NEET 2023 Chemistry Question Paper with Answer and Solution

(A) When both nitrogen and sulphur are present in an organic compound,sodium thiocyanate $(NaSCN)$ is formed during the preparation of Lassaigne's extract.
$Na + C + N + S \rightarrow NaSCN$
This $NaSCN$ reacts with $Fe^{3+}$ ions to form a blood-red coloured complex,$[Fe(SCN)]^{2+}$.
$Fe^{3+} + SCN^{-} \rightarrow [Fe(SCN)]^{2+}$
Since all nitrogen is consumed in the formation of $SCN^-$,no free cyanide ions $(CN^-)$ are available to form the prussian blue precipitate.

(C) The weight of impure limestone is $20 \ g$.
Since the limestone is $20 \%$ pure,the weight of pure $CaCO_3$ is $\frac{20}{100} \times 20 \ g = 4 \ g$.
The molar mass of $CaCO_3$ is $40 + 12 + (3 \times 16) = 100 \ g/mol$.
The number of moles of $CaCO_3$ is $n = \frac{4 \ g}{100 \ g/mol} = 0.04 \ mol$.
According to the reaction $CaCO_3 \rightarrow CaO + CO_2$,$1 \ mol$ of $CaCO_3$ produces $1 \ mol$ of $CO_2$.
Therefore,$0.04 \ mol$ of $CaCO_3$ produces $0.04 \ mol$ of $CO_2$.
The molar mass of $CO_2$ is $12 + (2 \times 16) = 44 \ g/mol$.
The mass of $CO_2$ produced is $0.04 \ mol \times 44 \ g/mol = 1.76 \ g$.

(D) Coke is largely used as a reducing agent in metallurgy.
In diamond,each carbon atom undergoes $sp^3$ hybridisation and is linked to four other carbon atoms by using hybridised orbitals in a tetrahedral fashion.
Buckminsterfullerene contains six-membered and five-membered rings and hence is a cage-like molecule.
Graphite is very soft and slippery. Hence,it is used as a dry lubricant in machines running at high temperature.

(D) Statement $A$ is correct: Hydrogen acts as a reducing agent for heavy metal oxides.
Statement $B$ is correct: Heavy water $(D_2O)$ is used as a tracer to study reaction mechanisms.
Statement $C$ is correct: Hydrogenation of oils produces saturated fats (vanaspati ghee).
Statement $D$ is incorrect: The $H-H$ bond dissociation enthalpy $(435.88 \ kJ \ mol^{-1})$ is actually the highest for a single bond between two atoms of any element.
Statement $E$ is incorrect: Hydrogen can only reduce oxides of metals that are less active than iron (e.g.,$Cu, Pb, Ag$). It cannot reduce oxides of highly active metals like $Na, K, Ca$.
Therefore,statements $D$ and $E$ are incorrect.

(B) $1$. All enzymes that utilize $ATP$ in phosphate transfer require $Mg$ as the cofactor.
$2$. Bone in the human body is not an inert and unchanging substance; it is continuously being solubilized and redeposited.
$3$. $Ca$ plays an important role in neuromuscular function,interneuronal transmission,cell membrane integrity,and blood coagulation.
$4$. The daily requirement of $Mg$ and $Ca$ in the human body is estimated to be $200-300 \ mg$ $(0.2-0.3 \ g)$.
Therefore,statement $B$ is correct.

(C) The reaction of sodium ethanoate $(CH_3COONa)$ with soda lime $(NaOH + CaO)$ is known as decarboxylation.
The chemical reaction is: $CH_3COONa + NaOH \xrightarrow{CaO} CH_4(g) + Na_2CO_3(s)$.
The organic compound obtained is methane $(CH_4)$.
The molar mass of $CH_4 = 12 + (4 \times 1) = 16 \, g/mol$.
The weight of $2$ moles of $CH_4 = 2 \, mol \times 16 \, g/mol = 32 \, g$.

(D) To achieve the nearest noble gas configuration,these elements form the following ions: $Na^+$,$O^{2-}$,$F^-$,and $N^{3-}$.
These ions are isoelectronic,meaning they all have the same number of electrons ($10$ electrons).
For isoelectronic species,the ionic size increases as the negative charge on the anion increases because the effective nuclear charge decreases.
Comparing the charges: $N^{3-}$ has the highest negative charge,therefore it has the largest ionic size.

(B) According to Molecular Orbital Theory,for homonuclear diatomic molecules like $B_2, C_2,$ and $N_2$,the $2s-2p$ mixing causes the $\sigma 2p_z$ orbital to have higher energy than the $\pi 2p_x$ and $\pi 2p_y$ orbitals.
Therefore,the correct increasing order of energy for $N_2$ is $\sigma 1s < \sigma^* 1s < \sigma 2s < \sigma^* 2s < (\pi 2p_x = \pi 2p_y) < \sigma 2p_z < (\pi^* 2p_x = \pi^* 2p_y) < \sigma^* 2p_z$.

(D) Intermolecular forces are the forces of attraction and repulsion between interacting molecules.
$A$. Dipole-dipole forces,$B$. Dipole-induced dipole forces,$C$. Hydrogen bonding,and $E$. Dispersion forces (London forces) are types of intermolecular forces.
$D$. Covalent bonding is an intramolecular force that holds atoms together within a molecule,not between molecules.
Therefore,$A, B, C,$ and $E$ are correct.

(D) Pyridine $(C_5H_5N)$ has a hexagonal ring structure similar to benzene where one $CH$ group is replaced by a nitrogen atom.
$1$. $\sigma$ bonds: There are $5$ $C-H$ bonds,$5$ $C-C/C-N$ single bonds,and $1$ $C-N$ double bond (which contains $1$ $\sigma$ bond). Total $\sigma$ bonds $= 5 + 5 + 1 = 11$.
$2$. $\pi$ bonds: There are $3$ double bonds in the ring,each containing $1$ $\pi$ bond. Total $\pi$ bonds $= 3$.
$3$. Lone pair of electrons: The nitrogen atom in pyridine has $1$ lone pair of electrons.
Thus,the number of $\sigma$ bonds,$\pi$ bonds,and lone pairs are $11, 3, 1$ respectively.

(A) - Atoms consist of three fundamental particles: electrons,protons,and neutrons. Statement $A$ is incorrect.
- The mass of the electron is $9.10939 \times 10^{-31} \ kg$. Statement $B$ is correct.
- All isotopes of a given element have the same number of electrons and protons,thus they show the same chemical properties. Statement $C$ is correct.
- Protons and neutrons present in the nucleus are collectively called nucleons. Statement $D$ is incorrect.
- Dalton's atomic theory regarded the atom as the ultimate particle of matter. Statement $E$ is correct.
Therefore,the correct statements are $B, C,$ and $E$.

(C) According to Boyle's law,for a fixed amount of gas at a constant temperature,the pressure is inversely proportional to the volume $(P \propto \frac{1}{V})$.
Using the ideal gas equation,$PV = nRT$,we can rearrange it as:
$P = (nRT) \times (\frac{1}{V})$
This equation is in the form of $y = mx$,where $y = P$,$x = \frac{1}{V}$,and the slope $m = nRT$.
Since the slope $(nRT)$ is directly proportional to the temperature $(T)$,a graph of $P$ versus $\frac{1}{V}$ will yield a straight line passing through the origin for each temperature. As the temperature increases,the slope $(nRT)$ also increases. Therefore,for $T_3 > T_2 > T_1$,the slope of the line for $T_3$ will be the steepest,followed by $T_2$ and then $T_1$.

(B) For a given azimuthal quantum number $l$,the magnetic quantum number $m$ can take values from $-l$ to $+l$,including zero.
The total number of permissible values of $m$,denoted as $n_m$,is given by the formula:
$n_m = 2l + 1$
Rearranging this formula to solve for $l$:
$n_m - 1 = 2l$
$l = \frac{n_m - 1}{2}$
Comparing this with the given options,option $B$ represents the correct relationship.

(B) To determine the number of species that do not follow the octet rule (i.e.,do not have $8$ electrons around the central atom),we analyze the valence electrons of each molecule:
$1$. $NH_3$: Nitrogen has $5$ valence electrons and forms $3$ bonds with Hydrogen,plus $1$ lone pair. Total electrons = $3 \times 2 + 2 = 8$ electrons.
$2$. $AlCl_3$: Aluminum has $3$ valence electrons and forms $3$ bonds with Chlorine. Total electrons = $3 \times 2 = 6$ electrons (Incomplete octet).
$3$. $BeCl_2$: Beryllium has $2$ valence electrons and forms $2$ bonds with Chlorine. Total electrons = $2 \times 2 = 4$ electrons (Incomplete octet).
$4$. $CCl_4$: Carbon has $4$ valence electrons and forms $4$ bonds with Chlorine. Total electrons = $4 \times 2 = 8$ electrons.
$5$. $PCl_5$: Phosphorus has $5$ valence electrons and forms $5$ bonds with Chlorine. Total electrons = $5 \times 2 = 10$ electrons (Expanded octet).
Species with $NOT$ $8$ electrons are $AlCl_3$,$BeCl_2$,and $PCl_5$.
Therefore,the total number of such species is $3$.

(D) Lewis acids are species that accept a lone pair of electrons due to the presence of a vacant orbital in their outermost shell.
$H_{2}\ddot{O}:$ acts as a Lewis base because it has lone pairs on the oxygen atom.
$BF_{3}$ acts as a Lewis acid because the boron atom has an incomplete octet (only $6$ electrons in its valence shell) and can accept a lone pair.
$OH^{-}$ acts as a Lewis base due to the presence of lone pairs on the oxygen atom.
$\ddot{N}H_{3}$ acts as a Lewis base because the nitrogen atom has a lone pair of electrons available for donation.

(D) When an alkali metal like sodium dissolves in liquid ammonia,it forms a deep blue solution due to the presence of ammoniated electrons,$[e(NH_3)_y]^-$.
These ammoniated electrons absorb energy in the visible region of light,which imparts the blue color.
Because of the presence of unpaired electrons,the solution is paramagnetic.
The chemical reaction is: $M + (x+y) NH_3 \rightarrow [M(NH_3)_x]^+ + [e(NH_3)_y]^-$.
Therefore,the Assertion $A$ is true,but the Reason $R$ is false because the blue color is not due to the formation of amide.

(C) The enthalpy change $(\Delta H)$ is defined as the change in internal energy $(\Delta U)$ plus the work done due to volume change at constant pressure.
For a chemical reaction involving gases,the relation is given by:
$\Delta H = \Delta U + \Delta n_{g} R T$
Where:
$\Delta H$ = Change in enthalpy
$\Delta U$ = Change in internal energy
$\Delta n_{g}$ = Change in the number of moles of gaseous products and reactants
$R$ = Universal gas constant
$T$ = Temperature in Kelvin

(A) Statement $I$ is false because eutrophication is caused by nutrient-enriched water bodies,not nutrient-deficient ones.
Statement $II$ is true because the excessive growth of plants and algae in eutrophic water bodies consumes dissolved oxygen during decomposition,leading to a decrease in oxygen levels and the death of aquatic life.
Therefore,Statement $I$ is incorrect but Statement $II$ is true.

(B) $H$ückel's rule states that a planar,cyclic,fully conjugated system is aromatic if it contains $(4n + 2) \pi$ electrons,where $n = 0, 1, 2, \dots$
Let's analyze each species:
$(i)$ Naphthalene: $10 \pi$ electrons $(n=2)$,aromatic.
$(ii)$ Cyclopentadienyl anion: $6 \pi$ electrons $(n=1)$,aromatic.
$(iii)$ Cyclobutadiene: $4 \pi$ electrons ($n=1$ for anti-aromatic),not $H$ückel's rule.
$(iv)$ Cyclopropenyl anion: $4 \pi$ electrons,anti-aromatic.
$(v)$ Cyclopropenyl cation: $2 \pi$ electrons $(n=0)$,aromatic.
$(vi)$ Cyclooctatetraene: $8 \pi$ electrons,non-planar,non-aromatic.
$(vii)$ Anthracene: $14 \pi$ electrons $(n=3)$,aromatic.
The species that obey $H$ückel's rule are $(i), (ii), (v),$ and $(vii)$.
Thus,the total number is $4$.

(D) The reaction is $A + B \rightleftharpoons C + D$.
At equilibrium,the concentrations are $[A] = 2 \, mol \, L^{-1}$,$[B] = 3 \, mol \, L^{-1}$,$[C] = 10 \, mol \, L^{-1}$,and $[D] = 6 \, mol \, L^{-1}$.
The equilibrium constant $K_{eq}$ is calculated as:
$K_{eq} = \frac{[C][D]}{[A][B]} = \frac{10 \times 6}{2 \times 3} = \frac{60}{6} = 10$.
The standard Gibbs free energy change $\Delta G^{\circ}$ is given by the formula:
$\Delta G^{\circ} = -RT \ln K_{eq} = -2.303 \, RT \log K_{eq}$.
Substituting the values $R = 2 \, cal \, mol^{-1} \, K^{-1}$,$T = 300 \, K$,and $K_{eq} = 10$:
$\Delta G^{\circ} = -2.303 \times 2 \times 300 \times \log(10)$.
Since $\log(10) = 1$:
$\Delta G^{\circ} = -2.303 \times 600 = -1381.8 \, cal \, mol^{-1}$.

(A) Using the ion-electron method:
Reduction half-reaction: $Cr_2O_7^{2-} + 14H^{+} + 6e^- \rightarrow 2Cr^{3+} + 7H_2O$
Oxidation half-reaction: $(SO_3^{2-} + H_2O \rightarrow SO_4^{2-} + 2H^{+} + 2e^-) \times 3$
Adding both half-reactions:
$Cr_2O_7^{2-} + 14H^{+} + 6e^- + 3SO_3^{2-} + 3H_2O \rightarrow 2Cr^{3+} + 7H_2O + 3SO_4^{2-} + 6H^{+} + 6e^-$
Simplifying the equation:
$Cr_2O_7^{2-} + 3SO_3^{2-} + 8H^{+} \rightarrow 2Cr^{3+} + 3SO_4^{2-} + 4H_2O$
Comparing this with the given equation $aCr_2O_7^{2-} + bSO_3^{2-} + cH^{+} \rightarrow 2aCr^{3+} + bSO_4^{2-} + \frac{c}{2}H_2O$:
$a = 1, b = 3, c = 8$.

(A) The process by which differentiated plant cells (like leaf mesophyll cells) regain the capacity to divide and form an undifferentiated mass of cells called callus is known as $Dedifferentiation$.
This is a characteristic feature of living plant cells under specific laboratory conditions in a culture medium containing growth regulators like auxins and cytokinins.

(A) In group $13$ elements,the stability of the $+1$ oxidation state increases down the group due to the inert pair effect.
For Thallium $(Tl)$,the $+1$ oxidation state is more stable than the $+3$ oxidation state.
Therefore,$TlI$ (where $Tl$ is $+1$) is more stable than $TlI_3$ (where $Tl$ is $+3$).

(C) The reaction proceeds in three steps:
$1$. Benzenediazonium chloride reacts with $Cu_2Br_2/HBr$ (Sandmeyer reaction) to form bromobenzene.
$2$. Bromobenzene reacts with $Mg$ in the presence of dry ether to form phenylmagnesium bromide (a Grignard reagent).
$3$. Phenylmagnesium bromide undergoes hydrolysis with $H_2O$ to yield benzene as the final product.

(D) The given structure is $C_6H_5-CH=CH-CH(X)-CH_2-CH_3$.
In this compound,the halogen atom $(X)$ is attached to an $sp^3$ hybridized carbon atom,which is adjacent to a carbon-carbon double bond $(C=C)$.
Such compounds,where the halogen is attached to an $sp^3$ hybridized carbon next to a double bond,are classified as allylic halides.
Therefore,the correct option is $D$.

(A) Assertion $A$ is false because a chemical reaction cannot have zero activation energy $(E_{a} > 0)$.
Reason $R$ is true because it correctly defines activation energy as the minimum extra energy required by reactant molecules to reach the threshold energy level.

(A) The reaction sequence is as follows:
$1$. Cyclohexanone $[A]$ reacts with $HCN$ to form a cyanohydrin,which is $1$-hydroxycyclohexanecarbonitrile $[B]$.
$2$. The treatment of $[B]$ with $conc. H_2SO_4$ and heat $(\Delta)$ leads to two simultaneous processes: dehydration of the tertiary alcohol group to form a double bond and hydrolysis of the nitrile group $(-CN)$ to a carboxylic acid group $(-COOH)$.
$3$. The final product $[C]$ is cyclohex-$1$-ene-$1$-carboxylic acid.

(D) Statement $I$ is correct because a nucleoside is formed by the attachment of a nitrogenous base to the $1'-$position of a pentose sugar.
Statement $II$ is incorrect because a nucleotide is formed when a nucleoside is linked to phosphoric acid $(H_3PO_4)$,not phosphorous acid $(H_3PO_3)$,at the $5'-$position of the sugar moiety.
Therefore,Statement $I$ is true but Statement $II$ is false.

(B) In a $CCP$ structure,let the number of atoms of $B$ be $n$.
The number of tetrahedral voids is $2n$.
Atoms of $A$ occupy $\frac{1}{3}$ of the tetrahedral voids,so the number of atoms of $A = \frac{1}{3} \times 2n = \frac{2n}{3}$.
The ratio of $A:B = \frac{2n}{3} : n = 2:3$.
Therefore,the formula is $A_2 B_3$.
Comparing with $A_x B_y$,we get $x = 2$ and $y = 3$.
Thus,$x + y = 2 + 3 = 5$.

(D) The stability of $Cu^{2+}_{(aq)}$ is more than $Cu^{+}_{(aq)}$ is due to the much more negative $\Delta_{hyd}H^{\circ}$ of $Cu^{2+}_{(aq)}$ than $Cu^{+}_{(aq)}$,which more than compensates for the second ionisation enthalpy of $Cu$.
$\Delta_{hyd}H^{\circ}$ of $Cu^{2+}_{(aq)} = -2121 \ kJ \ mol^{-1}$
$\Delta_{i}H_{1}^{\circ}$ of $Cu = +745 \ kJ \ mol^{-1}$
$\Delta_{i}H_{2}^{\circ}$ of $Cu = +1960 \ kJ \ mol^{-1}$

(D) Assertion $A$ is true because helium is used as a diluent for oxygen in diving apparatus to avoid the toxic effects of nitrogen at high pressures.
Reason $R$ is false because helium has very low solubility in blood,which is the actual reason it is preferred,not its solubility in $O_2$.
Therefore,$A$ is true but $R$ is false.

(A) Veronal is a derivative of barbituric acid and is classified as a barbiturate.
$Meprobamate$,$valium$,and $chlordiazepoxide$ are other types of tranquilizers used for stress and mild to severe mental diseases.

(D) The initial rate is given by $r = k[A]^2[B]$.
When the concentration of $A$ is tripled,the new concentration $[A'] = 3[A]$.
The new rate $r'$ is calculated as:
$r' = k[A']^2[B] = k(3[A])^2[B] = k(9[A]^2)[B] = 9k[A]^2[B]$.
Comparing the new rate with the initial rate,$r' = 9r$.
Therefore,the rate increases by a factor of $9$.

(A) The combination of $N_2$ and $H_2$ to form $NH_3$ in the presence of finely divided $Fe$ is an example of heterogeneous catalysis because the catalyst $(Fe_{(s)})$ is in a different physical state than the reactants ($N_2(g)$ and $H_2(g)$).
$N_{2(g)} + 3H_{2(g)} \xrightarrow{Fe_{(s)}} 2NH_{3(g)}$
All other options are examples of homogeneous catalysis where the catalyst and reactants are in the same phase:
$C_{12}H_{22}O_{11(aq)} + H_2O_{(l)} \xrightarrow{H^+_{(aq)}} \text{Glucose}_{(aq)} + \text{Fructose}_{(aq)}$
$2SO_{2(g)} + O_{2(g)} \xrightarrow{NO_{(g)}} 2SO_{3(g)}$

(B) homoleptic complex is a coordination compound in which the central metal atom or ion is bound to only one type of donor atom or ligand.
- Triamminetriaquachromium $(III)$ chloride: $[Cr(NH_3)_3(H_2O)_3]Cl_3$ (Heteroleptic)
- Potassium trioxalatoaluminate $(III)$: $K_3[Al(C_2O_4)_3]$ (Homoleptic,as all ligands are oxalate ions)
- Diamminechloridonitrito-$N$-platinum $(II)$: $[Pt(NH_3)_2Cl(NO_2)]$ (Heteroleptic)
- Pentaamminecarbonatocobalt $(III)$ chloride: $[Co(NH_3)_5(CO_3)]Cl$ (Heteroleptic)
Therefore,the correct option is $B$.

(B) The given reaction uses $Zn-Hg$ and $conc. HCl$,which are the reagents for the Clemmensen reduction.
Clemmensen reduction converts carbonyl groups $(>C=O)$ into methylene groups $(-CH_2-)$.
In the given reactant,there are two ketone groups: one attached to the cyclohexane ring and one attached to the benzene ring.
Both ketone groups are reduced to methylene groups.
Specifically,the acetyl group $(-COCH_3)$ is reduced to an ethyl group $(-CH_2CH_3)$.
Therefore,the product $(A)$ will have two ethyl groups attached to the respective rings.

(C) The equation $\Delta_r G = -nFE_{\text{cell}}$ relates the Gibbs energy change to the number of moles of electrons $(n)$ transferred,the Faraday constant $(F)$,and the cell potential $(E_{\text{cell}})$.
Since $\Delta_r G$ is directly proportional to $n$,the value of $\Delta_r G$ depends on $n$. Thus,Assertion $A$ is true.
$E_{\text{cell}}$ is an intensive property because it does not depend on the amount of matter,whereas $\Delta_r G$ is an extensive property because it depends on the amount of matter (number of moles $n$). Thus,Reason $R$ is true.
However,the fact that $E_{\text{cell}}$ is intensive and $\Delta_r G$ is extensive explains why $\Delta_r G$ is extensive,but it is not the direct reason why $\Delta_r G$ depends on $n$ in the equation. The dependence on $n$ is a mathematical consequence of the stoichiometry of the cell reaction. Therefore,$R$ is not the correct explanation of $A$.

(D) The relationship between conductivity $(k)$,conductance $(G)$,and cell constant $(G^*)$ is given by:
$k = G \times G^*$
Since conductance $G = \frac{1}{R}$,where $R$ is the resistance:
$k = \frac{1}{R} \times G^*$
Rearranging to solve for the cell constant $(G^*)$:
$G^* = k \times R$
Given:
$k = 0.0210 \, \Omega^{-1} cm^{-1}$
$R = 60 \, \Omega$
Calculation:
$G^* = 0.0210 \, \Omega^{-1} cm^{-1} \times 60 \, \Omega = 1.26 \, cm^{-1}$

(B) The reaction of $3\text{-methylbutan-2-ol}$ with $HBr$ proceeds via an $S_N1$ mechanism.
$1$. Protonation of the alcohol group followed by the loss of a water molecule generates a secondary carbocation: $CH_{3}-CH(CH_{3})-CH^+-CH_{3}$.
$2$. This secondary carbocation undergoes a $1,2\text{-hydride shift}$ to form a more stable tertiary carbocation: $CH_{3}-C^+(CH_{3})-CH_{2}-CH_{3}$.
$3$. Finally,the bromide ion $(Br^-)$ attacks the tertiary carbocation to form the major product: $2\text{-bromo-2-methylbutane}$,which is $CH_{3}-C(Br)(CH_{3})-CH_{2}-CH_{3}$.

(A) Neoprene is a synthetic rubber formed by the free radical polymerization of chloroprene $(2\text{-chloro-}1,3\text{-butadiene})$. The chemical reaction is as follows:
$nCH_2=C(Cl)-CH=CH_2 \xrightarrow{\text{Polymerization}} [CH_2-C(Cl)=CH-CH_2]_n$
Thus,the monomer required for the production of neoprene is chloroprene,which corresponds to option $A$.

(D) Let us analyze each reaction:
$(A)$ $CH_3CN + (i) LiAlH_4 / (ii) H_3O^{\oplus} \rightarrow CH_3CH_2NH_2$ (Ethylamine,a primary amine).
$(B)$ $CH_3CONH_2 + Br_2/KOH \rightarrow CH_3NH_2$ (Methylamine,a primary amine) via the Hoffmann bromamide degradation reaction.
$(C)$ $CH_3CONH_2 + (i) LiAlH_4 / (ii) H_3O^{\oplus} \rightarrow CH_3CH_2NH_2$ (Ethylamine,a primary amine).
$(D)$ $CH_3NC + (i) LiAlH_4 / (ii) H_3O^{\oplus} \rightarrow CH_3NHCH_3$ (Dimethylamine,a secondary amine).
Thus,reaction $(D)$ does not yield a primary amine.

(D) Statement $A$ is correct as transition metals form ionic $MO$ oxides.
Statement $B$ is correct as the highest oxidation state equals the group number from $Sc$ $(Group \ 3)$ to $Mn$ $(Group \ 7)$.
Statement $C$ is incorrect because the acidic character increases as the oxidation state of the metal increases $(V_2O_3 < V_2O_4 < V_2O_5)$. Therefore,the basic character decreases.
Statement $D$ is incorrect because $V_2O_4$ dissolves in acids to form $VO^{2+}$ ions,not $VO_4^{3-}$ salts.
Statement $E$ is correct as $CrO$ is basic and $Cr_2O_3$ is amphoteric.
Thus,statements $C$ and $D$ are incorrect.

(D) The reaction of benzyl phenyl ether with $HI$ involves the protonation of the ether oxygen atom.
Following protonation,the $C-O$ bond between the benzyl carbon and the oxygen atom breaks to form a stable benzyl carbocation $(C_6H_5CH_2^+)$ and phenol $(C_6H_5OH)$.
The benzyl carbocation is stabilized by resonance with the phenyl ring.
Finally,the iodide ion $(I^-)$ attacks the benzyl carbocation to form benzyl iodide $(C_6H_5CH_2I)$.
Thus,the products are $A = C_6H_5CH_2I$ and $B = C_6H_5OH$.

(D) In an $fcc$ unit cell,octahedral voids are located at the body centre and at the centre of each edge.
There are $12$ edges in a cube,and each edge is shared by $4$ adjacent unit cells.
Therefore,the contribution of an octahedral void located at the edge centre to a single unit cell is $\frac{1}{4}$.

(A) The dehydration of alcohols under acidic conditions proceeds via the formation of a carbocation intermediate. The rate of dehydration depends on the stability of the carbocation formed. Among the given options,the molecule that can form the most stable carbocation or has a structural feature facilitating elimination will dehydrate most readily. However,looking at the provided options and standard chemical behavior,$\beta$-hydroxy carbonyl compounds (aldols) are known to undergo dehydration very easily due to the formation of a conjugated $\alpha,\beta$-unsaturated system. If we re-examine the structures,the question likely intends to identify the compound that forms a stable conjugated system upon dehydration. Given the options provided,the most stable product is formed by the dehydration of $\beta$-hydroxy compounds. Based on standard competitive chemistry problems of this type,the correct answer is $A$.

(C) The structures of the given oxoacids of sulphur are as follows:
$A$. Peroxodisulphuric acid $(H_2S_2O_8)$: Contains two $S-OH$ groups,four $S=O$ bonds,and one peroxide linkage $(S-O-O-S)$. This matches $III$.
$B$. Sulphuric acid $(H_2SO_4)$: Contains two $S-OH$ groups and two $S=O$ bonds. This matches $IV$.
$C$. Pyrosulphuric acid $(H_2S_2O_7)$: Contains two $S-OH$ groups,four $S=O$ bonds,and one $S-O-S$ linkage. This matches $I$.
$D$. Sulphurous acid $(H_2SO_3)$: Contains two $S-OH$ groups and one $S=O$ bond. This matches $II$.
Therefore,the correct matching is $A-III, B-IV, C-I, D-II$.

(A) The given reaction involves the oxidation of an aldehyde group using Tollen's reagent $([Ag(NH_3)_2]^+)$ in a basic medium $(OH^-)$.
Aldehydes are oxidized to carboxylate ions $(COO^-)$ by Tollen's reagent,while ketones remain unaffected under these conditions.
In the given reactant,$2$-acetylbenzaldehyde,the aldehyde group $(-CHO)$ is oxidized to a carboxylate group $(-COO^-)$,while the acetyl group $(-COCH_3)$ remains unchanged.
Thus,the major product is the $2-$acetylbenzoate ion.

(B) The reaction sequence is as follows:
$1$. Reduction of acetaldehyde $(CH_3CHO)$ with $LiAlH_4$ followed by $H_3O^+$ gives ethanol $(CH_3CH_2OH)$ as $[A]$.
$2$. Dehydration of ethanol with $H_2SO_4$ at high temperature gives ethene $(CH_2=CH_2)$ as $[B]$.
$3$. Addition of $HBr$ to ethene gives bromoethane $(CH_3CH_2Br)$ as $[C]$.
$4$. The reaction of bromoethane with bromobenzene in the presence of $Na$ and dry ether is a Wurtz-Fittig reaction,which yields ethylbenzene as the final product $[D]$.

(B) In the blast furnace,the temperature range of $900 \ K$ to $1500 \ K$ is the lower part of the furnace.
The reactions occurring in this range are:
$1. C + CO_2 \rightarrow 2CO$
$2. FeO + CO \rightarrow Fe + CO_2$
$3. CaO + SiO_2 \rightarrow CaSiO_3$ (Slag formation)
The reaction $Fe_2O_3 + CO \rightarrow 2FeO + CO_2$ occurs at a lower temperature range of $500 \ K$ to $800 \ K$.
Therefore,the reaction that does not take place in the $900 \ K$ to $1500 \ K$ range is $Fe_2O_3 + CO \rightarrow 2FeO + CO_2$.

(D) Pumice stone is a type of colloid known as a solid sol.
In this system,the dispersed phase is $Gas$ and the dispersion medium is $Solid$.

(D) The stability of coordination complexes is significantly enhanced by the chelate effect,which occurs when polydentate ligands (chelating agents) form ring structures with the central metal ion.
In the given options,the complex $[CoCl_2(en)_2]NO_3$ contains ethylenediamine $(en)$,which is a bidentate ligand.
Bidentate ligands form more stable complexes compared to monodentate ligands like $NH_3$,$H_2O$,$Br^-$,or $NO_3^-$.
Therefore,the complex containing the chelating ligand $(en)$ is the most stable.