रसायनशास्त्र Chemistry

थोडक्यात सारांश

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.

महत्त्वाचे मुद्दे

• 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.
• Applications: detecting conjugation, finding concentration (colorimetry), distinguishing isomers, studying reaction kinetics, purity checks.

महत्त्वाच्या संज्ञा

• λmax – wavelength of maximum absorption.
• 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.

सूत्रे

• 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.

५ महत्त्वाचे प्रश्न

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.

• σ → σ*: highest energy, below 200 nm (vacuum UV); saturated hydrocarbons like methane.
• n → σ*: compounds with lone pairs on N, O, S, halogens (CH₃OH, CH₃NH₂).
• π → π*: alkenes, aromatics, conjugated systems; strong (ε about 10⁴).
• 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*.

परीक्षेसाठी टीप

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.

थोडक्यात सारांश

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.

महत्त्वाचे मुद्दे

• 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.

महत्त्वाच्या संज्ञा

• 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).

सूत्रे

• 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
• Vibrational modes: 3N − 6 (non-linear), 3N − 5 (linear)

५ महत्त्वाचे प्रश्न

Q1. Why are N₂ and O₂ IR inactive?

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).

परीक्षेसाठी टीप

Learn the table of 8–10 group frequencies; questions often give a spectrum's main peaks and ask you to identify the compound.

थोडक्यात सारांश

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).

महत्त्वाचे मुद्दे

• 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…).
• Relative intensities follow Pascal's triangle: doublet 1:1, triplet 1:2:1, quartet 1:3:3:1.
• Coupling constant (J), in Hz, is the distance between split peaks; it does not change with field strength.
• Protons of −OH and −NH often appear as broad singlets and usually do not split neighbours (fast exchange).
• Electronegative atoms and π systems (anisotropy) deshield protons: aldehyde H about 9–10 ppm, aromatic H about 6.5–8 ppm.

महत्त्वाच्या संज्ञा

• Shielding (आवरण) – reduction of the effective field at a nucleus by surrounding electrons.
• Downfield – towards higher δ (left side of spectrum), deshielded protons.
• Upfield – towards lower δ (right side), shielded protons.
• Equivalent protons – protons in identical chemical environments.
• Multiplicity – number of peaks in a split signal.

सूत्रे

• δ (ppm) = [(ν_sample − ν_TMS) / ν_spectrometer (Hz)] × 10⁶
• Multiplicity = n + 1 (n = equivalent neighbouring H)
• Resonance condition: ν = γB₀/2π

५ महत्त्वाचे प्रश्न

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).

• Two sets of protons → 2 signals.
• CH₃ (3H): neighbours CH₂ (2H) → triplet (1:2:1), δ about 1.7.
• CH₂ (2H): neighbours CH₃ (3H) → quartet (1:3:3:1), δ about 3.4 (deshielded by Br).
• Integration ratio 3:2.

Q5. Explain spin–spin coupling with the n + 1 rule.

• Spins of protons on neighbouring carbons can align with or against the field, slightly changing the field felt by the observed proton.
• n equivalent neighbours give n + 1 combinations of field effects → n + 1 lines.
• Intensities from Pascal's triangle.
• Equivalent protons do not split each other (e.g. CH₃ in acetone is a singlet).
• J value is the same in both coupled signals; it helps match partners.

परीक्षेसाठी टीप

For structure questions, always state: number of signals, δ of each, integration ratio and multiplicity. That four-point answer gets full marks.

थोडक्यात सारांश

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.

महत्त्वाचे मुद्दे

• 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)
• benzoyl compounds → C₆H₅CO⁺ (m/z 105) → C₆H₅⁺ (m/z 77)
• alcohols lose water (M − 18)
• Common fragmentations:
• 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.

महत्त्वाच्या संज्ञा

• 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.

सूत्रे

• 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

५ महत्त्वाचे प्रश्न

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.

Condition: C=O with a γ-H. Six-membered cyclic transition state; γ-H moves to O; β-bond breaks; neutral alkene lost; enol radical cation formed. Example: butanal (M = 72) → m/z 44 + ethene.

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

परीक्षेसाठी टीप

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).

थोडक्यात सारांश

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.

महत्त्वाचे मुद्दे

• 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.
• Spectrochemical series (increasing field strength): I⁻ < Br⁻ < S²⁻ < SCN⁻ < Cl⁻ < F⁻ < OH⁻ < C₂O₄²⁻ < H₂O < NCS⁻ < NH₃ < en < NO₂⁻ < CN⁻ < CO.
• 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.

महत्त्वाच्या संज्ञा

• 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).

सूत्रे

• 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

५ महत्त्वाचे प्रश्न

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.

• Low spin (CN⁻ strong): t2g⁶ eg⁰ → CFSE = 6 × (−0.4) Δo = −2.4 Δo (+ 2P extra pairing, relative to free ion). Unpaired = 0 → μ = 0, diamagnetic.
• High spin (H₂O weak): t2g⁴ eg² → CFSE = (4 × −0.4 + 2 × 0.6) Δo = (−1.6 + 1.2) Δo = −0.4 Δo. Unpaired = 4 → μ = √24 = 4.90 BM.

Q4. [Ti(H₂O)₆]³⁺ absorbs at 500 nm. Calculate Δo in kJ/mol.

• E per ion = hc/λ = (6.626 × 10⁻³⁴ × 3.0 × 10⁸)/(500 × 10⁻⁹)
• = 1.988 × 10⁻²⁵ / 5.0 × 10⁻⁷ = 3.976 × 10⁻¹⁹ J
• Per mole = 3.976 × 10⁻¹⁹ × 6.022 × 10²³ = 2.394 × 10⁵ J/mol
• Δo ≈ 239 kJ/mol

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.

परीक्षेसाठी टीप

For every complex, write: metal oxidation state → d count → weak or strong ligand → filling → CFSE → unpaired electrons → μ. This chain answers most CFT questions.

थोडक्यात सारांश

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.

महत्त्वाचे मुद्दे

• 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.

महत्त्वाच्या संज्ञा

• 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.

सूत्रे

• 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

५ महत्त्वाचे प्रश्न

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).

परीक्षेसाठी टीप

Always state d-count, field strength, configuration, n and then μ — this five-step chain earns full marks in magnetic moment problems.

थोडक्यात सारांश

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.

महत्त्वाचे मुद्दे

• 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.

महत्त्वाच्या संज्ञा

• 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.

सूत्रे

५ महत्त्वाचे प्रश्न

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.

परीक्षेसाठी टीप

"Lanthanide contraction: cause and consequences" is the most frequent long question. Learn the Zr/Hf example — examiners expect it.

थोडक्यात सारांश

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.

महत्त्वाचे मुद्दे

• 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.
• Schrödinger equation (time-independent): Ĥψ = Eψ, i.e. ∇²ψ + (8π²m/h²)(E − V)ψ = 0.
• ψₙ = √(2/L) sin(nπx/L)
• 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.

महत्त्वाच्या संज्ञा

• 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.

सूत्रे

• λ = h/mv; Δx·Δp ≥ h/4π
• Ĥ = −(h²/8π²m)∇² + V
• Eₙ = n²h²/(8mL²); ΔE (n → n + 1) = (2n + 1)h²/(8mL²)
• ψₙ = √(2/L) sin(nπx/L)

५ महत्त्वाचे प्रश्न

Q1. What are the conditions for an acceptable wave function?

ψ must be single-valued, finite everywhere, continuous (with continuous first derivative), and square-integrable so that it can be normalised.

Q2. Show that ψ = e^(ikx) is an eigenfunction of d/dx and find the eigenvalue.

d/dx (e^(ikx)) = ik·e^(ikx). The function comes back multiplied by a constant, so it is an eigenfunction with eigenvalue ik.

Q3. Derive the energy expression for a particle in a one-dimensional box.

• Set V = 0 inside; equation d²ψ/dx² + (8π²mE/h²)ψ = 0.
• General solution ψ = A sin kx + B cos kx, k² = 8π²mE/h².
• Boundary condition ψ(0) = 0 → B = 0; ψ(L) = 0 → kL = nπ.
• E = n²h²/(8mL²); normalise to get A = √(2/L).
• Comment on quantisation and zero-point energy.

Q4. Find the energy difference between n = 1 and n = 2 for an electron in a box of length 1 nm.

• h²/(8mL²) = (6.626 × 10⁻³⁴)²/(8 × 9.109 × 10⁻³¹ × (10⁻⁹)²) = 6.02 × 10⁻²⁰ J
• E₂ − E₁ = (4 − 1) × 6.02 × 10⁻²⁰ = 1.81 × 10⁻¹⁹ J (about 1.13 eV)

Q5. State the postulates of quantum mechanics.

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.

परीक्षेसाठी टीप

Practise the particle-in-a-box derivation until you can write it in five clear steps; it appears in almost every paper.

थोडक्यात सारांश

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.

महत्त्वाचे मुद्दे

• 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).

महत्त्वाच्या संज्ञा

• 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.

सूत्रे

• I = μr²; B = h/(8π²Ic) cm⁻¹
• E_J = BJ(J + 1); line positions ν̄ = 2B(J + 1); spacing = 2B
• ω̄ = (1/2πc)√(k/μ) → k = 4π²c²ω̄²μ
• E_v = (v + ½)ω̄ − (v + ½)²ω̄xₑ (anharmonic, xₑ = anharmonicity constant)

५ महत्त्वाचे प्रश्न

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.

• k = 4π²c²ω̄²μ
• = 4 × 9.87 × (2.998 × 10¹⁰ × 2886)² × 1.627 × 10⁻²⁷
• = 39.48 × 7.49 × 10²⁷ × 1.627 × 10⁻²⁷ ≈ 481 N m⁻¹ (use μ = 1.627 × 10⁻²⁷ kg)

Q5. Explain the Raman effect with a diagram of Stokes and anti-Stokes lines.

Energy-level diagram with virtual state; Rayleigh, Stokes (molecule gains energy, scattered light has lower frequency) and anti-Stokes lines (molecule loses energy, higher frequency, weaker because fewer molecules are excited); conditions for Raman activity; advantages (water as solvent, non-polar molecules).

परीक्षेसाठी टीप

Keep units straight: B and ω̄ are in cm⁻¹, so use c in cm s⁻¹ (2.998 × 10¹⁰) inside the formulas, and masses in kg.