Where are transition elements in the periodic table?
The d-block is the wide block in the middle of the periodic table. It has groups 3 to 12. It sits between the s-block (groups 1, 2) and the p-block (groups 13 to 18).
There are four rows (series): 3d (Sc to Zn), 4d (Y to Cd), 5d (La, Hf to Hg) and 6d (Ac, Rf onwards).
A transition element is one that has a partly filled d-subshell in the atom or in one of its common ions. So Zn, Cd and Hg (full d¹⁰ in atom and ion) are d-block elements but are not called true transition elements.
The name "transition" means "in between": their properties change from the very reactive s-block metals to the p-block elements.
Electronic configuration of d-block elements
General configuration: (n−1)d¹⁻¹⁰ ns¹⁻². "n−1" means the d-shell is one shell inside the outer shell. For the first series n = 4, so electrons go into 3d.
The 4s fills before 3d, but when the atom forms an ion, 4s electrons leave first.
Two special cases
A half-filled (d⁵) or full (d¹⁰) set is extra stable. So:
- Cr is [Ar] 3d⁵ 4s¹ (not 3d⁴ 4s²)
- Cu is [Ar] 3d¹⁰ 4s¹ (not 3d⁹ 4s²)
Electrons first go one per box with the same spin, then pair up (Hund's rule). Story step 2 shows this.
General properties of transition elements
All transition elements are metals. They are hard, shiny, have high melting and boiling points (many unpaired d-electrons make strong metallic bonds) and conduct heat and electricity well. Below are the main properties the syllabus asks for.
Atomic and ionic radii
Across a series the radius first gets smaller, then stays almost the same in the middle, and rises a little at the end. Why? Each new electron goes into an inner d-shell. It shields the outer electrons from the growing nuclear charge. So the pull on the outside hardly changes.
Down a group, 4d is bigger than 3d. But 4d and 5d (for example Zr 160 pm and Hf 159 pm) are almost the same, because of the lanthanoid contraction (see the lanthanoids lesson).
Ionisation enthalpy
Ionisation enthalpy is the energy needed to pull one electron out of a gas atom. Across the series it rises slowly (much less than across a p-block row), because shielding by d-electrons cancels much of the extra nuclear charge.
Some jumps are special. Zn has a high first value because 3d¹⁰ 4s² is fully filled. The second ionisation of Cr and Cu is very high because removing a second electron breaks a stable d⁵ or d¹⁰ set.
Variable oxidation states
The 4s and 3d electrons have nearly the same energy. So the metal can lose different numbers of them. That gives many oxidation states, usually changing in steps of one (Fe²⁺, Fe³⁺).
- The middle element Mn shows the most: +2 to +7.
- The ends show few: Sc only +3, Zn only +2.
- The highest state often equals the group number up to Mn (Mn is group 7 → +7 in KMnO₄).
- High states are found with oxygen or fluorine (small, very electronegative), like MnO₄⁻, CrO₄²⁻.
Stability in water: Mn²⁺ (d⁵) and Fe³⁺ (d⁵) are extra stable; Cu⁺ usually changes into Cu²⁺ and Cu in water, because Cu²⁺ has a very large hydration energy.
Why are transition metal ions coloured?
In an ion surrounded by water (or other groups), the five d-orbitals split into two energy levels. A d-electron can jump from the lower level to the upper level by absorbing some visible light. We see the remaining (complementary) colour. This jump is called a d–d transition.
So an ion needs a partly filled d-set to be coloured. d⁰ (Sc³⁺, Ti⁴⁺) and d¹⁰ (Zn²⁺, Cu⁺) ions are colourless.
Examples: Cu²⁺ blue, Ni²⁺ green, Co²⁺ pink, Mn²⁺ pale pink, Fe³⁺ yellow.
Magnetic properties and magnetic moment
An electron spins, so it acts like a tiny magnet. Two paired electrons cancel. Unpaired electrons do not cancel.
- Paramagnetic: has unpaired electrons, pulled into a magnet (Fe³⁺, Cu²⁺).
- Diamagnetic: all paired, weakly pushed away (Zn²⁺, Sc³⁺).
- Ferromagnetic: strongly magnetic solids like Fe, Co, Ni.
Spin-only magnetic moment: μ = √(n(n+2)) BM, where n = number of unpaired electrons and BM = Bohr magneton. More unpaired electrons → bigger μ.
Catalytic properties
A catalyst speeds up a reaction without being used up. Transition metals are good catalysts for two reasons:
- Variable oxidation states: the metal can take an electron and give it back (Fe³⁺ ⇌ Fe²⁺), making an easy path.
- Surface: reacting molecules stick to the metal surface, come close and react faster.
Examples: Fe in the Haber process (ammonia), V₂O₅ in the Contact process (sulphuric acid), Ni in hydrogenation of oils (vanaspati ghee), Pt/Pd/Rh in car catalytic converters.
Interstitial compounds and alloys
Interstitial compounds form when very small atoms (H, C, N) sit in the gaps of the metal crystal, like TiC, Mn₄N, Fe₃H. They are very hard, have high melting points and still conduct electricity.
Alloys form easily because transition metal atoms have similar sizes (radius within about 15%), so one atom can take the place of another in the crystal. Examples: steel (Fe + C + Cr/Ni), brass (Cu + Zn), bronze (Cu + Sn).
Other properties: complexes and enthalpy of atomisation
Transition metals form many complexes (like [Cu(NH₃)₄]²⁺) because their ions are small, highly charged and have empty d-orbitals to accept electron pairs.
They have high enthalpy of atomisation (energy to break the metal into atoms), because many unpaired electrons form strong metal–metal bonds. It is highest near the middle of each series.
Try it: predict, then check
1. Before you move the picker in the last 3D step, write down how many unpaired electrons you expect in Fe²⁺ (d⁶) and Cu²⁺ (d⁹). Then pick Fe and Cu and check the μ value.
2. At home: look at a blue copper sulphate crystal (from a school lab or garden shop) and a pale green iron (ferrous) sulphate crystal. Both colours come from d–d jumps. Zinc sulphate crystals are white: Zn²⁺ is d¹⁰.
3. Hold a strong fridge magnet near an iron nail and a copper wire. Only iron is pulled strongly (ferromagnetic).
Key formulas and definitions
- General configuration: (n−1)d¹⁻¹⁰ ns¹⁻²
- Spin-only magnetic moment: μ = √(n(n+2)) BM (n = unpaired electrons)
- Cr = [Ar] 3d⁵ 4s¹, Cu = [Ar] 3d¹⁰ 4s¹
- Ions: 4s electrons are lost before 3d electrons
- Coloured if d¹ to d⁹; colourless if d⁰ or d¹⁰
Worked examples
1. Write the electronic configuration of Mn (Z = 25) and Mn²⁺.
Step 1: Ar has 18 electrons, so 7 are left. Step 2: 4s takes 2, 3d takes 5 → Mn = [Ar] 3d⁵ 4s². Step 3: for the ion, remove the two 4s electrons first → Mn²⁺ = [Ar] 3d⁵.
2. How many unpaired electrons does Fe³⁺ (Z of Fe = 26) have? Find its spin-only magnetic moment.
Fe = [Ar] 3d⁶ 4s². Fe³⁺: remove 2 from 4s and 1 from 3d → 3d⁵. Five boxes, one electron each → n = 5. μ = √(5 × 7) = √35 = 5.92 BM.
3. Calculate the spin-only magnetic moment of Cu²⁺ (Z = 29).
Cu = [Ar] 3d¹⁰ 4s¹. Cu²⁺: remove 4s¹ and one 3d → 3d⁹. Nine electrons in five boxes: four pairs and one single → n = 1. μ = √(1 × 3) = √3 = 1.73 BM.
4. Which is coloured: Ti⁴⁺, V³⁺, Zn²⁺ or Sc³⁺? (Z: Ti 22, V 23, Zn 30, Sc 21)
Find d-count: Ti⁴⁺ = 3d⁰, V³⁺ = 3d², Zn²⁺ = 3d¹⁰, Sc³⁺ = 3d⁰. Only a partly filled d-set can do a d–d jump. So only V³⁺ is coloured (green).
5. An ion of a 3d metal has μ = 3.87 BM. How many unpaired electrons does it have? Suggest one such ion.
Set √(n(n+2)) = 3.87. Square it: n(n+2) = 15. Try n = 3: 3 × 5 = 15 ✔. So n = 3. A d³ ion like Cr³⁺ (or d⁷ Co²⁺) fits.
6. Arrange Mn²⁺, Fe²⁺, Ni²⁺ and Zn²⁺ in increasing order of magnetic moment. (Z: Mn 25, Fe 26, Ni 28, Zn 30)
d-counts: Mn²⁺ d⁵ (n = 5), Fe²⁺ d⁶ (n = 4), Ni²⁺ d⁸ (n = 2), Zn²⁺ d¹⁰ (n = 0). μ grows with n: Zn²⁺ (0) < Ni²⁺ (2.83) < Fe²⁺ (4.90) < Mn²⁺ (5.92 BM).
7. Why is the second ionisation enthalpy of Cu much higher than that of Zn?
Cu⁺ is 3d¹⁰: a full, very stable set. Taking the second electron must break it, so it costs a lot. Zn⁺ is 3d¹⁰ 4s¹: the second electron comes from 4s, which is easy. So IE₂(Cu) > IE₂(Zn).
Common mistakes
- Removing 3d electrons before 4s when making ions. The 4s electrons always leave first: Fe²⁺ is 3d⁶, not 3d⁴ 4s².
- Writing Cr as 3d⁴ 4s² and Cu as 3d⁹ 4s². The half-full and full d-sets win: 3d⁵ 4s¹ and 3d¹⁰ 4s¹.
- Saying Zn is a typical transition element. Zn has 3d¹⁰ in the atom and in Zn²⁺, so it is a d-block element but not a true transition element.
- Putting total electrons instead of unpaired electrons into μ = √(n(n+2)). Count only the single (unpaired) electrons in the boxes.