NEET / State Class 12 Board (e.g. ISC) Chemistry Syllabus

Calculate E°(cell) for Zn|Zn²⁺||Cu²⁺|Cu given E°(Zn²⁺/Zn)=−0.76 V, E°(Cu²⁺/Cu)=+0.34 V.

E°(cell) = E°(cathode) − E°(anode) = 0.34 − (−0.76) = +1.10 V.

Why does N show a maximum covalency of 4 while P can show 5 or 6?

Nitrogen has no vacant d-orbitals (only 2s, 2p), limiting it to 4 bonds. Phosphorus has accessible 3d orbitals, allowing expansion of the octet (e.g. PCl₅, PF₆⁻).

Why is the boiling point of HF higher than HCl despite F being smaller?

HF forms strong intermolecular hydrogen bonds due to fluorine's high electronegativity, raising its boiling point above the otherwise expected order.

Why does nitrogen exist as N₂ (gas) while phosphorus exists as P₄ (solid)?

Nitrogen forms a strong pπ–pπ triple bond (N≡N) due to its small size; phosphorus cannot form effective pπ–pπ bonds, so it forms single-bonded P₄ tetrahedra.

What is the trend in oxidising power of halogens and the strongest oxidiser?

Oxidising power decreases down the group: F₂ > Cl₂ > Br₂ > I₂. Fluorine is the strongest oxidising agent due to low bond dissociation energy and high hydration enthalpy.

Why is the +3 oxidation state more stable than +5 for Bi (inert pair effect)?

Down group 15, the 6s electrons (inert pair) are reluctant to participate in bonding due to poor shielding by d/f electrons, making the lower (+3) state more stable for heavy elements like Bi.

What are interhalogen compounds and give an example with shape.

Compounds between two different halogens, e.g. ClF₃ (T-shaped, sp³d), IF₇ (pentagonal bipyramidal, sp³d³). More reactive than halogens (except F₂).

Why do transition metals show variable oxidation states?

Because the energies of (n−1)d and ns orbitals are very close, electrons from both can participate in bonding, giving multiple stable oxidation states.

Why are most transition metal compounds coloured?

Due to d-d electronic transitions: an electron absorbs visible light to jump between split d-orbitals, and the complementary colour is observed. Requires partially filled d-orbitals.

What are lanthanoid contraction and its consequence?

The steady decrease in atomic/ionic radii across the lanthanoids due to poor shielding by 4f electrons. Consequence: Zr and Hf have nearly identical sizes and similar properties.

Why is Mn²⁺ more stable than Mn³⁺, while Fe³⁺ is more stable than Fe²⁺?

Mn²⁺ has a half-filled stable 3d⁵ configuration; Fe³⁺ also has stable 3d⁵, so Fe³⁺ is favoured over Fe²⁺ (3d⁶).

Why do Zn, Cd, and Hg not show typical transition metal properties?

They have completely filled d¹⁰ configurations in their common (+2) states, so they lack partially filled d-orbitals — hence no variable valency, colour, or strong catalysis.

Write the spin-only magnetic moment formula.

μ = √[n(n+2)] BM, where n is the number of unpaired electrons. (e.g. n=5 gives μ ≈ 5.92 BM.)

State Werner's definition of primary and secondary valency in coordination compounds.

Primary valency = oxidation number (ionisable, satisfied by anions); secondary valency = coordination number (non-ionisable, satisfied by ligands, fixed and directional).

What is a chelate ligand? Give an example.

A polydentate (bidentate or higher) ligand that binds the central metal at two or more points forming a ring, e.g. ethylenediamine (en) or EDTA. Chelation increases complex stability.

State the IUPAC name of [Co(NH₃)₆]Cl₃.

Hexaamminecobalt(III) chloride.

What does the crystal field splitting energy (Δ₀) determine in octahedral complexes?

Whether a complex is high-spin or low-spin. If Δ₀ > pairing energy (strong field ligand), electrons pair → low spin; if Δ₀ < pairing energy (weak field), → high spin.

Arrange these ligands by increasing field strength (spectrochemical series): I⁻, H₂O, CN⁻, NH₃.

I⁻ < H₂O < NH₃ < CN⁻. CN⁻ is a strong-field ligand, I⁻ is weak-field.

How many unpaired electrons does [Fe(CN)₆]³⁻ have and what is its geometry?

Fe is +3 (d⁵); CN⁻ is strong field → low spin (t₂g⁵) → 1 unpaired electron. Octahedral, d²sp³ hybridisation.

Which gives faster SN1 reactions: 3° or 1° haloalkanes, and why?

3° (tertiary) react faster by SN1 because the intermediate carbocation is stabilised by hyperconjugation and inductive effects from the alkyl groups.

Why are haloarenes less reactive than haloalkanes toward nucleophilic substitution?

Resonance gives the C–X bond partial double-bond character (shorter, stronger), plus sp² carbon holds electrons tightly and the aryl ring repels nucleophiles.

What is Markovnikov vs anti-Markovnikov addition of HBr to alkenes?

Markovnikov: H adds to the carbon with more H's (ionic mechanism). Anti-Markovnikov (peroxide effect/Kharasch): with peroxides, HBr adds opposite via a free-radical mechanism.

What is the Williamson ether synthesis?

Reaction of a sodium alkoxide (R–O⁻Na⁺) with a primary alkyl halide (SN2) to give an ether: R–O⁻ + R'–X → R–O–R' + X⁻. Use primary halides to avoid elimination.

Why are phenols more acidic than alcohols?

The phenoxide ion formed after losing H⁺ is stabilised by resonance delocalisation into the aromatic ring, whereas an alkoxide ion is not resonance-stabilised.

What does the Lucas test distinguish, and which alcohol reacts fastest?

It distinguishes 1°, 2°, and 3° alcohols using conc. HCl + ZnCl₂ (turbidity). 3° reacts immediately, 2° in ~5 min, 1° only on heating.

Why are aldehydes more reactive than ketones toward nucleophilic addition?

Aldehydes have less steric hindrance and fewer electron-donating alkyl groups, so the carbonyl carbon is more electrophilic and more accessible.

What is the aldol condensation reaction?

Two molecules of an aldehyde/ketone with α-hydrogens combine under dilute base to form a β-hydroxy carbonyl (aldol), which on heating loses water to give an α,β-unsaturated carbonyl.

Why are carboxylic acids more acidic than phenols and alcohols?

The carboxylate ion is stabilised by resonance over two equivalent oxygen atoms, delocalising the negative charge more effectively than phenoxide or alkoxide.

Why are aliphatic amines more basic than ammonia in aqueous solution (for 1°/2°)?

Alkyl groups are electron-donating (+I effect), increasing electron density on nitrogen so it donates its lone pair more readily; solvation of the cation also stabilises the conjugate acid.

What is the Hinsberg test used to distinguish?

It distinguishes 1°, 2°, and 3° amines using benzenesulphonyl chloride: 1° gives a base-soluble product, 2° gives a base-insoluble product, 3° does not react.

Name the bond linking amino acids in proteins and the bond linking monosaccharides in carbohydrates.

Amino acids are linked by peptide (amide) bonds; monosaccharides are linked by glycosidic bonds.