AIIMS 2018 Chemistry Question Paper with Answer and Solution

(A) For all three processes,the initial state $A$ and final state $B$ are the same.
Since internal energy $U$ is a state function,the change in internal energy $\Delta U = U_B - U_A$ is the same for all three paths.
According to the First Law of Thermodynamics,$\Delta Q = \Delta U + \Delta W$,where $\Delta W$ is the work done by the gas.
In a $P-V$ diagram,the work done $\Delta W$ is equal to the area under the curve with the volume axis.
From the figure,the area under the curves is such that $(Area)_1 < (Area)_2 < (Area)_3$.
Since $\Delta U$ is constant,$\Delta Q$ is directly proportional to $\Delta W$.
Therefore,${Q_1} < {Q_2} < {Q_3}$.

(B) In a unit cell,the number of $W$ atoms at the corners $= \frac{1}{8} \times 8 = 1$.
The number of $O$ atoms at the centre of edges $= \frac{1}{4} \times 12 = 3$.
The number of $Na$ atoms at the centre of the cube $= 1$.
Thus,the ratio of atoms is $Na:W:O = 1:1:3$.
Therefore,the formula for the compound is $NaWO_3$.

(D) The circuit is an $LR$ series circuit connected to a $DC$ source.
The time constant of the circuit is $\tau = \frac{L}{R} = \frac{10 \, H}{5 \, \Omega} = 2 \, s$.
The current in an $LR$ circuit at time $t$ is given by $I(t) = I_0(1 - e^{-t/\tau})$,where $I_0 = \frac{V}{R} = \frac{5 \, V}{5 \, \Omega} = 1 \, A$.
Given $t = 2 \, s$,we have $I(2) = 1 \times (1 - e^{-2/2}) = 1 - e^{-1} \, A$.

(A) The current in an $LR$ circuit during growth is given by the formula: $I(t) = I_0(1 - e^{-\frac{R}{L}t})$
Here,$I_0 = \frac{E}{R}$ is the maximum steady-state current.
Given values are: $L = 10\,H$,$R = 5\,\Omega$,$E = 5\,V$,and $t = 2\,s$.
First,calculate $I_0 = \frac{5\,V}{5\,\Omega} = 1\,A$.
Now,substitute the values into the growth equation:
$I(2) = 1 \times (1 - e^{-\frac{5}{10} \times 2})$
$I(2) = 1 - e^{-1} \,A$.

(D) Carbohydrates,fats,and proteins can all be used as respiratory substrates in cellular respiration.
During aerobic respiration,these molecules are broken down through various pathways and eventually converge at the formation of Acetyl $CoA$ before entering the Krebs' cycle ($TCA$ cycle).
Therefore,Acetyl $CoA$ acts as a common intermediate for the breakdown of all three major biomolecules.

(B) $XO$ type of sex chromosomes determine male sex in grasshoppers. This type of sex-determination comes under $XX-XO$ type.
It is common in cockroaches,grasshoppers,and bugs.
The female has two homomorphic sex chromosomes $XX$ and is homogametic.
It produces similar eggs,each with one $X$ chromosome.
The male has only one $X$ chromosome and is heterogametic.
It produces two types of sperms: gynosperms with $X$ and androsperms without $X$.
Other statements can be corrected as follows:
In birds,the females are heterogametic and represented as $ZW$.
Female sex in humans is represented as $XX$.
In Drosophila,males are heterogametic,i.e.,$XY$.

(B) In the grafting technique,parts of two plants are joined to form a composite plant.
The plant providing the root system is called the $Stock$,while the plant providing the shoot system (which bears flowers and fruits) is called the $Scion$.
Since the $Scion$ contains the genetic material that dictates the characteristics of the shoot,branches,flowers,and fruits,the quality of the fruits produced is determined by the genotype of the $Scion$.

(B) The circuit consists of parallel branches.
In the $1^{st}$ branch,there is $1$ capacitor of $1 \mu F$. Its equivalent capacitance is $C_1 = 1 \mu F$.
In the $2^{nd}$ branch,there are $2$ capacitors in series,each of $1 \mu F$. Its equivalent capacitance is $C_2 = \frac{1 \mu F}{2} = 0.5 \mu F$.
In the $3^{rd}$ branch,there are $4$ capacitors in series,each of $1 \mu F$. Its equivalent capacitance is $C_3 = \frac{1 \mu F}{4} = 0.25 \mu F$.
In the $4^{th}$ branch,there are $8$ capacitors in series,each of $1 \mu F$. Its equivalent capacitance is $C_4 = \frac{1 \mu F}{8} = 0.125 \mu F$.
Since all these branches are connected in parallel,the total equivalent capacitance $C_{eq}$ is the sum of the capacitances of each branch:
$C_{eq} = C_1 + C_2 + C_3 + C_4 + \dots$
$C_{eq} = 1 + \frac{1}{2} + \frac{1}{4} + \frac{1}{8} + \dots$
This is an infinite geometric progression with first term $a = 1$ and common ratio $r = \frac{1}{2}$.
The sum of an infinite geometric series is given by $S = \frac{a}{1 - r}$.
$C_{eq} = \frac{1}{1 - 1/2} = \frac{1}{1/2} = 2 \mu F$.

(D) The growth of current in an $L-R$ circuit is given by the formula:
$I = I_{0}(1 - e^{-t/\tau})$
where $I_{0}$ is the steady-state current and $\tau$ is the time constant.
First,calculate the steady-state current $I_{0}$:
$I_{0} = \frac{E}{R} = \frac{5\, V}{5\, \Omega} = 1\, A$
Next,calculate the time constant $\tau$:
$\tau = \frac{L}{R} = \frac{10\, H}{5\, \Omega} = 2\, s$
Given the time $t = 2\, s$,substitute the values into the growth equation:
$I = 1 \cdot (1 - e^{-2/2})$
$I = (1 - e^{-1})\, A$

(C) For free expansion,the external pressure $P_{ext} = 0$,therefore work done $W = -P_{ext} \Delta V = 0$.
For an adiabatic process,the heat exchange $q = 0$.
According to the first law of thermodynamics,$\Delta U = q + W$.
Since $q = 0$ and $W = 0$,the change in internal energy $\Delta U = 0$.
For an ideal gas,internal energy is a function of temperature only $(U = f(T))$,so $\Delta U = 0$ implies $\Delta T = 0$.
Thus,for free expansion of an ideal gas under adiabatic conditions,$q=0, \Delta T=0, W=0$.

(C) The work function $\Phi = 4.2 \, eV = 4.2 \times 1.602 \times 10^{-19} \, J \approx 6.73 \times 10^{-19} \, J$.
The energy of the incident radiation $E = \frac{hc}{\lambda} = \frac{6.626 \times 10^{-34} \times 3 \times 10^8}{2000 \times 10^{-10}} \, J = 9.939 \times 10^{-19} \, J$.
According to Einstein's photoelectric equation,$K_{max} = E - \Phi$.
$K_{max} = (9.939 - 6.73) \times 10^{-19} \, J = 3.209 \times 10^{-19} \, J$.
Thus,the kinetic energy is approximately $3.2 \times 10^{-19} \, J$.

(B) Given,$[N_2] = 3.0 \times 10^{-3} \ M$,$[O_2] = 4.2 \times 10^{-3} \ M$,and $[NO] = 2.8 \times 10^{-3} \ M$.
For the reaction $N_{2(g)} + O_{2(g)} \rightleftharpoons 2NO_{(g)}$,the equilibrium constant $K_C$ is given by:
$K_C = \frac{[NO]^2}{[N_2][O_2]}$
Substituting the values:
$K_C = \frac{(2.8 \times 10^{-3})^2}{(3.0 \times 10^{-3})(4.2 \times 10^{-3})} = \frac{7.84 \times 10^{-6}}{12.6 \times 10^{-6}} \approx 0.622$
The relationship between $K_p$ and $K_C$ is $K_p = K_C(RT)^{\Delta n}$.
Here,$\Delta n = (\text{moles of gaseous products}) - (\text{moles of gaseous reactants}) = 2 - (1 + 1) = 0$.
Since $\Delta n = 0$,$K_p = K_C(RT)^0 = K_C$.
Therefore,$K_p = 0.622$.

(C) $C^{4-}$,$N^{3-}$,and $O^{2-}$ are isoelectronic species,all having $10$ electrons.
The ionic radius of isoelectronic species decreases as the nuclear charge (atomic number) increases.
The atomic numbers are: $C (6)$,$N (7)$,and $O (8)$.
Therefore,the order of ionic radii is: $C^{4-} > N^{3-} > O^{2-}$.
Given the values $2.60 \ \mathring{A}$ for $C^{4-}$ and $1.40 \ \mathring{A}$ for $O^{2-}$,the value for $N^{3-}$ must lie between $1.40$ and $2.60$.
Among the given options,$1.71 \ \mathring{A}$ is the correct value.

(A) The stability of the given carbocations is determined by electronic effects:
$I.$ $CH_3-CH^{+}-CH_3$: The carbocation is stabilized by the $+I$ effect of two methyl groups.
$II.$ $CH_3-CH^{+}-OCH_3$: The carbocation is strongly stabilized by the $+R$ (resonance) effect of the $-OCH_3$ group,which is more effective than the $+I$ effect.
$III.$ $CH_3-CH^{+}-CH_2-OCH_3$: The $-OCH_3$ group is separated by a methylene group,so its $-I$ effect destabilizes the carbocation.
Comparing these,the resonance effect in $II$ provides maximum stability,followed by the inductive effect in $I$,while the $-I$ effect in $III$ makes it the least stable.
Therefore,the correct order of decreasing stability is $II > I > III$.

(D) For any chemical reaction,the relationship between enthalpy change $(\Delta H)$ and internal energy change $(\Delta E)$ is given by:
$\Delta H = \Delta E + \Delta n_{g}RT$
Where $\Delta n_{g}$ is the change in the number of moles of gaseous species.
$\Delta n_{g} = (\text{moles of gaseous products}) - (\text{moles of gaseous reactants})$
For the reaction: $Fe_{2}O_{3(s)} + 3H_{2(g)} \rightarrow 2Fe_{(s)} + 3H_{2}O_{(l)}$
Note that $Fe_{2}O_{3}$ and $Fe$ are solids,and $H_{2}O$ is a liquid. Only $H_{2}$ is in the gaseous state.
$\Delta n_{g} = 0 - 3 = -3$
Substituting this value into the equation:
$\Delta H = \Delta E + (-3)RT$
$\Delta H = \Delta E - 3RT$

(D) To determine the $IUPAC$ name,we identify the principal functional group,which is the ketone group $(-one)$.
We number the ring starting from the ketone carbon as $C-1$.
We then number the ring to give the lowest possible locants to the double bond and the hydroxyl group.
Starting from the ketone carbon $(C-1)$,we proceed towards the double bond to give it the lowest locant ($C-3$ to $C-4$ double bond).
Thus,the hydroxyl group is at position $5$.
The correct $IUPAC$ name is $5-$hydroxycyclohex$-3-$en$-1-$one.

(C) $AX_{2}$ is ionized as follows:
$AX_{2} \rightleftharpoons A^{2+} + 2X^{-}$
Solubility product of $AX_{2}$ is given by:
$K_{sp} = [A^{2+}][X^{-}]^{2} = (S) \times (2S)^{2} = 4S^{3}$
Given $K_{sp} = 3.2 \times 10^{-11}$
$4S^{3} = 3.2 \times 10^{-11}$
$S^{3} = 0.8 \times 10^{-11} = 8 \times 10^{-12}$
$S = \sqrt[3]{8 \times 10^{-12}} = 2 \times 10^{-4} \, mol / L$
Thus,the solubility is $2 \times 10^{-4} \, mol / L$.

(D) The central iodine atom $(I)$ in $IF_{7}$ has $7$ valence electrons. It forms $7$ bonds with $7$ fluorine atoms.
The steric number is calculated as:
$\text{Steric Number} = \frac{1}{2} \times (V + M - C + A) = \frac{1}{2} \times (7 + 7) = 7$.
$A$ steric number of $7$ corresponds to $sp^{3}d^{3}$ hybridization,which results in a pentagonal bipyramidal geometry.
In this structure,five fluorine atoms lie in a pentagonal plane with $I-F$ bond angles of $72^{\circ}$,and two fluorine atoms are positioned above and below the plane at $90^{\circ}$ to the equatorial plane.

(A) Apical dominance is a phenomenon in plants where the main central stem (apical bud) grows more strongly than other side stems.
This process is primarily regulated by the plant hormone $Auxin$.
$Auxin$ is synthesized at the shoot apex and transported downwards,where it inhibits the growth of lateral buds,thereby maintaining the dominance of the apical bud.

(A) The $Bt$ toxin is produced by specific genes known as $Cry$ genes.
These proteins are highly specific to certain insect pests.
The gene $Cry$ $I$ Ab controls the corn borer.
The gene $Cry$ $II$ Ab controls the cotton bollworms.
Therefore,$Cry$ $II$ Ab and $Cry$ $I$ Ab control cotton bollworms and corn borer respectively.

(B) In grasshoppers,the sex determination mechanism is of the $XO$ type.
In this system,males possess only one $X$ chromosome along with autosomes $(XO)$,while females possess a pair of $X$ chromosomes $(XX)$.
Therefore,the presence of a single $X$ chromosome determines the male sex in grasshoppers.
Option $A$ is incorrect because $ZZ$ chromosomes in birds determine the male sex (homogametic).
Option $C$ is incorrect because Klinefelter's syndrome is $XXY$,not $XO$,and $XO$ in humans results in Turner's syndrome (female).
Option $D$ is incorrect because in Drosophila,$XX$ chromosomes produce females,while $XY$ produces males.

(A) The Wurtz-Fittig reaction involves the reaction of an aryl halide with an alkyl halide in the presence of sodium metal to form an alkyl-substituted aromatic compound.
Specifically,$C_{6}H_{5}I + CH_{3}I + 2Na \rightarrow C_{6}H_{5}CH_{3} + 2NaI$ represents the Wurtz-Fittig reaction.
Option $A$ is the correct representation.

(A) The ease of nucleophilic substitution depends upon the nature of the leaving group. When the leaving tendency of a group in a compound is high,the compound is more reactive towards nucleophilic substitution.
The nucleophilic acyl substitution proceeds in two steps as shown in the mechanism:
$1$. Nucleophilic attack on the carbonyl carbon to form a tetrahedral intermediate.
$2$. Elimination of the leaving group $(A^-)$ to regenerate the carbonyl group.
The order of leaving tendency is $Cl^{-} > RCOO^{-} > RO^{-} > NH_{2}^{-}$.
Therefore,the order of reactivity of acyl compounds is:
$RCOCl$ (Acyl chloride) $ > (RCO)_2O$ (Acid anhydride) $ > RCOOR$ (Ester) $ > RCONH_2$ (Amide).

(D) The chemicals which are used to protect food from microbial action,i.e.,which arrest the process of fermentation,acidification,and any other decomposition of food,are known as food preservatives.
Table salt,sugar,vegetable oil,vinegar,sodium benzoate $(C_6H_5COONa)$,sodium metabisulphite $(Na_2S_2O_5)$,and vitamin $E$ are common examples of food preservatives.

(A) For a first order reaction,the integrated rate law is given by $k = \frac{2.303}{t} \log \frac{[A]_0}{[A]_t}$.
At half-life,$t = t_{1/2}$ and $[A]_t = \frac{[A]_0}{2}$.
Substituting these values into the equation:
$k = \frac{2.303}{t_{1/2}} \log \frac{[A]_0}{[A]_0 / 2} = \frac{2.303}{t_{1/2}} \log 2$.
Since $\log 2 \approx 0.3010$,we get $k = \frac{2.303 \times 0.3010}{t_{1/2}} = \frac{0.693}{t_{1/2}}$.
Therefore,$t_{1/2} = \frac{0.693}{k}$.
This expression shows that the half-life of a first order reaction is independent of the initial concentration of the reactant.

(D) The complex $\left[ Co(CN)_6 \right]^{3-}$ contains $Co^{3+}$ as the central metal ion.
The atomic number of $Co$ is $27$. The electronic configuration of $Co$ is $[Ar] 3d^7 4s^2$.
For $Co^{3+}$,the electronic configuration is $[Ar] 3d^6$.
$CN^-$ is a strong field ligand,which causes the pairing of electrons in the $d$-orbitals.
In an octahedral field,the $d$-orbitals split into $t_{2g}$ and $e_g$ sets.
Since $CN^-$ is a strong field ligand,all $6$ electrons of $Co^{3+}$ $(3d^6)$ will occupy the $t_{2g}$ orbitals,resulting in the configuration $t_{2g}^6 e_g^0$.
Therefore,the correct option is $D$.

(C) The reaction sequence is as follows:
$R-Br + Mg \xrightarrow{\text{Dry ether}} R-MgBr (Y)$
$R-MgBr + HCHO \xrightarrow{\text{Hydrolysis}} R-CH_{2}OH$
Given the product is $C_{4}H_{10}O$,which is a primary alcohol $(R-CH_{2}OH)$,we have $R+CH_{2} = C_{3}H_{7}$.
Thus,$R$ is a propyl group $(C_{3}H_{7}-)$.
Therefore,$X$ is $1-bromopropane$ $(CH_{3}CH_{2}CH_{2}Br)$:
$CH_{3}CH_{2}CH_{2}Br + Mg \xrightarrow{\text{Dry ether}} CH_{3}CH_{2}CH_{2}MgBr$
$CH_{3}CH_{2}CH_{2}MgBr + HCHO \rightarrow CH_{3}CH_{2}CH_{2}CH_{2}OMgBr$
$CH_{3}CH_{2}CH_{2}CH_{2}OMgBr + H_{2}O \rightarrow CH_{3}CH_{2}CH_{2}CH_{2}OH + Mg(OH)Br$

(A) The coordination number of the $Mn^{2+}$ ion in the complex ion $[MnBr_{4}]^{2-}$ is $4$.
This implies it can have either tetrahedral ($sp^{3}$ hybridization) or square planar ($dsp^{2}$ hybridization) geometry.
The given spin-only magnetic moment is $5.9 \ BM$.
Using the formula $\mu = \sqrt{n(n+2)}$,where $n$ is the number of unpaired electrons:
$5.9 \approx \sqrt{n(n+2)} \implies n \approx 5$.
Since $Mn^{2+}$ has a $d^{5}$ configuration,having $5$ unpaired electrons indicates a high-spin tetrahedral complex where the $d$-orbitals are not forced to pair up.
Therefore,the geometry is tetrahedral.

(C) Diamond is a giant molecule in which constituent atoms are held together by covalent bonds. Hence,it is a network solid.
$SO_{2}$ (solid),$H_{2}O$ (ice),and $I_{2}$ are examples of molecular solids.

(A) As the size of the halogen atom increases from $F$ to $I$,the $H-X$ bond length in halogen acids also increases from $HF$ to $HI$.
Bond length order: $HF < HCl < HBr < HI$.
Due to the increase in bond length,the strength of the $H-X$ bond decreases from $HF$ to $HI$.
Consequently,the bond dissociation enthalpy decreases from $HF$ to $HI$.
The order of $H-X$ bond dissociation enthalpy $(kJ/mol)$ is: $HF (574.0) > HCl (428.1) > HBr (362.5) > HI (294)$.
Therefore,$HF$ has the highest bond dissociation enthalpy.

(C) Let the oxidation state of $Pt$ be $x$ in $[Pt(Py)_{4}][PtCl_{4}]$.
For the cationic part $[Pt(Py)_{4}]^{2+}$,$x + 4(0) = +2$,so $x = +2$.
For the anionic part $[PtCl_{4}]^{2-}$,$x + 4(-1) = -2$,so $x = +2$.
Since the oxidation state of $Pt$ is $+2$ in both parts,option $C$ is correct.

(D) The given Arrhenius equation is $\ln k = \ln A - \frac{E_a}{RT}$.
Comparing this with the given equation $\ln k = 14.34 - \frac{1.25 \times 10^4 \ K}{T}$,we get $\frac{E_a}{R} = 1.25 \times 10^4 \ K$.
Given the gas constant $R \approx 2 \ cal \ K^{-1} \ mol^{-1}$,the activation energy $E_a$ is calculated as:
$E_a = (1.25 \times 10^4 \ K) \times (2 \ cal \ K^{-1} \ mol^{-1}) = 2.50 \times 10^4 \ cal \ mol^{-1}$.

(C) The correct increasing order of basic strength for group $15$ hydrides is: $NH_3 > PH_3 > AsH_3 > SbH_3$.
$NH_3$ is the most basic because of its small size,where the electron density of the lone pair is concentrated over a small region. As the size of the central atom increases,the electron density gets diffused over a larger surface area,decreasing the ability to donate the electron pair.
Thus,the arrangement in option $C$ is incorrect.
Option $A$ is correct as oxidising power increases down the group due to the inert pair effect.
Option $B$ is correct as acidic strength increases with the increase in bond length down the group.
Option $D$ is correct based on the electronic configuration and stability of half-filled $p$-orbitals in $N$.

(C) The net cell reaction is $Zn_{(s)} + Ag_{2}O_{(s)} \rightarrow ZnO_{(s)} + 2Ag_{(s)}$.
The standard Gibbs free energy change for the reaction is given by $\Delta G^{o} = \Delta G_{f}^{o}(ZnO) - \Delta G_{f}^{o}(Ag_{2}O)$.
Substituting the values: $\Delta G^{o} = -318.3 - (-11.21) = -307.09 \ kJ \ mol^{-1} = -307.09 \times 10^{3} \ J \ mol^{-1}$.
Using the relation $\Delta G^{o} = -n F E^{o}_{cell}$,where $n = 2$ (number of electrons transferred) and $F = 96500 \ C \ mol^{-1}$.
$-307.09 \times 10^{3} = -2 \times 96500 \times E^{o}_{cell}$.
$E^{o}_{cell} = \frac{307.09 \times 10^{3}}{2 \times 96500} = 1.591 \ V$.

(D) The structure of hypophosphoric acid $(H_4P_2O_6)$ is $(HO)_2P(=O)-P(=O)(OH)_2$,which contains a direct $P-P$ bond and no $P-O-P$ linkage.
In contrast,pyrophosphorus acid $(H_4P_2O_5)$,diphosphoric acid $(H_4P_2O_7)$,and isohypophosphoric acid ($H_4P_2O_6$ isomer) all contain a $P-O-P$ linkage.
Therefore,the correct option is $(D)$.

(B) Given:
Density $(\rho) = 8.55 \ g \ cm^{-3}$
Atomic mass $(M) = 93 \ g \ mol^{-1}$
For $bcc$ structure, number of atoms per unit cell $(Z) = 2$.
Avogadro's number $(N_A) = 6.022 \times 10^{23} \ mol^{-1}$.
Using the formula: $\rho = \frac{Z \times M}{a^3 \times N_A}$
$8.55 = \frac{2 \times 93}{a^3 \times 6.022 \times 10^{23}}$
$a^3 = \frac{186}{8.55 \times 6.022 \times 10^{23}} = 3.614 \times 10^{-23} \ cm^3$
$a = (3.614 \times 10^{-23})^{1/3} \ cm = 3.306 \times 10^{-8} \ cm = 330.6 \ pm$
For $bcc$ structure, the relation between radius $(r)$ and edge length $(a)$ is: $r = \frac{\sqrt{3}}{4} a$
$r = \frac{1.732}{4} \times 330.6 \ pm = 0.433 \times 330.6 \ pm \approx 143 \ pm$.

(B) The oxidation state of the central metal atom $Pt$ is calculated as follows:
Let the oxidation state of $Pt$ be $x$.
The sum of oxidation states of all ligands and the central metal equals the charge on the complex ion.
$(x) + 3(0) + (-1) + (-1) + (-1) = +1$
$x - 3 = +1$
$x = +4$
According to $IUPAC$ rules,ligands are named in alphabetical order: $ammine$ $(NH_3)$,$bromido$ $(Br^-)$,$chlorido$ $(Cl^-)$,and $nitro$ $(NO_2^-)$.
Thus,the name is triamminebromidochloridonitroplatinum $(IV)$ chloride.

(D) When $KI$ is added in excess to $AgNO_{3}$,the $AgI$ particles preferentially adsorb $I^{-}$ ions from the dispersion medium to form a negatively charged colloidal sol,represented as $AgI/I^{-}$.
This primary layer of $I^{-}$ ions attracts counter ions $(K^{+})$ from the medium to form the Helmholtz double layer,which is represented as $AgI/I^{-}:K^{+}$.

(A) The reaction of ammonia with chlorine depends on the relative amounts of the reactants.
When ammonia is in excess,the reaction is: $8 NH_{3} + 3 Cl_{2} \rightarrow 6 NH_{4}Cl + N_{2}$. Thus,$X = 6 NH_{4}Cl + N_{2}$.
When chlorine is in excess,the reaction is: $NH_{3} + 3 Cl_{2} \rightarrow NCl_{3} + 3 HCl$. Thus,$Y = NCl_{3} + 3 HCl$.

(B) The products of the reaction of copper with $HNO_{3}$ depend upon the concentration of $HNO_{3}$ used.
Copper metal reacts with dilute $HNO_{3}$ to form nitrogen $(II)$ oxide $(NO)$:
$3Cu + 8HNO_{3} \text{ (dil.)} \rightarrow 3Cu(NO_{3})_{2} + 2NO + 4H_{2}O$
Copper metal reacts with concentrated $HNO_{3}$ to form nitrogen $(IV)$ oxide or nitrogen dioxide $(NO_{2})$:
$Cu + 4HNO_{3} \text{ (conc.)} \rightarrow Cu(NO_{3})_{2} + 2NO_{2} + 2H_{2}O$
Thus,depending on the concentration,both $NO$ and $NO_{2}$ can be obtained.

(D) $Helium$ is used in filling balloons for meteorological observations because it is light and non-inflammable.
It is also used in gas-cooled nuclear reactors as a coolant.
Liquid $Helium$ is used as a cryogenic agent for carrying out various experiments at low temperature.

(B) In a unit cell:
Number of $W$ atoms at the corners $= 8 \times \frac{1}{8} = 1$
Number of $O$ atoms at the centre of edges $= 12 \times \frac{1}{4} = 3$
Number of $Na$ atoms at the centre of the cube $= 1 \times 1 = 1$
Therefore,the ratio of atoms is $Na:W:O = 1:1:3$.
Hence,the formula of the compound is $NaWO_{3}$.

(C) The dimerisation reaction is: $2 C_{6}H_{5}COOH \rightleftharpoons (C_{6}H_{5}COOH)_{2}$
Initially,we have $1$ mole of benzoic acid.
After association,the moles are:
Benzoic acid: $(1-x)$
Dimer: $x/2$
Total moles after association $= (1-x) + (x/2) = 1 - x/2$
The van't Hoff factor $(i)$ is defined as the ratio of total moles after association to the initial moles:
$i = \frac{1 - x/2}{1} = 1 - \frac{x}{2}$

(A) The reaction of methylenecyclohexane with $H_3O^+$ proceeds via acid-catalyzed hydration,which follows Markownikoff's rule. The carbocation formed at the tertiary carbon is more stable,leading to the formation of $1$-methylcyclohexanol $(Y)$.
The reaction with $(i) B_2H_6/THF$ and $(ii) H_2O_2/OH^-$ is hydroboration-oxidation,which follows anti-Markownikoff's rule. The hydroxyl group adds to the less substituted carbon,resulting in the formation of cyclohexylmethanol $(X)$.

(C) According to Kohlrausch's law of independent migration of ions,the molar conductivity of an electrolyte at infinite dilution is the sum of the molar conductivities of its constituent ions.
$\lambda_{m}^{o} (Ba(OH)_{2}) = \lambda_{m}^{o} (Ba^{2+}) + 2\lambda_{m}^{o} (OH^{-})$
Given values:
$\lambda_{m}^{o} (NaOH) = \lambda_{m}^{o} (Na^{+}) + \lambda_{m}^{o} (OH^{-}) = 248 \times 10^{-4} \ S \ m^{2} \ mol^{-1}$
$\lambda_{m}^{o} (NaCl) = \lambda_{m}^{o} (Na^{+}) + \lambda_{m}^{o} (Cl^{-}) = 126 \times 10^{-4} \ S \ m^{2} \ mol^{-1}$
$\lambda_{m}^{o} (BaCl_{2}) = \lambda_{m}^{o} (Ba^{2+}) + 2\lambda_{m}^{o} (Cl^{-}) = 280 \times 10^{-4} \ S \ m^{2} \ mol^{-1}$
To obtain $\lambda_{m}^{o} (Ba(OH)_{2})$,we perform the operation:
$\lambda_{m}^{o} (Ba(OH)_{2}) = \lambda_{m}^{o} (BaCl_{2}) + 2\lambda_{m}^{o} (NaOH) - 2\lambda_{m}^{o} (NaCl)$
Substituting the values:
$\lambda_{m}^{o} (Ba(OH)_{2}) = 280 \times 10^{-4} + 2(248 \times 10^{-4}) - 2(126 \times 10^{-4})$
$= (280 + 496 - 252) \times 10^{-4} \ S \ m^{2} \ mol^{-1}$
$= 524 \times 10^{-4} \ S \ m^{2} \ mol^{-1}$

(B) The given reaction sequence represents the synthesis of $Nylon-6$.
Cyclohexanone oxime undergoes a Beckmann rearrangement in the presence of $H_2SO_4$ to form $Caprolactam$ $(X)$.
$Caprolactam$ is then heated at $540 \ K$ to undergo ring-opening polymerization to form $Nylon-6$.
Thus,$X$ is $Caprolactam$.

(C) The spin-only magnetic moment $(\mu)$ depends on the number of unpaired electrons $(n)$ and is given by the formula: $\mu = \sqrt{n(n+2)} \ BM$.
The electronic configurations and number of unpaired electrons for the ions are:
$1$. For $Mn^{2+}$ $(Z=25)$: $[Ar] 3d^5$. Number of unpaired electrons $(n)$ = $5$. $\mu = \sqrt{5(5+2)} = \sqrt{35} \ BM$.
$2$. For $Cr^{2+}$ $(Z=24)$: $[Ar] 3d^4$. Number of unpaired electrons $(n)$ = $4$. $\mu = \sqrt{4(4+2)} = \sqrt{24} \ BM$.
$3$. For $Ti^{2+}$ $(Z=22)$: $[Ar] 3d^2$. Number of unpaired electrons $(n)$ = $2$. $\mu = \sqrt{2(2+2)} = \sqrt{8} \ BM$.
Comparing the values,the correct order of spin-only magnetic moment is $Mn^{2+} > Cr^{2+} > Ti^{2+}$.

(A) Aldol condensation requires the presence of at least one $\alpha$-hydrogen atom in the aldehyde or ketone.
In $CH_3-C(CH_3)_2-CHO$ ($2,2$-dimethylpropanal),the $\alpha$-carbon atom is bonded to three methyl groups and has no hydrogen atoms attached to it.
Therefore,it does not undergo aldol condensation.

(B) The greenish-yellow gas is $Cl_{2}$.
When $Cl_{2}$ reacts with a hot concentrated alkali metal hydroxide like $KOH$,it forms a chlorate (a type of halate),specifically $KClO_{3}$.
$KClO_{3}$ is widely used in the manufacturing of fireworks and safety matches.
The balanced chemical equation is:
$3Cl_{2} + 6KOH \rightarrow KClO_{3} + 5KCl + 3H_{2}O$

(A) On hydrolysis with dilute aqueous sulphuric acid,sucrose undergoes cleavage of the glycosidic bond to form an equimolar mixture of $D-(+)$-glucose and $D-(-)$-fructose.
The chemical reaction is:
$C_{12}H_{22}O_{11} + H_2O$ $\xrightarrow{H_2SO_4} C_6H_{12}O_6 (D-(+)\text{-glucose}) + C_6H_{12}O_6 (D-(-)\text{-fructose})$
Sucrose is dextrorotatory (specific rotation $= +66.1^{\circ}$),but the resulting mixture is levorotatory (specific rotation $= -20.0^{\circ}$) because the levorotation of fructose is greater than the dextrorotation of glucose. This process is known as the inversion of cane sugar.

(A) $[Cr(H_2O)_4Cl_2]^+$ shows geometrical isomerism because it is a $MA_4B_2$ type coordination compound which contains two sets of equivalent ligands,four $H_2O$ and two $Cl^-$.
Hence,the possible geometrical isomers are as shown in the figure.
Thus,the correct option is $(A)$.

(A) In the dehydrohalogenation of $2$-bromobutane with alcoholic $KOH$,the major product is the more substituted alkene $(but-2-ene)$ according to Saytzeff's rule.
Saytzeff's rule states that in dehydrohalogenation reactions,the alkene with the greater number of alkyl groups attached to the doubly bonded carbon atoms is the major product.