NEET 2017 Chemistry Question Paper with Answer and Solution

(C) The Doppler shift formula for light is given by $\frac{\Delta \lambda}{\lambda} = \frac{v}{c}$,where $\Delta \lambda$ is the shift in wavelength,$\lambda$ is the original wavelength,$v$ is the velocity of the source,and $c$ is the speed of light.
Given: $\Delta \lambda = 0.1 \ \mathring{A}$,$\lambda = 6000 \ \mathring{A}$,and $c = 3 \times 10^5 \ km/s$.
Rearranging the formula for velocity $v$: $v = \frac{\Delta \lambda}{\lambda} \times c$.
Substituting the values: $v = \frac{0.1}{6000} \times 3 \times 10^5 \ km/s$.
$v = \frac{1}{60000} \times 3 \times 10^5 = \frac{30}{6} = 5 \ km/s$.
Thus,the velocity of recession of the star is $5 \ km/s$.

(C) For hydrogen-like atoms (single-electron species),the energy of orbitals depends only on the principal quantum number $n$.
Therefore,the energy of $2s$ orbital is equal to the energy of $2p$ orbital $(E_{2s} = E_{2p})$.
Thus,the statement that the energy of $2s$ orbital is less than $2p$ orbital is incorrect.

(A) The atomic number of the element is $Z = 114$.
The electronic configuration is determined by filling orbitals in the order of increasing energy: $[Rn] \ 5f^{14} \ 6d^{10} \ 7s^2 \ 7p^2$.
Since the valence shell configuration is $ns^2 np^2$ (where $n=7$),the element belongs to the $14^{th}$ group,which is the carbon family.

(D) To be isoelectronic,the total number of valence electrons must be the same.
For $IBr_{2}^{-}$: $7 (I) + 2 \times 7 (Br) + 1 (\text{charge}) = 22$ valence electrons.
For $XeF_{2}$: $8 (Xe) + 2 \times 7 (F) = 22$ valence electrons.
Both species have $22$ valence electrons,so they are isoelectronic.
Both $IBr_{2}^{-}$ and $XeF_{2}$ have $3$ lone pairs and $2$ bond pairs around the central atom,resulting in a linear geometry ($sp^3d$ hybridization),making them isostructural.

(C) To determine the species with a bond angle of $120^{\circ}$,we analyze the hybridization and geometry of each molecule:
$1$. $ClF_3$: The central atom $Cl$ has $7$ valence electrons. It forms $3$ bonds with $F$ atoms and has $2$ lone pairs. The hybridization is $sp^3d$,and the geometry is $T$-shaped with bond angles less than $90^{\circ}$.
$2$. $NCl_3$: The central atom $N$ has $5$ valence electrons. It forms $3$ bonds with $Cl$ atoms and has $1$ lone pair. The hybridization is $sp^3$,and the geometry is pyramidal with bond angles approximately $107^{\circ}$.
$3$. $BCl_3$: The central atom $B$ has $3$ valence electrons. It forms $3$ bonds with $Cl$ atoms and has no lone pairs. The hybridization is $sp^2$,and the geometry is trigonal planar with bond angles of $120^{\circ}$.
$4$. $PH_3$: The central atom $P$ has $5$ valence electrons. It forms $3$ bonds with $H$ atoms and has $1$ lone pair. The geometry is pyramidal with bond angles close to $93^{\circ}$.
Therefore,$BCl_3$ is the correct species.

(B) The bond order of a species can be determined by its total number of electrons using Molecular Orbital Theory.
For $CN^{-}$: Total electrons = $6 (C) + 7 (N) + 1 = 14$. Bond order = $\frac{1}{2}(10 - 4) = 3$.
For $CO$: Total electrons = $6 (C) + 8 (O) = 14$. Bond order = $\frac{1}{2}(10 - 4) = 3$.
Since both $CN^{-}$ and $CO$ have $14$ electrons,they have the same bond order of $3$.

(A) The condition for spontaneity is $\Delta G < 0$.
Given $\Delta G = \Delta H - T \Delta S$.
For the reaction to be spontaneous,$\Delta H - T \Delta S < 0$,which implies $T \Delta S > \Delta H$ or $T > \frac{\Delta H}{\Delta S}$.
Given $\Delta H = 35.5 \ kJ \ mol^{-1} = 35500 \ J \ mol^{-1}$ and $\Delta S = 83.6 \ J \ K^{-1} \ mol^{-1}$.
Calculating the equilibrium temperature: $T = \frac{35500}{83.6} \approx 424.64 \ K \approx 425 \ K$.
Since $\Delta H > 0$ and $\Delta S > 0$,the reaction is spontaneous at temperatures higher than the equilibrium temperature,i.e.,$T > 425 \ K$.

(B) For an irreversible expansion process,the work done is given by $W = -P_{ext} \Delta V$.
Given $P_{ext} = 2.5 \ atm$,$V_i = 2.50 \ L$,and $V_f = 4.50 \ L$.
$W = -2.5 \ atm \times (4.50 \ L - 2.50 \ L) = -2.5 \times 2.0 = -5.0 \ L \ atm$.
Since $1 \ L \ atm = 101.3 \ J$,$W = -5.0 \times 101.3 \ J = -505 \ J$.
As the container is well-insulated,the process is adiabatic,so $q = 0$.
According to the First Law of Thermodynamics,$\Delta U = q + W$.
Therefore,$\Delta U = 0 + (-505 \ J) = -505 \ J$.

(A) Given reactions:
$(1) \ N_2 + 3H_2 \rightleftharpoons 2NH_3 \quad K_1$
$(2) \ N_2 + O_2 \rightleftharpoons 2NO \quad K_2$
$(3) \ H_2 + \frac{1}{2} O_2 \rightleftharpoons H_2O \quad K_3$
Target reaction:
$(4) \ 2NH_3 + \frac{5}{2} O_2 \rightleftharpoons 2NO + 3H_2O \quad K$
To obtain equation $(4)$,we perform the operation: $(2) + 3 \times (3) - (1)$.
Applying the rules of equilibrium constants:
$K = \frac{K_2 \times (K_3)^3}{K_1} = \frac{K_2 K_3^3}{K_1}$

(C) The dissociation of $Ag_2C_2O_4$ is given by: $Ag_2C_2O_4(s) \rightleftharpoons 2Ag^{+}(aq) + C_2O_4^{2-}(aq)$.
Given $[Ag^{+}] = 2.2 \times 10^{-4} \ mol \ L^{-1}$.
From the stoichiometry,$[C_2O_4^{2-}] = \frac{1}{2} [Ag^{+}] = \frac{2.2 \times 10^{-4}}{2} = 1.1 \times 10^{-4} \ mol \ L^{-1}$.
The solubility product $K_{sp}$ is defined as: $K_{sp} = [Ag^{+}]^2 [C_2O_4^{2-}]$.
Substituting the values: $K_{sp} = (2.2 \times 10^{-4})^2 \times (1.1 \times 10^{-4})$.
$K_{sp} = (4.84 \times 10^{-8}) \times (1.1 \times 10^{-4}) = 5.324 \times 10^{-12} \approx 5.3 \times 10^{-12}$.

(D) The equilibrium reaction is: $SrCO_{3(s)} \rightleftharpoons SrO_{(s)} + CO_{2(g)}$
The equilibrium constant $K_p$ is given by: $K_p = P_{CO_2} = 1.6 \, atm$.
This means the maximum pressure $CO_2$ can exert at equilibrium is $1.6 \, atm$.
Using Boyle's Law $(P_1V_1 = P_2V_2)$ for the gas phase,where $P_1 = 0.4 \, atm$,$V_1 = 20 \, L$,and $P_2 = 1.6 \, atm$:
$0.4 \times 20 = 1.6 \times V_2$
$V_2 = \frac{0.4 \times 20}{1.6} = \frac{8}{1.6} = 5 \, L$.
Thus,the volume of the container when the pressure reaches $1.6 \, atm$ is $5 \, L$.

(C) Ionic mobility is inversely proportional to the size of the hydrated ion: $\text{Ionic mobility} \propto \frac{1}{\text{size of hydrated ion}}$.
In aqueous solution,the extent of hydration depends on the charge density of the ion.
Smaller alkali metal ions have higher charge density,leading to greater hydration.
The order of hydrated ionic size is: $Li^{+}_{(aq)} > Na^{+}_{(aq)} > K^{+}_{(aq)} > Rb^{+}_{(aq)}$.
Since $Li^{+}$ has the largest hydrated size,it experiences the most drag in an electric field.
Therefore,$Li^{+}$ has the lowest ionic mobility.

(C) $o$-Nitrophenol and $p$-nitrophenol can be separated by steam distillation.
$o$-Nitrophenol is steam volatile due to the presence of intramolecular hydrogen bonding,which reduces its intermolecular attraction.
In contrast,$p$-nitrophenol exhibits intermolecular hydrogen bonding,leading to the association of molecules,which makes it less volatile.
Therefore,$o$-nitrophenol distills over with steam,while $p$-nitrophenol remains in the distillation flask.

(D) $1$. Identify the principal functional group: The aldehyde $(-CHO)$ group has higher priority than the ketone $(>C=O)$ group. Thus,the parent chain must include the aldehyde carbon as $C-1$.
$2$. Number the chain: Start numbering from the aldehyde carbon $(C-1)$. The chain is $6$ carbons long $(hex)$.
$3$. Identify substituents and unsaturation: There is a methyl group at $C-2$,a ketone group at $C-3$,and a double bond starting at $C-4$.
$4$. Combine the parts: The name is $2-$methyl$-3-$oxohex$-4-$enal.

(C) An electrophile is an electron-deficient species that can be either neutral (e.g.,$BF_3$,$AlCl_3$) or positively charged (e.g.,$H^+$,$NO_2^+$).
It acts as a Lewis acid and forms a chemical bond by accepting a pair of electrons from a nucleophile.

(A) The acidity of hydrocarbons depends on the hybridization of the carbon atom attached to the acidic hydrogen.
The order of electronegativity of carbon is $sp > sp^2 > sp^3$.
$CH \equiv CH$ ($sp$ hybridized) is the most acidic.
$CH_3-C \equiv CH$ ($sp$ hybridized) is less acidic than $CH \equiv CH$ due to the $+I$ effect of the methyl group.
$CH_2=CH_2$ ($sp^2$ hybridized) is less acidic than alkynes.
$CH_3-CH_3$ ($sp^3$ hybridized) is the least acidic.
Therefore,the correct order is $CH \equiv CH > CH_3-C \equiv CH > CH_2=CH_2 > CH_3-CH_3$.

(C) In the presence of $HgSO_4$ and $H_2SO_4$,alkynes undergo hydration following Markovnikov's rule.
Propyne $(CH_3-C \equiv CH)$ reacts with water to form an unstable enol intermediate,$CH_3-C(OH)=CH_2$ $(A)$.
This enol then undergoes tautomerization to form the more stable keto form,propanone $CH_3-C(=O)-CH_3$ $(B)$.
$CH_3-C \equiv CH + H_2O \xrightarrow{Hg^{2+}/H^{+}} [CH_3-C(OH)=CH_2] \to CH_3-C(=O)-CH_3$

(C) Conformers are different spatial arrangements of atoms in a molecule that can be interconverted by rotation about a single bond.
During this rotation,the bond lengths and bond angles of the molecule remain unchanged because the covalent bonds are not broken or distorted.

(D) Micro-organisms present in the soil act as a sink for $CO$ (carbon monoxide) by oxidizing it to $CO_2$.

(B) When $HgCl_2$ is dissolved in a solution containing $I^{-}$ ions,it first forms $HgI_2$ precipitate,which then reacts with excess $I^{-}$ to form the soluble complex $[HgI_4]^{2-}$.
$HgCl_2 + 2I^{-} \longrightarrow HgI_2(s) + 2Cl^{-}$
$HgI_2(s) + 2I^{-} \longrightarrow [HgI_4]^{2-}(aq)$
When $I_2$ is dissolved in a solution containing $I^{-}$ ions,it forms the triiodide ion,which is water-soluble.
$I_2 + I^{-} \longrightarrow I_3^{-}(aq)$
Thus,the final species formed are $[HgI_4]^{2-}$ and $I_3^{-}$.

(A) Archaebacteria are a unique group of bacteria that live in some of the most harsh habitats such as extreme salty areas (halophiles),hot springs (thermoacidophiles),and marshy areas (methanogens).
Archaebacteria differ from other bacteria in having a different cell wall structure,which is responsible for their survival in extreme conditions.
Therefore,the correct option is $A$.

(C) The opening and closing of the stomatal aperture are primarily controlled by the turgor pressure of the guard cells.
When guard cells become turgid due to the influx of water,they expand.
The cell walls of guard cells contain cellulose microfibrils that are oriented radially rather than longitudinally.
This radial orientation makes the cell wall more extensible in the transverse direction,causing the guard cells to bulge outwards and pull the stomatal aperture open.
Therefore,the radial arrangement of cellulose microfibrils is the key structural feature facilitating stomatal opening.

(D) The $Glycocalyx$ is the outermost layer of the bacterial cell envelope.
It is composed of polysaccharides and sometimes proteins.
This layer provides a sticky or slimy character to the bacterial cell,which helps in attachment to surfaces and protects the bacterium from the host's immune system.

(C) Sulphonation is an electrophilic aromatic substitution reaction. The rate of this reaction depends on the electron density of the benzene ring.
Groups that donate electrons to the ring (activating groups) increase the electron density and make the ring more susceptible to electrophilic attack.
$1$. $CH_3$ group in Toluene is an electron-donating group ($+I$ and hyperconjugation effect),which activates the benzene ring.
$2$. $Cl$ in Chlorobenzene is deactivating due to its strong $-I$ effect.
$3$. $NO_2$ in Nitrobenzene is a strongly deactivating group due to its $-I$ and $-M$ effects.
$4$. Benzene has no substituent.
Therefore,Toluene is the most reactive towards electrophilic substitution among the given options.

(B) Entropy is a measure of the randomness or disorder of a system. For gaseous molecules,entropy generally increases with an increase in the complexity and size of the molecule,as there are more degrees of freedom for rotation and vibration.
Comparing the given molecules:
$H_2$ (diatomic,small)
$CH_4$ (polyatomic,$5$ atoms)
$C_2H_2$ (polyatomic,$4$ atoms)
$C_2H_6$ (polyatomic,$8$ atoms)
Since $C_2H_6$ has the largest number of atoms and the most complex structure among the choices,it possesses the highest degree of randomness and thus the largest entropy value.
The standard molar entropy values $(J \ K^{-1} \ mol^{-1})$ at $298 \ K$ are:

$C_2H_2$	$200.8$
$H_2$	$130.58$
$C_2H_6$	$229.5$
$CH_4$	$186.2$

(C) The bond angle in hydrides of group $16$ elements depends on the electronegativity of the central atom.
As we move down the group,the electronegativity of the central atom decreases,which causes the bond pair electrons to move further away from the central atom.
This results in a decrease in bond pair-bond pair repulsion,leading to a smaller bond angle.
Therefore,$H_2O$ has the largest bond angle $(104.5^\circ)$ compared to $H_2S$ $(92.1^\circ)$,$H_2Se$ $(91^\circ)$,and $H_2Te$ $(90^\circ)$.

(B) $1$. Calculate the molar mass of the hydrocarbon:
Given $42 \, mg = 0.042 \, g$ contains $3.01 \times 10^{20}$ molecules.
Number of moles $n = \frac{3.01 \times 10^{20}}{6.022 \times 10^{23}} \approx 0.5 \times 10^{-3} \, mol = 5 \times 10^{-4} \, mol$.
Molar mass $M = \frac{\text{mass}}{\text{moles}} = \frac{0.042 \, g}{5 \times 10^{-4} \, mol} = 84 \, g/mol$.
$2$. Determine the number of atoms of $C$ and $H$:
Mass of $C = 85.7 \% \text{ of } 84 = 0.857 \times 84 \approx 72 \, g$.
Number of $C$ atoms $= \frac{72 \, g}{12 \, g/mol} = 6$.
Mass of $H = 84 - 72 = 12 \, g$.
Number of $H$ atoms $= \frac{12 \, g}{1 \, g/mol} = 12$.
$3$. Conclusion:
The molecular formula is $C_6H_{12}$.

(D) $1$. The bond angle order $CH_{4} (109.5^{\circ}) > NH_{3} (107^{\circ}) > H_{2}O (104.5^{\circ}) > H_{2}S (92^{\circ})$ is correct due to the presence of lone pairs.
$2$. The bond order for $O_{2}$ species is $O_{2}^{+} (2.5) > O_{2} (2.0) > O_{2}^{-} (1.5) > O_{2}^{2-} (1.0)$,which is correct.
$3$. The strength of $H$-bond depends on electronegativity. $HF > H_{2}O > NH_{3}$ is correct,but $HCl$ does not form significant $H$-bonds,so the order is correct.
$4$. In $CO_{2}$,the structure $O=C=O$ is more stable than $\overline{O}-C \equiv \stackrel{+}{O}$ because $O=C=O$ has no formal charges and satisfies the octet rule for all atoms. Therefore,option $D$ is incorrect.

(B) Potassium superoxide $(KO_{2})$ reacts with carbon dioxide $(CO_{2})$ to produce potassium carbonate $(K_{2}CO_{3})$ and oxygen gas $(O_{2})$.
The balanced chemical equation is:
$2 KO_{2} + CO_{2} \rightarrow K_{2}CO_{3} + \frac{3}{2} O_{2}$
Due to this property,it is used in breathing apparatus for submarines and space travel.

(C) The total number of orbitals in a shell with principal quantum number $n$ is given by the formula $n^2$.
For $n = 4$,the total number of orbitals $= 4^2 = 16$.

(A) In thermodynamics,the change in Gibbs free energy $( \Delta G )$ under isothermal and reversible conditions is equal to the maximum non-expansion work (or useful work) done by the system. Mathematically,$ \Delta G = W_{\text{non-expansion}} $.

(C) The relationship between $K_p$ and $K_c$ is given by the formula: $K_p = K_c(RT)^{\Delta n_g}$.
For the reaction $N_{2(g)} + 3H_{2(g)} \rightleftharpoons 2NH_{3(g)}$,the change in the number of moles of gaseous species is $\Delta n_g = 2 - (1 + 3) = -2$.
Given $K_p = 5.8 \times 10^5$,$T = 298 \, K$,and $R = 0.08314 \, L \, bar \, K^{-1} \, mol^{-1}$.
Rearranging the formula to solve for $K_c$: $K_c = K_p(RT)^{-\Delta n_g} = K_p(RT)^2$.
Substituting the values: $K_c = (5.8 \times 10^5) \times (0.08314 \times 298)^2$.
$K_c = (5.8 \times 10^5) \times (24.77572)^2$.
$K_c = (5.8 \times 10^5) \times 613.836$.
$K_c \approx 3.56 \times 10^8$.

(A) Tetrachloroethene $(Cl_{2}C=CCl_{2})$ was previously used as a solvent for dry cleaning,but it is a suspected carcinogen and pollutes groundwater.
Liquid $CO_{2}$ is now used as an environmentally friendly alternative for dry cleaning,as it is non-toxic and does not pollute the environment.

(D) According to the $Le \ Chatelier$ principle,a catalyst provides an alternative pathway with lower activation energy for both the forward and backward reactions. It increases the rate of both reactions equally,thus reaching the equilibrium state faster without changing the position of the equilibrium or the equilibrium constant. Therefore,the presence of a catalyst does not affect the equilibrium.

(A) The relationship between $K_p$ and $K_c$ is given by $K_p = K_c(RT)^{\Delta n_{(g)}}$.
Therefore,$\frac{K_p}{K_c} = (RT)^{\Delta n_{(g)}}$.
For the reaction $CO_{(g)} + Cl_{2(g)} \rightleftharpoons COCl_{2(g)}$,the change in the number of moles of gaseous species is $\Delta n_{(g)} = n_{p(g)} - n_{r(g)} = 1 - (1 + 1) = 1 - 2 = -1$.
Substituting this value,we get $\frac{K_p}{K_c} = (RT)^{-1} = \frac{1}{RT}$.

(B) To determine if species are isostructural,we check their hybridization and geometry:
$1$. $ICl_4^-$ ($sp^3d^2$,square planar) and $XeF_4$ ($sp^3d^2$,square planar) are isostructural.
$2$. $ClO_3^-$ ($sp^3$,pyramidal) and $CO_3^{2-}$ ($sp^2$,trigonal planar) are not isostructural.
$3$. $IBr_2^-$ ($sp^3d$,linear) and $XeF_2$ ($sp^3d$,linear) are isostructural.
$4$. $BrO_3^-$ ($sp^3$,pyramidal) and $XeO_3$ ($sp^3$,pyramidal) are isostructural.
Thus,the pair $ClO_3^-$ and $CO_3^{2-}$ is not isostructural.

(C) In an ethane molecule $(CH_3-CH_3)$,the carbon-carbon bond is a single sigma $(\sigma)$ bond.
Because the sigma $(\sigma)$ bond is formed by the head-on overlap of orbitals,it possesses cylindrical symmetry along the internuclear axis.
This cylindrical symmetry allows for free rotation around the carbon-carbon single bond,leading to different conformational isomers.

(D) The inability of $ns^{2}$ electrons of the valence shell to participate in bonding when moving down the group in heavier $p$-block elements is known as the inert pair effect.
As a result,$Pb(II)$ is more stable than $Pb(IV)$,and $Sn(IV)$ is more stable than $Sn(II)$.
Since $Pb(IV)$ is unstable,it is easily reduced to $Pb(II)$,making $Pb(IV)$ a strong oxidizing agent.
Since $Sn(II)$ is unstable,it is easily oxidized to $Sn(IV)$,making $Sn(II)$ a strong reducing agent.

(D) In $Schottky$ defect,ions leave the lattice,so density decreases. This is a correct statement.
$FeO_{0.98}$ shows metal deficiency defect because the ratio of $Fe$ to $O$ is less than $1:1$. This is a correct statement.
$NaCl_{(s)}$ is an insulator,$Si$ is a semiconductor,$Ag$ is a conductor,and quartz is a piezoelectric crystal. This is a correct statement.
Frenkel defect occurs in ionic compounds where there is a large difference in the size of the cation and anion (cation is much smaller than the anion). Therefore,the statement that it is favoured when sizes are almost equal is incorrect.

(C) The molal depression constant,also known as the cryoscopic constant $(K_f)$,is a characteristic property of the solvent.
It depends solely on the nature of the solvent and is independent of the concentration of the solute or the molality of the solution.
Therefore,if the molality of the dilute solution is doubled,the value of $(K_f)$ will remain unchanged.

(A) Molarity is defined as the number of moles of solute dissolved per liter of solution $(M = \frac{n}{V(L)})$.
Since the volume of a solution changes with temperature due to thermal expansion or contraction,molarity is temperature-dependent.
In contrast,mole fraction,weight percentage,and molality are based on mass,which remains constant regardless of temperature changes.

(B) The cell reaction for the Daniell cell is $Zn(s) + Cu^{2+}(aq) \longrightarrow Zn^{2+}(aq) + Cu(s).$
Using the Nernst equation $E = E^{\circ} - \frac{0.059}{n} \log \frac{[Zn^{2+}]}{[Cu^{2+}]}$ where $n = 2.$
For $E_1$: $[Zn^{2+}] = 0.01 \, M$ and $[Cu^{2+}] = 1.0 \, M.$
$E_1 = E^{\circ} - \frac{0.059}{2} \log \frac{0.01}{1} = E^{\circ} - \frac{0.059}{2} \log(10^{-2}) = E^{\circ} + 0.059.$
For $E_2$: $[Zn^{2+}] = 1.0 \, M$ and $[Cu^{2+}] = 0.01 \, M.$
$E_2 = E^{\circ} - \frac{0.059}{2} \log \frac{1}{0.01} = E^{\circ} - \frac{0.059}{2} \log(10^{2}) = E^{\circ} - 0.059.$
Comparing the two,$E_1 = E^{\circ} + 0.059$ and $E_2 = E^{\circ} - 0.059.$
Therefore,$E_1 > E_2.$

(C) The rate of the reaction is determined by the slow step $(ii)$:
$r = k[X][Y_2]$ $---$ $(1)$
From the fast equilibrium step $(i)$,we have:
$K_{eq} = \frac{[X]^2}{[X_2]}$
$[X]^2 = K_{eq}[X_2]$
$[X] = K_{eq}^{1/2}[X_2]^{1/2}$ $---$ $(2)$
Substituting equation $(2)$ into equation $(1)$:
$r = k \cdot K_{eq}^{1/2}[X_2]^{1/2}[Y_2]^1$
$r = k'[X_2]^{1/2}[Y_2]^1$
The overall order of the reaction is the sum of the exponents of the concentration terms in the rate law:
Order $= 0.5 + 1 = 1.5$

(A) For a first order reaction,the integrated rate equation is given by:
$t = \frac{2.303}{k} \log \frac{[A]_0}{[A]_t}$
Given:
$k = 10^{-2} \ sec^{-1}$
$[A]_0 = 20 \ g$
$[A]_t = 5 \ g$
Substituting the values:
$t = \frac{2.303}{10^{-2}} \log \frac{20}{5}$
$t = 230.3 \times \log(4)$
$t = 230.3 \times 0.6021$
$t \approx 138.6 \ sec$

(A) catalyst decreases the activation energies of both the forward and backward reactions by the same amount.
Therefore,it speeds up both the forward and backward reactions at the same rate.
As a result,the equilibrium constant $(K_{eq})$ remains unaffected by the presence of a catalyst at a given temperature.
Thus,the statement that the equilibrium constant changes is incorrect.

(C) The extraction of silver and gold involves the cyanide process (MacArthur-Forrest process).
First,the ore is leached with $NaCN$ in the presence of air $(O_2)$ to form a soluble complex:
$Ag_{2}S + 4 NaCN \xrightarrow{O_{2}} 2 Na[Ag(CN)_{2}] + Na_{2}SO_{4}$
Then,the silver is recovered from the solution by displacement using a more electropositive metal like zinc $(Zn)$:
$2 Na[Ag(CN)_{2}] + Zn \longrightarrow Na_{2}[Zn(CN)_{4}] + 2Ag(\downarrow)$
Thus,silver is recovered by displacement with $Zn$.

(A) $A-iii$: $XX'$ has linear geometry as only two atoms are present.
$B-i$: $XX'_3$ has $T$-shaped geometry as $3$ bond pairs and $2$ lone pairs of electrons are present. The electron pair geometry is trigonal bipyramidal.
$C-iv$: $XX'_5$ has square-pyramidal geometry as $5$ bond pairs and $1$ lone pair of electrons are present. The electron pair geometry is octahedral.
$D-ii$: $XX'_7$ has pentagonal bipyramidal geometry as $7$ bond pairs and $0$ lone pairs of electrons are present.

(A) The tetrathionate ion $(S_4O_6^{2-})$ has a structure with an $S-S-S-S$ chain,thus it contains $S-S$ bonds.
The thiosulfate ion $(S_2O_3^{2-})$ has a structure where one sulfur atom is bonded to another sulfur atom via a double bond ($S=S$ bond),thus it contains an $S-S$ bond.
The disulfate ion $(S_2O_7^{2-})$ has an $S-O-S$ linkage and no $S-S$ bond.
The peroxodisulfate ion $(S_2O_8^{2-})$ has an $S-O-O-S$ linkage and no $S-S$ bond.
Therefore,both $S_4O_6^{2-}$ and $S_2O_3^{2-}$ contain $S-S$ bonds.

(A) Sulfur dioxide $(SO_2)$ acts as a strong reducing agent.
When $SO_2$ gas is passed through an acidified solution of potassium permanganate $(KMnO_4)$,it reduces the purple-colored $Mn^{7+}$ ion to the colorless $Mn^{2+}$ ion.
The balanced chemical equation is:
$2KMnO_4 + 5SO_2 + 2H_2O \rightarrow K_2SO_4 + 2MnSO_4 + 2H_2SO_4$
Since $MnSO_4$ is a colorless solution,the purple color of $KMnO_4$ disappears.

(B) The actinoids exhibit a greater range of oxidation states compared to lanthanoids because the $5f$,$6d$,and $7s$ energy levels are very close in energy.
This small energy gap allows electrons from these subshells to participate in bonding,making excitation easier.

(A) Grignard's reagent,represented as $RMgX$,contains a direct $\sigma$-bond between the carbon atom of the alkyl group and the magnesium metal atom.
Therefore,it is classified as a $\sigma$-bonded organometallic compound.
In contrast,ferrocene,cobaltocene,and ruthenocene are examples of $\pi$-bonded organometallic compounds (metallocenes).

(B) The reaction of coordination complexes with excess $AgNO_3$ precipitates the chloride ions present outside the coordination sphere (ionizable chlorides).
$1$. $[Co(NH_3)_6]Cl_3$ contains $3$ ionizable $Cl^-$ ions,yielding $3 \ mol \ AgCl$.
$2$. $[Co(NH_3)_5Cl]Cl_2$ contains $2$ ionizable $Cl^-$ ions,yielding $2 \ mol \ AgCl$.
$3$. $[Co(NH_3)_4Cl_2]Cl$ contains $1$ ionizable $Cl^-$ ion,yielding $1 \ mol \ AgCl$.
Thus,the correct order is $3 \ AgCl, 2 \ AgCl, 1 \ AgCl$.

(B) The energy of absorbed light $(\Delta_o)$ is inversely proportional to the wavelength of absorption $(\lambda)$: $\Delta_o \propto \frac{1}{\lambda}$.
Stronger ligands cause greater crystal field splitting $(\Delta_o)$,which corresponds to higher energy and shorter wavelength of absorption.
The spectrochemical series for the given ligands is: $H_2O < NH_3 < en$.
As the ligand strength increases,the splitting energy $(\Delta_o)$ increases,and therefore the absorbed wavelength $(\lambda)$ decreases.
Thus,the increasing order of wavelength is: $[Co(H_2O)_6]^{3+} < [Co(NH_3)_6]^{3+} < [Co(en)_3]^{3+}$.

(B) $1$. The oxidation state of $Mn$ in $[Mn(CN)_6]^{3-}$ is calculated as: $x + 6(-1) = -3 \implies x = +3$.
$2$. The electronic configuration of $Mn$ $(Z=25)$ is $[Ar] 3d^5 4s^2$. Thus,$Mn^{3+}$ is $[Ar] 3d^4$.
$3$. $CN^-$ is a Strong Field Ligand $(SFL)$,which causes pairing of electrons in the $3d$ orbitals.
$4$. After pairing,two $3d$ orbitals,one $4s$ orbital,and three $4p$ orbitals are available for hybridization,resulting in $d^2sp^3$ hybridization.
$5$. Since the coordination number is $6$,the geometry is octahedral.

(A) The reaction of $3\text{-bromoanisole}$ with $NaNH_2$ proceeds via the formation of a benzyne intermediate.
$1$. The strong base $NH_2^-$ abstracts the ortho-hydrogen relative to the $Br$ atom,leading to the elimination of $Br^-$ and the formation of a benzyne intermediate.
$2$. The nucleophile $NH_2^-$ can then attack the benzyne intermediate at two different positions.
$3$. Attack at the meta-position (relative to the $OCH_3$ group) leads to a more stable carbanion intermediate because the negative charge is closer to the electron-withdrawing $OCH_3$ group.
$4$. Protonation of this intermediate yields $3\text{-aminoanisole}$ as the major product.
$5$. Since the incoming nucleophile ends up on the same carbon atom where the leaving group was originally present,this is not a cine substitution reaction; it is an elimination-addition reaction.

(B) When phenyl methyl ether $(Ph-O-CH_3)$ is heated with $HI$,the reaction proceeds via the protonation of the ether oxygen followed by a nucleophilic attack by the iodide ion $(I^-)$ on the less sterically hindered methyl group via an $S_N2$ mechanism.
The reaction is as follows:
$Ph-O-CH_3 + HI \rightarrow Ph-OH + CH_3I$
The products formed are phenol $(Ph-OH)$ and methyl iodide $(CH_3I)$. Among the given options,phenol is the correct product.

(C) The acidity of phenols depends on the stability of the phenoxide ion formed after the loss of a proton. Electron-withdrawing groups ($-I$ and $-M$ effects) increase acidity by stabilizing the phenoxide ion,while electron-donating groups (like $-CH_3$) decrease acidity.
$1$. $2,4,6$-Trinitrophenol has three $-NO_2$ groups,which exert strong $-I$ and $-M$ effects,making it the most acidic.
$2$. $p$-Nitrophenol has one $-NO_2$ group,which is acidic but less than picric acid.
$3$. Phenol has no substituent.
$4$. $p$-Cresol has a $-CH_3$ group,which is electron-donating ($+I$ and hyperconjugation),making it the least acidic.
Therefore,the order of acidity is: $2,4,6$-Trinitrophenol > $p$-Nitrophenol > Phenol > $p$-Cresol.

(B) $X$ is $C_2H_6O$. Since it reacts with $Cu$ at $573 \ K$ to give $A$,and $A$ gives a silver mirror test,$X$ must be a primary alcohol and $A$ must be an aldehyde.
$X = CH_3CH_2OH$ (Ethanol).
$A = CH_3CHO$ (Ethanal).
$2CH_3CHO \xrightarrow{^-OH, \Delta} CH_3CH=CHCHO$ ($Y$,But$-2-$enal) via Aldol condensation.
$CH_3CHO + NH_2NHCONH_2 \rightarrow CH_3CH=NNHCONH_2$ ($Z$,Semicarbazone).
Thus,$A = \text{Ethanal}, X = \text{Ethanol}, Y = \text{But-2-enal}, Z = \text{Semicarbazone}$.

(A) Cyclohexanone undergoes self-aldol condensation in the presence of a base $(OH^-)$ to form a $\beta$-hydroxy ketone intermediate. Upon heating $(\Delta)$,this intermediate undergoes dehydration (elimination of $H_2O$) to form an $\alpha,\beta$-unsaturated ketone. The final product is $2$-cyclohexylidenecyclohexanone.

(C) The basic strength of aromatic amines depends on the electron density on the nitrogen atom.
Electron-donating groups (like $-CH_3$) increase the electron density on the nitrogen atom,thereby increasing the basic strength.
Electron-withdrawing groups (like $-NO_2$) decrease the electron density on the nitrogen atom,thereby decreasing the basic strength.
In the given compounds:
$(I)$ is Aniline.
$(II)$ is $p$-Nitroaniline,where $-NO_2$ is a strong electron-withdrawing group,making it the least basic.
$(III)$ is $p$-Toluidine,where $-CH_3$ is an electron-donating group,making it the most basic.
Therefore,the increasing order of basic strength is: $II < I < III$.

(A) The conversion of acetamide $(CH_3CONH_2)$ to methanamine $(CH_3NH_2)$ involves the removal of a carbonyl group,which is characteristic of the Hoffmann bromamide degradation reaction.
The chemical equation is:
$CH_3CONH_2 + Br_2 + 4KOH \rightarrow CH_3NH_2 + K_2CO_3 + 2KBr + 2H_2O$
This reaction is known as the Hoffmann hypobromamide reaction.

(C) Denaturation is a process in which the secondary and tertiary structures of proteins are destroyed due to physical or chemical changes,such as change in $pH$ or temperature. During this process,the globules unfold and the helix becomes uncoiled,causing the protein to lose its biological activity. Therefore,the statement that denaturation makes proteins more active is incorrect.

(A) mixture of chloroxylenol and terpineol is commonly known as Dettol,which acts as an antiseptic.

(B) In coordination complexes,the electron distribution is determined by the competition between the crystal field splitting energy $(\Delta_{0})$ and the pairing energy $(P).$
If $\Delta_{0} > P$,the energy required to pair electrons is less than the energy required to promote an electron to the higher energy $e_{g}$ orbitals. This leads to the formation of low spin complexes.
If $\Delta_{0} < P$,the energy required to pair electrons is greater than the energy required to promote an electron to the higher energy $e_{g}$ orbitals. This leads to the formation of high spin complexes.
Therefore,the condition that favours the formation of high spin complexes is $\Delta_{0} < P$.

(B) Let $A$ be toluene and $B$ be benzene.
Given: $X_{A} = 0.50$,$X_{B} = 1 - 0.50 = 0.50$.
$P_{A}^{0} = 37.0 \ torr$,$P_{B}^{0} = 119 \ torr$.
Total vapour pressure $P_{total} = P_{A}^{0} X_{A} + P_{B}^{0} X_{B} = (37.0 \times 0.50) + (119 \times 0.50) = 18.5 + 59.5 = 78.0 \ torr$.
According to Dalton's Law,the mole fraction of toluene in the vapour phase $(y_{A})$ is given by $y_{A} = \frac{P_{A}}{P_{total}}$.
$P_{A} = P_{A}^{0} X_{A} = 37.0 \times 0.50 = 18.5 \ torr$.
$y_{A} = \frac{18.5}{78.0} \approx 0.237$.

(C) The reaction of benzene diazonium chloride $(ArN_2Cl)$ with copper powder $(Cu)$ in the presence of hydrochloric acid $(HCl)$ to form aryl chloride $(ArCl)$ is known as the Gattermann reaction.
Note: If $Cu_2Cl_2 / HCl$ were used instead of $Cu / HCl$,it would be the Sandmeyer reaction.

(C) In $[Cu(NH_3)_4]^{2+}$,the $Cu^{2+}$ ion undergoes $dsp^2$ hybridization,which results in a square planar geometry.
In contrast,the ions $NH_4^+$,$BF_4^-$,and $[NiCl_4]^{2-}$ all exhibit $sp^3$ hybridization,which corresponds to a tetrahedral geometry.

(B) When phenol is oxidized with chromic acid $(H_2CrO_4)$,it undergoes oxidation to form $p$-benzoquinone.
$p$-Benzoquinone is a cyclic conjugated diketone.
Therefore,the correct description of the product is a conjugated diketone.

(C) The reaction involves two steps:
$1$. The first step uses $Br_2/NaOH$,which is the condition for the Hofmann bromamide degradation reaction. This reaction converts an amide group $(-CONH_2)$ into a primary amine group $(-NH_2)$.
$2$. The starting material is methyl $2-(carbamoylmethyl)benzoate$. The $-CONH_2$ group is converted to $-CH_2NH_2$.
$3$. The intermediate formed is methyl $2-(aminomethyl)benzoate$.
$4$. In the second step,heating $(\Delta)$ causes an intramolecular cyclization (nucleophilic acyl substitution) where the $-NH_2$ group attacks the ester carbonyl group $(-COOCH_3)$,resulting in the loss of a methanol molecule $(-CH_3OH)$ and the formation of a cyclic amide (lactam).
$5$. The final product $(P)$ is $isoindolin-1-one$.

(A) For a general reaction $XA + YB \rightarrow ZC$,the rate of reaction is given by:
Rate $= -\frac{1}{X} \frac{d[A]}{dt} = -\frac{1}{Y} \frac{d[B]}{dt} = \frac{1}{Z} \frac{d[C]}{dt}$
Given that $\frac{-d[A]}{dt} = \frac{-d[B]}{dt} = \frac{d[C]}{dt}$,we can equate these to the rate expression:
$\frac{1}{X} (\text{Rate}) = \frac{1}{Y} (\text{Rate}) = \frac{1}{Z} (\text{Rate})$
This implies that $X = Y = Z$. Since the coefficients in a balanced chemical equation must be equal for the rates of disappearance and appearance to be equal,the simplest integer ratio is $X = Y = Z = 1$. However,looking at the options provided,the statement $X = Y = Z$ is implied by the equality of the rates. Given the options,the most consistent choice is that the coefficients are equal.

(A) The given reaction is a nucleophilic acyl substitution reaction. The rate of this reaction depends on the leaving group ability of $Z^-$. $A$ better leaving group makes the reaction faster. The leaving group ability is inversely proportional to the basicity of the group. Among the given options,$Cl^-$ is the weakest base and thus the best leaving group. Therefore,the reaction is fastest when $Z = Cl$.

(C) Let the number of oxide ions $(O^{2-})$ in the $CCP$ lattice be $N = 4$.
Number of octahedral voids $(OV)$ = $N = 4$.
Number of tetrahedral voids $(TV)$ = $2N = 8$.
Given that $Mg^{2+}$ ions occupy $\frac{1}{8}^{th}$ of the tetrahedral voids:
Number of $Mg^{2+} = \frac{1}{8} \times 8 = 1$.
Given that $Al^{3+}$ ions occupy $\frac{1}{2}$ of the octahedral voids:
Number of $Al^{3+} = \frac{1}{2} \times 4 = 2$.
Thus,the ratio of $Mg : Al : O$ is $1 : 2 : 4$.
The formula of the compound is $MgAl_2O_4$.

(C) The reaction of alcohols with conc. $HCl$ and anhydrous $ZnCl_{2}$ is known as the Lucas test.
This reaction proceeds via an $S_{N}1$ mechanism,where the rate-determining step is the formation of a carbocation.
Therefore,the reactivity of alcohols follows the order: $3^{\circ} > 2^{\circ} > 1^{\circ}$.
Among the given options:
$A$. Butan$-1-$ol is a $1^{\circ}$ alcohol.
$B$. Butan$-2-$ol is a $2^{\circ}$ alcohol.
$C$. $2-$methylpropan$-2-$ol is a $3^{\circ}$ alcohol.
$D$. $2-$methylpropan$-1-$ol is a $1^{\circ}$ alcohol.
Since $3^{\circ}$ alcohols form the most stable carbocation,$2-$methylpropan$-2-$ol reacts fastest.

(C) The atomic number of $Co$ is $27$. The electronic configuration of $Co^{2+}$ is $[Ar] 3d^7$.
Since $H_2O$ is a weak field ligand,it does not cause pairing of electrons in the $3d$ orbitals.
Thus,the $3d$ orbitals have $3$ unpaired electrons.
The hybridization involves the use of one $4s$,three $4p$,and two $4d$ orbitals to accommodate the six ligands.
Therefore,the hybridization is $sp^3d^2$.

(B) The central metal ion is $Mn^{2+}$.
The atomic number of $Mn$ is $25$, so the electronic configuration of $Mn^{2+}$ is $[Ar] 3d^5$.
In the tetrahedral complex $[MnBr_4]^{2-}$, $Br^-$ is a weak field ligand, so the electrons remain unpaired.
Number of unpaired electrons $(n)$ = $5$.
The spin-only magnetic moment $(\mu)$ is calculated using the formula $\mu = \sqrt{n(n+2)} \ BM$.
$\mu = \sqrt{5(5+2)} = \sqrt{35} \approx 5.92 \ BM$.

(B) According to the Hardy-Schulze rule,the flocculating power of an ion increases with the increase in the magnitude of the charge on the ion.
For a positive sol,the coagulating ions must be negative.
The correct order of flocculating power for a positive sol is $[Fe(CN)_{6}]^{4-} > PO_{4}^{3-} > SO_{4}^{2-} > Cl^{-}$.
Therefore,the statement in option $B$ is incorrect as it presents the reverse order.

(A) $Aspartame$ is an artificial sweetener that is unstable at cooking temperatures. It decomposes upon heating,which limits its use to cold foods and soft drinks.

(D) $1$. Among halogens,the bond dissociation enthalpy order is $Cl_{2} > Br_{2} > F_{2} > I_{2}$. Thus,$Cl_{2}$ has the highest bond dissociation enthalpy.
$2$. Among hydrogen halides,the bond dissociation enthalpy decreases down the group due to an increase in bond length: $HF > HCl > HBr > HI$. Thus,$HI$ has the lowest bond dissociation enthalpy.
$3$. Therefore,the correct pair is $(Cl_{2}, HI)$.

(A) The correct matching is as follows:
$(a)$ Nylon-$6$ is formed from $(v)$ Caprolactam.
$(b)$ Dacron is formed from $(i)$ Ethylene glycol and terephthalic acid.
$(c)$ Glyptal is formed from $(iii)$ Ethylene glycol and phthalic acid.
$(d)$ Novolac is formed from $(iv)$ Phenol and formaldehyde.
Therefore,the correct sequence is $(a)-(v), (b)-(i), (c)-(iii), (d)-(iv)$.

(A) The electronic configuration of $Ce$ $(Z=58)$ is $[Xe] 4f^1 5d^1 6s^2$.
By losing four electrons,$Ce$ attains the stable noble gas configuration of Xenon $([Xe])$,resulting in the $+4$ oxidation state.
Therefore,$Ce^{4+}$ is a stable ion.

(B) $1$. The first step is the Gattermann-Koch reaction where benzene reacts with $CO$ and $HCl$ in the presence of anhydrous $AlCl_3/CuCl$ to form $A$,which is benzaldehyde $(C_6H_5CHO)$.
$2$. The second step is the Claisen-Schmidt condensation reaction between benzaldehyde $(A)$ and a ketone $(B)$ in the presence of a base $(OH^-)$ to form the product,which is chalcone $(C_6H_5-CH=CH-CO-C_6H_5)$.
$3$. For the formation of chalcone,the ketone $B$ must be acetophenone $(C_6H_5COCH_3)$.
$4$. Therefore,the substance $B$ is acetophenone.

(B) To determine if a complex is diamagnetic,we check for the presence of unpaired electrons in the central metal ion.
$A$. In $[Ti(en)_{2}(NH_{3})_{2}]^{4+}$,$Ti$ is in the $+4$ oxidation state ($d^0$ configuration). It has no unpaired electrons,so it is diamagnetic.
$B$. In $[Cr(NH_{3})_{6}]^{3+}$,$Cr$ is in the $+3$ oxidation state ($d^3$ configuration). The $d$-orbitals have $3$ unpaired electrons $(t_{2g}^3)$,making it paramagnetic.
$C$. In $[Zn(NH_{3})_{6}]^{2+}$,$Zn$ is in the $+2$ oxidation state ($d^{10}$ configuration). It has no unpaired electrons,so it is diamagnetic.
$D$. In $[Sc(H_{2}O)_{3}(NH_{3})_{3}]^{3+}$,$Sc$ is in the $+3$ oxidation state ($d^0$ configuration). It has no unpaired electrons,so it is diamagnetic.
Therefore,$[Cr(NH_{3})_{6}]^{3+}$ is not diamagnetic.

(C) Fluorine $(F_2)$ is the strongest oxidizing agent among the halogens due to its high electronegativity and high standard reduction potential. It reacts vigorously with water to oxidize it to oxygen $(O_2)$ according to the following reaction:
$2F_2(g) + 2H_2O(l) \rightarrow 4HF(aq) + O_2(g)$

(B) According to Kohlrausch's law of independent migration of ions,the molar conductivity at infinite dilution for $AgCl$ can be calculated as:
$\Lambda_{m}^{\infty}(AgCl) = \Lambda_{m}^{\infty}(Ag^{+}) + \Lambda_{m}^{\infty}(Cl^{-})$
Using the given values:
$\Lambda_{m}^{\infty}(AgCl) = \Lambda_{m}^{\infty}(AgNO_{3}) + \Lambda_{m}^{\infty}(KCl) - \Lambda_{m}^{\infty}(KNO_{3})$
Substituting the values:
$= 133.4 + 149.9 - 144.9$
$= 283.3 - 144.9$
$= 138.4 \, S \, cm^{2} \, mol^{-1}$
Rounding to the nearest integer,the value is $138 \, S \, cm^{2} \, mol^{-1}$.

(C) Let the rate law be $Rate = k[Cl_2]^x[NO]^y$.
Given that doubling $[Cl_2]$ increases the rate by a factor of $2$,we have $2^x = 2$,which implies $x = 1$.
When both concentrations are doubled,the rate increases by a factor of $8$,so $2^x \times 2^y = 8$.
Substituting $x = 1$,we get $2^1 \times 2^y = 8$,which simplifies to $2^y = 4$.
Therefore,$y = 2$.
The order of the reaction with respect to $NO$ is $2$.

(B) The structure of hypophosphorous acid $(H_3PO_2)$ contains one $P=O$ bond,two $P-H$ bonds,and one $P-OH$ bond.
The presence of two $P-H$ bonds is responsible for its strong reducing character,as these bonds can easily release hydrogen atoms.

(C) The strength of an acid is determined by the stability of its conjugate base. Electron-withdrawing groups $(EWG)$ stabilize the carboxylate anion $(RCOO^-)$ through the inductive effect ($-I$ effect),thereby increasing the acidity of the parent carboxylic acid.
Comparing the substituents:
$1$. $NCCH_2-$: Contains a cyano group $(-CN)$,which has a $-I$ effect.
$2$. $O_2NCH_2-$: Contains a nitro group $(-NO_2)$,which has a strong $-I$ effect.
$3$. $F_3C-$: Contains three fluorine atoms. Fluorine is the most electronegative element,and the $-I$ effect of three fluorine atoms is significantly stronger than that of a single $-NO_2$ or $-CN$ group.
$4$. $Cl_3C-$: Contains three chlorine atoms. Chlorine is less electronegative than fluorine,so the $-I$ effect of $Cl_3C-$ is weaker than that of $F_3C-$.
Since fluorine is the most electronegative,$F_3CCOOH$ exhibits the strongest $-I$ effect,making its conjugate base the most stable and the acid the strongest among the given options.