General Properties of d-Block Elements
Questions 1–4
IUPAC defines a transition metal as an element whose atom has an incomplete d subshell or which can give rise to cations with an incomplete d subshell. By this definition, why are Zn, Cd, and Hg sometimes excluded from the transition metals?
Zinc, cadmium, and mercury have the ground-state configuration [noble gas] d10s2. When they form their most common cations (M2+), the two s electrons are lost, leaving a complete d10 shell. Since neither the neutral atom nor the common ion has a partially filled d subshell, they do not satisfy the IUPAC definition of a transition metal. While they do lack coloured ions, that is a consequence rather than the defining criterion.
Transition metals characteristically exhibit variable oxidation states. Which of the following best explains why this occurs?
In transition metals, the (n−1)d and ns orbitals lie close in energy. This means that successive ionisation energies do not show a dramatic jump as electrons are removed, unlike s- or p-block elements where core electrons are far more tightly held. Consequently, transition metals can lose varying numbers of d and s electrons without prohibitive energy cost, leading to multiple accessible oxidation states.
Examine the trend in atomic (metallic) radii across the first row transition metals from Sc to Zn. Which statement best describes the observed trend?
Across the first transition series, increasing nuclear charge tends to contract the atom, but electrons are added to inner (3d) orbitals that shield the outer 4s electrons. The initial contraction from Sc to about Cr/Mn reflects the increasing Zeff, but by mid-row the d–d electron repulsion partially offsets further contraction, leading to an almost flat region. At Cu and Zn, the d shell is nearly or fully filled and screens less effectively, producing a slight increase.
The lanthanide contraction refers to the steady decrease in ionic radii across the 4f series (La–Lu). What is its most important consequence for the chemistry of the 5d transition metals?
As the 4f shell fills from La to Lu, the poor shielding of 4f electrons causes a cumulative contraction in atomic size. By the time the 5d series begins (Hf onward), this contraction almost exactly cancels the expected size increase from adding another electron shell. The result is that pairs such as Zr/Hf, Nb/Ta, and Mo/W have nearly identical radii. This makes second- and third-row congeners far more similar in chemistry than first- and second-row pairs.
First Row Transition Metal Chemistry
Questions 5–8
For the early first-row metals Ti through Mn, the maximum oxidation state equals the total number of valence electrons (ns + (n−1)d). Beyond Mn, the maximum attainable oxidation state decreases. Why?
For Ti(+4), V(+5), Cr(+6), and Mn(+7), the maximum oxidation state corresponds to losing all valence electrons. Beyond Mn, Zeff is sufficiently large that removing all valence electrons becomes prohibitively expensive in terms of ionisation energy. The d electrons become increasingly core-like and harder to remove. Thus Fe reaches at most +6 (rarely), Co at most +5 (rare), and Ni and Cu rarely exceed +3 or +4 under normal conditions.
Cr3+ complexes are well known for being kinetically inert — they undergo ligand substitution extremely slowly. Which explanation accounts for this exceptional inertness?
Cr3+ has a d3 configuration. In an octahedral crystal field, all three d electrons occupy the lower t2g set with one electron in each orbital (half-filled t2g). This gives a large crystal field stabilisation energy (CFSE) of −1.2Δoct. Any ligand substitution mechanism (associative or dissociative) requires passing through a geometry that has lower CFSE, creating a substantial activation energy barrier. This makes Cr3+ complexes characteristically inert.
In aqueous solution, both Fe2+ (d6) and Fe3+ (d5) exist. Which ion is more thermodynamically stable in water under standard conditions, and why?
Fe3+ with its d5 high-spin configuration has one electron in each of the five d orbitals, maximising exchange energy (a quantum mechanical stabilisation from parallel electron spins). Under ambient aerobic aqueous conditions, Fe2+ is readily oxidised to Fe3+. The high third ionisation energy of Fe is offset by the large hydration enthalpy of the smaller, more highly charged Fe3+ ion and by the exchange stabilisation of the d5 configuration.
Copper has the ground-state configuration [Ar] 3d104s1, yet its most common oxidation state is +2 (d9) rather than +1 (d10). Why is Cu2+ favoured in aqueous chemistry?
Although Cu+ (d10) might seem electronically favourable, it is a large, singly charged ion with relatively weak interactions with water. Cu2+ is smaller and carries twice the charge, so its hydration enthalpy is far larger (about −2100 kJ/mol versus −580 kJ/mol for Cu+). This enormous gain in solvation energy far exceeds the cost of removing the second electron. Additionally, Cu+ readily disproportionates in water: 2Cu+ → Cu + Cu2+.
Second & Third Row Transition Metals
Questions 9–12
Higher oxidation states are generally more stable for the 4d and 5d transition metals compared to their 3d analogues (e.g. MoO3 is stable but CrO3 is a powerful oxidant). What is the primary reason?
The 4d and 5d orbitals extend further from the nucleus than 3d orbitals. This greater spatial extent improves overlap with ligand orbitals, forming stronger covalent bonds. The bond energy released when forming multiple strong M–O or M–F bonds compensates for the cost of reaching high oxidation states. For 3d metals, poorer orbital overlap means weaker bonds, so high oxidation states are often strongly oxidising or inaccessible altogether.
CrO42− (chromate), MoO42− (molybdate), and WO42− (tungstate) are all tetrahedral d0 species with the metal in the +6 oxidation state. How does their oxidising power compare?
CrO42− is a strong oxidising agent, particularly in acidic solution (as Cr2O72−, E° ≈ +1.33 V). In contrast, MoO42− and WO42− are essentially non-oxidising under normal conditions. This reflects the general trend: high oxidation states become progressively more stable descending a group because the larger 4d/5d orbitals form stronger bonds with oxygen, lowering the driving force for reduction.
Pd2+ and Pt2+ (both d8) almost invariably adopt square planar geometry, whereas Ni2+ (also d8) is often tetrahedral or octahedral. Why do the heavier congeners strongly prefer square planar?
For d8 ions, square planar geometry leaves the high-energy dx2−y2 orbital empty, but requires all eight electrons to be paired in four lower orbitals. This is favourable only when Δ is large enough to exceed the pairing energy. The 4d and 5d metals have inherently larger Δ values (roughly 1.5× that of 3d) due to better metal–ligand overlap. Thus Pd2+ and Pt2+ are almost always square planar, while Ni2+ adopts square planar geometry only with strong-field ligands such as CN−.
Metal–metal multiple bonds (double, triple, and even quadruple bonds) are an important feature of heavier transition metal chemistry. Which elements are most prominently associated with forming these bonds, and why?
Metal–metal multiple bonds require face-to-face overlap of d orbitals to form δ bonds, in addition to σ and π components. The spatially extended 4d and 5d orbitals of Mo, W, and Re overlap far more effectively than the compact 3d orbitals. The classic example is [Re2Cl8]2−, which contains a quadruple bond (σ2π4δ2). Mo–Mo quadruple bonds in Mo2(OAc)4 are another landmark example.
Special Topics: f-Block & Bioinorganic
Questions 13–15
The f-block elements comprise the lanthanides (4f series) and actinides (5f series). Which of the following correctly compares their chemistry?
The 4f orbitals of the lanthanides are deeply buried inside the atom and do not participate significantly in bonding, leading to predominantly ionic +3 chemistry across the whole series. In contrast, the 5f orbitals of the early actinides (U, Np, Pu) are more radially extended and energetically accessible. This allows the early actinides to access oxidation states from +3 to +7, resembling the variable oxidation state chemistry of d-block metals. Later actinides (Am onward) become more lanthanide-like, favouring +3.
The +3 oxidation state is overwhelmingly dominant for the lanthanides. Why is this the case?
Lanthanide atoms have the configuration [Xe] 4fn 5d0–1 6s2. The first three ionisation energies (removing the 6s and a 5d/4f electron) are moderate and readily compensated by lattice or hydration enthalpy. However, the fourth ionisation energy involves removing a deeply buried 4f electron and is very large. The 4f orbitals have almost no radial extension beyond the xenon core, so they are effectively shielded from chemical influence. Only at special configurations (Ce4+ = f0, Tb4+ = f7) is +4 accessible.
In biological systems, the most commonly utilised transition metals are Fe, Cu, Zn, and Mn. What factors explain why nature has selected these particular metals?
The bioavailability of a metal is the first requirement — Fe, Mn, Zn, and Cu are among the most abundant first-row transition metals in the environment. Beyond abundance, each plays a specific role: Fe and Cu provide accessible redox couples (Fe2+/Fe3+, Cu+/Cu2+) at potentials suitable for electron transfer and O2 chemistry. Zn2+ (d10, redox-inactive) serves as a Lewis acid catalyst. Mn provides multiple oxidation states for the oxygen-evolving complex in photosystem II. Their kinetic lability ensures rapid ligand exchange needed for catalytic turnover.