IIT JEE 2005 Chemistry Question Paper with Answer and Solution

(A) The number of radial nodes is calculated using the formula: $\text{Radial nodes} = (n - l - 1)$.
For $3s$ orbital: $n = 3, l = 0$.
$\text{Radial nodes} = 3 - 0 - 1 = 2$.
For $2p$ orbital: $n = 2, l = 1$.
$\text{Radial nodes} = 2 - 1 - 1 = 0$.
Therefore,the number of radial nodes for $3s$ and $2p$ orbitals are $2$ and $0$ respectively.

(D) To determine the number of lone pairs on the central atom,we calculate the valence electrons and subtract the electrons involved in bonding:
$1$. In $[ClO_3]^-$,the central $Cl$ atom has $7$ valence electrons + $1$ (negative charge) = $8$. It forms $3$ double bonds with $O$ atoms ($6$ electrons used). Lone pairs = $(8-6)/2 = 1$.
$2$. In $XeF_4$,the central $Xe$ atom has $8$ valence electrons. It forms $4$ single bonds with $F$ atoms ($4$ electrons used). Lone pairs = $(8-4)/2 = 2$.
$3$. In $SF_4$,the central $S$ atom has $6$ valence electrons. It forms $4$ single bonds with $F$ atoms ($4$ electrons used). Lone pairs = $(6-4)/2 = 1$.
$4$. In $[I_3]^-$,the central $I$ atom has $7$ valence electrons + $1$ (negative charge) = $8$. It forms $2$ single bonds with $I$ atoms ($2$ electrons used). Lone pairs = $(8-2)/2 = 3$.
Thus,$[I_3]^-$ has the maximum number of lone pairs $(3)$. The correct option is $(D)$.

(B) According to Graham's Law of diffusion,the rate of diffusion $(r)$ is inversely proportional to the square root of the molar mass $(M)$: $r \propto \frac{1}{\sqrt{M}}$.
For helium $(He)$ and methane $(CH_4)$,the ratio is given by: $\frac{r_{He}}{r_{CH_4}} = \sqrt{\frac{M_{CH_4}}{M_{He}}}$.
The molar mass of methane $(CH_4)$ is $16 \ g/mol$ and helium $(He)$ is $4 \ g/mol$.
Substituting the values: $\frac{r_{He}}{r_{CH_4}} = \sqrt{\frac{16}{4}} = \sqrt{4} = 2$.

(B) The reaction is: $CH_3NH_2 + HCl \rightarrow CH_3NH_3^+ + Cl^-$
Initial moles: $CH_3NH_2 = 0.1 \ mol$,$HCl = 0.08 \ mol$.
After reaction: $CH_3NH_2 = 0.02 \ mol$,$CH_3NH_3^+ = 0.08 \ mol$.
This forms a basic buffer solution.
Using the Henderson-Hasselbalch equation for a basic buffer:
$pOH = pK_b + \log \frac{[salt]}{[base]}$
$pK_b = -\log(5 \times 10^{-4}) = 4 - \log(5) = 4 - 0.699 = 3.301$.
$pOH = 3.301 + \log \frac{0.08}{0.02} = 3.301 + \log(4) = 3.301 + 0.602 = 3.903$.
$pH = 14 - pOH = 14 - 3.903 = 10.097$.
$[H^{+}] = 10^{-pH} = 10^{-10.097} \approx 8 \times 10^{-11} \ M$.

(A) For an adiabatic process,the first law of thermodynamics states that $\Delta U = q + w$. Since the process is adiabatic,$q = 0$,so $\Delta U = w$.
For an ideal gas,$\Delta U = n C_v \Delta T$. For a monoatomic gas,$C_v = \frac{3}{2} R$.
Thus,$n \times \frac{3}{2} R \times (T_f - T) = -P_{ext} \times \Delta V$.
Given $n = 1 \, \text{mole}$,$P_{ext} = 1 \, \text{atm}$,$\Delta V = (2 - 1) \, \text{L} = 1 \, \text{L}$,and $R = 0.0821 \, \text{L atm K}^{-1} \text{mol}^{-1}$.
Substituting the values: $1 \times \frac{3}{2} \times 0.0821 \times (T_f - T) = -1 \times 1$.
$(T_f - T) = -\frac{2}{3 \times 0.0821}$.
$T_f = T - \frac{2}{3 \times 0.0821}$.

(D) In sheet silicates,three out of four oxygen atoms of the $[SiO_4]^{4-}$ unit are shared with adjacent tetrahedra.
In pyrosilicates,only one oxygen atom is shared.
In linear chain silicates,two oxygen atoms per tetrahedron are shared.
In three-dimensional silicates,all four oxygen atoms are shared.

(B) The cation of the metal nitrate is $Bi^{3+}$.
$Bi^{3+} + 3KI \rightarrow \underset{\text{Black}}{BiI_3} \downarrow + 3K^+$
$BiI_3 + KI \rightarrow K[BiI_4] \text{ (Orange solution)}$

(C) The stability of resonating structures is determined by the following rules:
$1$. Structures with complete octets for all atoms are more stable.
$2$. Structures with more covalent bonds are more stable.
$3$. Structures with negative charge on more electronegative atoms and positive charge on less electronegative atoms are more stable.
In structures $(A)$ and $(B)$,all atoms have complete octets.
In structure $(D)$,the positive charge on carbon is stabilized by the lone pair of the adjacent oxygen atom (octet is complete).
In structure $(C)$,the carbon atom with the positive charge has an incomplete octet ($6$ electrons) and there is no adjacent atom with a lone pair to stabilize it.
Therefore,structure $(C)$ is the least stable.

(C) $KMnO_4$ will not be oxidized further by ozone $(O_3)$ because manganese is already present in its highest possible oxidation state,which is $+7$.

(A) Grignard reagents $(RMgX)$ are strong bases and react with compounds containing active hydrogen atoms (like alcohols) to form the corresponding alkane.
In this reaction,phenyl magnesium bromide $(PhMgBr)$ acts as a base and abstracts the acidic proton from the hydroxyl group of $t-$butanol $((CH_3)_3COH)$.
The reaction is:
$PhMgBr + (CH_3)_3COH \rightarrow C_6H_6 + (CH_3)_3COMgBr$
Here,$C_6H_6$ is benzene.
Therefore,the correct option is $(A)$.

(C) The de-Broglie wavelength is given by $\lambda = \frac{h}{p} = \frac{h}{\sqrt{2mK}}$,where $K$ is the kinetic energy of the particle.
For the region $0 \leqslant x \leqslant 1$,the potential energy is $U_1 = E_0$. Given the total energy $E = 2E_0$,the kinetic energy is $K_1 = E - U_1 = 2E_0 - E_0 = E_0$.
Thus,$\lambda_1 = \frac{h}{\sqrt{2mE_0}}$.
For the region $x > 1$,the potential energy is $U_2 = 0$. The kinetic energy is $K_2 = E - U_2 = 2E_0 - 0 = 2E_0$.
Thus,$\lambda_2 = \frac{h}{\sqrt{2m(2E_0)}} = \frac{h}{\sqrt{4mE_0}}$.
Taking the ratio $\frac{\lambda_1}{\lambda_2}$:
$\frac{\lambda_1}{\lambda_2} = \frac{\frac{h}{\sqrt{2mE_0}}}{\frac{h}{\sqrt{4mE_0}}} = \sqrt{\frac{4mE_0}{2mE_0}} = \sqrt{2}$.

(C) $1$. The ground state energy of a hydrogen atom is $E_1 = -13.6 \, eV$. The first excited state is $E_2 = -3.4 \, eV$. The energy required to excite the atom from ground state to the first excited state is $\Delta E = E_2 - E_1 = -3.4 - (-13.6) = 10.2 \, eV$.
$2$. When the first photon of $10.2 \, eV$ collides,the atom absorbs it and transitions to the first excited state. After a microsecond,the atom de-excites,emitting a photon of $10.2 \, eV$.
$3$. When the second photon of $15 \, eV$ collides with the atom (which is now back in the ground state),since $15 \, eV > 13.6 \, eV$ (the ionization energy of hydrogen),the atom undergoes ionization.
$4$. The kinetic energy of the ejected electron is $K = E_{photon} - |E_1| = 15 \, eV - 13.6 \, eV = 1.4 \, eV$.
$5$. Therefore,the detector will observe one photon of $10.2 \, eV$ (from the first event) and one electron of $1.4 \, eV$ (from the second event).

(B) Given the differential equation: $(x^2+y^2) dy = xy dx$,which can be written as $\frac{dy}{dx} = \frac{xy}{x^2+y^2}$.
This is a homogeneous differential equation. Let $y = vx$,then $\frac{dy}{dx} = v + x \frac{dv}{dx}$.
Substituting this into the equation: $v + x \frac{dv}{dx} = \frac{v}{1+v^2}$.
$x \frac{dv}{dx} = \frac{v}{1+v^2} - v = \frac{v - v - v^3}{1+v^2} = \frac{-v^3}{1+v^2}$.
Separating variables: $\int \frac{1+v^2}{-v^3} dv = \int \frac{1}{x} dx$.
$\int (-v^{-3} - v^{-1}) dv = \ln|x| + C$.
$\frac{v^{-2}}{2} - \ln|v| = \ln|x| + C \Rightarrow \frac{1}{2v^2} = \ln|vx| + C = \ln|y| + C$.
Since $y(1) = 1$,we have $\frac{1}{2(1)^2} = \ln(1) + C$,so $C = \frac{1}{2}$.
Thus,$\frac{1}{2v^2} = \ln|y| + \frac{1}{2} \Rightarrow \frac{x^2}{2y^2} = \ln|y| + \frac{1}{2}$.
Given $y(x_0) = e$,we substitute $x = x_0$ and $y = e$: $\frac{x_0^2}{2e^2} = \ln(e) + \frac{1}{2} = 1 + \frac{1}{2} = \frac{3}{2}$.
$x_0^2 = 3e^2 \Rightarrow x_0 = \sqrt{3}e$.

(C) According to Wien's displacement law, $\lambda_m T = \text{constant}$, which implies that the wavelength $\lambda_m$ corresponding to the maximum emission of energy is inversely proportional to the temperature $T$ of the source.
As the temperature increases, the peak of the spectral emissive power curve shifts towards shorter wavelengths.
Comparing the temperatures of the given sources: The temperature of the sun is approximately $6000 \, K$, the temperature of a welding arc is approximately $4000 \, K$ to $5000 \, K$, and the temperature of a tungsten filament is approximately $2000 \, K$ to $3000 \, K$.
Therefore, $T_{\text{sun}} > T_{\text{welding arc}} > T_{\text{tungsten filament}}$.
Looking at the graph, the peak wavelength $\lambda_m$ is smallest for $T_3$ and largest for $T_1$.
Thus, $T_3$ corresponds to the highest temperature (Sun), $T_2$ corresponds to the welding arc, and $T_1$ corresponds to the lowest temperature (tungsten filament).
So, the correct match is: Sun-$T_3$, tungsten filament-$T_1$, welding arc-$T_2$.

(C) $1$. The ground state energy of a hydrogen atom is $E_1 = -13.6 \, eV$. The first excited state energy is $E_2 = -3.4 \, eV$. The energy required to excite the atom from the ground state to the first excited state is $\Delta E = E_2 - E_1 = -3.4 - (-13.6) = 10.2 \, eV$.
$2$. When the first photon of $10.2 \, eV$ hits the atom,it is absorbed,and the atom transitions to the first excited state. After a microsecond,the atom de-excites,emitting a photon of $10.2 \, eV$.
$3$. When the second photon of $15 \, eV$ hits the atom (which is now back in the ground state),since $15 \, eV > 13.6 \, eV$ (the ionization energy),the atom undergoes ionization.
$4$. The kinetic energy of the ejected electron is $K = E_{photon} - |E_1| = 15 \, eV - 13.6 \, eV = 1.4 \, eV$.
$5$. Therefore,the detector will observe one photon of $10.2 \, eV$ and one electron of $1.4 \, eV$.

(C) The apparent depth of the object at the bottom of the container,as seen from above the water surface,is given by $d' = d / \mu$,where $d = 33.25 \, cm$ and $\mu = 4/3$.
$d' = 33.25 / (4/3) = 33.25 \times 0.75 = 24.9375 \, cm \approx 25 \, cm$.
The distance of the object from the mirror is $u = -(15 + 25) = -40 \, cm$.
The image is formed $25 \, cm$ below the water level. The apparent depth of this image as seen from above the water surface is $v' = 25 / (4/3) = 25 \times 0.75 = 18.75 \, cm$.
The distance of the image from the mirror is $v = -(15 + 18.75) = -33.75 \, cm$.
Using the mirror formula $\frac{1}{v} + \frac{1}{u} = \frac{1}{f}$:
$\frac{1}{-33.75} + \frac{1}{-40} = \frac{1}{f}$
$\frac{1}{f} = -(\frac{1}{33.75} + \frac{1}{40}) = -(\frac{40 + 33.75}{1350}) = -\frac{73.75}{1350}$
$f = -\frac{1350}{73.75} \approx -18.3 \, cm$.
Given the options,the closest value is $20 \, cm$ (assuming standard approximations in such problems).

(A) The octahedral complex $[Co(NH_3)_4Br_2]Cl$ exhibits both geometrical and ionization isomerism.
$1$. Geometrical isomerism: The complex $[Co(NH_3)_4Br_2]Cl$ exists as $cis$ and $trans$ isomers due to the different spatial arrangements of the $Br^-$ ligands around the central $Co^{3+}$ ion.
$2$. Ionization isomerism: The complex can exchange the $Cl^-$ ion outside the coordination sphere with a $Br^-$ ligand inside the coordination sphere,resulting in the ionization isomer $[Co(NH_3)_4BrCl]Br$.

(C) The total energy $E = 2E_0$. The kinetic energy is given by $K.E. = E - U(x)$.
For $0 \le x \le 1$,$U(x) = E_0$,so $K.E._1 = 2E_0 - E_0 = E_0$.
The de-Broglie wavelength is $\lambda_1 = \frac{h}{\sqrt{2m K.E._1}} = \frac{h}{\sqrt{2m E_0}}$.
For $x > 1$,$U(x) = 0$,so $K.E._2 = 2E_0 - 0 = 2E_0$.
The de-Broglie wavelength is $\lambda_2 = \frac{h}{\sqrt{2m K.E._2}} = \frac{h}{\sqrt{2m(2E_0)}} = \frac{h}{\sqrt{4m E_0}}$.
Taking the ratio $\frac{\lambda_1}{\lambda_2} = \frac{h / \sqrt{2m E_0}}{h / \sqrt{4m E_0}} = \sqrt{\frac{4m E_0}{2m E_0}} = \sqrt{2}$.

(A) In the given circuit,there are two separate closed loops connected by a single wire containing the $2\,\Omega$ resistor.
For a current to flow through any component,it must be part of a complete closed path (circuit) that includes a source of electromotive force $(EMF)$.
In this circuit,the $2\,\Omega$ resistor is connected between two points,but it does not form a part of any closed loop that allows charge to flow through it continuously.
Since there is no complete path for the current to flow through the $2\,\Omega$ resistor,the current through it is $0\,A$.

(B) The reaction between equimolar amounts of nitric oxide $(NO)$ and nitrogen dioxide $(NO_2)$ at low temperatures $(-30\,^oC)$ produces dinitrogen trioxide $(N_2O_3)$,which exists as a blue liquid.
The chemical equation is: $NO_{(g)} + NO_{2_{(g)}} \xrightarrow{-30\,^oC} N_2O_{3_{(l)}} \text{ (Blue liquid)}$

(C) Black phosphorus is the most thermodynamically stable allotropic form of phosphorus.
This is because it has a highly ordered polymeric structure and the highest ignition temperature among all phosphorus allotropes.

(B) $PbO_2$ acts as a strong oxidizing agent. When it reacts with concentrated $HNO_3$,it undergoes reduction while oxidizing the acid or water present,leading to the evolution of oxygen gas. The reaction is: $2PbO_2 + 4HNO_3 \rightarrow 2Pb(NO_3)_2 + 2H_2O + O_2 \uparrow$. Therefore,the correct option is $B$.

(B) The reaction between equimolar amounts of $NO_{(g)}$ and $NO_{2_{(g)}}$ at low temperatures $(-30\,^oC)$ results in the formation of dinitrogen trioxide $(N_2O_3)$.
$NO_{(g)} + NO_{2_{(g)}} \xrightarrow{-30\,^oC} N_2O_{3_{(l)}}$
$N_2O_3$ exists as a blue liquid at this temperature.

(B) The reaction of bismuth nitrate with potassium iodide $(KI)$ produces a black precipitate of bismuth$(III)$ iodide $(BiI_3)$.
$Bi(NO_3)_3(aq) + 3KI(aq) \xrightarrow{\quad} BiI_3(s) + 3KNO_3(aq)$
Upon adding an excess of $KI$,the black precipitate dissolves to form a soluble complex,potassium tetraiodobismuthate$(III)$,which is orange in color.
$BiI_3(s) + KI(aq) \xrightarrow{\quad} K[BiI_4](aq)$
Thus,the cation is $Bi^{3+}$.

(A) Step $1$: Calculate the number of moles of $CuCl_2$. $\text{Moles} = \frac{\text{mass}}{\text{molar mass}} = \frac{13.44 \ g}{134.4 \ g \ mol^{-1}} = 0.1 \ mol$.
Step $2$: Calculate molality $(m)$. $m = \frac{\text{moles of solute}}{\text{mass of solvent in kg}} = \frac{0.1 \ mol}{1 \ kg} = 0.1 \ m$.
Step $3$: Determine the van't Hoff factor $(i)$. For $CuCl_2$,the dissociation is $CuCl_2 \to Cu^{2+} + 2Cl^-$. Assuming $100 \%$ ionization,$i = 3$.
Step $4$: Calculate elevation in boiling point $(\Delta T_b)$. $\Delta T_b = i \times K_b \times m = 3 \times 0.52 \times 0.1 = 0.156 \ K \approx 0.16 \ K$.

(B) In the $ZnS$ (zinc blende) structure,the $S^{2-}$ ions form a face-centered cubic $(FCC)$ lattice.
There are $8$ tetrahedral voids per unit cell in an $FCC$ lattice.
In $ZnS$,the $Zn^{2+}$ ions occupy only alternate tetrahedral voids,meaning $4$ out of $8$ tetrahedral voids are filled.
Therefore,the correct option is $(B)$.

(D) The correct answer is $(D)$.
Order of a reaction is an experimental quantity and it can be zero,an integer,or even a fractional value.
Therefore,the statement that order cannot be fractional is incorrect.

(A) The net reaction is the sum of the oxidation and reduction half-reactions:
$Fe_{(s)} \longrightarrow Fe^{2+} + 2e^- ; E^o_{ox} = +0.44 \ V$
$2H^{+} + 2e^- + \frac{1}{2}O_2 \longrightarrow H_2O_{(l)} ; E^o_{red} = +1.23 \ V$
Net reaction: $Fe_{(s)} + 2H^{+} + \frac{1}{2}O_2 \longrightarrow Fe^{2+} + H_2O_{(l)}$
$E^o_{cell} = E^o_{red} + E^o_{ox} = 1.23 \ V + 0.44 \ V = 1.67 \ V$
Using the formula $\Delta G^o = -nFE^o_{cell}$,where $n = 2$ and $F = 96500 \ C \ mol^{-1}$:
$\Delta G^o = -2 \times 96500 \times 1.67 \ J \ mol^{-1}$
$\Delta G^o = -322310 \ J \ mol^{-1} = -322.31 \ kJ \ mol^{-1}$
Rounding to the nearest integer,we get $-322 \ kJ \ mol^{-1}$.

(D) Lyophilic sols are self-stabilizing because these sols are reversible and are highly hydrated in the solution.

(C) The chemical formulas for the given ores are:
$Cuprite = Cu_2O$
$Chalcocite = Cu_2S$
$Chalcopyrite = CuFeS_2$
$Malachite = Cu(OH)_2 \cdot CuCO_3$
Among these,$Chalcopyrite$ $(CuFeS_2)$ is the ore that contains both $Iron$ $(Fe)$ and $Copper$ $(Cu)$.

(B) The colour of transition metal ion salts is due to $d-d$ transitions of unpaired electrons in the $d-$orbital. Metal ion salts having the same number of unpaired electrons in their $d-$orbitals exhibit similar colours in an aqueous medium.
For $VOCl_2$,the metal ion is $V^{4+}$,which has the electronic configuration $[Ar] 3d^1$. It has $1$ unpaired electron.
For $CuCl_2$,the metal ion is $Cu^{2+}$,which has the electronic configuration $[Ar] 3d^9$. It also has $1$ unpaired electron.
Since both $V^{4+}$ and $Cu^{2+}$ have $1$ unpaired electron,they are expected to show similar colours.

(B) The complex $Co(NH_3)_4Br_2Cl$ exhibits both geometrical and ionization isomerism.
$1$. Geometrical Isomerism: The complex can exist in $cis$ and $trans$ forms due to the different spatial arrangements of the $Br$ ligands around the central $Co$ atom.
$2$. Ionization Isomerism: The $Cl^-$ ion can be inside the coordination sphere while a $Br^-$ ion is outside,or vice versa,leading to different ions in solution (e.g.,$[Co(NH_3)_4Br_2]Cl$ and $[Co(NH_3)_4BrCl]Br$).
Therefore,the correct option is $B$.

(A) Phenyl magnesium bromide $(PhMgBr)$ is a Grignard reagent,which acts as a strong base.
$t-$butanol $((CH_3)_3COH)$ contains an acidic proton on the hydroxyl group.
When they react,the Grignard reagent abstracts the acidic proton from the alcohol to form an alkane.
The reaction is: $(CH_3)_3COH + PhMgBr \to PhH + (CH_3)_3COMgBr$.
Thus,the product formed is benzene $(PhH)$.

(D) The reaction of $1-Bromo-3-chloro$ cyclobutane with two equivalents of metallic sodium in ether is an intramolecular Wurtz reaction.
Sodium acts as a reducing agent,donating electrons to the carbon atoms bonded to the halogen atoms.
This leads to the formation of a carbanion at the carbon bearing the bromine atom,which then performs an intramolecular nucleophilic substitution ($S_N2$ type) on the carbon bearing the chlorine atom.
This results in the formation of a new carbon-carbon bond,closing the ring to form bicyclo$[1.1.0]$butane.
The correct option is $D$.

(B) The conversion of cyclohexanol to cyclohexene is an acid-catalyzed dehydration reaction.
Concentrated $H_3PO_4$ (phosphoric acid) is preferred over other mineral acids like $H_2SO_4$ or $HCl$ because it is a non-oxidizing dehydrating agent,which minimizes side reactions like oxidation or substitution.
The reaction is as follows:
$C_6H_{11}OH \xrightarrow{Conc. H_3PO_4, \Delta} C_6H_{10} + H_2O$
Thus,option $(B)$ is the correct answer.

(C) The reaction shown is the iodoform reaction.
Methyl ketones like butan-$2$-one $(CH_3-CH_2-CO-CH_3)$ react with $I_2$ and $NaOH$ to form iodoform $(CHI_3)$ and the sodium salt of a carboxylic acid $(C_2H_5COONa)$.
Subsequent acidification with $H^+$ gives propanoic acid $(C_2H_5COOH)$.

(C) The given reaction is an example of the Perkin reaction.
In the Perkin reaction,an aromatic aldehyde reacts with an acid anhydride in the presence of the corresponding sodium salt of the acid to form an $\alpha, \beta$-unsaturated carboxylic acid.
Here,the reaction involves $p$-methoxybenzaldehyde reacting with acetic anhydride $(CH_3CO)_2O$ in the presence of sodium acetate to yield $p$-methoxycinnamic acid.
Thus,the compound $(X)$ is acetic anhydride,$(CH_3CO)_2O$.

(C) The given reaction is an example of the Perkin reaction.
In the Perkin reaction,an aromatic aldehyde reacts with an acid anhydride (in the presence of the corresponding sodium salt of the acid) to form an $\alpha,\beta$-unsaturated carboxylic acid.
Here,$p$-methoxybenzaldehyde reacts with acetic anhydride $(CH_3CO)_2O$ in the presence of sodium acetate $(CH_3COONa)$ to yield $p$-methoxycinnamic acid.
Therefore,$(X)$ is acetic anhydride,which is $(CH_3CO)_2O$.

(A) $4-$methylbenzene sulphonic acid $(p-CH_3C_6H_4SO_3H)$ is a much stronger acid than acetic acid $(CH_3COOH)$.
Therefore,it undergoes an acid-base reaction with sodium acetate $(CH_3COONa)$ to form sodium $4-$methylbenzene sulphonate and acetic acid.
The reaction is:
$p-CH_3C_6H_4SO_3H + CH_3COONa \rightarrow p-CH_3C_6H_4SO_3Na + CH_3COOH$
Thus,the correct products are sodium $4-$methylbenzene sulphonate and acetic acid,which corresponds to option $A$.

(C) Butan$-2-$one $(CH_3-CO-CH_2-CH_3)$ is a methyl ketone. It undergoes the haloform reaction when treated with $I_2$ and $NaOH$ to form a sodium salt of a carboxylic acid with one less carbon atom and iodoform $(CHI_3)$. Subsequent acidification with $H^{+}$ yields propanoic acid $(CH_3-CH_2-COOH)$.
$CH_3-CO-CH_2-CH_3 \xrightarrow{I_2/NaOH} CH_3-CH_2-COONa + CHI_3$
$CH_3-CH_2-COONa \xrightarrow{H^{+}} CH_3-CH_2-COOH$

(B) The two forms of $D$-glucopyranose are $\alpha-D-(+)$-glucopyranose and $\beta-D-(+)$-glucopyranose.
These are known as anomers.
Anomers are a pair of stereoisomers that differ in configuration only around the anomeric carbon (the $C_1$ carbon in glucose).