AIIMS 2002 Chemistry Question Paper with Answer and Solution

(A) The ladder is placed against a smooth vertical wall,meaning there is no friction at the wall contact point.
The weight of the ladder acts vertically downwards,$W = 250 \ N$.
The normal reaction force $R$ exerted by the floor on the ladder must balance the weight of the ladder in the vertical direction.
Therefore,$R = W = 250 \ N$.
The maximum force of static friction $f_{max}$ available at the contact point between the ladder and the floor is given by the formula $f_{max} = \mu R$,where $\mu$ is the coefficient of friction.
Given $\mu = 0.3$ and $R = 250 \ N$,we have:
$f_{max} = 0.3 \times 250 \ N = 75 \ N$.

(A) The de Broglie wavelength $\lambda$ is given by the formula $\lambda = \frac{h}{p} = \frac{h}{mv}$.
Rearranging for velocity $v$,we get $v = \frac{h}{m\lambda}$.
Substituting the values: $h = 6.63 \times 10^{-34} \ J \cdot s$,$m = 9.11 \times 10^{-31} \ kg$,and $\lambda = 10^{-10} \ m$.
$v = \frac{6.63 \times 10^{-34}}{(9.11 \times 10^{-31}) \times (10^{-10})} = \frac{6.63}{9.11} \times 10^{7} \ m/s \approx 0.727 \times 10^{7} \ m/s = 7.27 \times 10^6 \ m/s$.
Rounding to the nearest provided option,the speed is $7.25 \times 10^6 \ m/s$.

(B) The atomic mass of sulphur is $32 \ g/mol$.
Given that the compound contains $8\%$ sulphur by mass,it means $8 \ g$ of sulphur is present in $100 \ g$ of the compound.
To find the least molecular mass,we assume there is at least one atom of sulphur in the molecule.
Therefore,$1 \ g$ of sulphur is present in $\frac{100}{8} \ g$ of the compound.
For $32 \ g$ of sulphur (one mole of atoms),the mass of the compound is $\frac{100}{8} \times 32 = 400 \ g/mol$.

(C) The azimuthal quantum number $(l)$ determines the shape of the orbital and the angular momentum of the electron in a given orbital. The magnitude of angular momentum is given by the formula $\sqrt{l(l+1)} \frac{h}{2\pi}$.

(C) Quantum numbers are derived from the solution of the $Schr\ddot{o}dinger$ wave equation for the hydrogen atom. These numbers describe the size,shape,and orientation of orbitals,as well as the spin of electrons. While the principles listed in the options (Hund's rule,Aufbau's principle,Pauli's exclusion principle) govern the filling of electrons in orbitals,the quantum numbers themselves are fundamental mathematical solutions to the wave equation,which is consistent with the quantum mechanical model of the atom.

(B) The energy of an electronic configuration is determined by the sum of the energies of all the electrons present in the orbitals.
For a given principal quantum number $n=3$,the energy of the orbitals follows the order $3s < 3p < 3d$.
To find the configuration with the maximum energy,we compare the total number of electrons in the higher energy orbitals ($3p$ and $3d$).
Option $A$: $2$ electrons in $3s$,$2$ in $3p$,$0$ in $3d$. Total electrons in $3p+3d = 2$.
Option $B$: $2$ electrons in $3s$,$3$ in $3p$,$2$ in $3d$. Total electrons in $3p+3d = 5$.
Option $C$: $2$ electrons in $3s$,$3$ in $3p$,$1$ in $3d$. Total electrons in $3p+3d = 4$.
Option $D$: $2$ electrons in $3s$,$3$ in $3p$,$1$ in $3d$. Total electrons in $3p+3d = 4$.
Comparing the total number of electrons in the higher energy orbitals,Option $B$ has the highest number of electrons in the $3p$ and $3d$ orbitals,thus representing the state with the maximum energy.

(C) The bond energy depends on the bond order.
$N_2$ has a triple bond $(N \equiv N)$ with a bond order of $3$.
$O_2$ has a double bond $(O=O)$ with a bond order of $2$.
$F_2$ has a single bond $(F-F)$ with a bond order of $1$.
$C_2$ has a bond order of $2$.
Since bond energy is directly proportional to bond order,$N_2$ has the highest bond energy.

(A) $H_2C=O$ (formaldehyde) has the highest dipole moment due to the presence of a highly polar $C=O$ bond.
Oxygen is significantly more electronegative than carbon,leading to a large bond dipole.
In trans-isomers like trans-but$-2-$ene and trans$-2,3-$dichloro$-2-$butene,the bond dipoles cancel each other out due to molecular symmetry,resulting in a net dipole moment of approximately zero.
$2$-Methylpropene has a small dipole moment,but it is much lower than that of formaldehyde.

(A) $BF_3$ has a trigonal planar structure,making it non-polar with a dipole moment of $0 \ D$.
In $NH_3$,the orbital dipole due to the lone pair and the bond dipoles of $N-H$ bonds are in the same direction,resulting in a large net dipole moment of $1.47 \ D$.
In $NF_3$,the orbital dipole due to the lone pair and the bond dipoles of $N-F$ bonds are in opposite directions,resulting in a smaller net dipole moment of $0.24 \ D$.
Therefore,the correct order of increasing dipole moments is $BF_3 < NF_3 < NH_3$.

(C) In distilled water,the auto-ionization reaction is $2H_2O(l) \rightleftharpoons H_3O^{+}(aq) + OH^{-}(aq).$
Since the water is neutral,the concentration of hydronium ions is equal to the concentration of hydroxide ions: $[H_3O^{+}] = [OH^{-}] = 1 \times 10^{-6} \, mol/L.$
The ionic product of water,$K_w,$ is defined as $K_w = [H_3O^{+}][OH^{-}].$
Substituting the given values: $K_w = (1 \times 10^{-6}) \times (1 \times 10^{-6}) = 1 \times 10^{-12}.$
Therefore,the correct option is $C.$

(A) The dissociation of $CuBr$ is given by: $CuBr(s) \rightleftharpoons Cu^{+}(aq) + Br^{-}(aq)$
Let $S$ be the solubility of $CuBr$ in $mol/L$.
Then,$[Cu^{+}] = S$ and $[Br^{-}] = S$.
The solubility product constant $K_{sp}$ is given by: $K_{sp} = [Cu^{+}][Br^{-}] = S \times S = S^2$.
Given $S = 2 \times 10^{-4} \ mol/L$,we have:
$K_{sp} = (2 \times 10^{-4})^2 = 4 \times 10^{-8} \ mol^2 \ L^{-2}$.

(B) The solution is a buffer solution consisting of a weak acid $(CH_3COOH)$ and its salt with a strong base $(CH_3COONa)$.
We use the Henderson-Hasselbalch equation: $pH = pK_a + \log \frac{[salt]}{[acid]}$.
Given: $pK_a = 4.57$,$[salt] = 0.10\,M$,$[acid] = 0.03\,M$.
Substituting the values: $pH = 4.57 + \log \frac{0.10}{0.03} = 4.57 + \log(3.33)$.
Since $\log(3.33) \approx 0.52$,we get $pH = 4.57 + 0.52 = 5.09$.

(B) The reaction between a strong acid $(HNO_3)$ and a strong base $(KOH)$ is represented as: $H^+ + OH^- \rightarrow H_2O$,$\Delta H = -57.0 \, kJ \, mol^{-1}$.
Since $0.2 \, mole$ of $KOH$ is the limiting reagent,it will react with $0.2 \, mole$ of $HNO_3$ to produce $0.2 \, mole$ of $H_2O$.
The heat released is calculated as: $\text{Heat} = \Delta H \times \text{moles of water formed} = 57.0 \, kJ \, mol^{-1} \times 0.2 \, mol = 11.4 \, kJ$.

(D) Let the average oxidation state of $Fe$ be $x$.
In $Fe_3O_4$,the oxidation state of oxygen is $-2$.
The sum of oxidation states of all atoms in a neutral molecule is $0$.
$3x + 4(-2) = 0$
$3x - 8 = 0$
$3x = 8$
$x = \frac{8}{3}$

(A) The electronic configuration of $Li$ is $1s^2 \, 2s^1$,so $Li^{2+}$ has the electronic configuration $1s^1$.
Bohr's model and the resulting spectra are applicable to hydrogen-like species,which contain only one electron.
Since $Li^{2+}$ has only one electron,its spectrum is similar to that of the hydrogen atom $(H)$,which also has the configuration $1s^1$.

(A) The reaction of magnesium $(Mg)$ with water $(H_2O)$ produces magnesium hydroxide $(Mg(OH)_2)$ and hydrogen gas $(H_2)$:
$Mg + 2H_2O \to Mg(OH)_2 + H_2 \uparrow$
The other reactions listed produce hydrogen peroxide $(H_2O_2)$ instead of hydrogen gas.

(B) $1$. Identify the longest carbon chain containing the functional group (nitrile group). The chain has $6$ carbons,so the parent alkane is hexane.
$2$. Number the chain starting from the nitrile carbon as $C-1$.
$3$. The nitrile group is at $C-1$,so the suffix is $nitrile$.
$4$. There is a methyl group at the $4^{th}$ carbon.
$5$. The correct $IUPAC$ name is $4-$methylhexanenitrile.

(A) The structure of $pent-4-en-1-yne$ is $HC \equiv C-CH_2-CH=CH_2$.
Number of $\sigma$ bonds = $4$ ($C-C$ bonds) + $6$ ($C-H$ bonds) = $10$.
Number of $\pi$ bonds = $2$ (from $C \equiv C$) + $1$ (from $C=C$) = $3$.

(D) chiral compound is one that contains at least one chiral carbon atom (a carbon atom bonded to four different groups).
In $2,3,4-$trimethyl hexane,the carbon atom at position $3$ is bonded to a hydrogen atom,a methyl group,an ethyl group,and a $sec-$butyl group.
Since all four groups attached to the $C-3$ atom are different,it is a chiral center.
Therefore,$2,3,4-$trimethyl hexane is a chiral compound.

(C) The hydration of acetylene in the presence of dilute $H_2SO_4$ and $Hg^{2+}$ ions proceeds as follows:
$CH \equiv CH + H_2O \xrightarrow{H_2SO_4, Hg^{2+}} [CH_2 = CH - OH]$
This intermediate,vinyl alcohol,is unstable and undergoes tautomerization (rearrangement) to form a more stable carbonyl compound:
$[CH_2 = CH - OH] \rightarrow CH_3 - CHO$ (Acetaldehyde).

(D) $P(0)$: Phosphorus is a $p-$block element (Group $15$). Its valence shell is the $3rd$ shell $(3s^2 3p^3)$,and it does not have electrons in the $3d-$subshell.
$Fe(III)$: The electronic configuration is $[Ar] 3d^5$. It has valence electrons in the $3d-$subshell.
$Mn(II)$: The electronic configuration is $[Ar] 3d^5$. It has valence electrons in the $3d-$subshell.
$Cr(I)$: The electronic configuration is $[Ar] 3d^5$. It has valence electrons in the $3d-$subshell.
Therefore,$P(0)$ is the correct answer.

(D) The total number of individuals of a species per unit area or volume in a given habitat is known as absolute density. It represents the actual count of the population size within a specific area.

(D) For destructive interference to occur,the two waves must arrive at a point in opposite phases.
This happens when the path difference between the two waves is an odd multiple of half the wavelength.
The condition for destructive interference is given by the path difference $\Delta x = (2n + 1) \frac{\lambda}{2}$,where $n = 0, 1, 2, 3, \dots$.

(B) The electric potential $V$ at a point due to a point charge $Q$ at a distance $r$ is given by $V = \frac{1}{4\pi \epsilon_0} \cdot \frac{Q}{r}$.
In the given isosceles triangle,the distance from vertex $A$ to vertex $B$ is $a$,and the distance from vertex $A$ to vertex $C$ is also $a$ (since it is an isosceles triangle with $AB = AC = a$).
The potential at vertex $A$ due to charge $+q$ at $B$ is $V_B = \frac{1}{4\pi \epsilon_0} \cdot \frac{+q}{a}$.
The potential at vertex $A$ due to charge $-q$ at $C$ is $V_C = \frac{1}{4\pi \epsilon_0} \cdot \frac{-q}{a}$.
The total potential at vertex $A$ is the algebraic sum of the individual potentials: $V_A = V_B + V_C = \frac{1}{4\pi \epsilon_0} \cdot \frac{q}{a} + \frac{1}{4\pi \epsilon_0} \cdot \frac{-q}{a} = 0$.

(A) The formula for magnetic flux linkage in mutual induction is given by $\Delta \phi = M \Delta I$,where $\Delta \phi$ is the change in magnetic flux,$M$ is the coefficient of mutual inductance,and $\Delta I$ is the change in current.
Given:
$\Delta \phi = 2 \times 10^{-2} \, Wb$
$\Delta I = 0.01 \, A = 10^{-2} \, A$
Substituting the values into the formula:
$2 \times 10^{-2} = M \times 10^{-2}$
$M = \frac{2 \times 10^{-2}}{10^{-2}}$
$M = 2 \, H$
Therefore,the coefficient of mutual inductance is $2 \, H$.

(A) The maximum force of friction $(f_{max})$ is given by the formula $f_{max} = \mu N$,where $\mu$ is the coefficient of friction and $N$ is the normal reaction force.
Since the ladder is on a horizontal floor,the normal reaction force $N$ exerted by the floor on the ladder is equal to the weight of the ladder $(W = 250\,N)$.
Given $\mu = 0.3$ and $N = 250\,N$.
Therefore,$f_{max} = 0.3 \times 250 = 75\,N$.

(D) The Assertion is incorrect because,according to nuclear reactions,atoms can be created or destroyed (e.g.,in nuclear fission or fusion).
The Reason is also incorrect because,according to Avogadro's Law,equal volumes of all gases under the same conditions of temperature and pressure contain an equal number of molecules,and consequently,an equal number of atoms if the gases are monatomic.

(C) The Balmer series corresponds to transitions where $n_1 = 2$ and $n_2 = 3, 4, 5, ...$. Thus,the Assertion is correct.
For the highest wavelength,the energy difference between the levels must be the minimum.
Since $\Delta E = 13.6 \times Z^2 \times (\frac{1}{n_1^2} - \frac{1}{n_2^2}) \text{ eV}$ and $\lambda = \frac{hc}{\Delta E}$,the minimum energy transition for the Balmer series is from $n_2 = 3$ to $n_1 = 2$.
Therefore,the Reason is incorrect because the highest wavelength corresponds to the transition $n_2 = 3$ to $n_1 = 2$,not $n = 4$ and $6$.

(D) The Assertion is incorrect because an absorption spectrum consists of dark lines separated by bright spaces,which occur when specific wavelengths are absorbed by a substance.
The Reason is also incorrect because an emission spectrum consists of bright lines on a dark background,representing the light emitted by excited atoms or molecules.
Therefore,both the Assertion and the Reason are incorrect.

(C) The Assertion is correct because a sigma $(\sigma)$ bond is formed by the axial overlap of atomic orbitals,which is more extensive than the lateral overlap that forms a pi $(\pi)$ bond. Therefore,the sigma $(\sigma)$ bond is stronger than the pi $(\pi)$ bond.
The Reason is incorrect because atoms cannot rotate freely around a pi $(\pi)$ bond. The presence of a pi $(\pi)$ bond restricts rotation,as rotation would require breaking the lateral overlap of the $p$-orbitals.

(A) According to the ideal gas equation,$PV = nRT = (m/M)RT$,where $m$ is the mass and $M$ is the molar mass of the gas.
Rearranging the equation,we get $P = (m/V) \times (RT/M)$.
Since density $\rho = m/V$,we can substitute this into the equation: $P = \rho \times (RT/M)$.
At a constant temperature $T$,$R$ and $M$ are constants,so $P \propto \rho$.

(C) An adiabatic process is defined as a process where there is no exchange of heat between the system and the surroundings,i.e.,$q = 0$. Thus,the Assertion is correct.
During an adiabatic expansion,the system does work on the surroundings at the expense of its internal energy. Since $dU = dq + dw$ and $dq = 0$,$dU = dw$. For expansion,$dw < 0$,so $dU < 0$. Since internal energy is a function of temperature for an ideal gas,a decrease in internal energy leads to a decrease in temperature. Therefore,the Reason is incorrect.

(C) The properties of a system that depend upon the quantity of matter contained in it are called extensive properties,e.g.,mass,volume,heat capacity,etc.
However,$\text{Density} = \frac{\text{Mass}}{\text{Volume}}$.
Density is an intensive property because it is independent of the quantity of matter in a system.
Therefore,the Assertion is correct,but the Reason is incorrect.

(A) The internal energy of a substance is the sum of all forms of energy present in the system,such as electronic,nuclear,vibrational,and rotational energies.
Because it is impossible to determine the exact values of these constituent energies,the absolute value of the total internal energy of a substance cannot be determined.
Therefore,the Assertion is correct,and the Reason provides the correct explanation for it.

(A) Potassium $(K)$ and caesium $(Cs)$ are alkali metals with very low ionisation enthalpy.
Due to this low ionisation enthalpy,they can easily emit electrons when exposed to light (photoelectric effect).
Because of this property,they are widely used in photoelectric cells.
Therefore,the Reason correctly explains the Assertion.

(C) $Sn^{2+}$ is a strong reducing agent and it gets oxidized to $Sn^{4+}$.
The reaction with mercuric chloride $(HgCl_2)$ proceeds as follows:
$2HgCl_2 + SnCl_2 \to Hg_2Cl_2 + SnCl_4$ (white precipitate)
$Hg_2Cl_2 + SnCl_2 \to 2Hg + SnCl_4$ (grey precipitate of metallic mercury)
Thus,the Assertion is correct because $SnCl_2$ acts as a reducing agent,not an oxidizing agent. Therefore,the Reason is incorrect.

(B) Diamond is a bad conductor because all $4$ valence electrons of carbon are involved in $sp^3$ covalent bonding,leaving no free electrons.
Graphite is a good conductor of electricity because each carbon atom is $sp^2$ hybridized,leaving one free electron per carbon atom in its lattice.
Both the Assertion and the Reason are correct statements,but the fact that graphite is a good conductor does not explain why diamond is a bad conductor. Therefore,the Reason is not the correct explanation of the Assertion.

(C) pure line refers to a population of individuals that are genetically identical and homozygous for the traits of interest.
When closely related individuals are mated within the same breed for $4-6$ generations,the progeny obtained exhibit increased homozygosity.
This process,known as inbreeding,leads to the development of a pure line by eliminating recessive deleterious alleles and fixing desirable traits.

(D) The correct answer is $(d)$.
In the manufacturing process of lead shots,molten lead is poured from a height through a sieve. As the liquid lead falls,it forms droplets. Due to the property of surface tension,the free surface of a liquid drop always tends to minimize its surface area for a given volume. Since a sphere has the minimum surface area for a given volume,the falling droplets naturally take on a spherical shape before solidifying,resulting in spherical lead shots.

(C) $H_3PO_4$ (orthophosphoric acid) is a tribasic acid because it contains three replaceable hydrogen atoms attached to oxygen atoms as $P-OH$ groups.
The structure is $O=P(OH)_3$.

(C) The solubility of noble gases in water increases with an increase in atomic size and polarizability.
As we move down the group from $He$ to $Xe$,the atomic size increases,leading to stronger van der Waals forces of attraction between the gas molecules and water molecules.
Therefore,the correct order of solubility is $Xe > Kr > Ar > Ne > He$.

(C) The $AB_2$ type of structure is characteristic of the fluorite structure,which is exhibited by $CaF_2$.
In this structure,the cation $Ca^{2+}$ occupies the face-centered cubic lattice sites,while the anions $F^-$ occupy all the tetrahedral voids.
The dissociation can be represented as: $CaF_2 \rightleftharpoons Ca^{2+} + 2F^-$.

(A) $Schottky$ defect is a type of point defect that occurs in the crystal lattice of a $solid$. It arises when oppositely charged ions leave their lattice sites,creating vacancies to maintain electrical neutrality.

(B) For a $bcc$ structure, the number of atoms per unit cell $(n)$ is $2$.
Given: Atomic mass $(M)$ = $100 \ g/mol$, edge length $(a)$ = $400 \ pm = 400 \times 10^{-10} \ cm$, Avogadro's number $(N_A)$ = $6.022 \times 10^{23} \ mol^{-1}$.
The formula for density $(\rho)$ is $\rho = \frac{n \times M}{a^3 \times N_A}$.
Substituting the values: $\rho = \frac{2 \times 100}{(400 \times 10^{-10})^3 \times 6.022 \times 10^{23}}$.
$\rho = \frac{200}{64 \times 10^{-24} \times 6.022 \times 10^{23}} = \frac{200}{38.54} \approx 5.189 \ g/cm^3$.
Thus, the correct option is $(B)$.

(D) . The rate of this photochemical reaction is independent of the concentration of the reactants.
Therefore,it is a zero order reaction.

(C) Enzymes that possess two distinct sites are known as allosteric enzymes.
These two sites are the $active \ site$ and the $allosteric \ site$.
The $active \ site$ is where the substrate binds to undergo the catalytic reaction.
The $allosteric \ site$ is a specific region on the enzyme where an effector or modulator molecule can bind.
This binding is reversible and regulates the enzyme's activity by inducing a conformational change.

(A) The size of colloidal particles is intermediate between true solutions and suspensions.
The range of particle size for colloids is typically $1 \ nm$ to $1000 \ nm$.
Converting this to meters: $1 \ nm = 10^{-9} \ m$ and $1000 \ nm = 10^{-6} \ m$.
Among the given options,the range $10^{-9} \ m$ to $10^{-7} \ m$ (which is $1 \ nm$ to $100 \ nm$) represents the standard colloidal range.
Therefore,the correct option is $A$.

(C) The chemical formulas for the given ores are as follows:
$1$. Haematite: $Fe_2O_3$ (Ore of iron)
$2$. Magnetite: $Fe_3O_4$ (Ore of iron)
$3$. Cassiterite: $SnO_2$ (Ore of tin)
$4$. Limonite: $Fe_2O_3 \cdot 3H_2O$ (Ore of iron)
Therefore,Cassiterite is the ore that does not represent an ore of iron.

(B) The reaction of an alcohol with thionyl chloride $(SOCl_2)$ in the presence of pyridine is a standard method for the preparation of alkyl chlorides.
This specific reaction is known as Darzen's procedure.
It is preferred because the by-products ($SO_2$ and $HCl$) are gases,which escape,leaving behind pure alkyl chloride.

(A) Lucas test is used to distinguish between primary,secondary,and tertiary alcohols based on the rate of reaction with Lucas reagent (a mixture of concentrated $HCl$ and anhydrous $ZnCl_2$).

(C) The reaction between an aldehyde $(C_6H_5CHO)$ and a primary amine $(C_6H_5NH_2)$ results in the formation of an imine,which is commonly referred to as a Schiff's base.
In this specific reaction,benzaldehyde reacts with aniline to form $N$-benzylideneaniline,which is a Schiff's base.

(D) Teflon is a polymer of $Tetrafluoroethene$.
It is obtained by the polymerisation of $Tetrafluoroethene$ $(CF_2 = CF_2)$ under high pressure in the presence of a peroxide or persulphate catalyst.
Therefore,the correct option is $(D)$.

(D) Amides generally exhibit high boiling and melting points compared to other compounds of similar molecular mass,but when compared to their corresponding carboxylic acids,they are generally lower or comparable depending on the specific structure.
However,the statement that amides have higher boiling and melting points than corresponding acids is scientifically incorrect,as carboxylic acids form stronger intermolecular hydrogen bonds (dimers) than amides.
Therefore,the Assertion is incorrect,and the Reason is correct regarding the presence of intermolecular hydrogen bonding in amides.

(A) The sky appears blue due to the scattering of sunlight by colloidal particles (dust and air molecules) present in the atmosphere.
According to Rayleigh scattering,the intensity of scattered light is inversely proportional to the fourth power of the wavelength $(I \propto 1/\lambda^4)$.
Since blue light has a shorter wavelength compared to red light,it is scattered more effectively by these particles,making the sky appear blue.

(D) The Assertion is incorrect because physical adsorption is a surface phenomenon,but the term used in the assertion is 'absorption',which is a bulk phenomenon.
Furthermore,the Reason is incorrect because physical adsorption involves weak van der Waals forces and does not involve the breaking of chemical bonds.
Therefore,both the Assertion and the Reason are incorrect.

(D) Fluorine $(F_2)$ is the most reactive halogen due to its extremely low bond dissociation energy,which makes the $F-F$ bond very easy to break.
Therefore,the Assertion that fluorine has lower reactivity is incorrect,while the Reason that the $F-F$ bond has low bond dissociation energy is correct.

(D) The Assertion is incorrect because the dinegative anion of sulphur $(S^{2-})$ is actually very common (e.g.,in metal sulphides like $ZnS$,$FeS$),whereas the formation of $O^{2-}$ is energetically demanding due to high electron-electron repulsion in the small oxygen atom.
The Reason is also incorrect because the covalency of oxygen is generally two,but this is not the reason for the stability or commonality of its anions.
Therefore,both the Assertion and the Reason are incorrect.

(A) All halogens are coloured because they absorb light in the visible region of the electromagnetic spectrum.
This absorption of visible light causes the excitation of electrons to higher energy levels,which results in the characteristic colours of the halogens.
Therefore,the Assertion is correct,and the Reason is the correct explanation for the Assertion.

(D) The electronic configuration of $Cu$ is $[Ar] 3d^{10} 4s^1$.
For $Cu^{+}$ (cuprous ion),the configuration is $[Ar] 3d^{10}$,which has no unpaired electrons.
For $Cu^{2+}$ (cupric ion),the configuration is $[Ar] 3d^9$,which has one unpaired electron.
Therefore,the Assertion is incorrect because it states the opposite.
$Cu^{+}$ is colourless due to the absence of unpaired electrons,while $Cu^{2+}$ is blue in aqueous solution due to $d-d$ transitions in the presence of water ligands. Thus,the Reason is correct.
The correct option is $D$.

(B) Phenol $(C_6H_5OH)$ is a stronger acid than ethanol $(C_2H_5OH)$ because the phenoxide ion $(C_6H_5O^-)$ formed after the loss of a proton is resonance-stabilized,whereas the ethoxide ion $(C_2H_5O^-)$ is not.
Groups with $+M$ effect (like $-OH$,$-OR$,$-NH_2$) increase electron density in the ring,which destabilizes the phenoxide ion and decreases the acidity of phenols when present at the $p-$position.
Both the Assertion and the Reason are scientifically correct statements,but the Reason does not explain why phenol is a stronger acid than ethanol.
Therefore,the correct option is $(b)$.

(D) Sucrose is a non-reducing sugar because it does not contain a free hemiacetal or hemiketal group.
Mutarotation is a property of reducing sugars that possess a free hemiacetal or hemiketal group,allowing them to exist in equilibrium between their $\alpha$ and $\beta$ anomeric forms.
Since sucrose lacks this free group,it does not undergo mutarotation.
Therefore,the Assertion is incorrect,while the Reason is correct.

(D) $DNA$ is found mainly in the nucleus of the cell,while $RNA$ is found mainly in the cytoplasm of the cell. Therefore,the assertion is incorrect.
Enzymes are biological catalysts that function within a specific temperature range. Upon heating,they undergo denaturation,which causes them to lose their specific biological activity. Therefore,the reason is also incorrect.