The d- and f-Block Elements: common doubts, answered
The questions students ask most often about The d- and f-Block Elements, each with a short answer. For the full chapter, read the The d- and f-Block Elements notes.
About the chapter
How is The d- and f-Block Elements chapter usually tested in NEET?
Expect questions on electronic configurations of ions, spin-only magnetic moments, why particular ions are coloured or colourless, and trends in E° values and oxidation states. The reactions of potassium dichromate and permanganate in different media come up often, as does lanthanoid contraction and its consequences. Removing 3d electrons before 4s, and calling the chromate-dichromate change redox, are the classic slips.
Position and electronic configuration
Read this section in the notes →Why are zinc, cadmium and mercury not counted as transition elements?
A transition element is one with a partly filled d subshell, either in the atom or in one of its common ions. Zinc, cadmium and mercury have a full d¹⁰ set both as atoms and in their usual +2 ions, so they do not qualify, though they are placed in the d-block. Copper does qualify, because Cu²⁺ is 3d⁹.
Why are chromium and copper configurations 3d⁵4s¹ and 3d¹⁰4s¹?
Half-filled and completely filled d subshells have extra stability, from symmetrical electron distribution and greater exchange energy. The 3d and 4s energies are very close, so moving one electron from 4s into 3d to reach d⁵ or d¹⁰ lowers the total energy. Chromium therefore has 3d⁵4s¹ instead of 3d⁴4s², and copper has 3d¹⁰4s¹ instead of 3d⁹4s².
Why are 4s electrons removed before 3d electrons when a transition metal ionises?
Once the 3d orbitals begin to fill, they drop below 4s in energy, and the outer 4s electrons are lost first. So Fe, 3d⁶4s², becomes Fe²⁺ as 3d⁶ and Fe³⁺ as 3d⁵. Removing 3d electrons first, to write Fe²⁺ as 3d⁴4s², is one of the most frequent errors in configuration questions.
Physical properties and atomic size
Read this section in the notes →Why do transition metals have high melting points?
Their metallic bonding is strong because, besides the ns electrons, many (n−1)d electrons also take part in bonding between atoms. More unpaired d electrons generally mean stronger bonding, so melting points rise to a maximum near the middle of each series, around d⁵, and then fall. Zinc, cadmium and mercury have no unpaired d electrons and melt at low temperatures.
What is lanthanoid contraction and why do Zr and Hf have almost the same radius?
Across the lanthanoids, the atomic and ionic radii shrink steadily, because 4f electrons shield one another poorly from the increasing nuclear charge. This cancels the size gain expected on moving from the 4d to the 5d series, so zirconium (160 pm) and hafnium (159 pm) end up nearly the same size. They behave so alike that separating them is very difficult.
Ionisation enthalpies
Read this section in the notes →Why does the first ionisation enthalpy change only slowly across a transition series?
Each step adds one proton to the nucleus but also one electron to the inner 3d subshell. The new d electron partly shields the outer 4s electrons from the extra nuclear charge, so the effective attraction on them rises only a little. That makes ionisation enthalpies increase gradually and irregularly across the series, compared with the steep rise across a period of the s and p blocks.
Why is the third ionisation enthalpy of manganese so high?
Mn²⁺ is 3d⁵, a half-filled subshell with extra stability. Removing a third electron would break that stable arrangement, so a great deal of energy is needed. For the same reason, the second ionisation enthalpies of chromium and copper are unusually high, because their singly charged ions are already d⁵ and d¹⁰.
Oxidation states
Read this section in the notes →Why does manganese show the most oxidation states in the 3d series?
Manganese, 3d⁵4s², has seven electrons in 4s and 3d that can be used in bonding, and the energies of these orbitals are close enough for any number of them to be involved. So it shows every state from +2 to +7, the highest in KMnO₄. Elements at the ends of the series have fewer usable electrons or too many paired d electrons, so they show fewer states.
Why does oxygen bring out higher oxidation states than fluorine in some metals?
Oxygen can form multiple bonds with the metal, so a few oxygen atoms can satisfy a very high oxidation state. Manganese reaches +7 in Mn₂O₇, but its highest fluoride is MnF₄, since each fluorine forms only a single bond. Similarly osmium reaches +8 in OsO₄. Fluorine still stabilises high states well, as in VF₅ and CrF₆, through its small size and high electronegativity.
Electrode potentials
Read this section in the notes →Why is the E° value for Cu²⁺/Cu positive while the others in the series are negative?
Copper's high enthalpy of atomisation and high sum of first two ionisation enthalpies are not paid back by the hydration enthalpy of Cu²⁺. So converting copper metal to aqueous Cu²⁺ is energetically uphill, giving E° of +0.34 V. That is why copper does not release hydrogen from dilute acids and dissolves only in oxidising acids such as nitric acid.
Higher oxidation states and reactivity
Read this section in the notes →Why is Cr²⁺ a reducing agent while Mn³⁺ is an oxidising agent, though both are d⁴?
Cr²⁺ readily loses an electron to become Cr³⁺, which is d³ with a half-filled t₂g set that is especially stable, so Cr²⁺ is reducing. Mn³⁺ readily gains an electron to become Mn²⁺, which is d⁵, a half-filled d subshell, so Mn³⁺ is oxidising. In each case the ion moves towards the more stable configuration available to it.
Magnetic properties
Read this section in the notes →How do you calculate the spin-only magnetic moment of a transition metal ion?
Use μ = √[n(n+2)] BM, where n is the number of unpaired electrons. Find n from the ion's d configuration after removing the 4s electrons first. Fe³⁺, d⁵, has 5 unpaired electrons and μ ≈ 5.92 BM; Cu²⁺, d⁹, has 1 and μ ≈ 1.73 BM. Zn²⁺, d¹⁰, has none, so it is diamagnetic.
Coloured ions and complexes
Read this section in the notes →Why are Sc³⁺, Ti⁴⁺ and Zn²⁺ colourless while Cu²⁺ is blue?
Colour in these ions comes from d-d transitions, in which visible light lifts an electron from a lower d orbital to a higher one. Sc³⁺ and Ti⁴⁺ are d⁰ with no d electron to promote, and Zn²⁺ is d¹⁰ with no vacant d orbital to receive one, so they absorb no visible light. Cu²⁺, d⁹, absorbs red-orange light and appears blue.
Catalysts, interstitial compounds and alloys
Read this section in the notes →Why are transition metals and their compounds good catalysts?
They can use their d electrons and vacant d orbitals to form temporary bonds with reactant molecules on their surface, which weakens bonds in the reactants and lowers the activation energy. They also switch easily between oxidation states, forming unstable intermediates. Iron in the Haber process, V₂O₅ in the contact process and nickel in hydrogenation are examples.
What are interstitial compounds and what are their properties?
They form when small atoms such as hydrogen, carbon or nitrogen get trapped in the spaces of a metal's crystal lattice, as in TiC, Fe₃H and Mn₄N. Their formulas are usually non-stoichiometric and do not follow normal valencies. They have higher melting points than the parent metal, are very hard, keep metallic conductivity and are chemically fairly inert.
Potassium dichromate and permanganate
Read this section in the notes →Why does the colour of potassium dichromate change from orange to yellow in alkali?
Orange dichromate, Cr₂O₇²⁻, and yellow chromate, CrO₄²⁻, are in an equilibrium that depends on pH. Adding alkali pushes it towards chromate, and adding acid pushes it back to dichromate. Chromium stays +6 in both ions, so this is an acid-base change, not a redox reaction, a point that trips many students.
How many electrons does permanganate gain in acidic, neutral and basic solutions?
In acidic solution MnO₄⁻ gains 5 electrons and becomes nearly colourless Mn²⁺, so manganese goes from +7 to +2. In neutral or faintly alkaline solution it gains 3 electrons and forms brown MnO₂, manganese +4. In strongly alkaline solution it can take just one electron to give green manganate, MnO₄²⁻. The electron count therefore depends on the medium.
Why is HCl not used to acidify permanganate in titrations?
Permanganate is strong enough to oxidise chloride ions to chlorine gas, so some of it would be consumed by the acid itself and the titre would come out too high. Dilute sulphuric acid is used instead, because sulphate is not oxidised by permanganate. Permanganate titrations also need no separate indicator, since the first excess drop leaves a permanent pink colour.
The lanthanoids
Read this section in the notes →Why is +3 the common oxidation state of lanthanoids, and why are Ce⁴⁺ and Eu²⁺ found?
All lanthanoids readily lose three electrons, two from 6s and one from 5d or 4f, and +3 is their typical state. Other states appear when they lead to an empty, half-filled or full f subshell. Ce⁴⁺ is 4f⁰, so it forms, though it is a strong oxidising agent. Eu²⁺ is 4f⁷, half-filled, but it is a strong reducing agent and readily goes to +3.
The actinoids and applications
Read this section in the notes →Why do actinoids show more oxidation states than lanthanoids?
In the actinoids the 5f, 6d and 7s orbitals lie very close in energy, so many electrons can take part in bonding, giving states such as +6 for uranium and +7 for neptunium. The 5f orbitals also extend further from the nucleus than 4f, so they bond more readily. Lanthanoids, with deeply buried 4f orbitals, stick mainly to +3.
What is the difference between lanthanoids and actinoids?
Lanthanoids fill the 4f subshell, mostly show +3, and except for promethium are not radioactive. Actinoids fill 5f, show a wider range of oxidation states, are all radioactive, and their contraction from one element to the next is larger because 5f electrons shield even more poorly. Actinoids also form complexes more readily, and many have short half-lives, which makes them harder to study.
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