NEET 2018 Chemistry Question Paper with Answer and Solution

(C) Secondary pollutants are not emitted directly into the atmosphere but are formed through chemical reactions between primary pollutants and other atmospheric components.
Ozone $(O_3)$ is a classic example of a secondary pollutant,formed in the troposphere by the photochemical reaction of nitrogen oxides $(NO_x)$ and volatile organic compounds $(VOCs)$ in the presence of sunlight.
$SO_2$,$CO$,and $NO_2$ are primary pollutants as they are emitted directly from sources like combustion processes.

(A) primary pollutant is an air pollutant emitted directly from a source,such as $CO$,$NO_2$,and $SO_2$.
Secondary pollutants are not emitted directly; they form in the atmosphere when primary pollutants react or interact.
$PAN$ (Peroxyacetyl nitrate) is a classic example of a secondary pollutant formed by the photochemical reaction of nitrogen oxides and volatile organic compounds in the presence of sunlight.
Therefore,the correct option is $A$.

(D) Let $V$ be the volume of the wires. Since $V = A \times L$,and both wires have the same volume and material,we have $A_1 L_1 = A_2 L_2$.
Given $A_1 = A$ and $A_2 = 3A$,we get $A L_1 = 3A L_2$,which implies $L_1 = 3 L_2$.
The Young's modulus $Y$ is given by $Y = \frac{F L}{A \Delta x}$.
For wire $1$: $Y = \frac{F L_1}{A \Delta x} \implies F = \frac{Y A \Delta x}{L_1}$.
For wire $2$: $Y = \frac{F' L_2}{3A \Delta x} \implies F' = \frac{Y (3A) \Delta x}{L_2}$.
Substitute $L_2 = \frac{L_1}{3}$ into the equation for $F'$:
$F' = \frac{Y (3A) \Delta x}{L_1 / 3} = 9 \left( \frac{Y A \Delta x}{L_1} \right)$.
Since $F = \frac{Y A \Delta x}{L_1}$,we get $F' = 9F$.

(D) The Young's modulus $Y$ is given by $Y = \frac{F L}{A \Delta x}$,where $L$ is the length,$A$ is the area,and $\Delta x$ is the extension.
Since the volume $V = A L$ is constant,we can write $L = \frac{V}{A}$.
Substituting $L$ into the formula,we get $F = \frac{Y A \Delta x}{L} = \frac{Y A \Delta x}{(V/A)} = \frac{Y A^2 \Delta x}{V}$.
Since $Y, V,$ and $\Delta x$ are the same for both wires,we have $F \propto A^2$.
Therefore,$\frac{F_2}{F_1} = \left(\frac{A_2}{A_1}\right)^2 = \left(\frac{3A}{A}\right)^2 = 3^2 = 9$.
Thus,$F_2 = 9 F_1 = 9F$.

(C) Formic acid $(HCOOH)$ reacts with conc. $H_2SO_4$ to produce $CO$ and $H_2O$: $HCOOH \xrightarrow{conc. H_2SO_4} CO + H_2O$.
Oxalic acid $(H_2C_2O_4)$ reacts with conc. $H_2SO_4$ to produce $CO$,$CO_2$,and $H_2O$: $H_2C_2O_4 \xrightarrow{conc. H_2SO_4} CO + CO_2 + H_2O$.
$KOH$ pellets absorb the acidic gas $CO_2$,leaving $CO$ as the remaining gaseous product.
Moles of $HCOOH = \frac{2.3 \ g}{46 \ g/mol} = 0.05 \ mol$. This produces $0.05 \ mol$ of $CO$.
Moles of $H_2C_2O_4 = \frac{4.5 \ g}{90 \ g/mol} = 0.05 \ mol$. This produces $0.05 \ mol$ of $CO$ and $0.05 \ mol$ of $CO_2$.
Total moles of $CO$ remaining = $0.05 \ mol + 0.05 \ mol = 0.1 \ mol$.
Mass of $CO = 0.1 \ mol \times 28 \ g/mol = 2.8 \ g$.

(B) In the alkaline earth metals group,as we move down the group,the metallic character increases,which leads to an increase in the basic nature of the oxides. $BeO$ is amphoteric in nature,while $MgO$,$CaO$,and $BaO$ are basic. Among the given options,$BeO$ is the most acidic (or least basic) because it has the smallest size and highest polarizing power.

(A) The ionic character of metal hydrides depends on the electronegativity difference between the metal and hydrogen,or alternatively,on the polarizing power of the cation.
According to Fajan's rule,smaller cations have higher polarizing power,which increases the covalent character of the bond.
As we move down the group $2$ ($Be$ to $Ba$),the size of the cation increases,and its polarizing power decreases.
Therefore,the covalent character decreases and the ionic character increases down the group.
The order of ionic character is $BeH_2 < CaH_2 < BaH_2$.

(A) To find the number of molecules,we calculate the number of moles $(n)$ in each case,as $Number \; of \; molecules = n \times N_A$.
$A$) $18 \; mL$ of water: Since density $d = 1 \; g/mL$,mass $W = 18 \; g$. Moles $n = \frac{18}{18} = 1 \; mol$. Molecules $= 1 \times N_A$.
$B$) $0.18 \; g$ of water: Moles $n = \frac{0.18}{18} = 0.01 \; mol$. Molecules $= 0.01 \times N_A$.
$C$) $0.00224 \; L$ of water vapour at $STP$: Moles $n = \frac{0.00224}{22.4} = 0.0001 \; mol$. Molecules $= 0.0001 \times N_A$.
$D$) $10^{-3} \; mol$ of water: Moles $n = 0.001 \; mol$. Molecules $= 0.001 \times N_A$.
Comparing the values,$1 \times N_A$ is the maximum. Thus,the correct option is $A$.

(D) The electronic configuration of element $(X)$ is $1s^{2} 2s^{2} 2p^{3}$.
This corresponds to Nitrogen $(N)$,which has $5$ valence electrons and needs $3$ electrons to complete its octet,forming an ion with a charge of $-3$ $(N^{3-})$.
Magnesium $(Mg)$ is an alkaline earth metal with $2$ valence electrons,forming an ion with a charge of $+2$ $(Mg^{2+})$.
To form a neutral ionic compound,the charges must balance: $3 \times (+2) + 2 \times (-3) = 0$.
Therefore,the simplest formula for the compound is $Mg_{3}X_{2}$.

(C) The electronic configuration of $N$ $(Z=7)$ is $1s^2 2s^2 2p_x^1 2p_y^1 2p_z^1$. According to Hund's rule of maximum multiplicity,electrons in degenerate orbitals (like $2p_x, 2p_y, 2p_z$) should be filled singly first with parallel spins. The provided image for option $C$ shows the last electron with an opposite spin,which violates Hund's rule. Thus,option $C$ is the wrong statement.

(B) The bond order of a diatomic species can be calculated using the formula: $\text{Bond Order} = \frac{N_b - N_a}{2}$,where $N_b$ is the number of bonding electrons and $N_a$ is the number of antibonding electrons.
For species with total electrons up to $14$,the configuration is $\sigma 1s^2, \sigma^* 1s^2, \sigma 2s^2, \sigma^* 2s^2, \pi 2p_x^2 = \pi 2p_y^2, \sigma 2p_z^2$.
$1$. $CN^{+}$ ($12$ electrons): $\text{Bond Order} = \frac{8-4}{2} = 2.0$
$2$. $CN$ ($13$ electrons): $\text{Bond Order} = \frac{9-4}{2} = 2.5$
$3$. $CN^{-}$ ($14$ electrons): $\text{Bond Order} = \frac{10-4}{2} = 3.0$
$4$. $NO$ ($15$ electrons): $\text{Bond Order} = \frac{10-5}{2} = 2.5$
Comparing the values,$CN^{-}$ has the highest bond order of $3.0$.

(C) The general formula for the complex ion is $MF_{6}^{3-}$.
Boron $(B)$ belongs to the $2^{nd}$ period and has a maximum covalency of $4$ because it lacks vacant $d$-orbitals in its valence shell.
Therefore,it cannot expand its octet to form the $BF_{6}^{3-}$ ion.
Other elements like $Al$,$Ga$,and $In$ have vacant $d$-orbitals and can form such complex ions.

(B) The central atom $Cl$ has $7$ valence electrons.
In $ClF_3$,$Cl$ forms $3$ single bonds with $3$ $F$ atoms.
This leaves $7 - 3 = 4$ electrons,which form $2$ lone pairs.
According to $VSEPR$ theory,these $2$ lone pairs occupy the equatorial positions in the trigonal bipyramidal geometry to minimize repulsion,resulting in a $T$-shaped molecular geometry.

(D) In group $13$,the atomic radii increase down the group from $B$ to $Al$.
However,due to the poor shielding effect of $d$-electrons in $Ga$ (transition contraction),the atomic radius of $Ga$ is slightly smaller than that of $Al$.
Thus,the correct order is $B < Ga < Al < In < Tl$.

(A) To find the decreasing order of oxidation states,we calculate the oxidation state of nitrogen $(N)$ in each compound:
$1$. In $HNO_{3}$: $1 + x + 3(-2) = 0 \implies x = +5$
$2$. In $NO$: $x + (-2) = 0 \implies x = +2$
$3$. In $N_{2}$: The oxidation state of an element in its free state is $0$.
$4$. In $NH_{4}Cl$: $x + 4(1) + (-1) = 0 \implies x + 3 = 0 \implies x = -3$
Comparing the values: $+5 > +2 > 0 > -3$.
Therefore,the correct order is $HNO_{3} > NO > N_{2} > NH_{4}Cl$.

(D) For a solution to have $pH = 1$,the concentration of $[H^+]$ must be $10^{-1} \; M = 0.1 \; M$.
Let us calculate the final $[H^+]$ for each case:
$a. \; [H^+] = \frac{(60 \times 0.1) - (40 \times 0.1)}{60 + 40} = \frac{6 - 4}{100} = 0.02 \; M \; (pH \approx 1.7)$
$b. \; [H^+] = \frac{(55 \times 0.1) - (45 \times 0.1)}{55 + 45} = \frac{5.5 - 4.5}{100} = 0.01 \; M \; (pH = 2)$
$c. \; [H^+] = \frac{(75 \times 0.2) - (25 \times 0.2)}{75 + 25} = \frac{15 - 5}{100} = \frac{10}{100} = 0.1 \; M \; (pH = 1)$
$d. \; [H^+] = 0 \; M \; (pH = 7, \text{neutral solution})$
Thus,the solution $c$ has $pH = 1$.

(A) The solubility $(s)$ in $mol \ L^{-1}$ is calculated by dividing the solubility in $g \ L^{-1}$ by the molar mass of $BaSO_{4}$.
$s = \frac{2.42 \times 10^{-3} \ g \ L^{-1}}{233 \ g \ mol^{-1}} \approx 1.0386 \times 10^{-5} \ mol \ L^{-1}$.
For a salt like $BaSO_{4}$,the dissociation is $BaSO_{4}(s) \rightleftharpoons Ba^{2+}(aq) + SO_{4}^{2-}(aq)$.
The solubility product $K_{sp} = [Ba^{2+}][SO_{4}^{2-}] = s \times s = s^{2}$.
$K_{sp} = (1.0386 \times 10^{-5})^{2} \approx 1.078 \times 10^{-10} \ mol^{2} \ L^{-2}$.
Rounding to two decimal places,we get $1.08 \times 10^{-10} \ mol^{2} \ L^{-2}$.

(A) The ease of liquefaction of a gas is directly proportional to the magnitude of the van der Waals constant $a$,which represents the intermolecular forces of attraction.
Greater value of $a$ implies stronger intermolecular forces,leading to a higher critical temperature $(T_{C})$.
Comparing the given values: $a(NH_{3}) = 4.17$,$a(H_{2}) = 0.244$,$a(O_{2}) = 1.36$,and $a(CO_{2}) = 3.59$.
Since $NH_{3}$ has the highest value of $a$ $(4.17)$,it has the strongest intermolecular forces and is the most easily liquefied gas among the given options.

(D) The hydrocarbon $(A)$ is methane $(CH_4)$.
$CH_4$ reacts with $Br_2$ in the presence of light $(hv)$ via free radical substitution to form methyl bromide $(CH_3Br)$.
$CH_4 + Br_2 \xrightarrow{hv} CH_3Br + HBr$
Methyl bromide $(CH_3Br)$ undergoes the Wurtz reaction in the presence of sodium $(Na)$ and dry ether to form ethane $(CH_3-CH_3)$.
$2CH_3Br + 2Na \xrightarrow{\text{dry ether}} CH_3-CH_3 + 2NaBr$
Ethane $(CH_3-CH_3)$ is a gaseous hydrocarbon containing two carbon atoms,which is less than four carbon atoms.

(A) Nitrogen oxides like $NO$ and $NO_2$ are common pollutants produced by both natural processes (e.g.,lightning,forest fires) and human activities (e.g.,combustion in automobile engines). $N_2O$ is primarily produced by natural biological processes in soil and oceans. $N_2O_5$ is not a common pollutant introduced into the atmosphere by these activities.

(B) The reduction half-reaction is: $MnO_{4}^{-} + 8H^{+} + 5e^{-} \longrightarrow Mn^{2+} + 4H_{2}O$ ... $(I)$
The oxidation half-reaction is: $C_{2}O_{4}^{2-} \longrightarrow 2CO_{2} + 2e^{-}$ ... $(II)$
To balance the electrons,multiply $(I)$ by $2$ and $(II)$ by $5$:
$2MnO_{4}^{-} + 16H^{+} + 10e^{-} \longrightarrow 2Mn^{2+} + 8H_{2}O$
$5C_{2}O_{4}^{2-} \longrightarrow 10CO_{2} + 10e^{-}$
Adding these two equations gives the balanced redox reaction:
$2MnO_{4}^{-} + 5C_{2}O_{4}^{2-} + 16H^{+} \longrightarrow 2Mn^{2+} + 10CO_{2} + 8H_{2}O$
Thus,the coefficients for $MnO_{4}^{-}$,$C_{2}O_{4}^{2-}$,and $H^{+}$ are $2$,$5$,and $16$ respectively.

(A) The given reaction is $A_{2(g)} + B_{2(g)} \rightleftharpoons X_{2(g)}$.
$1$. The enthalpy change $\Delta_{r} H = -X \ kJ$ indicates that the reaction is exothermic $(\Delta H < 0)$. According to Le Chatelier's principle,for an exothermic reaction,a low temperature favours the formation of the product.
$2$. The change in the number of gaseous moles is $\Delta n_{g} = n_{p} - n_{r} = 1 - (1 + 1) = -1$. Since $\Delta n_{g} < 0$,an increase in pressure will shift the equilibrium towards the side with fewer moles,which is the product side.
Therefore,low temperature and high pressure favour the maximum formation of the product.

(D) The van der Waals equation for real gases is given by $(P + \frac{an^2}{V^2})(V - nb) = nRT$.
In this equation,the term $\frac{an^2}{V^2}$ represents the pressure correction.
The constant '$a$' is a measure of the magnitude of the intermolecular forces of attraction between the gas molecules.

(C) Let the bond dissociation energies ($B$.$E$.) of $X_2$,$Y_2$,and $XY$ be $x \ kJ \ mol^{-1}$,$0.5x \ kJ \ mol^{-1}$,and $x \ kJ \ mol^{-1}$ respectively.
The chemical equation for the formation of $XY$ is: $\frac{1}{2}X_2(g) + \frac{1}{2}Y_2(g) \rightarrow XY(g)$.
The enthalpy of reaction is given by: $\Delta H = \Sigma(B.E.)_{\text{reactants}} - \Sigma(B.E.)_{\text{products}}$.
Substituting the values: $-200 = [\frac{1}{2} \times (x) + \frac{1}{2} \times (0.5x)] - [1 \times (x)]$.
$-200 = [0.5x + 0.25x] - x$.
$-200 = 0.75x - x$.
$-200 = -0.25x$.
$x = \frac{200}{0.25} = 800 \ kJ \ mol^{-1}$.
Therefore,the bond dissociation energy of $X_2$ is $800 \ kJ \ mol^{-1}$.

(D) zwitterion is a molecule that contains both a positive and a negative charge,making it electrically neutral overall.
Amino acids,such as glycine $(NH_{2}CH_{2}COOH)$,contain both an acidic carboxyl group $(-COOH)$ and a basic amino group $(-NH_{2})$.
In an aqueous solution,the proton from the carboxyl group is transferred to the amino group,resulting in the formation of a zwitterion:
$HOOC-CH_{2}-NH_{2} \rightleftharpoons ^{-}OOC-CH_{2}-NH_{3}^{+}$

(A) The inductive effect ($-I$ effect) is directly proportional to the electronegativity of the atom attached to the carbon chain.
Electronegativity order of the atoms is: $N < O < F$.
Therefore,the $-I$ effect order for the given substituents is: $-NH_2 < -OR < -F$ and $-NR_2 < -OR < -F$.
Both options $A$ and $B$ represent the correct order of the $-I$ effect.

(C) The $-NO_2$ group is a strong electron-withdrawing group ($-I$ and $-M$ effects).
In electrophilic aromatic substitution,the stability of the intermediate carbocation (arenium ion) determines the orientation.
When the positive charge is at the ortho or para positions,it comes into direct conjugation with the $-NO_2$ group,leading to significant destabilization due to the electron-withdrawing effect.
When the positive charge is at the meta position,it is not in direct conjugation with the $-NO_2$ group,making it relatively more stable than the ortho and para intermediates.
Therefore,the meta-substituted carbocation is the most stable among the options.

(B) The correct molecule is $CH_{2}=CH-C \equiv CH$.
$1$. The first carbon $(CH_{2}=)$ is bonded to two hydrogens and one carbon via a double bond,so it has $3$ sigma bonds and $0$ lone pairs,resulting in $sp^2$ hybridisation.
$2$. The second carbon $(-CH-)$ is bonded to one hydrogen,one carbon via a double bond,and one carbon via a single bond,resulting in $sp^2$ hybridisation.
$3$. The third carbon $(-C \equiv)$ is bonded to one carbon via a single bond and one carbon via a triple bond,resulting in $sp$ hybridisation.
$4$. The fourth carbon $(\equiv CH)$ is bonded to one carbon via a triple bond and one hydrogen,resulting in $sp$ hybridisation.
Thus,the order is $sp^2, sp^2, sp, sp$.

(B) The inheritance of $ABO$ blood groups in humans is governed by the gene $I$.
$1$. Dominance: The alleles $I^A$ and $I^B$ are dominant over $i$,as $I^A$ and $I^B$ produce sugars,while $i$ does not produce any sugar.
$2$. Co-dominance: When both $I^A$ and $I^B$ are present together,they both express their own types of sugars,resulting in blood group $AB$. This is an example of co-dominance.
$3$. Multiple Alleles: Since there are three alleles $(I^A, I^B, i)$ for a single gene locus,this is an example of multiple allelism.
Therefore,$a, b,$ and $c$ are the correct characteristics.

(B) The $Limbic$ $system$ (or $Limbic$ $lobe$) along with the $Hypothalamus$ is involved in the regulation of sexual behavior,expression of emotional reactions (e.g.,excitement,pleasure,rage,and fear),and motivation. It does not control voluntary movement; that function is primarily associated with the $Cerebellum$ and the $Motor$ $cortex$ of the $Cerebrum$. Therefore,the pairing in option $B$ is incorrect. The $Medulla$ $oblongata$ contains centers that control respiration,cardiovascular reflexes,and gastric secretions. The $Hypothalamus$ contains a number of centers which control body temperature,urge for eating and drinking,and produces releasing hormones. The $Corpus$ $callosum$ is a tract of nerve fibers that connects the left and right cerebral hemispheres.

(D) In a strong acidic medium,the lone pair of electrons on the nitrogen atom of aniline is protonated by the acid to form the anilinium ion $(C_6H_5NH_3^+)$.
The $-NH_3^+$ group is strongly electron-withdrawing due to its positive charge and exerts a strong $-I$ effect.
This deactivates the benzene ring and makes it meta-directing for electrophilic substitution reactions.
Therefore,the nitration of aniline in a strong acidic medium yields a significant amount of $m$-nitroaniline.

(A) Amylose is a long unbranched chain of $\alpha-D$-glucose units held by $C_1-C_4$ glycosidic linkages.
Amylopectin is a branched-chain polymer of $\alpha-D$-glucose units in which the main chain is formed by $C_1-C_4$ glycosidic linkages,while branching occurs by $C_1-C_6$ glycosidic linkages.
Therefore,the correct statement is that amylopectin has both $1 \rightarrow 4$ $\alpha$-linkage and $1 \rightarrow 6$ $\alpha$-linkage.

(D) Cross-linked or network polymers are formed from bi-functional and tri-functional monomers.
They contain strong covalent bonds between various linear polymer chains.
Examples include $Bakelite$ and $Melamine$.
Since all statements $A$,$B$,$C$,and $D$ are technically correct descriptions of cross-linked polymers,there is no incorrect statement provided in the options. However,in the context of standard chemistry curriculum,all these statements are considered true properties of network polymers.

(D) The given reaction is the Reimer-Tiemann reaction,where phenol reacts with chloroform $(CHCl_3)$ in the presence of aqueous sodium hydroxide $(NaOH)$ to form salicylaldehyde.
In this reaction,the electrophile is generated from chloroform.
First,$CHCl_3$ reacts with $OH^-$ to form the trichloromethyl anion $(\stackrel{\ominus}{C}Cl_3)$.
This anion then undergoes $\alpha$-elimination by losing a chloride ion $(Cl^-)$ to form dichlorocarbene $(:CCl_2)$.
Dichlorocarbene is an electron-deficient species with a sextet of electrons,making it a strong electrophile.

(D) Carboxylic acids have higher boiling points than aldehydes,ketones,and even alcohols of comparable molecular mass.
This is due to the more extensive association of carboxylic acid molecules through intermolecular $H$-bonding,which often results in the formation of stable dimers in the vapor phase or non-polar solvents.

(C) The haloform reaction is given by compounds containing the $CH_3-CO-$ group or the $CH_3-CH(OH)-$ group.
Compound $A$ with molecular formula $C_8H_{10}O$ that contains the $CH_3-CH(OH)-$ group is $1-phenylethanol$ $(C_6H_5-CH(OH)-CH_3)$.
$NaOI$ is prepared by the reaction of $I_2$ with $NaOH$.
Therefore,$A$ is $1-phenylethanol$ and $Y$ is $I_2$.

(B) For a $1^{st}$-order reaction,the half-life is given by $t_{1/2} = \frac{0.693}{k}$,which is independent of the initial concentration $[A]_0$.
For a $2^{nd}$-order reaction,the half-life is given by $t_{1/2} = \frac{1}{k[A]_0}$,which is inversely proportional to the initial concentration $[A]_0$.

(D) species undergoes disproportionation if its standard reduction potential $(SRP)$ to a lower oxidation state is greater than its $SRP$ to a higher oxidation state.
For $HBrO$:
$HBrO + H^{+} + e^{-} \rightarrow \frac{1}{2} Br_2 + H_2O$ $(E^{\circ} = 1.595 \ V)$
$HBrO + 2H_2O \rightarrow BrO_3^{-} + 5H^{+} + 4e^{-}$ $(E^{\circ} = -1.5 \ V)$
Adding these,the overall reaction is:
$2HBrO \rightarrow Br_2 + BrO_3^{-} + H_2O$
$E^{\circ}_{cell} = E^{\circ}_{cathode} - E^{\circ}_{anode} = 1.595 \ V - 1.5 \ V = 0.095 \ V$
Since $E^{\circ}_{cell} > 0$,the reaction is spontaneous,and $HBrO$ undergoes disproportionation.

(C) For $bcc$ structure,the number of atoms per unit cell $Z_{bcc} = 2$ and the relation between edge length $a$ and atomic radius $r$ is $4r = \sqrt{3}a$,so $a_{bcc} = \frac{4r}{\sqrt{3}}$.
For $fcc$ structure,the number of atoms per unit cell $Z_{fcc} = 4$ and the relation between edge length $a$ and atomic radius $r$ is $4r = \sqrt{2}a$,so $a_{fcc} = \frac{4r}{\sqrt{2}}$.
The density $d$ is given by $d = \frac{Z \times M}{N_A \times a^3}$.
Taking the ratio: $\frac{d_{bcc}}{d_{fcc}} = \frac{Z_{bcc}}{Z_{fcc}} \times (\frac{a_{fcc}}{a_{bcc}})^3$.
Substituting the values: $\frac{d_{bcc}}{d_{fcc}} = \frac{2}{4} \times (\frac{4r/\sqrt{2}}{4r/\sqrt{3}})^3 = \frac{1}{2} \times (\frac{\sqrt{3}}{\sqrt{2}})^3 = \frac{1}{2} \times \frac{3\sqrt{3}}{2\sqrt{2}} = \frac{3\sqrt{3}}{4\sqrt{2}}$.

(A) The statement '$All$ form monobasic oxyacids' is incorrect because $fluorine$ does not form a series of oxyacids like other halogens due to its high electronegativity and small size. It only forms $HOF$ ($Fluoric(I)$ acid).
All halogens act as oxidizing agents.
$Chlorine$ has the highest electron-gain enthalpy among all halogens.
Except for $fluorine$,all halogens show positive oxidation states $(+1, +3, +5, +7)$ in their oxyacids.

(C) According to the Ellingham diagram,a metal can reduce the oxide of another metal if its own oxide formation curve lies below the curve of the metal oxide being reduced.
This is because the metal with the curve lower in the diagram has a more negative Gibbs free energy of formation $(-\Delta G)$ for its oxide.
Magnesium $(Mg)$ forms $MgO$,and its formation curve lies below the formation curve of alumina $(Al_2O_3)$ at temperatures below $1623 \ K$.
Therefore,$Mg$ can reduce $Al_2O_3$ to $Al$.

(C) According to the $Hardy-Schulze$ rule,the coagulating power of an ion depends on both the magnitude and the sign of the charge on the ion.
Greater the magnitude of the charge on the ion,the higher is its coagulating power for a given colloidal solution.

(D) The reaction sequence is as follows:
$1$. Compound $A$ is ethanol $(C_{2}H_{5}OH)$.
$2$. When $A$ $(C_{2}H_{5}OH)$ reacts with $Na$,it forms sodium ethoxide $(B = C_{2}H_{5}ONa)$.
$3$. When $A$ $(C_{2}H_{5}OH)$ reacts with $PCl_{5}$,it forms ethyl chloride $(C = C_{2}H_{5}Cl)$.
$4$. $B$ $(C_{2}H_{5}ONa)$ and $C$ $(C_{2}H_{5}Cl)$ react via Williamson's synthesis to produce diethyl ether $(C_{2}H_{5}-O-C_{2}H_{5})$.
Therefore,the correct order is $A = C_{2}H_{5}OH, B = C_{2}H_{5}ONa, C = C_{2}H_{5}Cl$.

(A) Step $1$: Toluene $(C_{7}H_{8})$ reacts with $3Cl_{2}$ in the presence of heat to undergo side-chain chlorination,forming benzotrichloride $(C_{6}H_{5}CCl_{3})$ as product '$A$'.
Step $2$: The $-CCl_{3}$ group is strongly electron-withdrawing and meta-directing. Therefore,electrophilic substitution with $Br_{2}/Fe$ occurs at the meta position,forming $m-$bromobenzotrichloride as product '$B$'.
Step $3$: Reduction of the $-CCl_{3}$ group with $Zn/HCl$ converts it back into a methyl group $(-CH_{3})$,yielding $m-$bromotoluene as the final product '$C$'.
Thus,the correct option is $A$.

(B) For a zero order reaction,the half-life period $(t_{1/2})$ is given by the formula: $t_{1/2} = \frac{[A]_0}{2k}$.
Here,$[A]_0$ is the initial concentration of the reactant and $k$ is the rate constant.
Since $t_{1/2} \propto [A]_0$,if the initial concentration $[A]_0$ is doubled,the half-life period $t_{1/2}$ will also be doubled.

(D) $1$. The reaction of benzene with $n$-propyl chloride in the presence of anhydrous $AlCl_3$ is a Friedel-Crafts alkylation. The primary carbocation formed $(CH_3CH_2CH_2^+)$ undergoes a $1,2$-hydride shift to form a more stable secondary carbocation $(CH_3CH^+CH_3)$.
$2$. This secondary carbocation attacks the benzene ring to form isopropylbenzene,also known as cumene $(P)$.
$3$. Cumene $(P)$ undergoes oxidation with $O_2$ to form cumene hydroperoxide,which upon treatment with dilute acid $(H_3O^+/\Delta)$ undergoes rearrangement to yield phenol $(Q)$ and acetone $(R)$.

(A) The complex $[CoCl_2(en)_2]$ contains two identical ligands $(Cl^-)$ and two bidentate ligands $(en)$.
It exists in two forms: $cis$ and $trans$.
In the $trans$ form,the two $Cl$ atoms are at $180^{\circ}$ to each other.
In the $cis$ form,the two $Cl$ atoms are at $90^{\circ}$ to each other.
Since the arrangement of ligands around the central metal atom differs,this is an example of geometrical isomerism.

(D) To exhibit $d-d$ transition and paramagnetism,the ion must have at least one unpaired electron in its $d$-orbitals.
$A$. $CrO_4^{2-}$: $Cr$ is in $+6$ oxidation state $(3d^0)$. It is diamagnetic and shows no $d-d$ transition.
$B$. $Cr_2O_7^{2-}$: $Cr$ is in $+6$ oxidation state $(3d^0)$. It is diamagnetic and shows no $d-d$ transition.
$C$. $MnO_4^{-}$: $Mn$ is in $+7$ oxidation state $(3d^0)$. It is diamagnetic and shows no $d-d$ transition.
$D$. $MnO_4^{2-}$: $Mn$ is in $+6$ oxidation state $(3d^1)$. It has one unpaired electron,making it paramagnetic,and it exhibits $d-d$ transition.

(B) The atomic number of $Ni$ is $28$. The electronic configuration of $Ni$ is $[Ar] 3d^8 4s^2$.
In the complex $[Ni(CO)_4]$,the oxidation state of $Ni$ is $0$.
$CO$ is a strong field ligand $(SFL)$,which causes the pairing of electrons in the $3d$ orbital.
The $4s$ electrons are promoted to the $3d$ orbital,and the $4s$ and $4p$ orbitals undergo $sp^3$ hybridisation.
Since all electrons are paired,the complex is diamagnetic.
The $sp^3$ hybridisation results in a tetrahedral geometry.

(B) The chemical formula of iron carbonyl is $Fe(CO)_5$.
In this complex,there is only one central metal atom $(Fe)$ present.
$A$ complex containing only one central metal atom is classified as mononuclear.
Therefore,$Fe(CO)_5$ is mononuclear.

(A) The spin-only magnetic moment $(\mu)$ is calculated using the formula $\mu = \sqrt{n(n+2)} \ B.M.$,where $n$ is the number of unpaired electrons.
$1. \ Co^{3+} (3d^6)$: $n = 4$,so $\mu = \sqrt{4(4+2)} = \sqrt{24} \ B.M.$ (Matches $iv$)
$2. \ Cr^{3+} (3d^3)$: $n = 3$,so $\mu = \sqrt{3(3+2)} = \sqrt{15} \ B.M.$ (Matches $v$)
$3. \ Fe^{3+} (3d^5)$: $n = 5$,so $\mu = \sqrt{5(5+2)} = \sqrt{35} \ B.M.$ (Matches $ii$)
$4. \ Ni^{2+} (3d^8)$: $n = 2$,so $\mu = \sqrt{2(2+2)} = \sqrt{8} \ B.M.$ (Matches $i$)
Therefore,the correct sequence is $(a)-(iv), (b)-(v), (c)-(ii), (d)-(i)$.