AIIMS 1999 Chemistry Question Paper with Answer and Solution

(B) Given,half-life $T_{1/2} = 20 \ minutes$.
We use the radioactive decay law: $\frac{N}{N_0} = \left( \frac{1}{2} \right)^{t/T_{1/2}}$.
For $20\%$ decay,the remaining amount is $N_1 = 80\%$ of $N_0$,so $\frac{N_1}{N_0} = 0.8 = \left( \frac{1}{2} \right)^{t_1/20}$ ... $(i)$.
For $80\%$ decay,the remaining amount is $N_2 = 20\%$ of $N_0$,so $\frac{N_2}{N_0} = 0.2 = \left( \frac{1}{2} \right)^{t_2/20}$ ... $(ii)$.
Dividing equation $(i)$ by $(ii)$:
$\frac{0.8}{0.2} = \frac{(1/2)^{t_1/20}}{(1/2)^{t_2/20}} = \left( \frac{1}{2} \right)^{(t_1-t_2)/20} = \left( \frac{1}{2} \right)^{-(t_2-t_1)/20} = 2^{(t_2-t_1)/20}$.
$4 = 2^{(t_2-t_1)/20}$.
Since $4 = 2^2$,we have $2 = \frac{t_2-t_1}{20}$.
Therefore,$t_2 - t_1 = 40 \ minutes$.

(A) The principal quantum number $(n)$ determines the size of the orbital.
The azimuthal quantum number $(l)$ determines the shape of the orbital.
The magnetic quantum number $(m)$ determines the orientation of the orbital in space.
Therefore,they are respectively related to size,shape,and orientation.

(C) According to Fajan's rule,covalent character in an ionic bond is favoured by:
$1$. Small size of the cation.
$2$. Large size of the anion.
$3$. High charge on the cation or anion.
$4$. Cation with pseudo-noble gas configuration (e.g.,$Cu^+$,$Zn^{2+}$).
Therefore,a small cation and a large anion favour covalent bond formation.

(A) $He$ is a noble gas (Group $18$ element).
Noble gases consist of individual atoms held together by very weak London dispersion forces (a type of Van der Waals force).
In contrast,$HCl$ exhibits dipole-dipole interactions,while $NH_3$ and $H_2O$ exhibit strong hydrogen bonding.
Therefore,$He$ exhibits the weakest intermolecular forces.

(C) The strength of intermolecular forces is determined by the nature of the particles.
$NH_3$ and $H_2O$ exhibit strong hydrogen bonding.
$HCl$ exhibits dipole-dipole interactions.
$He$ is a noble gas and only exhibits very weak London dispersion forces (van der Waals forces).
Therefore,$He$ has the weakest intermolecular forces among the given options.

(D) The equilibrium constant $(K_c)$ is defined as the ratio of the rate constant of the forward reaction $(K_f)$ to the rate constant of the backward reaction $(K_b)$.
Given: $K_f = 1.1 \times 10^{-2} \text{ min}^{-1}$ and $K_b = 1.5 \times 10^{-3} \text{ min}^{-1}$.
$K_c = \frac{K_f}{K_b} = \frac{1.1 \times 10^{-2}}{1.5 \times 10^{-3}} = \frac{11 \times 10^{-3}}{1.5 \times 10^{-3}} = \frac{11}{1.5} = 7.33$.
Therefore,the correct option is $(D)$.

(B) The relationship between $K_p$ and $K_c$ is given by the equation $K_p = K_c(RT)^{\Delta n}$.
For the reaction $2H_2S_{(g)} \rightleftharpoons 2H_{2(g)} + S_{2(g)}$,the change in the number of moles of gaseous products and reactants is $\Delta n = (2 + 1) - 2 = 1$.
Substituting $\Delta n = 1$ into the equation,we get $K_p = K_c(RT)^1$,which implies $K_c = K_p / (RT)$.
Since the temperature $T = 106.5 + 273.15 = 379.65 K$ is greater than $1$,the value of $(RT)$ is much greater than $1$.
Therefore,$K_c = K_p / (RT)$ will be less than $K_p$.
Thus,$K_c < 1.2 \times 10^{-2}$.

(D) The heat of neutralisation is defined as the amount of heat released when $1 \ gram$ equivalent of an acid is neutralised by $1 \ gram$ equivalent of a base.
For the reaction between a strong acid and a strong base,the enthalpy change is constant at approximately $-57.1 \ kJ \ mol^{-1}$ or $-13.7 \ kcal \ mol^{-1}$,because the reaction is essentially the formation of water from $H^+$ and $OH^-$ ions.
When a weak acid or a weak base is involved,some of the heat released is consumed in the dissociation of the weak electrolyte.
Since $NaOH$ is a strong base and $HCl$ is a strong acid,their neutralisation reaction releases the maximum amount of heat compared to reactions involving weak acids $(CH_3COOH)$ or weak bases $(NH_4OH)$.
Therefore,the correct option is $(D)$.

(B) The chemical equation for the formation of ferric oxide $(Fe_2O_3)$ is: $2Fe(s) + \frac{3}{2}O_2(g) \rightarrow Fe_2O_3(s)$.
Given that $4 \ g$ of iron releases $29.28 \ kJ$ of heat.
The molar mass of iron $(Fe)$ is $56 \ g \ mol^{-1}$.
For $2 \ moles$ of iron,the mass is $2 \times 56 = 112 \ g$.
Since heat is evolved,the enthalpy change $(\Delta H)$ is negative.
For $4 \ g$ of $Fe$,$\Delta H = -29.28 \ kJ$.
For $112 \ g$ of $Fe$ (which corresponds to $1 \ mole$ of $Fe_2O_3$),the enthalpy of formation is:
$\Delta H_f = \frac{-29.28 \ kJ}{4 \ g} \times 112 \ g = -819.8 \ kJ \ mol^{-1}$.

(D) The standard free energy change is calculated using the Gibbs-Helmholtz equation: $\Delta G^\circ = \Delta H^\circ - T\Delta S^\circ$.
Given: $\Delta H^\circ = 58.04 \, kJ = 58040 \, J$,$\Delta S^\circ = 176.7 \, J/K$,and $T = 25 + 273 = 298 \, K$.
Substituting the values: $\Delta G^\circ = 58040 \, J - (298 \, K \times 176.7 \, J/K)$.
$\Delta G^\circ = 58040 \, J - 52656.6 \, J = 5383.4 \, J$.
Converting to $kJ$: $\Delta G^\circ = 5.3834 \, kJ \approx 5.39 \, kJ$.

(D) Let the oxidation number of $Os$ be $x$.
In $OsO_4$,the oxidation number of oxygen is $-2$.
Applying the rule that the sum of oxidation numbers in a neutral molecule is zero:
$x + 4(-2) = 0$
$x - 8 = 0$
$x = +8$
Therefore,the oxidation number of $Os$ in $OsO_4$ is $+8$.

(D) The molecular weight of $H_3PO_4$ is $3 \times 1 + 1 \times 31 + 4 \times 16 = 98 \ g/mol$.
In the given reaction,$NaOH + H_3PO_4 \to NaH_2PO_4 + H_2O$,one molecule of $H_3PO_4$ reacts with one molecule of $NaOH$ to replace one $H^+$ ion.
Therefore,the n-factor (basicity) of $H_3PO_4$ in this reaction is $1$.
The equivalent weight is calculated as: $\text{Equivalent weight} = \frac{\text{Molecular weight}}{\text{n-factor}} = \frac{98}{1} = 98$.

(A) $1$. In a period,atomic radius decreases from left to right. Therefore,the atomic radius of $O > F$.
$2$. $O^{2-}$ and $F^{-}$ are isoelectronic species,both having $10$ electrons.
$3$. For isoelectronic species,the ionic radius increases as the nuclear charge (atomic number) decreases.
$4$. Since the atomic number of $O$ $(Z=8)$ is less than $F$ $(Z=9)$,the ionic radius of $O^{2-} > F^{-}$.
$5$. Combining these,the overall order is $O^{2-} > F^{-} > O > F$.

(B) Hydrogen has only one orbital which is strongly attracted by the nucleus.
Thus,it has a very small size compared to alkali metals,and hence it has a high ionization enthalpy.

(C) is calcium hydroxide,$Ca(OH)_2$,also known as slaked lime.
$Ca(OH)_2$ is used in the Clark's process for the removal of temporary hardness of water.
It reacts with sodium carbonate $(Na_2CO_3)$ to produce caustic soda $(NaOH)$: $Ca(OH)_2 + Na_2CO_3 \to CaCO_3 + 2NaOH$.
When $CO_2$ is passed through $Ca(OH)_2$ solution,it forms $CaCO_3$,which makes the solution cloudy: $Ca(OH)_2 + CO_2 \to CaCO_3 + H_2O$.

(D) When sodium is heated in moist air,it first reacts with oxygen to form sodium oxide $(Na_2O)$:
$2Na + \frac{1}{2}O_2 \rightarrow Na_2O$
Then,the sodium oxide reacts with the moisture (water vapor) present in the air to form sodium hydroxide $(NaOH)$:
$Na_2O + H_2O \rightarrow 2NaOH$
Thus,the final mixture contains both $Na_2O$ and $NaOH$.

(A) Alkaline earth metals are denser than alkali metals because they have two valence electrons,which leads to stronger metallic bonding.
This allows the atoms to be packed more tightly due to their greater nuclear charge and smaller atomic radii compared to alkali metals.

(D) Anhydrous $CuSO_4$ is used to test the presence of water in any liquid.
It is white in its anhydrous form and changes its colour to blue upon hydration due to the formation of $CuSO_4 \cdot 5H_2O$.

(A) . As the number of branches increases,the surface area of the molecule decreases.
This reduction in surface area leads to weaker $Vander \ Waals$ forces of attraction between the molecules.
Consequently,the boiling point $(B.P.)$ of branched-chain alkanes is lower than that of their corresponding straight-chain isomers.

(A) Alkenes are unsaturated hydrocarbons containing a carbon-carbon double bond $(C=C)$.
Due to the presence of the $\pi$-bond,they are electron-rich and generally undergo electrophilic addition reactions.

(D) The reaction of propene $(CH_3-CH=CH_2)$ with $HBr$ follows Markovnikov's rule.
According to this rule,the negative part of the addendum $(Br^-)$ attaches to the carbon atom having a lesser number of hydrogen atoms.
Thus,the major product formed is $2-$bromopropane $(CH_3-CH(Br)-CH_3)$.

(D) When benzene reacts with chlorine in the presence of sunlight ($UV$ light),an addition reaction occurs instead of substitution.
The delocalized $\pi$-electron system of the benzene ring is broken,and one chlorine atom adds to each of the six carbon atoms.
The product formed is $1,2,3,4,5,6-$hexachlorocyclohexane,commonly known as benzene hexachloride $(BHC)$ or gammexane,with the formula $C_6H_6Cl_6$.

(B) $1$. The reaction of benzene $(C_6H_6)$ with nitrating mixture $(HNO_3 + H_2SO_4)$ is an electrophilic aromatic substitution reaction that yields nitrobenzene $(X)$.
$2$. Nitrobenzene $(X)$ contains a $-NO_2$ group,which is a strong electron-withdrawing group and a meta-directing group.
$3$. When nitrobenzene reacts with chlorine $(Cl_2)$ in the presence of a Lewis acid catalyst $(FeCl_3)$,the incoming electrophile $(Cl^+)$ is directed to the meta-position.
$4$. Therefore,the product $Y$ is $3-$nitrochlorobenzene (also known as $m-$nitrochlorobenzene).

(B) $Graphite$ is an allotrope of carbon where each carbon atom is $sp^2$ hybridized,leaving one free electron per atom. These delocalized electrons allow $Graphite$ to conduct electricity,unlike $Diamond$ which has all four valence electrons involved in covalent bonding.

(D) The $Endoplasmic \text{ } reticulum$ $(ER)$ is a network of membranous tubules that extends from the plasma membrane to the nuclear membrane, providing mechanical support to the cell.
$Rough \text{ } Endoplasmic \text{ } Reticulum$ $(RER)$ has ribosomes attached to its surface, which are the sites of protein synthesis.
Furthermore, the $ER$ system acts as a transport network for various enzymes and proteins within the cell.

(A) Hormones are classified based on their chemical nature into peptide,steroid,iodothyronines,and amino-acid derivatives.
$1$. Epinephrine (also known as adrenaline) is derived from the amino acid tyrosine.
$2$. Progesterone and Estrogen are steroid hormones derived from cholesterol.
$3$. Prostaglandins are lipid compounds derived from fatty acids (arachidonic acid).
Therefore,Epinephrine is the correct answer as it is an amino acid derivative.

(A) The dot product of two vectors $\vec{P}$ and $\vec{Q}$ is defined as $\vec{P} \cdot \vec{Q} = PQ \cos \theta$,where $\theta$ is the angle between the vectors.
Given that $\vec{P} \cdot \vec{Q} = PQ$.
Substituting this into the formula: $PQ = PQ \cos \theta$.
Dividing both sides by $PQ$ (assuming $P, Q \neq 0$),we get $\cos \theta = 1$.
Since $\cos 0^\circ = 1$,the angle $\theta = 0^\circ$.

(C) $Ca(OH)_2$ (slaked lime) is used to remove temporary hardness from water.
Reaction with sodium carbonate:
$Na_2CO_3 + Ca(OH)_2 \rightarrow 2NaOH + CaCO_3$
Reaction with $CO_2$:
$Ca(OH)_2 + CO_2 \rightarrow CaCO_3(s) + H_2O$
The formation of $CaCO_3$ makes the solution milky.

(D) The given reaction is: $NaOH + H_3PO_4 \to NaH_2PO_4 + H_2O$.
In this reaction,only one $H^+$ ion from $H_3PO_4$ is replaced by one $Na^+$ ion.
Therefore,the n-factor (basicity) of $H_3PO_4$ in this specific reaction is $1$.
The molecular weight $(MW)$ of $H_3PO_4$ is $(3 \times 1) + 31 + (4 \times 16) = 98 \ g/mol$.
The equivalent weight $(EW)$ is calculated as: $EW = \frac{MW}{n\text{-factor}} = \frac{98}{1} = 98$.

(D) Internal energy $(U)$ is the sum of all microscopic forms of energy in a system,including translational,rotational,vibrational,electronic,and nuclear energy.
It does not include macroscopic forms of energy such as kinetic energy due to the motion of the system as a whole or potential energy due to the gravitational pull on the system.
Therefore,the correct answer is $D$.

(C) The empirical formula mass of $CH_2O$ is $(1 \times 12) + (2 \times 1) + (1 \times 16) = 30 \ g/mol$.
The value of $n$ is calculated as: $n = \frac{\text{Molecular weight}}{\text{Empirical formula mass}} = \frac{180}{30} = 6$.
The molecular formula is given by: $(\text{Empirical formula})_n = (CH_2O)_6 = C_6H_{12}O_6$.

(A) The principal quantum number $(n)$ describes the size of the orbital and the energy level.
The azimuthal quantum number $(l)$ describes the shape of the orbital.
The magnetic quantum number $(m_l)$ describes the orientation of the orbital in space.
Therefore,the principal,azimuthal,and magnetic quantum numbers are respectively related to size,shape,and orientation.

(A) The principal quantum number $(n)$ determines the size and energy of the orbital.
The azimuthal quantum number $(l)$ determines the shape of the orbital.
The magnetic quantum number $(m_l)$ determines the spatial orientation of the orbital.
Therefore,the correct sequence is size,shape,and orientation.

(C) The chemical $A$ is calcium hydroxide,$Ca(OH)_2$,also known as slaked lime.
$1$. $Ca(OH)_2$ is used in the Clark's process for removing temporary hardness of water by converting soluble bicarbonates into insoluble carbonates: $Ca(HCO_3)_2 + Ca(OH)_2 \rightarrow 2CaCO_3 \downarrow + 2H_2O$.
$2$. It reacts with sodium carbonate $(Na_2CO_3)$ to produce caustic soda $(NaOH)$: $Ca(OH)_2 + Na_2CO_3 \rightarrow CaCO_3 + 2NaOH$.
$3$. When $CO_2$ is passed through lime water $(Ca(OH)_2)$,it turns milky due to the formation of insoluble calcium carbonate: $Ca(OH)_2 + CO_2 \rightarrow CaCO_3 \downarrow + H_2O$.

(B) The thermal decomposition of ammonium nitrate at $250\,^oC$ produces nitrous oxide and water vapor.
The chemical equation is:
$NH_4NO_3 \xrightarrow{250\,^oC} N_2O + 2H_2O$

(C) The average kinetic energy $(E)$ of a gas molecule is directly proportional to its absolute temperature $(T)$,given by the relation $E = \frac{3}{2} k_B T$,where $k_B$ is the Boltzmann constant.
Therefore,$E \propto T$,which implies $\frac{E_1}{E_2} = \frac{T_1}{T_2}$.
Given temperatures in Kelvin:
$T_1 = 27 + 273 = 300\,K$
$T_2 = 227 + 273 = 500\,K$
Given energy $E_1 = 6.21 \times 10^{-21}\,J$.
Substituting the values into the ratio:
$\frac{6.21 \times 10^{-21}}{E_2} = \frac{300}{500}$
$\frac{6.21 \times 10^{-21}}{E_2} = \frac{3}{5}$
$E_2 = \frac{6.21 \times 10^{-21} \times 5}{3}$
$E_2 = 2.07 \times 10^{-21} \times 5 = 10.35 \times 10^{-21}\,J$.

(C) The chemical $A$ is calcium hydroxide,$Ca(OH)_2$,also known as slaked lime.
$1$. $Ca(OH)_2$ is used in the Clark's process to remove temporary hardness of water by converting soluble bicarbonates into insoluble carbonates: $Ca(HCO_3)_2 + Ca(OH)_2 \rightarrow 2CaCO_3 \downarrow + 2H_2O$.
$2$. It reacts with sodium carbonate $(Na_2CO_3)$ to produce caustic soda $(NaOH)$: $Ca(OH)_2 + Na_2CO_3 \rightarrow CaCO_3 + 2NaOH$.
$3$. When $CO_2$ is passed through lime water $(Ca(OH)_2)$,it turns milky due to the formation of insoluble $CaCO_3$: $Ca(OH)_2 + CO_2 \rightarrow CaCO_3 \downarrow + H_2O$.

(B) When ammonium nitrate $(NH_4NO_3)$ is heated,it undergoes thermal decomposition.
At temperatures around $250\,^oC$,the reaction is as follows:
$NH_4NO_3 \xrightarrow{\Delta} N_2O + 2H_2O$
Here,$N_2O$ is nitrous oxide (also known as dinitrogen monoxide).

(D) Atoms are electrically neutral because the number of electrons (negatively charged) and protons (positively charged) are equal in a neutral atom. Therefore,both the Assertion and the Reason are incorrect.

(B) The Assertion is correct: Water $(H_2O)$ exists as a liquid at room temperature due to the presence of intermolecular hydrogen bonding,whereas $H_2S$ does not form hydrogen bonds and exists as a gas.
The Reason is correct: Oxygen $(O_2)$ is paramagnetic due to the presence of two unpaired electrons in its antibonding molecular orbitals $(\pi^*)$.
However,the paramagnetism of oxygen is not the reason for the liquid state of water. Therefore,the Reason is not the correct explanation for the Assertion.

(A) Diffusion is the process by which matter is transported in small quantities in the absence of bulk flow.

(D) The molar mass of water vapor $(H_2O)$ is $18 \ g/mol$,while the average molar mass of dry air (mostly $N_2$ and $O_2$) is approximately $29 \ g/mol$.
Since water vapor replaces heavier air molecules,wet air is actually lighter than dry air,making the Assertion incorrect.
The density of dry air is much lower than the density of liquid water,making the Reason also incorrect.

(C) The chemical $A$ is calcium hydroxide,$Ca(OH)_2$,also known as slaked lime.
$1$. It is used for water softening to remove temporary hardness by reacting with calcium bicarbonate: $Ca(HCO_3)_2 + Ca(OH)_2 \to 2CaCO_3 \downarrow + 2H_2O$.
$2$. It reacts with sodium carbonate $(Na_2CO_3)$ to produce caustic soda $(NaOH)$: $Ca(OH)_2 + Na_2CO_3 \to 2NaOH + CaCO_3$.
$3$. When $CO_2$ is passed through lime water $(Ca(OH)_2)$,it turns milky due to the formation of insoluble calcium carbonate: $Ca(OH)_2 + CO_2 \to CaCO_3 \downarrow + H_2O$.

(C) The purification of alumina is carried out by the Baeyer's process,which involves the digestion of bauxite ore with a concentrated solution of $NaOH$ at high temperature and pressure to form soluble sodium aluminate.

(C) The windscreen of automobiles is made up of $Safety \ glass$.
$Safety \ glass$ is a type of glass that is processed by thermal or chemical treatments to increase its strength compared to normal glass.
It is designed to shatter into small,blunt pieces rather than sharp shards when broken,making it less likely to cause injury,which is essential for automotive safety.

(B) The thermal decomposition of ammonium nitrate $(NH_4NO_3)$ at $250 \, ^oC$ yields nitrous oxide $(N_2O)$ and water vapor.
The balanced chemical equation is:
$NH_4NO_3(s) \xrightarrow{\Delta} N_2O(g) + 2H_2O(g)$
Thus,the correct option is $B$.

(D) Toluene $(C_6H_5CH_3)$ can be oxidized to benzoic acid $(C_6H_5COOH)$ using strong oxidizing agents such as alkaline potassium permanganate $(KMnO_4)$ or acidic potassium dichromate $(K_2Cr_2O_7)$.
Both reagents are capable of oxidizing the alkyl side chain attached to the benzene ring to a carboxylic acid group.
Therefore,the correct option is $(d)$.

(C) Colligative properties are those properties of a solution that depend only on the number of solute particles present in the solution,regardless of their chemical nature.
Examples of colligative properties include:
$1$. Relative lowering of vapour pressure
$2$. Elevation of boiling point
$3$. Depression in freezing point
$4$. Osmotic pressure
Among the given options,$Osmotic \ pressure$ is a colligative property.

(A) For a body-centered cubic $(bcc)$ lattice,the relationship between the edge length $(a)$ and the atomic radius $(r)$ is given by: $4r = \sqrt{3} a$.
Given $a = 4.29 \ \mathring{A}$.
$r = \frac{\sqrt{3} \times 4.29}{4} \ \mathring{A}$.
$r = \frac{1.732 \times 4.29}{4} \ \mathring{A} = 1.8574 \ \mathring{A}$.
Since $1 \ \mathring{A} = 10^{-8} \ cm$,the radius is $1.8574 \times 10^{-8} \ cm$.

(C) The relationship between average life $(\tau)$ and half-life $(t_{1/2})$ is given by the formula: $\tau = 1.44 \times t_{1/2}$.
Given that $t_{1/2} = 1580 \ yrs$.
Therefore,$\tau = 1.44 \times 1580 \ yrs$.
$\tau = 2275.2 \ yrs = 2.275 \times 10^{3} \ yrs$.

(D) For a first order reaction,the rate law is given by: $Rate = K[A]$.
Given that the rate constant $(K) = 3 \times 10^{-6} \ s^{-1}$ and the initial concentration $[A] = 0.10 \ M$.
Substituting these values into the rate equation:
$Rate = (3 \times 10^{-6} \ s^{-1}) \times (0.10 \ M) = 3 \times 10^{-7} \ M \ s^{-1}$.
Therefore,the initial rate of reaction is $3 \times 10^{-7} \ M \ s^{-1}$.

(B) The reaction at the anode is: $Cl^{-} \to \frac{1}{2}Cl_2 + e^-$
The equivalent mass of $Cl_2$ is: $E_{Cl_2} = \frac{71}{2} = 35.5 \ g/eq$
Using Faraday's law of electrolysis: $W = \frac{E \times I \times t}{96500}$
Given: $I = 2 \ A$,$t = 30 \ min = 1800 \ s$,$E = 35.5 \ g/eq$
$W = \frac{35.5 \times 2 \times 1800}{96500} \approx 1.32 \ g$.

(C) For a galvanic cell,the cathode is the electrode with the higher reduction potential and the anode is the electrode with the lower reduction potential.
Here,$E^o_{Ag^+/Ag} = +0.80 \ V$ (cathode) and $E^o_{Cu^{2+}/Cu} = +0.34 \ V$ (anode).
The $emf$ of the cell is calculated as:
$E^o_{cell} = E^o_{cathode} - E^o_{anode}$
$E^o_{cell} = 0.80 \ V - 0.34 \ V = +0.46 \ V$.

(D) Flux is a substance added to the ore to remove gangue (impurities) by forming a fusible material called slag.
Depending on the nature of the impurity,different types of flux are used.
If the impurity is acidic (e.g.,$SiO_2$),a basic flux (e.g.,$CaO$) is used.
If the impurity is basic (e.g.,$FeO$),an acidic flux (e.g.,$SiO_2$) is used.
Therefore,flux is used to remove both acidic and basic impurities.

(D) $Mn$ (Manganese) belongs to group $7$ of the periodic table. Its electronic configuration is $[Ar] 3d^5 4s^2$. It can lose all $7$ valence electrons to exhibit a maximum oxidation state of $+7$,as seen in compounds like $KMnO_4$.

(C) $Cu$ is present in brass,bronze,and German silver.
All these substances are the alloys of copper.

(D) The reaction of copper sulphate with potassium cyanide proceeds in steps:
$1$. $CuSO_4 + 2KCN \to Cu(CN)_2 + K_2SO_4$
$2$. The unstable $Cu(CN)_2$ decomposes to give $Cu_2(CN)_2$ and cyanogen gas: $2Cu(CN)_2 \to Cu_2(CN)_2 + (CN)_2$
$3$. $Cu_2(CN)_2$ then reacts with excess $KCN$ to form the stable complex potassium tetracyanocuprate$(I)$: $Cu_2(CN)_2 + 6KCN \to 2K_3[Cu(CN)_4]$
Thus,the final product is $K_3[Cu(CN)_4]$.

(C) Vinegar is a dilute solution of acetic acid $(CH_3COOH)$.
It is produced by the fermentation of ethanol,which can be derived from cane sugar.
Vinegar typically contains $8-10\%$ acetic acid by volume.

(C) Bakelite is a thermosetting polymer.
It becomes infusible on heating and cannot be remoulded.

(D) Disaccharides are carbohydrates that,upon hydrolysis,yield two same or different monosaccharide units.
Sucrose is the most common disaccharide,commonly known as table sugar.
It is naturally occurring and found in many plants.
The molecular formula of sucrose is $C_{12}H_{22}O_{11}$.

(A) gene is defined as a specific segment of a $DNA$ molecule that contains the instructions to code for a particular protein or functional $RNA$ molecule.

(A) The relationship between specific conductivity $(K)$,resistance $(R)$,and cell constant $(G^*)$ is given by the formula: $K = \frac{1}{R} \times G^*$.
Rearranging the formula to solve for the cell constant: $G^* = K \times R$.
Given values: $K = 0.0212\,ohm^{-1}\,cm^{-1}$ and $R = 55\,ohm$.
Substituting the values: $G^* = 0.0212 \times 55 = 1.166\,cm^{-1}$.

(A) Water is a covalent compound,hence pure water is a weak electrolyte and feebly ionised,making it a poor conductor of electricity.
Addition of a small amount of acid or alkali provides ions that facilitate the flow of current,thereby increasing the conductivity of water for electrolysis.
Thus,both Assertion and Reason are correct,and the Reason is the correct explanation of the Assertion.

(D) Iodine $(I_2)$ is a non-polar covalent molecule.
According to the principle of "like dissolves like",non-polar substances are more soluble in non-polar solvents like $CCl_4$ than in polar solvents like water $(H_2O)$.
Therefore,the assertion that iodine is more soluble in water than in $CCl_4$ is incorrect.
Additionally,the reason that iodine is a polar compound is also incorrect,as $I_2$ is non-polar.
Thus,both the assertion and the reason are incorrect.

(A) Inert gases (Noble gases) have a stable electronic configuration with a complete octet ($ns^2 np^6$,except $He$ which is $1s^2$).
Because of this stable configuration,they do not need to form bonds with other atoms to achieve stability.
Therefore,they exist as monoatomic gases.
Thus,the Reason correctly explains the Assertion.

(A) Amines are organic derivatives of ammonia $(NH_3)$.
The nitrogen atom in amines possesses one lone pair of electrons.
This lone pair can be donated to a Lewis acid (like $H^+$),which makes amines basic in nature.
Therefore,the presence of a lone pair of electrons on the nitrogen atom is the correct explanation for the basic nature of amines.

(A) The Lassaigne's test (sodium fusion test) is used to detect nitrogen in organic compounds.
Benzene diazonium chloride $(C_6H_5N_2Cl)$ is unstable and decomposes upon heating to release $N_2$ gas before the fusion process is complete.
Consequently,it does not give the characteristic Prussian blue color test for nitrogen.
Therefore,both the Assertion and the Reason are correct,and the Reason is the correct explanation for the Assertion.