AP EAMCET 2019 Chemistry Question Paper with Answer and Solution

(D) The condition for a circle $S_1 = x^2 + y^2 + 2g_1x + 2f_1y + c_1 = 0$ to bisect the circumference of another circle $S_2 = x^2 + y^2 + 2g_2x + 2f_2y + c_2 = 0$ is that the common chord must pass through the center of the second circle.
The common chord is given by $S_1 - S_2 = 0$,which is $(6-2)x + (-2+6)y + (k+15) = 0$,or $4x + 4y + k + 15 = 0$.
The center of the second circle $x^2 + y^2 + 2x - 6y - 15 = 0$ is $(-g_2, -f_2) = (-1, 3)$.
Substituting $(-1, 3)$ into the equation of the common chord: $4(-1) + 4(3) + k + 15 = 0$.
$-4 + 12 + k + 15 = 0$.
$8 + k + 15 = 0$.
$k + 23 = 0 \Rightarrow k = -23$.

(C) The magnetic force on a charged particle moving in a magnetic field is given by $F = Bqv \sin \theta$. Since the electron enters the field transversely,$\theta = 90^\circ$,so $F = Bqv$.
The kinetic energy gained by an electron accelerated through a potential difference $V$ is given by $\frac{1}{2}mv^2 = eV$,which implies $v = \sqrt{\frac{2eV}{m}}$.
Substituting this into the force equation: $F = B e \sqrt{\frac{2eV}{m}} = B e \sqrt{\frac{2e}{m}} \sqrt{V}$.
From this expression,we see that $F \propto \sqrt{V}$.
If the potential is increased to $V' = 2V$,the new force $F'$ will be $F' \propto \sqrt{2V}$.
Taking the ratio: $\frac{F'}{F} = \frac{\sqrt{2V}}{\sqrt{V}} = \sqrt{2}$.
Therefore,$F' = \sqrt{2} F$.

(C) The kinetic energy of the electron is given by $K = eV$,where '$e$' is the charge of the electron and '$V$' is the accelerating potential.
In the first case,$K_1 = eV_1 = eV$.
In the second case,$K_2 = eV_2 = e(2V) = 2eV$.
Since $K = \frac{1}{2}mv^2$,we have $v = \sqrt{\frac{2K}{m}}$.
Thus,the ratio of velocities is $\frac{v_2}{v_1} = \sqrt{\frac{K_2}{K_1}} = \sqrt{\frac{2eV}{eV}} = \sqrt{2}$.
The magnetic force on a moving charge is given by $F = evB \sin(\theta)$. Since the electron enters the transverse magnetic field,$\theta = 90^\circ$ and $\sin(90^\circ) = 1$,so $F = evB$.
Since '$e$' and '$B$' are constant,$F \propto v$.
Therefore,$\frac{F_2}{F_1} = \frac{v_2}{v_1} = \sqrt{2}$.
Hence,$F_2 = \sqrt{2}F$.

(B) In Wurtz reaction, two molecules of bromoethane react with $2$ molecules of sodium to form $n$-butane:
$2CH_3CH_2Br + 2Na \rightarrow CH_3CH_2CH_2CH_3 + 2NaBr$
Decarboxylation of the sodium salt of a carboxylic acid with sodalime $(NaOH + CaO)$ removes one carbon atom as $Na_2CO_3$ to produce an alkane with one less carbon atom than the original acid.
To obtain $n$-butane ($4$ carbons), the sodium salt must be derived from a pentanoic acid ($5$ carbons).
The reaction is:
$CH_3CH_2CH_2CH_2COONa + NaOH \xrightarrow{CaO, \Delta} CH_3CH_2CH_2CH_3 + Na_2CO_3$
Thus, $X$ is pentanoic acid, $CH_3(CH_2)_3COOH$.

(B) Hydrogen atoms do not undergo hybridization. The number of hybrid orbitals is calculated as follows:
$(A)$ $C_6H_6$: All six $C$-atoms are $sp^2$-hybridized (each has $3$ hybrid orbitals). Total $= 6 \times 3 = 18$.
$(B)$ $(CH_3)_4C$: All five $C$-atoms are $sp^3$-hybridized (each has $4$ hybrid orbitals). Total $= 5 \times 4 = 20$.
$(C)$ $(CH_3)_2C=O$: Two $C$-atoms in $CH_3$ groups are $sp^3$-hybridized ($2 \times 4 = 8$ orbitals). The carbonyl $C$-atom is $sp^2$-hybridized ($3$ orbitals). The $O$-atom is $sp^2$-hybridized ($3$ orbitals). Total $= 8 + 3 + 3 = 14$.
$(D)$ $CH_3-CH=CH-CN$: $C_4$ is $sp^3$ ($4$ orbitals),$C_3$ and $C_2$ are $sp^2$ ($2 \times 3 = 6$ orbitals),$C_1$ is $sp$ ($2$ orbitals),and $N$ is $sp$ ($2$ orbitals). Total $= 4 + 6 + 2 + 2 = 14$.
Thus,$(CH_3)_4C$ has the maximum number of hybrid orbitals. The correct option is $(B)$.

(D) To determine the number of lone pairs,we draw the Lewis structures:
$1$. $CO$: The structure is $:C \equiv O:$. It has $1$ lone pair on $C$ and $1$ lone pair on $O$,total = $2$ lone pairs.
$2$. $NO_2^-$: The structure is $[:O-N=O:]^-$. $N$ has $1$ lone pair,one $O$ has $3$ lone pairs,and the other $O$ has $2$ lone pairs,total = $6$ lone pairs.
$3$. $CO_3^{2-}$: The structure has $3$ oxygen atoms each with $3$ lone pairs (for single-bonded $O$) or $2$ lone pairs (for double-bonded $O$). Total lone pairs = $8$.
$4$. $NF_3$: $N$ has $1$ lone pair and each of the $3$ $F$ atoms has $3$ lone pairs,total = $1 + 3 \times 3 = 10$ lone pairs.
Comparing the counts: $CO (2) < NO_2^- (6) < CO_3^{2-} (8) < NF_3 (10)$.
Thus,the increasing order is $A < B < D < C$. The correct option is $(D)$.

(C) The arrangement of electron pairs around the central atom is octahedral when the steric number is $6$ (i.e.,$sp^3d^2$ hybridization). $A$ molecule has a non-octahedral shape if it contains one or more lone pairs of electrons.
$(i)$ $SF_6$: Steric number = $6$ ($6$ bond pairs,$0$ lone pairs). Shape is octahedral.
$(ii)$ $XeF_6$: Steric number = $7$ ($6$ bond pairs,$1$ lone pair). The electron pair arrangement is pentagonal bipyramidal,not octahedral.
$(iii)$ $BrF_5$: Steric number = $6$ ($5$ bond pairs,$1$ lone pair). The electron pair arrangement is octahedral,but the shape is square pyramidal (not octahedral).
$(iv)$ $XeO_2F_4$: Steric number = $6$ ($6$ bond pairs,$0$ lone pairs). Shape is octahedral.
Therefore,$BrF_5$ is the molecule where the electron pair arrangement is octahedral but the shape is not.

(C) To determine the geometry,we calculate the hybridization and the number of lone pairs for each molecule:
$1$. $PCl_5$: $sp^3d$ hybridization,trigonal bipyramidal geometry.
$2$. $BrF_5$: $sp^3d^2$ hybridization with $1$ lone pair,square pyramidal geometry.
$3$. $ClF_5$: $sp^3d^2$ hybridization with $1$ lone pair,square pyramidal geometry.
$4$. $PF_5$: $sp^3d$ hybridization,trigonal bipyramidal geometry.
$5$. $ClF_3$: $sp^3d$ hybridization with $2$ lone pairs,$T$-shaped geometry.
$6$. $XeF_4$: $sp^3d^2$ hybridization with $2$ lone pairs,square planar geometry.
$7$. $XeF_2$: $sp^3d$ hybridization with $3$ lone pairs,linear geometry.
$8$. $IF_5$: $sp^3d^2$ hybridization with $1$ lone pair,square pyramidal geometry.
Thus,the molecules with square pyramidal geometry are $BrF_5, ClF_5$,and $IF_5$.
The total count is $3$.

(D) In the set $SF_4, XeF_4, CF_4$,the molecules have different geometries and their central atoms exhibit different hybridisations as shown below:

Molecule	Hybridisation	Geometry
$SF_4$	$sp^3d$	See-saw
$XeF_4$	$sp^3d^2$	Square planar
$CF_4$	$sp^3$	Tetrahedral

(C) The molecular geometries are determined by $VSEPR$ theory:

Molecule	Type	Shape
$BrF_5$	$AB_5E$	Square pyramidal
$SF_4$	$AB_4E$	See-saw
$XeF_4$	$AB_4E_2$	Square planar
$ClF_3$	$AB_3E_2$	$T$-shape

Matching the items:
$A$ $(BrF_5)$ matches $III$ ($AB_5E$,square pyramidal).
$B$ $(SF_4)$ matches $I$ ($AB_4E$,see-saw).
$C$ $(XeF_4)$ matches $II$ ($AB_4E_2$,square planar).
$D$ $(ClF_3)$ matches $IV$ ($AB_3E_2$,$T$-shape).
Therefore,the correct match is $A-III, B-I, C-II, D-IV$.

(B) Statement $(i)$ is correct: $ClF_3$ has $3$ bond pairs and $2$ lone pairs $(AB_3E_2)$,and $SO_2$ has $2$ bond pairs and $1$ lone pair $(AB_2E)$.
Statement $(ii)$ is correct: $SF_4$ has $4$ bond pairs and $1$ lone pair,resulting in a "See-saw" geometry.
Statement $(iii)$ is incorrect: $HgCl_2$ has $2$ bond pairs and $0$ lone pairs,resulting in a linear shape. $PbCl_2$ has $2$ bond pairs and $1$ lone pair,resulting in a bent shape.
Therefore,only statement $(iii)$ is incorrect.

(C) The molecular orbital configurations and bond orders $(BO)$ are calculated as follows:
$He_2^{+}: (\sigma 1s)^2 (\sigma^* 1s)^1$; $BO = \frac{2-1}{2} = 0.5$. It is paramagnetic due to one unpaired electron.
$Li_2^{+}: (\sigma 1s)^2 (\sigma^* 1s)^2 (\sigma 2s)^1$; $BO = \frac{3-2}{2} = 0.5$. It is paramagnetic.
$B_2: (\sigma 1s)^2 (\sigma^* 1s)^2 (\sigma 2s)^2 (\sigma^* 2s)^2 (\pi 2p_x)^1 (\pi 2p_y)^1$; $BO = \frac{6-4}{2} = 1.0$. It is paramagnetic due to two unpaired electrons.
$C_2: (\sigma 1s)^2 (\sigma^* 1s)^2 (\sigma 2s)^2 (\sigma^* 2s)^2 (\pi 2p_x)^2 (\pi 2p_y)^2$; $BO = \frac{8-4}{2} = 2.0$. It is diamagnetic.
Thus,the molecule with a bond order of $1.0$ and paramagnetic nature is $B_2$.

(A) $N_2$ ($14$ electrons): Bond order $= \frac{10-4}{2} = 3$.
$N_2^{+}$ ($13$ electrons): Bond order $= \frac{9-4}{2} = 2.5$.
Since bond order decreases,the bond length of $N-N$ increases.
$O_2$ ($16$ electrons): Bond order $= \frac{10-6}{2} = 2$.
$O_2^{+}$ ($15$ electrons): Bond order $= \frac{10-5}{2} = 2.5$.
Since bond order increases,the bond length of $O-O$ decreases.
Therefore,the changes are increases and decreases respectively.

(B) Based on Molecular Orbital Theory $(MOT)$:
$H_2$ $(1s^2)$: No unpaired electrons $\rightarrow$ Diamagnetic
$N_2$ $(KK(\sigma 2s)^2(\sigma^* 2s)^2(\pi 2p_x)^2(\pi 2p_y)^2(\sigma 2p_z)^2)$: No unpaired electrons $\rightarrow$ Diamagnetic
$O_2$ $(KK(\sigma 2s)^2(\sigma^* 2s)^2(\sigma 2p_z)^2(\pi 2p_x)^2(\pi 2p_y)^2(\pi^* 2p_x)^1(\pi^* 2p_y)^1)$: $2$ unpaired electrons $\rightarrow$ Paramagnetic
$N_2^+$: $1$ unpaired electron $\rightarrow$ Paramagnetic
$O_2^+$: $1$ unpaired electron $\rightarrow$ Paramagnetic
$O_2^-$: $1$ unpaired electron $\rightarrow$ Paramagnetic
$F_2$: No unpaired electrons $\rightarrow$ Diamagnetic
Thus,$O_2, O_2^+, O_2^-, N_2^+$ are all paramagnetic.

(B) The correct statement is $(b)$.
All these molecules $(C_2, N_2, O_2, F_2)$ are formed by the combination of atomic orbitals of the second period elements.
In each case,the total number of atomic orbitals involved is $10$ ($5$ from each atom),which results in the formation of $5$ bonding molecular orbitals $(BMOs)$ and $5$ antibonding molecular orbitals $(ABMOs)$ in the molecular orbital diagram (considering the $1s$ shell).
Therefore,they all possess the same number of bonding and antibonding molecular orbitals.

(B) $(I)$ Incorrect: In $NH_3$,the orbital dipole due to the lone pair and the net dipole moment of the three $N-H$ bonds are in the same direction,whereas in $NF_3$,they are in opposite directions. Thus,$\mu(NH_3) > \mu(NF_3)$. $BF_3$ is symmetrical and has zero dipole moment. The correct order is $NH_3 > NF_3 > BF_3$.
$(II)$ Correct: Bond length increases with the increase in the size of the atoms. Atomic radii follow the order $C > N > O > H$,so the bond length order $C-O > N-O > O-H$ is correct.
$(III)$ Correct: According to Molecular Orbital Theory,the bond order of $C_2$ is $2$,$B_2$ is $1$,and $He_2$ is $0$. The order $C_2 > B_2 > He_2$ is correct.

(B) The overall reaction is the sum of the two given equilibrium steps:
Step $1$: $Ag^{+} + NH_3 \rightleftharpoons [Ag(NH_3)]^{+}$,$K_1 = 1.6 \times 10^3$
Step $2$: $[Ag(NH_3)]^{+} + NH_3 \rightleftharpoons [Ag(NH_3)_2]^{+}$,$K_2 = 6.8 \times 10^3$
Adding these two equations gives the net reaction:
$Ag^{+} + 2 NH_3 \rightleftharpoons [Ag(NH_3)_2]^{+}$
The equilibrium constant $(K_{net})$ for the net reaction is the product of the equilibrium constants of the individual steps:
$K_{net} = K_1 \times K_2$
$K_{net} = (1.6 \times 10^3) \times (6.8 \times 10^3)$
$K_{net} = 10.88 \times 10^6 = 1.088 \times 10^7$

(A) Given reaction: $N_{2(g)} + O_{2(g)} \rightleftharpoons 2 NO_{(g)}$
Equilibrium constant,$K_C = 0.5625$
Equilibrium concentration,$[NO] = 3.0 \times 10^{-3} \ M$
The expression for equilibrium constant is $K_C = \frac{[NO]^2}{[N_2][O_2]}$
Since $[N_2] = [O_2]$,we can write $K_C = \frac{[NO]^2}{[N_2]^2}$
Substituting the values: $0.5625 = \frac{(3.0 \times 10^{-3})^2}{[N_2]^2}$
$[N_2]^2 = \frac{9.0 \times 10^{-6}}{0.5625} = 16 \times 10^{-6}$
$[N_2] = \sqrt{16 \times 10^{-6}} = 4.0 \times 10^{-3} \ M$
Therefore,the correct option is $(A)$.

(C) The equilibrium constant expression for the reaction $CO_{2(g)} + C_{(s)} \rightleftharpoons 2CO_{(g)}$ is given by:
$K_c = \frac{[CO]^2}{[CO_2]}$
Given $K_c = 0.036$ and $[CO_2] = 0.004 \ M$.
Substituting the values:
$0.036 = \frac{[CO]^2}{0.004}$
$[CO]^2 = 0.036 \times 0.004 = 0.000144$
$[CO] = \sqrt{0.000144} = 0.012 \ M = 1.2 \times 10^{-2} \ mol \ L^{-1}$.

(B) For the reaction,$CO_{2(g)} + C_{(s)} \rightleftharpoons 2 CO_{(g)}$
$K_p = \frac{(p_{CO})^2}{p_{CO_2}} = \frac{(0.6)^2}{0.15} = \frac{0.36}{0.15} = 2.4$
Using the relation $K_p = K_c(RT)^{\Delta n}$,where $\Delta n = 2 - 1 = 1$ and $R = 0.08314 \ L \ bar \ K^{-1} \ mol^{-1}$.
$K_c = \frac{K_p}{RT} = \frac{2.4}{0.08314 \times 1000} = \frac{2.4}{83.14} \approx 0.02887 \approx 2.89 \times 10^{-2}$.
Thus,the correct option is $(B)$.

(A) Given,$K_p$ for the conversion of oxygen to ozone at $400 \ K$ is $1.0 \times 10^{-30}$.
Standard Gibbs energy change $\Delta G^{\circ} = -RT \ln K_p = -2.303 RT \log_{10} K_p$.
Here,$R = 8.314 \ J \ K^{-1} \ mol^{-1}$ and $T = 400 \ K$.
Substituting the values:
$\Delta G^{\circ} = -2.303 \times 8.314 \times 400 \times \log_{10} (1.0 \times 10^{-30})$
$\Delta G^{\circ} = -2.303 \times 8.314 \times 400 \times (-30)$
$\Delta G^{\circ} = 229765.7 \ J \ mol^{-1}$
Converting to $kJ \ mol^{-1}$:
$\Delta G^{\circ} \approx 229.8 \ kJ \ mol^{-1}$.

(B) Initial pressure $(p)$ of $N_2O_4$ can be calculated using the ideal gas equation $pV = nRT$:
$p = \frac{nRT}{V} = \frac{(18.4 \ g / 92 \ g \ mol^{-1}) \times 0.083 \ L \ bar \ K^{-1} \ mol^{-1} \times 400 \ K}{1 \ L} = 0.2 \ mol \times 0.083 \times 400 = 6.64 \ bar$.
For the equilibrium reaction $N_2O_{4(g)} \rightleftharpoons 2NO_{2(g)}$:
Initial: $p$ atm,$0$
At equilibrium: $(p - p_i)$ atm,$2p_i$ atm
Total pressure $(p_T)$ at equilibrium:
$p_T = (p - p_i) + 2p_i = p + p_i$
$10.64 = 6.64 + p_i$
$p_i = 10.64 - 6.64 = 4.00 \ bar$.
Partial pressures at equilibrium:
$p_{N_2O_4} = p - p_i = 6.64 - 4.00 = 2.64 \ bar$
$p_{NO_2} = 2p_i = 2 \times 4.00 = 8.00 \ bar$
Equilibrium constant $K_p$:
$K_p = \frac{(p_{NO_2})^2}{p_{N_2O_4}} = \frac{(8.00)^2}{2.64} = \frac{64}{2.64} \approx 24.24$.

(C) For the reaction $N_{2(g)} + 3H_{2(g)} \longrightarrow 2NH_{3(g)}$,the rate expression is given by: $\text{Rate} = -\frac{d[N_2]}{dt} = -\frac{1}{3} \frac{d[H_2]}{dt} = \frac{1}{2} \frac{d[NH_3]}{dt}$.
Given that $-\frac{d[N_2]}{dt} = 0.02 \ mol \ L^{-1} \ s^{-1}$.
Equating the terms: $-\frac{d[N_2]}{dt} = -\frac{1}{3} \frac{d[H_2]}{dt}$.
Therefore,$-\frac{d[H_2]}{dt} = 3 \times (-\frac{d[N_2]}{dt}) = 3 \times 0.02 \ mol \ L^{-1} \ s^{-1} = 0.06 \ mol \ L^{-1} \ s^{-1}$.

(D) For the reaction $N_{2(g)} + 3H_{2(g)} \longrightarrow 2NH_{3(g)}$,the rate of reaction is given by the expression:
Rate $= -\frac{d[N_2]}{dt} = -\frac{1}{3} \frac{d[H_2]}{dt} = \frac{1}{2} \frac{d[NH_3]}{dt}$.
Equating the terms for $NH_3$ and $H_2$:
$-\frac{1}{3} \frac{d[H_2]}{dt} = \frac{1}{2} \frac{d[NH_3]}{dt}$.
Multiplying both sides by $6$,we get:
$-2 \frac{d[H_2]}{dt} = 3 \frac{d[NH_3]}{dt}$.
Rearranging gives $3 \frac{d[NH_3]}{dt} = -2 \frac{d[H_2]}{dt}$.

(B) The correct matching is as follows:
$A$. Insecticide $\rightarrow III$. $Na_3AsO_3$ (Sodium arsenite is used as an insecticide).
$B$. $K_2Cr_2O_7 / 50\% H_2SO_4$ $\rightarrow I$. $COD$ (This mixture is used to determine the Chemical Oxygen Demand).
$C$. Bleaching of clothes and paper $\rightarrow V$. $H_2O_2$ (Hydrogen peroxide is a common bleaching agent).
$D$. Eye irritant $\rightarrow II$. $PAN$ (Peroxyacetyl nitrate is a well-known eye irritant in photochemical smog).
Thus,the correct sequence is $A-III, B-I, C-V, D-II$. The correct option is $B$.

(C) Dry ice is solid $CO_2$.
It is widely used as a refrigerant for ice-cream and frozen foods because it sublimes at $-78.5 \ ^{\circ}C$ at atmospheric pressure,providing intense cooling without leaving any liquid residue.

(D) Acidic oxides $\rightarrow Mn_2O_7, CrO_3, V_2O_5$ (due to higher oxidation state of the central metal).
Basic oxides $\rightarrow Na_2O, CaO, BaO$ (due to lower oxidation state of the central metal).
Alkali and alkaline earth metal oxides are generally basic.
Neutral oxides $\rightarrow CO, NO, N_2O$.
Amphoteric oxides $\rightarrow ZnO, PbO, BeO$ (exhibit both acidic and basic properties).
Hence,option $(D)$ is correct.

(C) $I$. The tendency to form halide hydrates decreases down the group as the size of the cation increases,which reduces the hydration energy.
$II$. As we move down the group,the size of the alkali metal cation increases,which stabilizes the large superoxide ion $(O_2^-)$ more effectively due to reduced polarizing power.
$III$. $LiF$ has a very high lattice energy due to the small size of both $Li^+$ and $F^-$ ions,which outweighs its hydration energy,leading to low solubility.
$IV$. The solubility of group-$2$ carbonates decreases down the group because the lattice energy decreases less rapidly than the hydration energy.
Therefore,statements $II$ and $III$ are correct.

(A) On moving down the group,the electron gain enthalpy generally decreases.
Oxygen,due to its small size,experiences significant electron-electron repulsions in its relatively small $2p$-subshell.
Consequently,an incoming electron is not accepted as easily as in the case of other elements in the same group,which is why oxygen has a lower electron gain enthalpy than sulfur.
Therefore,the correct order of electron gain enthalpy for the group $16$ elements is $S > Se > Te > O$.

(B) $I$ and $II$ are correct: The properties of elements are periodic functions of their atomic numbers,and non-metals are fewer in number compared to metallic elements.
$III$ is correct: In the periodic table,the first ionisation energy of elements along a period does not vary in a regular manner due to variations in effective nuclear charge and shielding effects.
$IV$ is incorrect: The ground state electronic configuration of $Pd (Z=46)$ is $[Kr] 4d^{10} 5s^0$.
Therefore,statements $I, II,$ and $III$ are correct.

(C) The process of electron gain is represented as: $Cl_{(g)} + e^{-} \xrightarrow{\Delta_{eg}H} Cl^{-}_{(g)}$
Energy of the product $Cl^{-}_{(g)}$ is equal to the sum of the energy of the reactant $Cl_{(g)}$ and the enthalpy change $\Delta_{eg}H$ of the reaction.
$\text{Energy of } Cl^{-}_{(g)} = \text{Energy of } Cl_{(g)} + \Delta_{eg}H$
Given that the energy of $Cl_{(g)}$ is $x \ kJ \ mol^{-1}$ and $\Delta_{eg}H = -349 \ kJ \ mol^{-1}$.
$\text{Energy of } Cl^{-}_{(g)} = x + (-349) = x - 349 \ kJ \ mol^{-1}$.

(B) In Lassaigne's test,the organic compound is fused with sodium metal to convert nitrogen into sodium cyanide $(NaCN)$.
$Na + C + N \rightarrow NaCN$
When the extract is treated with ferrous sulphate $(FeSO_4)$,sodium ferrocyanide is formed:
$6NaCN + FeSO_4 \rightarrow Na_4[Fe(CN)_6] + Na_2SO_4$
Some $Fe^{2+}$ ions are oxidized to $Fe^{3+}$ ions by the atmosphere or concentrated $H_2SO_4$ added during acidification.
These $Fe^{3+}$ ions react with the ferrocyanide ions to form ferric ferrocyanide,which is Prussian blue in colour:
$4Fe^{3+} + 3[Fe(CN)_6]^{4-} \rightarrow Fe_4[Fe(CN)_6]_3$
Thus,the complex responsible for the Prussian blue colour is $Fe_4[Fe(CN)_6]_3$.

(C) For a balanced Wheatstone bridge,the condition is $\frac{A}{B} = \frac{C}{D}$.
Case $1$: When $C = 100 \Omega$,the bridge is balanced,so $\frac{A}{B} = \frac{100}{D}$ ... $(i)$
Case $2$: When $A$ and $B$ are interchanged,the new resistances in the arms are $B$ and $A$ respectively. The bridge balances for $C = 121 \Omega$,so $\frac{B}{A} = \frac{121}{D}$ ... (ii)
From equation $(i)$,we have $\frac{A}{B} = \frac{100}{D}$. Therefore,$\frac{B}{A} = \frac{D}{100}$.
Substituting this into equation (ii),we get $\frac{D}{100} = \frac{121}{D}$.
$D^2 = 100 \times 121 = 12100$.
$D = \sqrt{12100} = 110 \Omega$.

(C) For a balanced Wheatstone bridge,the ratio of resistances is given by $\frac{A}{B} = \frac{C}{D}$.
Let $D = x$. Initially,$\frac{A}{B} = \frac{100}{x}$.
When $A$ and $B$ are interchanged,the new condition is $\frac{B}{A} = \frac{121}{x}$.
From the first equation,$\frac{B}{A} = \frac{x}{100}$.
Equating the two expressions for $\frac{B}{A}$,we get $\frac{x}{100} = \frac{121}{x}$.
$x^2 = 121 \times 100 = 12100$.
$x = \sqrt{12100} = 110 \ \Omega$.
Thus,the value of $D$ is $110 \ \Omega$.

(A) The reaction $2 CuSO_{4(aq)} + 4 KI_{(aq)} \longrightarrow 2 CuI_2 + 2 K_2SO_4$ does not occur as written because $CuI_2$ is unstable and spontaneously decomposes into $Cu_2I_2$ and $I_2$.
The correct reaction is: $2 CuSO_{4(aq)} + 4 KI_{(aq)} \longrightarrow Cu_2I_2(s) + 2 K_2SO_4(aq) + I_2(s)$.

(C) Gypsum $(CaSO_4 \cdot 2H_2O)$ is a significant non-biodegradable solid waste byproduct generated during the production of phosphate fertilizers.

(D) In environmental chemistry,Chemical Oxygen Demand $(COD)$ is the total amount of oxygen (in $ppm$) required to chemically oxidize all organic and inorganic pollutants in a given water sample. It is not a measure of bacteria.
Thus,option $(D)$ is incorrect.
- $NO_2$ is known to be a lung irritant.
- Municipal sewage typically has a $BOD$ (Biological Oxygen Demand) value ranging from $100-4000 \ ppm$.
- The primary source of $CO$ (carbon monoxide) in the atmosphere is automobile exhaust fumes.

(A) Acid rain is primarily caused by the emission of sulphur dioxide $(SO_2)$ and nitrogen oxides $(NO_x)$,specifically nitrogen dioxide $(NO_2)$.
These gases react with water vapor,oxygen,and other chemicals in the atmosphere to form sulphuric acid $(H_2SO_4)$ and nitric acid $(HNO_3)$.
These acids then fall to the ground as acid rain.

(D) Statement $(i)$ is correct because $CF_2Cl_2$ (Freon-$12$) undergoes photolysis in the presence of $UV$ light to produce chlorine free radicals,which catalyze the depletion of ozone $(O_3)$ to oxygen $(O_2)$.
Statement $(ii)$ is incorrect because fluoride concentration in drinking water should be around $1 \ ppm$. Excess fluoride (e.g.,$10 \ ppm$) causes harmful effects like mottled teeth and skeletal fluorosis.
Statement $(iii)$ is correct as the maximum permissible concentration of lead in drinking water is indeed $50 \ ppb$ $(0.05 \ ppm)$.

It is given that $(N + 1)$ divisions on vernier scale $(VSD) = N$ divisions on main scale $(MSD)$.
$\Rightarrow 1 \ VSD = \frac{N}{N + 1} \ MSD$
Least count of vernier callipers $(LC) = 1 \ MSD - 1 \ VSD$
$\Rightarrow LC = a - a \left( \frac{N}{N + 1} \right) = \frac{a}{N + 1} \ unit$

(D) $1$. In cyclic compounds containing a double bond,the double bond is given priority in numbering.
$2$. The carbon atoms of the double bond are assigned positions $1$ and $2$ such that the substituent gets the lowest possible locant.
$3$. Starting from the double bond,if we number in the direction that gives the methoxy group position $4$,we get $4-$methoxycyclohexene.
$4$. Therefore,the correct $IUPAC$ name is $4-$methoxycyclohexene.

(D) Key Idea: The number of no-bond resonance (hyperconjugation) structures is equal to the number of $\alpha$-hydrogen atoms present on the carbon atom directly attached to the benzene ring.
$(a)$ tert-Butylbenzene: The $\alpha$-carbon has $0$ $H$-atoms. Number of hyperconjugation structures = $0$.
$(b)$ Ethylbenzene: The $\alpha$-carbon $(CH_2)$ has $2$ $H$-atoms. Number of hyperconjugation structures = $2$.
$(c)$ Isopropylbenzene: The $\alpha$-carbon $(CH)$ has $1$ $H$-atom. Number of hyperconjugation structures = $1$.
$(d)$ Toluene: The $\alpha$-carbon $(CH_3)$ has $3$ $H$-atoms. Number of hyperconjugation structures = $3$.
Since toluene has the maximum number of $\alpha$-hydrogen atoms $(3)$,it has the maximum number of no-bond resonance structures. Hence,option $(d)$ is the correct answer.

(D) In structure $(A)$,the curved arrow shows the movement of a lone pair of electrons from the oxygen atom to the adjacent bond position to form a $\pi$ bond between oxygen and carbon.
In structure $(B)$,the curved arrow shows the movement of electrons from the $\pi$ bond between the carbon and nitrogen atoms to the adjacent nitrogen atom.

(A) $(I)$ is a neutral structure with a complete octet for all atoms,making it the most stable resonance structure.
Between $(II)$ and $(III)$,structure $(II)$ places a negative charge on the more electronegative oxygen atom,whereas structure $(III)$ places a positive charge on the oxygen atom.
Since a more electronegative atom is more stable with a negative charge,$(II)$ is more stable than $(III)$.
Therefore,the stability order is $III < II < I$.
Hence,option $(A)$ is correct.

(A) The pressure of dry $N_2$ gas is $P_{dry} = P_{total} - P_{aqueous} = 740 - 15 = 725 \ mm \ Hg$.
Using the ideal gas equation at $STP$ $(P_2 = 760 \ mm \ Hg, T_2 = 273 \ K)$:
$\frac{P_1 V_1}{T_1} = \frac{P_2 V_2}{T_2} \Rightarrow \frac{725 \times 50}{300} = \frac{760 \times V_2}{273}$.
$V_2 = \frac{725 \times 50 \times 273}{300 \times 760} \approx 43.47 \ mL$.
Since $22400 \ mL$ of $N_2$ at $STP$ weighs $28 \ g$,the mass of $N_2$ is:
$\text{Mass} = \frac{28 \times 43.47}{22400} \approx 0.0543 \ g$.
Percentage of $N_2 = \frac{\text{Mass of } N_2}{\text{Mass of compound}} \times 100 = \frac{0.0543}{1} \times 100 = 5.43 \%$.
Rounding to the nearest provided option,$5.42 \%$ is the correct value.

(A) Key Idea: Kjeldahl's method is not applicable to compounds containing nitrogen in nitro $(-NO_2)$,azo $(-N=N-)$ groups,or nitrogen present in a heterocyclic ring (like pyridine).
$I$. Aniline $(C_6H_5NH_2)$: Nitrogen is in the amino group $(-NH_2)$,which can be estimated by Kjeldahl's method.
$II$. Azobenzene $(C_6H_5-N=N-C_6H_5)$: Contains an azo $(-N=N-)$ group,so it cannot be estimated.
$III$. Nitrobenzene $(C_6H_5NO_2)$: Contains a nitro $(-NO_2)$ group,so it cannot be estimated.
$IV$. Pyridine $(C_5H_5N)$: Nitrogen is part of a heterocyclic aromatic ring,so it cannot be estimated.
Therefore,compounds $II, III,$ and $IV$ cannot be estimated by Kjeldahl's method.
Thus,option $(A)$ is the correct answer.

(A) In an $S_N2$ reaction,the nucleophile attacks from the side opposite to the leaving group,resulting in a Walden inversion (inversion of configuration). Thus,$X$ is the inverted alcohol.
In the reaction of a primary alcohol with $Conc. \ HCl$,the reaction typically proceeds via an $S_N2$ mechanism (or $S_N1$ depending on the substrate,but here the primary alcohol structure is given). However,looking at the options and the standard transformation of a primary alcohol to a primary alkyl chloride,the reaction involves the substitution of $-OH$ by $-Cl$. Since the chiral center is not involved in the bond-breaking/forming step for the primary alcohol,the configuration at the chiral center remains unchanged. Thus,$Y$ is the product with retained configuration.

(C) Kolbe electrolysis of $CH_3COONa$ produces $C_2H_6$ and $CO_2$ at the anode.
$2CH_3COONa + 2H_2O \xrightarrow{\text{electrolysis}} CH_3-CH_3 (A) + 2CO_2 (B) + H_2 + 2NaOH$
When $B$ $(CO_2)$ is heated with $O_2$ in the presence of $(CH_3COO)_2Mn$,it is not the standard reaction; however,the question implies the oxidation of $CH_3-CH_3$ $(A)$. The combustion of ethane is:
$2C_2H_6 + 7O_2 \xrightarrow{(CH_3COO)_2Mn} 4CO_2 (C) + 6H_2O (D)$
$C$ $(CO_2)$ reacts with $NaOH$ to form sodium carbonate $(Na_2CO_3)$ or sodium bicarbonate $(NaHCO_3)$,which are salts.
Thus,$A$ is $C_2H_6$ and $D$ is $H_2O$.

(D) The reaction sequence is as follows:
$1$. $C_2H_4 + HBr \rightarrow CH_3CH_2Br$ $(W)$
$2$. $W + Mg / \text{dry ether} \rightarrow CH_3CH_2MgBr$
$3$. $CH_3CH_2MgBr + CO_2 \xrightarrow{H_3O^+} CH_3CH_2COOH$ $(X)$
$4$. $CH_3CH_2COOH$ undergoes Kolbe's electrolysis $(Y)$ to form $CH_3CH_2CH_2CH_3$ $(Z)$.
$5$. $W + Na / \text{dry ether}$ (Wurtz reaction) also forms $CH_3CH_2CH_2CH_3$ $(Z)$.
Thus,$Y$ is electrolysis and $Z$ is $CH_3CH_2CH_2CH_3$.

(C) Reductive ozonolysis of $Hex-2-ene$ $(CH_3-CH_2-CH_2-CH=CH-CH_3)$ involves the addition of ozone across the double bond followed by cleavage with $Zn/H_2O$ to produce two carbonyl compounds.
Cleavage of the $C=C$ bond in $Hex-2-ene$ occurs at the $2^{nd}$ position.
The reaction is: $CH_3-CH_2-CH_2-CH=CH-CH_3 \xrightarrow{O_3, Zn/H_2O} CH_3-CH_2-CH_2-CHO + CH_3-CHO$.
Thus,$X = CH_3-CH_2-CH_2-CHO$ $(Butanal)$ and $Y = CH_3-CHO$ $(Ethanal)$.

(D) $LiAlH_4$ acts as a source of hydride ion $(H^-)$,which attacks the electrophilic carbonyl carbon of cyclohexanone to form an alkoxide intermediate.
In the second step,$D_2O$ acts as a source of deuterium $(D^+)$,which protonates (deuterates) the alkoxide oxygen to form the final product,which is a cyclohexanol derivative with an $-OD$ group and a hydrogen atom attached to the alpha carbon.

(C) $1$. The starting material is propan$-1-$ol $(CH_3CH_2CH_2OH)$.
$2$. Oxidation with $CrO_3/H_2SO_4$ (Jones reagent) converts the primary alcohol to propanoic acid $(CH_3CH_2COOH)$.
$3$. Treatment with $SOCl_2$ converts the carboxylic acid to propanoyl chloride $(CH_3CH_2COCl)$,which is $X$.
$4$. Reaction of $X$ with dimethylcadmium $((CH_3)_2Cd)$ is a standard method to prepare ketones from acid chlorides,yielding butan$-2-$one $(CH_3CH_2COCH_3)$. Thus,$Y = (CH_3)_2Cd$.
$5$. The final step is the protection of the ketone as a cyclic acetal using ethylene glycol $(HOCH_2CH_2OH)$ in the presence of dry $HCl$ gas. Thus,$Z = HOCH_2CH_2OH/HCl \text{ (gas)}$.
$6$. Therefore,option $C$ is correct.

(A) The dehydration of alcohols depends on the degree of substitution of the alcohol:
$1$. Primary $(1^{\circ})$ alcohols like ethanol undergo dehydration using concentrated $H_2SO_4$ at $443 \ K$.
$2$. Secondary $(2^{\circ})$ alcohols like propan$-2-$ol undergo dehydration using $85 \% \ H_3PO_4$ at $440 \ K$.
$3$. Tertiary $(3^{\circ})$ alcohols like tert-butyl alcohol undergo dehydration using $20 \% \ H_3PO_4$ at $358 \ K$.
Therefore,$X = H_2SO_4, 443 \ K$; $Y = 85 \% \ H_3PO_4, 440 \ K$; $Z = 20 \% \ H_3PO_4, 358 \ K$.

(A) $C_6H_7N$ corresponds to aniline $(A)$.
Aniline undergoes diazotisation reaction with $NaNO_2 / HCl$ at $273-278 \ K$ to form benzene diazonium chloride,which on warming with water produces phenol $(B)$.
Phenol reacts with concentrated nitric acid $(HNO_3)$ to undergo electrophilic aromatic substitution,yielding $2,4,6-$trinitrophenol $(C)$,which is commonly known as picric acid.

(C) The reaction sequence is as follows:
$1$. $p-chloronitrobenzene$ reacts with $NaOH$ at $443 \ K$ to form $p-nitrophenol$ $(X)$.
$2$. $p-nitrophenol$ is reduced using $Sn + HCl$ to form $p-aminophenol$ $(Y)$.
$3$. $p-aminophenol$ reacts with $NaNO_2 / HCl$ at $0-5^{\circ}C$ to form a diazonium salt intermediate.
$4$. Hydrolysis of this diazonium salt with $H_3O^{+}$ at $10^{\circ}C$ replaces the diazonium group with an $-OH$ group,resulting in the formation of $hydroquinone$ $(Z)$.

(C) The given reaction is the industrial preparation of phenol from cumene (isopropylbenzene).
$1$. Cumene reacts with oxygen $(O_2)$ to form cumene hydroperoxide $(A)$:
$C_6H_5-CH(CH_3)_2 + O_2 \rightarrow C_6H_5-C(CH_3)_2-O-OH$
$2$. Cumene hydroperoxide $(A)$ on treatment with dilute acid $(H_3O^+)$ undergoes rearrangement to produce phenol $(B)$ and acetone $(C)$:
$C_6H_5-C(CH_3)_2-O-OH \xrightarrow{H_3O^+} C_6H_5OH + (CH_3)_2CO$
Thus,$A$ is cumene hydroperoxide,$B$ is phenol,and $C$ is acetone. The correct option is $C$.

(B) The reaction involves a tertiary alkyl halide,$(CH_3)_3C-Cl$,and a strong base,$C_2H_5ONa$ (sodium ethoxide).
Since the alkyl halide is sterically hindered (tertiary),the $S_N2$ pathway is unfavorable.
Instead,the base acts as a proton acceptor,leading to an $E2$ elimination reaction.
The base abstracts a proton from one of the $\beta$-carbon atoms,resulting in the formation of an alkene.
The major product is $2$-methylpropene,which is $(CH_3)_2C=CH_2$.

(C) The first reaction is: $(CH_3)_3 C-O^- Na^+ + CH_3 CH_2 Br \longrightarrow (CH_3)_3 C-OCH_2 CH_3 (X) + NaBr$.
This is a $S_N2$ reaction where a primary halide reacts with a bulky alkoxide to form an ether.
The second reaction is: $(CH_3)_3 C-Br + CH_3 CH_2 O^- Na^+ \longrightarrow CH_3-C(CH_3)=CH_2 (Y) + CH_3 CH_2 OH (Z) + NaBr$.
This is an elimination reaction $(E2)$ where a tertiary alkyl halide reacts with a strong base (alkoxide) to form an alkene and an alcohol.

(A) $(i)$ The reaction of an ether with $HI$ produces an alcohol and an alkyl iodide. In the presence of a $3^{\circ}$ carbon atom in one of the alkyl groups of the ether,the iodide ion $(-I)$ attacks the more substituted carbon atom (the $3^{\circ}$ carbon) via the $S_N1$ mechanism.
$(ii)$ This occurs because the $3^{\circ}$ carbocation formed during the reaction is highly stable compared to $1^{\circ}$ or $2^{\circ}$ carbocations,and the $S_N2$ mechanism is hindered by significant steric hindrance at the $3^{\circ}$ carbon.
$(iii)$ In the given reaction,the ether is $2-ethoxy-2-methylbutane$. The $HI$ cleaves the bond between the oxygen and the $3^{\circ}$ carbon,leading to the formation of ethanol $(CH_3CH_2OH)$ and $2-iodo-2-methylbutane$ $((CH_3)_2(CH_3CH_2)C-I)$ via the $S_N1$ mechanism.
Thus,option $A$ is the correct answer.

(C) When an aldehyde $(R-CHO)$ reacts with Tollen's reagent,the following reaction takes place:
$R-CHO + 2[Ag(NH_3)_2]^+ + 3OH^- \longrightarrow R-COO^- + 2Ag + 4NH_3 + 2H_2O$
Thus,the products formed are $Ag, H_2O, R-COO^-$ and $NH_3$,and option $(C)$ is the correct answer.

(D) The reaction sequence is as follows:
$1$. Reductive ozonolysis of ethene $(CH_2=CH_2)$ gives formaldehyde $(HCHO)$ as product $A$.
$CH_2=CH_2 \xrightarrow[2. Zn/H_2O]{1. O_3} 2HCHO (A)$
$2$. Formaldehyde $(HCHO)$ lacks $\alpha$-hydrogen atoms,so it undergoes a self-oxidation-reduction reaction (disproportionation) in the presence of concentrated $NaOH$,known as the Cannizzaro reaction.
$2HCHO + NaOH (conc.) \rightarrow CH_3OH (B) + HCOONa (C)$
Here,$B$ is methanol (alcohol) and $C$ is sodium formate (sodium salt of carboxylic acid).
Therefore,the reaction of $A$ to give $B$ and $C$ is an example of the Cannizzaro reaction.

(A) When ethanal $(CH_3CHO)$ and propanone $(CH_3COCH_3)$ undergo crossed-aldol condensation,the nucleophilic enolate of propanone attacks the carbonyl carbon of ethanal.
$1$. The initial addition product is $4$-hydroxypentan-$2$-one $(CH_3-CH(OH)-CH_2-COCH_3)$.
$2$. Upon dehydration (heating),this yields the unsaturated product,pent-$3$-en-$2$-one $(CH_3-CH=CH-COCH_3)$.

(A) Acidic character $\propto$ stability of conjugate base (phenoxide ion).
The stability of the phenoxide ion is increased by electron-withdrawing groups ($-M$ and $-I$ effects) and decreased by electron-donating groups ($+M$,$+H$,and $+I$ effects).
The substituents are:
$(I)$ $p$-methylphenol: $-CH_3$ group shows $+H$ and $+I$ effects,which destabilize the phenoxide ion.
$(II)$ $m$-nitrophenol: $-NO_2$ group shows only $-I$ effect ($-M$ effect does not operate at $m$-position).
$(III)$ $p$-nitrophenol: $-NO_2$ group shows both $-M$ and $-I$ effects,providing strong stabilization.
$(IV)$ Phenol: No substituent.
Comparing these,the order of stability of the conjugate base is: $p$-methylphenoxide < phenoxide < $m$-nitrophenoxide < $p$-nitrophenoxide.
Therefore,the increasing order of acidic character is: $I < IV < II < III$.

(A) The reaction is a Claisen-Schmidt condensation,which is a type of crossed aldol condensation between benzaldehyde (which has no $\alpha$-hydrogens) and propanal (which has $\alpha$-hydrogens).
In the presence of dilute $NaOH$,the $\alpha$-carbon of propanal acts as a nucleophile and attacks the carbonyl carbon of benzaldehyde to form an aldol intermediate.
Subsequent dehydration (loss of $H_2O$) under heating $(\Delta)$ yields the $\alpha,\beta$-unsaturated aldehyde,$C_6H_5CH=C(CH_3)CHO$.
Thus,the correct product is $C_6H_5CH=C(CH_3)CHO$.

(C) When benzaldehyde is heated with concentrated $NaOH$,it undergoes the Cannizzaro reaction.
In this reaction,aldehydes that do not contain an $\alpha$-hydrogen atom undergo self-oxidation and reduction (disproportionation).
Benzaldehyde $(C_6H_5CHO)$ lacks an $\alpha$-hydrogen,so it is reduced to benzyl alcohol $(C_6H_5CH_2OH)$ and oxidized to sodium benzoate $(C_6H_5COO^-Na^+)$.
The overall reaction is:
$2C_6H_5CHO + NaOH \xrightarrow{\Delta} C_6H_5CH_2OH + C_6H_5COO^-Na^+$

(A) $NaOCl$ acts as an oxidizing agent and is used in the haloform reaction. Compounds that undergo the haloform reaction with $NaOCl$ are:
$1$. Secondary alcohols with a methyl group attached to the carbinol carbon $(RCH(OH)CH_3)$.
$2$. Methyl ketones $(R-COCH_3)$.
$3$. Acetaldehyde $(CH_3CHO)$.
$4$. Compounds containing a $CH_3CO-$ group,such as $V$ $( (CH_3)_2C=C(CH_3)COCH_3 )$.
Analyzing the given compounds:
$I. RCH(OH)CH_3$ is a $2^{\circ}$ alcohol with a $CH_3$ group,so it reacts.
$II. RCH_2CH_2-CO-CH_2CH_3$ is a ketone but lacks a $CH_3$ group attached to the carbonyl carbon,so it does not react.
$III. R-COCH_3$ is a methyl ketone,so it reacts.
$IV. CH_3CHO$ is acetaldehyde,which reacts.
$V. (CH_3)_2C=C(CH_3)COCH_3$ contains a $CH_3CO-$ group,so it reacts.
Therefore,$I, III, IV,$ and $V$ are oxidized by $NaOCl$.

(C) $1$. Aniline reacts with $Br_2/H_2O$ to form $2,4,6$-tribromoaniline $(X)$.
$2$. $2,4,6$-tribromoaniline reacts with $NaNO_2/HCl$ at $273-278 \ K$ to form the diazonium salt,$2,4,6$-tribromobenzenediazonium chloride $(Y)$.
$3$. The diazonium salt $(Y)$ reacts with ethanol $(C_2H_5OH)$ to undergo reduction,where the diazonium group is replaced by a hydrogen atom,yielding $1,3,5$-tribromobenzene $(Z)$.

(A) $(i)$ Aniline is highly reactive towards electrophilic substitution. When treated with $Br_2$ in $H_2O$,it undergoes rapid poly-substitution to form $2, 4, 6-$tribromoaniline as the major product $(X)$.
$(ii)$ To control the reactivity of the $-NH_2$ group,it is first acetylated using acetic anhydride $(CH_3CO)_2O$ to form acetanilide. The $-NHCOCH_3$ group is less activating than the $-NH_2$ group,which restricts the bromination to the $p-$position,yielding $p-$bromoacetanilide as the major product $(Y)$.

(B) $(X) \longrightarrow$ Sucrose on hydrolysis gives glucose and fructose.
Sucrose $\longrightarrow$ glucose + fructose
$(Y) \longrightarrow$ Lactose on hydrolysis gives $D$-glucose and $D$-galactose.
Lactose $\longrightarrow$ $D$-glucose + $D$-galactose
$(Z) \longrightarrow$ Maltose on hydrolysis gives $2$ units of glucose.
Maltose $\longrightarrow$ glucose + glucose
Thus,the correct option is $(B)$.

(C) The explanations of the given statements are as follows:
$(A)$ $A$ tripeptide contains $3$ amino acids linked by $2$ peptide bonds. Thus,statement $(A)$ is correct.
$(B)$ $A$ pentapeptide contains $5$ amino acids linked by $4$ peptide bonds. Thus,statement $(B)$ is correct.
$(C)$ $A$ nucleoside is a product of a base and a sugar. $A$ nucleotide is a product of a base,a sugar,and a phosphate group. Thus,statement $(C)$ is incorrect.
$(D)$ In cellulose,$\beta-D$-glucose units are joined by $\beta$-glycosidic linkages between $C-1$ of one glucose and $C-4$ of the next glucose unit. Thus,statement $(D)$ is correct.
Therefore,the correct statements are $(A)$,$(B)$,and $(D)$. Hence,option $(C)$ is correct.

(A) The general structure of an $\alpha$-amino acid is $NH_2-CH(R)-COOH$.
In tryptophan,the $R$ group is an indolyl-methyl group (specifically,a $3$-indolylmethyl group).
In histidine,the $R$ group is an imidazolyl-methyl group (specifically,a $4$-imidazolylmethyl group).
Therefore,the correct representation for $R$ in tryptophan $(X)$ and histidine $(Y)$ is the indolyl-methyl group and the imidazolyl-methyl group respectively.

(C) Both $Pepsin$ and $Trypsin$ are proteolytic enzymes that break down proteins into amino acids. $Pepsin$ acts in the stomach,while $Trypsin$ acts in the small intestine. Given the options,$Trypsin$ is a standard answer for pancreatic protein digestion. $Pepsin$ is also a correct answer in biological contexts. However,in many academic contexts,$Trypsin$ is highlighted for its role in the final breakdown of proteins into amino acids in the small intestine.

(C) $I$. $DNA$ has a double helix structure.
$II$. Adenine forms hydrogen bonds with thymine $(A=T)$ and cytosine forms hydrogen bonds with guanine $(C \equiv G)$.
$III$. The two strands in $DNA$ molecules are complementary to each other.
$IV$. $DNA$ strand contains pentose $2$-deoxyribose sugar.
Thus,option $(C)$ is incorrect.

(D) $I)$ According to Chargaff's rule,in $DNA$,the amount of adenine equals thymine $(A = T)$. This is correct.
$II)$ In $RNA$,adenine pairs with uracil,but there is no rule stating their amounts must be equal. This is incorrect.
$III)$ Amylose is a linear polymer of $\alpha-D$-glucose units linked by $\alpha-1 \rightarrow 4$ glycosidic linkages. Amylopectin is the branched polymer. This is incorrect.
$IV)$ Addison disease is caused by the hypofunctioning of the adrenal cortex,leading to a deficiency of glucocorticoids and mineralocorticoids. This is correct.
Therefore,statements $I$ and $IV$ are correct.

(A) Let us analyze the given sets of vitamins,their sources,and deficiency diseases:
$(a)$ Vitamin $B_6$ (pyridoxine) is found in milk,cereals,and fish. Its deficiency can lead to convulsions. This is a correct set.
$(b)$ Vitamin $K$ is found in green leafy vegetables. Its deficiency leads to increased blood clotting time,not anaemia. This is an incorrect set.
$(c)$ Vitamin $C$ (ascorbic acid) is found in citrus fruits and green leafy vegetables,not primarily in fish. Its deficiency causes scurvy. This is an incorrect set.
$(d)$ Vitamin $D$ is obtained from exposure to sunlight and egg yolk,not citrus fruits. Its deficiency causes rickets. This is an incorrect set.
Therefore,the correct set is $(a)$.

(B) $1$. Acetic acid $(CH_3COOH)$ reacts with $NH_3$ to form ammonium acetate,which upon heating undergoes dehydration to form acetamide $(A)$: $CH_3COOH + NH_3$ $\rightarrow CH_3COONH_4$ $\xrightarrow{\Delta} CH_3CONH_2 (A) + H_2O$.
$2$. Acetamide $(A)$ is reduced by $LiAlH_4$ to form ethylamine $(B)$: $CH_3CONH_2 \xrightarrow{LiAlH_4} CH_3CH_2NH_2 (B)$.
$3$. Ethylamine $(B)$ reacts with chloroform $(CHCl_3)$ in the presence of $KOH$ (Carbylamine reaction) to form ethyl isocyanide $(C)$: $CH_3CH_2NH_2 + CHCl_3 + 3KOH \rightarrow CH_3CH_2NC (C) + 3KCl + 3H_2O$.
$4$. Therefore,$B$ is $CH_3CH_2NH_2$ and $C$ is $CH_3CH_2NC$.

(D) $1$. $CH_3COOH + SOCl_2 \rightarrow CH_3COCl$ (Compound $A$ is acetyl chloride).
$2$. $CH_3COCl + C_6H_6 \xrightarrow{AlCl_3} C_6H_5COCH_3$ (Friedel-Crafts acylation,Compound $B$ is acetophenone).
$3$. $C_6H_5COCH_3 + HCN \rightarrow C_6H_5C(OH)(CH_3)CN$ (Nucleophilic addition,Compound $C$ is acetophenone cyanohydrin).
$4$. $C_6H_5C(OH)(CH_3)CN + H_2O \rightarrow C_6H_5C(OH)(CH_3)COOH$ (Hydrolysis of nitrile to carboxylic acid,Compound $D$ is $2-$hydroxy$-2-$phenylpropanoic acid).

(D) The mechanism of esterification (Fischer esterification) involves the following steps:
$1$. Protonation of the carbonyl oxygen of the carboxylic acid: The carboxylic acid $(R-COOH)$ reacts with $H^+$ to form a protonated species,which is represented by structure $4$.
$2$. Nucleophilic attack by the alcohol: The alcohol $(R'-OH)$ attacks the electrophilic carbonyl carbon of the protonated carboxylic acid,leading to the formation of the tetrahedral intermediate,which is represented by structure $2$.
$3$. Proton transfer: $A$ proton is transferred from the oxygen of the alcohol group to one of the hydroxyl groups,resulting in the formation of an oxonium ion,represented by structure $1$.
$4$. Elimination of water and deprotonation: The water molecule is eliminated,and the loss of a proton leads to the formation of the protonated ester,which is represented by structure $3$.
Thus,the correct sequence is $4$ $\rightarrow 2$ $\rightarrow 1$ $\rightarrow 3$.

(A) Key Idea: The Fries rearrangement is the rearrangement of a phenolic ester to a hydroxy aryl ketone in the presence of a Lewis acid catalyst.
In this reaction,the acyl group $(RCO-)$ migrates from the oxygen atom of the ester to the ortho or para position of the benzene ring.
Thus,it represents the conversion of an $O$-acylated phenol (phenolic ester) to a $C$-acylated phenol (hydroxy aryl ketone).
Therefore,option $(A)$ is correct.
The reaction is as follows:
$C_6H_5OCOR \xrightarrow{AlCl_3} o-HOC_6H_4COR + p-HOC_6H_4COR$

(A) In alcohols,the oxygen atom is $sp^3$ hybridized.
Due to the presence of two lone pairs on the oxygen atom,the bond angle is slightly less than the ideal tetrahedral angle of $109.5^{\circ}$ $(109^{\circ} 28')$.
Therefore,the bond angle in alcohols is close to $109^{\circ}$.

(A) The overall stability constant $\beta_n$ for a complex formation reaction is the product of the stepwise stability constants $K_1, K_2, ..., K_n$.
Given the stepwise constants:
$K_1 = 1.0 \times 10^4$
$K_2 = 1.0 \times 10^3$
$K_3 = 1.0 \times 10^3$
$K_4 = 1.0 \times 10^2$
The overall stability constant $\beta_4$ is calculated as:
$\beta_4 = K_1 \times K_2 \times K_3 \times K_4$
$\beta_4 = (1.0 \times 10^4) \times (1.0 \times 10^3) \times (1.0 \times 10^3) \times (1.0 \times 10^2)$
$\beta_4 = 1.0 \times 10^{(4+3+3+2)} = 1.0 \times 10^{12}$

(C) For the reaction $N_{2(g)} + 3H_{2(g)} \longrightarrow 2NH_{3(g)}$,the rate of reaction is given by:
Rate $= -\frac{d[N_2]}{dt} = -\frac{1}{3} \frac{d[H_2]}{dt} = \frac{1}{2} \frac{d[NH_3]}{dt}$.
Given that $-\frac{d[N_2]}{dt} = 0.02 \ mol \ L^{-1} \ s^{-1}$.
Equating the terms for $N_2$ and $H_2$:
$-\frac{d[N_2]}{dt} = -\frac{1}{3} \frac{d[H_2]}{dt}$.
Therefore,$-\frac{d[H_2]}{dt} = 3 \times (-\frac{d[N_2]}{dt})$.
$-\frac{d[H_2]}{dt} = 3 \times 0.02 = 0.06 \ mol \ L^{-1} \ s^{-1}$.

(D) For the reaction $N_{2(g)} + 3H_{2(g)} \longrightarrow 2NH_{3(g)}$,the rate of reaction is expressed as:
$Rate = -\frac{d[N_2]}{dt} = -\frac{1}{3} \frac{d[H_2]}{dt} = \frac{1}{2} \frac{d[NH_3]}{dt}$
Equating the terms for $NH_3$ and $H_2$:
$-\frac{1}{3} \frac{d[H_2]}{dt} = \frac{1}{2} \frac{d[NH_3]}{dt}$
Multiplying both sides by $6$:
$-2 \frac{d[H_2]}{dt} = 3 \frac{d[NH_3]}{dt}$
Rearranging gives $3 \frac{d[NH_3]}{dt} = -2 \frac{d[H_2]}{dt}$.

(A) For a zero order reaction,the integrated rate equation is given by:
$[A]_t = [A]_0 - kt$
Given:
Rate constant,$k = 1 \times 10^{-3} \ mol \ L^{-1} \ s^{-1}$
Initial concentration,$[A]_0 = 0.1 \ mol \ L^{-1}$
Time,$t = 10 \ s$
Substituting the values into the equation:
$[A]_t = 0.1 - (1 \times 10^{-3} \times 10)$
$[A]_t = 0.1 - 10^{-2}$
$[A]_t = 0.1 - 0.01$
$[A]_t = 0.09 \ mol \ L^{-1}$

(A) For a first order reaction,the rate constant $K$ is given by:
$K = \frac{2.303}{t} \log \frac{[A]_0}{[A]_t}$
Given $K = 3.3 \times 10^{-4} \ s^{-1}$ and the reaction is $90 \%$ complete,so $[A]_t = [A]_0 - 0.90[A]_0 = 0.10[A]_0$.
Substituting the values:
$3.3 \times 10^{-4} = \frac{2.303}{t} \log \frac{[A]_0}{0.10[A]_0}$
$3.3 \times 10^{-4} = \frac{2.303}{t} \log(10)$
Since $\log(10) = 1$:
$t = \frac{2.303}{3.3 \times 10^{-4}} \ s$
$t \approx 0.6978 \times 10^4 \ s = 6978 \ s$
To convert time into minutes:
$t = \frac{6978}{60} \ min \approx 116.3 \ min$
Rounding to the nearest provided option,the answer is $116.67 \ min$.

(A) For a first-order reaction: $[R] = [R]_0 e^{-kt}$.
Taking natural logarithm on both sides: $\ln [R] = \ln [R]_0 - kt$.
Rearranging the terms: $\ln [R]_0 - \ln [R] = kt$,which gives $\ln \frac{[R]_0}{[R]} = kt$.
Converting to base $10$ logarithm: $2.303 \log \frac{[R]_0}{[R]} = kt$.
Therefore,$\log \frac{[R]_0}{[R]} = \frac{k}{2.303} t$.
Comparing this with the equation of a straight line $y = mx + c$,a plot of $\log \frac{[R]_0}{[R]}$ vs $t$ gives a straight line passing through the origin with a positive slope $m = \frac{k}{2.303}$.

(A) For a first-order reaction,the half-life is given by $t_{1/2} = \frac{0.693}{k}$.
At $T_1 = 310 \ K$,$t_{1/2} = 12 \ min$,so $k_1 = \frac{0.693}{12} \ min^{-1}$.
At $T_2 = 300 \ K$,$t_{1/2} = 2 \ hrs = 120 \ min$,so $k_2 = \frac{0.693}{120} \ min^{-1}$.
Using the Arrhenius equation: $\ln\left(\frac{k_1}{k_2}\right) = \frac{E_a}{R} \left(\frac{T_1 - T_2}{T_1 T_2}\right)$.
$\ln\left(\frac{0.693/12}{0.693/120}\right) = \ln(10) = 2.303$.
$2.303 = \frac{E_a}{8.3} \left(\frac{310 - 300}{310 \times 300}\right)$.
$2.303 = \frac{E_a}{8.3} \left(\frac{10}{93000}\right) = \frac{E_a}{8.3 \times 9300}$.
$E_a = 2.303 \times 8.3 \times 9300 \approx 177760 \ J \ mol^{-1} = 177.76 \ kJ \ mol^{-1}$.

(C) $(II)$,$(III)$,and $(IV)$ are broad spectrum antibiotics.
$\text{Penicillin G}$ is a narrow spectrum antibiotic.
Broad spectrum antibiotics are effective against a wide range of Gram-positive and Gram-negative bacteria.

(A) Ofloxacin is an antibiotic,whereas paracetamol,morphine,and codeine act as analgesics.
Thus,option $(A)$ is the correct answer.

(A) $Codeine$ is an analgesic,while $Equanil$ is a tranquilizer. Thus,this pair is correct.
$Chloramphenicol$ is an antibiotic,while $Nardil$ is an antidepressant.
$Histamine$ is a chemical messenger involved in inflammatory responses,while $Salvarsan$ is an antibiotic used to treat syphilis.
$Norethindrone$ is a synthetic progesterone derivative used as an antifertility drug,while $Alitame$ is a high-potency artificial sweetener.

(C) Antihistamines are drugs that treat allergic reactions by blocking the action of histamine.
$2. \text{Dimetane}$ (Brompheniramine) and $4. \text{Seldane}$ (Terfenadine) are well-known antihistamines.
$1. \text{Serotonin}$ is a neurotransmitter.
$3. \text{Phenelzine}$ is an antidepressant ($MAO$ inhibitor).
Therefore,the correct option is $C$.

(C) $(i)$ Binding of inhibitor at allosteric site changes the shape of the active site in such a way that the substrate cannot recognize it. This statement is correct.
$(ii)$ The shape of the receptor changes after the attachment of a chemical messenger to it. This statement is incorrect.
$(iii)$ $A$ chemical messenger gives a message to the cell without entering the cell. This statement is incorrect.
$(iv)$ Erythromycin is a bacteriostatic antibiotic used for the treatment of a number of bacterial infections. This statement is correct.
Hence,option $(C)$ is the correct answer.

(C) The structures of the given compounds are as follows:
$(a)$ Ranitidine: An antacid.
$(b)$ Saccharin: An artificial sweetener.
$(c)$ Salvarsan: An antimicrobial drug that contains an $-As=As-$ linkage in its structure.
$(d)$ Seldane: An antihistamine.
Therefore,Salvarsan contains the $-As=As-$ linkage,and option $(c)$ is the correct answer.

$(A)$ In the $4f$-series, as the atomic number increases from $La$ to $Lu$, the atomic and ionic radii decrease gradually.
This decrease in atomic and ionic size from lanthanum to lutetium is known as lanthanide contraction.
Since the atomic number of $Pr$ $(59)$, $Gd$ $(64)$, and $Tm$ $(69)$ increases in that order, their ionic radii decrease accordingly.
Therefore, the correct order of ionic radii is $Pr^{3+} (101 \text{ pm}) > Gd^{3+} (94 \text{ pm}) > Tm^{3+} (87 \text{ pm})$.

(C) The chemical formula of copper sulphate pentahydrate is $[Cu(H_2O)_4]SO_4 \cdot H_2O$.
Copper sulphate pentahydrate contains $5$ molecules of water of crystallisation.
In the crystal structure,four water molecules are coordinated to the $Cu^{2+}$ ion,while the fifth water molecule is hydrogen-bonded to the $SO_4^{2-}$ ion.
Thus,the correct representation is $[Cu(H_2O)_4]SO_4 \cdot H_2O$.

(D) To find the magnetic moment,we calculate the number of unpaired electrons $(n)$ in each complex. The magnetic moment is given by $\mu = \sqrt{n(n+2)} \ BM$.
$(I)$ $\left[Cr(CN)_6\right]^{3-}$: $Cr^{3+}$ is $3d^3$. With $CN^-$ as a strong field ligand,all $3$ electrons remain unpaired in the $t_{2g}$ orbitals. $n = 3$,$\mu = \sqrt{3(3+2)} = \sqrt{15} \ BM$.
$(II)$ $\left[Mn(CN)_6\right]^{3-}$: $Mn^{3+}$ is $3d^4$. With $CN^-$ as a strong field ligand,electrons pair up,leaving $n = 2$ unpaired electrons. $\mu = \sqrt{2(2+2)} = \sqrt{8} \ BM$.
$(III)$ $\left[Fe(CN)_6\right]^{3-}$: $Fe^{3+}$ is $3d^5$. With $CN^-$ as a strong field ligand,electrons pair up,leaving $n = 1$ unpaired electron. $\mu = \sqrt{1(1+2)} = \sqrt{3} \ BM$.
$(IV)$ $\left[Co(CN)_6\right]^{3-}$: $Co^{3+}$ is $3d^6$. With $CN^-$ as a strong field ligand,all electrons pair up in the $t_{2g}$ orbitals $(t_{2g}^6 e_g^0)$. $n = 0$,$\mu = 0 \ BM$.
Thus,$\left[Co(CN)_6\right]^{3-}$ has the lowest magnetic moment.

(C) The wavelength of light absorbed by a complex is inversely proportional to the crystal field splitting energy $(\Delta_o)$,which in turn depends on the strength of the ligands.
According to the spectrochemical series,the strength of the ligands is $en > H_2O$.
Therefore,the crystal field splitting energy follows the order: $[Ni(en)_3]^{2+} > [Ni(H_2O)_4en]^{2+} > [Ni(H_2O)_6]^{2+}$.
Since $E = \frac{hc}{\lambda}$,the energy is inversely proportional to the wavelength.
Thus,the order of wavelengths absorbed is: $[Ni(H_2O)_6]^{2+} > [Ni(H_2O)_4en]^{2+} > [Ni(en)_3]^{2+}$.
This corresponds to $\lambda_1 > \lambda_3 > \lambda_2$.

(B) Given,$\Delta_0$ for the coordination complex is $1000 \ kJ \ mol^{-1}$.
Energy of $t_{2g}$ orbitals,$E(t_{2g}) = -400 \ kJ \ mol^{-1}$.
We know that the crystal field splitting energy $\Delta_0$ is defined as the difference between the energies of $e_g$ and $t_{2g}$ orbitals:
$\Delta_0 = E(e_g) - E(t_{2g})$
Substituting the given values:
$1000 = E(e_g) - (-400)$
$1000 = E(e_g) + 400$
$E(e_g) = 1000 - 400$
$E(e_g) = 600 \ kJ \ mol^{-1}$
Thus,the energy of $e_g$ orbitals is $600 \ kJ \ mol^{-1}$.
Therefore,option $(B)$ is the correct answer.

(A) The crystal field splitting energy in a tetrahedral complex $(\Delta_t)$ is related to the octahedral complex $(\Delta_o)$ by the expression: $\Delta_t = \frac{4}{9} \Delta_o$.
Given that $\Delta_t = x \ eV$,we can rearrange the formula to solve for $\Delta_o$:
$\Delta_o = \frac{9}{4} \Delta_t = \frac{9x}{4} \ eV$.

(B) The stability of a complex ion is influenced by the nature of the ligand and the chelate effect.
$CN^-$ is a strong field ligand that forms a very stable complex with $Co^{3+}$.
Additionally,the stability of coordination complexes is often determined by the chelate effect,where polydentate ligands like oxalate $(C_2O_4^{2-})$ form more stable complexes than monodentate ligands.
However,among the given options,$[Co(CN)_6]^{3-}$ is exceptionally stable due to the strong field nature of the $CN^-$ ligand,which causes large crystal field splitting energy $(CFSE)$.
Thus,$[Co(CN)_6]^{3-}$ is the most stable complex.