Build orbital populations, test magnetic approximations and connect electronic gaps to observed colour.
JEE Main · NEET UG52 min study guide9 indexed sections
BY THE END, YOU CAN
Calculate correct ionic d counts and spin-only moments.
Distinguish discrete-ion response from collective magnetic ordering.
Compare high-spin, low-spin and geometry-dependent occupancies.
Explain why f-block orbital contributions require more than spin counting.
Calculate absorption energies and distinguish d–d from charge-transfer colour.
BEFORE YOU START
Oxidation states and electronic configurations.
Octahedral and tetrahedral splitting.
Photon energy and wavelength units.
Original educational lesson. Work through the examples before revealing the practice answers. Check the current official syllabus separately; historical question frequency does not establish eligibility or predict future questions.
CONNECT THE CONCEPT TO THE EXAM
Selected exam questions & worked solutions
These selected examples are separate from the original practice below. Each identifies its exam, year and wording. They do not represent every paper or every selected year.
No source-verified exam questions have been published for this lesson yet. The worked examples and practice below are original teaching material, not past-exam questions.
01 / UNDERSTAND & APPLY
1. Identify the ion before counting spins
Magnetic behaviour connects an electron configuration to a measurable response to a field. For a transition-metal ion, first obtain oxidation state from the complete formula, then determine its d count. Only after that should the ligand geometry and splitting pattern be used to distribute electrons. Neutral-atom configurations are not automatically the configurations of the ions.
The outer ns electrons are removed before the (n−1)d electrons in the familiar first-row cations. Fe is commonly written [Ar]3d⁶4s², Fe²⁺ is 3d⁶ and Fe³⁺ is 3d⁵. Cu²⁺ is 3d⁹, whereas Zn²⁺ is 3d¹⁰. The unpaired-electron count is obtained from the occupied orbitals, not from oxidation state alone.
The shortcut d count=group number−oxidation state works for the usual transition-metal ions in this lesson. It does not say that a complex with six d electrons must have six unpaired electrons: each orbital accommodates at most two, and ligand splitting may favour pairing.
An isolated ion model is also different from a bulk metal or a magnetically coupled solid. Start with local electronic structure, then ask whether collective interactions or orbital contributions invalidate a simple spin-only prediction.
ORIGINAL WORKED EXAMPLE
Find the right starting configuration
Determine the d counts in [FeF₆]³⁻ and [Fe(CN)₆]⁴⁻.
Each F⁻ or CN⁻ contributes −1 to the charge equation.
For [FeF₆]³⁻, x−6=−3 gives Fe(III); group 8−3=5.
For [Fe(CN)₆]⁴⁻, x−6=−4 gives Fe(II); group 8−2=6.
Answer The first is d⁵; the second d⁶. Spin populations still require the ligand-field information.
02 / UNDERSTAND & APPLY
2. Diamagnetism, paramagnetism and collective ordering
All electron clouds have a diamagnetic response: the induced magnetization opposes the applied field and produces a weak repulsion from stronger-field regions. For closed-shell molecules or ions, this is usually the dominant response. In the elementary isolated-ion classification, all electrons paired implies diamagnetic behaviour.
Unpaired electrons can provide permanent magnetic moments. Thermal motion randomizes their directions without a field; an applied field biases the population toward alignment, giving paramagnetism. A paramagnetic sample is attracted into a stronger field. It is not necessarily a permanent magnet after the field is removed.
Ferromagnetism is collective parallel ordering of magnetic moments below an ordering temperature, and can give strong spontaneous magnetization. Antiferromagnetic ordering has oppositely aligned sublattices with cancellation in the simplest case. Ferrimagnetism also has opposing sublattices but unequal moments leave a net magnetization. These are properties of interacting ensembles, not synonyms for having one unpaired electron.
A salt containing paramagnetic ions can display antiferromagnetic coupling, while a bulk metal can show band magnetism not captured by isolated free-ion counts. State whether the problem asks about a discrete complex, a free ion or a solid. Temperature, coupling and impurities can change the observed response.
ORIGINAL PRACTICE · 1
A complex has one unpaired electron. Must it retain permanent magnetization after removing a field?
Reveal answer & reasoning
No.
An isolated paramagnetic population can randomize thermally when the field is removed. Permanent magnetization requires appropriate collective ordering and material behaviour.
ORIGINAL PRACTICE · 2
Two equal spin sublattices point oppositely and their magnetizations cancel. Which idealized ordering is this?
Reveal answer & reasoning
Antiferromagnetic ordering.
Opposite alignment with unequal magnitudes would instead give a net ferrimagnetic moment.
03 / UNDERSTAND & APPLY
3. Derive and use the spin-only expression
Each unpaired electron contributes spin 1/2. If n unpaired electrons are aligned according to Hund’s rule within the relevant occupancy, their total spin is S=n/2. Taking the electron spin g factor approximately as 2 gives the spin-only effective moment 2√[S(S+1)] μB, equivalent to √[n(n+2)] μB.
The Bohr magneton μB is a unit of magnetic moment, approximately 9.274×10⁻²⁴ J T⁻¹. A reported value such as 4.90 BM means 4.90 Bohr magnetons; it is not a number of electrons or a field strength in tesla. The effective moment extracted from susceptibility is also not simply the maximum projected moment nμB.
For n=0,1,2,3,4,5 the spin-only values are 0,1.73,2.83,3.87,4.90,5.92 μB respectively. These landmarks help check arithmetic. A measured value near a landmark can support an electron assignment only when the spin-only approximation and sample model are suitable.
Orbital angular momentum, spin–orbit coupling, magnetic exchange, temperature-dependent populations and covalency can all produce deviations. The approximation often works reasonably for many first-row transition-metal ions with largely quenched orbital contributions, but it is not an exact universal experimental law.
μ_so = g√[S(S+1)] μB ≈ √[n(n+2)] μB
Use g≈2 and S=n/2; neglect orbital contributions and collective interactions.
ORIGINAL WORKED EXAMPLE
Compute and interpret a moment
An isolated complex has three unpaired electrons and negligible orbital contribution. Estimate its effective moment.
n=3 gives n(n+2)=3×5=15.
√15=3.87, so the spin-only prediction is 3.87 μB.
The value is an approximation to an effective moment; it does not mean the complex contains 3.87 unpaired electrons.
Answer Approximately 3.87 μB.
ORIGINAL PRACTICE · 1
Under a valid spin-only model, a moment near 4.90 μB corresponds to how many unpaired electrons?
Reveal answer & reasoning
Four.
Solve n(n+2)≈4.90²≈24; n=4 gives 4×6=24.
ORIGINAL PRACTICE · 2
What is the spin-only moment for an all-paired d⁶ complex?
Reveal answer & reasoning
0 μB.
n=0. The substance still has ordinary diamagnetic response; zero spin-only moment does not mean zero interaction with every magnetic field.
04 / UNDERSTAND & APPLY
4. Octahedral high spin and low spin
In an octahedral field, the three t₂g orbitals lie below the two e_g orbitals, separated by Δₒ. Electrons occupy separate orbitals within a degenerate set before pairing. When lower orbitals are singly occupied, additional electrons may either pair there or enter the upper set; the competition is between splitting and pairing costs.
For d⁴–d⁷ octahedral ions, weak fields can favour high spin and strong fields low spin. In the simple model, Δₒ<P favours the high-spin alternative and Δₒ>P favours low spin. These terms compare alternative distributions at the same d count, not two different oxidation states.
For octahedral d⁶, high spin t₂g⁴e_g² has four unpaired electrons, whereas low spin t₂g⁶e_g⁰ has none. For d⁵, high spin t₂g³e_g² has five unpaired electrons and low spin t₂g⁵ has one. Draw each orbital to avoid equating the exponent with the number of unpaired electrons.
Octahedral d¹, d² and d³ fill the lower set without a pairing-versus-upper-level choice; d⁸, d⁹ and d¹⁰ also lack distinct ordinary high/low-spin alternatives in this elementary octahedral scheme. Geometry still matters: square-planar d⁸ can be all-paired, while a common tetrahedral d⁸ complex has two unpaired electrons.
Full-size scientific diagram. Swipe or scroll horizontally; keyboard users can focus the diagram and use arrow keys.
Idealized octahedral occupancy. Spin-only predictions neglect orbital and collective contributions; zero spin-only moment does not remove diamagnetic response.
ORIGINAL WORKED EXAMPLE
Same metal ion, different ligands
Compare spin-only predictions for octahedral [Fe(H₂O)₆]²⁺, treated as high spin, and [Fe(CN)₆]⁴⁻, treated as low spin.
Both contain Fe²⁺ and are d⁶.
The aqua complex has t₂g⁴e_g²: one pair and four unpaired electrons. Its moment is √24=4.90 μB.
The cyanide complex has t₂g⁶: three pairs and no unpaired electrons.
Answer Approximately 4.90 μB versus zero spin-only moment.
ORIGINAL PRACTICE · 1
What are n and μ_so for octahedral high-spin and low-spin d⁵?
Reveal answer & reasoning
High spin: n=5, 5.92 μB. Low spin: n=1, 1.73 μB.
High spin occupies all five d orbitals singly; low-spin t₂g⁵ contains two pairs and one unpaired electron.
05 / UNDERSTAND & APPLY
5. Geometry cannot be read from one number alone
Tetrahedral splitting is usually smaller than octahedral splitting for comparable metal and ligands. Most familiar tetrahedral first-row complexes are therefore high spin. Avoid the absolute assertion that no low-spin tetrahedral complex could ever exist; the school-level trend is broad but not a universal law of nature.
Square-planar d⁸ complexes commonly have a sufficiently high d(x²−y²) level that the eight electrons occupy the four lower orbitals in pairs. In contrast, tetrahedral Ni²⁺ is d⁸ with two unpaired electrons. Thus [Ni(CN)₄]²⁻ is a standard diamagnetic square-planar reference, and [NiCl₄]²⁻ a paramagnetic tetrahedral reference.
A measured moment can eliminate incompatible assignments under stated assumptions, but does not uniquely establish every bond angle or ligand arrangement. For example, several d counts can share one unpaired electron. Combine magnetic evidence with composition, ligand field, spectra and structural information.
Count ligand charge rather than assuming that a familiar ligand always produces one oxidation state. Then keep the geometry-specific orbital ordering explicit. Applying an octahedral t₂g/e_g diagram to a square-planar complex is a change of model, not a harmless notation choice.
ORIGINAL PRACTICE · 1
Why does “one unpaired electron” not uniquely identify Cu²⁺?
Reveal answer & reasoning
Other configurations and spin states can also have one unpaired electron.
Low-spin octahedral d⁵ and d¹ ions are examples. Composition and charge are needed to distinguish them.
06 / UNDERSTAND & APPLY
6. f-block moments and the orbital contribution
For lanthanoid ions, the 4f electrons are relatively shielded from the ligands by filled outer shells. The orbital contribution is often retained more strongly than in familiar 3d complexes. Spin–orbit coupling combines total orbital angular momentum L and total spin S into total angular momentum J. Consequently, counting unpaired electrons alone usually does not predict the measured moment.
For an isolated well-defined ground J multiplet in the free-ion approximation, μ_eff≈g_J√[J(J+1)] μB. The Landé factor is g_J=1+[J(J+1)+S(S+1)−L(L+1)]/[2J(J+1)]. This expression assumes a suitable angular-momentum coupling model and negligible thermal population of higher multiplets; it must not be applied mechanically when J=0 or when nearby states matter.
Gd³⁺ is 4f⁷ with S=7/2 and L=0 in its ground state. Then J=7/2, g≈2, and the result is √63≈7.94 μB, matching the spin-only estimate well. Ce³⁺ is 4f¹ with a ground-state orbital contribution; using L=3, S=1/2 and J=5/2 gives g_J=6/7 and a free-ion estimate about 2.54 μB, rather than 1.73 μB from one unpaired spin alone.
Sm³⁺ and Eu³⁺ illustrate additional limits because nearby levels and temperature-dependent effects matter. In particular, a J=0 ground state does not justify claiming zero observed susceptibility at every temperature. For actinoids, stronger spin–orbit effects, covalency and more complicated level structures further restrict simple free-ion or spin-only models.
The broad d- and f-block comparison is therefore causal: ligand-induced orbital quenching often makes spin-only useful for 3d complexes; shielded 4f electrons retain substantial orbital character; 5f systems can add stronger covalency and coupling complexity. It is a hierarchy of approximations, not a claim that one formula fits all metals.
ORIGINAL WORKED EXAMPLE
A supplied f-ion model
For a hypothetical ground multiplet specified by L=3, S=1/2 and J=5/2, calculate g_J and μ_eff in the free-ion approximation.
J(J+1)=35/4=8.75, S(S+1)=3/4=0.75, and L(L+1)=12.
g_J=1+(8.75+0.75−12)/(2×8.75)=1−2.5/17.5=6/7.
μ_eff=(6/7)√8.75 μB≈2.54 μB. The supplied quantum numbers, not only the unpaired count, determine this result.
Answer g_J=6/7; μ_eff≈2.54 μB.
ORIGINAL PRACTICE · 1
Why is Gd³⁺ a relatively good spin-only f-block example?
Reveal answer & reasoning
Its 4f⁷ ground state has L=0.
The half-filled shell gives no resultant orbital angular momentum in that ground term, so the leading moment is spin-derived: √63≈7.94 μB.
07 / UNDERSTAND & APPLY
7. Colour begins with the absorbed photon
Ligands split metal d levels. If an allowed or weakly allowed transition between relevant states requires a visible photon, selective absorption can colour the complex. The absorbed colour is removed from the incident white-light mixture; the transmitted or reflected appearance is associated with the remaining light, often described approximately as the complementary colour.
Photon energy is E=hν=hc/λ. A larger energy gap corresponds to shorter absorbed wavelength. For a simple d¹ octahedral model, the main d–d absorption can directly estimate Δₒ. Multi-electron ions have electron–electron repulsion and multiple terms, so not every observed band equals a single Δₒ without further analysis.
In centrosymmetric complexes, d–d transitions are Laporte-forbidden in the simplest electric-dipole approximation, but vibrations and distortion can lend them intensity. Spin-changing transitions are also weak under the spin selection rule. Thus partially filled d orbitals do not guarantee intense visible colour. Tetrahedral complexes lack an inversion centre, which can make their d–d bands more intense.
Charge-transfer transitions move electron density substantially between ligand and metal and can be intense. MnO₄⁻ has Mn(VII), formally d⁰, yet is strongly coloured; its colour cannot be assigned to a d–d promotion from an occupied d orbital. Similarly, ligand-centred transitions may colour a compound even when the metal is d⁰ or d¹⁰.
Ligand identity, oxidation state, geometry and solvent can alter energy gaps and therefore colour. Colour alone is not a reliable unique identification of a metal, and a diamagnetic complex can still be coloured through allowed electronic transitions. Magnetism and optical absorption interrogate related but different aspects of electronic structure.
E_photon=hc/λ; E_molar=N_Ahc/λ
Use λ in metres with h in J s and c in m s⁻¹; multiply by Avogadro constant for J mol⁻¹.
Full-size scientific diagram. Swipe or scroll horizontally; keyboard users can focus the diagram and use arrow keys.
Conceptual optical-path diagram; no particular absorption spectrum or observed hue is implied. A single wavelength is not a complete colour model.
ORIGINAL WORKED EXAMPLE
Convert a band to energy
A simple d¹ model absorbs at 600 nm. Estimate the photon energy and molar gap using h=6.626×10⁻³⁴ J s, c=3.00×10⁸ m s⁻¹ and N_A=6.022×10²³ mol⁻¹.
Convert 600 nm to 6.00×10⁻⁷ m.
E=(6.626×10⁻³⁴)(3.00×10⁸)/(6.00×10⁻⁷)=3.313×10⁻¹⁹ J.
Multiply by N_A and divide by 1000: about 199.5 kJ mol⁻¹. The observed solution colour is not simply the absorbed orange colour.
Answer 3.31×10⁻¹⁹ J per photon; about 200 kJ mol⁻¹.
ORIGINAL PRACTICE · 1
A d⁰ oxometal anion is intensely coloured. Must the oxidation-state calculation be wrong?
Reveal answer & reasoning
No.
An intense ligand-to-metal charge-transfer transition can produce colour without a d–d transition. Permanganate is a familiar example.
08 / UNDERSTAND & APPLY
8. Combine magnetic and optical evidence
Suppose a correctly assigned d⁶ complex is diamagnetic and has a visible absorption. An octahedral low-spin configuration is compatible with the magnetic evidence, and visible absorption is compatible with excitations from its filled lower levels into higher levels. “All paired” describes its ground state, not the absence of excited electronic states.
Suppose a sample has a moment larger than the elementary spin-only prediction. First check composition, oxidation state, geometry, purity and the convention used to extract the moment. Only then consider orbital contributions or exchange. A discrepancy is evidence that a model or assignment needs examination, not permission to invent a fractional unpaired electron.
In qualitative multiple-statement problems, distinguish “can” from “must.” A partially filled d subshell can support d–d transitions; their energy may be outside the visible range. Unpaired electrons can cause paramagnetism; a coupled solid may have a cancelling ground state. A ligand can shift colour without changing oxidation state.
09 / UNDERSTAND & APPLY
9. Revision decision path
For a discrete transition-metal complex: solve ligand-charge algebra, obtain dⁿ, select the actual geometry, fill its orbitals with the stated field assumptions, count unpaired electrons, and only then use the spin-only expression if justified. For f-block ions, explicitly ask whether orbital angular momentum must be retained.
For colour: name the candidate transition, relate the absorption energy to wavelength, and distinguish absorbed from observed light. Check charge-transfer and ligand-centred alternatives before concluding that d⁰ or d¹⁰ necessarily means colourless.
For solids: keep collective ordering separate from an isolated-ion moment. A complete explanation identifies both the microscopic source of moments and the interactions that align or cancel them.
The number under the square root is n(n+2), not n².
An effective moment is reported in μB or BM, not tesla.
High/low spin depends on d count, geometry and splitting versus pairing.
Spin-only is a model; f-block orbital contributions are substantive, not minor bookkeeping.
KEEP THE BIG IDEAS
Quick revision
Count the correct ion before counting unpaired electrons.
μ_so is useful when orbital and collective effects are negligible.
f-block moments often retain orbital contributions.
Absorbed and observed colours differ; charge transfer can colour d⁰ compounds.
The teaching notes, worked explanations and original practice are written for this lesson. Separately labeled selected exam questions identify their exam, year, source and whether wording is paraphrased or an exact excerpt. Educational references do not certify current exam coverage.