AIIMS 2016 Chemistry Question Paper with Answer and Solution

(D) In a series $LCR$ circuit,the impedance $Z$ is given by $Z = \sqrt{R^2 + (\omega L - \frac{1}{\omega C})^2}$.
The peak current is $I_{\max} = \frac{\varepsilon_{\max}}{Z}$.
As the angular frequency $\omega$ changes,the impedance $Z$ changes. Specifically,at resonance $(\omega = \frac{1}{\sqrt{LC}})$,$Z$ is minimum $(Z = R)$,and the current is maximum.
If $\omega$ increases away from the resonance frequency,$Z$ increases,causing the current to decrease.
Therefore,the current does not monotonically increase with the increase in angular frequency. The Assertion is incorrect.
The Reason provides the correct formula for peak current,but the Assertion itself is false.

(A) Mass of $6.022 \times 10^{23}$ atoms of oxygen $= 16 \ g$.
Mass of one atom of oxygen $= \frac{16}{6.022 \times 10^{23}} \approx 2.66 \times 10^{-23} \ g$.
Mass of $6.022 \times 10^{23}$ atoms of nitrogen $= 14 \ g$.
Mass of one atom of nitrogen $= \frac{14}{6.022 \times 10^{23}} \approx 2.32 \times 10^{-23} \ g$.
Mass of $1 \times 10^{-10}$ mole of oxygen $= 16 \times 10^{-10} \ g$.
Mass of $1 \times 10^{-10}$ mole of copper $= 63 \times 10^{-10} \ g$.
Comparing the values:
$2.32 \times 10^{-23} (II) < 2.66 \times 10^{-23} (I) < 16 \times 10^{-10} (III) < 63 \times 10^{-10} (IV)$.
Thus,the order of increasing mass is $II < I < III < IV$.

(D) The Rydberg formula for a hydrogen-like species is given by $\frac{1}{\lambda} = R Z^2 \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right)$.
For $He^{+}$ $(Z = 2)$,the transition from $n_2 = 4$ to $n_1 = 2$ gives $\frac{1}{\lambda} = R (2)^2 \left( \frac{1}{2^2} - \frac{1}{4^2} \right) = 4R \left( \frac{1}{4} - \frac{1}{16} \right) = 4R \left( \frac{3}{16} \right) = \frac{3R}{4}$.
For a hydrogen atom $(Z = 1)$,we need to find $n_1$ and $n_2$ such that $\frac{1}{\lambda} = R \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right) = \frac{3R}{4}$.
Comparing the two,we get $\frac{1}{n_1^2} - \frac{1}{n_2^2} = \frac{3}{4}$.
This is satisfied when $n_1 = 1$ and $n_2 = 2$ because $\frac{1}{1^2} - \frac{1}{2^2} = 1 - \frac{1}{4} = \frac{3}{4}$.

(C) The Assertion is based on Heisenberg's Uncertainty Principle,which states that it is impossible to determine simultaneously the exact position and exact momentum of a microscopic particle like an electron. This is a fundamental law of quantum mechanics.
The Reason states that the path of an electron in an atom is clearly defined. This is incorrect because,according to quantum mechanics,electrons do not move in well-defined circular orbits (as proposed in the Bohr model). Instead,they exist in orbitals,which are regions of space where the probability of finding an electron is high. Therefore,the path of an electron cannot be defined.

(C) The general trend for first ionization enthalpy across a period is that it increases from left to right due to an increase in effective nuclear charge and a decrease in atomic size.
However,there are exceptions due to stable electronic configurations.
The electronic configurations of the second-period elements are:
$Li (1s^2 2s^1)$,$Be (1s^2 2s^2)$,$B (1s^2 2s^2 2p^1)$,$C (1s^2 2s^2 2p^2)$,$N (1s^2 2s^2 2p^3)$,$O (1s^2 2s^2 2p^4)$,$F (1s^2 2s^2 2p^5)$,$Ne (1s^2 2s^2 2p^6)$.
$Be$ has a fully filled $2s$ orbital,which is more stable than the $2p^1$ configuration of $B$.
$N$ has a half-filled $2p$ orbital,which is more stable than the $2p^4$ configuration of $O$.
Thus,the correct decreasing order for the elements $Be, B, C, N, F$ is $F > N > C > Be > B$.

(C) The bond order $(B.O.)$ is calculated using the formula: $B.O. = \frac{1}{2} (N_b - N_a)$,where $N_b$ is the number of bonding electrons and $N_a$ is the number of antibonding electrons.
$1$. For $CN^{-}$ ($14$ electrons): Configuration is $\sigma 1s^2, \sigma^* 1s^2, \sigma 2s^2, \sigma^* 2s^2, \pi 2p_x^2 = \pi 2p_y^2, \sigma 2p_z^2$. $B.O. = \frac{10 - 4}{2} = 3$.
$2$. For $O_2^-$ ($17$ electrons): Configuration is $\sigma 1s^2, \sigma^* 1s^2, \sigma 2s^2, \sigma^* 2s^2, \sigma 2p_z^2, \pi 2p_x^2 = \pi 2p_y^2, \pi^* 2p_x^2 = \pi^* 2p_y^1$. $B.O. = \frac{10 - 7}{2} = 1.5$.
$3$. For $NO^{+}$ ($14$ electrons): Configuration is $\sigma 1s^2, \sigma^* 1s^2, \sigma 2s^2, \sigma^* 2s^2, \sigma 2p_z^2, \pi 2p_x^2 = \pi 2p_y^2$. $B.O. = \frac{10 - 4}{2} = 3$.
$4$. For $CN^{+}$ ($12$ electrons): Configuration is $\sigma 1s^2, \sigma^* 1s^2, \sigma 2s^2, \sigma^* 2s^2, \pi 2p_x^2 = \pi 2p_y^2$. $B.O. = \frac{8 - 4}{2} = 2$.
Thus,$CN^{-}$ and $NO^{+}$ both have a bond order of $3$.

(A) According to the $VSEPR$ theory,lone pairs are localized on the central atom,whereas bond pairs are shared between two atoms.
Because lone pairs are attracted by only one nucleus,they occupy more space around the central atom compared to bond pairs,which are attracted by two nuclei.
This increased spatial requirement leads to greater repulsive interactions between lone pairs than between lone pair-bond pair or bond pair-bond pair interactions.
Thus,the Reason correctly explains the Assertion. Therefore,the correct option is $A$.

(A) Given,$P_1 = 15 \, atm$,$P_2 = 60 \, atm$.
$V_1 = 76 \, cm^3$,$V_2 = 20.5 \, cm^3$.
If the gas is an ideal gas,then according to Boyle's law,it must follow the equation $P_1V_1 = P_2V_2$.
$P_1 \times V_1 = 15 \times 76 = 1140$.
$P_2 \times V_2 = 60 \times 20.5 = 1230$.
Since $P_1V_1 \neq P_2V_2$,the gas behaves non-ideally.
Additionally,processes like dimerization or adsorption on vessel walls would also lead to a deviation from the expected ideal volume change,making all three statements possible explanations for the observed behavior.

(A) The critical temperature $(T_c)$ of a gas is the temperature above which it cannot be liquefied,no matter how much pressure is applied.
This is because,above $T_c$,the kinetic energy of the gas molecules is so high that the intermolecular forces of attraction are insufficient to hold the molecules together in the liquid state.
Therefore,both the Assertion and the Reason are correct,and the Reason is the correct explanation of the Assertion.

(B) The relationship between enthalpy change $\Delta H$ and internal energy change $\Delta U$ is given by the equation: $\Delta H = \Delta U + \Delta n_g RT$.
If $\Delta n_g = 0$,then $\Delta H = \Delta U$.
If $\Delta n_g \neq 0$,then $\Delta H \neq \Delta U$.
Let us calculate $\Delta n_g$ (change in the number of moles of gaseous species) for each reaction:
$A$: $\Delta n_g = 1 - 1 = 0$
$B$: $\Delta n_g = 1 - (1 + 1) = -1$
$C$: $\Delta n_g = 1 - 1 = 0$
$D$: $\Delta n_g = 2 - (1 + 1) = 0$
Since $\Delta n_g \neq 0$ for option $B$,$\Delta H \neq \Delta U$ in this reaction.

(B) We are given the following combustion reactions:
$C_6H_{6(l)} + \frac{15}{2}O_{2(g)} \to 6CO_{2(g)} + 3H_2O_{(l)}$; $\Delta H = -3270 \ kJ \ mol^{-1} \dots (i)$
$C(gr) + O_{2(g)} \to CO_{2(g)}$; $\Delta H = -394 \ kJ \ mol^{-1} \dots (ii)$
$H_{2(g)} + \frac{1}{2}O_{2(g)} \to H_2O_{(l)}$; $\Delta H = -286 \ kJ \ mol^{-1} \dots (iii)$
We need to find the enthalpy of formation of $C_6H_{6(l)}$:
$6C(gr) + 3H_{2(g)} \to C_6H_{6(l)}$; $\Delta H_f = ? \dots (iv)$
By performing the operation $6 \times (ii) + 3 \times (iii) - (i)$,we get:
$\Delta H_f = [6 \times (-394) + 3 \times (-286)] - (-3270)$
$\Delta H_f = [-2364 - 858] + 3270$
$\Delta H_f = -3222 + 3270 = +48 \ kJ \ mol^{-1}$

(D) To determine the most and least soluble compounds,we calculate the molar solubility $(S)$ for each:
$1$. For $AgCl$ ($1:1$ type): $K_{sp} = S^2 \Rightarrow S = \sqrt{1.1 \times 10^{-10}} \approx 1.05 \times 10^{-5} \, M$
$2$. For $AgI$ ($1:1$ type): $K_{sp} = S^2 \Rightarrow S = \sqrt{1.0 \times 10^{-16}} = 1.0 \times 10^{-8} \, M$
$3$. For $PbCrO_4$ ($1:1$ type): $K_{sp} = S^2 \Rightarrow S = \sqrt{4.0 \times 10^{-14}} = 2.0 \times 10^{-7} \, M$
$4$. For $Ag_2CO_3$ ($2:1$ type): $K_{sp} = 4S^3$ $\Rightarrow S = \sqrt[3]{K_{sp}/4} = \sqrt[3]{8.0 \times 10^{-12} / 4} = \sqrt[3]{2.0 \times 10^{-12}} \approx 1.26 \times 10^{-4} \, M$
Comparing the solubilities: $1.26 \times 10^{-4} > 1.05 \times 10^{-5} > 2.0 \times 10^{-7} > 1.0 \times 10^{-8}$.
Thus,$Ag_2CO_3$ is the most soluble and $AgI$ is the least soluble.

(D) For the first equilibrium: $AB \rightleftharpoons A^{+} + B^{-}$,$K_1 = \frac{[A^{+}][B^{-}]}{[AB]}$
For the second equilibrium: $AB + B^{-} \rightleftharpoons AB_2^-$,$K_2 = \frac{[AB_2^-]}{[AB][B^{-}]}$
Dividing $K_1$ by $K_2$,we get:
$\frac{K_1}{K_2} = \frac{[A^{+}][B^{-}]}{[AB]} \times \frac{[AB][B^{-}]}{[AB_2^-]} = \frac{[A^{+}][B^{-}]^2}{[AB_2^-]}$
Rearranging the expression to find the ratio $\frac{[A^{+}]}{[AB_2^-]}$:
$\frac{[A^{+}]}{[AB_2^-]} = \frac{K_1}{K_2} \times \frac{1}{[B^{-}]^2}$
Thus,the ratio $\frac{[A^{+}]}{[AB_2^-]}$ is inversely proportional to the square of $[B^{-}]$.

(C) The solubility product constant $(K_{sp})$ values at $25^{\circ}C$ are approximately $K_{sp}(AgCl) \approx 1.8 \times 10^{-10}$ and $K_{sp}(AgBr) \approx 5.0 \times 10^{-13}$.
Since $K_{sp}(AgBr) < K_{sp}(AgCl)$,$AgBr$ requires a lower concentration of $Ag^+$ ions to exceed its solubility product compared to $AgCl$.
Therefore,$AgBr$ will precipitate first.
The Assertion is correct,but the Reason is incorrect because it states $K_{sp}(AgCl) < K_{sp}(AgBr)$,which is false.

(A) The precipitation of a salt occurs when the ionic product exceeds its solubility product constant $(K_{sp})$.
Since $K_{sp}$ of $AgBr$ $(5.0 \times 10^{-13})$ is significantly lower than the $K_{sp}$ of $AgCl$ $(1.8 \times 10^{-10})$,the ionic product of $AgBr$ reaches its $K_{sp}$ value first upon the addition of $Ag^+$ ions.
Therefore,$AgBr$ precipitates first.
Both the Assertion and the Reason are correct,and the Reason correctly explains the Assertion.

(C) species can act as both an oxidizing and a reducing agent if the central atom is in an intermediate oxidation state,allowing it to either increase (oxidation) or decrease (reduction) its oxidation number.

Species	Oxidation State of Central Atom
$Cl^{-}$	$-1$ (Minimum)
$ClO_4^{-}$	$+7$ (Maximum)
$ClO^{-}$	$+1$ (Intermediate)
$MnO_4^{-}$	$+7$ (Maximum)

In $ClO^{-}$,the chlorine atom is in the $+1$ oxidation state. Since its oxidation state can increase (e.g.,to $+3, +5, +7$) or decrease (e.g.,to $-1$),it can function as both an oxidizing and a reducing agent.

(C) Ionic hydrides are stoichiometric compounds of dihydrogen formed with most of the $s$-block elements. They are crystalline,non-volatile,and non-conducting in the solid state. However,when in a molten state,they conduct electricity and liberate dihydrogen gas at the $anode$,not the $cathode$. Therefore,statement $(C)$ is incorrect.

(A) The reaction $2M_2O \xrightarrow[673 \ K]{\text{heat}} M_2O_2 + 2M$ is generally not a standard method for preparing peroxides of alkali metals.
Specifically,$Li$ is the smallest alkali metal and its oxide $Li_2O$ is very stable.
$Li$ does not form a peroxide $(Li_2O_2)$ under these conditions; it primarily forms the monoxide $Li_2O$ upon combustion in excess air.
Therefore,the reaction shown in option $A$ is chemically incorrect.

(B) $BF_3$ has a planar trigonal structure with bond angles of $120^{\circ}$.
Due to its symmetrical geometry,the resultant dipole moment of two $B-F$ bonds is equal and opposite to the third $B-F$ bond.
Therefore,the net dipole moment of the $BF_3$ molecule is zero.

(B) Borax is a compound of boron,which is chemically known as sodium tetraborate decahydrate.
Its structural formula is represented as $Na_2[B_4O_5(OH)_4] \cdot 8 H_2O$.

(C) The assertion is correct because,due to the 'inert pair effect',the stability of the $+4$ oxidation state decreases down the group $14$ $(C > Si > Ge > Sn > Pb)$.
Consequently,$Pb^{4+}$ is highly unstable and acts as a strong oxidizing agent to reduce itself to the more stable $+2$ state.
The reason is incorrect because the higher oxidation states (like $+4$) become less stable,not more stable,for the heavier members of the group due to the 'inert pair effect'.

(B) If the organic compound contains nitrogen or sulphur,the Lassaigne's extract is boiled with $HNO_3$ to decompose sodium cyanide $(NaCN)$ or sodium sulphide $(Na_2S)$ formed during fusion.
$NaCN + HNO_3 \rightarrow NaNO_3 + HCN \uparrow $
$Na_2S + 2HNO_3 \rightarrow 2NaNO_3 + H_2S \uparrow $
If cyanide and sulphide ions are not removed,they react with silver nitrate $(AgNO_3)$ and interfere with the halogen test by forming precipitates:
$NaCN + AgNO_3 \rightarrow AgCN \text{ (White ppt.)} + NaNO_3$
$Na_2S + 2AgNO_3 \rightarrow Ag_2S \text{ (Black ppt.)} + 2NaNO_3$

(A) Enantiomerism is exhibited by molecules that are chiral,meaning they possess at least one chiral carbon atom (a carbon atom bonded to four different groups).
$1.$ Diphenyl methanol: The central carbon is bonded to one $H$ atom,one $OH$ group,and two identical phenyl groups. Since it has two identical groups,it is achiral and cannot exhibit enantiomerism.
$2.$ $1-$Bromo$-2-$chlorobutane: The carbon at position $2$ is bonded to $H$,$Cl$,$CH_3$,and $CH_2Br$. All four groups are different,so it is chiral.
$3.$ $2-$Butanol: The carbon at position $2$ is bonded to $H$,$OH$,$CH_3$,and $CH_2CH_3$. All four groups are different,so it is chiral.
$4.$ Tartaric acid: It contains two chiral carbon atoms and can exist as enantiomers (e.g.,$(+)-$ and $(-)-$tartaric acid).
Therefore,Diphenyl methanol cannot exhibit enantiomerism.

(D) $1$. Identify the longest carbon chain containing both double bonds,which consists of $10$ carbon atoms (decadiene).
$2$. Number the chain from the end that gives the lowest locants to the double bonds and substituents.
$3$. Numbering from left to right: $CH_3(1)-CH(CH_3)(2)-CH_2(3)-CH(4)=CH(5)-CH(6)=CH(7)-CH(CH_3)(8)-CH_2(9)-CH_3(10)$.
$4$. The double bonds are at positions $4$ and $6$,and methyl groups are at positions $2$ and $8$.
$5$. Therefore,the $IUPAC$ name is $2,8-$ dimethyl $-4,6-$ decadiene.

(A) The addition of $HBr$ to unsymmetrical alkenes follows anti-Markovnikov's rule in the presence of peroxide (peroxide effect or Kharasch effect).
However,for symmetrical alkenes,the product formed remains the same regardless of the presence or absence of peroxide because the molecule is symmetric.
$2-$butene $(CH_3-CH=CH-CH_3)$ is a symmetrical alkene.
Therefore,the addition of $HBr$ to $2-$butene yields $2-$bromobutane in both cases.

(D) The Assertion is incorrect because the energy of the resonance hybrid is always lower than the energy of any of the individual canonical structures. This difference in energy is known as resonance energy.
The Reason is correct because a resonance hybrid is a theoretical construct representing the weighted average of all contributing canonical structures,and no single Lewis structure can fully describe the actual electron distribution in the molecule.
Therefore,the Assertion is incorrect and the Reason is correct.

(C) The Tropylium cation $(C_7H_7^+)$ is a seven-membered ring with $6\pi$ electrons ($n=1$ in $H$ückel's rule $(4n+2)\pi$).
It is planar and fully conjugated,making it aromatic.
Therefore,the Assertion is correct.
However,the Reason is incorrect because aromaticity is determined by multiple factors,including cyclic structure,planarity,complete conjugation,and the presence of $(4n+2)\pi$ electrons,not just planarity alone.

(A) In mitosis,a single cell divides to form two daughter cells.
So,the number of cells produced after $n$ divisions is given by the formula $2^n$.
We are given that the total number of cells required is $128$.
Therefore,$2^n = 128$.
Since $128 = 2^7$,we have $2^n = 2^7$.
By comparing the exponents,we get $n = 7$.
Thus,$7$ mitotic divisions are needed for a single cell to produce $128$ cells.

(B) Net primary productivity $(NPP)$ is the rate at which producers store organic matter in their bodies per unit time and area.
It is calculated as Gross primary productivity $(GPP)$ minus the energy lost due to respiration $(R)$,expressed as $NPP = GPP - R$.
Secondary productivity is defined as the rate of formation of new organic matter by heterotrophs (consumers) per unit time and area.
Both statements are scientifically correct,but the Reason does not explain why $NPP$ is defined as $GPP - R$.

(C) The irritant red haze observed in traffic and congested urban areas is primarily caused by the presence of oxides of nitrogen $(NO_x)$.
These oxides,particularly nitrogen dioxide $(NO_2)$,contribute to the formation of photochemical smog,which appears as a brownish-red haze.

(A) Statement $(i)$ is incorrect because the special membranous structure formed by the extension of the plasma membrane into the cell in prokaryotes is called a mesosome,not a polysome. $A$ polysome is a chain of ribosomes attached to a single $mRNA$ molecule.
Statement $(ii)$ is incorrect because the smooth endoplasmic reticulum $(SER)$ is the major site for lipid synthesis,whereas the rough endoplasmic reticulum $(RER)$ is the site for protein and glycoprotein synthesis.
Statement $(iii)$ is correct as $RuBisCO$ (Ribulose bisphosphate carboxylase-oxygenase) is the most abundant protein in the biosphere.
Statement $(iv)$ is correct because the endomembrane system includes the endoplasmic reticulum,Golgi complex,lysosomes,and vacuoles. Mitochondria,chloroplasts,and peroxisomes are not part of this system because their functions are not coordinated with the endomembrane components.
Therefore,statements $(iii)$ and $(iv)$ are correct.

(C) palindromic sequence in $DNA$ is a sequence of base pairs that reads the same on the two strands when the orientation of reading is kept the same (e.g.,$5' \rightarrow 3'$).
Among the given options,the sequence $5' - GAATTC - 3'$ / $3' - CTTAAG - 5'$ is a well-known palindromic sequence recognized by the restriction enzyme $EcoRI$.
This enzyme cuts the $DNA$ between the $G$ and $A$ bases,which is near the middle of the recognition site,producing sticky ends.

(C) In $XeO_4$,the central atom $Xe$ is in its $8^{th}$ oxidation state.
It undergoes $sp^{3}$ hybridization involving one $5s$ and three $5p$ orbitals.
These four $sp^{3}$ hybrid orbitals form four $\sigma$ bonds with four oxygen atoms.
Additionally,the four unpaired electrons in the $5d$ orbitals of $Xe$ form four $p\pi-d\pi$ bonds with the $2p$ orbitals of the oxygen atoms.
Therefore,the molecule contains four $p\pi-d\pi$ bonds.

(A) Gallium $(Ga)$ is a soft,silvery metal.
Its melting point is $30\,^oC$.
This metal expands by $3.1\%$ when it solidifies and hence,it should not be stored in glass or metal containers.

(A) $S_N1$ reactions proceed via the formation of a carbocation intermediate. The rate of $S_N1$ reaction depends on the stability of the carbocation formed.
$(A)$ forms a $1^o$ carbocation $(CH_3CH_2CH_2CH_2^+)$,which is the least stable.
$(B)$ forms an allylic carbocation $(CH_2=CH-CH^+(CH_3))$,which is resonance-stabilized and therefore the most stable.
$(C)$ forms a $2^o$ carbocation $(CH_3CH_2CH^+(CH_3))$,which is more stable than a $1^o$ carbocation but less stable than an allylic carbocation.
Thus,the stability order of the carbocations is $B > C > A$. Consequently,the order of $S_N1$ reactivity is $B > C > A$.

(C) $C_6H_6$ is diamagnetic $(i-5)$.
$CrO_2$ is ferromagnetic $(ii-3)$.
$MnO$ is antiferromagnetic $(iii-1)$.
$Fe_3O_4$ is ferrimagnetic $(iv-2)$.
$Fe^{3+}$ is paramagnetic with $5$ unpaired electrons $(v-4)$.

(C) In $hcp$ and $ccp$ arrangements,the packing efficiency is $74\%$.
Therefore,the void space (empty space) is $100\% - 74\% = 26\%$.
Statement $C$ claims the void space in $hcp$ is $32\%$,which is incorrect.

(D) The depression in freezing point is given by $\Delta T_f = T_f^o - T_f = 0 - (-0.465) = 0.465 \ K$.
Using the formula for molar mass $M$ of the solute:
$M = \frac{1000 \times K_f \times w_2}{\Delta T_f \times w_1} = \frac{1000 \times 1.86 \times 1.8}{0.465 \times 40} = 180 \ g \ mol^{-1}$.
The empirical formula mass of $CH_2O$ is $12 + (2 \times 1) + 16 = 30 \ g \ mol^{-1}$.
Calculating the value of $n$:
$n = \frac{\text{Molecular mass}}{\text{Empirical formula mass}} = \frac{180}{30} = 6$.
Therefore,the molecular formula is $(CH_2O)_6 = C_6H_{12}O_6$.

(C) According to Raoult's Law,the total vapour pressure of a solution containing two volatile components is given by $p_s = p_1 + p_2 = x_1 p_1^o + x_2 p_2^o$.
If the solute (component $2$) is more volatile than the solvent (component $1$),then $p_2^o > p_1^o$.
In this case,the total vapour pressure $p_s$ can be greater than the pure solvent vapour pressure $p_1^o$ depending on the mole fraction.
However,the Reason is incorrect because both components contribute to the total vapour pressure,and both form vapours,with the vapour phase being richer in the more volatile component.

(B) The standard cell potential is calculated using the formula:
$E_{cell}^o = E_{cathode}^o - E_{anode}^o$
Here,the cathode is the $N^{+}/N$ electrode and the anode is the $M^{+}/M$ electrode.
$E_{cell}^o = E_{N^{+}/N}^o - E_{M^{+}/M}^o$
$E_{cell}^o = 0.25 \ V - 0.52 \ V = -0.27 \ V$
Since $E_{cell}^o < 0$,the cell reaction is non-spontaneous.

(D) Specific conductance (conductivity,$\kappa$) is defined as the conductance of $1 \ cm^3$ of the solution.
On increasing dilution,the number of ions per unit volume decreases,which leads to a decrease in specific conductance.
Therefore,the Assertion is incorrect.
Although the degree of ionization and ionic mobility increase with dilution,the Assertion itself is false,making the correct choice $D$.

(D) According to collision theory,the rate of a chemical reaction depends on three main factors:
$1$. The frequency of collisions between reactant molecules,which is influenced by temperature and concentration.
$2$. The energy of the collisions,which must exceed the activation energy $(E_a)$.
$3$. The orientation or geometry of the colliding molecules (steric factor).
Since temperature affects the kinetic energy (and thus velocity) and collision frequency,and orientation is a fundamental requirement for effective collisions,all the factors listed in options $A$,$B$,and $C$ influence the reaction rate. Therefore,none of the options provided are independent of the rate.

(A) The rate law is given as $\text{Rate} = k [NO_2]^2$.
This indicates that the reaction is zero order with respect to $CO_{(g)}$.
The rate constant $k$ depends only on temperature and remains constant as long as the temperature is constant.
Since the reaction is zero order with respect to $CO$,adding $0.1 \ mol$ of $CO_{(g)}$ does not change the rate of the reaction.
Therefore,both $k$ and the reaction rate remain the same.

(A) Colloidal antimony is used in the treatment of the disease kala azar.

(D) The enthalpy of chemisorption is high,typically in the range of $40-400 \ kJ \ mol^{-1}$,because it involves the formation of chemical bonds.
In contrast,physisorption involves weak van der Waal's forces,resulting in a low enthalpy of adsorption,typically in the range of $20-40 \ kJ \ mol^{-1}$.
Therefore,the enthalpy of physisorption is lower than that of chemisorption.
Thus,the Assertion is false,and the Reason is true.

(C) In the blast furnace,the reduction of iron oxides $(Fe_2O_3)$ occurs primarily by carbon monoxide $(CO)$ to form iron $(Fe)$ and carbon dioxide $(CO_2)$,represented by reaction $(i)$.
Additionally,the removal of silica $(SiO_2)$ impurity as slag $(CaSiO_3)$ occurs via reaction $(iv)$,where calcium oxide $(CaO)$ acts as a flux.
Reaction $(ii)$ is not a primary reaction in the blast furnace for iron extraction,and reaction $(iii)$ is not the main reduction mechanism compared to the $CO$ reduction.
Therefore,the correct reactions are $(i)$ and $(iv)$.

(C) The first reaction represents the oxidation of water by fluorine,where $F_2$ acts as a strong oxidizing agent: $2F_{2(g)} + 2H_2O_{(l)} \to 4H^{+}_{(aq)} + 4F^{-}_{(aq)} + O_{2(g)}$. Thus,$X = F$.
The second reaction represents the disproportionation of halogens (like chlorine) in water: $Cl_{2(g)} + H_2O_{(l)} \to HCl_{(aq)} + HOCl_{(aq)}$. Thus,$Y = Cl$.

(A) In $SF_6$,the sulphur atom is sterically protected by $six$ fluorine atoms,which prevents the attack of $H_2O$ molecules.
Additionally,the sulphur atom in $SF_6$ is coordinatively saturated.
In contrast,$SF_4$ has a lone pair on the sulphur atom and is not sterically hindered to the same extent,allowing $H_2O$ to attack the sulphur atom and undergo hydrolysis.
Thus,both the assertion and the reason are correct,and the reason correctly explains the assertion.

(B) The magnetic moment values of actinides are lower than the theoretically predicted values because the $5f$ electrons are less effectively shielded by the outer shells. This leads to the quenching of the orbital contribution to the magnetic moment,making the observed values smaller than those calculated using the spin-only formula. While actinides are paramagnetic,the reason provided does not explain why the magnetic moments are lower than predicted.

(B) When excess ammonia is added to a copper sulphate solution,it forms a deep blue complex.
The reaction is: $CuSO_4(aq) + 4NH_3(aq) \to [Cu(NH_3)_4]SO_4(aq)$.
The deep blue colour is due to the formation of the tetraamminecopper$(II)$ ion,$[Cu(NH_3)_4]^{2+}$.

(D) When excess of $AgNO_3$ and $BaCl_2$ are added to solution $X$:
$1.$ Reaction with $AgNO_3$:
$[Co(NH_3)_5Br]Cl_2 + 2AgNO_3 \rightarrow [Co(NH_3)_5Br](NO_3)_2 + 2AgCl(s) (Y)$
Since $1 \, \text{mole}$ of $[Co(NH_3)_5Br]Cl_2$ provides $2 \, \text{moles}$ of $Cl^-$ ions,$0.02 \, \text{mole}$ of the complex yields $0.02 \times 2 = 0.04 \, \text{mole}$ of $AgCl$ precipitate $(Y)$.
$2.$ Reaction with $BaCl_2$:
$[Co(NH_3)_5Cl]SO_4 + BaCl_2 \rightarrow [Co(NH_3)_5Cl]Cl_2 + BaSO_4(s) (Z)$
Since $1 \, \text{mole}$ of $[Co(NH_3)_5Cl]SO_4$ provides $1 \, \text{mole}$ of $SO_4^{2-}$ ions,$0.02 \, \text{mole}$ of the complex yields $0.02 \, \text{mole}$ of $BaSO_4$ precipitate $(Z)$.
Therefore,the number of moles of $Y$ and $Z$ are $0.04$ and $0.02$ respectively.

(D) The Assertion is incorrect because alkylbenzene $CAN$ be prepared by Friedel-Crafts alkylation,although it has limitations such as polyalkylation and rearrangement.
The Reason is also incorrect because alkyl halides are generally more reactive or comparable in reactivity to acyl halides in the context of Friedel-Crafts reactions,but the primary issue with alkylation is the formation of polyalkylated products due to the activating nature of the alkyl group added to the benzene ring.
Therefore,both the Assertion and the Reason are incorrect.

(D) The acidity of an alcohol depends on the stability of the conjugate base (alkoxide ion) formed after the loss of a proton $(H^+)$.
In $ClCH_2CH_2OH$,the chlorine atom exerts a $-I$ (negative inductive) effect,which is electron-withdrawing.
This $-I$ effect helps in dispersing the negative charge on the oxygen atom of the alkoxide ion $(ClCH_2CH_2O^-)$,thereby increasing its stability compared to the ethoxide ion $(CH_3CH_2O^-)$.
Since the conjugate base of $ClCH_2CH_2OH$ is more stable,it is a stronger acid than $CH_3CH_2OH$.

(A) Alkyl aryl ethers are cleaved at the alkyl-oxygen bond because the phenyl-oxygen bond has partial double bond character due to resonance,making it stronger and more stable.
Therefore,the reaction of ethyl phenyl ether $(C_6H_5-O-C_2H_5)$ with $HBr$ proceeds as follows:
$C_6H_5-O-C_2H_5 + HBr \rightarrow C_6H_5OH + C_2H_5Br$
Both the Assertion and the Reason are correct,and the Reason correctly explains why the cleavage occurs at the alkyl-oxygen bond.

(D) Aldol condensation occurs in aldehydes and ketones that possess at least one $\alpha$-hydrogen atom.
$1$. Propanal $(CH_3CH_2CHO)$ has $2$ $\alpha$-hydrogen atoms.
$2$. Trichloroethanal $(Cl_3CCHO)$ has no $\alpha$-hydrogen atoms.
$3$. Methanal $(HCHO)$ has no $\alpha$-hydrogen atoms.
$4$. Ethanal $(CH_3CHO)$ has $3$ $\alpha$-hydrogen atoms.
$5$. Benzaldehyde $(C_6H_5CHO)$ has no $\alpha$-hydrogen atoms.
Therefore,trichloroethanal $(2)$,methanal $(3)$,and benzaldehyde $(5)$ do not undergo aldol condensation.

(A) Aldehydes and ketones possess a polar carbonyl group $(C=O)$,which creates a permanent dipole moment.
Due to this polarity,there exist dipole-dipole interactions between the molecules of aldehydes and ketones.
These intermolecular dipole-dipole interactions are stronger than the weak van der Waals forces present in hydrocarbons and ethers of comparable molecular masses.
Consequently,more energy is required to overcome these forces,leading to higher boiling points for aldehydes and ketones compared to hydrocarbons and ethers.
Thus,both the Assertion and the Reason are correct,and the Reason is the correct explanation of the Assertion.

(D) The benzene diazonium cation acts as a weak electrophile.
It undergoes electrophilic aromatic substitution (azo coupling) only with electron-rich aromatic compounds that contain strong electron-donating groups like $-OH$,$-NH_2$,or $-OCH_3$.
Nitrobenzene contains a strong electron-withdrawing group $(-NO_2)$,which deactivates the benzene ring towards electrophilic attack.
Therefore,nitrobenzene does not undergo the azo coupling reaction.

(C) The assertion is correct: Acylation of amines (using acid chlorides or anhydrides) results in a monosubstituted product because the electron-withdrawing nature of the acyl group reduces the nucleophilicity of the nitrogen atom,preventing further acylation. In contrast,alkylation of amines (using alkyl halides) leads to a mixture of primary,secondary,tertiary amines,and quaternary ammonium salts because the alkyl group is electron-donating,which increases the nucleophilicity of the nitrogen atom,making it more reactive toward further alkylation.
The reason is incorrect: The limitation of acylation is due to the electronic effect (electron-withdrawing nature of the acyl group) rather than steric hindrance. Therefore,the assertion is correct,but the reason is incorrect.

(C) Except for glycine,all other naturally occurring amino acids contain a chiral center at the $\alpha$-carbon atom.
Glycine has the structure $H_2N-CH_2-COOH$,where the $\alpha$-carbon is bonded to two identical hydrogen atoms,making it achiral.
Therefore,all amino acids except glycine are optically active.

(D) Vitamin $D$ is a fat-soluble vitamin.
Fat-soluble vitamins $(A, D, E, K)$ are stored in the liver and adipose tissues of the body.
Therefore,the Assertion is incorrect because Vitamin $D$ can be stored in the body.
The Reason is also incorrect because fat-soluble vitamins are not excreted in urine; only water-soluble vitamins ($B$ complex and $C$) are excreted in urine.

(B) Nylon $6$ is synthesized from caprolactam.
When caprolactam is heated with water at high temperatures,it undergoes ring-opening polymerization to form Nylon $6$.
The repeating unit is $[CO-(CH_2)_5-NH]$.

(C) Bakelite is a cross-linked polymer formed by the condensation reaction of phenol and formaldehyde.
It is a thermosetting polymer,which means it becomes hard and infusible on heating and cannot be remelted or reshaped.
Therefore,the Assertion is correct,but the Reason is incorrect.

(A) Broad spectrum antibiotics are those medicines which are effective against a wide range of Gram-positive and Gram-negative bacteria.
$Tetracycline$ is a well-known broad spectrum antibiotic.
It is effective against a number of types of bacteria,large viruses,and typhus fever.
Since the reason correctly explains why $Tetracycline$ is classified as a broad spectrum antibiotic,both statements are correct and the reason is the correct explanation of the assertion.