Why Copper Sulphate Is Blue but Zinc Sulphate Is White

Iron in a bridge, copper in wiring, platinum in a car's exhaust system and the blue of copper sulphate crystals all come from the d-block. Partly filled d orbitals give these metals variable oxidation states, colour, magnetism and remarkable catalytic power.

This part covers the position and configurations of the d- and f-block elements, the physical properties of transition metals, their oxidation states, colour and magnetism, and their catalysis, alloys and complexes.

**The d-block spans groups $3$ to $12$, where $(n-1)d$ orbitals fill with general configuration $(n-1)d^{1-10}n{s}^{1-2}$; the f-block — lanthanoids and actinoids — sits below the main table filling $4f$ and $5f$ orbitals; and chromium and copper break the pattern to gain stable half-filled or full d subshells.

The series:

- First transition series** ($3d$) — scandium to zinc
- Second series ($4d$) — yttrium to cadmium
- Third series ($5d$) — lanthanum, then hafnium to mercury
- f-block — lanthanoids ($4f$) and actinoids ($5f$)

The exceptions:

$$\text{Cr: }[\text{Ar}]3d^{5}4s^1,\text{Cu: }[\text{Ar}]3d^{10}4s^1$$

The $3d$ and $4s$ energies are very close, and half-filled and completely filled subshells gain extra stability from exchange energy.

Worked example. Iron ($Z=26$) is $[\text{Ar}]3d^{6}4s^2$. Forming Fe${}^{3+}$ removes both $4s$ electrons and then one $3d$ electron, giving $[\text{Ar}]3d^5$ — a stable half-filled configuration.

An everyday example. Steel utensils in an Indian kitchen are mostly iron alloyed with chromium and nickel — three members of the first transition series.

The substance. Zinc, cadmium and mercury are not treated as transition elements, because their d orbitals are completely filled in both the atoms and their common ions.

Transition elements are hard, lustrous metals with high melting points and densities, because unpaired d electrons strengthen metallic bonding; across a series their radii shrink only slightly, ionisation enthalpies rise gradually and density increases.

Trends in the first series:

- Metallic character — all are metals and good conductors of heat and electricity
- Melting points — high, peaking near the middle of the series where unpaired d electrons are most numerous
- Radii — decrease slowly, because each added electron enters an inner d orbital that shields the growing nuclear charge poorly
- Ionisation enthalpy — rises gradually, less steeply than across the s and p blocks
- Density — increases from scandium to copper as mass rises and radius shrinks

Heavier series. The $4d$ and $5d$ elements are larger than the $3d$ ones, but the $4d$ and $5d$ members of a group, such as zirconium and hafnium, have almost the same radii because of the lanthanoid contraction.

Worked example. Copper's molar mass is $\frac{63.5}{47.9}\approx 1.33$ times titanium's, yet its density, about $8.9$ g cm${}^{-3}$ against $4.5$ g cm${}^{-3}$, is $\frac{8.9}{4.5}\approx 1.98$ times as great — the extra comes from copper's smaller atoms packing more mass into each cubic centimetre.

An everyday example. Tungsten filaments in old light bulbs survived glowing white-hot because tungsten has an exceptionally high melting point.

The substance. Manganese has a surprisingly low melting point for the middle of the series, because its stable $3d^5$ configuration holds its electrons tightly.

**Because $(n-1)d$ and $ns$ electrons have similar energies, transition metals can lose different numbers of them, giving variable oxidation states; electrons moving between split d orbitals absorb visible light and cause colour; and unpaired electrons make compounds paramagnetic, with spin-only moment $\mu =\sqrt{n(n+2)}$ BM.

Oxidation states:**

- Manganese shows every state from $+2$ to $+7$
- The highest states occur in oxides and fluorides, such as Mn${}_2$O${}_7$
- The number of states is greatest near the middle of the series

Colour. In a compound, the d orbitals split into sets of slightly different energy. An electron absorbing visible light to jump between them — a d–d transition — removes that colour, and we see the complementary colour. Ions with $d^0$ or $d^{10}$ configurations, such as Sc${}^{3+}$ and Zn${}^{2+}$, are colourless.

Magnetic moment:

$$\mu =\sqrt{n(n+2)}\text{ BM}$$

where $n$ is the number of unpaired electrons.

Worked example. Fe${}^{2+}$ is $3d^6$ with $4$ unpaired electrons; Cu${}^{2+}$ is $3d^9$ with $1$:

$$\mu ({\text{Fe}}^{2+})=\sqrt{4\times 6}\approx 4.90\text{ BM},\mu ({\text{Cu}}^{2+})=\sqrt{1\times 3}\approx 1.73\text{ BM}$$

An everyday example. Copper sulphate crystals are blue while zinc sulphate is white — Cu${}^{2+}$ has a d–d transition available, but Zn${}^{2+}$, with a full $3d^{10}$ subshell, does not.

The substance. Magnetic moment reveals the number of unpaired electrons, so a measured value near $5.92$ BM points to a $d^5$ ion such as Mn${}^{2+}$.

Transition metals catalyse reactions because they change oxidation state easily and adsorb reactants on their surfaces; small atoms such as hydrogen, carbon and nitrogen fit into gaps in their lattices to form interstitial compounds; similar atomic sizes let them form alloys; and their small, highly charged ions with vacant d orbitals readily form complexes.

Catalytic activity:

- Variable oxidation states let the metal accept and return electrons
- Surface adsorption holds reactants close together and weakens their bonds
- Examples: iron in making ammonia, vanadium(V) oxide in making sulphuric acid, nickel in hydrogenating oils

Interstitial compounds such as TiC, Mn${}_4$N and Fe${}_3$H are hard, have high melting points and keep metallic conductivity; their formulas are often non-stoichiometric.

Alloys form readily between metals whose atomic radii are within about $15$ percent of each other, as in brass, bronze and stainless steel.

Complexes include $[{\text{Cu(NH}}_3{)}_4{]}^{2+}$ and $[{\text{Fe(CN)}}_6{]}^{3-}$.

Worked example. Fe${}^{3+}$ catalyses $2{\text{I}}^{-}+{\text{S}}_2{\text{O}}_8^{2-}\to {\text{I}}_2+2{\text{SO}}_4^{2-}$ by cycling between two oxidation states:

$$2{\text{Fe}}^{3+}+2{\text{I}}^{-}\to 2{\text{Fe}}^{2+}+{\text{I}}_2$$

$$2{\text{Fe}}^{2+}+{\text{S}}_2{\text{O}}_8^{2-}\to 2{\text{Fe}}^{3+}+2{\text{SO}}_4^{2-}$$

An everyday example. Vanaspati is made by hydrogenating vegetable oils over a nickel catalyst.

The substance. The catalyst ends unchanged — iron(III) is regenerated in the second step, ready to react again.