NEET 2026 Chemistry Question Paper with Answer and Solution

(B) For a buffer of weak acid $HX$ and its conjugate base $X^-$,the Henderson-Hasselbalch equation for $pOH$ is given by $pOH = pK_b + log \frac{[X^-]}{[HX]}$.
Given that the concentrations are equal,$[X^-] = [HX]$,so $log \frac{[X^-]}{[HX]} = log(1) = 0$.
Therefore,$pOH = pK_b = -log(K_b) = -log(10^{-10}) = 10$.
Using the relation $pH + pOH = 14$ at $298 \text{ K}$,we get $pH = 14 - pOH = 14 - 10 = 4$.

(A) The relationship between $K_p$ and $K_c$ is given by the equation $K_p = K_c(RT)^{\Delta n_g}$,where $\Delta n_g$ is the change in the number of moles of gaseous products and reactants.
For $K_p = K_c$,the condition is $\Delta n_g = 0$.
Let us calculate $\Delta n_g$ for each reaction:
$(A)$ $N_2(g) + O_2(g) \rightleftharpoons 2NO(g) \implies \Delta n_g = 2 - (1+1) = 0$.
$(B)$ $H_2O(g) + CO(g) \rightleftharpoons H_2(g) + CO_2(g) \implies \Delta n_g = (1+1) - (1+1) = 0$.
$(C)$ $N_2(g) + 3H_2(g) \rightleftharpoons 2NH_3(g) \implies \Delta n_g = 2 - (1+3) = -2$.
$(D)$ $H_2(g) + I_2(g) \rightleftharpoons 2HI(g) \implies \Delta n_g = 2 - (1+1) = 0$.
Reactions $(A)$,$(B)$,and $(D)$ all satisfy the condition $\Delta n_g = 0$. In standard chemistry problems of this type,if multiple options are correct,the question may be flawed or require selecting the most representative example.

(A) For the reaction $2A(g) + B(g) \rightarrow 2D(g)$,the change in the number of gaseous moles is $\Delta n_g = 2 - (2 + 1) = -1$.
We know the relation $\Delta H^{\circ} = \Delta U^{\circ} + \Delta n_g RT$.
Substituting the values: $\Delta H^{\circ} = -10 \times 10^3 \text{ J mol}^{-1} + (-1) \times 8.31 \text{ J K}^{-1} \text{ mol}^{-1} \times 298 \text{ K} = -10000 - 2476.38 = -12476.38 \text{ J mol}^{-1} = -12.476 \text{ kJ mol}^{-1}$.
Now,we use the Gibbs free energy equation: $\Delta G^{\circ} = \Delta H^{\circ} - T\Delta S^{\circ}$.
$\Delta G^{\circ} = -12.476 \text{ kJ mol}^{-1} - 298 \text{ K} \times (-44 \times 10^{-3} \text{ kJ K}^{-1} \text{ mol}^{-1}) = -12.476 + 13.112 = +0.636 \text{ kJ mol}^{-1}$.
Since $\Delta G^{\circ} > 0$,the reaction is non-spontaneous.

(B) In Lassaigne's test,the organic compound is fused with metallic sodium.
This process converts the covalently bonded elements (like $N$,$S$,and Halogens) present in the organic compound into their corresponding water-soluble sodium salts,which are ionic in nature (e.g.,$NaCN$,$Na_2S$,$NaX$).

(B) In $\text{ClF}_3$,the central chlorine atom has $7$ valence electrons.
It forms $3$ covalent bonds with fluorine atoms,leaving $2$ lone pairs on the chlorine atom.
According to the $VSEPR$ theory,the steric number is $3 + 2 = 5$,which corresponds to $sp^3d$ hybridization.
The presence of $2$ lone pairs in the equatorial positions of the trigonal bipyramidal electron geometry results in a $T$-shaped molecular geometry.

(B) $1$. In the first reaction,benzene reacts with $6Cl_2$ in the presence of anhydrous $AlCl_3$ (a Lewis acid) under dark and cold conditions. This is an electrophilic substitution reaction where all hydrogen atoms of benzene are replaced by chlorine atoms,resulting in hexachlorobenzene $(C_6Cl_6)$. Thus,$X$ contains $6$ chlorine atoms.
$2$. In the second reaction,benzene reacts with $3Cl_2$ in the presence of $UV$ light at $500 \text{ K}$. This is a free radical addition reaction. Benzene undergoes addition to form benzene hexachloride $(C_6H_6Cl_6)$,which contains $6$ chlorine atoms. Thus,$Y$ contains $6$ chlorine atoms.
$3$. Therefore,the number of chlorine atoms in $X$ and $Y$ are $6$ and $6$,respectively.

According to the first law of thermodynamics, $\Delta U = q + w$. Here, heat absorbed by the system $q = +500 \text{ J}$, and work done by the system $w = -200 \text{ J}$. Thus, $\Delta U = 500 - 200 = 300 \text{ J}$.

(A) $1$. Identify the longest carbon chain: The longest continuous chain contains $7$ carbon atoms,so the parent alkane is heptane.
$2$. Numbering the chain: Number the chain from the end that gives the substituents the lowest possible locants. Numbering from left to right gives substituents at positions $3$ and $5$. Numbering from right to left also gives substituents at positions $3$ and $5$.
$3$. Alphabetical order: When locants are the same from both ends,the substituent that comes first alphabetically gets the lower number. Ethyl $(E)$ comes before methyl $(M)$. Therefore,the ethyl group is assigned position $3$ and the methyl group is assigned position $5$.
$4$. Final Name: $3$-ethyl-$5$-methylheptane.

(D) The chemical reaction is: $CO_{2}(g) + C(s) \rightarrow 2CO(g)$.
Let the initial volume of $CO_{2}$ be $1 \ dm^{3}$.
Let $x$ be the volume of $CO_{2}$ that reacts with carbon.
According to the stoichiometry of the reaction,$1 \ mole$ of $CO_{2}$ produces $2 \ moles$ of $CO$. Thus,$x \ dm^{3}$ of $CO_{2}$ will produce $2x \ dm^{3}$ of $CO$.
The remaining volume of $CO_{2}$ is $(1 - x) \ dm^{3}$.
The total volume of the gaseous mixture is the sum of the remaining $CO_{2}$ and the produced $CO$: $(1 - x) + 2x = 1 + x$.
Given that the final volume is $1.4 \ dm^{3}$,we have $1 + x = 1.4$,which gives $x = 0.4 \ dm^{3}$.
Therefore,the volume of $CO$ produced is $2x = 2(0.4) = 0.8 \ dm^{3}$.
The volume of remaining $CO_{2}$ is $1 - 0.4 = 0.6 \ dm^{3}$.

(D) The formula for formal charge is: $\text{Formal charge} = (\text{Valence electrons}) - (\text{Non-bonding electrons}) - \frac{1}{2}(\text{Bonding electrons})$.
For oxygen atom $2$ (terminal oxygen with double bond): Valence electrons = $6$,Non-bonding electrons = $4$,Bonding electrons = $4$. Formal charge = $6 - 4 - \frac{1}{2}(4) = 6 - 4 - 2 = 0$.
For oxygen atom $1$ (central oxygen): Valence electrons = $6$,Non-bonding electrons = $2$,Bonding electrons = $6$. Formal charge = $6 - 2 - \frac{1}{2}(6) = 6 - 2 - 3 = +1$.
For oxygen atom $3$ (terminal oxygen with single bond): Valence electrons = $6$,Non-bonding electrons = $6$,Bonding electrons = $2$. Formal charge = $6 - 6 - \frac{1}{2}(2) = 6 - 6 - 1 = -1$.
Thus,the formal charges on oxygen atoms $2, 1$ and $3$ are $0, +1$ and $-1$ respectively.

(C) Metallic character is defined as the tendency of an element to lose electrons.
Metallic character increases as we move down a group and decreases as we move from left to right across a period.
The positions of the given elements in the periodic table are:
$P$ (Group $15$,Period $3$)
$Si$ (Group $14$,Period $3$)
$Mg$ (Group $2$,Period $3$)
$Na$ (Group $1$,Period $3$)
$Be$ (Group $2$,Period $2$)
Comparing these,$P$ is the least metallic (non-metal),followed by $Si$ (metalloid). Among the metals,$Be$ is less metallic than $Mg$ (since $Be$ is in Period $2$ and $Mg$ is in Period $3$ of the same group). $Na$ is the most metallic as it is in Group $1$.
Thus,the correct order of increasing metallic character is: $P < Si < Be < Mg < Na$.

(B) The molar mass of urea $(CO(NH_{2})_{2})$ is $60 \ g \ mol^{-1}$.
Number of moles of urea = $\frac{\text{Given mass}}{\text{Molar mass}} = \frac{5.4 \ g}{60 \ g \ mol^{-1}} = 0.09 \ mol$.
Each molecule of urea contains $4$ hydrogen atoms.
Therefore,the number of moles of hydrogen atoms = $0.09 \ mol \times 4 = 0.36 \ mol$.
Number of hydrogen atoms = $\text{Number of moles} \times N_{A} = 0.36 \times 6.022 \times 10^{23} = 2.168 \times 10^{23}$ atoms.

(D) The equilibrium is $BiO(OH)(s) \rightleftharpoons BiO^{+}(aq) + OH^{-}(aq)$.
The solubility product constant is $K = [BiO^{+}][OH^{-}] = 4 \times 10^{-10}$.
Assuming $[BiO^{+}] = [OH^{-}] = s$,then $s^2 = 4 \times 10^{-10}$,which gives $s = 2 \times 10^{-5} \text{ M}$.
Thus,$[OH^{-}] = 2 \times 10^{-5} \text{ M}$.
Calculating $pOH$: $pOH = -\log(2 \times 10^{-5}) = 5 - \log 2 = 5 - 0.3010 = 4.699$.
Finally,$pH = 14 - pOH = 14 - 4.699 = 9.301$.

(D) The total power of the bulb is $P_{\text{total}} = 150 \text{ W}$.
The power converted into light is $P_{\text{light}} = 150 \times 0.08 = 12 \text{ J/s}$.
The energy of a single photon is given as $E_{\text{photon}} = 4.42 \times 10^{-19} \text{ J}$.
The number of photons emitted per second $(n)$ is calculated by dividing the power of light by the energy of one photon:
$n = \frac{P_{\text{light}}}{E_{\text{photon}}} = \frac{12}{4.42 \times 10^{-19}} \approx 2.715 \times 10^{19}$ photons per second.
Therefore,the correct option is $D$.

(D) Metamerism arises due to the difference in the nature of alkyl groups attached to the same polyvalent functional group (like $-O-$,$-S-$,$-NH-$,$-CO-$).
In option $(D)$,both molecules are ethers with the same molecular formula $C_{4}H_{10}O$.
$CH_{3}CH_{2}OCH_{2}CH_{3}$ is diethyl ether,where the oxygen atom is attached to two ethyl groups.
$CH_{3}OCH_{2}CH_{2}CH_{3}$ is methyl propyl ether,where the oxygen atom is attached to one methyl group and one propyl group.
Since the distribution of alkyl groups around the oxygen atom is different,they are metamers.

(A) The bonding in the given molecules is as follows:
$(A)$ $C_{2}H_{4}$ (Ethene): Structure is $CH_{2}=CH_{2}$. It contains $5\sigma$ bonds and $1\pi$ bond.
$(B)$ $C_{2}H_{2}$ (Ethyne): Structure is $CH \equiv CH$. It contains $3\sigma$ bonds and $2\pi$ bonds.
$(C)$ $CH_{4}$ (Methane): Structure is $CH_{4}$ with $sp^{3}$ hybridization. It contains $4\sigma$ bonds.
$(D)$ $NH_{3}$ (Ammonia): Structure is $NH_{3}$ with $sp^{3}$ hybridization. It contains $3\sigma$ bonds and $1$ lone pair on the nitrogen atom.
Therefore,the correct matching is: $A-IV, B-I, C-III, D-II$.

(A) The orbital is represented as $nl$,where $n$ is the principal quantum number and $l$ is the azimuthal quantum number.
The values of $l$ correspond to orbitals: $l=0$ $(s)$,$l=1$ $(p)$,$l=2$ $(d)$,$l=3$ $(f)$.
$(A)$ $n=2, l=1$ corresponds to $2p$.
$(B)$ $n=4, l=0$ corresponds to $4s$.
$(C)$ $n=5, l=3$ corresponds to $5f$.
$(D)$ $n=3, l=2$ corresponds to $3d$.
Therefore,the correct matching is $A-II, B-III, C-IV, D-I$.

(A) The reaction of methane with steam is known as steam reforming.
The chemical equation for this reaction is:
$CH_4(g) + H_2O(g) \xrightarrow{Ni, 1273 \ K} CO(g) + 3H_2(g)$
In this process,methane reacts with steam at $1273 \ K$ in the presence of a nickel catalyst to produce carbon monoxide $(CO)$ and hydrogen gas $(H_2)$.

(A) Oxygen commonly exhibits an oxidation state of $-2$,but it also shows $-1$ (in peroxides),$-1/2$ (in superoxides),and even positive states when combined with fluorine (e.g.,$OF_2$). Therefore,the statement that oxygen exhibits only $-2$ oxidation state is incorrect.

(A) $1$. Comparing the sizes: $Mg$ $(160 \text{ pm})$ > $Al$ $(143 \text{ pm})$ > $Mg^{2+}$ $(72 \text{ pm})$ > $Al^{3+}$ $(54 \text{ pm})$. Thus,$Mg$ is the largest and $Al^{3+}$ is the smallest. Statement $A$ is incorrect.
$2$. The element with atomic number $107$ is Bohrium $(Bh)$,but its systematic $IUPAC$ name is Unnilseptium. Statement $B$ is correct.
$3$. $Li$ and $Mg$ show diagonal relationship due to similar charge-to-size ratios. Statement $C$ is correct.
$4$. In $[AlCl(H_2O)_5]^{2+}$,$Al$ is in $+3$ oxidation state and forms $6$ coordinate bonds (covalency $6$). Statement $D$ is correct.

(D) The reaction of benzene with methyl chloride via Friedel-Crafts alkylation produces toluene $(W)$.
Nitration of toluene with dilute $HNO_3/H_2SO_4$ produces a mixture of ortho-nitrotoluene $(X)$ and para-nitrotoluene $(Y)$.
These isomers have significantly different boiling points due to differences in polarity and intermolecular forces,making fractional distillation the ideal method for their separation.

(C) $[Pt(Cl)_2(NH_3)_2]$ is a $d^8$ complex,which exhibits square planar geometry $(III)$.
$[Co(NH_3)_6]Cl_3$ contains the $[Co(NH_3)_6]^{3+}$ ion,which is an octahedral complex $(I)$.
$[NiCl_4]^{2-}$ is a $d^8$ tetrahedral complex $(IV)$.
$[Fe(CO)_5]$ is a $d^8$ complex with trigonal bipyramidal geometry $(II)$.
Therefore,the correct matching is $A-III, B-I, C-IV, D-II$.

(B) The half-cell reaction is: $H_2(g) \to 2H^+(aq) + 2e^-$.
Using the Nernst equation for the oxidation potential: $E = E^\circ - \frac{0.059}{n} \log Q$.
Here,$n = 2$,$E^\circ = 0 \text{ V}$,$[H^+] = 0.02 \text{ M}$,and $P_{H_2} = 2 \text{ atm}$.
$Q = \frac{[H^+]^2}{P_{H_2}} = \frac{(0.02)^2}{2} = \frac{0.0004}{2} = 0.0002 = 2 \times 10^{-4}$.
$E = 0 - \frac{0.059}{2} \log(2 \times 10^{-4})$.
$E = -0.0295 \times (\log 2 + \log 10^{-4}) = -0.0295 \times (0.3010 - 4) = -0.0295 \times (-3.699) \approx 0.109 \text{ V}$.

(D) For a zero-order reaction,the rate of reaction is independent of the concentration of the reactant.
The integrated rate equation for a zero-order reaction is given by $[R]_t = -kt + [R]_0$.
Comparing this with the equation of a straight line $y = mx + c$,where $y = [R]_t$,$x = t$,$m = -k$ (slope),and $c = [R]_0$ (intercept).
$A$ linear plot of concentration $[R]$ vs time with a constant negative slope signifies a zero-order reaction.

(D) Comparing the given equation $\ln k = 14.34 - \frac{1.25 \times 10^4}{T}$ with the Arrhenius equation $\ln k = \ln A - \frac{E_a}{RT}$:
$-\frac{E_a}{R} = -1.25 \times 10^4$
$E_a = 1.25 \times 10^4 \times R$
$E_a = 1.25 \times 10^4 \times 1.987 \text{ cal mol}^{-1} \text{ K}^{-1} = 2.48375 \times 10^4 \text{ cal mol}^{-1}$
$E_a = 24.8375 \text{ kcal mol}^{-1} \approx 24.84 \text{ kcal mol}^{-1}$.

(C) Nitriles $(R-C\equiv N)$ are reduced to primary amines $(R-CH_2-NH_2)$ using strong reducing agents.
$1$. $\text{LiAlH}_4$ (Lithium Aluminium Hydride) is a strong reducing agent that reduces nitriles to primary amines.
$2$. $\text{H}_2/\text{Ni}$ (Catalytic hydrogenation) also reduces nitriles to primary amines.
$3$. $\text{Na(Hg)}/\text{C}_2\text{H}_5\text{OH}$ is known as the Mendius reduction,which reduces nitriles to primary amines.
$4$. $\text{Sn} + \text{HCl}$ is primarily used for the reduction of nitro compounds to amines.
$5$. $\text{Br}_2/\text{aq. NaOH}$ is used for the Hofmann Bromamide degradation of amides to primary amines with one carbon atom less.
Therefore,the correct reagents are $A, C$,and $D$.

(A) $DNA$ generally exists as a double-stranded helix and contains Adenine $(A)$,Guanine $(G)$,Cytosine $(C)$,and Thymine $(T)$ as its nitrogenous bases.
$RNA$ is typically single-stranded and contains Uracil $(U)$ instead of Thymine $(T)$.

(B) Chloroform $(CHCl_3)$ and acetone $(CH_3COCH_3)$ form a solution that shows a negative deviation from Raoult's law.
This occurs because the intermolecular forces,specifically the formation of hydrogen bonding between chloroform and acetone molecules,are stronger than the individual intermolecular forces present in the pure components.
This increased attraction reduces the escaping tendency of the molecules,thereby decreasing the total vapour pressure of the solution compared to the ideal behavior.

(C) The smell of vinegar is a characteristic property of acetic acid $(\text{CH}_3\text{COOH})$.
When dilute $\text{H}_2\text{SO}_4$ is added to a salt containing the acetate anion $(\text{CH}_3\text{COO}^-)$,the following reaction occurs:
$2\text{CH}_3\text{COO}^- + \text{H}_2\text{SO}_4 \rightarrow 2\text{CH}_3\text{COOH} + \text{SO}_4^{2-}$.
Acetic acid is liberated as colourless vapours,which are acidic in nature and turn blue litmus paper red.

(C) . $[Pt(NH_3)_2Cl_2]$ shows geometrical isomerism (cis-trans).
$B$. $[Co(en)_3]^{3+}$ shows optical isomerism due to the lack of a plane of symmetry.
$C$. $[Co(NH_3)_5NO_2]Cl_2$ exhibits linkage isomerism because of the ambidentate $NO_2^-$ ligand.
$D$. $[Cr(H_2O)_6]Cl_3$ exhibits solvate (hydrate) isomerism.
Therefore,the correct matching is $A-III, B-I, C-IV, D-II$.

(C) Nitrogen has a small atomic size and high electronegativity,which allows it to form stable $p\pi$-$p\pi$ multiple bonds with itself (as in $N_{2}$).
Nitrogen lacks vacant $d$-orbitals in its valence shell,hence it cannot form $d\pi$-$p\pi$ bonds.
Therefore,the statement that nitrogen can form $d\pi$-$p\pi$ bonds with oxygen is incorrect.

(B) $1$. Compound $P$ $(C_{8}H_{8}O)$ reacts with $2,4-DNP$ reagent,which confirms the presence of a carbonyl group (aldehyde or ketone).
$2$. It does not reduce Fehling's reagent,which indicates that $P$ is a ketone,not an aldehyde.
$3$. The molecular formula $C_{8}H_{8}O$ corresponds to acetophenone $(C_{6}H_{5}COCH_{3})$.
$4$. On drastic oxidation with chromic acid $(H_{2}CrO_{4})$,the alkyl group attached to the benzene ring is oxidized to a carboxylic acid group.
$5$. Acetophenone $(C_{6}H_{5}COCH_{3})$ on oxidation yields benzoic acid $(C_{6}H_{5}COOH)$ as product $Q$.
$6$. Benzoic acid $(Q)$ reacts with aqueous $NaHCO_{3}$ to produce effervescence due to the evolution of $CO_{2}$ gas.
$7$. Therefore,$P$ is acetophenone and $Q$ is benzoic acid.

(B) $1$. The reaction of propan$-1-$ol $(CH_{3}CH_{2}CH_{2}OH)$ with phosphorus pentachloride $(PCl_{5})$ is given by:
$CH_{3}CH_{2}CH_{2}OH + PCl_{5} \rightarrow CH_{3}CH_{2}CH_{2}Cl + POCl_{3} + HCl$
Thus,$X = POCl_{3}$.
$2$. The product $CH_{3}CH_{2}CH_{2}Cl$ reacts with alcoholic $KOH$ upon heating to undergo dehydrohalogenation,forming propene $(Y = CH_{3}CH=CH_{2})$.
$3$. Propene $(CH_{3}CH=CH_{2})$ reacts with $HBr$ in the presence of peroxide $((C_{6}H_{5}CO)_{2}O_{2})$ via the anti-Markownikoff addition mechanism to yield $1-$bromopropane $(Z = CH_{3}CH_{2}CH_{2}Br)$.
Therefore,$X = POCl_{3}$ and $Z = CH_{3}CH_{2}CH_{2}Br$.

(B) The general unit for the rate constant $k$ for a reaction of order $n$ is given by the formula: $k = (mol \text{ } L^{-1})^{1-n} s^{-1}$.
For $n=0$ (Zero order): $k = (mol \text{ } L^{-1})^{1-0} s^{-1} = mol \text{ } L^{-1} s^{-1}$ (Matches $IV$).
For $n=1$ (First order): $k = (mol \text{ } L^{-1})^{1-1} s^{-1} = s^{-1}$ (Matches $III$).
For $n=2$ (Second order): $k = (mol \text{ } L^{-1})^{1-2} s^{-1} = mol^{-1} L s^{-1}$ (Matches $I$).
For $n=3$ (Third order): $k = (mol \text{ } L^{-1})^{1-3} s^{-1} = mol^{-2} L^{2} s^{-1}$ (Matches $II$).
Therefore,the correct matching is $A-IV, B-III, C-I, D-II$.

(D) The correct matching is as follows:
$A$. $H_{3}C-CH(OH)-CH_{3}$ is formed by the reaction of $CH_{3}COOH$ with $CH_{3}OH$ followed by reduction (Grignard or similar sequence). This matches with $III$.
$B$. $CH_{3}COOH \rightarrow CH_{3}CH_{2}OH$ is a reduction reaction. This matches with $II$.
$C$. $CH_{3}CH_{2}OH \rightarrow H_{3}C-CH(OH)-CH_{3}$ involves dehydration to ethene followed by hydration. This matches with $IV$.
$D$. Benzene $\rightarrow$ Phenol is the Cumene process. This matches with $I$.
Therefore,the correct sequence is $A-III, B-II, C-IV, D-I$. Note: The provided options in the prompt were slightly inconsistent with standard chemical pathways; based on the logic,the correct mapping is $A-III, B-II, C-IV, D-I$.

(A) Molality $(m)$ = $\frac{\text{moles of solute}}{\text{mass of solvent in kg}} = \frac{2.5 / 60}{75 / 1000} = \frac{0.04167}{0.075} \approx 0.556 \text{ m}$. Statement $(A)$ is correct.
$(B)$ Molarity $(M)$ = $\frac{\text{moles of solute}}{\text{volume of solution in L}} = \frac{5 / 40}{450 / 1000} = \frac{0.125}{0.45} \approx 0.278 \text{ M}$. Statement $(B)$ is correct.
$(C)$ According to Henry's law,the solubility of gases in liquids increases with a decrease in temperature. Thus,aquatic species are more comfortable in cold water. Statement $(C)$ is correct.
$(D)$ According to Henry's law,the solubility of a gas in a liquid is directly proportional to the partial pressure of the gas. Thus,solubility increases with an increase in pressure. Statement $(D)$ is incorrect.
$(E)$ The mole fraction of component $B$ is defined as $x_{B} = \frac{n_{B}}{n_{A} + n_{B}}$. Statement $(E)$ is incorrect.
Therefore,statements $(A)$,$(B)$,and $(C)$ are correct.

(C) An ambidentate ligand is a ligand that can coordinate through two different atoms.
Thiocyanate $(SCN^{-})$ is an ambidentate ligand because it can coordinate through either the sulphur atom (thiocyanato-$S$) or the nitrogen atom (isothiocyanato-$N$).
Oxalate,ethane$-1,2-$diamine,and ethylenediaminetetraacetate ion are polydentate ligands but are not ambidentate.

(D) The phthalein dye test is a characteristic chemical test used to identify the presence of phenolic groups.
In this test,a phenol reacts with phthalic anhydride in the presence of concentrated sulphuric acid $(H_2SO_4)$ as a dehydrating agent.
This reaction results in the formation of a phthalein dye (such as phenolphthalein),which exhibits a distinct color change in an alkaline medium.
Therefore,the correct functional group identified by this test is the phenolic group.

(A) The mass of the substance deposited during electrolysis is given by Faraday's law: $m = (I \times t \times M) / (n \times F)$.
Here,the current $I = 1.5 \text{ A}$,time $t = 10 \text{ minutes} = 600 \text{ s}$,molar mass $M = 63 \text{ g mol}^{-1}$,and $n = 2$ (since $Cu^{2+} + 2e^{-} \rightarrow Cu$).
Substituting the values: $m = (1.5 \times 600 \times 63) / (2 \times 96487)$.
$m = 56700 / 192974 \approx 0.2938 \text{ g}$.
Thus,the mass of copper deposited is $0.2938 \text{ g}$.

(B) $1$. $C_2H_6 \xrightarrow{Cl_2, UV \text{ light}} C_2H_5Cl$ ($X$ is ethyl chloride).
$2$. $C_2H_5Cl \xrightarrow{NH_3} C_2H_5NH_2$ ($Y$ is ethylamine).
$3$. $C_2H_5NH_2 \xrightarrow{(i) NaNO_2/HCl, (ii) H_2O} C_2H_5OH$.
Diazotization of ethylamine followed by hydrolysis yields ethanol $(Z)$.

(C) Cerium $(Ce)$ has an atomic number of $58$.
Its electronic configuration is $[Xe] 4f^1 5d^1 6s^2$.
By losing four electrons,it reaches the stable noble gas configuration of Xenon $(4f^0)$,which is why it readily exhibits the $+4$ oxidation state.

(A) Phenolphthalein is a synthetic acid-base indicator.
In an acidic medium,phenolphthalein remains colourless,whereas in a basic (alkaline) medium,it turns pink.
In the titration of $NaOH$ (a strong base) against oxalic acid (a weak acid),the $NaOH$ solution is taken in the conical flask with phenolphthalein,making the initial solution pink.
As the acid is added from the burette,the $pH$ of the solution decreases.
At the equivalence point,the solution transitions from a basic state to a neutral/slightly acidic state,causing the pink colour to disappear.
Therefore,the observed colour change is pink to colourless.

(B) Reaction $1$: $C_2H_5Cl + AgCN \rightarrow C_2H_5NC$ (Ethyl isocyanide,which is a foul-smelling compound).
Reaction $2$: This involves the Hofmann bromamide degradation followed by the carbylamine reaction.
Step $1$: $C_2H_5CONH_2 \xrightarrow{Br_2, NaOH} C_2H_5NH_2$ ($Y$ is ethylamine).
Step $2$: $C_2H_5NH_2 + CHCl_3 + 3KOH \xrightarrow{\Delta} C_2H_5NC + 3KCl + 3H_2O$ (Carbylamine reaction).
Thus,$Z$ is ethyl isocyanide $(C_2H_5NC)$ and $X$ is $AgCN$.

(C) The formula for the spin-only magnetic moment is $\mu = \sqrt{n(n+2)}$ $BM$,where $n$ is the number of unpaired electrons.
For the $Ti^{2+}$ ion with electronic configuration $3d^2$,the number of unpaired electrons $n = 2$.
Substituting the value of $n$ into the formula:
$\mu = \sqrt{2(2+2)} = \sqrt{2 \times 4} = \sqrt{8} \approx 2.83$ $BM$.
Rounding to two decimal places,we get $2.84$ $BM$.
Thus,the correct option is $C$.

(B) The correct matches are as follows:
$A$. $V_2O_5$ is used as a catalyst in the Contact Process for the industrial preparation of $H_2SO_4$ from $SO_2$. Thus,$A-III$.
$B$. $Fe$ is used as a catalyst in the Haber process for the preparation of ammonia from a $N_2/H_2$ mixture. Thus,$B-I$.
$C$. $PdCl_2$ is used as the Wacker catalyst for the oxidation of ethyne to ethanal. Thus,$C-IV$.
$D$. $Ni$ complexes are used to catalyze the polymerization of alkynes. Thus,$D-II$.
Therefore,the correct matching is $A-III, B-I, C-IV, D-II$.