NEET 2019 Chemistry Question Paper with Answer and Solution

(A) An amphoteric substance is one that can react with both acids and bases.
$Be(OH)_2$ is amphoteric in nature because it reacts with both acids and bases to form salts and water.
For example:
$Be(OH)_2 + 2HCl \rightarrow BeCl_2 + 2H_2O$ (acts as a base)
$Be(OH)_2 + 2NaOH \rightarrow Na_2[Be(OH)_4]$ (acts as an acid)
Other hydroxides like $Mg(OH)_2$,$Ca(OH)_2$,and $Sr(OH)_2$ are basic in nature.

(A) The work done during expansion against a constant external pressure is given by the formula:
$W = -P_{ext} \times \Delta V$
Given:
$P_{ext} = 2 \; bar$
$V_{1} = 0.1 \; L$
$V_{2} = 0.25 \; L$
Change in volume:
$\Delta V = V_{2} - V_{1} = 0.25 \; L - 0.1 \; L = 0.15 \; L$
Calculating work done:
$W = -2 \; bar \times 0.15 \; L = -0.30 \; bar \cdot L$
Since $1 \; L \cdot bar = 100 \; J$:
$W = -0.30 \times 100 \; J = -30 \; J$
The work done by the gas is $-30 \; J$.

(A) The dissociation of $Ca(OH)_{2}$ is given by: $Ca(OH)_{2(s)} \rightleftharpoons Ca^{2+}_{(aq)} + 2OH^{-}_{(aq)}$
Given $pH = 9$,we know $pOH = 14 - pH = 14 - 9 = 5$.
Therefore,$[OH^{-}] = 10^{-pOH} = 10^{-5} \ M$.
From the stoichiometry,$[OH^{-}] = 2S$,where $S$ is the solubility of $Ca(OH)_{2}$.
So,$2S = 10^{-5} \implies S = 0.5 \times 10^{-5} \ M$.
The solubility product expression is $K_{sp} = [Ca^{2+}][OH^{-}]^{2} = (S)(2S)^{2} = 4S^{3}$.
Substituting the value of $S$: $K_{sp} = 4 \times (0.5 \times 10^{-5})^{3} = 4 \times 0.125 \times 10^{-15} = 0.5 \times 10^{-15}$.

(C) The balanced chemical equation for Haber's process is:
$N_{2(g)} + 3H_{2(g)} \rightarrow 2NH_{3(g)}$
From the stoichiometry of the reaction,$2 \ \text{moles}$ of $NH_{3(g)}$ are produced from $3 \ \text{moles}$ of $H_{2(g)}$.
Therefore,to produce $20 \ \text{moles}$ of $NH_{3(g)}$,the required moles of $H_{2(g)}$ are:
$= \frac{3 \ \text{mol } H_2}{2 \ \text{mol } NH_3} \times 20 \ \text{mol } NH_3 = 30 \ \text{moles of } H_{2(g)}$.

(D) Greenhouse gases are gases that trap heat in the atmosphere. The primary greenhouse gases include carbon dioxide $(CO_2)$,methane $(CH_4)$,nitrous oxide $(N_2O)$,water vapour,chlorofluorocarbons $(CFCs)$,and ozone $(O_3)$.
Sulphur dioxide $(SO_2)$ is primarily responsible for acid rain and is not considered a greenhouse gas.

(A) The structural formula of pent$-2-$en$-4-$yne is $CH_3-CH=CH-C \equiv CH$.
Number of $\sigma$ bonds:
$C-H$ bonds = $6$
$C-C$ bonds = $4$
Total $\sigma$ bonds = $6 + 4 = 10$.
Number of $\pi$ bonds:
$1$ (from $C=C$) + $2$ (from $C \equiv C$) = $3$.

(C) According to $M.O.T.$,the electronic configuration of the $C_2$ molecule is:
$\sigma 1s^2 < \sigma^* 1s^2 < \sigma 2s^2 < \sigma^* 2s^2 < \pi 2p_x^2 = \pi 2p_y^2$
In the $C_2$ molecule,the bond order is $2$,and both bonds are $\pi$ bonds due to the presence of electrons only in the $\pi 2p$ orbitals.

(A) disproportionation reaction is a redox reaction in which the same element is simultaneously oxidized and reduced.
$(a)$ $2 Cu^{+} \rightarrow Cu^{2+} + Cu^{0}$
Here,$Cu^{+}$ (oxidation state $+1$) is oxidized to $Cu^{2+}$ $(+2)$ and reduced to $Cu^{0}$ $(0)$. This is a disproportionation reaction.
$(b)$ $3 MnO_{4}^{2-} + 4 H^{+} \rightarrow 2 MnO_{4}^{-} + MnO_{2} + 2 H_{2}O$
Here,$Mn$ in $MnO_{4}^{2-}$ (oxidation state $+6$) is oxidized to $MnO_{4}^{-}$ $(+7)$ and reduced to $MnO_{2}$ $(+4)$. This is a disproportionation reaction.
$(c)$ $2 KMnO_{4} \xrightarrow{\Delta} K_{2}MnO_{4} + MnO_{2} + O_{2}$
This is a decomposition reaction,not a disproportionation reaction.
$(d)$ $2 MnO_{4}^{-} + 3 Mn^{2+} + 2 H_{2}O \rightarrow 5 MnO_{2} + 4 H^{+}$
This is a comproportionation reaction where $Mn(+7)$ and $Mn(+2)$ combine to form $Mn(+4)$.
Therefore,only $(a)$ and $(b)$ are disproportionation reactions.

(C) The conjugate base of a Bronsted acid is formed by the removal of a proton $(H^{+})$ from the acid.
For $H_{2}O$: $H_{2}O - H^{+} \rightarrow OH^{-}$.
For $HF$: $HF - H^{+} \rightarrow F^{-}$.
Therefore,the conjugate bases are $OH^{-}$ and $F^{-}$ respectively.

(C) basic buffer is formed by a mixture of a weak base and its salt with a strong acid.
In option $C$,we have $NH_{4}OH$ (weak base) and $HCl$ (strong acid).
Initial milli-moles of $HCl = 100 \times 0.1 = 10 \ mmol$.
Initial milli-moles of $NH_{4}OH = 200 \times 0.1 = 20 \ mmol$.
The reaction is: $HCl + NH_{4}OH \rightarrow NH_{4}Cl + H_{2}O$.
After the reaction,$10 \ mmol$ of $NH_{4}OH$ remains,and $10 \ mmol$ of $NH_{4}Cl$ (salt) is formed.
Since the mixture contains a weak base and its salt,it forms a basic buffer.

(B) The conversion of an internal alkyne (but$-2-$yne) to a $cis$-alkene (cis$-2-$butene) requires a partial reduction that involves $syn$-addition of hydrogen.
This is achieved using Lindlar's catalyst,which is $Pd$ supported on $CaCO_3$ or $BaSO_4$ poisoned with quinoline or lead acetate $(H_2, Pd / C, \text{quinoline})$.
$Na / \text{liquid } NH_3$ (Birch reduction) would result in the $trans$-alkene.
Therefore,the correct reagent is $H_2, Pd / C, \text{quinoline}$.

(D) $[SiCl_{6}]^{2-}$ does not exist because:
$(I)$ The size of the $Cl^{-}$ ion is large,so it cannot be accommodated around the small $Si^{4+}$ ion due to steric hindrance.
$(II)$ The interaction between the lone pair of the chloride ion and the $Si^{4+}$ ion is not strong enough to stabilize the complex.

(D) $Be(OH)_{2}$ is an amphoteric hydroxide because it reacts with both acids and bases.
All other hydroxides listed,i.e.,$Mg(OH)_{2}$,$Ca(OH)_{2}$,and $Sr(OH)_{2}$,are basic in nature.

(B) The first ionisation enthalpy generally increases across a period due to an increase in effective nuclear charge.
However,there are deviations due to stable electronic configurations.
The electronic configurations for second period elements are:
$Li (2s^1)$,$Be (2s^2)$,$B (2s^2 2p^1)$,$C (2s^2 2p^2)$,$N (2s^2 2p^3)$,$O (2s^2 2p^4)$,$F (2s^2 2p^5)$,$Ne (2s^2 2p^6)$.
$Be$ has a fully filled $2s$ orbital,making its ionisation energy higher than $B$.
$N$ has a half-filled $2p$ subshell,making its ionisation energy higher than $O$.
The correct order is: $Li < B < Be < C < O < N < F < Ne$.

(D) The $PCl_{5}$ molecule has a trigonal bipyramidal geometry.
In this structure,the three equatorial $P-Cl$ bonds are at $120^{\circ}$ to each other.
The two axial $P-Cl$ bonds are at $180^{\circ}$ to each other.
Due to greater repulsion from equatorial bond pairs,the axial $P-Cl$ bonds are longer than the equatorial bonds.
$PCl_{5}$ is a highly reactive molecule because it easily undergoes hydrolysis and acts as a chlorinating agent. Therefore,the statement that it is non-reactive is incorrect.

(A) To determine the energy order,we use the $(n+l)$ rule:
For $4d$: $n=4, l=2$,so $(n+l) = 4+2 = 6$.
For $5p$: $n=5, l=1$,so $(n+l) = 5+1 = 6$.
For $5f$: $n=5, l=3$,so $(n+l) = 5+3 = 8$.
For $6p$: $n=6, l=1$,so $(n+l) = 6+1 = 7$.
Comparing the $(n+l)$ values: $8 (5f) > 7 (6p) > 6 (4d, 5p)$.
For orbitals with the same $(n+l)$ value,the one with the higher $n$ value has higher energy. Thus,$5p (n=5) > 4d (n=4)$.
The final decreasing order of energy is $5f > 6p > 5p > 4d$.

(C) $1$. Ozonolysis of alkene $A$ yields propanone $(CH_{3}COCH_{3})$ and ethanal $(CH_{3}CHO)$.
$2$. By reversing the ozonolysis reaction,we join the carbonyl carbons with a double bond: $(CH_{3})_{2}C=CH-CH_{3}$. This is $2-methylbut-2-ene$.
$3$. Addition of $HCl$ to $2-methylbut-2-ene$ follows Markovnikov's rule,which proceeds via the formation of the most stable carbocation.
$4$. Protonation of the double bond at the $C-3$ position leads to a tertiary carbocation: $(CH_{3})_{2}C^{+}-CH_{2}-CH_{3}$.
$5$. Subsequent attack by the chloride ion $(Cl^{-})$ on the tertiary carbocation yields $2-chloro-2-methylbutane$,which is $CH_{3}-CH_{2}-C(Cl)(CH_{3})_{2}$.

(B) All enzymes that utilize $ATP$ in phosphate transfer require magnesium $(Mg^{2+})$ as the cofactor.
Magnesium ions play a crucial role in stabilizing the negative charges on the phosphate groups of $ATP$,facilitating the transfer of the phosphate group.

(B) In the spectrum of the hydrogen atom,the spectral lines of the $Balmer$ series lie in the visible region.
The $Lyman$ series lies in the ultraviolet region,while the $Paschen$,$Brackett$,and $Pfund$ series lie in the infrared region.

(D) For the reaction $2H_{(g)} \rightarrow H_{2(g)}$,two moles of gaseous atoms combine to form one mole of a gaseous molecule.
Since the number of moles of gas decreases,the randomness or disorder of the system decreases.
Therefore,the change in entropy $(\Delta S)$ is negative.

(A) The statement '$PbF_4$ is covalent in nature' is incorrect.
$PbF_4$ is an ionic compound because the large size of the $Pb^{4+}$ cation and the small size of the $F^-$ anion favor ionic bonding.
$SiCl_4$ undergoes easy hydrolysis due to the presence of vacant $d$-orbitals.
$GeX_4$ is more stable than $GeX_2$ due to the inert pair effect being less pronounced in $Ge$ compared to $Pb$.
$SnF_4$ is ionic in nature.

(C) The compressibility factor $Z$ is defined as the ratio of the molar volume of a real gas $(V_{m})_{real}$ to the molar volume of an ideal gas $(V_{m})_{ideal}$ at the same temperature and pressure: $Z = \frac{(V_{m})_{real}}{(V_{m})_{ideal}}$.
Given that $(V_{m})_{real}$ is $20$ percent smaller than $(V_{m})_{ideal}$,we have $(V_{m})_{real} = 0.8 \times (V_{m})_{ideal}$.
Therefore,$Z = \frac{0.8 \times (V_{m})_{ideal}}{(V_{m})_{ideal}} = 0.8$.
Since $Z < 1$,the gas shows negative deviation from ideal behavior,which indicates that attractive intermolecular forces are dominant.

(B) Clark's method is used to remove temporary hardness of water by adding lime $(Ca(OH)_2)$.
The chemical reaction is:
$Ca(HCO_3)_2 + Ca(OH)_2 \rightarrow 2 CaCO_3 \downarrow + 2 H_2O$

(C) The shoot tips of plants contain high concentrations of $Auxins$,which are responsible for apical dominance. $Apical$ $dominance$ inhibits the growth of lateral buds. By removing the shoot tips,the source of $Auxins$ is eliminated. This removes the inhibitory effect on the lateral buds,allowing them to grow and produce more branches,which in turn increases the production of tea-leaves.

(D) Liquid $CO_2$ is used as a solvent in dry cleaning because it is environmentally friendly and effective at dissolving grease and stains when combined with a suitable detergent. This process is known as supercritical fluid extraction or liquid $CO_2$ cleaning.

(D) $Na_{2}O$ is a metallic oxide,hence it is $Basic$.
$Al_{2}O_{3}$ is an $Amphoteric$ oxide as it reacts with both acids and bases.
$N_{2}O$ is a $Neutral$ oxide.
$Cl_{2}O_{7}$ is a non-metallic oxide,hence it is $Acidic$.
Therefore,the correct matching is $(i)-(b), (ii)-(d), (iii)-(a), (iv)-(c)$.

(A) The stability of carbocations is determined by the inductive effect $(+I)$ and hyperconjugation (number of $\alpha-H$ atoms).
$1$. $(CH_{3})_{3}C^{+}$ is a tertiary $(3^{\circ})$ carbocation with $9$ $\alpha-H$ atoms.
$2$. $CH_{3}CH_{2}CH^{+}CH_{2}CH_{3}$ is a secondary $(2^{\circ})$ carbocation with $4$ $\alpha-H$ atoms.
$3$. $CH_{3}CH^{+}CH_{2}CH_{2}CH_{3}$ is a secondary $(2^{\circ})$ carbocation with $3$ $\alpha-H$ atoms.
$4$. $CH_{3}CH_{2}CH_{2}^{+}$ is a primary $(1^{\circ})$ carbocation with $2$ $\alpha-H$ atoms.
Since the stability order is $3^{\circ} > 2^{\circ} > 1^{\circ}$, the tertiary carbocation $(CH_{3})_{3}C^{+}$ is the most stable.

(B) For an alkane to produce only one mono-chloro product,all hydrogen atoms in the molecule must be equivalent.
Neopentane ($2,2-$dimethylpropane) has the structure $C(CH_3)_4$.
In neopentane,all $12$ hydrogen atoms are attached to equivalent primary carbon atoms.
Therefore,substitution of any of these $12$ hydrogen atoms by a chlorine atom results in the same product,$1-$chloro$-2,2-$dimethylpropane.
Other options like $n-$pentane,isopentane,and $2,2-$dimethylbutane contain non-equivalent hydrogen atoms,leading to a mixture of isomeric mono-chloro products.

(A) The reaction of propyne $(CH_3-C \equiv CH)$ with a red hot iron tube at $873 \ K$ is a cyclic polymerization reaction.
Three molecules of propyne undergo cyclization to form $1,3,5$-trimethylbenzene (mesitylene) as the product $A$.
The structure of $1,3,5$-trimethylbenzene is a benzene ring with three methyl groups at positions $1, 3,$ and $5$.
To calculate the number of $\sigma$ bonds in $1,3,5$-trimethylbenzene $(C_9H_{12})$:
- There are $6$ $\sigma$ bonds in the benzene ring ($C$-$C$ bonds).
- There are $3$ $\sigma$ bonds between the ring carbons and the methyl carbons.
- Each of the $3$ methyl groups has $3$ $C$-$H$ $\sigma$ bonds,totaling $3 \times 3 = 9$ $\sigma$ bonds.
- There are $3$ $\sigma$ bonds between the ring carbons and the remaining $3$ hydrogens,totaling $3$ $\sigma$ bonds.
Total $\sigma$ bonds = $6 \text{ (ring C-C)} + 3 \text{ (ring-methyl C-C)} + 9 \text{ (methyl C-H)} + 3 \text{ (ring C-H)} = 21$.
Thus,the product $A$ contains $21$ $\sigma$ bonds.

(C) Silicones are synthetic polymers containing $Si-O-Si$ linkages. They are used as sealants,greases,electrical insulators,and for waterproofing fabrics. Due to their biocompatibility,they are also widely used in surgical and cosmetic implants.

(A) The reaction of calcium phosphide $(Ca_{3}P_{2})$ with water produces phosphine gas $(PH_{3})$,which has a characteristic rotten fish smell.
$Ca_{3}P_{2} + 6H_{2}O \rightarrow 3Ca(OH)_{2} + 2PH_{3} (Y)$
Phosphine gas $(PH_{3})$ reacts with copper sulfate $(CuSO_{4})$ to form copper phosphide $(Cu_{3}P_{2})$.
$3CuSO_{4} + 2PH_{3} \rightarrow Cu_{3}P_{2} + 3H_{2}SO_{4}$
Therefore,the compound $X$ is $Ca_{3}P_{2}$.

(C) In neutral or faintly alkaline medium,$KMnO_4$ acts as an oxidizing agent and reduces to $MnO_2$.
The balanced chemical equation for the reaction between $KMnO_4$ and potassium iodide $(KI)$ is:
$2KMnO_4 + KI + H_2O \longrightarrow 2MnO_2 + KIO_3 + 2KOH$.
Here,the iodide ion $(I^-)$ is oxidized to the iodate ion $(IO_3^-)$.
Therefore,'$X$' is $IO_3^-$.

(D) According to Molecular Orbital Theory $(MOT)$,the electronic configuration of $O_2$ ($16$ electrons) is: $\sigma 1s^2, \sigma^* 1s^2, \sigma 2s^2, \sigma^* 2s^2, \sigma 2p_z^2, \pi 2p_x^2 = \pi 2p_y^2, \pi^* 2p_x^1 = \pi^* 2p_y^1$.
Since $O_2$ contains two unpaired electrons in the $\pi^*$ antibonding orbitals,it exhibits paramagnetic behavior.
In contrast,$N_2$ ($14$ electrons),$H_2$ ($2$ electrons),and $Li_2$ ($6$ electrons) have all paired electrons,making them diamagnetic.

(A) The dipole moment $(\mu)$ depends on the polarity of bonds and the geometry of the molecule.
$1$. $BF_{3}$ is a planar molecule with a symmetric structure,so its net dipole moment is $\mu = 0$.
$2$. $NF_{3}$ has a pyramidal structure with a lone pair. The bond dipoles of $N-F$ bonds are in the opposite direction to the lone pair dipole,resulting in a small net dipole moment $(\mu \approx 0.24 \ D)$.
$3$. $NH_{3}$ has a pyramidal structure where the bond dipoles of $N-H$ bonds and the lone pair dipole are in the same direction,resulting in a higher net dipole moment $(\mu \approx 1.47 \ D)$.
$4$. $H_{2}O$ has a bent structure with two $O-H$ bonds and two lone pairs,resulting in a high net dipole moment $(\mu \approx 1.85 \ D)$.
Therefore,the correct order is $BF_{3} < NF_{3} < NH_{3} < H_{2}O$.

(A) Crude $NaCl$ obtained by crystallisation of brine solution typically contains impurities like $Na_2SO_4$,$CaCl_2$,$MgCl_2$,and $CaSO_4$.
$MgSO_4$ is not typically present as an impurity in this process.

(A) Among alkali metal ions,$Li^{+}$ has the smallest ionic size and the highest charge density.
Due to this,$Li^{+}$ has the maximum hydration enthalpy and the strongest tendency to form hydrated salts.
Therefore,$LiCl$ readily forms its dihydrate salt,$LiCl \cdot 2H_2O$.

(C) $NaOH_{(aq)}$ is a strong base solution.
Since $NaOH$ dissociates completely,$[OH^-] = 0.01 \ M = 10^{-2} \ M$.
$pOH = -\log[OH^-] = -\log(10^{-2}) = 2$.
We know that $pH + pOH = 14$ at $25^{\circ}C$.
Therefore,$pH = 14 - pOH = 14 - 2 = 12$.

(C) Bronsted acid is a proton donor,and a Bronsted base is a proton acceptor.
Species that can act as both are called amphoteric.
$HCO_3^-$,$NH_3$,and $HSO_4^-$ possess both a proton to donate and a lone pair/negative charge to accept a proton.
$HCl$ can only donate a proton $(HCl \rightarrow H^+ + Cl^-)$ but cannot accept a proton to form $H_2Cl^+$,thus it cannot act as a Bronsted base.

(C) $CaF_2$ dissociates as: $CaF_{2(s)} \rightleftharpoons Ca^{2+}_{(aq)} + 2F^{-}_{(aq)}$
$NaF$ is a strong electrolyte: $NaF_{(aq)} \rightarrow Na^{+}_{(aq)} + F^{-}_{(aq)}$
Given $[NaF] = 0.1 \ M$,so $[F^-] = 0.1 \ M$ from $NaF$.
Let $s$ be the molar solubility of $CaF_2$. Then $[Ca^{2+}] = s$ and total $[F^-] = (2s + 0.1) \approx 0.1 \ M$ (since $s$ is very small).
$K_{sp} = [Ca^{2+}][F^-]^2$
$5.3 \times 10^{-11} = s \times (0.1)^2$
$s = \frac{5.3 \times 10^{-11}}{0.01} = 5.3 \times 10^{-9} \ mol \ L^{-1}$

(C) The compound $CrO_6$ is a peroxide derivative of chromium.
In this structure,$Cr$ is bonded to four oxygen atoms via two peroxide linkages $(-O-O-)$ and one double-bonded oxygen atom,or more accurately,it is a butterfly structure where $Cr$ is in the $+6$ oxidation state.
Since the maximum oxidation state of $Cr$ is $+6$,it maintains this state in $CrO_6$.

(B) In $CuSO_4 \cdot 5H_2O$,the structure consists of four water molecules coordinated directly to the $Cu^{2+}$ ion.
The fifth water molecule is held by hydrogen bonds between the $SO_4^{2-}$ ion and the coordinated water molecules.
Therefore,only $1$ water molecule is involved in hydrogen bonding.

(C) The total pressure of the saturated air is the sum of the partial pressure of dry air and the partial pressure of water vapour.
Given,the mole fraction of water vapour,$X_{H_{2}O} = 0.02$.
The mole fraction of dry air,$X_{dry\;air} = 1 - X_{H_{2}O} = 1 - 0.02 = 0.98$.
The partial pressure of a component is given by $P_{i} = X_{i} \times P_{total}$.
Therefore,the partial pressure of dry air $= X_{dry\;air} \times P_{total} = 0.98 \times 1.2 \; atm = 1.176 \; atm$.

(C) $2 \ M$ solution of $NaOH$ means $2 \ mol$ of $NaOH$ is present in $1 \ L$ of solution.
Density of solution $= 1.28 \ g \ mL^{-1}$.
Mass of solution $=$ Volume $\times$ Density $= 1000 \ mL \times 1.28 \ g \ mL^{-1} = 1280 \ g$.
Mass of solute $(NaOH)$ $= 2 \ mol \times 40 \ g \ mol^{-1} = 80 \ g$.
Mass of solvent (water) $=$ Mass of solution $-$ Mass of solute $= 1280 \ g - 80 \ g = 1200 \ g = 1.2 \ kg$.
Molality $(m) = \frac{\text{moles of solute}}{\text{mass of solvent in kg}} = \frac{2 \ mol}{1.2 \ kg} = 1.67 \ m$.

(C) The number of angular nodes is given by the azimuthal quantum number $\ell$. Given $\ell = 3$.
The total number of nodes is given by the formula $n - 1$,where $n$ is the principal quantum number.
Given total nodes $= 3$,we have $n - 1 = 3$,which implies $n = 4$.
Since $n = 4$ and $\ell = 3$,the orbital corresponds to the $4f$ subshell.

(B) According to Bohr's postulate,the circumference of the orbit is an integral multiple of the de Broglie wavelength: $n \lambda = 2 \pi r_n$.
For a hydrogen atom,the radius of the $n^{th}$ orbit is given by $r_n = a_0 \times n^2$,where $a_0 = 52.9 \; pm$ and $Z = 1$.
Substituting $n = 2$ for the second orbit: $r_2 = a_0 \times (2)^2 = 4 a_0$.
Now,substituting this into the wavelength formula: $2 \lambda = 2 \pi (4 a_0)$.
Therefore,$\lambda = 4 \pi a_0$.
Calculating the value: $\lambda = 4 \times \pi \times 52.9 \; pm = 211.6 \pi \; pm$.

(D) Using the ideal gas equation: $PV = nRT$
First,calculate the number of moles $(n)$ of water vapour:
$n = \frac{\text{mass}}{\text{molar mass}} = \frac{1.8 \; g}{18 \; g \; mol^{-1}} = 0.1 \; mol$
Convert the temperature $(T)$ from Celsius to Kelvin:
$T = 374 + 273 = 647 \; K$
Given pressure $(P)$ is $1 \; bar$ and gas constant $(R)$ is $0.083 \; bar \; L \; K^{-1} \; mol^{-1}$.
Now,calculate the volume $(V)$:
$V = \frac{nRT}{P} = \frac{0.1 \; mol \times 0.083 \; bar \; L \; K^{-1} \; mol^{-1} \times 647 \; K}{1 \; bar} = 5.37 \; L$

(B) The work done on the gas during an expansion against a constant external pressure is given by the formula:
$W = -P_{ext} \times \Delta V$
Given:
$P_{ext} = 10^{5} \; Nm^{-2}$
$V_i = 10^{-3} \; m^{3}$
$V_f = 10^{-2} \; m^{3}$
Change in volume $\Delta V = V_f - V_i = 10^{-2} - 10^{-3} = 10^{-2} - 0.1 \times 10^{-2} = 0.9 \times 10^{-2} \; m^{3}$
Substituting the values:
$W = -10^{5} \; Nm^{-2} \times (0.9 \times 10^{-2} \; m^{3})$
$W = -0.9 \times 10^{3} \; J$
$W = -900 \; J$
Since the work done is negative,it indicates that work is done by the gas on the surroundings.

(D) In an adiabatic expansion,the system does work at the expense of its internal energy,leading to a decrease in temperature. Thus,$T_C < T_A$.
For an adiabatic process:
$q = 0$,$\Delta U = W$
Since it is an expansion,$W < 0$,therefore $\Delta U < 0$.
Since $\Delta U = nC_{vm} \Delta T$,we have $nC_{vm} \Delta T < 0$,which implies $\Delta T < 0$.
Therefore,$T_C - T_A < 0$,or $T_C < T_A$.
Option $D$ states $T_C > T_A$,which is incorrect.

(B) In $XO$ type of sex determination (e.g.,grasshoppers),males produce two types of gametes: $50 \%$ with an $X$ chromosome and $50 \%$ without any sex chromosome. Thus,statement $A$ is correct.
In birds (including domesticated fowls),sex determination is of $ZW-ZZ$ type. Here,females are heterogametic $(ZW)$ and males are homogametic $(ZZ)$. Therefore,the sex of the progeny depends on the type of egg,not the sperm. Thus,statement $B$ is incorrect.
In humans,males are $XY$. The $Y$ chromosome is significantly shorter than the $X$ chromosome. Thus,statement $C$ is correct.
In fruit flies $(Drosophila)$,males are $XY$ (heterogametic) and females are $XX$ (homogametic). Thus,statement $D$ is correct.

(C) The genetic code is nearly universal,meaning that the same codons specify the same amino acids in almost all organisms,from bacteria to humans.
Because of this universality,a human gene inserted into a bacterial cell can be correctly transcribed and translated by the bacterial machinery to produce the same human protein (insulin).
Therefore,the ability of bacteria to produce human insulin is a direct consequence of the genetic code being nearly universal.

(C) In a hexagonal close packed $(hcp)$ lattice,the number of anions $(A)$ per unit cell is $6$.
The number of octahedral voids in an $hcp$ lattice is equal to the number of atoms,which is $6$.
The cations $(C)$ occupy $75 \%$ of these octahedral voids.
Number of cations $(C) = 6 \times \frac{75}{100} = 6 \times \frac{3}{4} = \frac{18}{4} = 4.5$ or $\frac{9}{2}$.
The ratio of cations $(C)$ to anions $(A)$ is $\frac{9}{2} : 6$.
Multiplying both sides by $2$,we get $9 : 12$.
Simplifying the ratio by dividing by $3$,we get $3 : 4$.
Therefore,the formula of the compound is $C_{3}A_{4}$.

(C) An ideal solution is defined as a solution that obeys Raoult's law over the entire range of concentration.
For an ideal solution,the enthalpy of mixing is zero,i.e.,$\Delta_{mix} H = 0$.
Also,the volume of mixing is zero,i.e.,$\Delta_{mix} V = 0$.
However,the entropy of mixing $(\Delta_{mix} S)$ is always greater than zero for any mixing process,and the Gibbs free energy of mixing $(\Delta_{mix} G)$ is always less than zero for a spontaneous process.

(C) The relationship between the standard cell potential and the equilibrium constant is given by the Nernst equation at equilibrium:
$E_{cell}^{\ominus} = \frac{2.303 \; RT}{nF} \log_{10} K_{eq}$
Given:
$E_{cell}^{\ominus} = 0.59 \; V$
$n = 1$
$\frac{2.303 \; RT}{F} = 0.059 \; V$
Substituting the values into the equation:
$0.59 = \frac{0.059}{1} \log_{10} K_{eq}$
$\log_{10} K_{eq} = \frac{0.59}{0.059} = 10$
$K_{eq} = 10^{10}$

(A) Antibiotics that are effective mainly against either Gram-positive or Gram-negative bacteria are classified as narrow spectrum antibiotics.
Penicillin $G$ is a classic example of a narrow spectrum antibiotic.
In contrast,ampicillin,amoxycillin,and chloramphenicol are broad spectrum antibiotics,meaning they are effective against a wide range of Gram-positive and Gram-negative bacteria.

(A) In aqueous solution,the basic strength of amines depends on three factors: inductive effect,solvation effect (solvation of protonated amine),and steric hindrance.
For methyl substituted amines,the order is determined by the combined effect of these factors:
$(CH_{3})_{2}NH > CH_{3}NH_{2} > (CH_{3})_{3}N$
Here,$(CH_{3})_{2}NH$ is the most basic due to the optimal balance of the $+I$ effect and solvation,while $(CH_{3})_{3}N$ is the least basic among these due to steric hindrance preventing effective solvation of the protonated cation.

(A) The formation of a negatively charged $[AgI]I^{-}$ sol occurs when iodide ions $(I^-)$ are preferentially adsorbed on the surface of $AgI$ particles.
This happens when $KI$ is present in excess relative to $AgNO_3$ in the reaction: $AgNO_3 + KI \rightarrow AgI + KNO_3$.
In option $A$,the moles of $KI$ are $50 \ mL \times 1.5 \ M = 75 \ mmol$,while the moles of $AgNO_3$ are $50 \ mL \times 1 \ M = 50 \ mmol$.
Since $KI$ is in excess,$I^-$ ions are adsorbed on the $AgI$ precipitate,resulting in a negatively charged $[AgI]I^-$ sol.

(D) Protonation involves the donation of a lone pair of electrons from the oxygen atom to a proton $(H^+)$.
Compounds with higher electron density on the oxygen atom are more basic and easier to protonate.
In $PhOH$ (phenol),the lone pair of electrons on the oxygen atom is delocalized into the benzene ring due to resonance.
This delocalization reduces the electron density on the oxygen atom,making it the least basic and therefore the most difficult to protonate among the given options.

(B) The reaction shown is the industrial preparation of phenol from cumene (isopropylbenzene).
In the first step,cumene is oxidized by atmospheric oxygen $(O_2)$ to form cumene hydroperoxide as the intermediate $A$.
In the second step,cumene hydroperoxide is treated with dilute acid $(H^+/H_2O)$ to yield phenol and acetone $(CH_3COCH_3)$ as products.

(A) $MnO_{4}^{2-}$ (manganate ion) and $MnO_{4}^{-}$ (permanganate ion) both have a tetrahedral geometry.
In these ions,the manganese atom is in a high oxidation state and uses its $d-$orbitals to form $p\pi-d\pi$ bonds with the oxygen atoms.
Specifically,the $\pi-$bonding involves the overlap of filled $2p-$orbitals of oxygen with empty $3d-$orbitals of manganese.

(C) For a first order reaction,the integrated rate equation is $t = \frac{2.303}{k} \log\left(\frac{[A]_0}{[A]_t}\right)$.
For $99\%$ completion,let $[A]_0 = 100$,then $[A]_t = 100 - 99 = 1$.
Substituting the values into the equation:
$t = \frac{2.303}{k} \log\left(\frac{100}{1}\right)$
$t = \frac{2.303}{k} \log(10^2)$
$t = \frac{2.303 \times 2}{k} \log(10)$
Since $\log(10) = 1$,we get $t = \frac{4.606}{k}$.

(B) Nylon $2$-nylon $6$ is the correct answer.
It is an alternating polyamide copolymer of glycine $(H_{2}N-CH_{2}-COOH)$ and aminocaproic acid $[H_{2}N(CH_{2})_{5}COOH]$.
This polymer is biodegradable.

(B) $XeF_{4}$ has $sp^{3}d^{2}$ hybridization and a square planar structure.
$XeF_{6}$ has $sp^{3}d^{3}$ hybridization and a distorted octahedral structure.
$XeOF_{4}$ has $sp^{3}d^{2}$ hybridization and a square pyramidal structure.
$XeO_{3}$ has $sp^{3}$ hybridization and a pyramidal structure.
Therefore,the correct matching is: $(a)-(ii), (b)-(iii), (c)-(iv), (d)-(i)$.

(C) The thermal stability of hydrides of group $16$ elements depends on the bond dissociation enthalpy of the $E-H$ bond.
As the size of the central atom $E$ increases down the group $(O < S < Se < Te < Po)$,the bond length increases and the bond dissociation enthalpy decreases.
Therefore,the thermal stability decreases down the group.
The correct order is: $H_2 O > H_2 S > H_2 Se > H_2 Te > H_2 Po$.

(A) Tribromooctaoxide has the chemical formula $Br_3O_8$.
In this structure,the central bromine atom is bonded to two other bromine atoms.
Each of the terminal bromine atoms is bonded to three oxygen atoms via double bonds and to the central bromine atom.
The central bromine atom is bonded to two oxygen atoms via double bonds.
This results in a neutral molecule where all bromine atoms achieve a stable oxidation state.
The other options represent anionic species,which do not correspond to the neutral $Br_3O_8$ molecule.

(D) Malachite is a green copper carbonate hydroxide mineral with the chemical formula $CuCO_3 \cdot Cu(OH)_2$.

(A) Maximum boiling azeotropes are formed by solutions that show large negative deviation from Raoult's law.
In these solutions,the solute-solvent interactions are stronger than the solute-solute and solvent-solvent interactions.
Water + Nitric acid $(H_2O + HNO_3)$ shows a large negative deviation from ideal behavior,resulting in a maximum boiling azeotrope.

(A) The given cell reaction is: $2 Fe^{3+}_{(aq)} + 2 I^{-}_{(aq)} \rightarrow 2 Fe^{2+}_{(aq)} + I_{2(aq)}$
Here,the number of electrons involved in the reaction is $n = 2$.
The formula for standard Gibbs energy is: $\Delta_r G^{\ominus} = -n F E^{\ominus}_{cell}$
Substituting the values: $\Delta_r G^{\ominus} = -2 \times 96500 \ C \ mol^{-1} \times 0.24 \ V$
$\Delta_r G^{\ominus} = -46320 \ J \ mol^{-1}$
Converting to $kJ \ mol^{-1}$: $\Delta_r G^{\ominus} = -46.32 \ kJ \ mol^{-1}$

(D) Pure nitrogen is obtained by the thermal decomposition of sodium azide or barium azide: $2NaN_3 \xrightarrow{\Delta} 2Na + 3N_2$ and $Ba(N_3)_2 \xrightarrow{\Delta} Ba + 3N_2$. $(a-iv)$
Haber process is used for the industrial manufacture of ammonia: $N_2 + 3H_2 \rightleftharpoons 2NH_3$. $(b-iii)$
Contact process is used for the manufacture of sulphuric acid $(H_2SO_4)$. $(c-ii)$
Deacon's process is used for the formation of chlorine gas: $4HCl + O_2 \xrightarrow{CuCl_2} 2H_2O + 2Cl_2$. $(d-i)$
Therefore,the correct matching is $a-iv, b-iii, c-ii, d-i$.

(C) Non-essential amino acids are those that can be synthesized by the human body.
Among the given options,$alanine$ is a non-essential amino acid.
$Valine$,$leucine$,and $lysine$ are essential amino acids that must be obtained through the diet.

(B) Electrophilic substitution reaction involves the replacement of a hydrogen atom on an aromatic ring by an electrophile.
In the reaction of benzene with chlorine in the presence of a Lewis acid like anhydrous $AlCl_3$,the electrophile $Cl^+$ is generated,which attacks the benzene ring to form chlorobenzene.
This is a classic example of electrophilic aromatic substitution.
Reaction: $C_6H_6 + Cl_2 \xrightarrow{Anhyd. AlCl_3} C_6H_5Cl + HCl$.

(B) When phthalic acid reacts with ammonia $(NH_3)$,it first forms ammonium phthalate.
Upon heating,ammonium phthalate loses two molecules of water to form phthalamide.
On further strong heating,phthalamide loses a molecule of ammonia $(NH_3)$ to form the cyclic imide known as phthalimide.

(C) For the reaction $N_{2(g)} + 3 H_{2(g)} \rightleftharpoons 2 NH_{3(g)}$,the rate of reaction is expressed as:
$-\frac{d[N_{2}]}{dt} = -\frac{1}{3} \frac{d[H_{2}]}{dt} = +\frac{1}{2} \frac{d[NH_{3}]}{dt}$.
Comparing this with the given options,we see that $-\frac{d[N_{2}]}{dt} = \frac{1}{2} \frac{d[NH_{3}]}{dt}$ is the correct relationship.

(B) In the complex $K_{4}[Fe(CN)_{6}]$,the oxidation state of $Fe$ is $+2$.
The electronic configuration of $Fe$ $(Z=26)$ is $[Ar] 3d^{6} 4s^{2}$.
For $Fe^{2+}$,the configuration is $3d^{6}$.
$CN^{-}$ is a strong field ligand $(SFL)$,which causes pairing of electrons in the $d$-orbitals.
According to crystal field theory for an octahedral complex,the $d$-orbitals split into $t_{2g}$ and $e_{g}$ sets.
Due to the strong field,all $6$ electrons pair up in the lower energy $t_{2g}$ orbitals,resulting in the configuration $t_{2g}^{6} e_{g}^{0}$.

(D) The rate of disintegration is given by $R = N \lambda$,where $\lambda$ is the decay constant.
The decay constant $\lambda$ is related to the half-life $T_{1/2}$ by the formula $\lambda = \frac{0.693}{T_{1/2}}$.
Substituting $\lambda$ into the rate equation: $R = N \times \frac{0.693}{T_{1/2}}$.
Rearranging to solve for the number of atoms $N$: $N = \frac{R \times T_{1/2}}{0.693}$.
Given $R = 10^{10} \ s^{-1}$ and $T_{1/2} = 2.2 \times 10^{9} \ s$:
$N = \frac{10^{10} \times 2.2 \times 10^{9}}{0.693} = \frac{2.2 \times 10^{19}}{0.693} \approx 3.17 \times 10^{19} \ atoms$.

(C) The oxidation of primary alcohols with strong oxidizing agents like $K_2Cr_2O_7$ or $KMnO_4/H^+$ yields carboxylic acids. Thus,phenylmethanol $(C_6H_5CH_2OH)$ gives benzoic acid $(C_6H_5COOH)$ in options $A$ and $D$.
The haloform reaction of acetophenone $(C_6H_5COCH_3)$ with $NaOCl$ followed by acid workup $(H_3O^+)$ is a standard method to produce benzoic acid,as shown in option $B$.
$PCC$ (Pyridinium chlorochromate) is a mild oxidizing agent that oxidizes primary alcohols only up to the aldehyde stage. Therefore,the reaction of phenylmethanol with $PCC$ yields benzaldehyde $(C_6H_5CHO)$ as the major product,not benzoic acid. Hence,option $C$ is the correct answer.

(A) Hinsberg's reagent (benzene sulfonyl chloride,$C_6H_5SO_2Cl$) is used to distinguish between primary,secondary,and tertiary amines.
$1$. Primary $(1^\circ)$ amines react to form $N$-alkylbenzene sulfonamides,which contain an acidic hydrogen on the nitrogen atom and are therefore soluble in alkali.
$2$. Secondary $(2^\circ)$ amines react to form $N,N$-dialkylbenzene sulfonamides,which lack an acidic hydrogen on the nitrogen atom and are insoluble in alkali.
$3$. Tertiary $(3^\circ)$ amines do not react with Hinsberg's reagent.
In the given options,$(CH_3)_2CH-NH-CH(CH_3)_2$ is a secondary amine. It reacts with Hinsberg's reagent to form an $N,N$-disubstituted sulfonamide that has no acidic hydrogen,making it insoluble in alkali.

(B) During the process of denaturation,the hydrogen bonds and other interactions that stabilize the $2^o$ and $3^o$ structures are disrupted,leading to their loss.
However,the peptide bonds that hold the amino acids together in the $1^o$ structure remain intact.

(C) Polyacrylonitrile is used as a substitute for wool in making commercial fibres,such as Orlon or Acrilan.

(C) Sucralose is a trichloro derivative of sucrose. It is stable at cooking temperature and does not provide calories to the body.

(A) The rate of nucleophilic substitution reaction $(NSR)$ depends on the stability of the transition state or the nature of the $C-X$ bond.
$A$. $C_{6}H_{5}Cl$ is an aryl halide. In aryl halides,the $C-Cl$ bond acquires partial double bond character due to resonance,making it very strong and difficult to break. Thus,it undergoes hydrolysis at the slowest rate.
$B$. $CH_{3}CH_{2}Cl$ is a primary alkyl halide,which undergoes $S_{N}2$ reaction.
$C$. $CH_{2}=CH-CH_{2}Cl$ is an allylic halide,which is highly reactive towards $S_{N}1$ or $S_{N}2$ due to resonance stabilization of the carbocation or transition state.
$D$. $C_{6}H_{5}CH_{2}Cl$ is a benzyl chloride,which is highly reactive due to resonance stabilization of the benzyl carbocation.
Therefore,the correct option is $A$.

(C) When vapours of a secondary alcohol are passed over heated copper at $573 \; K$,dehydrogenation occurs to form a ketone.
Reaction: $CH_3-CH(OH)-CH_3 \xrightarrow{Cu / 573 \; K} CH_3-CO-CH_3 + H_2$
Thus,a $2^o$ alcohol yields a ketone.

(A) The reaction of an ether with excess $HI$ involves the protonation of the ether oxygen followed by nucleophilic attack by $I^-$.
$CH_{3}CH_{2}CH_{2}OC(CH_{3})_{3} + HI \rightarrow CH_{3}CH_{2}CH_{2}OH + (CH_{3})_{3}CI$.
Since $HI$ is in excess,the alcohol formed $(CH_{3}CH_{2}CH_{2}OH)$ further reacts with $HI$ to form an alkyl iodide.
$CH_{3}CH_{2}CH_{2}OH + HI \rightarrow CH_{3}CH_{2}CH_{2}I + H_{2}O$.
The tertiary carbocation formed from the cleavage of the $C-O$ bond in the tert-butyl group is highly stable,leading to the formation of $(CH_{3})_{3}CI$.
Thus,the final products are $CH_{3}CH_{2}CH_{2}I$ and $(CH_{3})_{3}CI$.

(A) $i. Na_{2}O$ is used in the oxidation of $SO_{2}$ to $SO_{3}$ in the contact process for the manufacture of $H_{2}SO_{4}$ $(i-c)$.
$ii. TiCl_{4} + Al(CH_{3})_{3}$ is the Ziegler-Natta catalyst used for the polymerisation of ethylene $(ii-d)$.
$iii. PdCl_{2}$ is used in the Wacker process for the oxidation of ethyne to ethanal $(iii-a)$.
$iv. \text{Nickel complexes}$ are used for the polymerisation of alkynes $(iv-b)$.
Thus,the correct matching is $i-c, ii-d, iii-a, iv-b$.

(A) $AlCl_{3}$ in an acidified aqueous solution undergoes hydration to form the octahedral complex $[Al(H_{2}O)_{6}]^{3+}$.
The electronic configuration of $Al^{3+}$ is $[Ne] \ 3s^{0} \ 3p^{0} \ 3d^{0}$.
To accommodate six water ligands,the $Al^{3+}$ ion utilizes one $3s$,three $3p$,and two $3d$ orbitals,resulting in $sp^{3}d^{2}$ hybridisation.
Therefore,$A = [Al(H_{2}O)_{6}]^{3+}$ and $B = sp^{3}d^{2}$.

(D) The correct definition of $Gangue$ is the earthy or undesired materials (such as sand,clay,or rocks) associated with the ore. Option $D$ is incorrect because $Gangue$ is not the ore itself,but the impurities present in the ore.

(C) The reducing property of phosphorus oxoacids is directly proportional to the number of $P-H$ bonds present in the molecule.
$H_3PO_2$ (hypophosphorous acid) contains two $P-H$ bonds.
$H_3PO_3$ (phosphorous acid) contains one $P-H$ bond.
$H_3PO_4$ (phosphoric acid) contains zero $P-H$ bonds.
$H_4P_2O_7$ (pyrophosphoric acid) contains zero $P-H$ bonds.
Since $H_3PO_2$ has the maximum number of $P-H$ bonds,it acts as the strongest reducing agent among the given options.

(A) Oleum is also known as pyrosulphuric acid.
Its chemical formula is $H_2S_2O_7$.

(D) The relationship between the crystal field splitting energy of tetrahedral complexes $(\Delta_{t})$ and octahedral complexes $(\Delta_{o})$ is given by the formula: $\Delta_{t} = \frac{4}{9} \times \Delta_{o}$.
Given that $\Delta_{o}$ for $[CoCl_{6}]^{4-}$ is $18000 \; cm^{-1}$.
Therefore,$\Delta_{t}$ for $[CoCl_{4}]^{2-}$ is calculated as: $\Delta_{t} = \frac{4}{9} \times 18000 \; cm^{-1} = 8000 \; cm^{-1}$.

(D) According to Kohlrausch's law,the limiting molar conductivity of an electrolyte is the sum of the limiting molar conductivities of its constituent ions.
$1. \lambda_{m(H_{2}SO_{4})}^{0} = 2\lambda_{H^{+}}^{0} + \lambda_{SO_{4}^{2-}}^{0} = x$
$2. \lambda_{m(K_{2}SO_{4})}^{0} = 2\lambda_{K^{+}}^{0} + \lambda_{SO_{4}^{2-}}^{0} = y$
$3. \lambda_{m(CH_{3}COOK)}^{0} = \lambda_{CH_{3}COO^{-}}^{0} + \lambda_{K^{+}}^{0} = z$
We need to find $\lambda_{m(CH_{3}COOH)}^{0} = \lambda_{CH_{3}COO^{-}}^{0} + \lambda_{H^{+}}^{0}$.
From the given equations:
$\lambda_{m(CH_{3}COOH)}^{0} = \lambda_{m(CH_{3}COOK)}^{0} + \frac{1}{2}\lambda_{m(H_{2}SO_{4})}^{0} - \frac{1}{2}\lambda_{m(K_{2}SO_{4})}^{0}$
Substituting the values:
$\lambda_{m(CH_{3}COOH)}^{0} = z + \frac{x}{2} - \frac{y}{2} = \frac{x-y}{2} + z \ S \ cm^{2} mol^{-1}$.

(D) For a first order reaction,the integrated rate equation is: $t = \frac{2.303}{k} \log \frac{[A]_0}{[A]_t}$
Given: $k = 2.303 \times 10^{-3} \; s^{-1}$,$[A]_0 = 40 \; g$,$[A]_t = 10 \; g$
Substituting the values: $t = \frac{2.303}{2.303 \times 10^{-3}} \log \frac{40}{10}$
$t = \frac{1}{10^{-3}} \log 4$
$t = 1000 \times \log(2^2) = 1000 \times 2 \times \log 2$
$t = 2000 \times 0.3010 = 602 \; s$

(B) According to the Arrhenius equation: $\log \left( \frac{K_{2}}{K_{1}} \right) = \frac{E_{a}}{2.303 \ R} \left( \frac{1}{T_{1}} - \frac{1}{T_{2}} \right)$.
Given that the activation energy $E_{a} = 0$.
Substituting $E_{a} = 0$ into the equation,we get: $\log \left( \frac{K_{2}}{K_{1}} \right) = \frac{0}{2.303 \ R} \left( \frac{1}{T_{1}} - \frac{1}{T_{2}} \right) = 0$.
This implies $\frac{K_{2}}{K_{1}} = 10^{0} = 1$,which means $K_{2} = K_{1}$.
Therefore,the rate constant at $400 \ K$ remains the same as at $200 \ K$,which is $1.6 \times 10^{6} \ s^{-1}$.

(A) The Freundlich adsorption isotherm is given by the equation $x/m = K(P)^{1/n}$.
In this equation,the exponent $1/n$ represents the fraction of the surface covered by the adsorbate,and its value typically lies between $0$ and $1$ (i.e.,$0 < 1/n < 1$).
Comparing the given options with the condition $0 < 1/n < 1$,the value $0.3$ satisfies this range.
Therefore,the correct option is $x/m = k p^{0.3}$.

(A) Let the amount of $Ni^{2+}$ ions in the crystal $Ni_{0.98}O$ be $x$.
Then,the amount of $Ni^{3+}$ ions in the crystal will be $(0.98 - x)$.
Since the crystal is electrically neutral,the total positive charge must equal the total negative charge (from $O^{2-}$).
$2(x) + 3(0.98 - x) = 2$
$2x + 2.94 - 3x = 2$
$-x = 2 - 2.94$
$-x = -0.94$
$x = 0.94$
Therefore,the fraction of nickel existing as $Ni^{2+}$ ions is $\frac{0.94}{0.98} \approx 0.96$.

(A) For a solution to exhibit positive deviation from Raoult's law:
$1$. The intermolecular attractive forces between $A-A$ and $B-B$ are stronger than those between $A-B$.
$2$. The enthalpy of mixing is positive,i.e.,$\Delta_{mix} H > 0$.
$3$. The volume of mixing is positive,i.e.,$\Delta_{mix} V > 0$.
Since options $A$,$B$,and $C$ are all correct descriptions of positive deviation,this question is a multiple-correct type. However,in standard single-choice contexts,the primary cause is the intermolecular force interaction.

(B) The reducing power of a metal is inversely proportional to its Standard Reduction Potential $(SRP)$ value.
Given $SRP$ values are:
$E^{\circ}_{K^{+}/K} = -2.93 \ V$
$E^{\circ}_{Al^{3+}/Al} = -1.66 \ V$
$E^{\circ}_{Cr^{3+}/Cr} = -0.74 \ V$
$E^{\circ}_{Ag^{+}/Ag} = 0.80 \ V$
Since the reducing power follows the order of decreasing $SRP$ values (more negative $SRP$ means stronger reducing agent),the order is:
$K > Al > Cr > Ag$.