AIIMS 2019 Chemistry Question Paper with Answer and Solution

(C) The reaction follows Markovnikov's addition of $HCl$ to the alkene.
$1$. The $H^+$ ion attacks the terminal carbon of the double bond to form the most stable carbocation.
$2$. The intermediate formed is a $3^{\circ}$ benzylic carbocation $(C_6H_5-CH^+-CH_2-CH_3)$,which is highly stabilized by resonance with the phenyl ring.
$3$. The $Cl^-$ ion then attacks this carbocation to form the final product: $1$-chloro-$1$-phenylpropane $(C_6H_5-CH(Cl)-CH_2-CH_3)$.

(A) Excess nitrate in drinking water can cause methemoglobinemia,which is also known as blue baby syndrome.

(A) The $S-S$ bond length depends on the oxidation state of sulfur and the repulsion between lone pairs on the sulfur atoms.
In $S_2O_4^{2-}$ (dithionite),each sulfur atom has one lone pair,leading to significant $L.P.-L.P.$ repulsion between the two sulfur atoms,which increases the $S-S$ bond length.
In $S_2O_5^{2-}$ (disulfite),one sulfur has a lone pair and the other does not,resulting in less repulsion compared to $S_2O_4^{2-}$.
In $S_2O_6^{2-}$ (dithionate),both sulfur atoms are in a higher oxidation state and have no lone pairs involved in $S-S$ repulsion,resulting in the shortest $S-S$ bond length.
Therefore,the correct order of $S-S$ bond length is $I > II > III$.

(C) For a $4p$ electron,the principal quantum number $n=4$ and the azimuthal quantum number $l=1$.
The magnetic quantum number $m$ can take values from $-l$ to $+l$,which means for $l=1$,$m$ can be $-1, 0, +1$.
The value $m=2$ is not possible for $l=1$ because $m$ must satisfy the condition $|m| \leq l$.
Therefore,the set $n=4, l=1, m=2, m_{s}=+\frac{1}{2}$ is not possible.

(C) The change in enthalpy is defined as $\Delta H = \Delta U + \Delta(PV)$.
Given $\Delta U = 30.0 \, L \, atm$.
Initial state: $P_1 = 2.0 \, atm$,$V_1 = 3.0 \, L$.
Final state: $P_2 = 4.0 \, atm$,$V_2 = 5.0 \, L$.
$\Delta(PV) = P_2 V_2 - P_1 V_1 = (4.0 \times 5.0) - (2.0 \times 3.0) = 20.0 - 6.0 = 14.0 \, L \, atm$.
Therefore,$\Delta H = 30.0 + 14.0 = 44.0 \, L \, atm$.

(B) The auto-ionization of water is an endothermic process.
According to Le Chatelier's principle,an increase in temperature shifts the equilibrium in the forward direction.
Therefore,the ionic product of water $(K_w = [H^+][OH^-])$ increases as the temperature increases.
At $298 \ K$,$K_w = 1 \times 10^{-14}$.
Since $393 \ K > 298 \ K$,the value of $K_w$ at $393 \ K$ will be greater than $1 \times 10^{-14}$.

(B) In $o-$nitrophenol,the $-OH$ group forms an intramolecular hydrogen bond with the adjacent nitro group.
This makes the $-OH$ group unavailable for intermolecular hydrogen bonding with water molecules.
Consequently,$o-$nitrophenol exhibits lower solubility in water compared to $m-$ and $p-$nitrophenols,which form intermolecular hydrogen bonds with water.

(A) For isoelectronic species,the ionic radius increases as the negative charge increases and decreases as the positive charge increases.
All the given species $(C^{4-}, N^{3-}, O^{2-}, Mg^{2+})$ are isoelectronic with $10$ electrons.
Among these,$C^{4-}$ has the highest negative charge,which results in the greatest inter-electronic repulsion and the weakest effective nuclear attraction.
Therefore,$C^{4-}$ has the maximum ionic radius.

(C) To find the lightest,we calculate the mass for each option:
$1$. Mass of $0.2 \ mol$ of $H_2 = 0.2 \ mol \times 2 \ g/mol = 0.4 \ g$.
$2$. $6.023 \times 10^{23}$ molecules of $N_2$ corresponds to $1 \ mol$. Mass of $1 \ mol$ of $N_2 = 1 \ mol \times 28 \ g/mol = 28 \ g$.
$3$. Mass of $0.1 \ g$ of silver is given as $0.1 \ g$.
$4$. Mass of $0.1 \ mol$ of $O_2 = 0.1 \ mol \times 32 \ g/mol = 3.2 \ g$.
Comparing the masses: $0.1 \ g < 0.4 \ g < 2.8 \ g < 3.2 \ g$.
Therefore,$0.1 \ g$ of silver is the lightest.

(B) According to Graham's Law of diffusion,the rate of diffusion $(r)$ is inversely proportional to the square root of the molar mass $(M)$: $r \propto \frac{1}{\sqrt{M}}$.
First,calculate the molar masses of the given compounds:
$M(CO_{2}) = 12 + 2 \times 16 = 44 \ g/mol$
$M(SO_{2}) = 32 + 2 \times 16 = 64 \ g/mol$
$M(SO_{3}) = 32 + 3 \times 16 = 80 \ g/mol$
$M(PCl_{3}) = 31 + 3 \times 35.5 = 31 + 106.5 = 137.5 \ g/mol$
Since the rate of diffusion is inversely proportional to the square root of the molar mass,the order of the rate of diffusion will be the reverse of the order of their molar masses:
$M(CO_{2}) < M(SO_{2}) < M(SO_{3}) < M(PCl_{3})$
Therefore,the rate of diffusion order is: $CO_{2} > SO_{2} > SO_{3} > PCl_{3}$.

(C) For geometrical isomerism to occur at a double bond,each carbon atom of the double bond must be attached to two different groups.
$I$: The carbon atom is attached to two identical methyl groups $(a, a)$,so geometrical isomerism is not possible.
$II$: Both carbons are attached to different groups,so it shows geometrical isomerism.
$III$: The carbon atom is attached to two identical methyl groups,so geometrical isomerism is not possible.
$IV$: Both carbons are attached to different groups,so it shows geometrical isomerism.
Therefore,geometrical isomerism is not possible at sites $I$ and $III$.

(A) The magnetic quantum number $m$ for an $s$-orbital is always $0$.
For $Na$ $(Z=11)$,the electronic configuration is $1s^{2} 2s^{2} 2p^{6} 3s^{1}$. The last electron enters the $3s$ orbital,which has $l=0$,therefore $m=0$.
For $O$ $(Z=8)$,the configuration is $1s^{2} 2s^{2} 2p^{4}$. The last electron enters a $2p$ orbital $(l=1)$,where $m$ can be $-1, 0, +1$.
For $Cl$ $(Z=17)$,the configuration is $1s^{2} 2s^{2} 2p^{6} 3s^{2} 3p^{5}$. The last electron enters a $3p$ orbital $(l=1)$,where $m$ can be $-1, 0, +1$.
For $N$ $(Z=7)$,the configuration is $1s^{2} 2s^{2} 2p^{3}$. The last electron enters a $2p$ orbital $(l=1)$,where $m$ can be $-1, 0, +1$.
Thus,only $Na$ has its last electron in an $s$-orbital where $m$ is strictly $0$.

(A) For a reaction to be spontaneous,the Gibbs free energy change $\Delta G^{\circ}$ must be less than $0$.
Given the equation: $\Delta G^{\circ} = \Delta H^{\circ} - T \Delta S^{\circ}$.
At equilibrium,$\Delta G^{\circ} = 0$,so $T = \frac{\Delta H^{\circ}}{\Delta S^{\circ}}$.
Given values: $\Delta H^{\circ} = 179.1 \ kJ \ mol^{-1} = 179100 \ J \ mol^{-1}$ and $\Delta S^{\circ} = 160.2 \ J \ K^{-1} \ mol^{-1}$.
Substituting these values into the equation:
$T = \frac{179100 \ J \ mol^{-1}}{160.2 \ J \ K^{-1} \ mol^{-1}} \approx 1117.97 \ K$.
Since the reaction is endothermic and entropy increases,the reaction becomes spontaneous at temperatures above $1118 \ K$.

(B) When aluminium is treated with dilute $H_{2}SO_{4}$:
$2Al + 3H_{2}SO_{4} \rightarrow Al_{2}(SO_{4})_{3} + 3H_{2}$
From the stoichiometry,$2 \text{ moles of } Al$ produce $3 \text{ moles of } H_{2}$.
When aluminium is treated with $NaOH$:
$2Al + 2NaOH + 6H_{2}O \rightarrow 2Na[Al(OH)_{4}] + 3H_{2}$
From the stoichiometry,$2 \text{ moles of } Al$ produce $3 \text{ moles of } H_{2}$.
Since the same mass of aluminium $(2 \ g)$ is used in both cases,the number of moles of $Al$ is the same.
Therefore,the ratio of the volumes of hydrogen evolved is $3:3$,which simplifies to $1:1$.

(C) Electrophilic substitution reactivity depends on the electron density of the aromatic ring.
Electron donating groups $(EDG)$ increase electron density,while electron withdrawing groups $(EWG)$ decrease it.
$(a)$ Toluene $(-CH_3)$: $+I$ and hyperconjugation effect (activating).
$(b)$ Anisole $(-OCH_3)$: $+M$ effect is very strong (strongly activating).
$(c)$ Chlorobenzene $(-Cl)$: $-I$ effect dominates over $+M$ effect (deactivating).
$(d)$ Benzaldehyde $(-CHO)$: $-M$ and $-I$ effects (strongly deactivating).
The order of activation is: Anisole $(b)$ > Toluene $(a)$ > Chlorobenzene $(c)$ > Benzaldehyde $(d)$.
Therefore,the correct order is $b > a > c > d$.

(D) The reaction is an electrophilic addition of $HBr$ to a conjugated diene $(2\text{-methylbuta-1,3-diene})$ at low temperature $(-80^{\circ}C)$,which favors the kinetic product $(1,2\text{-addition})$.
$1$. The proton $(H^+)$ from $HBr$ attacks the terminal double bond to form the most stable carbocation.
$2$. Protonation at the $C_1$ position leads to a tertiary allylic carbocation,which is more stable than the secondary allylic carbocation formed by protonation at the $C_4$ position.
$3$. The bromide ion $(Br^-)$ then attacks the tertiary carbocation at the $C_2$ position to form the $1,2\text{-addition}$ product.
$4$. The final product $A$ is $2\text{-bromo-2-methylbut-3-ene}$.

(C) In general,the anti-staggered conformer is the most stable for most molecules due to minimal steric hindrance. However,for ethane-$1,2$-diol,the gauche conformer is more stable than the anti-staggered form. This is due to the formation of an intramolecular hydrogen bond between the two hydroxyl $(-OH)$ groups,which stabilizes the gauche conformation. The structure showing this intramolecular hydrogen bonding is given in the reference image.

(D) The solubility of ionic compounds in water depends on their lattice energy and hydration energy.
According to Fajan's rule,compounds with higher covalent character exhibit lower solubility in polar solvents like water.
$MgS$ has a high degree of covalent character due to the small size and high charge density of the $Mg^{2+}$ ion,which polarizes the $S^{2-}$ ion significantly.
Therefore,$MgS$ is the least soluble among the given options.

(D) The formation reaction of $C_{2}H_{5}OH_{(l)}$ is:
$2C_{(s)} + 3H_{2(g)} + \frac{1}{2} O_{2(g)} \rightarrow C_{2}H_{5}OH_{(l)}$
To obtain this,we perform the operation: $2 \times (I) + 3 \times (II) + (III)$.
$\Delta H_f = 2 \times (-393) + 3 \times (-287.3) + 1366.8$
$\Delta H_f = -786 - 861.9 + 1366.8$
$\Delta H_f = -1647.9 + 1366.8 = -281.1 \, kJ \, mol^{-1}$.

(C) An isolated system is defined as a system that cannot exchange energy or matter with its surroundings.
For an isolated system,the total internal energy remains constant,so $ \Delta U = 0 $.
According to the second law of thermodynamics,for a spontaneous process in an isolated system,the total entropy of the system must increase,so $ \Delta S_{total} > 0 $.
Note that $ \Delta G $ is typically defined for systems at constant temperature and pressure,which is not applicable to an isolated system in the same way; however,the primary criterion for spontaneity in an isolated system is $ \Delta S > 0 $.

(D) The energy of a single photon is given by $E_{photon} = \frac{hc}{\lambda}$.
The total energy $E$ is given by $E = n \times E_{photon} = \frac{nhc}{\lambda}$,where $n$ is the number of photons.
Given:
$E = 3 \times 10^{-18} \, J$
$\lambda = 660 \, nm = 660 \times 10^{-9} \, m$
$h = 6.6 \times 10^{-34} \, J \cdot s$
$c = 3 \times 10^{8} \, m/s$
Rearranging for $n$:
$n = \frac{E \times \lambda}{h \times c}$
Substituting the values:
$n = \frac{3 \times 10^{-18} \times 660 \times 10^{-9}}{6.6 \times 10^{-34} \times 3 \times 10^{8}}$
$n = \frac{3 \times 660 \times 10^{-27}}{6.6 \times 3 \times 10^{-26}}$
$n = \frac{1980 \times 10^{-27}}{19.8 \times 10^{-26}} = 100 \times 10^{-1} = 10$
Thus,the number of emitted photons is $10$.

(C) Dimethylamine is a weak base: $(CH_3)_2NH + H_2O \rightleftharpoons (CH_3)_2NH_2^+ + OH^-$
Given: $C = 0.05 \ M$,$K_b = 5 \times 10^{-4}$,and $[OH^-]_{NaOH} = 0.1 \ M$.
Due to the common ion effect,the concentration of $OH^-$ is dominated by $NaOH$,so $[OH^-] \approx 0.1 \ M$.
The expression for $K_b$ is: $K_b = \frac{[(CH_3)_2NH_2^+][OH^-]}{[(CH_3)_2NH]} = \frac{C\alpha \times 0.1}{C(1-\alpha)} \approx \frac{0.05 \times \alpha \times 0.1}{0.05} = 0.1 \alpha$.
$5 \times 10^{-4} = 0.1 \alpha \implies \alpha = 5 \times 10^{-3}$.
Percentage dissociation $= \alpha \times 100 = 5 \times 10^{-3} \times 100 = 0.5 \% = 5 \times 10^{-1} \%$.

(C) The limiting reagent is the reactant that is completely consumed in a reaction and limits the amount of product formed.
For the reaction $A + B \rightarrow AB$,the stoichiometric ratio is $1:1$.
To find the limiting reagent,we compare the number of atoms of $A$ and $B$.
The reactant with the smaller number of atoms (when stoichiometric coefficients are $1$) is the limiting reagent.
In option $C$,we have $50$ atoms of $A$ and $30$ atoms of $B$.
Since $30 < 50$,$B$ is the limiting reagent.
Therefore,the correct option is $C$.

(A) $K_{2}Cr_{2}O_{7}$ acts as a strong oxidizing agent in acidic medium.
It can oxidize sulfite ions $(SO_{3}^{2-})$ to sulfate ions $(SO_{4}^{2-})$.
The balanced ionic reaction is:
$Cr_{2}O_{7}^{2-} + 3SO_{3}^{2-} + 8H^{+} \rightarrow 2Cr^{3+} + 3SO_{4}^{2-} + 4H_{2}O$
Other ions like $CO_{3}^{2-}$,$SO_{4}^{2-}$,and $NO_{3}^{-}$ are either already in their highest oxidation state or do not undergo redox reaction with dichromate under standard conditions.

(D) The reaction is a solvolysis reaction of $1-$iodo$-2-$methylcyclohexane in water.
$1$. The leaving group $I^-$ departs to form a secondary carbocation at the $C-2$ position.
$2$. This secondary carbocation undergoes a $1,2-$hydride shift to form a more stable tertiary carbocation at the $C-1$ position.
$3$. Water acts as a nucleophile and attacks the tertiary carbocation to form an oxonium ion intermediate.
$4$. Finally,deprotonation yields $1-$methylcyclohexanol.
Comparing the options:
- Option $A$ is the oxonium ion intermediate.
- Option $B$ is the initial secondary carbocation.
- Option $C$ is the tertiary carbocation.
- Option $D$ is a primary carbocation,which is highly unstable and not expected to be formed in this mechanism.

(D) $1$. Identify the longest carbon chain containing the principal functional group $(-OH)$. The chain has $6$ carbons,so the parent alkane is hexane.
$2$. Number the chain starting from the end closer to the principal functional group $(-OH)$. Thus,the $-OH$ group is at position $2$.
$3$. Identify the substituents: an amino group $(-NH_2)$ at position $4$ and an ethyl group $(-CH_2CH_3)$ at position $3$.
$4$. Arrange the substituents alphabetically: amino comes before ethyl.
$5$. Combining these,the $IUPAC$ name is $4-$amino$-3-$ethylhexan$-2-$ol.

(B) The reaction proceeds as follows:
$1$. Addition of $2 \text{ moles}$ of $HBr$ to propyne $(CH_{3}-C \equiv CH)$ follows Markovnikov's rule,resulting in the formation of a geminal dibromide: $CH_{3}-C(Br)_{2}-CH_{3}$.
$2$. Hydrolysis of the geminal dibromide with $H_{2}O$ replaces the bromine atoms with hydroxyl groups,forming a geminal diol (gem-diol): $CH_{3}-C(OH)_{2}-CH_{3}$.
$3$. Gem-diols are unstable and lose a molecule of water to form a ketone: $CH_{3}-C(OH)_{2}-CH_{3} \rightarrow CH_{3}-C(=O)-CH_{3} + H_{2}O$.
Therefore,the final product is acetone $(CH_{3}-C(=O)-CH_{3})$.

(A) $H_2O_2$ is prepared by the action of dilute sulfuric acid on hydrated barium peroxide $(BaO_2 \cdot 8H_2O)$.
The chemical reaction is:
$BaO_2 \cdot 8H_2O_{(s)} + H_2SO_{4(aq)} \rightarrow BaSO_{4(s)} + H_2O_{2(aq)} + 8H_2O_{(l)}$.

(D) Below the critical temperature $(T_c)$,the isotherms show a distinct region where the gas and liquid phases coexist. In this region,the pressure remains constant as the volume decreases during the condensation process. This results in a characteristic horizontal or nearly horizontal segment in the $P-V$ isotherm,as shown in the provided figure (option $D$).

(A) The flame colors of the given alkali metals are:
$Li$ (Lithium): Crimson red flame $(\lambda \approx 670 \ nm)$
$Na$ (Sodium): Golden yellow flame $(\lambda \approx 589 \ nm)$
$K$ (Potassium): Violet flame $(\lambda \approx 404 \ nm)$
$Cs$ (Cesium): Blue flame $(\lambda \approx 455 \ nm)$
Since the wavelength $(\lambda)$ is inversely proportional to the energy of the emitted light,the color with the lowest energy corresponds to the maximum wavelength.
Among the given options,the crimson red color of $Li$ has the longest wavelength.

(B) Electronegativity $(E.N.)$ increases across a period and decreases down a group. Fluorine $(F)$ is the most electronegative element $(E.N. = 4.0)$,while alkali metals have the lowest $E.N.$ values.
Comparing the given pairs:
$1$. $Li$ $(1.0)$ and $F$ $(4.0)$: Difference = $3.0$
$2$. $Na$ $(0.9)$ and $F$ $(4.0)$: Difference = $3.1$
$3$. $Na$ $(0.9)$ and $Br$ $(2.8)$: Difference = $1.9$
$4$. $Na$ $(0.9)$ and $Cl$ $(3.0)$: Difference = $2.1$
Thus,the pair with the maximum electronegativity difference is $Na \& F$.

(A) The structure of $P_4O_{10}$ consists of a tetrahedral arrangement of four phosphorus atoms.
Each phosphorus atom is bonded to three oxygen atoms via $P-O-P$ bridges,and one terminal oxygen atom via a $P=O$ double bond.
Counting the bonds in the structure:
$1$. Total $P-O-P$ bonds = $6$.
$2$. Total $P=O$ bonds = $4$.
$3$. Total $P-P$ bonds = $0$.
Therefore,the order is $P-O-P (6) > P=O (4) > P-P (0)$.

(A) The given reaction is $A_{2} \rightleftharpoons 2A$ with $\Delta H > 0$ (endothermic).
According to Le Chatelier's principle,for an endothermic reaction,an increase in temperature shifts the equilibrium in the forward direction,thereby increasing the yield of the product $A$.
Additionally,increasing the concentration of the reactant $A_{2}$ shifts the equilibrium in the forward direction to counteract the change,further increasing the yield of $A$.
Therefore,an increase in both temperature and concentration of the reactant will increase the yield of the monomer $A$.

(A) The $pH$ of a salt of a weak acid and a weak base is given by the formula:
$pH = \frac{1}{2}(pK_w + pK_a - pK_b)$
Given:
$pK_w = 14$
$pK_a = 4$
$pK_b = 5$
Substituting the values:
$pH = \frac{1}{2}(14 + 4 - 5)$
$pH = \frac{1}{2}(13)$
$pH = 6.5$

(A) The radius of the $n^{th}$ orbit is given by $r_n = 0.529 \times \frac{n^2}{Z} \ \mathring{A}$.
For $H$ atom,$Z = 1$ and $n = 1$,so $r_H = 0.529 \times \frac{1^2}{1} = 0.529 \ \mathring{A}$.
For $Be^{3+}$ ion,$Z = 4$. Let the orbit number be $n$.
Given $r_H = r_{Be^{3+}}$,so $0.529 \times \frac{1^2}{1} = 0.529 \times \frac{n^2}{4}$.
$1 = \frac{n^2}{4} \implies n^2 = 4 \implies n = 2$.
The energy of an orbit is given by $E_n = -13.6 \times \frac{Z^2}{n^2} \ \text{eV}$.
For $Be^{3+}$,$Z = 4$ and $n = 2$,so $E = -13.6 \times \frac{4^2}{2^2} = -13.6 \times \frac{16}{4} = -13.6 \times 4 = -54.4 \ \text{eV}$.

(C) $PCl_5$: Bond pairs $(bp)$ = $5$,Lone pairs $(lp)$ = $0$. Total = $5$. Hybridization = $sp^3d$. Shape = Trigonal bipyramidal.
$BrF_5$: Bond pairs $(bp)$ = $5$,Lone pairs $(lp)$ = $1$. Total = $6$. Hybridization = $sp^3d^2$. Shape = Square pyramidal.
$IF_7$: Bond pairs $(bp)$ = $7$,Lone pairs $(lp)$ = $0$. Total = $7$. Hybridization = $sp^3d^3$. Shape = Pentagonal bipyramidal.
Thus,the correct statement is that one of the following $(BrF_5)$ is square pyramidal.

(D) The conversion $CH_2=CH-CH_2-CHO \to CH_3-CH_2-CH_2-CH_2OH$ involves the reduction of both the carbon-carbon double bond $(C=C)$ and the aldehyde group $(-CHO)$.
$LiAlH_4$,$NaBH_4$,and $Na + C_2H_5OH$ are selective reducing agents that reduce the aldehyde group to a primary alcohol but do not reduce isolated $C=C$ double bonds.
Catalytic hydrogenation using $H_2 / Ni$ (or $Pt/Pd$) is the appropriate method to reduce both the alkene and the aldehyde functional groups to an alkane and a primary alcohol,respectively.

(D) The covalent radii for the given transition elements are as follows:
$Ni = 125 \text{ pm}$
$Cu = 128 \text{ pm}$
$Zn = 137 \text{ pm}$
$Mn = 137 \text{ pm}$
Comparing these values, $Ni$ has the smallest covalent radius among the given options.

(D) Nitric acid $(HNO_3)$ is a strong oxidizing agent. Among the given non-metals,$P$,$S$,and $I_2$ are oxidized by concentrated $HNO_3$ to their respective oxoacids ($H_3PO_4$,$H_2SO_4$,and $HIO_3$). Chlorine $(Cl_2)$ does not react with nitric acid because $HNO_3$ acts as an oxidizing agent and $Cl_2$ is already in a state where it is not easily oxidized further by $HNO_3$ under standard conditions. Thus,chlorine reacts most slowly (or effectively not at all) compared to the others.

(C) Carbon black,charcoal,coke,and graphene are forms or allotropes of carbon. Among these,$Coke$ is primarily used in the blast furnace for the reduction of iron ore to produce steel.

(A) The molecular formula of cyanogen gas is $(CN)_{2}$.
Cyanogen gas has a linear structure with the formula $N \equiv C - C \equiv N$.
In this molecule,both carbon atoms are $sp$ hybridized.
It is classified as a pseudohalogen because its chemical behavior resembles that of halogens.
Therefore,the statement that it has a bent structure is incorrect.

(A) complex is optically inactive if it possesses a plane of symmetry or a center of inversion.
$1$. $[RhCl(CO)(PPh_3)(NH_3)]$ is a square planar complex. All square planar complexes are optically inactive because they possess a plane of symmetry (the molecular plane).
$2$. $[Fe(C_2O_4)_3]^{3-}$ is an octahedral complex with three bidentate ligands,which is chiral and optically active.
$3$. $[Fe(en)_2Cl_2]$ can exist as cis and trans isomers; the cis-isomer is optically active.
$4$. $[Pd(en)_2Cl_2]$ is a square planar complex,but in the context of coordination chemistry,$[RhCl(CO)(PPh_3)(NH_3)]$ is the most classic example of an inherently achiral square planar geometry.

(A) The crystal field splitting energy $(\Delta_{0})$ increases as the field strength of the ligand increases.
The spectrochemical series order for the given ligands is: $CN^{-} > NH_3 > H_2O > Cl^{-}$.
Since the energy of the absorbed light is inversely proportional to the wavelength $(\lambda \propto \frac{1}{\Delta_{0}})$,a stronger ligand field results in a larger $\Delta_{0}$ and a shorter wavelength of absorption.
Thus,the increasing order of wavelength is: $[Co(CN)_6]^{3-} < [Co(NH_3)_6]^{3+} < [Co(NH_3)_5(H_2O)]^{3+} < [Co(NH_3)_5Cl]^{2+}$.

(D) The elevation in boiling point is given by the formula: $\Delta T_{b} = i \times K_{b} \times m$.
For $NaCl$,the van't Hoff factor $i = 2$.
The molality $m$ is defined as: $m = \frac{W_{NaCl} \times 1000}{M_{NaCl} \times W_{H_2O(g)}}$.
Given $\Delta T_{b} = 1 \,K$,$K_{b} = 0.52 \,K \cdot kg/mol$,$M_{NaCl} = 58.5 \,g/mol$,and $W_{H_2O} = 500 \,g$.
Substituting the values: $1 = 2 \times 0.52 \times \frac{W_{NaCl} \times 1000}{58.5 \times 500}$.
Solving for $W_{NaCl}$: $W_{NaCl} = \frac{1 \times 58.5 \times 500}{2 \times 0.52 \times 1000} = 28.125 \,g$.

(A) The reaction at the hydrogen electrode is: $H_{2}(g) \rightarrow 2H^{+}(aq) + 2e^{-}$.
Given $pH = 10$,so $[H^{+}] = 10^{-10} \, M$.
Pressure of $H_{2}$ gas,$P_{H_{2}} = 1 \, atm$.
Using the Nernst equation for the oxidation potential:
$E_{ox} = E^{\circ}_{ox} - \frac{0.0591}{n} \log \frac{[H^{+}]^{2}}{P_{H_{2}}}$
Since $E^{\circ}_{ox}$ for $S.H.E$ is $0 \, V$ and $n = 2$:
$E_{ox} = 0 - \frac{0.0591}{2} \log \frac{(10^{-10})^{2}}{1}$
$E_{ox} = -0.02955 \times \log(10^{-20})$
$E_{ox} = -0.02955 \times (-20) = 0.591 \, V$.
Rounding to two decimal places,the potential is $0.59 \, V$.

(C) The decomposition reaction of $NH_3$ is: $2NH_3 \rightarrow N_2 + 3H_2$.
Since it is a zero-order reaction,the rate of reaction is equal to the rate constant $k = 2 \times 10^{-4} \ mol \ L^{-1} \ s^{-1}$.
The rate of reaction is given by: $\text{Rate} = -\frac{1}{2} \frac{d[NH_3]}{dt} = \frac{d[N_2]}{dt} = \frac{1}{3} \frac{d[H_2]}{dt} = k$.
Therefore,the rate of appearance of $N_2$ is $\frac{d[N_2]}{dt} = k = 2 \times 10^{-4} \ mol \ L^{-1} \ s^{-1}$.
The rate of appearance of $H_2$ is $\frac{d[H_2]}{dt} = 3k = 3 \times (2 \times 10^{-4}) = 6 \times 10^{-4} \ mol \ L^{-1} \ s^{-1}$.

(D) Given that the rate constant doubles,$K_{2} = 2 K_{1}$.
The Arrhenius equation in logarithmic form is:
$\log \frac{K_{2}}{K_{1}} = \frac{E_{a}}{2.303 \times R} \left(\frac{T_{2} - T_{1}}{T_{1} \times T_{2}}\right)$
Substituting the given values:
$\log(2) = \frac{E_{a}}{2.303 \times 8.314} \left(\frac{400 - 300}{300 \times 400}\right)$
$0.3010 = \frac{E_{a}}{19.147} \times \left(\frac{100}{120000}\right)$
$0.3010 = \frac{E_{a}}{19.147} \times \frac{1}{1200}$
$E_{a} = 0.3010 \times 19.147 \times 1200 \approx 6914 \, J \, mol^{-1} \approx 6.91 \, kJ \, mol^{-1}$.
Rounding to the nearest provided option,the correct value is $6.88 \, kJ \, mol^{-1}$.

(D) The colour of transition metal ions is due to the presence of unpaired electrons which allow $d-d$ transitions.
Electronic configurations:
$Zn^{2+} (Z=30): [Ar] 3d^{10} 4s^0$ (All electrons are paired,so it is colourless).
$Ni^{2+} (Z=28): [Ar] 3d^8 4s^0$ (Has $2$ unpaired electrons,so it is coloured).
$Cr^{3+} (Z=24): [Ar] 3d^3 4s^0$ (Has $3$ unpaired electrons,so it is coloured).
Therefore,only $Zn^{2+}$ is colourless,while $Ni^{2+}$ and $Cr^{3+}$ are coloured.

(D) $1$. Identify the ligands: There are $5$ ammine $(NH_3)$ ligands and $1$ nitrito$-N$ $(NO_2^-)$ ligand.
$2$. Determine the oxidation state of the central metal atom $(Co)$: Let the oxidation state be $x$. The charge on $NH_3$ is $0$,$NO_2^-$ is $-1$,and $Cl^-$ is $-1$. The total charge of the complex is $0$. Thus,$x + 5(0) + 1(-1) + 2(-1) = 0$,which gives $x - 3 = 0$,so $x = +3$.
$3$. Name the complex: The ligands are named in alphabetical order (ammine before nitrito$-N$). The metal is cobalt followed by its oxidation state in Roman numerals in parentheses. The counter ion is chloride.
$4$. The correct name is Pentaammine nitrito$-N$-cobalt$(III)$ chloride.

(D) The molar ionic conductance is directly proportional to the number of ions produced in the solution.
$(i)$ $[Pt(NH_3)_5 Cl]Cl_3 \rightarrow [Pt(NH_3)_5 Cl]^{3+} + 3Cl^-$ ($4$ ions)
$(ii)$ $[Pt(NH_3)_4 Cl_2]Cl_2 \rightarrow [Pt(NH_3)_4 Cl_2]^{2+} + 2Cl^-$ ($3$ ions)
$(iii)$ $[Pt(NH_3)_3 Cl_3]Cl \rightarrow [Pt(NH_3)_3 Cl_3]^+ + Cl^-$ ($2$ ions)
$(iv)$ $[Pt(NH_3)_2 Cl_4]$ is a neutral complex and does not dissociate into ions ($0$ ions).
Thus,the order of molar ionic conductance is $(iv) < (iii) < (ii) < (i)$.

(D) The reactivity towards $S_{N}1$ reaction depends on the stability of the carbocation intermediate formed after the loss of the leaving group $(Cl^-)$.
$1$. For $MeCOCH_{2}Cl$,the carbocation $MeCOCH_{2}^+$ is destabilized by the electron-withdrawing carbonyl group.
$2$. For cyclopropyl chloride,the cyclopropyl cation is highly unstable due to significant angle strain.
$3$. For $C_{6}H_{5}CH_{2}CH_{2}Cl$,the carbocation $C_{6}H_{5}CH_{2}CH_{2}^+$ is a primary carbocation,which is relatively unstable.
$4$. For $MeOCH_{2}Cl$,the carbocation formed is $MeOCH_{2}^+$. This carbocation is resonance-stabilized by the lone pair on the oxygen atom: $Me-O-CH_{2}^+ \leftrightarrow Me-O^+=CH_{2}$. This structure is highly stable because every atom has a complete octet.
Therefore,$MeOCH_{2}Cl$ is the most reactive towards $S_{N}1$ reaction.

(B) $DIBAL-H$ (Diisobutylaluminium hydride) acts as a selective reducing agent. It is capable of reducing both esters $(-COOR)$ and nitriles $(-CN)$ to aldehydes $(-CHO)$ at low temperatures. In the given substrate,both the ester group $(-COOCH_3)$ and the nitrile group $(-CN)$ are reduced to aldehyde groups $(-CHO)$. Therefore,the product is a cyclohexane$-1,4-$dicarbaldehyde. Thus,option $(B)$ is the correct answer.

(C) The rate of $S_{N}1$ reaction depends on the stability of the carbocation intermediate formed after the departure of the leaving group $(Cl^-)$.
$(i)$ Forms an allylic carbocation,which is resonance stabilized.
(ii) Forms a secondary $(2^{\circ})$ cyclohexyl carbocation.
(iii) Forms a tertiary $(3^{\circ})$ carbocation.
Comparing the stability of these carbocations:
$1$. The allylic carbocation (from $i$) is highly stable due to resonance.
$2$. The tertiary carbocation (from $iii$) is stable due to inductive effect and hyperconjugation.
$3$. The secondary carbocation (from $ii$) is the least stable among the three.
Thus,the order of stability of carbocations is $i > iii > ii$. Consequently,the order of $S_{N}1$ reactivity is $i > iii > ii$. However,based on standard textbook problems of this type,the resonance-stabilized allylic carbocation is significantly more reactive. Given the options,$i > iii > ii$ is the correct chemical trend.

(C) $(i)$ $N$-methylaniline: The $+I$ effect of the methyl group increases basicity compared to aniline.
$(ii)$ $p$-methoxyaniline: The $+M$ effect of the $-OCH_3$ group significantly increases basicity,making it more basic than $N$-methylaniline.
$(iii)$ Benzylamine: The $-NH_2$ group is not directly attached to the benzene ring,so the lone pair is not involved in resonance. It is the most basic.
$(iv)$ Aniline: The lone pair on the $-NH_2$ group is involved in resonance with the benzene ring,making it the least basic.
Comparing these,the order is: Benzylamine $(iii) > p$-methoxyaniline $(ii) > N$-methylaniline $(i) >$ Aniline $(iv)$.
Therefore,the correct order is $iii > ii > i > iv$.

(A) The reaction between phthalic acid and ethylamine $(C_2H_5NH_2)$ under heating $(\Delta)$ involves the formation of an amide linkage.
Initially,the acid-base reaction forms a salt,which upon further heating undergoes dehydration to form the cyclic imide,$N$-ethylphthalimide.
The reaction is:
Phthalic acid $C_2H_5NH_2 \xrightarrow{\Delta} N$-ethylphthalimide $2H_2O$.

(A) Starch is a common polysaccharide.
When it reacts with iodine solution,it forms a complex that produces a characteristic blue-black color.
Therefore,the iodine test is used to detect the presence of polysaccharides like starch.

(B) Given the equation $\ln k = 10 - \frac{69 \, kJ}{RT} \cdots (i)$.
Comparing this with the Arrhenius equation $\ln k = \ln A - \frac{E_a}{RT}$,we get $E_a = 69 \, kJ/mol = 69000 \, J/mol$.
For two temperatures $T_1 = 300 \, K$ and $T_2$,the rate constant $k_2 = 2k_1$.
Using the Arrhenius equation: $\ln \frac{k_2}{k_1} = \frac{E_a}{R} \left( \frac{1}{T_1} - \frac{1}{T_2} \right)$.
Substituting the values: $\ln 2 = \frac{69000}{8.314} \left( \frac{1}{300} - \frac{1}{T_2} \right)$.
$0.693 = 8300 \left( \frac{T_2 - 300}{300 T_2} \right)$.
Solving for $T_2$: $\frac{0.693}{8300} = \frac{T_2 - 300}{300 T_2}$.
$8.349 \times 10^{-5} = \frac{T_2 - 300}{300 T_2}$.
$0.02505 T_2 = T_2 - 300$.
$0.97495 T_2 = 300$.
$T_2 \approx 307.7 \, K$.

(D) The $CFSE$ (Crystal Field Stabilization Energy) depends on the oxidation state of the metal,the nature of the ligand,and the $d$-electron configuration.
$1$. In $K_{3}[Fe(CN)_{6}]$,$Fe$ is in $+3$ oxidation state ($d^{5}$ configuration). $CN^{-}$ is a strong field ligand,resulting in a low-spin $t_{2g}^{5}e_{g}^{0}$ configuration. $CFSE = -2.0 \Delta_{o}$.
$2$. In $K_{3}[Co(Ox)_{3}]$,$Co$ is in $+3$ oxidation state ($d^{6}$ configuration). $Ox^{2-}$ is a moderate field ligand,resulting in a low-spin $t_{2g}^{6}e_{g}^{0}$ configuration. $CFSE = -2.4 \Delta_{o}$.
$3$. In $K_{3}[CoF_{6}]$,$Co$ is in $+3$ oxidation state ($d^{6}$ configuration). $F^{-}$ is a weak field ligand,resulting in a high-spin $t_{2g}^{4}e_{g}^{2}$ configuration. $CFSE = -0.4 \Delta_{o}$.
$4$. In $K_{3}[Co(CN)_{6}]$,$Co$ is in $+3$ oxidation state ($d^{6}$ configuration). $CN^{-}$ is a strong field ligand,resulting in a low-spin $t_{2g}^{6}e_{g}^{0}$ configuration. $CFSE = -2.4 \Delta_{o}$.
Comparing the magnitudes,$[Co(CN)_{6}]^{3-}$ and $[Co(Ox)_{3}]^{3-}$ have the highest $CFSE$ values. However,since $CN^{-}$ is a stronger field ligand than $Ox^{2-}$,$\Delta_{o}$ for $[Co(CN)_{6}]^{3-}$ is significantly larger,making its $CFSE$ the maximum.

(A) When ammonia $(NH_3)$ reacts with excess bleaching powder $(CaOCl_2)$,it undergoes oxidation to produce dinitrogen gas $(N_2)$.
The balanced chemical equation is:
$3 CaOCl_2 + 2 NH_3 \rightarrow 3 CaCl_2 + N_2 + 3 H_2O$

(A) For a general reaction $aA + bB \rightarrow cC + dD$,the rate of reaction is given by:
Rate $= -\frac{1}{a} \frac{d[A]}{dt} = -\frac{1}{b} \frac{d[B]}{dt} = \frac{1}{c} \frac{d[C]}{dt} = \frac{1}{d} \frac{d[D]}{dt}$.
For the given reaction $A + 2 B \rightarrow C + D$,the stoichiometric coefficients are $a=1$ and $b=2$.
Substituting these values,the rate of reaction is:
Rate $= -\frac{1}{1} \frac{d[A]}{dt} = -\frac{1}{2} \frac{d[B]}{dt}$.
Thus,the correct expression is $\frac{-d[A]}{dt} = \frac{-1}{2} \frac{d[B]}{dt}$.

(A) In an $FCC$ unit cell,there are $8$ tetrahedral voids located at a distance of $\frac{a}{4}$ from each corner along the body diagonal.
Each body diagonal has a length of $\sqrt{3}a$.
The distance between two nearest tetrahedral voids is the distance between two voids located on the same body diagonal,which is $\frac{a}{4} + \frac{a}{4} = \frac{a}{2}$.

(A) The $CO$ ligand acts as a $2$-electron donor,while the $NO$ ligand acts as a $3$-electron donor.
For $Fe(CO)_5$: Total valence electrons = $8 + (5 \times 2) = 18$. To replace $CO$ with $NO$ while maintaining $18$ electrons,we replace $3$ $CO$ ($6$ $e^-$) with $2$ $NO$ ($6$ $e^-$),resulting in $Fe(CO)_2(NO)_2$.
For $Cr(CO)_6$: Total valence electrons = $6 + (6 \times 2) = 18$. To replace $CO$ with $NO$ while maintaining $18$ electrons,we replace $6$ $CO$ ($12$ $e^-$) with $4$ $NO$ ($12$ $e^-$),resulting in $Cr(NO)_4$.
Thus,the number of $CO$ ligands replaced are $3$ and $6$ respectively.

(B) The iron content in various forms of iron is as follows:
$1$. Cast iron: Contains about $93-95 \%$ iron.
$2$. Pig iron: Contains about $93-95 \%$ iron.
$3$. Wrought iron: Contains about $99.5-99.9 \%$ iron,making it the purest form of commercial iron.
$4$. Stainless steel: Contains about $70-80 \%$ iron along with chromium and nickel.
Therefore,Wrought iron has the maximum iron content.

(A) The molarity $(M)$ of a solution is given by the formula: $M = \frac{\% \ w/w \times d \times 10}{M_{solute}}$.
Here,the percentage by weight $(\% \ w/w)$ is $63$,the density $(d)$ is $5.4 \ g/mL$,and the molar mass of $HNO_3$ $(M_{solute})$ is $1 + 14 + (3 \times 16) = 63 \ g/mol$.
Substituting these values into the formula:
$M = \frac{63 \times 5.4 \times 10}{63} = 54 \ M$.

(C) $K_{2}Cr_{2}O_{7}$ is a strong oxidizing agent used in redox titrations.
It is prepared from $K_{2}CrO_{4}$ by adding acid.
It is orange in color.
However,it is not stable in both acid and base; it exists as $Cr_{2}O_{7}^{2-}$ (dichromate) in acidic medium and converts to $CrO_{4}^{2-}$ (chromate) in basic medium.
Thus,the statement that it is stable in both acid and base is incorrect.

(A) First,calculate the molar conductivity $(\lambda_{m})$ using the formula: $\lambda_{m} = \frac{1000 \times \kappa}{M}$
$\lambda_{m} = \frac{1000 \times 10^{-3}}{0.05} = 20 \, S \, cm^{2} \, mol^{-1}$
Next,calculate the degree of dissociation $(\alpha)$: $\alpha = \frac{\lambda_{m}}{\lambda_{m}^{\infty}} = \frac{20}{500} = 0.04$
Finally,calculate the dissociation constant $(K_{a})$ using the formula: $K_{a} = C \alpha^{2}$
$K_{a} = 0.05 \times (0.04)^{2} = 0.05 \times 0.0016 = 8 \times 10^{-5}$

(B) The spin-only magnetic moment is given by the formula $\mu = \sqrt{n(n+2)} \ BM$,where $n$ is the number of unpaired electrons.
Given $\mu = 1.73 \ BM$,we have $1.73 = \sqrt{n(n+2)}$.
Squaring both sides,$3 = n(n+2)$,which gives $n^2 + 2n - 3 = 0$.
Solving for $n$,we get $(n+3)(n-1) = 0$,so $n = 1$ (since $n$ cannot be negative).
Vanadium $(V)$ has the atomic number $23$ and electronic configuration $[Ar] 3d^3 4s^2$.
For $n=1$,the vanadium ion must have one unpaired electron.
In $VCl_4$,vanadium is in the $+4$ oxidation state $(V^{4+})$.
The configuration of $V^{4+}$ is $[Ar] 3d^1$,which contains $1$ unpaired electron.
Thus,the formula is $VCl_4$.

(A) The isoelectric point $(pI)$ is the $pH$ at which the amino acid exists as a zwitterion,i.e.,$H_3N^{+}-CH(R)-COO^{-}$.
At $pH$ values lower than the $pI$,the concentration of $H^{+}$ ions is high.
This leads to the protonation of the carboxylate group $(COO^{-})$,converting it into a carboxylic acid group $(COOH)$.
Since the given $pH$ $(1.0)$ is significantly lower than the $pI$ $(6.0)$,the amino acid will exist predominantly in its cationic form,$H_3N^{+}-CH(R)-COOH$.

(D) $NaBH_4$ (Sodium borohydride) is a selective reducing agent that specifically reduces aldehydes and ketones to their corresponding alcohols.
It does not reduce esters or carbon-carbon double bonds $(C=C)$ under standard conditions.
In the given substrate,there is a ketone group on the cyclohexene ring and an ester group $(-COOCH_3)$ on the side chain.
Therefore,$NaBH_4$ will selectively reduce the ketone to a secondary alcohol $(-OH)$ while leaving the ester and the double bond intact.
The product $X$ is $2\text{-hydroxycyclohex-3-en-1-yl} \text{ acetate derivative}$ (specifically,the structure shown in option $D$).

(A) The reaction involves the acid-catalyzed ring opening of an epoxide.
In the presence of an acid catalyst like $H_2SO_4$,the epoxide oxygen is protonated,making the epoxide ring more susceptible to nucleophilic attack.
Since the reaction occurs under acidic conditions,the nucleophile $(CH_3OH)$ attacks the more substituted carbon atom of the epoxide due to the development of significant carbocation character at that position in the transition state.
Therefore,the reagent $X$ is $CH_3OH$ in the presence of an acid catalyst $(H_2SO_4)$.

(B) When calcium acetate is heated (dry distillation),it undergoes thermal decomposition to form acetone and calcium carbonate.
$(CH_3COO)_2Ca \xrightarrow{\Delta} CH_3COCH_3 + CaCO_3$

(B) The diazo-coupling reaction is an electrophilic aromatic substitution reaction. The reactivity of the diazonium ion towards coupling depends on the electrophilicity of the diazonium cation.
Greater electron-withdrawing character of the substituent on the benzene ring increases the electrophilicity of the diazonium group,thereby increasing the reactivity.
The substituent effects are as follows:
$(I)$ $-NMe_2$: Strong $+M$ effect (strongly electron-donating).
$(II)$ $-NO_2$: Strong $-M$ effect (strongly electron-withdrawing).
$(III)$ $-OCH_3$: $+M$ effect (electron-donating).
$(IV)$ $-CH_3$: $+I$ effect (weakly electron-donating).
Thus,the electron-withdrawing order is: $-NO_2 > -CH_3 > -OCH_3 > -NMe_2$.
Therefore,the order of reactivity is: $(I) < (III) < (IV) < (II)$.

(B) The nitration of acetanilide $(C_6H_5NHCOCH_3)$ is an electrophilic aromatic substitution reaction. The acetamido group $(-NHCOCH_3)$ is strongly activating and ortho/para-directing. Due to the steric hindrance of the bulky acetamido group,the para-isomer is the major product.
$1$. Nitration of acetanilide with $conc. HNO_3 / H_2SO_4$ yields $p-$nitroacetanilide as the major product.
$2$. Subsequent alkaline hydrolysis of $p-$nitroacetanilide removes the acetyl group to yield $p-$nitroaniline.
The reaction sequence is:
$C_6H_5NHCOCH_3$ $\xrightarrow{HNO_3/H_2SO_4} p-NO_2-C_6H_4-NHCOCH_3$ $\xrightarrow{OH^-/H_2O} p-NO_2-C_6H_4-NH_2$.

(D) During vigorous exercise,the demand for energy in muscles exceeds the supply of oxygen.
Under these anaerobic conditions,the body converts pyruvic acid into lactic acid to regenerate $NAD^{+}$ and continue glycolysis for $ATP$ production.
This accumulation of $L^{-}$-Lactic acid in the muscle tissues leads to muscle fatigue and soreness.
The process is represented as: $\text{Glucose}$ $\rightarrow \text{Pyruvic acid}$ $\rightarrow \text{Lactic acid}$.

(C) In the primary structure of proteins,amino acids are arranged in a specific sequence.
These amino acids are linked to each other by covalent bonds known as peptide bonds,which are formed by the condensation reaction between the carboxyl group $(-COOH)$ of one amino acid and the amino group $(-NH_2)$ of the next amino acid.

(D) The given structure represents the repeating unit of $Nylon-6,6$.
It is formed by the condensation polymerization of hexamethylenediamine $(H_2N(CH_2)_6NH_2)$ and adipic acid $(HOOC(CH_2)_4COOH)$.
Since it is formed from two different monomers,it is classified as a copolymer.

(A) The given polymer is polyisobutylene,which is formed by the polymerization of $2$-methylpropene (isobutylene).
The structure of the monomer is $CH_3-C(CH_3)=CH_2$.

(B) The structure shown is $2$-acetoxybenzoic acid,commonly known as $aspirin$.
$Aspirin$ is a well-known drug that acts as an $analgesic$ (pain reliever) and an $antipyretic$ (fever reducer).
It also possesses anti-inflammatory properties,but among the given options,$analgesic$ is the most standard classification for its primary use.

(A) The method of concentrating ore based on the difference in densities of the ore and the gangue (impurities) is known as $Levigation$ or $Gravity$ $Separation$ or $Hydraulic$ $Washing$. In this process,the powdered ore is washed with a stream of water,where lighter gangue particles are carried away,leaving behind the heavier ore particles.