Punjab Board Class 12 Chemistry Question Paper 2025 Set A with Solution PDF

The paragraph explicitly mentions the two types of nucleic acids found in the nucleus of a living cell.


DNA stands for Deoxyribonucleic acid.


RNA stands for Ribonucleic acid.
Quick Tip: Remember that the difference in names comes from the sugar moiety: Deoxyribose in DNA and Ribose in RNA.

According to the provided text, RNA molecules are of three types.


These types are listed as messenger RNA (m-RNA), ribosomal RNA (r-RNA), and transfer RNA (t-RNA).
Quick Tip: The three types of RNA play distinct roles in protein synthesis: m-RNA carries the code, t-RNA brings amino acids, and r-RNA forms the ribosome structure.

The first sentence of the passage states that "It has been observed that nucleus of a living cell is responsible for this transmission of inherent characters, also called heredity."


Therefore, the nucleus is the organelle responsible for heredity.
Quick Tip: While the nucleus is the organelle, the specific molecular carriers of heredity within the nucleus are the chromosomes and DNA.

Based on the text, heredity is defined as "this transmission of inherent characters".


It refers to the passing of genetic traits from parents to offspring.
Quick Tip: Heredity is the biological process responsible for the resemblance between parents and offspring.

The passage states that nucleic acids "are mainly of two types".


These two types are identified as deoxyribonucleic acids (DNA) and ribonucleic acid (RNA).
Quick Tip: DNA is primarily the genetic material, while RNA is involved in the expression of that genetic information.

The strength of a ligand is determined by its position in the spectrochemical series.


Carbon monoxide (CO) is a strong field ligand and acts as a \(\pi\)-acceptor, placed at the high end of the series.


Chloride ion (Cl\(^{-1}\)) is a weak field ligand, placed at the lower end of the series.


Therefore, CO causes greater crystal field splitting than Cl\(^{-1}\), making it a stronger ligand.
Quick Tip: Recall the spectrochemical series order: Halogens \(<\) OH \(<\) H\(_2\)O \(<\) NH\(_3\) \(<\) CN \(<\) CO. Ligands with Carbon donors (CO, CN) are generally very strong.

In the Victor Meyer test, different classes of alcohols produce distinct colours.


Primary alcohols react to produce a blood-red colouration (due to the formation of nitrolic acid).


Secondary alcohols produce a deep blue colouration (due to the formation of pseudonitrol).


Therefore, the statement that primary alcohols produce a deep blue colour is incorrect.
Quick Tip: Mnemonic for Victor Meyer Test colours: RBC (Red, Blue, Colourless) corresponds to Primary (\(1^\circ\)), Secondary (\(2^\circ\)), and Tertiary (\(3^\circ\)) alcohols respectively.

The name "formic acid" (methanoic acid) is derived from the Latin word "formica", which means ant.


It was historically first isolated by the distillation of the bodies of red ants.


Thus, the statement is factually correct.
Quick Tip: Methanoic acid is the IUPAC name for formic acid. It is also responsible for the stinging sensation in ant bites.

Acetone is the common name for the ketone with the formula CH\(_3\)COCH\(_3\).


Its IUPAC name is Propanone, as it contains a three-carbon chain.


Butanone is the IUPAC name for Methyl Ethyl Ketone (CH\(_3\)COCH\(_2\)CH\(_3\)), which has a four-carbon chain.


Therefore, Acetone is not Butanone.
Quick Tip: "Prop-" indicates 3 carbons, while "But-" indicates 4 carbons. Acetone is the simplest ketone with 3 carbons.

In Nickel Tetracarbonyl [Ni(CO)\(_4\)], the oxidation state of Nickel is 0 (since CO is a neutral ligand).


The electronic configuration of Ni(0) is [Ar] 3d\(^8\) 4s\(^2\).


CO is a strong field ligand, which causes the two electrons from the 4s orbital to pair up with the electrons in the 3d orbital.


This results in a 3d\(^{10}\) configuration with no unpaired electrons.


Since there are no unpaired electrons, the complex is diamagnetic.
Quick Tip: Species with no unpaired electrons are diamagnetic and are weakly repelled by magnetic fields.

The ebullioscopic constant (\(K_b\)) relates the elevation in boiling point (\(\Delta T_b\)) to the molality (\(m\)) of the solution: \(\Delta T_b = K_b \cdot m\).


Rearranging for \(K_b\), we get \(K_b = \frac{\Delta T_b}{m}\).


The unit of temperature change (\(\Delta T_b\)) is Kelvin (K).


The unit of molality (\(m\)) is mol kg\(^{-1}\).


Therefore, the unit of \(K_b\) is \(\frac{K}{mol kg^{-1}} = K kg mol^{-1}\).
Quick Tip: Both ebullioscopic (\(K_b\)) and cryoscopic (\(K_f\)) constants have the same units: K kg mol\(^{-1}\).

Galvanisation is a specific metallurgical process used to protect steel or iron from rusting.


This involves applying a protective layer of Zinc (Zn) to the metal.


Zinc acts as a sacrificial anode, corroding in preference to the underlying iron.


Therefore, the correct coating material is Zinc.
Quick Tip: Think of "Galvanisation" and "Zinc" together. It is the most common method for rust prevention on iron sheets.

For a reaction of \(n\)-th order, the rate law is Rate \(= k[A]^n\).


The units of the rate constant \(k\) are given by the general formula \((mol L^{-1})^{1-n} s^{-1}\).


For a first order reaction, \(n = 1\).


Substituting \(n=1\) into the formula: \((mol L^{-1})^{1-1} s^{-1} = (mol L^{-1})^0 s^{-1} = s^{-1}\).


Alternatively, since Rate (mol L\(^{-1}\) s\(^{-1}\)) \(= k \times\) Concentration (mol L\(^{-1}\)), \(k\) must have units of s\(^{-1}\).
Quick Tip: The unit of the rate constant for a first-order reaction is strictly time\(^{-1}\) and is independent of concentration units.

Let the total mass of the solution be 100 g.


Mass of Benzene (C\(_6\)H\(_6\)) = 30 g. Mass of Carbon tetrachloride (CCl\(_4\)) = 100 - 30 = 70 g.


Molar mass of Benzene (C\(_6\)H\(_6\)) = \(12(6) + 1(6) = 78\) g/mol.


Molar mass of CCl\(_4\) = \(12 + 35.5(4) = 12 + 142 = 154\) g/mol.


Moles of Benzene (\(n_B\)) = \(\frac{30}{78} \approx 0.385\) mol.


Moles of CCl\(_4\) (\(n_A\)) = \(\frac{70}{154} \approx 0.455\) mol.


Total moles = \(0.385 + 0.455 = 0.840\) mol.


Mole fraction of Benzene (\(X_B\)) = \(\frac{n_B}{n_{total}} = \frac{0.385}{0.840} \approx 0.4583\).


This value is closest to 0.459.
Quick Tip: Mole fraction is a dimensionless quantity. Always ensure you calculate moles using the correct molar masses (\(C=12, H=1, Cl=35.5\)).

Reverse osmosis requires a semi-permeable membrane that allows solvent to pass but blocks solute.


The most common material used for this membrane is Cellulose Acetate.


Option (A) says "Cellulose acetone", which appears to be a typo for Cellulose Acetate found in the original exam paper (confirmed by Hindi version "Cellulose Acetate").


Options B, C, and D are chemical salts or carbohydrates not typically used as membrane films for RO.


Therefore, (A) is the intended correct answer.
Quick Tip: In exam questions, if you see a slight typo like "Cellulose Acetone" instead of "Cellulose Acetate", choose it if it's clearly the intended answer among unrelated distractors.

DDT (Dichlorodiphenyltrichloroethane) is a synthetic organochlorine insecticide.


It is chemically stable and does not break down easily in the environment by natural biological processes.


Because it persists in the environment and accumulates in the food chain (biomagnification), it is classified as a non-biodegradable pollutant.
Quick Tip: DDT is the classic example of a non-biodegradable pollutant that led to the decline of bird populations due to eggshell thinning.

We compare the acid strengths based on the stability of their conjugate bases and pK\(_a\) values.


Acetic acid (CH\(_3\)COOH) has a pK\(_a\) of 4.76. Propanoic acid (CH\(_3\)CH\(_2\)COOH) has a pK\(_a\) of 4.88.


Benzoic acid (C\(_6\)H\(_5\)COOH) has a pK\(_a\) of 4.20. The phenyl group exerts an electron-withdrawing effect (due to sp\(^2\) carbons) which stabilizes the carboxylate ion relative to alkyl groups.


Phenylacetic acid (C\(_6\)H\(_5\)CH\(_2\)COOH) has a pK\(_a\) of 4.31. The -I effect of the phenyl ring is diminished by the intervening CH\(_2\) group compared to direct attachment in Benzoic acid.


Since a lower pK\(_a\) indicates a stronger acid, Benzoic acid (pK\(_a\) 4.20) is the strongest among the given options.
Quick Tip: General order of acidity: Formic Acid \(>\) Benzoic Acid \(>\) Acetic Acid. Aromatic carboxylic acids are generally stronger than aliphatic ones (except Formic).

Lithium Aluminium Hydride (LiAlH\(_4\)) is a strong reducing agent.


It reduces carboxylic acids (-COOH) directly to primary alcohols (-CH\(_2\)OH).


The reaction is: C\(_6\)H\(_5\)COOH \(\xrightarrow{LiAlH_4}\) C\(_6\)H\(_5\)CH\(_2\)OH.


The product C\(_6\)H\(_5\)CH\(_2\)OH is named Benzyl Alcohol.
Quick Tip: LiAlH\(_4\) is strong enough to reduce acids, esters, and ketones to alcohols. It does not reduce the benzene ring itself under normal conditions.

Vinegar is a common household item used for cooking and preservation.


Chemically, it consists of a dilute solution of acetic acid in water (typically 5-8%).


The IUPAC name for acetic acid is Ethanoic acid.
Quick Tip: The word vinegar comes from French "vin aigre" meaning sour wine, which is formed when ethanol in wine oxidizes to ethanoic acid.

Let the oxidation state of Iron (Fe) be \(x\).


The cyanide ligand (CN) has a charge of -1. There are 6 cyanide ions.


The overall charge on the complex ion is -3.


The equation for the sum of oxidation states is: \(x + 6(-1) = -3\).

\(x - 6 = -3 \implies x = +3\).


Therefore, the oxidation state of Fe is +3.
Quick Tip: Always equate the sum of oxidation states of the central metal and ligands to the total charge on the coordination sphere. CN is always -1.

1. Dissociation: Double salts (e.g., Mohr's salt) completely dissociate into their constituent simple ions when dissolved in water. Coordination compounds (e.g., Potassium ferrocyanide) do not completely dissociate; the complex ion remains intact.


2. Properties: The properties of a double salt are essentially the same as those of its constituent compounds. The properties of a coordination compound are different from its constituents.


3. Valency: In double salts, the metal exhibits normal valency. In coordination compounds, the metal satisfies two types of valencies: primary (oxidation state) and secondary (coordination number).
Quick Tip: Remember: Double salts exist only in solid state, while coordination compounds exist in both solid and solution states.

Definition: A monodentate (or unidentate) ligand is a ligand that has only one donor atom capable of forming a coordinate bond with the central metal ion.


Examples:

1. Ammonia (:NH\(_3\)) - The Nitrogen atom is the donor.

2. Water (H\(_2\)O:) - The Oxygen atom is the donor.

3. Chloride ion (Cl\(^-\)) - Acts as a donor.
Quick Tip: "Mono" means one and "dentate" refers to teeth/binding sites. So, "one-toothed" ligands bind at one site only.

1. Alcohols (R-OH) contain a highly electronegative oxygen atom attached to a hydrogen atom, which allows them to form intermolecular hydrogen bonds.


2. These hydrogen bonds are strong attractive forces that require significant energy to break during boiling.


3. Haloalkanes (R-X) are polar but cannot form hydrogen bonds with each other; they only have weaker dipole-dipole interactions and Van der Waals forces.


4. Consequently, the boiling point of alcohols is much higher than that of haloalkanes of comparable molecular mass.
Quick Tip: Hydrogen bonding is the key factor for the unusually high boiling points of alcohols, water, and carboxylic acids compared to other organic compounds.

Step 1: Dehydration: Heat Propan-1-ol with concentrated H\(_2\)SO\(_4\) at 443 K to form Propene.

CH\(_3\)CH\(_2\)CH\(_2\)OH \(\xrightarrow[443 K]{conc. H_2SO_4}\) CH\(_3\)CH=CH\(_2\) + H\(_2\)O


Step 2: Hydration: Treat Propene with water in the presence of acid (Acid-catalyzed hydration). The addition follows Markovnikov's rule, where the OH group attaches to the more substituted carbon (the middle one).

CH\(_3\)CH=CH\(_2\) + H\(_2\)O \(\xrightarrow{H^+}\) CH\(_3\)CH(OH)CH\(_3\) (Propan-2-ol)
Quick Tip: Conversion of \(1^\circ\) alcohol to \(2^\circ\) alcohol always involves an alkene intermediate. Dehydration \(\to\) Alkene \(\to\) Markovnikov Hydration.

1. Definition: Rosenmund reduction is a chemical reaction used to convert acid chlorides (acyl chlorides) into aldehydes.


2. Reagents: Hydrogen gas (H\(_2\)) is passed through the solution in the presence of a catalyst, which is Palladium (Pd) deposited on Barium Sulphate (BaSO\(_4\)). The catalyst is often poisoned with sulfur or quinoline to prevent further reduction to alcohols.


3. Reaction:

R-COCl + H\(_2\) \(\xrightarrow{Pd/BaSO_4}\) R-CHO + HCl


4. Example: Benzoyl chloride is reduced to Benzaldehyde.

C\(_6\)H\(_5\)COCl + H\(_2\) \(\xrightarrow{Pd/BaSO_4}\) C\(_6\)H\(_5\)CHO + HCl
Quick Tip: BaSO\(_4\) acts as a catalyst poison here. Without it, the highly active Pd catalyst would reduce the aldehyde further into a primary alcohol.

1. Lower carboxylic acids (like Methanoic acid, Ethanoic acid) have a polar carboxyl group (-COOH).


2. The oxygen atoms in the carboxyl group and the hydrogen atom of the -OH group can form hydrogen bonds with water molecules.

(R-COOH \(\cdots\) OH\(_2\))


3. This extensive hydrogen bonding allows the acid molecules to mix freely with water.


4. As the size of the hydrophobic alkyl chain (R) increases, the solubility decreases because the hydrophobic interaction overcomes the hydrogen bonding.
Quick Tip: "Like dissolves like." Since water is polar and H-bonded, it dissolves other polar, H-bonding substances like lower carboxylic acids.

Step 1: Diazotization: Aniline is treated with Sodium Nitrite (NaNO\(_2\)) and Hydrochloric acid (HCl) at a low temperature (0-5\(^\circ\)C) to form Benzene Diazonium Chloride.

C\(_6\)H\(_5\)NH\(_2\) + NaNO\(_2\) + 2HCl \(\xrightarrow{0-5^\circC}\) C\(_6\)H\(_5\)N\(_2^+\)Cl\(^-\) + NaCl + 2H\(_2\)O


Step 2: Sandmeyer Reaction: The Benzene Diazonium Chloride is treated with Cuprous Chloride (Cu\(_2\)Cl\(_2\) or CuCl) in HCl.

C\(_6\)H\(_5\)N\(_2^+\)Cl\(^-\) \(\xrightarrow{CuCl/HCl}\) C\(_6\)H\(_5\)Cl + N\(_2\)


The product C\(_6\)H\(_5\)Cl is Chlorobenzene.
Quick Tip: The diazonium salt is a versatile intermediate. Changing the copper salt allows you to make Bromobenzene (CuBr) or Benzonitrile (CuCN) as well.

1. Basicity depends on the availability of the lone pair on the nitrogen atom to accept a proton.


2. In Methylamine (CH\(_3\)NH\(_2\)), the methyl group (-CH\(_3\)) is an electron-releasing group (has a +I effect).


3. This effect increases the electron density on the nitrogen atom, making the lone pair more available for donation compared to Ammonia (NH\(_3\)).


4. Additionally, the methylammonium ion formed after accepting a proton is stabilized by the +I effect of the methyl group, shifting the equilibrium forward.


Thus, Methylamine is more basic than Ammonia.
Quick Tip: In gas phase, basicity order is \(3^\circ > 2^\circ > 1^\circ > NH_3\). In aqueous phase, solvation effects change this order, but all alkyl amines remain stronger than ammonia.

1. Sugar: DNA contains the sugar 2-deoxyribose, whereas RNA contains the sugar ribose.


2. Nitrogenous Bases: DNA contains the pyrimidine base Thymine (T). RNA contains Uracil (U) instead of Thymine. (Both contain Adenine, Guanine, and Cytosine).


3. Structure: DNA usually exists as a double-stranded helix. RNA usually exists as a single-stranded molecule.
Quick Tip: An easy way to remember: DNA = Double stranded, Deoxyribose. RNA = Ribose, and "U" comes after "T" alphabetically? No, just remember U replaces T.

1. Raoult's Law: Ideal solutions obey Raoult's law (\(P_A = P_A^0 X_A\)) over the entire range of concentration. Non-ideal solutions deviate (positively or negatively) from Raoult's law.


2. Thermodynamics of Mixing: For an ideal solution, the enthalpy of mixing (\(\Delta H_{mix}\)) is zero, and the volume of mixing (\(\Delta V_{mix}\)) is zero. For a non-ideal solution, \(\Delta H_{mix} \neq 0\) and \(\Delta V_{mix} \neq 0\).
Quick Tip: Ideal solutions are rare (e.g., Benzene + Toluene). Most real solutions are non-ideal due to differences in intermolecular forces between solute and solvent.

Given:

Molality of solution, \(m = 0.1\) molal.

Boiling point of solution, \(T_b = 100.052^\circ\)C.

Boiling point of pure water, \(T_b^0 = 100^\circ\)C.


Calculate elevation in boiling point (\(\Delta T_b\)):
\(\Delta T_b = T_b - T_b^0 = 100.052 - 100 = 0.052^\circ\)C (or 0.052 K).


Formula for elevation in boiling point:
\(\Delta T_b = K_b \times m\)


Rearrange to solve for \(K_b\):
\(K_b = \frac{\Delta T_b}{m}\)
\(K_b = \frac{0.052}{0.1}\)
\(K_b = 0.52\) K kg mol\(^{-1}\).
Quick Tip: The value 0.52 K kg/mol is a standard constant for water. If you get a calculation result far from this, check your math!

Given:

Mass of solute (Glucose), \(w_2 = 18\) g.

Molar mass of Glucose (C\(_6\)H\(_{12}\)O\(_6\)), \(M_2 = 180\) g/mol.

Mass of solvent (Water), \(w_1 = 1\) kg.
\(K_b = 0.52\) K kg mol\(^{-1}\).


Step 1: Calculate moles of glucose (\(n_2\)):
\(n_2 = \frac{18}{180} = 0.1\) mol.


Step 2: Calculate Molality (\(m\)):
\(m = \frac{moles of solute}{mass of solvent in kg} = \frac{0.1}{1} = 0.1\) m.


Step 3: Calculate Elevation in boiling point (\(\Delta T_b\)):
\(\Delta T_b = K_b \times m = 0.52 \times 0.1 = 0.052\) K.


Step 4: Calculate Boiling Point of solution (\(T_b\)):

Pure water boils at 373.15 K (100\(^\circ\)C) at 1.013 bar.
\(T_b = T_b^0 + \Delta T_b = 373.15 + 0.052 = 373.202\) K.

(In Celsius: \(100 + 0.052 = 100.052^\circ\)C).
Quick Tip: Always check the pressure units. 1.013 bar is standard atmospheric pressure, so pure water boils at 100\(^\circ\)C.

1. Energy Conversion: A Galvanic cell converts chemical energy into electrical energy (spontaneous reaction). An Electrolytic cell converts electrical energy into chemical energy (non-spontaneous reaction).


2. Polarity of Electrodes: In a Galvanic cell, the anode is negative and the cathode is positive. In an Electrolytic cell, the anode is positive and the cathode is negative.
Quick Tip: In both cells, oxidation always occurs at the anode and reduction at the cathode. Only the sign changes.

The oxidation reaction of water is:

H\(_2\)O \(\to\) 2H\(^+\) + \(\frac{1}{2}\)O\(_2\) + 2e\(^-\)


1. Moles of Water:

Molar mass of H\(_2\)O = 18 g/mol.

Moles of H\(_2\)O = \(\frac{90}{18} = 5\) moles.


2. Electrons required:

From the equation, oxidation of 1 mole of H\(_2\)O releases 2 moles of electrons.

So, 5 moles of H\(_2\)O will release \(5 \times 2 = 10\) moles of electrons.


3. Charge Calculation:

Charge of 1 mole of electrons (1 Faraday) \(\approx\) 96487 C.

Total Charge = \(10 \times 96487 = 964870\) C.
Quick Tip: Always start by writing the balanced half-reaction to determine 'n' (number of electrons transferred per mole).

1. Definition/Determination: Order is the sum of powers of concentration terms in the experimental rate law; it is an experimental quantity. Molecularity is the number of reacting species colliding simultaneously in an elementary step; it is a theoretical concept.


2. Values: Order can be zero, fractional, or an integer. Molecularity can only be a positive integer (1, 2, 3) and cannot be zero or fractional.
Quick Tip: Molecularity has no meaning for complex (multi-step) reactions, whereas order applies to both elementary and complex reactions.

For a first-order reaction, the integrated rate equation is:
\(t = \frac{2.303}{k} \log \frac{[A]_0}{[A]}\)


Given:
\(k = 60\) s\(^{-1}\).

We want the final concentration \([A]\) to be \(\frac{1}{10}\) of initial \([A]_0\).

So, \(\frac{[A]_0}{[A]} = 10\).


Substitute values:
\(t = \frac{2.303}{60} \log(10)\)

Since \(\log(10) = 1\):
\(t = \frac{2.303}{60}\)
\(t \approx 0.03838\) s.


Rounding to significant figures, \(t \approx 3.84 \times 10^{-2}\) s.
Quick Tip: Memorize \(\ln(10) \approx 2.303\). This conversion factor is ubiquitous in kinetics problems involving base-10 logs.

For a first-order reaction, the half-life (\(t_{1/2}\)) is given by the formula:
\(t_{1/2} = \frac{0.693}{k}\)


Given:
\(k = 8\) min\(^{-1}\).


Substitute the value:
\(t_{1/2} = \frac{0.693}{8}\)
\(t_{1/2} = 0.086625\) min.


Rounding to reasonable precision: \(t_{1/2} \approx 0.087\) min (or approx 5.2 seconds).
Quick Tip: The half-life of a first-order reaction is constant and independent of the initial concentration.

1. Zirconium (Zr) belongs to the 4d series and Hafnium (Hf) belongs to the 5d series of transition elements.


2. Usually, atomic size increases down a group. However, the filling of 4f orbitals before Hf results in "Lanthanoid Contraction" (poor shielding by f-electrons).


3. This contraction causes the atomic radius of Hf (159 pm) to be virtually identical to that of Zr (160 pm).


4. Because they have similar atomic radii and the same valence shell configuration, they possess very similar physical and chemical properties.
Quick Tip: Zr and Hf are often called "chemical twins" because separating them is extremely difficult due to this similarity.

"Two-third life" (\(t_{2/3}\)) is the time required for the reaction to be 2/3 complete.

This means the amount consumed is \(2/3\), so the amount remaining \([A]\) is \(1 - 2/3 = 1/3\) of \([A]_0\).

So, \(\frac{[A]_0}{[A]} = 3\).


Formula: \(t = \frac{2.303}{k} \log \frac{[A]_0}{[A]}\)


Substitute values (\(k = 5.48 \times 10^{-14}\) s\(^{-1}\)):
\(t_{2/3} = \frac{2.303}{5.48 \times 10^{-14}} \log(3)\)


We know \(\log(3) \approx 0.4771\).
\(t_{2/3} = \frac{2.303 \times 0.4771}{5.48 \times 10^{-14}}\)
\(t_{2/3} = \frac{1.0987}{5.48} \times 10^{14}\)
\(t_{2/3} \approx 0.2005 \times 10^{14}\)
\(t_{2/3} \approx 2.0 \times 10^{13}\) s.
Quick Tip: Make sure you understand if "2/3 life" means remaining is 2/3 or consumed is 2/3. In kinetics contexts, it usually refers to completion, so remaining = 1 - completion.

1. The acidity of alcohols depends on the stability of the alkoxide ion (R-O\(^-\)) formed after releasing a proton.


2. Alkyl groups are electron-releasing (+I effect). They increase electron density on the oxygen atom, destabilizing the negative charge on the alkoxide ion.


3. Primary alcohols (\(1^\circ\)): Have one alkyl group. Least +I effect. Alkoxide is most stable. Strongest acid.


4. Secondary alcohols (\(2^\circ\)): Have two alkyl groups. Moderate +I effect.


5. Tertiary alcohols (\(3^\circ\)): Have three alkyl groups. Maximum +I effect. Alkoxide is least stable. Weakest acid.


Order of Acidity: \(1^\circ > 2^\circ > 3^\circ\).
Quick Tip: Acidity is inversely proportional to the +I effect of the attached groups. More alkyl groups = Less acidic.

1. Reaction: When phenol is treated with chloroform (CHCl\(_3\)) in the presence of aqueous sodium hydroxide (NaOH) at 340K, followed by acid hydrolysis.


2. Mechanism: The reaction involves an electrophilic substitution where the electrophile is dichlorocarbene (:CCl\(_2\)).


3. Product: An aldehyde group (-CHO) is introduced, predominantly at the ortho position to the phenolic -OH group.


4. Equation:

Phenol + CHCl\(_3\) + 3NaOH \(\to\) Salicylaldehyde (2-Hydroxybenzaldehyde) + 3NaCl + 2H\(_2\)O.
Quick Tip: This is a specific method to synthesize Salicylaldehyde. If CCl\(_4\) is used instead of CHCl\(_3\), the product is Salicylic Acid (Kolbe-like product).

1. Resonance Effect: In aniline (C\(_6\)H\(_5\)NH\(_2\)), the lone pair of electrons on the nitrogen atom is in conjugation with the benzene ring. It participates in resonance, delocalizing the negative charge over the ortho and para positions of the ring. This makes the lone pair less available for donation to an acid.


2. Inductive Effect: The phenyl group exerts a -I (electron withdrawing) effect, further reducing electron density on nitrogen. In contrast, the ethyl group in ethylamine has a +I (electron releasing) effect, increasing electron density on nitrogen.


3. Stability of Conjugate Acid: The anilinium ion formed by protonation is less stable because resonance is disrupted (to an extent), whereas the ethylammonium ion is stabilized by the +I effect of the ethyl group.
Quick Tip: Key phrase: "Delocalization of lone pair." Whenever an amine's N is attached to a benzene ring, basicity drops drastically.

According to Kohlrausch's Law of independent migration of ions, the limiting molar conductivity of an electrolyte is the sum of the limiting ionic conductivities of the cation and anion, multiplied by the number of ions per formula unit.


Formula for CaCl\(_2\):
\(\Lambda_m^\circ(CaCl_2) = \lambda^\circ(Ca^{2+}) + 2 \times \lambda^\circ(Cl^-)\)


Substitute values:
\(\Lambda_m^\circ = 119.5 + 2 \times (76.3)\)
\(\Lambda_m^\circ = 119.5 + 152.6\)
\(\Lambda_m^\circ = 272.1\) S cm\(^2\) mol\(^{-1}\).
Quick Tip: Don't forget the stoichiometry! CaCl\(_2\) produces two chloride ions, so you must multiply the chloride conductivity by 2.

Cell Reaction: Fe(s) + 2H\(^+\) \(\to\) Fe\(^{2+}\) + H\(_2\)(g)

Here, \(n=2\) electrons transferred.


E\(^\circ_{cell} = E^\circ_{cathode} - E^\circ_{anode}\)

Cathode is SHE (H\(^+\)/H\(_2\)), so \(E^\circ = 0.00\) V.

Anode is Fe/Fe\(^{2+}\), so \(E^\circ = -0.44\) V.

E\(^\circ_{cell} = 0 - (-0.44) = +0.44\) V.


Nernst Equation:
\(E_{cell} = E^\circ_{cell} - \frac{0.059}{n} \log \frac{[Products]}{[Reactants]}\)
\(E_{cell} = 0.44 - \frac{0.059}{2} \log \frac{[Fe^{2+}]}{[H^+]^2}\)


Substitute values (\([Fe^{2+}]=10^{-3}\), \([H^+]=1\)):
\(E_{cell} = 0.44 - 0.0295 \log \frac{10^{-3}}{1^2}\)
\(E_{cell} = 0.44 - 0.0295 \times (-3)\)
\(E_{cell} = 0.44 + 0.0885\)
\(E_{cell} = 0.5285\) V.
Quick Tip: Pay attention to the signs. \(\log(10^{-3})\) is \(-3\), so the term adds to the standard potential. Lower product concentration increases voltage.

(i) Wurtz-Fittig Reaction: Reaction of an aryl halide with an alkyl halide and sodium metal to form substituted benzene.

C\(_6\)H\(_5\)X + 2Na + R-X \(\xrightarrow{Dry Ether}\) C\(_6\)H\(_5\)-R + 2NaX


(ii) Ullmann Reaction: Heating an aryl iodide with copper powder to form biaryl.

2 C\(_6\)H\(_5\)I + 2Cu \(\xrightarrow{\Delta}\) C\(_6\)H\(_5\)-C\(_6\)H\(_5\) (Biphenyl) + 2CuI


(iii) Gattermann Reaction: Modification of Sandmeyer reaction using Copper powder instead of CuCl.

C\(_6\)H\(_5\)N\(_2^+\)Cl\(^-\) \(\xrightarrow{Cu/HCl}\) C\(_6\)H\(_5\)Cl + N\(_2\)


(iv) Hunsdiecker Reaction: Reaction of silver salt of carboxylic acid with Bromine to form alkyl bromide (Decarboxylation).

R-COOAg + Br\(_2\) \(\xrightarrow{CCl_4, \Delta}\) R-Br + CO\(_2\) + AgBr


(v) Balz-Schiemann Reaction: Preparation of fluorobenzene from diazonium salt using HBF\(_4\).

C\(_6\)H\(_5\)N\(_2^+\)Cl\(^-\) + HBF\(_4\) \(\to\) C\(_6\)H\(_5\)N\(_2^+\)BF\(_4^-\) \(\xrightarrow{\Delta}\) C\(_6\)H\(_5\)F + N\(_2\) + BF\(_3\)
Quick Tip: Name reactions are high-yield questions. Create a chart: Reactant \(\to\) Reagent \(\to\) Product for each.

1. Steps: SN\(^1\) is a two-step reaction (formation of carbocation, then nucleophile attack). SN\(^2\) is a single-step concerted reaction (bond breaking and making happen simultaneously).


2. Kinetics: SN\(^1\) is first order (Unimolecular), Rate \(\propto\) [Alkyl Halide]. SN\(^2\) is second order (Bimolecular), Rate \(\propto\) [Alkyl Halide][Nucleophile].


3. Stereochemistry: SN\(^1\) leads to racemization (retention + inversion). SN\(^2\) leads to complete inversion of configuration (Walden inversion).


4. Reactivity Order: For SN\(^1\): \(3^\circ > 2^\circ > 1^\circ\) (due to carbocation stability). For SN\(^2\): \(1^\circ > 2^\circ > 3^\circ\) (due to steric hindrance).
Quick Tip: SN1 = "1" stands for Unimolecular, but it has 2 steps. SN2 = "2" stands for Bimolecular, but it has 1 step.

Optical activity is the property of certain organic substances to rotate the plane of plane-polarized light when it is passed through their solution.

If the compound rotates light to the right (clockwise), it is dextrorotatory (\(d\) or +).

If it rotates light to the left (counter-clockwise), it is levorotatory (\(l\) or -).

This property arises due to molecular chirality (lack of a plane of symmetry).
Quick Tip: A carbon atom attached to four different groups is a chiral center, which is the primary source of optical activity in organic molecules.

1. Oxidation States: Lanthanoids show +3 as the dominant oxidation state, with occasional +2 or +4. Actinoids show a wider range of oxidation states (+3, +4, +5, +6, +7) due to comparable energies of 5f, 6d, and 7s shells.


2. Binding Energy: 4f electrons (Lanthanoids) have higher binding energy and are more deeply buried. 5f electrons (Actinoids) have lower binding energy and are more available for bonding (covalent character).


3. Radioactivity: Only Promethium is radioactive among Lanthanoids. All Actinoids are radioactive.
Quick Tip: Actinoid contraction is greater than Lanthanoid contraction due to poorer shielding by 5f electrons compared to 4f electrons.

1. Cu\(^{+2}\) Configuration: Copper (atomic number 29) has configuration [Ar] 3d\(^{10}\) 4s\(^1\). Cu\(^{+2}\) is [Ar] 3d\(^9\).

It has one unpaired electron in the d-orbital. This allows for d-d transitions (movement of electron between split d-orbitals) by absorbing visible light, resulting in colour.


2. Zn\(^{+2}\) Configuration: Zinc (atomic number 30) has configuration [Ar] 3d\(^{10}\) 4s\(^2\). Zn\(^{+2}\) is [Ar] 3d\(^{10}\).

The d-subshell is completely filled. There are no empty d-orbitals for electron transition. Hence, no visible light is absorbed, and the salts appear colourless (white).
Quick Tip: Colour in transition metals usually requires an incomplete d-subshell (d\(^1\) to d\(^9\)). d\(^0\) and d\(^{10}\) are generally colourless.

1. Transition metals have very similar atomic sizes (radii) within a series (e.g., Cr, Mn, Fe, Co, Ni).


2. Because the atoms are of similar size, atoms of one metal can easily replace the atoms of another metal in its crystal lattice without disrupting the structure significantly.


3. This ready substitution allows the formation of solid solutions or alloys (like Brass, Bronze, Steel) which are hard and have high melting points.
Quick Tip: Interstitial compounds are different: they form when small atoms (H, C, N) get trapped in the lattice gaps. Alloys involve metal-metal substitution.

1. Fe\(^{+3}\) (Iron, Z=26):

Neutral Fe: [Ar] 3d\(^6\) 4s\(^2\).

Fe\(^{+3}\) (remove 2 from s, 1 from d): [Ar] 3d\(^5\).

Hund's rule: 5 electrons in 5 d-orbitals \(\implies\) 5 unpaired electrons.


2. Zn\(^{+2}\) (Zinc, Z=30):

Neutral Zn: [Ar] 3d\(^{10}\) 4s\(^2\).

Zn\(^{+2}\) (remove 2 from s): [Ar] 3d\(^{10}\).

All electrons paired \(\implies\) 0 unpaired electrons.


3. Mn\(^{+2}\) (Manganese, Z=25):

Neutral Mn: [Ar] 3d\(^5\) 4s\(^2\).

Mn\(^{+2}\) (remove 2 from s): [Ar] 3d\(^5\).

Hund's rule: 5 electrons in 5 d-orbitals \(\implies\) 5 unpaired electrons.
Quick Tip: Half-filled (d\(^5\)) and fully-filled (d\(^{10}\)) shells are extra stable. Both Fe\(^{+3}\) and Mn\(^{+2}\) are d\(^5\).