The yellow color in potassium chromate solution is due to the chromate ion (\(CrO_4^{2-}\)).


The orange color is due to the dichromate ion (\(Cr_2O_7^{2-}\)).


These two ions exist in an equilibrium that is dependent on the pH of the solution.


The equilibrium reaction is: \(2CrO_4^{2-} (yellow) + 2H^+ \rightleftharpoons Cr_2O_7^{2-} (orange) + H_2O\).


When the solution turned from yellow to orange, it indicates that the equilibrium shifted to the right.


According to Le Chatelier's principle, this shift is caused by an increase in the concentration of \(H^+\) ions, which means the medium became acidic.


Among the given options, HCl (hydrochloric acid) is a strong acid that provides \(H^+\) ions.


Therefore, the most likely substance present in the test tube was HCl solution.
Quick Tip: Remember that the chromate (\(CrO_4^{2-}\)) and dichromate (\(Cr_2O_7^{2-}\)) equilibrium is a classic example of a pH-dependent color change. Acidic conditions favor the orange dichromate ion, while basic (alkaline) conditions favor the yellow chromate ion.

A catalyst is a substance that increases the rate of a chemical reaction without being consumed in the process.


It achieves this by providing an alternative reaction pathway with a lower activation energy (\(E_a\)).


Activation energy is the minimum amount of energy required for reactants to transform into products.


By lowering this energy barrier, more reactant molecules can successfully form products in a given amount of time, thus increasing the reaction rate.


A catalyst does not alter the thermodynamic quantities of a reaction, such as the equilibrium constant (\(K\)), enthalpy of reaction (\(\Delta H\)), or Gibbs free energy (\(\Delta G\)).


Therefore, the role of a catalyst is to change the activation energy of the reaction.
Quick Tip: A catalyst accelerates both the forward and reverse reactions by the same factor. This is why it doesn't shift the position of equilibrium but simply helps in attaining it faster.

Let's match each cell type with its common application:


i. Lead storage cell: This is a secondary (rechargeable) battery used in high-current applications like automobiles and power backup systems (inverters). So, i matches with d.


ii. Mercury cell: This primary cell provides a very constant voltage, making it ideal for small devices like wrist watches and hearing aids. So, ii matches with c.


iii. Dry cell (Leclanché cell): This is a common primary cell used in low-drain devices like wall clocks, flashlights, and radios. So, iii matches with a.


iv. Fuel cell: This cell generates electricity from a fuel. \(H_2-O_2\) fuel cells were famously used in the Apollo Space Programme. So, iv matches with b.


The correct combination is: i-d, ii-c, iii-a, iv-b.


This corresponds to option (B).
Quick Tip: Differentiate between cell types: Primary cells (e.g., Dry cell, Mercury cell) are non-rechargeable. Secondary cells (e.g., Lead storage cell) are rechargeable. Fuel cells operate continuously as long as fuel is supplied.

The reaction is the conversion of ethanol (\(CH_3CH_2OH\)), a primary alcohol, to ethanal (\(CH_3CHO\)), an aldehyde.


This transformation is an oxidation.


Let's analyze the reagents:


(A) Catalytic hydrogenation (\(H_2\)/Pd) is a reduction process.


(B) \(LiAlH_4\) (Lithium aluminium hydride) is a strong reducing agent.


(C) PCC (Pyridinium chlorochromate) is a mild oxidizing agent that selectively oxidizes primary alcohols to aldehydes.


(D) \(KMnO_4\) (Potassium permanganate) is a strong oxidizing agent that would oxidize a primary alcohol all the way to a carboxylic acid (\(CH_3COOH\)).


Therefore, to stop the oxidation at the aldehyde stage, the mild reagent PCC must be used.
Quick Tip: The choice of oxidizing agent for primary alcohols is crucial. Use mild agents like PCC or PDC to get aldehydes. Use strong agents like acidified \(KMnO_4\) or \(K_2Cr_2O_7\) to get carboxylic acids.

This compound is a tertiary amine.


To name it using IUPAC rules, we first identify the longest carbon chain attached to the nitrogen atom.


The alkyl groups attached to nitrogen are ethyl, methyl, and propyl.


The longest chain is the propyl group (3 carbons), so the parent name is propanamine.


Since the nitrogen is attached to carbon-1 of the propyl chain, the parent is propan-1-amine.


The other two groups, ethyl and methyl, are treated as substituents on the nitrogen atom.


We use the prefix 'N-' to denote their position and list them alphabetically.


So, the name becomes N-ethyl-N-methylpropan-1-amine.
Quick Tip: When naming tertiary amines, always select the longest alkyl chain attached to the nitrogen as the parent chain. The other alkyl groups are named as N-substituents and are listed in alphabetical order.

The graph shows a plot of the concentration of reactant, [R], versus time, t.


The plot is a straight line with a negative slope.


For a zero-order reaction, the integrated rate law is \([R] = -kt + [R]_0\).


This equation is in the form of a straight line, \(y = mx + c\).


Here, \(y = [R]\), \(x = t\), the slope \(m = -k\), and the y-intercept \(c = [R]_0\).


The graph perfectly matches the linear relationship expected for a zero-order reaction.
Quick Tip: To determine the order of a reaction from a graph, remember the linear plots: Zero Order: [R] vs. time gives a straight line with slope = -k. First Order: ln[R] vs. time gives a straight line with slope = -k. Second Order: 1/[R] vs. time gives a straight line with slope = k.

The nitrite ion (\(NO_2^-\)) is an ambidentate nucleophile.


Silver nitrite (\(AgNO_2\)) has a significant covalent character in the Ag-O bond.


This makes the lone pair on the nitrogen atom more available for nucleophilic attack.


The nitrogen atom attacks the electrophilic carbon of ethyl bromide, forming a C-N bond.


The reaction is: \(CH_3CH_2Br + AgNO_2 \rightarrow CH_3CH_2NO_2 + AgBr\).


The product, \(CH_3CH_2NO_2\), is nitroethane.


In contrast, an ionic nitrite like \(KNO_2\) would favor attack through the oxygen atom, forming ethyl nitrite (\(CH_3CH_2ONO\)).
Quick Tip: Remember the role of the cation with ambidentate nucleophiles. Covalent nitrites (\(AgNO_2\)) favor N-attack leading to nitroalkanes. Ionic nitrites (\(KNO_2\)) favor O-attack leading to alkyl nitrites.

The depression in freezing point (\(\Delta T_f\)) is a colligative property given by \(\Delta T_f = i \cdot K_f \cdot m\).


The highest freezing point corresponds to the smallest depression in freezing point (\(\Delta T_f\)).


This will be the solution with the lowest effective concentration of particles, i.e., the smallest van 't Hoff factor (\(i\)).


Let's find the value of \(i\) for each solute:


(A) \(KCl \rightarrow K^+ + Cl^-\) ; \(i = 2\).


(B) \(Na_2SO_4 \rightarrow 2Na^+ + SO_4^{2-}\) ; \(i = 3\).


(C) Glucose (\(C_6H_{12}O_6\)) is a non-electrolyte and does not dissociate; \(i = 1\).


(D) \(AlCl_3 \rightarrow Al^{3+} + 3Cl^-\) ; \(i = 4\).


Since Glucose has the lowest van 't Hoff factor (\(i=1\)), it will cause the least depression in freezing point, resulting in the highest freezing point.
Quick Tip: Be careful with the wording. "Highest freezing point" means "smallest depression in freezing point". This corresponds to the solute that produces the fewest particles in solution, determined by the van 't Hoff factor (\(i\)).

The Cannizzaro reaction is a disproportionation reaction given by aldehydes that do not have any \(\alpha\)-hydrogen atoms.


An \(\alpha\)-hydrogen is a hydrogen atom attached to the \(\alpha\)-carbon (the carbon adjacent to the carbonyl group).


Let's examine each option:


(A) 2-methylpropanal: The \(\alpha\)-carbon has one hydrogen atom. It will not undergo Cannizzaro.


(B) 2,2-dimethylpropanal: The \(\alpha\)-carbon has zero \(\alpha\)-hydrogen atoms. It will undergo the Cannizzaro reaction.


(C) Propanal: The \(\alpha\)-carbon has two hydrogen atoms. It will not undergo Cannizzaro.


(D) 2,3-dimethylbutanal: The \(\alpha\)-carbon has one hydrogen atom. It will not undergo Cannizzaro.


Therefore, only 2,2-dimethylpropanal lacks an \(\alpha\)-hydrogen and will undergo the Cannizzaro reaction.
Quick Tip: The key condition for the Cannizzaro reaction is the absence of \(\alpha\)-hydrogen atoms in the aldehyde. Aldehydes with one or more \(\alpha\)-hydrogens will undergo aldol condensation in the presence of a base.

The color of transition metal ions is typically due to d-d electronic transitions, which require partially filled d-orbitals.


Let's determine the electronic configuration of each ion:


I. \(Cu^+\) (Z=29): Configuration is [Ar] \(3d^{10}\). The d-orbital is completely filled. Hence, colorless.


II. \(Ti^{4+}\) (Z=22): Configuration is [Ar] \(3d^0\). The d-orbital is empty. Hence, colorless.


III. \(Co^{2+}\) (Z=27): Configuration is [Ar] \(3d^7\). The d-orbital is partially filled. Hence, coloured (pink).


IV. \(Fe^{2+}\) (Z=26): Configuration is [Ar] \(3d^6\). The d-orbital is partially filled. Hence, coloured (pale green).


Therefore, the group in which both ions are coloured is III (\(Co^{2+}\)) and IV (\(Fe^{2+}\)).
Quick Tip: An aqueous transition metal ion is typically colored if its d-orbital is partially filled (\(d^1\) to \(d^9\)). Ions with empty (\(d^0\)) or completely filled (\(d^{10}\)) d-orbitals are generally colorless.

A molecule is chiral if it contains a chiral center (a carbon atom bonded to four different groups).


Let's examine the options:


(A) 1-chloropropane (\(CH_2Cl-CH_2-CH_3\)): No chiral center.


(B) 2-chloropropane (\(CH_3-CHCl-CH_3\)): The central carbon is bonded to two identical methyl groups. Not chiral.


(C) 1-chlorobutane (\(CH_2Cl-CH_2-CH_2-CH_3\)): No chiral center.


(D) 2-chlorobutane (\(CH_3-CHCl-CH_2-CH_3\)): Carbon-2 is bonded to four different groups: a hydrogen atom (-H), a chlorine atom (-Cl), a methyl group (\(-CH_3\)), and an ethyl group (\(-CH_2CH_3\)).


Since Carbon-2 in 2-chlorobutane is a chiral center, the molecule is chiral.
Quick Tip: To quickly find a chiral carbon, look for a carbon atom with four single bonds and check if all four attached atoms or groups are different from each other.

We need to distinguish between propanal (an aldehyde) and propanoic acid (a carboxylic acid).


(A) Sodium bicarbonate test: Carboxylic acids react with sodium bicarbonate (\(NaHCO_3\)) to produce brisk effervescence of carbon dioxide gas.
\(CH_3CH_2COOH + NaHCO_3 \rightarrow CH_3CH_2COONa + H_2O + CO_2 \uparrow\)

Aldehydes do not give this test. This can distinguish between them.


(B) Hinsberg test: This test is used for amines.


(C) Iodoform test: This test is given by compounds containing a \(CH_3-C=O\) group. Neither propanal nor propanoic acid has this structure.


(D) Lucas test: This test is used for alcohols.


Therefore, the sodium bicarbonate test is the correct choice.
Quick Tip: The sodium bicarbonate test is a reliable and simple chemical test to confirm the presence of the carboxylic acid functional group. The evolution of \(CO_2\) gas is a positive result.

For a given alkyl group (R), the boiling point of alkyl halides depends on intermolecular van der Waals forces.


The magnitude of van der Waals forces increases with the size and mass of the halogen atom.


The order of size and mass of halogens is I > Br > Cl > F.


This leads to stronger van der Waals forces in the same order: RI > RBr > RCl > RF. Thus, Reason (R) is true.


Stronger intermolecular forces require more energy to overcome, resulting in higher boiling points.


Therefore, the boiling points also decrease in the order: RI > RBr > RCl > RF. Thus, Assertion (A) is true.


The reason correctly explains the assertion.
Quick Tip: For isomeric compounds or compounds with similar polarity, the boiling point generally increases with increasing molar mass due to stronger London dispersion forces (a type of van der Waals force).

Using a direct current (DC) source to measure the resistance of an ionic solution causes electrolysis.


Electrolysis involves chemical reactions at the electrodes, which alters the concentration of ions in the solution.


A change in concentration changes the resistance, making the measurement inaccurate. Thus, Reason (R) is true.


To prevent this, an alternating current (AC) source is used. The rapid reversal of polarity prevents any net electrolysis, keeping the concentration constant.


This allows for an accurate measurement of resistance. Thus, Assertion (A) is true.


The reason is the correct explanation for the assertion.
Quick Tip: In conductivity experiments, use AC to measure the intrinsic property of resistance/conductance without changing the solution. Use DC to intentionally drive a chemical change (electrolysis).

The dissolution of most gases in liquids is an exothermic process.


According to Le Chatelier's principle, increasing the temperature shifts the equilibrium in the reverse (endothermic) direction, decreasing the solubility of the gas. Therefore, Reason (R) is true.


Henry's law is given by \(p = K_H \cdot x\).


This can be rearranged to \(K_H = p/x\), where x is the mole fraction (solubility).


This shows that \(K_H\) is inversely proportional to solubility (\(x\)).


Since solubility (\(x\)) decreases with an increase in temperature, the value of \(K_H\) must increase with an increase in temperature.


Therefore, Assertion (A) is false.
Quick Tip: Remember the inverse relationship between Henry's constant (\(K_H\)) and gas solubility. A gas with a higher \(K_H\) value is less soluble at a given temperature.

Lower aldehydes and ketones are soluble in water due to hydrogen bonding between their polar carbonyl oxygen and water molecules.


As the size of the non-polar alkyl group (hydrophobic part) increases, it hinders this hydrogen bonding.


The larger hydrophobic part outweighs the polar part, causing a decrease in solubility. So, Assertion (A) is true.


Aldehydes and ketones have a polar C=O bond, which creates a dipole moment. This leads to dipole-dipole interactions between their molecules. So, Reason (R) is true.


However, the reason (dipole-dipole interaction) explains their boiling points, not their solubility trend in water.


The correct explanation for the assertion is the increasing size of the hydrophobic alkyl part.


Therefore, both statements are true, but the reason is not the correct explanation for the assertion.
Quick Tip: When explaining solubility in water, the primary factor is the ability to form hydrogen bonds with water. As the non-polar (hydrophobic) part of a molecule increases, its solubility in a polar solvent like water generally decreases.

Variation of Conductivity (\(\kappa\)):


Conductivity decreases as the concentration decreases (on dilution).


This is because conductivity is defined as the conductance of ions present in a unit volume of the solution.


Upon dilution, the number of current-carrying ions per unit volume decreases, which leads to a decrease in conductivity.


Variation of Molar Conductivity (\(\Lambda_m\)):


Molar conductivity increases as the concentration decreases (on dilution).


Molar conductivity is the conducting power of all the ions produced by one mole of an electrolyte in a given volume V. (\(\Lambda_m = \kappa V\))


On dilution, the volume V increases. Although conductivity (\(\kappa\)) decreases, the increase in volume V is much greater.


This is also because at lower concentrations, the inter-ionic attractions decrease, allowing ions to move more freely, which increases molar conductivity.
Quick Tip: Remember the key difference: Conductivity (\(\kappa\)) depends on the concentration of ions in a fixed unit volume, so it decreases on dilution. Molar conductivity (\(\Lambda_m\)) considers the total ions from one mole, and it increases on dilution due to greater ionic mobility.

The formula for elevation in boiling point is \(\Delta T_b = i \cdot K_b \cdot m\).


First, determine the van't Hoff factor (\(i\)). \(CaCl_2\) dissociates completely as:
\(CaCl_2 \rightarrow Ca^{2+} + 2Cl^{-}\)

The number of ions produced is \(1 + 2 = 3\). So, \(i = 3\).


Next, calculate the moles of solute (\(CaCl_2\)):

Moles (\(n_2\)) = Mass / Molar Mass = 3 g / 111 g mol\(^{-1}\) \(\approx\) 0.0270 mol.


Next, calculate the molality (\(m\)) of the solution:

Molality (\(m\)) = Moles of solute / Mass of solvent in kg.

Mass of solvent (water) = 260 g = 0.260 kg.
\(m\) = 0.0270 mol / 0.260 kg \(\approx\) 0.1039 mol kg\(^{-1}\).


Finally, calculate the elevation in boiling point (\(\Delta T_b\)):
\(\Delta T_b = 3 \times (0.52 K kg mol^{-1}) \times (0.1039 mol kg^{-1})\).
\(\Delta T_b = 1.56 \times 0.1039\) K.
\(\Delta T_b \approx 0.162\) K.
Quick Tip: For problems involving colligative properties of electrolytes, never forget to include the van't Hoff factor (\(i\)). Its value depends on the number of ions the solute dissociates into.

According to Raoult's law for an ideal solution, the total vapour pressure (\(P_{total}\)) is given by:
\(P_{total} = P_X^\circ x_X + P_Y^\circ x_Y\)

where \(P_X^\circ\) and \(P_Y^\circ\) are the vapour pressures of pure components, and \(x_X\) and \(x_Y\) are their mole fractions.


Since the solution contains equal moles of 'X' and 'Y', their mole fractions are equal.

Mole fraction of X, \(x_X = 0.5\).

Mole fraction of Y, \(x_Y = 0.5\).


Now, substitute the values into Raoult's law:
\(P_{total} = (120 mm Hg) \times 0.5 + (160 mm Hg) \times 0.5\).
\(P_{total} = 60 mm Hg + 80 mm Hg\).
\(P_{total} = 140 mm Hg\).
Quick Tip: For a binary ideal solution with equal moles of two components, the mole fraction of each component is always 0.5. The total vapour pressure will be the simple average of the pure component vapour pressures.

The rate constant, denoted by \(k\), is a proportionality constant that relates the rate of a chemical reaction to the concentration of the reactants.


For a general reaction, Rate = \(k[Reactant]^n\).


The rate constant is defined as the rate of the reaction when the molar concentration of each reactant is unity (1 mol L\(^{-1}\)).


It is also known as the specific reaction rate.
Quick Tip: The value of the rate constant (\(k\)) is specific to a particular reaction, depends on temperature (as described by the Arrhenius equation), and is independent of the reactant concentrations.

The unit of the rate constant (\(k\)) for a reaction of order 'n' is given by:

Unit of \(k\) = (mol L\(^{-1}\))\(^{1-n}\) s\(^{-1}\).


The given unit is L mol\(^{-1}\) s\(^{-1}\), which can be rewritten as (mol L\(^{-1}\))\(^{-1}\) s\(^{-1}\).


By comparing the general formula with the given unit, we equate the powers of the concentration term:
\(1 - n = -1\).


Solving for n:
\(n = 1 + 1 = 2\).


Therefore, the reaction is of the second order.
Quick Tip: You can quickly identify the reaction order from the units of k: 0 order: mol L\(^{-1}\) s\(^{-1}\) 1st order: s\(^{-1}\) 2nd order: L mol\(^{-1}\) s\(^{-1}\)

At 413 K, the dehydration of ethanol forms diethyl ether (intermolecular dehydration) via an \(S_N2\) mechanism.


Step 1: Protonation of alcohol

An ethanol molecule is protonated by the acid to form a protonated alcohol (ethyloxonium ion).
\(CH_3CH_2OH + H^+ \rightleftharpoons CH_3CH_2\overset{+}{O}H_2\)


Step 2: Nucleophilic attack

A second molecule of ethanol acts as a nucleophile and attacks the carbon atom of the protonated alcohol, displacing a water molecule in a single step.
\(CH_3CH_2OH + CH_3CH_2\overset{+}{O}H_2 \rightarrow CH_3CH_2-\overset{+}{O}(H)-CH_2CH_3 + H_2O\)


Step 3: Deprotonation

The resulting protonated ether loses a proton to form the final product, diethyl ether, and regenerate the acid catalyst.
\(CH_3CH_2-\overset{+}{O}(H)-CH_2CH_3 \rightarrow CH_3CH_2-O-CH_2CH_3 + H^+\)
Quick Tip: Remember the temperature dependence of alcohol dehydration with acid. At lower temperatures (\(\sim\)413 K), intermolecular dehydration occurs, forming an ether. At higher temperatures (\(\sim\)443 K), intramolecular dehydration (elimination) occurs, forming an alkene.

The iodoform test can be used to distinguish between 2-pentanone and 3-pentanone.


Principle: This test is given by compounds containing a methyl ketone group (\(CH_3-C=O\)-) or a group that can be oxidized to it (\(CH_3-CH(OH)\)-).


Procedure and Observation:

2-pentanone (\(CH_3COCH_2CH_2CH_3\)) possesses a methyl ketone group. When warmed with iodine and sodium hydroxide solution, it will give a yellow precipitate of iodoform (\(CHI_3\)).
\(CH_3COCH_2CH_2CH_3 + 3I_2 + 4NaOH \rightarrow CHI_3 \downarrow (yellow ppt) + CH_3CH_2CH_2COONa + 3NaI + 3H_2O\)


3-pentanone (\(CH_3CH_2COCH_2CH_3\)) does not have a methyl ketone group. Therefore, it will not give a positive iodoform test, and no yellow precipitate will be formed.
Quick Tip: To check if a compound will give a positive iodoform test, look for the \(CH_3-C=O-\) group (methyl ketone) or the \(CH_3-CH(OH)-\) group (methyl secondary alcohol). Acetaldehyde is the only aldehyde that gives this test.

Clemmensen reduction is a reaction that reduces the carbonyl group of an aldehyde or a ketone to a methylene group (\(-CH_2-\)).


The reagents used are zinc amalgam (Zn-Hg) and concentrated hydrochloric acid (HCl).


The general equation is:
\(R-CO-R' + 4[H] \xrightarrow{Zn-Hg, conc. HCl} R-CH_2-R' + H_2O\)

(Aldehyde/Ketone) \(\hspace{3.5cm}\) (Alkane)


For example, the reduction of propanone (acetone) to propane:
\(CH_3-CO-CH_3 + 4[H] \xrightarrow{Zn-Hg, conc. HCl} CH_3-CH_2-CH_3 + H_2O\)
Quick Tip: Clemmensen reduction is performed in a strongly acidic medium and is not suitable for compounds that are sensitive to acids. The Wolff-Kishner reduction (using hydrazine and a strong base) achieves the same transformation under basic conditions.

Let the rate law for the reaction be: Rate = \(k[A]^x[B]^y\).


Order with respect to A (x):

Compare experiments 1 and 2, where [B] is constant.

As [A] is doubled (0.1 M \(\rightarrow\) 0.2 M), the rate also doubles (0.05 \(\rightarrow\) 0.10).

So, Rate \(\propto\) [A]\(^1\). The order with respect to A is 1 (\(x=1\)).


Order with respect to B (y):

Compare experiments 1 and 3, where [A] is constant.

As [B] is doubled (0.1 M \(\rightarrow\) 0.2 M), the rate remains unchanged (0.05).

So, the rate is independent of [B]. The order with respect to B is 0 (\(y=0\)).


Overall order of the reaction:

Overall order = \(x + y = 1 + 0 = 1\).

The overall order of the reaction is 1.

The rate law is: Rate = \(k[A]\).
Quick Tip: To find the order with respect to a specific reactant using the method of initial rates, choose two experiments where the concentration of only that reactant changes, while the concentrations of all other reactants are kept constant.

The final volume (20.2 mL) is greater than the sum of the initial volumes (10 mL + 10 mL = 20 mL).


This means the solution shows a positive deviation from Raoult's law.


(i) The volume change (\(\Delta V_{mix} > 0\)) occurred because the new intermolecular forces of attraction between A and B molecules are weaker than the original forces between A-A and B-B molecules. This allows the molecules to be farther apart, thus increasing the total volume.


(ii) There will be a decrease in temperature. The process is endothermic (\(\Delta H_{mix} > 0\)) because more energy is required to break the stronger A-A and B-B bonds than is released when forming the weaker A-B bonds. This absorption of heat from the surroundings results in cooling.


(iii) An example of this type of solution is a mixture of ethanol and acetone, or carbon disulphide and acetone.
Quick Tip: For non-ideal solutions: Positive Deviation: Weaker A-B forces, \(\Delta V_{mix} > 0\) (volume increases), \(\Delta H_{mix} > 0\) (endothermic, cools down). Negative Deviation: Stronger A-B forces, \(\Delta V_{mix} < 0\) (volume decreases), \(\Delta H_{mix} < 0\) (exothermic, heats up).

(i) Sprinkling salt (like \(NaCl\) or \(CaCl_2\)) on snow depresses the freezing point of water. The salt dissolves in the thin layer of water on the ice, forming a solution with a freezing point lower than 0°C. If the ambient temperature is above this new freezing point, the snow melts.


(ii) The fluid inside red blood cells (RBCs) is isotonic with a 0.9% (mass/vol) \(NaCl\) solution. The external 0.5% \(NaCl\) solution is hypotonic (less concentrated). Due to osmosis, water moves from the hypotonic solution (outside the cell) into the RBCs. This causes the cells to swell and eventually burst (a process called hemolysis).


(iii) A major application of reverse osmosis is the desalination of seawater. By applying a pressure greater than the osmotic pressure, pure water is forced out of the sea water through a semipermeable membrane, leaving the salts behind.
Quick Tip: Osmosis is the net movement of solvent from a less concentrated (hypotonic) solution to a more concentrated (hypertonic) solution. In reverse osmosis, external pressure is used to force the solvent in the opposite direction.

1. Oxidation State of Ni: Let the oxidation state of Ni be x. \(x + 4(-1) = -2 \implies x = +2\).


2. Electronic Configuration: Ni (Z=28): \([Ar] 3d^8 4s^2\). For \(Ni^{2+}\), the configuration is \([Ar] 3d^8\).


3. Hybridization and Geometry: \(Cl^-\) is a weak field ligand and does not cause pairing of electrons. The \(3d^8\) configuration has two unpaired electrons. For bonding with four \(Cl^-\) ligands, the \(Ni^{2+}\) ion uses its vacant outer orbitals: one \(4s\) and three \(4p\) orbitals. This leads to \(sp^3\) hybridization. The geometry for \(sp^3\) hybridization is tetrahedral.


4. Magnetic Character: Since the complex has two unpaired electrons in its 3d orbitals, it is paramagnetic.
Quick Tip: For coordination number 4, remember the common geometries: \(sp^3\) hybridization (usually with weak field ligands) leads to tetrahedral geometry, while \(dsp^2\) hybridization (usually with strong field ligands like \(CN^-\)) leads to square planar geometry.

This complex exhibits geometrical isomerism (cis-trans). 'en' (ethylenediamine) is a bidentate ligand.


1. cis-isomer: The two chloro ligands are adjacent (90° apart). This isomer is optically active.




2. trans-isomer: The two chloro ligands are opposite (180° apart). This isomer is optically inactive.


Quick Tip: For octahedral complexes of the type \([M(AA)_2X_2]\), where AA is a symmetric bidentate ligand, cis-trans isomerism is possible. The cis isomer is always optically active, while the trans isomer is optically inactive.

Inversion of configuration is characteristic of the \(S_N2\) (bimolecular nucleophilic substitution) mechanism.


(i) 1-chloropropane is a primary alkyl halide. Primary halides are sterically unhindered and favor the \(S_N2\) mechanism, where the nucleophile (\(OH^-\)) attacks from the side opposite to the leaving group, causing an inversion of configuration.


(ii) 2-chloro-2-methylpropane is a tertiary alkyl halide. It is sterically hindered, preventing \(S_N2\) attack. It reacts via the \(S_N1\) mechanism, which proceeds through a planar carbocation intermediate, leading to racemization (a mix of inversion and retention).


Therefore, 1-chloropropane will undergo inversion of configuration.
Quick Tip: Reactivity order for nucleophilic substitution: \(S_N1\): \(3^\circ > 2^\circ > 1^\circ\) (favored by stable carbocations) \(S_N2\): \(1^\circ > 2^\circ > 3^\circ\) (favored by less steric hindrance)

The reaction is the dehydrohalogenation of 2-bromobutane, an elimination reaction.


The formation of the major product is governed by Saytzeff's (or Zaitsev's) rule.


Saytzeff's rule states that the more substituted alkene (the alkene with more alkyl groups on the double-bonded carbons) is the more stable and thus the major product.


(A) But-1-ene (\(CH_3-CH_2-CH=CH_2\)) is a monosubstituted alkene.


(B) But-2-ene (\(CH_3-CH=CH-CH_3\)) is a disubstituted alkene.


Since But-2-ene (B) is more substituted, it is more stable.


Therefore, according to Saytzeff's rule, But-2-ene (B) will be the major product.
Quick Tip: In elimination reactions, remember the two main rules: \textbf{Saytzeff's Rule:} Forms the more substituted alkene (major product with small bases like alc. KOH). \textbf{Hofmann's Rule:} Forms the less substituted alkene (major product with bulky bases like potassium tert-butoxide).

Williamson's synthesis is a method to prepare ethers by the reaction of a sodium alkoxide with a primary alkyl halide.


The reaction follows an \(S_N2\) mechanism.


General Equation:
\(R-O^-Na^+ + R'-X \rightarrow R-O-R' + NaX\)

(Sodium alkoxide + Alkyl halide \(\rightarrow\) Ether + Sodium halide)


Example:
\(CH_3CH_2O^-Na^+ + CH_3-Br \rightarrow CH_3CH_2-O-CH_3 + NaBr\)

(Sodium ethoxide + Bromomethane \(\rightarrow\) Methoxyethane)
Quick Tip: For good yields in Williamson's synthesis, the alkyl halide must be primary. If a secondary or tertiary alkyl halide is used, elimination becomes the major reaction, producing an alkene instead of an ether.

Method: The mixture is separated by steam distillation.


Principle: The separation is based on the difference in volatility due to different types of hydrogen bonding.


o-Nitrophenol: Exhibits intramolecular hydrogen bonding (within the molecule). This prevents association with other molecules, making it more volatile and thus steam volatile.


p-Nitrophenol: Exhibits intermolecular hydrogen bonding (between different molecules). This causes molecular association, making it less volatile and not steam volatile.


When steam is passed through the mixture, the volatile o-nitrophenol vaporizes and distills over with the steam, while the less volatile p-nitrophenol is left behind.
Quick Tip: Remember the effect of hydrogen bonding on physical properties. Intramolecular H-bonding decreases the boiling point and water solubility. Intermolecular H-bonding increases the boiling point and water solubility.

1. Identify P: P reacts with ammonia (\(NH_3\)) to form ammonium acetate (\(CH_3COO^{-}NH_4^{+}\)). This is an acid-base reaction, so P must be acetic acid.

P = \(CH_3COOH\) (Acetic acid)


2. Identify Q: Ammonium acetate, on heating (\(\Delta\)), undergoes dehydration to form an amide.

Q = \(CH_3CONH_2\) (Acetamide)


3. Identify R: Acetamide (Q) reacts with phosphorus pentachloride (\(PCl_5\)), a strong dehydrating agent, which removes water from the amide to form a nitrile.

R = \(CH_3CN\) (Acetonitrile)
Quick Tip: Remember these key transformations: Carboxylic Acid + Ammonia \(\rightarrow\) Ammonium Salt \(\xrightarrow{Heat}\) Amide. Amide + Dehydrating agent (\(P_2O_5\), \(PCl_5\), \(SOCl_2\)) \(\rightarrow\) Nitrile.

Pernicious anaemia is a type of megaloblastic anemia caused by the body's inability to absorb vitamin B\(_{12}\).


Therefore, the vitamin is Vitamin B\(_{12}\) (also known as Cobalamin).
Quick Tip: It's helpful to remember the names and deficiency diseases of key vitamins: Vitamin A (Night blindness), Vitamin C (Scurvy), Vitamin D (Rickets), Vitamin B\(_{12}\) (Pernicious anaemia).

\begin{tabular{|l|l|l|
\hline
Property & Globular Proteins & Fibrous Proteins

\hline
Shape & Polypeptide chains coil to give a & Polypeptide chains run parallel to

& spherical or spheroidal shape. & form a thread-like or fibre-like structure.

\hline
Solubility & Generally soluble in water. & Generally insoluble in water.

\hline
Function & Involved in metabolic functions like & Provide structural strength and support, e.g.,

& catalysis (enzymes) and transport. & keratin (in hair) and collagen (in tendons).

\hline
\end{tabular
Quick Tip: Think of "globular" like "globe" (spherical) and "fibrous" like "fibre" (thread-like). Their shapes are closely related to their functions: compact spheres are good for transport and catalysis, while long fibres are excellent for building structures.

(a)

(i) Since one mole of the compound precipitates one mole of AgCl, there must be one chloride ion outside the coordination sphere. The remaining ligands (4 \(H_2O\) molecules and 2 \(Cl^-\) ions) are inside the sphere.

The structural formula is \([Cr(H_2O)_4Cl_2]Cl\).


(ii) The secondary valency is the coordination number of the central metal ion. Here, Cr is coordinated to four water molecules and two chloride ions.

Secondary valency = 4 + 2 = 6.


(b)

A double salt (e.g., Mohr's salt, \(FeSO_4 \cdot (NH_4)_2SO_4 \cdot 6H_2O\)) exists only in the solid state and dissociates completely into its constituent ions in solution, losing its identity.

A coordination compound (e.g., \(K_4[Fe(CN)_6]\)) retains its identity in solution. The complex ion (\([Fe(CN)_6]^{4-}\)) does not dissociate further into its constituent ions.


(c) (i)

Conductivity depends on the number of ions produced in solution.

- \([Cr(NH_3)_3Cl_3]\): Non-electrolyte, produces 0 ions.

- \([Cr(NH_3)_5Cl]Cl_2\): Dissociates into \([Cr(NH_3)_5Cl]^{2+}\) and \(2Cl^-\), producing 3 ions.

- \([Cr(NH_3)_6]Cl_3\): Dissociates into \([Cr(NH_3)_6]^{3+}\) and \(3Cl^-\), producing 4 ions.

The increasing order of conductivity is:

\([Cr(NH_3)_3Cl_3] < [Cr(NH_3)_5Cl]Cl_2 < [Cr(NH_3)_6]Cl_3\)


OR


(c) (ii)

\begin{tabular{|l|l|
\hline
Primary Valency & Secondary Valency

\hline
1. Corresponds to the oxidation state. & Corresponds to the coordination number.

\hline
2. It is ionisable. & It is non-ionisable.

\hline
3. Satisfied only by negative ions. & Satisfied by negative ions or neutral molecules.

\hline
4. It is non-directional. & It is directional and determines the geometry.

\hline
\end{tabular
Quick Tip: Remember Werner's key concepts: Primary valency = Oxidation state (ionisable, non-directional). Secondary valency = Coordination number (non-ionisable, directional, determines geometry).

(a)

Presence of aldehydic group (-CHO):

Glucose on reaction with a mild oxidizing agent like bromine water gets oxidized to gluconic acid. This confirms the presence of an aldehydic group.
\(CHO-(CHOH)_4-CH_2OH \xrightarrow{Br_2/H_2O} COOH-(CHOH)_4-CH_2OH\)

(Glucose) \(\hspace{3.5cm}\) (Gluconic acid)


Presence of a straight chain:

Glucose on prolonged heating with hydroiodic acid (HI) and red phosphorus undergoes complete reduction to form n-hexane, confirming that all six carbon atoms are in a straight chain.
\(CHO-(CHOH)_4-CH_2OH \xrightarrow{HI, \Delta} CH_3-CH_2-CH_2-CH_2-CH_2-CH_3\)

(Glucose) \(\hspace{4.5cm}\) (n-Hexane)


(b) (i) Anomers:

Anomers are a pair of cyclic stereoisomers of a sugar that differ only in the configuration of the hydroxyl group (-OH) at the anomeric carbon (the hemiacetal or hemiketal carbon, which is C-1 in glucose).


OR


(b) (ii) Structure of \(\beta\)-D-Glucopyranose:

In the Haworth projection of glucose, the \(\beta\)-anomer is the one where the -OH group on the anomeric carbon (C-1) is on the same side as the terminal \(-CH_2OH\) group (C-6), i.e., pointing 'up'.



(c) Invert Sugar:

Sucrose is dextrorotatory (specific rotation = +66.5°).

On hydrolysis, sucrose yields an equimolar mixture of D-glucose and D-fructose.
\(C_{12}H_{22}O_{11} (Sucrose) + H_2O \rightarrow C_6H_{12}O_6 (D-Glucose) + C_6H_{12}O_6 (D-Fructose)\)

D-glucose is dextrorotatory (+52.5°), but D-fructose is strongly laevorotatory (-92.4°).

The resulting mixture is net laevorotatory. Because the sign of optical rotation "inverts" from positive to negative during hydrolysis, the product mixture is called "invert sugar".
Quick Tip: For D-sugars in Haworth projections: \(\alpha\)-anomer has the C1-OH group pointing down ('trans' to \(-CH_2OH\)). \(\beta\)-anomer has the C1-OH group pointing up ('cis' to \(-CH_2OH\)).

(i) The reagent \(C_6H_5SO_2Cl\) is Hinsberg's reagent, used to test for primary, secondary, and tertiary amines.

A primary amine reacts with Hinsberg's reagent to form N-alkylbenzenesulphonamide, which is acidic due to the hydrogen attached to the nitrogen atom. This makes it soluble in alkali (NaOH).

A secondary amine forms an N,N-dialkylbenzenesulphonamide, which has no acidic hydrogen and is insoluble in alkali.

A tertiary amine does not react.

Since the product is soluble in NaOH, amine 'X' must be a primary amine.

The only primary amine with the formula \(C_2H_7N\) is ethylamine (\(CH_3CH_2NH_2\)).


(ii) A higher \(pK_b\) value corresponds to a weaker base. The order of basic strength in aqueous phase is: \((CH_3)_2NH > CH_3NH_2 > (CH_3)_3N > NH_3 > C_6H_5NH_2\).

Therefore, the increasing order of \(pK_b\) values (decreasing basic strength) is:

\((CH_3)_2NH < CH_3NH_2 < (CH_3)_3N < NH_3 < C_6H_5NH_2\)


(iii) Nitration is carried out in a strongly acidic medium. In this medium, aniline gets protonated to form the anilinium ion (\(C_6H_5NH_3^+\)). The \(-NH_3^+\) group is a meta-directing and strongly deactivating group. Due to the presence of the anilinium ion, a significant amount of m-nitroaniline is formed.


(iv)

(I) Aniline to p-bromoaniline: The amino group is first protected by acetylation. Then bromination is carried out, followed by hydrolysis.
\(C_6H_5NH_2 \xrightarrow{(CH_3CO)_2O} C_6H_5NHCOCH_3 \xrightarrow{Br_2/CH_3COOH} p-Br-C_6H_4NHCOCH_3 \xrightarrow{H^+/H_2O} p-Br-C_6H_4NH_2\)


(II) Aniline to phenol: Aniline is first converted to benzene diazonium chloride by diazotization, which is then hydrolyzed by warming with water.
\(C_6H_5NH_2 \xrightarrow{NaNO_2 + HCl, 273-278K} C_6H_5N_2^+Cl^- \xrightarrow{H_2O, warm} C_6H_5OH + N_2 + HCl\)
Quick Tip: Hinsberg's Test is a key method to distinguish amine types. Remember: \(1^\circ\) amine product dissolves in alkali, \(2^\circ\) amine product does not, and \(3^\circ\) amine does not react.

(i) This is the carbylamine test (or isocyanide test), which is given by primary amines. Ethylamine, being a primary amine, reacts with chloroform (\(CHCl_3\)) and alcoholic potassium hydroxide (KOH) to form ethyl isocyanide, a foul-smelling compound.
\(CH_3CH_2NH_2 + CHCl_3 + 3KOH (alc.) \xrightarrow{\Delta} CH_3CH_2NC + 3KCl + 3H_2O\)


(ii)

Compound A is reduced by \(H_2/Pd\) to form aniline (\(C_6H_5NH_2\)). This is a standard reduction of a nitro group. So, A is Nitrobenzene (\(C_6H_5NO_2\)).

Aniline reacts with bromine. The question gives \(Br_2/NaOH\), but typically aqueous bromine is used. The \(-NH_2\) group is highly activating, directing bromination to the ortho and para positions. Aniline reacts with bromine water to give a white precipitate of 2,4,6-tribromoaniline. We will assume this is the intended reaction.

So, B is 2,4,6-tribromoaniline.
\(C_6H_5NH_2 + 3Br_2 \rightarrow C_6H_2Br_3NH_2 + 3HBr\)


(iii)

(I) Aniline to benzene: Aniline is diazotized to form benzene diazonium chloride, which is then reduced using hypophosphorous acid (\(H_3PO_2\)).
\(C_6H_5NH_2 \xrightarrow{NaNO_2 + HCl, 273K} C_6H_5N_2^+Cl^- \xrightarrow{H_3PO_2 + H_2O} C_6H_6 + N_2 + H_3PO_3 + HCl\)


(II) Aniline to sulphanilic acid: Aniline is treated with concentrated sulfuric acid. The anilinium hydrogensulphate formed rearranges on heating to form p-aminobenzenesulphonic acid (sulphanilic acid), which exists as a zwitterion.
\(C_6H_5NH_2 + H_2SO_4 (conc.) \rightarrow C_6H_5NH_3^+HSO_4^- \xrightarrow{453-473K} p-H_2N-C_6H_4-SO_3H\)
Quick Tip: The carbylamine reaction is a specific confirmatory test for primary amines (both aliphatic and aromatic). The formation of a foul-smelling isocyanide is a positive result.

(i)

(I) Pyrolusite is \(MnO_2\). Fusion with KOH in air produces potassium manganate, which is green. So, A is \(K_2MnO_4\).

In acidic medium, the green manganate ion disproportionates into the purple permanganate ion and manganese dioxide. So, B is \(KMnO_4\).


(II) Permanganate ion (\(MnO_4^-\) from B) oxidizes \(Fe^{2+}\) to \(Fe^{3+}\) in acidic medium, while it gets reduced to \(Mn^{2+}\).

The balanced ionic equation is:
\(MnO_4^- + 5Fe^{2+} + 8H^+ \rightarrow Mn^{2+} + 5Fe^{3+} + 4H_2O\)


(ii)

(I) \(Ce^{4+}\) is a good oxidizing agent because it has a high tendency to gain an electron to form the very stable \(Ce^{3+}\) ion. The reduction potential \(E^\circ(Ce^{4+}/Ce^{3+})\) is high and positive (+1.74 V), indicating a strong driving force for the reduction process.


(II) The actinoid contraction is greater because 5f electrons have a much poorer shielding effect than 4f electrons. This poor shielding results in a stronger effective nuclear charge experienced by the outer electrons, leading to a more significant pull and a greater contraction in size across the series.


(III) The \(E^\circ(Zn^{2+}/Zn)\) value is negative because the energy required to convert solid Zn to gaseous atoms (atomization) and then ionize it is more than compensated by the very high hydration enthalpy of the small \(Zn^{2+}\) ion. For copper, the high second ionization enthalpy required to form \(Cu^{2+}\) is not fully compensated by its hydration enthalpy, resulting in a positive electrode potential.
Quick Tip: The stability of half-filled (\(d^5\)) and completely-filled (\(d^{10}\)) d-orbitals plays a crucial role in determining the electrode potentials and oxidation states of d-block elements.

(i)

(I) The phenomenon responsible is the Lanthanoid Contraction.

Definition: The Lanthanoid Contraction is the steady and regular decrease in the atomic and ionic radii of the lanthanoid elements with increasing atomic number. This is caused by the poor shielding effect of the inner 4f electrons, which leads to an increase in the effective nuclear charge.


(II) A major consequence is that the elements of the second (4d) and third (5d) transition series in the same group have very similar atomic radii and chemical properties. For example, Zr and Hf, or Nb and Ta, are known as "chemical twins" and are very difficult to separate.


(ii)

(I) Transition metals exhibit catalytic properties due to two main reasons: (1) Their ability to show variable oxidation states allows them to form unstable intermediates. (2) They can provide a large surface area for the adsorption of reactant molecules.


(II) Transition metals have high enthalpies of atomisation because of the strong metallic bonding in their crystal lattices. This strong bonding is due to the involvement of a large number of electrons from both (n-1)d and ns orbitals in forming metallic bonds.


(III) An element is considered a transition element if it has an incompletely filled d-orbital in its ground state or in any of its common oxidation states.

- Scandium (Sc, Z=21) has the configuration \([Ar] 3d^1 4s^2\). It has an incomplete d-orbital in its ground state, so it is a transition element.

- Zinc (Zn, Z=30) has the configuration \([Ar] 3d^{10} 4s^2\). Its only common oxidation state is +2, with the configuration \([Ar] 3d^{10}\). Since it does not have an incomplete d-orbital in either its ground state or its common oxidation state, it is not considered a transition element.
Quick Tip: The definition of a transition element is strict: it must have a partially filled d subshell in its elemental form or its common ions. This is why Group 12 (Zn, Cd, Hg) is often excluded from the typical transition metals.

(i) In a galvanic cell, the half-reaction with the higher (less negative) standard reduction potential (\(E^\circ\)) occurs as reduction at the cathode. The half-reaction with the lower (more negative) \(E^\circ\) is reversed and occurs as oxidation at the anode.

Since \(E^\circ(Fe^{2+}/Fe) = -0.44\) V is higher than \(E^\circ(Cr^{3+}/Cr) = -0.74\) V.

- Reduction Reaction: The reaction \(Fe^{2+} + 2e^- \rightarrow Fe(s)\) will remain as the reduction half-reaction (at the cathode).

- Oxidation Reaction: The reaction \(Cr^{3+} + 3e^- \rightarrow Cr(s)\) will be reversed to become \(Cr(s) \rightarrow Cr^{3+} + 3e^-\) (at the anode).


(ii)

Cell Representation: The oxidation half-reaction (\(Mg \rightarrow Mg^{2+}\)) is written on the left (anode), and the reduction half-reaction (\(Ag^+ \rightarrow Ag\)) is written on the right (cathode).

\(Mg(s) | Mg^{2+}(0.100 M) || Ag^+(0.001 M) | Ag(s)\)


Calculation of \(E_{cell}\):

Using the Nernst equation: \(E_{cell} = E^\circ_{cell} - \frac{0.0591}{n} \log Q\).

From the overall reaction, the number of electrons transferred, \(n=2\).

The reaction quotient, \(Q = \frac{[Mg^{2+}]}{[Ag^+]^2} = \frac{0.100}{(0.001)^2} = \frac{10^{-1}}{(10^{-3})^2} = \frac{10^{-1}}{10^{-6}} = 10^5\).

Substituting the values:
\(E_{cell} = 3.17 V - \frac{0.0591}{2} \log(10^5)\)
\(E_{cell} = 3.17 - (0.02955) \times (5 \log 10)\)
\(E_{cell} = 3.17 - (0.02955 \times 5)\)
\(E_{cell} = 3.17 - 0.14775\)

\(E_{cell} \approx 3.022 V\)
Quick Tip: Remember the cell notation: Anode | Anode Ion (\(C_1\)) || Cathode Ion (\(C_2\)) | Cathode. The Nernst equation is essential for calculating cell potential under non-standard conditions.

(i) Kohlrausch's Law of Independent Migration of Ions:

The law states that the limiting molar conductivity of an electrolyte can be represented as the sum of the individual contributions of the anion and cation of the electrolyte.

Mathematically, \(\Lambda^\circ_m (A_xB_y) = x \lambda^\circ_{A^{y+}} + y \lambda^\circ_{B^{x-}}\).


Two Applications:

1. Calculation of the limiting molar conductivity (\(\Lambda^\circ_m\)) for weak electrolytes, which cannot be determined experimentally by extrapolation.

2. Calculation of the degree of dissociation (\(\alpha\)) and the dissociation constant (\(K_a\) or \(K_b\)) for a weak electrolyte at a given concentration.


(ii) Calculation of Degree of Dissociation (\(\alpha\)):

The formula for the degree of dissociation is \(\alpha = \frac{\Lambda_m}{\Lambda^\circ_m}\).

We are given \(\Lambda_m (NH_4OH) = 9.33\) S cm\(^2\) mol\(^{-1}\).

First, we must calculate the limiting molar conductivity for \(NH_4OH\) using Kohlrausch's law:
\(\Lambda^\circ_m(NH_4OH) = \Lambda^\circ_m(NH_4Cl) + \Lambda^\circ_m(NaOH) - \Lambda^\circ_m(NaCl)\)
\(\Lambda^\circ_m(NH_4OH) = (129.8 + 217.4 - 108.9)\) S cm\(^2\) mol\(^{-1}\)
\(\Lambda^\circ_m(NH_4OH) = 238.3\) S cm\(^2\) mol\(^{-1}\).

Now, calculate \(\alpha\):
\(\alpha = \frac{9.33 S cm^2 mol^{-1}}{238.3 S cm^2 mol^{-1}}\)

\(\alpha \approx 0.03915\)

(or 3.915%)
Quick Tip: Kohlrausch's law is like an algebraic puzzle. To find the \(\Lambda^\circ_m\) of a weak electrolyte, combine the \(\Lambda^\circ_m\) values of strong electrolytes in a way that the desired ions remain and the unwanted ions cancel out.