UV-Visible spectroscopy studies the absorption of ultraviolet (200–400 nm) and visible (400–800 nm) light by molecules. The absorbed energy promotes electrons from lower to higher energy orbitals (electronic transitions). This topic covers the Beer–Lambert law, types of electronic transitions, chromophores and auxochromes, spectral shifts, and Woodward–Fieser rules for calculating λmax.
Key points
Energy and wavelength: E = hν = hc/λ. Shorter wavelength means higher energy.
Beer–Lambert law: absorbance is proportional to concentration and path length, A = εcl.
Electronic transitions (in order of decreasing energy): σ → σ* (below 200 nm, e.g. alkanes), n → σ* (e.g. alcohols, amines, around 150–250 nm), π → π* (alkenes, conjugated systems; strong, high ε), n → π* (carbonyl compounds; weak, longer wavelength, around 270–300 nm for simple ketones).
Chromophore: group responsible for absorption, usually with π electrons: C=C, C=O, N=N, NO₂.
Auxochrome: a group with lone pairs (−OH, −NH₂, −OR, −Cl) that does not absorb by itself above 200 nm but shifts the absorption of a chromophore to longer wavelength and raises intensity.
Bathochromic (red) shift: λmax moves to longer wavelength. Hypsochromic (blue) shift: to shorter wavelength.
Hyperchromic effect: increase in intensity (ε). Hypochromic effect: decrease in intensity.
Conjugation lowers the HOMO–LUMO gap, so λmax increases: ethene about 171 nm, 1,3-butadiene about 217 nm. Long conjugated molecules like β-carotene absorb in the visible region and are coloured.
Polar solvents shift n → π* bands to shorter wavelength (blue shift) and π → π* bands often to longer wavelength.
Woodward–Fieser rules estimate λmax of conjugated dienes and α,β-unsaturated carbonyl compounds.
Molar absorptivity (ε) – absorbance of a 1 mol/L solution in a 1 cm cell; unit L mol⁻¹ cm⁻¹.
Chromophore (वर्णधारी गट) – group that causes absorption of UV/visible light.
Auxochrome – colour-enhancing group with lone pairs.
Transmittance (T) – fraction of light that passes through, I/I₀.
Formulas
E = hν = hc/λ
A = log (I₀/I) = −log T = εcl (c in mol/L, l in cm)
Woodward–Fieser (conjugated dienes): base value acyclic diene (butadiene) 217 nm, heteroannular diene 214 nm, homoannular diene 253 nm; add 30 nm per extra conjugated C=C (double bond extending conjugation), 5 nm per alkyl substituent or ring residue, 5 nm per exocyclic double bond.
α,β-unsaturated ketone (acyclic or six-membered ring): base 215 nm; α-alkyl +10, β-alkyl +12 nm.
5 important questions
Q1. Define chromophore and auxochrome with examples.
A chromophore is a group that absorbs UV/visible light because of its π or n electrons, like C=C, C=O and NO₂. An auxochrome is a group with lone pairs like −OH or −NH₂ which, when attached to a chromophore, shifts λmax to longer wavelength and increases intensity.
Q2. Why does λmax increase with conjugation?
Conjugation spreads π electrons over more atoms. This brings the highest occupied and lowest unoccupied π orbitals closer in energy. A smaller energy gap means absorption at longer wavelength.
Q3. A solution in a 1 cm cell has absorbance 0.60. If ε = 1.2 × 10⁴ L mol⁻¹ cm⁻¹, find the concentration.
A = εcl → c = A/(εl)
c = 0.60/(1.2 × 10⁴ × 1)
c = 5.0 × 10⁻⁵ mol/L
Q4. Calculate λmax for 2,3-dimethyl-1,3-butadiene using Woodward–Fieser rules.
Acyclic diene (butadiene) base value: 217 nm
Alkyl substituents on the diene carbons: 2 methyl groups × 5 = 10 nm
No ring residues, no exocyclic bonds, no extra conjugation
λmax = 217 + 10 = 227 nm (observed value is close, about 226 nm)
Q5. Explain the types of electronic transitions with examples.
n → π*: carbonyl, nitro groups; weak (ε small, under 100 for simple ketones), longest wavelength.
Show an energy level diagram: σ < π < n < π* < σ*.
Exam tip
Woodward–Fieser numericals are easy if you write the base value and each addition on separate lines. Learn the four transitions with the energy diagram.
Infrared (IR) spectroscopy studies the absorption of infrared radiation, which makes bonds in a molecule vibrate. Each functional group absorbs at a characteristic frequency, so the IR spectrum works like a fingerprint of the molecule. This topic covers the IR region, Hooke's law, types of vibrations, selection rule, factors affecting frequency, and characteristic group frequencies.
Key points
IR positions are given as wavenumber (cm⁻¹). The useful mid-IR range is 4000–400 cm⁻¹ (2.5–25 μm).
A molecule absorbs IR only if the vibration changes its dipole moment (selection rule). So homonuclear molecules like O₂, N₂ and H₂ are IR inactive.
The bond is treated as a spring (Hooke's law): frequency rises with bond strength and falls with heavier atoms.
Types of vibrations: stretching (symmetric and asymmetric; change in bond length) and bending (scissoring, rocking, wagging, twisting; change in bond angle). Stretching needs more energy than bending.
Number of fundamental vibrations: 3N − 6 for non-linear and 3N − 5 for linear molecules. CO₂ has 4; H₂O has 3.
Functional group region (4000–1500 cm⁻¹): peaks of O–H, N–H, C–H, C≡N, C=O, C=C.
Fingerprint region (1500–400 cm⁻¹): complex pattern unique to each molecule; used to confirm identity by comparison.
Hydrogen bonding lowers the O–H frequency and broadens the band (alcohols in concentrated solution; carboxylic acids give a very broad band about 3300–2500 cm⁻¹).
Conjugation lowers C=O frequency (e.g. conjugated ketones absorb lower than saturated ketones).
Bond order effect: C≡C (about 2100–2260) > C=C (about 1620–1680) > C–C (about 1200).
Applications: identifying functional groups, checking purity, following reactions (disappearance of C=O), studying hydrogen bonding, distinguishing cis/trans isomers.
Important terms
Wavenumber (ν̄) – number of waves per cm, 1/λ; unit cm⁻¹.
Stretching vibration (ताणन कंपन) – vibration that changes bond length.
Bending vibration – vibration that changes bond angle.
Fingerprint region – 1500–400 cm⁻¹; unique to each compound.
Reduced mass (μ) – m₁m₂/(m₁ + m₂).
Formulas
ν̄ = 1/λ (cm⁻¹); λ (μm) = 10⁴ / ν̄ (cm⁻¹)
Hooke's law: ν̄ = (1/2πc) √(k/μ), k = force constant, μ = m₁m₂/(m₁ + m₂), c = speed of light
A molecule absorbs IR only if its dipole moment changes during vibration. N₂ and O₂ are symmetric diatomic molecules with zero dipole moment, and stretching keeps it zero. So they do not absorb IR.
Q2. How will you distinguish ethanol and acetone by IR?
Ethanol shows a broad O–H stretching band near 3200–3550 cm⁻¹ and no C=O band. Acetone shows a strong C=O band near 1715 cm⁻¹ and no O–H band.
Q3. Calculate the number of fundamental vibrations in CO₂, H₂O and benzene.
CO₂ (linear, N = 3): 3 × 3 − 5 = 4
H₂O (non-linear, N = 3): 3 × 3 − 6 = 3
C₆H₆ (non-linear, N = 12): 3 × 12 − 6 = 30
Not all may appear (some are IR inactive or degenerate).
Q4. Convert a wavelength of 5.85 μm to wavenumber and identify the likely group.
ν̄ = 10⁴/λ(μm) = 10⁴/5.85
ν̄ ≈ 1709 cm⁻¹
Strong band near 1710 cm⁻¹ → C=O stretching of a ketone (or carboxylic acid).
Q5. Explain the factors affecting vibrational frequency.
Bond strength (force constant k): stronger bond, higher frequency (C≡C > C=C > C–C).
Mass of atoms (reduced mass): heavier atoms, lower frequency (C–H about 3000 vs C–D about 2200).
Hydrogen bonding: lowers frequency, broadens O–H and N–H bands.
Conjugation/resonance: lowers C=O frequency.
Inductive effect: electron-withdrawing groups raise C=O frequency (acid chlorides about 1800).
Exam tip
Learn the table of 8–10 group frequencies; questions often give a spectrum's main peaks and ask you to identify the compound.
Nuclear Magnetic Resonance (NMR) spectroscopy studies the absorption of radio-frequency radiation by certain nuclei (mainly ¹H) placed in a strong magnetic field. Proton (¹H) NMR tells us how many kinds of hydrogen a molecule has, their chemical environment, how many of each, and which neighbours they have. This topic covers the principle, shielding, chemical shift, TMS, equivalent protons, peak area (integration) and spin–spin splitting (n + 1 rule).
Key points
Nuclei with spin (I ≠ 0) like ¹H and ¹³C behave like tiny magnets. ¹²C and ¹⁶O have I = 0 and give no NMR signal.
In a magnetic field, a ¹H nucleus (I = ½) has two energy states (aligned and opposed). Absorption of radio waves flips the spin: this is resonance.
Electrons around a proton create a small opposing field. This shielding means a higher external field is needed. Less electron density = deshielding = signal at higher δ (downfield).
Tetramethylsilane (TMS), (CH₃)₄Si, is the reference (δ = 0). Reasons: 12 equivalent protons give one sharp signal; it is highly shielded so its signal lies away from most others; it is inert, volatile and soluble in organic solvents.
Chemical shift (δ) in ppm is independent of the instrument's frequency.
Number of signals = number of sets of chemically equivalent protons.
Area (integration) under each signal is proportional to the number of protons of that type.
Spin–spin splitting: a signal is split by protons on adjacent carbons. n + 1 rule: n equivalent neighbouring protons split a signal into n + 1 peaks (singlet, doublet, triplet, quartet…).
Multiplicity = n + 1 (n = equivalent neighbouring H)
Resonance condition: ν = γB₀/2π
Proton type
Approx. δ (ppm)
R–CH₃
0.9
CH₃–C=O
2.1
CH₃–O
3.3–3.8
C=C–H
4.5–6.5
Ar–H
6.5–8.0
R–CHO
9–10
R–COOH
10–12
5 important questions
Q1. Why is TMS used as the reference in NMR?
TMS has 12 equivalent protons, giving one strong sharp signal. Its protons are highly shielded, so the signal appears upfield of most organic protons. It is chemically inert, volatile (easy to remove) and soluble in common NMR solvents.
Q2. How many signals will be seen in the ¹H NMR of ethanol, acetone and toluene?
Ethanol (CH₃CH₂OH): 3 signals. Acetone (CH₃COCH₃): 1 signal (6 equivalent H). Toluene: 2 main signals at low resolution (CH₃ and aromatic H; aromatic H may split into more at high resolution).
Q3. A signal appears 1260 Hz from TMS on a 400 MHz instrument. Find δ.
δ = (Δν / spectrometer frequency) × 10⁶
= (1260 / 400 × 10⁶) × 10⁶
= 1260/400
δ = 3.15 ppm
Q4. Predict the ¹H NMR spectrum of ethyl bromide (CH₃CH₂Br).
Mass spectrometry measures the mass-to-charge ratio (m/z) of ions made from a molecule. It gives the molecular mass and, from the fragments, clues about structure. This topic covers the principle and instrument, the molecular ion and base peak, isotope peaks (M+1, M+2), the nitrogen rule, common fragmentation patterns, and simple problems.
Key points
In electron impact (EI) ionisation, the sample vapour is hit by a beam of electrons (usually 70 eV). One electron is knocked out to give the molecular ion M⁺• (a radical cation).
Excess energy breaks M⁺• into smaller fragment ions and neutral pieces. Only charged particles are detected.
Instrument: inlet → ionisation chamber → accelerating plates → magnetic (or quadrupole) analyser → detector → recorder. Ions are separated by m/z.
Mass spectrum: bar graph of relative abundance against m/z.
Base peak: tallest peak, given 100 % intensity (most stable ion).
Molecular ion peak (M⁺): gives molecular mass; strong for aromatic compounds, weak or absent for alcohols and highly branched alkanes.
Isotope peaks: ¹³C (1.1 %) gives an M+1 peak; Cl gives M : M+2 ≈ 3 : 1 (³⁵Cl/³⁷Cl); Br gives M : M+2 ≈ 1 : 1 (⁷⁹Br/⁸¹Br).
Nitrogen rule: a molecule with an odd molecular mass contains an odd number of N atoms; even mass means zero or even number of N.
alkanes: loss of CH₂ units → peaks 14 apart (m/z 29, 43, 57…)
α-cleavage in ketones → acylium ions (CH₃CO⁺, m/z 43)
aromatic compounds with CH₂ side chain → tropylium ion (C₇H₇⁺, m/z 91)
McLafferty rearrangement: carbonyl compounds with a γ-hydrogen transfer it to oxygen through a six-membered transition state and split off an alkene. Butanal gives m/z 44.
Metastable ions appear as broad peaks at m* = (m₂)²/m₁.
Important terms
m/z – mass-to-charge ratio; for singly charged ions equal to mass.
Base peak (मूळ शिखर) – most intense peak, set at 100 %.
Fragmentation – breaking of the molecular ion into smaller ions.
Formulas
m/z = B²r²e/(2V) (magnetic analyser: B = field, r = radius of path, V = accelerating voltage) — for a fixed B and V, heavier ions follow a larger radius.
Number of C atoms ≈ (height of M+1 / height of M) × 100 / 1.1
Metastable peak m* = m₂²/m₁
m/z
Ion
Typical source
15
CH₃⁺
methyl groups
29
C₂H₅⁺ / CHO⁺
ethyl, aldehydes
43
CH₃CO⁺ / C₃H₇⁺
methyl ketones, propyl
77
C₆H₅⁺
benzene ring
91
C₇H₇⁺ (tropylium)
toluene, benzyl compounds
105
C₆H₅CO⁺
acetophenone, benzoates
5 important questions
Q1. What is the base peak? Is it always the molecular ion peak?
The base peak is the tallest peak in the spectrum, given 100 % intensity. It is not always the molecular ion; often it is a stable fragment (m/z 43 in 2-butanone, m/z 91 in toluene).
Q2. A compound shows two molecular ion peaks at m/z 156 and 158 of nearly equal height. What does it suggest?
An M : M+2 ratio of about 1 : 1 shows one bromine atom (⁷⁹Br and ⁸¹Br are almost equally abundant). Mass 156 = C₆H₅ (77) + ⁷⁹Br, so the compound is bromobenzene. A 3 : 1 ratio would show one chlorine atom instead.
Q3. Explain the principle and working of a mass spectrometer.
Draw the block diagram. Explain vaporisation, ionisation by 70 eV electrons, acceleration, deflection in a magnetic field depending on m/z, detection and recording. Mention the high vacuum needed.
Q4. Explain the McLafferty rearrangement with an example.
Q5. Predict the main peaks in the mass spectrum of acetophenone (C₆H₅COCH₃, M = 120).
M⁺ at 120
Loss of CH₃ (α-cleavage) → C₆H₅CO⁺ at 105 (base peak)
Loss of CO from 105 → C₆H₅⁺ at 77
Loss of C₆H₅ → CH₃CO⁺ at 43
Ring fragment C₄H₃⁺ at 51
Exam tip
For every problem, first find M⁺, then check the nitrogen rule and isotope pattern, then subtract to name the lost fragments (M − 15 = CH₃, M − 18 = H₂O, M − 29 = CHO/C₂H₅, M − 43 = CH₃CO).
Crystal Field Theory (CFT), developed by Bethe and Van Vleck, explains bonding in coordination complexes as purely electrostatic attraction between the metal ion and the ligands (treated as point charges). Its main idea is the splitting of the five d orbitals. This topic covers splitting in octahedral and tetrahedral fields, the spectrochemical series, high-spin and low-spin complexes, crystal field stabilisation energy (CFSE), colour, magnetism, and limitations.
Key points
In a free metal ion, the five d orbitals have equal energy (degenerate). Ligands approaching the metal break this degeneracy.
Octahedral field: ligands come along the x, y, z axes. dx²−y² and dz² point at ligands and rise in energy (eg set); dxy, dyz, dxz lie between axes and fall (t2g set).
The energy gap is Δo (or 10 Dq). t2g is lowered by 0.4 Δo and eg is raised by 0.6 Δo.
Tetrahedral field: the splitting is reversed — e set lower, t2 set higher — and smaller: Δt ≈ (4/9) Δo. So tetrahedral complexes are almost always high spin.
Square planar (e.g. d⁸ Ni²⁺, Pt²⁺ with strong ligands): dx²−y² is very high in energy, so the 8 electrons pair in the lower four orbitals → diamagnetic.
If Δo < pairing energy (P): electrons enter eg before pairing → high spin (weak field). If Δo > P: electrons pair in t2g first → low spin (strong field). This choice matters only for d⁴ to d⁷ in octahedral complexes.
Colour: due to d–d transitions; the energy absorbed equals Δo. [Ti(H₂O)₆]³⁺ (d¹) absorbs around 500 nm and looks purple.
Δo increases with higher oxidation state of the metal and down a group (3d < 4d < 5d).
Jahn–Teller distortion: unevenly filled eg orbitals (e.g. Cu²⁺ d⁹) cause octahedral complexes to distort.
Limitations: treats ligands as point charges, ignores covalent character, cannot explain why CO (neutral) is a strong ligand. Ligand field / MO theory improves on it.
Important terms
Crystal field splitting (Δ) – energy gap between split d orbital sets.
Pairing energy (P) – energy needed to pair two electrons in one orbital.
CFSE (स्फटिक क्षेत्र स्थिरीकरण ऊर्जा) – energy gained by placing electrons in the lower split orbitals.
Strong field ligand – causes large Δ, gives low-spin complexes (CN⁻, CO).
Weak field ligand – small Δ, gives high-spin complexes (I⁻, F⁻, H₂O).
Formulas
Octahedral CFSE = (−0.4 × n(t2g) + 0.6 × n(eg)) Δo + (number of extra pairs) P
Tetrahedral: e lowered by 0.6 Δt, t2 raised by 0.4 Δt; Δt ≈ (4/9) Δo
Δo = hc/λ (from absorption wavelength); per mole multiply by Nₐ
μ (spin only) = √(n(n + 2)) BM
5 important questions
Q1. What is the spectrochemical series?
It is the arrangement of ligands in increasing order of the splitting (Δ) they produce. Part of it: I⁻ < Br⁻ < Cl⁻ < F⁻ < H₂O < NH₃ < en < CN⁻ < CO. Ligands at the right end are strong field ligands.
Q2. Why are tetrahedral complexes generally high spin?
In a tetrahedral field only four ligands approach and none point directly at the d orbitals, so Δt is small (about 4/9 of Δo). Δt is almost always smaller than the pairing energy, so electrons occupy higher orbitals before pairing.
Q3. Calculate the CFSE and magnetic moment of [Fe(CN)₆]⁴⁻ (low spin) and [Fe(H₂O)₆]²⁺ (high spin). Both have Fe²⁺, d⁶.
Q5. Explain crystal field splitting in an octahedral complex.
Six ligands approach along ±x, ±y, ±z.
dx²−y² and dz² lobes point at ligands → more repulsion → raised (eg).
dxy, dyz, dxz lobes point between ligands → lowered (t2g).
Barycentre rule: t2g at −0.4 Δo, eg at +0.6 Δo.
Draw the diagram: one level for the free ion, a spherical-field level, then the split levels with Δo marked.
Exam tip
For every complex, write: metal oxidation state → d count → weak or strong ligand → filling → CFSE → unpaired electrons → μ. This chain answers most CFT questions.
Electronic Spectra and Magnetic Properties of Complexes
Quick summary
Transition metal complexes are often coloured and many are paramagnetic. This topic explains both properties. It covers types of magnetic behaviour, the spin-only formula and its use to decide high-spin or low-spin, measurement by the Gouy method, d–d and charge-transfer spectra, selection rules, the Orgel diagram for d¹ ions, and the Jahn–Teller effect.
Key points
Diamagnetic substances have all electrons paired and are weakly repelled by a magnetic field. Paramagnetic substances have unpaired electrons and are attracted. Ferromagnetic substances (Fe, Co, Ni) show strong, permanent magnetism due to aligned domains.
Spin-only magnetic moment μs = √(n(n + 2)) BM, where n = number of unpaired electrons. Orbital contribution is mostly "quenched" in first-row complexes.
Measured μ tells n, and n tells whether the complex is high spin (weak-field ligand) or low spin (strong-field ligand). Example: [Fe(H₂O)₆]²⁺ (d⁶, high spin, n = 4, μ ≈ 4.90 BM) but [Fe(CN)₆]⁴⁻ (d⁶, low spin, n = 0, diamagnetic).
Gouy method: sample in a tube hung from a balance between the poles of a magnet; a paramagnetic sample appears heavier when the field is switched on.
Colour arises because the complex absorbs some wavelengths of visible light; we see the complementary colour.
d–d transitions: an electron moves from t₂g to eg; the energy absorbed equals Δo. [Ti(H₂O)₆]³⁺ (d¹) absorbs at about 20 300 cm⁻¹ (about 493 nm, green-yellow) and looks purple.
Laporte rule: transitions between orbitals of the same parity (d → d) are forbidden in centrosymmetric complexes, so octahedral d–d bands are weak. Tetrahedral complexes (no centre of symmetry) are more intensely coloured.
Spin rule: transitions that change the number of unpaired electrons are forbidden. [Mn(H₂O)₆]²⁺ (high-spin d⁵) has only spin-forbidden bands, so it is very pale pink.
Selection rules:
Charge-transfer (CT) bands are very intense: ligand-to-metal (MnO₄⁻ deep purple, CrO₄²⁻ yellow) or metal-to-ligand.
Orgel diagram: for d¹ (and d⁶ high spin) one band ²T₂g → ²Eg; d⁹ (Cu²⁺) also gives one broad band (the inverse diagram).
Jahn–Teller effect: a non-linear molecule with a degenerate electronic state distorts to remove the degeneracy. Strong in d⁹ (Cu²⁺) and high-spin d⁴ — octahedral Cu(II) complexes are elongated along the z-axis.
Complexes with d⁰ (TiO₂, Sc³⁺) and d¹⁰ (Zn²⁺) ions are colourless because no d–d transition is possible.
Important terms
Bohr magneton (BM) – unit of magnetic moment, 9.27 × 10⁻²⁴ J T⁻¹.
Complementary colour (पूरक रंग) – colour seen when one part of white light is absorbed.
Charge-transfer transition – electron moves between metal-based and ligand-based orbitals.
Formulas
μs = √(n(n + 2)) BM
Δo (cm⁻¹) = 10⁷/λ (λ in nm); E = hcν̄
Effective magnetic moment from susceptibility: μeff = 2.83 √(χM·T) BM
n (unpaired)
μs (BM)
Example
0
0
[Fe(CN)₆]⁴⁻
1
1.73
[Cu(H₂O)₆]²⁺
2
2.83
[Ni(NH₃)₆]²⁺
3
3.87
[Cr(H₂O)₆]³⁺
4
4.90
[Fe(H₂O)₆]²⁺
5
5.92
[Mn(H₂O)₆]²⁺
5 important questions
Q1. Calculate the spin-only magnetic moment of [CoF₆]³⁻.
Co³⁺ is d⁶. F⁻ is a weak-field ligand, so the complex is high spin: t₂g⁴ eg², n = 4. μs = √(4 × 6) = √24 = 4.90 BM.
Q2. Why is [Ti(H₂O)₆]³⁺ coloured but [Sc(H₂O)₆]³⁺ colourless?
Ti³⁺ is d¹; its electron can jump from t₂g to eg by absorbing visible light (about 493 nm), so it looks purple. Sc³⁺ is d⁰ and has no d electron to excite.
Q3. Explain the selection rules for electronic transitions in complexes.
Laporte (parity) rule and spin rule with examples; why octahedral d–d bands are weak (ε about 1–10), tetrahedral stronger, and CT bands very strong (ε about 10³–10⁴); relaxation by vibronic coupling.
Q4. Describe the Gouy method of measuring magnetic susceptibility.
Diagram of tube, balance and electromagnet; weighing with field off and on; Δw relation to χ; calculation of μeff and n; conclusion about high/low spin.
Q5. What is the Jahn–Teller effect? Explain with Cu(II) complexes.
Statement; d⁹ configuration t₂g⁶ eg³ with unequal filling of dz² and dx²–y²; tetragonal elongation (two long axial bonds, four short equatorial); effect on spectra (broad, split band).
Exam tip
Always state d-count, field strength, configuration, n and then μ — this five-step chain earns full marks in magnetic moment problems.
Lanthanides and actinides are the two series of f-block (inner transition) elements. In them the 4f (lanthanides) or 5f (actinides) orbitals are being filled. This topic covers their electronic configuration, oxidation states, lanthanide contraction and its consequences, colour and magnetic properties, separation of lanthanides, and a comparison of lanthanides with actinides.
Key points
Lanthanides: the 14 elements after lanthanum, cerium (Ce, Z = 58) to lutetium (Lu, Z = 71). General configuration [Xe] 4f¹⁻¹⁴ 5d⁰⁻¹ 6s². (La is often studied with them.)
Actinides: the 14 elements after actinium, thorium (Th, 90) to lawrencium (Lr, 103). Configuration [Rn] 5f¹⁻¹⁴ 6d⁰⁻¹ 7s². Elements after uranium (Z > 92) are man-made transuranic elements.
All actinides are radioactive.
The common and most stable oxidation state of lanthanides is +3. Some show +2 (Eu²⁺, Yb²⁺, giving f⁷ and f¹⁴) or +4 (Ce⁴⁺ f⁰, Tb⁴⁺ f⁷) because empty, half-filled or full f shells are stable.
Actinides show more varied oxidation states (+3 to +7; U commonly +6, Np and Pu up to +7), because 5f, 6d and 7s orbitals have similar energies.
Lanthanide contraction: steady decrease in atomic and ionic (M³⁺) radii from La to Lu. Cause: 4f electrons shield each other poorly from the increasing nuclear charge.
Consequences: (1) 4d and 5d elements of the same group have nearly equal size (Zr 160 pm ≈ Hf 159 pm) and similar properties, making them hard to separate; (2) basicity of Ln(OH)₃ decreases from La(OH)₃ to Lu(OH)₃; (3) Y³⁺ has a size similar to Ho³⁺ and occurs with heavy lanthanides; (4) similar chemical properties make separation of lanthanides difficult.
Colour: many Ln³⁺ ions are coloured because of f–f transitions. f–f bands are sharp and weak because 4f orbitals are buried and hardly affected by ligands. La³⁺ (f⁰) and Lu³⁺ (f¹⁴) are colourless.
Magnetism: Ln³⁺ ions with unpaired f electrons are paramagnetic; orbital contribution is large, so spin-only formula does not fit well.
Separation of lanthanides: modern method is ion exchange chromatography (smaller ions bind more strongly to complexing agents, so heavy lanthanides come out first). Older methods: fractional crystallisation, solvent extraction.
Uses: misch metal (Ce-based alloy) in lighter flints; lanthanide oxides as phosphors in screens; Nd in strong magnets; U and Pu as nuclear fuels.
Important terms
Inner transition elements – f-block elements; their last electron enters the (n − 2)f subshell.
Lanthanide contraction (लँथेनाइड आकुंचन) – steady decrease in size across the lanthanide series.
Transuranic elements – elements with Z greater than 92.
Actinide contraction – similar but larger decrease in size across the actinides (5f shielding is even poorer).
Misch metal – alloy of lanthanides (about 50% Ce) with a little iron.
Formulas
Point
Lanthanides
Actinides
Orbitals filled
4f
5f
Main oxidation state
+3
+3, many others (+4 to +7)
Radioactivity
only Pm radioactive
all radioactive
Complex formation
less
more
Oxocations
not formed
formed (UO₂²⁺, PuO₂²⁺)
Spectra
sharp f–f bands
broader bands
5 important questions
Q1. What is lanthanide contraction?
It is the steady decrease in the atomic and ionic radii of lanthanides from La to Lu as atomic number increases. It happens because each added 4f electron shields the outer electrons poorly, so the effective nuclear charge rises.
Q2. Why is +3 the common oxidation state of lanthanides?
Removing the two 6s electrons and one 5d (or one 4f) electron needs moderate energy, and the M³⁺ ion is stabilised by high hydration and lattice energies. So +3 is the most stable state for all of them.
Q3. Why do Ce⁴⁺ and Eu²⁺ exist though +3 is the common state?
Ce = [Xe] 4f¹ 5d¹ 6s². Ce⁴⁺ = [Xe] 4f⁰, a stable noble-gas configuration. Ce⁴⁺ is a strong oxidising agent (tends to go to Ce³⁺).
Eu = [Xe] 4f⁷ 6s². Eu²⁺ = 4f⁷, a stable half-filled shell. Eu²⁺ is a reducing agent.
So empty, half-filled and full f shells explain these extra states.
Q4. Explain the consequences of lanthanide contraction.
Zr and Hf (and Nb–Ta, Mo–W) have nearly the same radii → similar chemistry, hard to separate.
Basic strength of Ln(OH)₃ decreases from La to Lu (smaller ion, more covalent hydroxide).
Yttrium is found with heavy lanthanides due to similar size.
Separation of lanthanides from each other is difficult.
Density and electronegativity rise slightly across the series.
Q5. Compare lanthanides and actinides.
Both are f-block; 4f vs 5f.
Lanthanides mainly +3; actinides show many oxidation states.
Actinides all radioactive; among lanthanides only promethium is.
Actinides form complexes and oxocations more easily (5f orbitals are less buried).
Both show contraction; both mostly form coloured, paramagnetic ions.
Exam tip
"Lanthanide contraction: cause and consequences" is the most frequent long question. Learn the Zr/Hf example — examiners expect it.
Quantum chemistry explains the behaviour of electrons in atoms and molecules using quantum mechanics. This topic covers the failures of classical mechanics, the wave function and its meaning, operators, the Schrödinger wave equation, postulates of quantum mechanics, and the particle in a one-dimensional box with its energy levels and applications.
Key points
Failures of classical mechanics: black-body radiation (Planck, quantised energy E = hν), photoelectric effect (Einstein, photons), atomic line spectra and heat capacity of solids could not be explained.
de Broglie: every moving particle has a wave nature, λ = h/mv.
Heisenberg uncertainty principle: position and momentum cannot both be known exactly at the same time: Δx·Δp ≥ h/4π.
Wave function (ψ): describes the state of a particle. ψ itself has no physical meaning; **ψ² (or ψψ*) is the probability density** of finding the particle (Born interpretation).
A well-behaved ψ must be single-valued, finite, continuous and normalisable (∫ψ*ψ dτ = 1).
Operators: a rule that changes one function into another. Important ones — Hamiltonian (energy) Ĥ, momentum p̂ₓ = (h/2πi) d/dx, Laplacian ∇².
Eigenvalue equation: Âψ = aψ; a is the eigenvalue (measurable value), ψ the eigenfunction.
Eₙ = n²h²/(8mL²), n = 1, 2, 3 … (n = 0 not allowed)
Particle in a 1-D box (length L, V = 0 inside, V = ∞ outside):
Results: energy is quantised; lowest energy (zero-point energy) is not zero; gaps grow as n² increases; a larger box gives lower and closer energy levels.
Application: π electrons of conjugated polyenes treated as particles in a box explain why longer conjugation shifts absorption to longer wavelength (butadiene → β-carotene).
Number of nodes in ψₙ = n − 1.
Important terms
Wave function (तरंग फलन) – mathematical function describing a quantum state.
Normalisation – making the total probability of finding the particle equal to 1.
Orthogonal functions – ∫ψᵢ*ψⱼ dτ = 0 for i ≠ j.
Zero-point energy – lowest possible energy, h²/(8mL²) for a box.
State of a system described by ψ; every observable has a linear Hermitian operator; measured values are eigenvalues; average (expectation) value ⟨a⟩ = ∫ψ*Âψ dτ; time evolution given by the Schrödinger equation.
Exam tip
Practise the particle-in-a-box derivation until you can write it in five clear steps; it appears in almost every paper.
Molecular Spectroscopy (Rotational, Vibrational and Raman)
Quick summary
Molecular spectroscopy studies how molecules absorb or scatter electromagnetic radiation, which tells us about bond lengths, bond strengths and molecular structure. This physical-chemistry topic covers the regions of the spectrum, rotational (microwave) spectra of diatomic molecules, vibrational spectra using the harmonic and anharmonic oscillator models, and Raman spectra.
Key points
Energy of a molecule = electronic + vibrational + rotational (Born–Oppenheimer approximation). Electronic gaps are largest (UV-visible), vibrational next (infrared), rotational smallest (microwave).
Rotational spectra are shown only by molecules with a permanent dipole moment (HCl, CO), not by H₂, N₂, O₂ or CO₂.
Rigid rotor: rotational energy levels E_J = BJ(J + 1) in cm⁻¹, with J = 0, 1, 2… Selection rule ΔJ = ±1.
Rotational lines appear at 2B, 4B, 6B …; spacing between lines = 2B. From B we get the moment of inertia and the bond length.
Vibrational spectra need a change in dipole moment during the vibration (HCl active; H₂ inactive).
Harmonic oscillator: E_v = (v + ½)ω̄ (cm⁻¹), v = 0, 1, 2…; selection rule Δv = ±1; one line at ω̄. Zero-point energy = ½ω̄.
Anharmonic oscillator (real bonds, Morse curve): levels come closer at higher v; overtones (Δv = ±2, ±3) appear weakly at nearly 2ω̄, 3ω̄.
Force constant (k) measures bond stiffness; a higher k means a stronger bond and higher vibration frequency.
Degrees of freedom: a molecule of N atoms has 3N − 6 vibrational modes (3N − 5 if linear). H₂O has 3; CO₂ has 4.
Raman effect (C. V. Raman, 1928; Nobel Prize 1930): when monochromatic light is scattered, most has the same frequency (Rayleigh), but a tiny part has lower (Stokes) and higher (anti-Stokes) frequencies.
Raman activity needs a change in polarisability, so homonuclear molecules like H₂, N₂ and O₂ are Raman active.
Rule of mutual exclusion: in molecules with a centre of symmetry (CO₂), vibrations that are IR active are Raman inactive and vice versa.
Rotational Raman lines follow ΔJ = 0, ±2 and are spaced 4B from each other (6B for the first line from the Rayleigh line).
Important terms
Wavenumber (ν̄) (तरंग संख्या) – number of waves per cm; ν̄ = 1/λ.
Reduced mass (μ) – m₁m₂/(m₁ + m₂).
Polarisability – ease with which an electron cloud is distorted by an electric field.
E_v = (v + ½)ω̄ − (v + ½)²ω̄xₑ (anharmonic, xₑ = anharmonicity constant)
Spectrum
Region
Condition
Gives
Rotational
microwave
permanent dipole
bond length
Vibrational
infrared
change in dipole
force constant
Raman
scattered visible/UV
change in polarisability
both, for non-polar too
5 important questions
Q1. Why does HCl show a rotational spectrum but H₂ does not?
A molecule must have a permanent dipole moment to interact with microwave radiation. HCl is polar; H₂ is non-polar, so it has no pure rotational spectrum.
Q2. State the rule of mutual exclusion.
In a molecule with a centre of symmetry, no vibration can be both IR active and Raman active. For CO₂, the symmetric stretch is Raman active only, while the asymmetric stretch and bending are IR active only.
Q3. The rotational constant of ¹H³⁵Cl is 10.59 cm⁻¹. Find its bond length (μ = 1.627 × 10⁻²⁷ kg).
I = h/(8π²cB) = 6.626 × 10⁻³⁴/(8 × 9.87 × 2.998 × 10¹⁰ × 10.59) = 2.64 × 10⁻⁴⁷ kg m²
r = √(I/μ) = √(2.64 × 10⁻⁴⁷/1.627 × 10⁻²⁷) = 1.27 × 10⁻¹⁰ m = 1.27 Å (accepted value 1.27 Å)
Q4. The fundamental vibration of HCl is at 2886 cm⁻¹. Calculate its force constant.
Nuclear chemistry deals with changes in the nucleus of an atom. This topic covers nuclear particles and forces, mass defect and binding energy, the n/p ratio and stability, radioactive decay (α, β, γ), the rate of decay and half-life, artificial transmutation, nuclear fission and fusion, and applications such as radiocarbon dating and radioisotopes in medicine and agriculture.
Key points
The nucleus contains protons and neutrons (nucleons). Mass number A = protons + neutrons; atomic number Z = protons.
Isotopes: same Z, different A (¹²C, ¹⁴C). Isobars: same A, different Z (⁴⁰Ar, ⁴⁰Ca). Isotones: same number of neutrons.
Nucleons are held by the very strong, short-range nuclear force.
Mass defect (Δm): the mass of a nucleus is less than the total mass of its separate nucleons. This missing mass appears as binding energy (E = Δmc²).
Binding energy per nucleon is highest near iron (Fe-56, about 8.8 MeV), so medium nuclei are the most stable. Heavy nuclei gain stability by fission, light nuclei by fusion.
n/p ratio: stable light nuclei have n/p ≈ 1; heavier stable nuclei need more neutrons (up to about 1.5). Nuclei above the stability belt emit β⁻; those below emit positrons or capture electrons.
α decay: loses ⁴₂He → A decreases by 4, Z by 2. β⁻ decay: a neutron changes to a proton → Z increases by 1, A unchanged. γ emission: energy only, no change in A or Z.
Group displacement law (Soddy and Fajans): α emission moves the element two places left in the periodic table; β emission moves it one place right.
Radioactive decay is a first-order process: rate does not depend on temperature or pressure.
Nuclear fission: a heavy nucleus (U-235) splits into two medium nuclei when hit by a slow neutron, releasing about 200 MeV and 2–3 neutrons; this can start a chain reaction (used in reactors and atom bombs).
Nuclear fusion: light nuclei (isotopes of hydrogen) combine to form a heavier nucleus; it powers the Sun and needs very high temperature.
Applications: C-14 dating of old organic material, Co-60 in cancer therapy, I-131 for thyroid study and treatment, P-32 as a tracer in agriculture, irradiation of food.
Important terms
Radioactivity (किरणोत्सर्ग) – spontaneous emission of radiation by unstable nuclei, discovered by Henri Becquerel (1896).
Half-life (t½) – time for half of the radioactive nuclei to decay.
Decay constant (λ) – fraction of nuclei decaying per unit time.
Chain reaction – a self-sustaining series of fissions caused by released neutrons.
Critical mass – minimum mass of fissile material needed to sustain a chain reaction.
Formulas
E = Δmc²; 1 amu (u) ≡ 931.5 MeV
Δm = [Z mₚ + (A − Z) mₙ] − M(nucleus)
N = N₀ e^(−λt); λ = (2.303/t) log (N₀/N)
t½ = 0.693/λ; average life τ = 1/λ
After n half-lives: N = N₀ (½)ⁿ
C-14 dating: t = (2.303/λ) log (N₀/N), t½ of C-14 ≈ 5730 years
5 important questions
Q1. Define mass defect and binding energy.
Mass defect is the difference between the total mass of free protons and neutrons and the actual mass of the nucleus. Binding energy is the energy equivalent of this mass defect (E = Δmc²); it is the energy needed to break the nucleus into its nucleons.
Q2. ²³⁸₉₂U emits one α and two β particles. Find the product.
α: A = 238 − 4 = 234, Z = 92 − 2 = 90. Two β⁻: Z = 90 + 2 = 92, A stays 234. Product is ²³⁴₉₂U, an isotope of uranium.
Q3. Calculate the binding energy of ⁴₂He. (mₚ = 1.00728 u, mₙ = 1.00867 u, mass of He nucleus = 4.00151 u)
Mass of nucleons = 2 × 1.00728 + 2 × 1.00867 = 2.01456 + 2.01734 = 4.03190 u
Δm = 4.03190 − 4.00151 = 0.03039 u
BE = 0.03039 × 931.5 = 28.31 MeV
BE per nucleon = 28.31/4 ≈ 7.08 MeV
Q4. A wooden sample has 25% of the C-14 activity of fresh wood. Find its age (t½ = 5730 years).
N/N₀ = 0.25 = (½)², so 2 half-lives have passed.
Age = 2 × 5730 = 11,460 years
Check: λ = 0.693/5730 = 1.209 × 10⁻⁴ per year; t = (2.303/λ) log 4 = (2.303 × 0.602)/1.209 × 10⁻⁴ ≈ 11,460 years ✓
Q5. Distinguish between nuclear fission and nuclear fusion.
Fission: heavy nucleus splits; fusion: light nuclei join.
Fission starts with slow neutrons at ordinary temperature; fusion needs millions of degrees.
Fission gives radioactive waste; fusion products are mostly not radioactive.
Fusion gives more energy per unit mass.
Fission is used in nuclear power stations; fusion powers the Sun and stars (controlled fusion is still under research).
Exam tip
For decay problems, balance both A and Z on each side. For half-life questions, first check if the fraction is a simple power of ½ — it saves time.
Bioinorganic chemistry studies the role of metal ions and other inorganic elements in living systems. This topic covers essential and trace elements, the sodium–potassium pump, metalloproteins such as haemoglobin and myoglobin, the role of Mg, Zn, Co, Cu and Mo in biology, toxicity of heavy metals and chelation therapy, and metal-based drugs like cisplatin.
Key points
Bulk (macro) elements: C, H, O, N, P, S, Na, K, Mg, Ca, Cl. Trace elements: Fe, Zn, Cu, Mn, Co, Mo, Se, I, Cr. Some elements are toxic at any level (Hg, Pb, Cd, As).
Na⁺/K⁺ pump: an enzyme (Na⁺/K⁺-ATPase) in the cell membrane uses one ATP to push 3 Na⁺ out and bring 2 K⁺ in. It keeps high K⁺ inside and high Na⁺ outside, needed for nerve impulses.
Ca²⁺: bones and teeth (hydroxyapatite), muscle contraction, blood clotting.
Haemoglobin (Hb): four subunits, each with a haem group — Fe(II) held by four N atoms of a porphyrin ring and a fifth N from a histidine. O₂ binds at the sixth position. Iron stays Fe(II) while carrying O₂.
Cooperative binding: when one subunit takes O₂, the others take it more easily, giving an S-shaped (sigmoidal) curve. Myoglobin (one haem, in muscle) gives a hyperbolic curve and stores O₂.
CO poisoning: CO binds to haem iron about 200–250 times more strongly than O₂, blocking oxygen transport.
Chlorophyll: Mg²⁺ in a porphyrin-type (chlorin) ring; absorbs red and blue light in photosynthesis.
Vitamin B₁₂ (cobalamin): contains cobalt in a corrin ring; needed to form red blood cells.
Cytochromes (Fe), ferredoxins (Fe–S clusters) and copper proteins (plastocyanin) carry electrons. Nitrogenase (Fe–Mo) fixes N₂ into NH₃ in root-nodule bacteria.
Toxicity: Pb (anaemia, brain damage), Hg (Minamata disease, nervous system), Cd (itai-itai disease, bones and kidneys), As (skin lesions, cancer). Excess Fe or Cu is also harmful.
Chelation therapy: a chelating agent binds the toxic metal into a soluble complex that is excreted. EDTA (as Ca-EDTA) for lead, BAL (British anti-Lewisite) for Hg and As, desferrioxamine for iron overload, D-penicillamine for copper (Wilson's disease).
Cisplatin, cis-[Pt(NH₃)₂Cl₂], is an anticancer drug that binds to DNA; the trans isomer is inactive.
Important terms
Metalloprotein – protein containing a metal ion as an essential part.
Porphyrin – large flat ring with four pyrrole units that binds metals through four N atoms.
Chelation (कीलेशन) – binding of a metal ion by a ligand through two or more donor atoms.
Formulas
Metal
Biomolecule
Function
Fe
haemoglobin, myoglobin
O₂ transport and storage
Fe, Cu
cytochromes
electron transfer
Mg
chlorophyll
photosynthesis
Co
vitamin B₁₂
RBC formation
Zn
carbonic anhydrase
CO₂ hydration
Mo, Fe
nitrogenase
nitrogen fixation
Na⁺/K⁺ pump: 3Na⁺(in) + 2K⁺(out) + ATP → 3Na⁺(out) + 2K⁺(in) + ADP + Pi
5 important questions
Q1. Why is CO poisonous?
CO binds to the Fe(II) of haemoglobin much more strongly than O₂, forming carboxyhaemoglobin. Hb can no longer carry oxygen, so tissues suffer oxygen shortage.
Q2. Name the metal in chlorophyll and vitamin B₁₂.
Q3. Explain the structure and function of haemoglobin.
Four globin chains, haem structure, Fe(II) coordination (four porphyrin N + histidine + O₂), oxy- and deoxy-Hb, cooperative effect and sigmoid curve, comparison with myoglobin, why Fe is not oxidised to Fe(III) inside the protein pocket.
Q4. Describe the sodium–potassium pump.
Concentration difference across the membrane, role of ATP, stoichiometry (3 Na⁺ out, 2 K⁺ in), importance in nerve conduction and osmotic balance; diagram.
Q5. Write a note on metal toxicity and chelation therapy.
Toxic effects of Pb, Hg, Cd and As with diseases; principle of chelation; agents (Ca-EDTA, BAL, desferrioxamine, D-penicillamine) matched to metals; why Ca-EDTA and not Na₂-EDTA is used (to avoid removing body calcium).
Exam tip
A table of metal → biomolecule → function covers most short questions; learn haemoglobin in detail for the long answer.
Carbohydrates are optically active polyhydroxy aldehydes or ketones, or compounds that give these on hydrolysis. They are the main energy source of living things. This topic covers classification, the open-chain and cyclic structures of glucose and fructose, mutarotation, important reactions, interconversion (chain lengthening and shortening), and an introduction to disaccharides and polysaccharides.
Key points
General formula of many carbohydrates: Cₓ(H₂O)ᵧ (hence "hydrates of carbon"), but this is not a strict rule.
Aldoses have a −CHO group (glucose is an aldohexose); ketoses have a C=O group (fructose is a ketohexose).
Reducing sugars reduce Tollens' and Fehling's reagents (all monosaccharides, maltose, lactose). Sucrose is non-reducing because the reducing groups of glucose and fructose are joined in the glycosidic bond.
Evidence for the open-chain structure of glucose: forms oxime with NH₂OH (C=O present); gives n-hexane with HI/red P (six-carbon straight chain); forms a penta-acetate (five −OH groups); oxidised by Br₂ water to gluconic acid (−CHO present); oxidised by HNO₃ to saccharic (glucaric) acid (a −CH₂OH at the end).
Objections to the open chain: glucose does not give Schiff's test or a bisulphite addition product; its penta-acetate does not react with NH₂OH; it exists as two forms (α and β); it shows mutarotation.
Cyclic (pyranose) structure: the C-5 −OH adds to the C-1 aldehyde, forming a six-membered ring (like pyran) and a new chiral centre at C-1 (anomeric carbon). This gives α-D-glucose and β-D-glucose (anomers).
Mutarotation: specific rotation of a fresh glucose solution changes slowly to an equilibrium value of +52.7° (α form +112°, β form +18.7°), because α and β forms interconvert through the open chain.
Glucose and fructose form the same osazone with excess phenylhydrazine, because only C-1 and C-2 react, and the rest of the molecule is identical.
Kiliani–Fischer synthesis lengthens an aldose by one carbon (via cyanohydrin). Ruff degradation shortens it by one carbon.
Oxidation: CHO → COOH (Br₂ water) gives gluconic acid
With HI/P: glucose → n-hexane (CH₃(CH₂)₄CH₃)
Number of stereoisomers of an aldohexose (4 chiral C in open chain) = 2⁴ = 16
Sugar
Units
Linkage
Reducing
Maltose
glucose + glucose
α-1,4
yes
Lactose
galactose + glucose
β-1,4
yes
Sucrose
glucose + fructose
α-1, β-2
no
5 important questions
Q1. What is mutarotation?
It is the change in specific rotation of a freshly prepared solution of a sugar until it reaches a constant value. Glucose changes from +112° (α) or +18.7° (β) to +52.7°, because α and β forms interconvert through the open-chain form and reach equilibrium.
Q2. Why is sucrose a non-reducing sugar?
In sucrose, C-1 of glucose and C-2 of fructose — the two reducing (carbonyl) carbons — are joined in the glycosidic bond. No free aldehyde or ketone group can form, so sucrose does not reduce Fehling's or Tollens' reagent.
Q3. Give evidence for the open-chain structure of glucose.
Molecular formula C₆H₁₂O₆; with HI/P gives n-hexane → straight 6-carbon chain.
With NH₂OH forms oxime; with HCN forms cyanohydrin → carbonyl group present.
Br₂ water oxidises it to gluconic acid (same carbon number) → aldehyde group.
Acetylation gives penta-acetate → five −OH groups.
HNO₃ gives saccharic acid → a primary alcohol group (−CH₂OH) also present.
Q4. Why do glucose and fructose give the same osazone?
Osazone formation involves only C-1 and C-2.
In glucose, C-1 (CHO) reacts first, then C-2 is oxidised and reacts.
In fructose, C-2 (C=O) reacts first, then C-1 is oxidised and reacts.
C-3 to C-6 are identical in both, so the products are the same.
Mannose also gives the same osazone (C-2 epimer).
Q5. Explain the Kiliani–Fischer synthesis (arabinose → glucose and mannose).
Aldopentose + HCN → cyanohydrin (new chiral C formed; two epimers).
Hydrolysis → aldonic acids.
Heating → lactones.
Reduction with Na/Hg → aldohexoses with one more carbon.
Arabinose gives glucose and mannose (epimers at C-2).
Exam tip
Learn glucose evidence as a list of reaction → conclusion. Draw the Haworth projection of α- and β-D-glucopyranose neatly; it is a frequent diagram question.
Amino acids are organic compounds containing both an amino group (−NH₂) and a carboxyl group (−COOH). α-Amino acids join through peptide bonds to form peptides and proteins. This topic covers classification of amino acids, zwitterions and the isoelectric point, synthesis, the peptide bond, structure of proteins (primary to quaternary), denaturation, and tests for proteins.
Key points
In α-amino acids, the −NH₂ is on the carbon next to −COOH. General formula: R−CH(NH₂)−COOH.
About 20 standard amino acids occur in proteins. All except glycine (R = H) are optically active; natural ones mainly have the L-configuration.
Essential amino acids cannot be made by the human body and must come from food (e.g. valine, leucine, lysine, tryptophan). Non-essential ones are made by the body (glycine, alanine).
Zwitterion: in water an amino acid exists as a dipolar ion, H₃N⁺−CHR−COO⁻. This explains its high melting point, solubility in water and low solubility in organic solvents.
Amino acids are amphoteric: they react with both acids and bases.
Isoelectric point (pI): the pH at which the amino acid has no net charge and does not move in an electric field.
Synthesis: Strecker synthesis (aldehyde + NH₃ + HCN, then hydrolysis); Gabriel phthalimide synthesis; amination of α-halo acids.
Peptide bond: −CO−NH− formed between the −COOH of one amino acid and the −NH₂ of another, with loss of water. The peptide bond is planar (partial double bond character).
Protein structure: primary (sequence of amino acids), secondary (α-helix and β-pleated sheet, held by hydrogen bonds), tertiary (overall 3D folding: H-bonds, disulphide bridges, ionic and hydrophobic interactions), quaternary (arrangement of two or more polypeptide chains, e.g. haemoglobin).
Denaturation: loss of biological activity when heat, pH change or chemicals disturb the secondary/tertiary structure; primary structure stays intact (boiling an egg).
Tests: Biuret test (violet colour with alkaline CuSO₄; for peptide bonds), Ninhydrin test (purple/blue colour with amino acids), Xanthoproteic test (yellow with conc. HNO₃, aromatic rings).
Important terms
Zwitterion (द्विध्रुवी आयन) – molecule with both positive and negative charges but neutral overall.
Isoelectric point (pI) – pH at which net charge is zero.
Peptide – chain of amino acids joined by peptide bonds (dipeptide, tripeptide, polypeptide).
Denaturation – destruction of protein's 3D structure and activity.
N-terminal / C-terminal – end with free −NH₂ / free −COOH of a peptide chain.
Isoelectric point for a simple neutral amino acid: pI = (pKa₁ + pKa₂)/2
Number of possible peptides from n different amino acids (each used once) = n!
5 important questions
Q1. What is a zwitterion? Why do amino acids have high melting points?
A zwitterion is a dipolar ion with both + and − charges, like H₃N⁺CH₂COO⁻. Because amino acids exist as zwitterions in the solid, strong ionic attractions hold them together, giving high melting points (they often decompose before melting).
Q2. Define isoelectric point. Calculate pI of glycine (pKa₁ = 2.34, pKa₂ = 9.60).
Isoelectric point is the pH at which an amino acid carries no net charge. pI = (2.34 + 9.60)/2 = 11.94/2 = 5.97.
Q3. How many tripeptides can be formed from glycine, alanine and valine, each used once? Name them.
Nucleic acids (DNA and RNA) are biopolymers that store and transfer genetic information. Their monomers are nucleotides. This topic covers the components of nucleic acids, nucleosides and nucleotides, the Watson–Crick double helix of DNA, Chargaff's rules, types of RNA, replication, and an outline of protein synthesis and the genetic code.
Key points
Nucleic acids were first isolated by Friedrich Miescher (1869) from cell nuclei.
Components: a pentose sugar, a nitrogenous base and phosphoric acid.
Sugars: DNA has 2-deoxy-D-ribose (no –OH at C-2′); RNA has D-ribose.
Bases: purines — adenine (A) and guanine (G); pyrimidines — cytosine (C), thymine (T, DNA only) and uracil (U, RNA only).
Nucleoside = base + sugar (joined by an N-glycosidic bond at C-1′). Nucleotide = nucleoside + phosphate (ester at C-5′).
Nucleotides join by 3′,5′-phosphodiester bonds to form a polynucleotide chain with a 5′ end and a 3′ end.
Watson–Crick model (1953), based on X-ray data of Rosalind Franklin and Maurice Wilkins: two antiparallel strands coiled into a right-handed double helix; sugar–phosphate backbone outside, bases inside.
Base pairing: A pairs with T by two H-bonds; G pairs with C by three H-bonds. So GC-rich DNA is more stable (higher melting temperature).
Dimensions (B-DNA): diameter about 2 nm, one turn about 3.4 nm, about 10 base pairs per turn, 0.34 nm between base pairs.
Chargaff's rules: in DNA, A = T and G = C, so purines = pyrimidines; the ratio (A + T)/(G + C) varies from species to species.
Replication is semi-conservative (Meselson–Stahl, 1958): each new DNA has one old and one new strand.
RNA is usually single-stranded. mRNA carries the code from DNA, tRNA brings amino acids (cloverleaf shape, anticodon), rRNA forms ribosomes.
Central dogma: DNA → (transcription) → RNA → (translation) → protein.
Genetic code: triplet code of 64 codons; 61 code for 20 amino acids, 3 are stop codons (UAA, UAG, UGA); AUG is the start codon (methionine). The code is nearly universal and degenerate.
Important terms
Nucleotide (न्यूक्लिओटाइड) – base + sugar + phosphate; monomer of nucleic acids.
Codon – three-base sequence on mRNA coding for one amino acid.
Denaturation of DNA – separation of strands by heat (melting).
Formulas
Feature
DNA
RNA
Sugar
2-deoxyribose
ribose
Bases
A, G, C, T
A, G, C, U
Strands
double
usually single
Function
stores genetic information
protein synthesis
Location
mainly nucleus
nucleus and cytoplasm
Example: if a DNA has 30 % A, then T = 30 %, G + C = 40 %, so G = C = 20 %.
5 important questions
Q1. Distinguish between a nucleoside and a nucleotide.
A nucleoside is a base joined to a pentose sugar (adenosine). A nucleotide is a nucleoside with a phosphate group on the sugar (adenosine monophosphate, AMP).
Q2. A DNA sample contains 18 % cytosine. Find the percentage of the other bases.
G = C = 18 %, so G + C = 36 %. A + T = 64 %, and A = T = 32 %.
Q3. Describe the double helical structure of DNA.
Antiparallel strands, right-handed helix, backbone and base positions, complementary base pairing with number of H-bonds, major and minor grooves, dimensions; draw a labelled diagram and a base-pair sketch.
Q4. Explain the replication of DNA.
Unwinding by helicase, replication fork, primer, DNA polymerase adding nucleotides in the 5′ → 3′ direction, leading and lagging strands (Okazaki fragments, joined by ligase), semi-conservative nature with the Meselson–Stahl experiment.
Q5. Write a note on the types of RNA and their role in protein synthesis.
mRNA (codons), tRNA (anticodon, amino-acid arm, cloverleaf), rRNA (ribosome); transcription in the nucleus and translation on ribosomes; start and stop codons; peptide bond formation.
Exam tip
Chargaff's rule numericals are quick marks: A = T, G = C, and all four add to 100 %.
Polymers are very large molecules (macromolecules) made by joining many small repeating units called monomers. This topic covers classification of polymers, addition and condensation polymerisation with mechanisms, molecular mass averages, natural and synthetic rubber, and important polymers like polythene, PVC, nylon-6,6, terylene and Bakelite, along with biodegradable polymers.
Key points
Monomer: small molecule that joins to form a polymer (ethene → polythene). The process is polymerisation.
By structure: linear (HDPE), branched (LDPE), cross-linked/network (Bakelite, melamine).
By monomers: homopolymer (one type of monomer: polythene) and copolymer (two or more types: Buna-S, nylon-6,6).
By forces/properties: elastomers (weak forces, rubber), fibres (strong H-bonds, nylon, terylene), thermoplastics (soften on heating and can be remoulded: PVC, polythene), thermosetting (become hard permanently on heating: Bakelite).
Addition (chain-growth) polymerisation: monomers with double bonds add without losing any molecule. Free-radical mechanism: initiation (by peroxide), propagation, termination.
Condensation (step-growth) polymerisation: monomers with two functional groups join with loss of small molecules like H₂O or HCl (nylon-6,6, terylene).
Degree of polymerisation = M̄n / mass of repeating unit
Polymer
Monomer(s)
Type
Use
Polythene
CH₂=CH₂
addition
bags, bottles
PVC
CH₂=CHCl
addition
pipes, cable coats
Teflon
CF₂=CF₂
addition
non-stick pans
Nylon-6,6
hexamethylenediamine + adipic acid
condensation
fibres, ropes
Terylene
ethylene glycol + terephthalic acid
condensation
fabrics
Bakelite
phenol + formaldehyde
condensation
switches, handles
5 important questions
Q1. Distinguish between thermoplastic and thermosetting polymers.
Thermoplastics are linear or slightly branched and soften on heating, so they can be remoulded many times (PVC, polythene). Thermosetting polymers form a 3D cross-linked network on heating and become permanently hard; they cannot be remoulded (Bakelite).
Q2. What is vulcanisation of rubber? Why is it done?
Vulcanisation is heating natural rubber with sulphur (about 3–5%) at 373–415 K. Sulphur forms cross-links between chains. This makes rubber stronger, more elastic, less sticky and usable over a wider temperature range (tyres).
Q3. A polymer sample has 100 molecules of molar mass 10,000 g/mol and 200 molecules of molar mass 20,000 g/mol. Calculate M̄n, M̄w and PDI.
Q5. Write the preparation of nylon-6,6 and Bakelite.
Nylon-6,6: n H₂N(CH₂)₆NH₂ + n HOOC(CH₂)₄COOH → −[NH(CH₂)₆NH−CO(CH₂)₄CO]ₙ− + 2n H₂O; heated under pressure; "6,6" = six carbons in each monomer.
Bakelite: phenol + formaldehyde (acid or base catalyst) → o- and p-hydroxymethylphenols → Novolac (linear) → heating with more HCHO gives cross-linked Bakelite.
Both are condensation polymers.
Exam tip
Classification questions are common: always give one example for each class. Learn monomers of six key polymers from the table.
Dyes are coloured organic compounds that can fix themselves to a fabric and give it a lasting colour. This topic covers the qualities of a good dye, Witt's theory of colour (chromophores and auxochromes), classification of dyes by structure and by method of application, and the synthesis and uses of methyl orange, Congo red, malachite green, phenolphthalein, fluorescein, alizarin and indigo.
Key points
A dye must: have a suitable colour, attach firmly to the fibre, and be fast (resist washing, light and sweat). Not every coloured substance is a dye.
W. H. Perkin made the first synthetic dye, mauveine, in 1856.
Witt's theory (1876): colour comes from chromophores — unsaturated groups such as –N=N– (azo), C=O, –NO₂, C=C, quinonoid rings. The compound containing a chromophore is a chromogen.
Auxochromes (–OH, –NH₂, –NHR, –NR₂) do not give colour by themselves but deepen it and help the dye bind to the fibre by salt formation.
Modern view: colour arises from absorption of visible light by an extended conjugated system; longer conjugation shifts absorption to longer wavelength (bathochromic shift).
Classification by structure: nitro and nitroso, azo (methyl orange, Congo red), triphenylmethane (malachite green), phthaleins and xanthenes (phenolphthalein, fluorescein), anthraquinone (alizarin), indigoid (indigo).
acid dyes (sulphonic acid groups; wool, silk) – e.g. orange II
basic dyes (amino groups; wool, silk, acrylic) – malachite green
direct dyes (applied straight to cotton from water) – Congo red
mordant dyes (need a metal ion, Al³⁺/Cr³⁺, as fixing agent) – alizarin
vat dyes (reduced to a soluble leuco form, applied, then oxidised in air) – indigo
disperse dyes (polyester), reactive dyes (form covalent bonds with cotton)
Classification by application:
Methyl orange: diazotised sulphanilic acid coupled with N,N-dimethylaniline. Used as an acid–base indicator (red below pH 3.1, yellow above pH 4.4).
Phenolphthalein: phthalic anhydride + 2 phenol with conc. H₂SO₄. Colourless in acid, pink in base (pH 8.3–10).
Fluorescein: phthalic anhydride + resorcinol with ZnCl₂; strong green fluorescence in alkaline solution; used in tracing water leaks.
Chromophore (रंगदायी गट) – group responsible for colour.
Auxochrome – group that intensifies colour and fixes the dye.
Mordant – metal salt that binds a dye to the fibre.
Leuco compound – colourless reduced form of a dye.
Formulas
Dye
Structural class
Application class
Main use
Methyl orange
azo
acid
indicator
Congo red
azo (bis-azo)
direct
cotton, indicator
Malachite green
triphenylmethane
basic
silk, wool, paper
Phenolphthalein
phthalein
–
indicator, laxative (old use)
Alizarin
anthraquinone
mordant
cotton (Turkey red)
Indigo
indigoid
vat
denim
5 important questions
Q1. What are chromophores and auxochromes? Give two examples of each.
Chromophores are unsaturated groups that cause colour, e.g. –N=N–, –NO₂. Auxochromes are groups that deepen colour and help binding, e.g. –OH, –NH₂.
Q2. Why does phenolphthalein turn pink in alkaline solution?
In base it loses protons and the lactone ring opens to give a quinonoid dianion with an extended conjugated system, which absorbs visible light and appears pink. In acid it stays in the colourless lactone (benzenoid) form.
Q3. Give the synthesis and uses of methyl orange.
Sulphanilic acid + Na₂CO₃ + NaNO₂ + HCl at 0–5 °C → diazonium salt.
Coupling with N,N-dimethylaniline at the para position → helianthin (red, acid form).
Treatment with NaOH → sodium salt, methyl orange.
Uses: indicator for strong acid–weak base titrations; dyeing wool.
Q4. Classify dyes on the basis of application with one example each.
Write acid, basic, direct, mordant, vat, disperse and reactive dyes with the fibre each suits and how it is fixed.
Q5. Write the synthesis of fluorescein and alizarin.
Fluorescein: phthalic anhydride + 2 resorcinol, ZnCl₂, about 200 °C. Alizarin: anthraquinone + fuming H₂SO₄ → anthraquinone-2-sulphonic acid; fusion with NaOH and KClO₃; acidification. State a use of each.
Exam tip
Learn each dye as a set: reactants, condition, structural class, application class and one use. That set answers both short and long questions.
Drugs are chemicals that are used to prevent, diagnose or cure disease. This topic covers basic terms, the classification of drugs by therapeutic action, and the synthesis and uses of important drugs: aspirin, paracetamol, sulphanilamide, isoniazid, chloroquine and an introduction to antibiotics, with a note on drug safety.
Key points
Chemotherapy: treatment of disease with chemicals that harm the disease-causing organism more than the patient. The term was introduced by Paul Ehrlich, who made salvarsan (an arsenic drug for syphilis, 1910).
Pharmacophore: the part of a drug molecule responsible for its biological action.
Therapeutic index = LD₅₀/ED₅₀; a larger value means a safer drug.
Aspirin (acetylsalicylic acid): salicylic acid + acetic anhydride (conc. H₂SO₄ or H₃PO₄ catalyst). Also prevents blood clotting; can irritate the stomach.
Paracetamol (p-acetamidophenol): p-aminophenol + acetic anhydride. p-Aminophenol is made from phenol (nitration → p-nitrophenol → reduction) or from nitrobenzene by electrolytic reduction.
Sulpha drugs (antibacterial): sulphanilamide from acetanilide → chlorosulphonation (ClSO₃H) → p-acetamidobenzenesulphonyl chloride → NH₃ → hydrolysis. They block bacterial folic acid synthesis by imitating p-aminobenzoic acid (PABA).
Antibiotics: substances made by micro-organisms that kill or stop other microbes. Penicillin was discovered by Alexander Fleming (1928). Broad-spectrum: chloramphenicol, tetracycline; narrow-spectrum: penicillin G.
Q1. What is the difference between an antiseptic and a disinfectant?
Antiseptics kill or stop microbes on living tissue (Dettol, 0.2 % phenol, iodine tincture). Disinfectants are used on non-living objects like floors and toilets (1 % phenol, chlorine). The same substance can be either, depending on concentration.
Q2. How do sulpha drugs act?
Sulphanilamide is structurally like p-aminobenzoic acid (PABA). Bacteria take it up instead of PABA, so they cannot make folic acid and stop multiplying. Humans get folic acid from food, so they are not affected.
Q3. Give the synthesis and uses of paracetamol.
Phenol → (dil. HNO₃) p-nitrophenol → (Sn/HCl or H₂/Ni) p-aminophenol → (acetic anhydride) paracetamol. Uses: antipyretic and analgesic. Overdose damages the liver.
Q4. Describe the synthesis of sulphanilamide from acetanilide.
Acetanilide + ClSO₃H → p-acetamidobenzenesulphonyl chloride (the –NHCOCH₃ group protects –NH₂ and directs para).
+ NH₃ → p-acetamidobenzenesulphonamide.
Acid hydrolysis removes the acetyl group → sulphanilamide.
Q5. Write notes on antibiotics and antitubercular drugs.
Definition, Fleming and penicillin, broad and narrow spectrum with examples, bactericidal vs bacteriostatic; synthesis of isoniazid in steps; why TB drugs are given in combination (to prevent resistance).
Exam tip
For each drug write: class, synthesis in arrows, and one use. Names of discoverers (Ehrlich, Fleming) are common one-mark questions.
Green chemistry is the design of chemical products and processes that reduce or eliminate the use and production of hazardous substances. It aims to prevent pollution at the source instead of cleaning it later. This topic covers the meaning and need of green chemistry, the twelve principles given by Paul Anastas and John Warner (1998), atom economy, green solvents and reagents, alternative energy sources like microwaves and ultrasound, and examples of green syntheses.
Key points
Green chemistry is also called sustainable chemistry. Its aim: less waste, safer chemicals, less energy, renewable raw materials.
Need: traditional processes use toxic solvents and reagents, produce large amounts of waste, use much energy, and cause pollution and health hazards.
Twelve principles (Anastas and Warner): 1. Prevent waste. 2. Maximise atom economy. 3. Less hazardous chemical syntheses. 4. Design safer chemicals. 5. Safer solvents and auxiliaries. 6. Design for energy efficiency. 7. Use renewable feedstocks. 8. Reduce derivatives (avoid unnecessary protecting groups). 9. Use catalysts rather than stoichiometric reagents. 10. Design for degradation. 11. Real-time analysis for pollution prevention. 12. Inherently safer chemistry for accident prevention.
Atom economy (Barry Trost) measures how many atoms of the reactants end up in the desired product. Addition and rearrangement reactions have 100% atom economy; substitution and elimination reactions are lower.
Green solvents: water, supercritical CO₂ (used for decaffeination of coffee), ionic liquids, and solvent-free reactions.
Catalysis: catalysts are used in small amounts and can be reused; biocatalysts (enzymes) work in water at mild conditions.
Examples: the Boots–Hoechst–Celanese (BHC) route to ibuprofen uses 3 catalytic steps instead of the older 6-step Boots process and gives much less waste; dry cleaning with liquid CO₂ instead of perchloroethylene; making adipic acid from glucose using enzymes instead of benzene; hydrogen peroxide as a clean oxidant (gives only water).
Green chemistry is linked to the idea of the E-factor (Roger Sheldon): kg of waste per kg of product. Lower is better.
Important terms
Green chemistry (हरित रसायनशास्त्र) – chemistry that avoids pollution at the source.
Atom economy – percentage of reactant mass that ends up in the desired product.
E-factor – mass of waste ÷ mass of product.
Supercritical fluid – substance above its critical temperature and pressure with properties of both gas and liquid.
Renewable feedstock – raw material from biomass that can be replaced naturally.
Formulas
% Atom economy = (Molar mass of desired product / Sum of molar masses of all reactants) × 100
Q1. Define green chemistry. Who gave the twelve principles?
Green chemistry is the design of chemical products and processes that reduce or eliminate hazardous substances. The twelve principles were given by Paul Anastas and John Warner in 1998.
Q2. Why is supercritical CO₂ called a green solvent?
It is non-toxic, non-flammable, cheap and easily available. After use it simply turns into gas and can be recycled, leaving no solvent residue. It is used for extracting caffeine from coffee and in dry cleaning.
Q3. Calculate the atom economy for the preparation of ethene from ethanol: C₂H₅OH → C₂H₄ + H₂O.
Molar masses: C₂H₅OH = 46 g/mol; C₂H₄ = 28 g/mol
% Atom economy = (28/46) × 100
= 60.9%
The rest (water) is a by-product, so elimination reactions have lower atom economy.
Q4. Calculate the atom economy for the addition reaction CH₂=CH₂ + Br₂ → CH₂Br−CH₂Br.
Reactants: C₂H₄ = 28, Br₂ = 159.8; total = 187.8
Product C₂H₄Br₂ = 187.8
% Atom economy = (187.8/187.8) × 100 = 100%
All atoms of the reactants are in the product; addition reactions are ideal from the atom-economy view.
Q5. Explain any five principles of green chemistry with examples.
Prevention of waste: better to avoid waste than treat it (BHC ibuprofen process).
Atom economy: prefer addition and rearrangement reactions.
Safer solvents: water, supercritical CO₂ instead of benzene or CCl₄.
Catalysis: catalysts and enzymes instead of excess reagents.
Energy efficiency: microwave reactions at room pressure, shorter time.
Design for degradation: biodegradable polymers like PHBV.
Exam tip
Atom economy numericals are easy marks. Write the balanced equation, molar masses of all reactants and the desired product, then use the formula.