given $pK_a$ for $CH_3COOH$ is approximately $4.76$ and $pK_w = 14$,the calculated $pH$ is greater than $7$.
Therefore,the $pH$ of the solution is $8$.

(D) An acidic buffer is a solution containing a mixture of a weak acid and its salt with a strong base.
$CH_3COOH$ is a weak acid,but $CH_3COONH_4$ is a salt of a weak acid $(CH_3COOH)$ and a weak base $(NH_4OH)$.
Therefore,the mixture of $CH_3COOH$ and $CH_3COONH_4$ does not form an acidic buffer.
Since the assertion is false and the definition provided in the reason is also incorrect (it should be a salt with a strong base),both the assertion and the reason are incorrect.

(C) The equilibrium constant ($K_c$ or $K_p$) has a fixed value for a given chemical reaction at a specific temperature.
However,the composition of the final equilibrium mixture (the concentrations of reactants and products) at a particular temperature is independent of the initial amounts of reactants added.
Therefore,the Assertion is true,but the Reason is false.

(B) Atomic radii increase as we move down the group due to the addition of an extra energy shell.
Although the nuclear charge also increases,the effect of the addition of a new energy shell predominates.
The increase in atomic radius is most significant when moving from $Na$ ($Z=11$,$3s^1$) to $K$ ($Z=19$,$4s^1$) because the transition from the $n=3$ shell to the $n=4$ shell involves a substantial increase in shielding and distance from the nucleus compared to subsequent transitions in the group.
Therefore,the largest difference in radii is observed for the pair $Na, K$.

(B) Silicones are organosilicon polymers containing $Si-O-Si$ linkages.
They are formed by the hydrolysis of alkyl or aryl substituted chlorosilanes (e.g.,$R_2SiCl_2$).
The general formula for linear silicones is $(R_2SiO)_n$,where the repeating unit is $-(R_2Si-O)-$

(A) Hydroboration-oxidation of propene follows anti-Markovnikov addition of water across the double bond.
$3CH_3-CH=CH_2 + BH_3 \rightarrow (CH_3CH_2CH_2)_3B$
$(CH_3CH_2CH_2)_3B + 3H_2O_2 + 3OH^- \rightarrow 3CH_3CH_2CH_2OH + BO_3^{3-}$

(A) Oxymercuration-demercuration is a reaction that adds $H$ and $OH$ across a double bond following Markovnikov's rule without rearrangement.
For allylbenzene $(C_6H_5CH_2CH=CH_2)$,the double bond is between the $C_2$ and $C_3$ carbons of the side chain.
According to Markovnikov's rule,the $OH$ group attaches to the more substituted carbon $(C_2)$ and the $H$ atom attaches to the less substituted carbon $(C_3)$.
The reaction proceeds as follows:
$1$. Mercuration: The alkene reacts with mercuric acetate $(CH_3COO)_2Hg$ in water to form an organomercurial intermediate.
$2$. Demercuration: The intermediate is reduced with $NaBH_4$ in $NaOH$ to replace the mercury group with a hydrogen atom.
The final product is $1$-phenylpropan-$2$-ol,which is $C_6H_5CH_2CH(OH)CH_3$.

(B) In the sulphonation of benzene,concentrated $H_2SO_4$ (or fuming sulphuric acid,$H_2SO_4 + SO_3$) is used.
The electrophile that attacks the benzene ring is sulphur trioxide,$SO_3$.
Even in concentrated $H_2SO_4$,the active electrophilic species is $SO_3$.

(D) The reaction of $trans-but-2-ene$ with $Br_2$ involves $anti$ addition.
$Anti$ addition of $Br_2$ to $trans-but-2-ene$ results in the formation of a $meso$ compound (specifically,$2,3-dibromobutane$).
Since the product is a $meso$ compound,it is optically inactive and not a racemic mixture.
Therefore,the Assertion is incorrect.
The Reason statement is also incorrect as $trans$ addition to a $trans$ alkene does not necessarily form two types of stereoisomers in the manner described.
Hence,both the Assertion and Reason are incorrect.

(A) Acetylene reacts with sodamide $(NaNH_2)$ to form sodium acetylide and ammonia.
$CH\equiv CH + NaNH_2 \rightarrow HC\equiv C^{-}Na^{+} + NH_3$
This reaction occurs because the $sp$ hybridized carbon atoms in acetylene have $50\%$ $s$-character.
Due to the high $s$-character,the electrons are held more tightly by the nucleus,making the carbon atom significantly electronegative.
This increased electronegativity makes the terminal hydrogen atom acidic,allowing it to be replaced by the sodium ion from the sodamide.
Therefore,both the Assertion and the Reason are correct,and the Reason is the correct explanation of the Assertion.

(B) All molecules exhibit short-range London dispersion forces,which are a type of van der Waals forces.
When inert gases are mixed with iodine vapours,these short-range London dispersion forces exist between them.

(C) The assertion is true because $F$ is the most electronegative element in the periodic table.
The reason is false because $F$ has a lower electron affinity than $Cl$.
This is due to the small size of the $F$ atom,which leads to strong inter-electronic repulsions among the electrons in its $2p$ subshell,making it less favorable to add an extra electron compared to $Cl$.

(C) Copper sulphate solution is not stored in a zinc vessel because $Zn$ is more reactive than $Cu$ and has a higher tendency to undergo oxidation,thereby displacing $Cu$ from its aqueous solution.
The chemical reaction is: $Zn_{(s)} + CuSO_{4(aq)} \to ZnSO_{4(aq)} + Cu_{(s)}$.
$Zn$ does not form a complex with $CuSO_4$. The reason provided is scientifically incorrect. Therefore,the Assertion is true,but the Reason is false.

$(A)$ The $N_2$ molecule contains a $N \equiv N$ triple bond, while the $O_2$ molecule contains an $O = O$ double bond.
The bond dissociation energy of the $N \equiv N$ triple bond is significantly higher than that of the $O = O$ double bond, making $N_2$ chemically inert or less reactive under normal conditions.
The bond length of $N_2$ $(109.8 \ pm)$ is indeed shorter than that of $O_2$ $(121 \ pm)$ due to the higher bond order.
Since the high bond dissociation energy (a consequence of the short, strong triple bond) is the reason for the low reactivity of $N_2$, the Reason correctly explains the Assertion.
Therefore, both $A$ and $R$ are true and $R$ is the correct explanation of $A$.

(C) In $NaCl$ crystal,$Na^{+}$ and $Cl^{-}$ ions are strongly held by electrostatic forces of attraction.
When the salt dissolves,these ions leave the crystal lattice and enter the solution,thereby acquiring greater freedom (increased entropy).
Thermodynamically,for a process to be spontaneous,the change in Gibbs free energy,$\Delta G = \Delta H - T\Delta S$,must be negative.
For the dissolution of $NaCl$,the enthalpy change $\Delta H$ is positive (endothermic),but the entropy change $\Delta S$ is positive,making the term $T\Delta S$ large and positive.
Since $T\Delta S > \Delta H$,the value of $\Delta G$ becomes negative,making the process spontaneous.
The reason statement claims $\Delta H$ is highly positive and $T\Delta S$ is a low negative value,which is incorrect as $T\Delta S$ is actually positive for this process.
Thus,the Assertion is correct,but the Reason is incorrect.

(B) The composition of Portland cement typically includes the following major constituents:
$1$. Tricalcium silicate $(Ca_3SiO_5)$: $50-70\%$
$2$. Dicalcium silicate $(Ca_2SiO_4)$: $20-30\%$
$3$. Tricalcium aluminate $(Ca_3Al_2O_6)$: $5-10\%$
$4$. Tetracalcium aluminoferrite $(Ca_4Al_2Fe_2O_{10})$: $5-15\%$
Comparing the given options,$Ca_3SiO_5$ (tricalcium silicate) is present in the highest amount.

(A) Chain transfer agents are substances that react with the growing polymer chain to terminate its growth,thereby initiating a new chain.
This process limits the average molecular mass of the polymer.
Among the given options,$CCl_4$ acts as a chain transfer agent in radical polymerization because the $C-Cl$ bond is relatively weak and can easily undergo homolytic cleavage to transfer a chlorine atom to the growing radical chain.

(B) In a hexagonal close-packed $(hcp)$ structure,each atom is in contact with $6$ atoms in its own layer,$3$ atoms in the layer above,and $3$ atoms in the layer below.
Therefore,the total number of nearest touching neighbours is $6 + 3 + 3 = 12$.
Thus,the coordination number is $12$.

(C) Given: $p_{benzene}^o = 0.850 \ bar$,$p_{solution} = 0.845 \ bar$,$W_{benzene} = 39.0 \ g$,$M_{benzene} = 78 \ g/mol$,$w_{solid} = 0.5 \ g$.
Using Raoult's law for non-volatile solutes: $\frac{p^o - p}{p^o} = \frac{n_{solid}}{n_{solid} + n_{benzene}} \approx \frac{n_{solid}}{n_{benzene}}$ (for dilute solutions).
$n_{benzene} = \frac{39.0 \ g}{78 \ g/mol} = 0.5 \ mol$.
$\frac{0.850 - 0.845}{0.850} = \frac{0.5 / M_{solid}}{0.5}$.
$\frac{0.005}{0.850} = \frac{0.5}{M_{solid} \times 0.5} = \frac{1}{M_{solid}}$.
$M_{solid} = \frac{0.850}{0.005} = 170 \ g/mol$.

(D) The elevation in boiling point is given by the formula $\Delta T_b = i \times K_b \times m$.
Since $1 \ mol$ of each compound is dissolved in $1 \ L$ of solution,the molality $(m)$ is the same for all.
Therefore,$\Delta T_b$ is directly proportional to the van't Hoff factor $(i)$.
The van't Hoff factor $(i)$ depends on the degree of dissociation $(\alpha)$.
Stronger acids dissociate more completely in water. The acid strength order for hydrogen halides is $HI > HBr > HCl > HF$.
This is because the bond dissociation energy follows the order $HI < HBr < HCl < HF$.
Since $HI$ has the lowest bond dissociation energy,it dissociates most readily,leading to the highest number of ions in the solution.
Thus,$i$ is maximum for $HI$,resulting in the largest $\Delta T_b$ value.

(B) catalyst provides an alternative reaction pathway with a lower activation energy $(E_a)$.
By lowering the activation energy,it increases the number of molecules that possess sufficient energy to cross the energy barrier,thereby increasing the rate of both forward and backward reactions equally.
It does not change the equilibrium constant or the overall mechanism of the reaction.

(D) In the process of smelting,metal oxides are reduced by carbon or carbon monoxide $(CO)$.
For $ZnO$: $ZnO + C \to Zn + CO$.
For $FeO$: $FeO + C \to Fe + CO$.
Metals like $Zn, Fe, Pb,$ and $Sn$ are typically extracted from their oxide ores using carbon reduction (smelting). Both $FeO$ and $ZnO$ are standard examples of oxides reduced by carbon in industrial metallurgy.

(C) Complete hydrolysis of $XeF_6$ yields $XeO_3$.
$XeF_6 + 3H_2O \to XeO_3 + 6HF$

(A) little more than $47\%$ of the earth's crust consists of oxygen. The most common rock constituents of the earth's crust are nearly all oxides.

(A) Among the given options,$CuSO_{4(aq)}$ and $KCN$ react to form an unstable copper $(II)$ cyanide,which rapidly decomposes to give copper $(I)$ cyanide and cyanogen gas.
The chemical equations are:
$2CuSO_{4(aq)} + 4KCN \to 2Cu(CN)_2 + 2K_2SO_4$
$2Cu(CN)_2 \to 2CuCN + (CN)_2 \uparrow$ (Cyanogen)

(C) The assertion is true because ozone $(O_3)$ is indeed an allotrope of oxygen.
The reason is false. While liquid oxygen is pale blue,it is paramagnetic in its triplet state due to the presence of two unpaired electrons in its antibonding molecular orbitals. In its singlet state,all electrons are paired,making it diamagnetic,not paramagnetic.

(D) $Co-60$ is used as an anticancerous agent among the given radioactive isotopes.
It emits $\beta$-particles and high-energy gamma rays,which is why it is widely used in radiation therapy for cancer treatment.

(A) $SnI_4$ is an orange-colored solid.
The colour arises due to the charge transfer phenomenon.
In $SnI_4$,an electron is transferred from the iodide ion $(I^-)$ to the tin center $(Sn^{4+})$ upon absorption of light,which is known as ligand-to-metal charge transfer $(LMCT)$.
This absorption of light in the blue region of the visible spectrum results in the compound appearing orange.

(A) In lanthanide ions,the $4f$ electrons are well-shielded from the external field by the overlying $5s$ and $5p$ electrons. Thus,the magnetic effect of the orbital motion of the electron is not quenched. Consequently,the magnetic moment must be calculated by taking into account both the spin and orbital contributions.
The magnetic moment is given by $\mu = g\sqrt{J(J+1)}$,where $g = 1 + \frac{S(S+1) - L(L+1) + J(J+1)}{2J(J+1)}$.
For $Dy^{3+}$ $([Xe]4f^9)$,the $4f$ subshell is more than half-filled. In such cases,the spin and orbital angular momenta couple to give a high total angular momentum $J$,resulting in the highest magnetic moment among lanthanoids.
Therefore,both the Assertion and Reason are correct,and the Reason is the correct explanation of the Assertion.

(A) In metal carbonyls,a $\pi$-bond arises from the overlap of filled $d$-orbitals on the metal with empty $\pi^*$-antibonding orbitals of the $CO$ ligand.
This back-bonding increases the electron density in the antibonding orbital of $CO$,which weakens the $C-O$ bond compared to free $CO$.
As a result,the $C-O$ bond order decreases and the bond length increases.
Therefore,both the Assertion and Reason are correct,and the Reason is the correct explanation of the Assertion.

(D) Ferrocene,$Fe(\eta^5-C_5H_5)_2$,is a classic example of an organometallic compound.
It possesses a sandwich-type structure where the $Fe$ atom is located between two parallel cyclopentadienyl rings.
Due to the delocalization of $\pi$-electrons in the cyclopentadienyl rings,all $Fe-C$ bond lengths are equal.
It is widely recognized as one of the earliest discovered sandwich compounds,significantly influencing organometallic chemistry.
Therefore,all the given statements are true.

(D) Nickel dimethylglyoxime,represented as $[Ni(DMG)_2]$,is a square planar coordination compound.
It is a neutral,non-ionic complex that is insoluble in water and is used for the gravimetric estimation of $Ni^{2+}$ ions.

(C) $Cu^{2+}$ ($3d^9$ configuration) forms complexes with the same magnetic moment and geometry irrespective of the nature of the ligand.
The electronic configuration of $Cu^{2+}$ is $[Ar] 3d^9$. The $3d$ subshell contains one unpaired electron.
Since the $3d$ subshell is almost completely filled,the incoming ligands cannot cause any pairing or redistribution of electrons in the $3d$ orbitals. Therefore,the number of unpaired electrons remains $1$ (magnetic moment $\mu = \sqrt{1(1+2)} = \sqrt{3} \approx 1.73 \ BM$) regardless of whether the ligand is strong or weak.
Furthermore,the geometry of $Cu^{2+}$ complexes is typically determined by the number of ligands (e.g.,square planar for $4$ ligands due to Jahn-Teller distortion),which remains consistent for a given coordination number.

(B) The color of a complex is complementary to the color of the light it absorbs.
Since red light is absorbed,the complex will appear blue-green.
Among the given options,$[Cu(NH_3)_4]^{2+}$ is a well-known deep blue-colored complex,which corresponds to the absorption of red light.

(A) The complex $[Cu(H_2O)_6]^{2+}$ is blue in aqueous solution.
When $HCl$ is added,the $Cl^-$ ligand replaces one $H_2O$ molecule to form $[CuCl(H_2O)_5]^+$.
This substitution leads to a change in the crystal field splitting energy,resulting in a color change from blue to green/yellowish-green depending on the concentration of $HCl$.

(C) The reaction sequence is as follows:
$1$. $Toluene$ undergoes oxidation $([O])$ to form $Benzoic \ acid$ $(A)$.
$2$. $Benzoic \ acid$ reacts with $SOCl_2$ to form $Benzoyl \ chloride$ $(B)$.
$3$. $Benzoyl \ chloride$ reacts with $NaN_3$ to form $Benzoyl \ azide$ $(C)$.
$4$. $Benzoyl \ azide$ undergoes $Curtius \ rearrangement$ upon heating to form $Phenyl \ isocyanate$ $(D)$ with the loss of $N_2$ gas.
Therefore,$D$ is $Phenyl \ isocyanate$.

(A) The reaction of chloral $(CCl_3CHO)$ with chlorobenzene in the presence of concentrated $H_2SO_4$ produces $DDT$ ($p,p'$-dichlorodiphenyltrichloroethane).
This reaction involves the condensation of two molecules of chlorobenzene with one molecule of chloral,resulting in the elimination of a water molecule.
The mechanism is an electrophilic aromatic substitution where the protonated chloral acts as an electrophile attacking the chlorobenzene ring.
Since both the assertion (that chloral reacts with chlorobenzene to form $DDT$) and the reason (that it is an electrophilic substitution reaction) are correct,and the reason correctly explains the mechanism of the reaction,option $A$ is the correct answer.

(C) The reaction of alkyl chlorides or alkyl bromides with $NaI$ in acetone to form alkyl iodides is known as the $Finkelstein$ reaction.
The chemical equation is: $R-X + NaI \xrightarrow{\text{acetone}} R-I + NaX$ (where $X = Cl, Br$).
In this reaction,$NaI$ is soluble in acetone,whereas $NaCl$ and $NaBr$ are insoluble in acetone.
Because $NaCl$ or $NaBr$ precipitate out of the solution,the equilibrium shifts in the forward direction according to $Le$ $Chatelier's$ principle,facilitating the formation of alkyl iodide.
Therefore,the Assertion is correct,but the Reason is incorrect because it states the solubility properties in reverse.

(A) The reaction of ethers with $HI$ involves the cleavage of the $C-O$ bond.
For the first ether,$CH_3-O-C_2H_5$,the reaction follows the $S_N2$ mechanism,where the iodide ion attacks the less sterically hindered alkyl group,forming $CH_3I + C_2H_5OH$.
For the second ether,$(CH_3)_3C-O-CH_3$,the reaction follows the $S_N1$ mechanism because the tert-butyl carbocation is highly stable,leading to the formation of $(CH_3)_3CI + CH_3OH$.

(C) The reaction shown is the Fries rearrangement.
When phenyl esters are treated with a Lewis acid like anhydrous $AlCl_3$,they undergo rearrangement to yield a mixture of $o-$hydroxyketone and $p-$hydroxyketone.
The reaction is as follows:
$Phenyl \ acetate \xrightarrow{AlCl_3, \Delta} o-hydroxyacetophenone + p-hydroxyacetophenone$.

(B) When benzoic acid $(C_6H_5COOH)$ is treated with lithium aluminium hydride $(LiAlH_4)$,it acts as a strong reducing agent.
It reduces the carboxylic acid group $(-COOH)$ to a primary alcohol group $(-CH_2OH)$.
The reaction is:
$C_6H_5COOH + 4[H] \xrightarrow{LiAlH_4} C_6H_5CH_2OH + H_2O$
Thus,the product obtained is benzyl alcohol.

(A) In acetamide $(CH_{3}CONH_{2})$,the nitrogen atom has a lone pair that participates in resonance with the $C=O$ group. However,the $-NH_{2}$ group is a stronger electron-donating group by resonance compared to the $-OC_{2}H_{5}$ group in ethyl acetoacetate $(CH_{3}COCH_{2}COOC_{2}H_{5})$.
Because nitrogen is less electronegative than oxygen,the lone pair on nitrogen is more readily donated into the carbonyl system,increasing the electron density on the oxygen atom of the $C=O$ bond,which makes the $C=O$ bond more polar.
Thus,both the Assertion and the Reason are correct,and the Reason is the correct explanation for the Assertion.

(C) Among the given statements,only $(c)$ is true. Methylamine reacts with nitrous acid to liberate $N_2$ gas:
$CH_3-NH_2 + HNO_2 \to CH_3OH + N_2 + H_2O$.
Trimethylamine does not react with Hinsberg reagent $(C_6H_5SO_2Cl)$ in the presence of $KOH$ because it lacks an acidic hydrogen,and it remains insoluble.
Dimethylamine reacts with Hinsberg reagent to form a sulfonamide which is insoluble in $KOH$.
Azo dyes are typically formed by the coupling reaction of diazonium salts with phenols or aromatic amines,not by simple secondary amines.

(D) All aliphatic primary amines react with nitrous acid $(NaNO_2 + HCl)$ to form unstable alkyl diazonium salts,which decompose to liberate $N_2$ gas.
$RNH_2 + HONO \longrightarrow R-OH + N_2 + H_2O$
Since all the given options ($C_2H_5NH_2$,$CH_3NH_2$,and $(CH_3)_2CHNH_2$) are aliphatic primary amines,they all will evolve $N_2$ gas.

(C) The Assertion is correct: Benzene diazonium salts react with water upon boiling to form phenol,releasing $N_2$ gas.
The Reason is incorrect: The $C-N$ bond in the diazonium salt is indeed polar,but the reaction proceeds because the $N_2$ group is an excellent leaving group,not simply due to the polarity of the bond.

(A) $\alpha$-amino acids contain both an acidic carboxyl group $(-COOH)$ and a basic amino group $(-NH_2)$.
In aqueous solution,the carboxyl group loses a proton $(H^{+})$ and the amino group accepts it,forming a dipolar ion known as a zwitterion or internal salt: $NH_2-CHR-COOH \rightleftharpoons NH_3^+-CHR-COO^-$.
This occurs because the acidic and basic groups are in close proximity.
Therefore,both Assertion and Reason are correct,and the Reason is the correct explanation.