Chapter 4: The d- and f-Block Elements

A transition element is defined as an element whose atom in the ground state, or one of whose simple ions, has a partially (incompletely) filled $$d$$ subshell.

Scandium ($$Z = 21$$): its ground-state electronic configuration is $$\mathrm{[Ar]\,3d^1\,4s^2}$$. The $$3d$$ subshell contains one electron, i.e. it is partially filled. Because the atom itself has an incompletely filled $$d$$ subshell, scandium qualifies as a transition element.

Zinc ($$Z = 30$$): its ground-state configuration is $$\mathrm{[Ar]\,3d^{10}\,4s^2}$$. Here the $$3d$$ subshell is completely filled. Moreover, in its only common oxidation state the ion formed is $$\mathrm{Zn^{2+}}$$, with configuration $$\mathrm{[Ar]\,3d^{10}}$$ — again a completely filled $$d$$ subshell.

Since neither the zinc atom nor its ion has a partially filled $$d$$ subshell, zinc is not regarded as a transition element (it is a $$d$$-block element only by position).

The classification of an element as a transition element does not require the $$d$$ subshell to be partially filled in the atom alone — it is sufficient that the atom or any one of its stable ions has an incompletely filled $$d$$ subshell.

Silver ($$Z = 47$$) has the ground-state configuration $$\mathrm{[Kr]\,4d^{10}\,5s^1}$$, in which the $$4d$$ subshell is indeed completely filled.

However, silver also exhibits the +2 oxidation state (for example in $$\mathrm{AgO}$$ and $$\mathrm{AgF_2}$$). The ion $$\mathrm{Ag^{2+}}$$ has the configuration $$\mathrm{[Kr]\,4d^9}$$, i.e. it has an incompletely filled $$4d$$ subshell.

Since one of its ions ($$\mathrm{Ag^{2+}}$$) possesses a partially filled $$d$$ subshell, silver is classified as a transition element.

The enthalpy of atomisation is the energy required to break the metallic bonds and convert one mole of the metal into free gaseous atoms. A high value therefore signals strong metallic bonding.

In transition elements the atoms have a large number of unpaired electrons in the $$(n-1)d$$ as well as the $$ns$$ orbitals. These unpaired electrons are available for the formation of metallic bonds.

The greater the number of unpaired electrons, the stronger the interatomic metallic bonding (which involves both the $$d$$ and the $$s$$ electrons), and hence the larger the force holding the atoms together in the metallic lattice.

Because such strong metallic bonds must be broken during atomisation, the transition elements exhibit high enthalpies of atomisation.

The enthalpy of atomisation depends on the strength of the metallic bonding, which in turn depends on the number of unpaired electrons available for bonding.

Zinc ($$Z = 30$$) has the ground-state configuration $$\mathrm{[Ar]\,3d^{10}\,4s^2}$$. The $$3d$$ subshell is completely filled, so there are no unpaired $$d$$ electrons; the $$4s$$ orbital is also fully paired.

Because the $$d$$ electrons (being fully paired) do not participate in metallic bonding, only weak metallic bonding results. Every other element in the series Sc to Zn has at least one unpaired $$d$$ electron contributing to bonding, but zinc has none.

Weak metallic bonds require little energy to break, so zinc has the lowest enthalpy of atomisation ($$126 \, \mathrm{kJ\,mol^{-1}}$$) in the series.

An element exhibits variable oxidation states when, in addition to the $$ns$$ electrons, a variable number of $$(n-1)d$$ electrons can also take part in bonding.

Consider scandium ($$Z = 21$$), configuration $$\mathrm{[Ar]\,3d^1\,4s^2}$$. After losing the two $$4s$$ electrons together with the single $$3d$$ electron, it attains the stable noble-gas configuration of argon ($$\mathrm{[Ar]}$$).

Since this is the only energetically favourable possibility, scandium shows only the $$+3$$ oxidation state and does not exhibit variable oxidation states.

Among the $$3d$$-series transition metals, manganese ($$Z = 25$$) exhibits the largest number of oxidation states.

Its ground-state configuration is $$\mathrm{[Ar]\,3d^5\,4s^2}$$. The number of oxidation states an element can show depends on the number of electrons it can use in bonding — i.e. the $$ns$$ electrons together with the $$(n-1)d$$ electrons.

Manganese has $$2$$ electrons in the $$4s$$ orbital and $$5$$ electrons in the $$3d$$ subshell, giving a total of $$7$$ electrons available for bonding. It can therefore lose them in steps and show every oxidation state from $$+2$$ up to $$+7$$, i.e. $$+2,\;+3,\;+4,\;+5,\;+6,\;+7$$.

No other element of the series has as many electrons available for bonding; hence manganese shows the maximum number of oxidation states.

Both $$\mathrm{Cr^{2+}}$$ and $$\mathrm{Mn^{3+}}$$ have the $$d^4$$ configuration, yet they behave oppositely. The behaviour is decided by the stability of the configuration produced when the ion is oxidised or reduced.

$$\mathrm{Cr^{2+}}$$ is reducing: a reducing agent is itself oxidised. When $$\mathrm{Cr^{2+}}$$ ($$d^4$$) loses one electron it becomes $$\mathrm{Cr^{3+}}$$ ($$d^3$$):

$$\mathrm{Cr^{2+} \longrightarrow Cr^{3+} + e^-}$$

The $$d^3$$ configuration corresponds to a half-filled $$t_{2g}$$ level, which is extra stable. Since oxidation leads to a more stable state, $$\mathrm{Cr^{2+}}$$ readily gives up an electron and therefore acts as a reducing agent.

$$\mathrm{Mn^{3+}}$$ is oxidising: an oxidising agent is itself reduced. When $$\mathrm{Mn^{3+}}$$ ($$d^4$$) gains one electron it becomes $$\mathrm{Mn^{2+}}$$ ($$d^5$$):

$$\mathrm{Mn^{3+} + e^- \longrightarrow Mn^{2+}}$$

The $$d^5$$ configuration is an exactly half-filled $$3d$$ subshell, which is exceptionally stable. Since reduction leads to a more stable state, $$\mathrm{Mn^{3+}}$$ readily accepts an electron and therefore acts as an oxidising agent.

The standard electrode potential $$E^{\circ}(\mathrm{M^{2+}/M})$$ for the process $$\mathrm{M(s) \longrightarrow M^{2+}(aq) + 2e^-}$$ is governed by the balance of three enthalpy terms:

• the enthalpy of atomisation, $$\Delta_a H^{\circ}$$ — energy required to convert $$\mathrm{M(s)}$$ into $$\mathrm{M(g)}$$;
• the ionisation enthalpy, $$\Delta_i H^{\circ}\,(=\mathrm{IE_1 + IE_2})$$ — energy required to convert $$\mathrm{M(g)}$$ into $$\mathrm{M^{2+}(g)}$$;
• the hydration enthalpy, $$\Delta_{\mathrm{hyd}} H^{\circ}$$ — energy released when $$\mathrm{M^{2+}(g)}$$ is hydrated to $$\mathrm{M^{2+}(aq)}$$.

For copper, the sum of the atomisation enthalpy and the ionisation enthalpy needed to reach $$\mathrm{Cu^{2+}(g)}$$ is high, whereas the hydration enthalpy of $$\mathrm{Cu^{2+}}$$ is comparatively low.

Hence the large energy required to take $$\mathrm{Cu(s)}$$ all the way to $$\mathrm{Cu^{2+}(g)}$$ is not sufficiently compensated by the energy released on hydration. The overall change $$\mathrm{Cu(s) \to Cu^{2+}(aq)}$$ is therefore energetically unfavourable, so copper is not readily oxidised by $$\mathrm{H^+}$$. This makes $$E^{\circ}(\mathrm{Cu^{2+}/Cu})$$ positive ($$+0.34\,\mathrm{V}$$).

In each of these oxoanions the central metal is in its highest oxidation state:

• $$\mathrm{VO_2^+}$$ — vanadium is in the $$+5$$ state;
• $$\mathrm{Cr_2O_7^{2-}}$$ — chromium is in the $$+6$$ state;
• $$\mathrm{MnO_4^-}$$ — manganese is in the $$+7$$ state.

An oxidising agent works by getting itself reduced, i.e. by allowing the central metal to drop from its highest oxidation state to a lower, more stable one.

On moving from V to Cr to Mn, the highest oxidation state ($$+5 \to +6 \to +7$$) becomes progressively less stable. The less stable the highest oxidation state, the greater the tendency of the species to accept electrons and be reduced.

Therefore the ease of reduction — and hence the oxidising power — increases in the order $$\mathrm{VO_2^+ < Cr_2O_7^{2-} < MnO_4^-}$$.

The ionisation enthalpies of the first transition series do not increase smoothly with atomic number. The irregularities arise from two opposing factors and from the extra stability of certain electronic configurations.

First ionisation enthalpy ($$\mathrm{IE_1}$$): as we move along the series the nuclear charge increases, which tends to raise $$\mathrm{IE_1}$$. At the same time, the electrons being added enter the inner $$3d$$ subshell and shield the $$4s$$ electrons from the nucleus, which tends to lower $$\mathrm{IE_1}$$. These two effects very nearly cancel, so $$\mathrm{IE_1}$$ rises only slowly and somewhat irregularly across the series instead of increasing steadily.

Second ionisation enthalpy ($$\mathrm{IE_2}$$): this shows sharp irregularities, especially at chromium and copper.

• For $$\mathrm{Cr}$$: the ion $$\mathrm{Cr^+}$$ has the configuration $$\mathrm{[Ar]\,3d^5}$$ — an exceptionally stable, exactly half-filled $$d$$ subshell. Removing a second electron from this stable arrangement requires unusually high energy, so $$\mathrm{IE_2}$$ of Cr is very high.
• For $$\mathrm{Cu}$$: the ion $$\mathrm{Cu^+}$$ has the configuration $$\mathrm{[Ar]\,3d^{10}}$$ — a stable, completely-filled $$d$$ subshell. Removing a second electron from this stable arrangement likewise requires unusually high energy, so $$\mathrm{IE_2}$$ of Cu is very high.

Thus the irregular variation reflects the shifting balance between increasing nuclear charge and increasing $$d$$-electron shielding, together with the special stability of half-filled and fully-filled $$d$$ configurations.

The standard electrode potential $$E^{\circ}(\mathrm{M^{2+}/M})$$ depends on the combined effect of three energy terms: the enthalpy of atomisation $$\Delta_a H^{\circ}$$, the ionisation enthalpy $$\Delta_i H^{\circ}\,(=\mathrm{IE_1+IE_2})$$, and the hydration enthalpy $$\Delta_{\mathrm{hyd}} H^{\circ}$$. Since these three quantities do not vary regularly across the series, the $$E^{\circ}$$ values are also irregular.

The general trend is that $$E^{\circ}$$ becomes less negative from left to right, mainly because the sum $$\mathrm{IE_1+IE_2}$$ increases. However, there are clear exceptions:

• Mn ($$E^{\circ}=-1.18\,\mathrm{V}$$) is much more negative than its neighbours Cr ($$-0.91$$) and Fe ($$-0.44$$). This is because the ion formed, $$\mathrm{Mn^{2+}}$$, has the extra-stable half-filled $$3d^5$$ configuration; the high stability of $$\mathrm{Mn^{2+}}$$ makes its formation easy, so $$E^{\circ}$$ is exceptionally negative.
• Ni ($$E^{\circ}=-0.25\,\mathrm{V}$$) is more negative than expected from the smooth trend; this is associated with the high (most negative) hydration enthalpy of $$\mathrm{Ni^{2+}}$$.
• Cu ($$E^{\circ}=+0.34\,\mathrm{V}$$) is the only positive value. For copper the high enthalpy of atomisation and high ionisation enthalpy are not compensated by its comparatively low hydration enthalpy, so forming $$\mathrm{Cu^{2+}(aq)}$$ from $$\mathrm{Cu(s)}$$ is unfavourable.

Hence the irregularity results from the irregular variation of $$\Delta_a H^{\circ}$$, $$\Delta_i H^{\circ}$$ and $$\Delta_{\mathrm{hyd}} H^{\circ}$$, modified by the special stability of the $$\mathrm{Mn^{2+}}$$ ($$d^5$$) configuration and the unusual enthalpy values of copper.

The electrode potential $$E^{\circ}(\mathrm{M^{3+}/M^{2+}})$$ measures the tendency of $$\mathrm{M^{3+}}$$ to be reduced to $$\mathrm{M^{2+}}$$. A large positive value means the change $$\mathrm{M^{3+} + e^- \to M^{2+}}$$ is highly favourable, i.e. $$\mathrm{M^{2+}}$$ is much more stable than $$\mathrm{M^{3+}}$$.

$$\mathrm{Mn^{3+}/Mn^{2+}}$$: reduction gives $$\mathrm{Mn^{2+}}$$, configuration $$\mathrm{[Ar]\,3d^5}$$ — an extra-stable, exactly half-filled $$d$$ subshell. Because the change $$\mathrm{Mn^{3+}\,(d^4) \to Mn^{2+}\,(d^5)}$$ produces this very stable configuration, it is strongly favoured, so $$E^{\circ}$$ is large and positive.

$$\mathrm{Cr^{3+}/Cr^{2+}}$$: here $$\mathrm{Cr^{3+}}$$ already has the stable configuration $$\mathrm{[Ar]\,3d^3}$$ (half-filled $$t_{2g}$$ level). Reduction to $$\mathrm{Cr^{2+}}$$ ($$d^4$$) destroys this stability, so the reduction is not favoured and $$E^{\circ}$$ is negative.

$$\mathrm{Fe^{3+}/Fe^{2+}}$$: here $$\mathrm{Fe^{3+}}$$ already has the stable configuration $$\mathrm{[Ar]\,3d^5}$$. Reduction to $$\mathrm{Fe^{2+}}$$ ($$d^6$$) destroys the half-filled stability, so the reduction is only weakly favoured and $$E^{\circ}$$ is only slightly positive.

Thus only for manganese does reduction create the stable $$d^5$$ configuration; for chromium and iron, reduction destroys a stable configuration. This is why $$E^{\circ}(\mathrm{Mn^{3+}/Mn^{2+}})$$ is much more positive than the other two.

A metal can be raised to its highest oxidation state only by an element that is itself a very strong oxidising agent and that can form a large number of stable bonds to the metal.

Oxygen and fluorine are the two elements best suited for this:

• Both are small in size and highly electronegative, hence strong oxidising agents capable of raising the metal to its highest oxidation state.
• Fluorine forms strong, stable single bonds, and its small size allows several fluorine atoms to pack around one metal atom.
• Oxygen has the additional ability to form multiple ($$p\pi$$-$$d\pi$$) bonds with the metal, which gives extra stability to very high oxidation states.

Examples: manganese shows its highest state $$+7$$ only in $$\mathrm{Mn_2O_7}$$ and in the oxoanion $$\mathrm{MnO_4^-}$$; chromium shows $$+6$$ in $$\mathrm{CrO_3}$$; vanadium shows $$+5$$ in $$\mathrm{V_2O_5}$$ and $$\mathrm{VF_5}$$.

Since no other element can both oxidise the metal so strongly and form so many stabilising bonds, the highest oxidation state of a metal is exhibited only in its oxide or fluoride.

A reducing agent is a species that is itself readily oxidised; the more readily it loses an electron, the stronger a reducing agent it is.

Consider the oxidation of each ion:

$$\mathrm{Cr^{2+} \longrightarrow Cr^{3+} + e^-}\qquad (d^4 \to d^3)$$

$$\mathrm{Fe^{2+} \longrightarrow Fe^{3+} + e^-}\qquad (d^6 \to d^5)$$

On oxidation $$\mathrm{Cr^{2+}}$$ gives $$\mathrm{Cr^{3+}}$$, which has the stable half-filled $$t_{2g}$$ ($$d^3$$) configuration. This strong driving force makes $$\mathrm{Cr^{2+}}$$ lose its electron very readily. Although $$\mathrm{Fe^{2+}}$$ on oxidation also reaches a stable configuration ($$d^5$$), $$\mathrm{Fe^{2+}}$$ ($$d^6$$) is itself reasonably stable, so it is oxidised much less readily.

The standard electrode potentials confirm this: $$E^{\circ}(\mathrm{Cr^{3+}/Cr^{2+}}) = -0.41\,\mathrm{V}$$, while $$E^{\circ}(\mathrm{Fe^{3+}/Fe^{2+}}) = +0.77\,\mathrm{V}$$. The more negative $$E^{\circ}$$ of the chromium couple shows that $$\mathrm{Cr^{2+}}$$ is much more easily oxidised.

Hence $$\mathrm{Cr^{2+}}$$ is the stronger reducing agent.

The element with atomic number $$25$$ is manganese, $$\mathrm{Mn}$$, whose ground-state configuration is $$\mathrm{[Ar]\,3d^5\,4s^2}$$.

A divalent ion is formed by removing the two $$4s$$ electrons:

$$\mathrm{Mn^{2+}: [Ar]\,3d^5}$$

In the $$3d^5$$ configuration the five electrons occupy the five $$3d$$ orbitals singly (Hund's rule of maximum multiplicity), so the number of unpaired electrons is $$n = 5$$.

The spin-only magnetic moment is given by

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

Substituting $$n = 5$$:

$$\mu = \sqrt{5(5+2)} = \sqrt{5 \times 7} = \sqrt{35}$$

$$\mu \approx 5.92\;\mathrm{BM}$$

The element with atomic number $$27$$ is cobalt, $$\mathrm{Co}$$, with ground-state configuration $$\mathrm{[Ar]\,3d^7\,4s^2}$$.

The divalent ion $$\mathrm{M^{2+}}$$ is formed by removing the two $$4s$$ electrons:

$$\mathrm{Co^{2+}: [Ar]\,3d^7}$$

Distributing $$7$$ electrons over the five $$3d$$ orbitals by Hund's rule: each orbital first receives one electron (using $$5$$ electrons), and the remaining $$2$$ electrons pair up. This leaves $$2$$ doubly-occupied orbitals and $$3$$ singly-occupied orbitals, so the number of unpaired electrons is $$n = 3$$.

The spin-only magnetic moment is

$$\mu = \sqrt{n(n+2)} = \sqrt{3(3+2)} = \sqrt{3 \times 5} = \sqrt{15}$$

$$\mu \approx 3.87\;\mathrm{BM}$$

Disproportionation is a special type of redox reaction in which a single species in an intermediate oxidation state is simultaneously oxidised and reduced — i.e. the same element goes partly to a higher oxidation state and partly to a lower one.

It occurs when a particular oxidation state becomes less stable relative to the oxidation states lying on either side of it.

Example: in manganese chemistry, the $$+6$$ state in manganate ion disproportionates in acidic solution into the $$+7$$ state (permanganate) and the $$+4$$ state ($$\mathrm{MnO_2}$$):

$$\mathrm{3MnO_4^{2-} + 4H^+ \longrightarrow 2MnO_4^- + MnO_2 + 2H_2O}$$

Here manganese in the $$+6$$ state is oxidised to $$+7$$ (in $$\mathrm{MnO_4^-}$$) and reduced to $$+4$$ (in $$\mathrm{MnO_2}$$).

In aqueous solution the $$\mathrm{Cu^+}$$ ion undergoes disproportionation:

$$\mathrm{2Cu^+(aq) \longrightarrow Cu^{2+}(aq) + Cu(s)}$$

That is, $$\mathrm{Cu^+}$$ is simultaneously oxidised to $$\mathrm{Cu^{2+}}$$ and reduced to metallic $$\mathrm{Cu}$$.

The reason lies in the hydration enthalpies. To form $$\mathrm{Cu^{2+}}$$ a second electron must be removed from $$\mathrm{Cu^+}$$, which costs the (large) second ionisation energy. However, $$\mathrm{Cu^{2+}}$$ is smaller in size and carries a higher charge than $$\mathrm{Cu^+}$$, so its hydration enthalpy is much greater (more negative).

The large hydration enthalpy released when $$\mathrm{Cu^{2+}}$$ is hydrated more than compensates for the energy needed for the second ionisation. As a result, in water $$\mathrm{Cu^{2+}}$$ is thermodynamically more stable than $$\mathrm{Cu^+}$$, and $$\mathrm{Cu^+}$$ spontaneously disproportionates.

Hence $$\mathrm{Cu^+}$$ ion is not stable in aqueous solution.

The most common and stable oxidation state of the lanthanoids is $$+3$$. A few members also show $$+4$$, and this is usually associated with the tendency to attain an empty ($$4f^0$$) or half-filled ($$4f^7$$) subshell.

Cerium ($$Z = 58$$), configuration $$\mathrm{[Xe]\,4f^1\,5d^1\,6s^2}$$, is the lanthanoid best known for the $$+4$$ state. On forming $$\mathrm{Ce^{4+}}$$ it attains the stable empty-subshell configuration $$\mathrm{[Xe]\,4f^0}$$ (the configuration of the noble gas xenon).

This is why $$\mathrm{Ce^{4+}}$$ (in ceric salts) is well known and is even used as a strong oxidising agent in volumetric analysis.

In the actinoids the differentiating electron enters the $$5f$$ subshell, whereas in the lanthanoids it enters the $$4f$$ subshell. Both series therefore show a steady decrease in atomic and ionic size with increasing atomic number — the lanthanoid contraction and the actinoid contraction.

The magnitude of the contraction depends on how effectively the $$f$$ electrons shield the outer electrons from the steadily increasing nuclear charge.

The $$5f$$ orbitals are more diffuse — they extend farther from the nucleus — than the more compact $$4f$$ orbitals. Because of this, the $$5f$$ electrons are poorer at shielding the nuclear charge than the $$4f$$ electrons.

Due to this poorer shielding by the $$5f$$ electrons, each additional proton produces a larger increase in the effective nuclear charge felt by the outer electrons in the actinoids. Consequently the decrease in size from one element to the next is greater in the actinoids than in the lanthanoids.

The stability of a $$+2$$ ion towards oxidation to the $$+3$$ state depends on the electronic configuration produced when the electron is lost.

Manganese: $$\mathrm{Mn^{2+}}$$ has the configuration $$\mathrm{[Ar]\,3d^5}$$ — an exactly half-filled $$3d$$ subshell, which is exceptionally stable. Oxidising it to $$\mathrm{Mn^{3+}}$$ ($$d^4$$) would destroy this stable half-filled arrangement, so $$\mathrm{Mn^{2+}}$$ strongly resists oxidation:

$$\mathrm{Mn^{2+}\;(d^5,\;stable) \longrightarrow Mn^{3+}\;(d^4) + e^-}$$  — unfavourable.

Iron: $$\mathrm{Fe^{2+}}$$ has the configuration $$\mathrm{[Ar]\,3d^6}$$. Oxidising it to $$\mathrm{Fe^{3+}}$$ ($$d^5$$) actually produces the stable half-filled $$3d^5$$ configuration, so $$\mathrm{Fe^{2+}}$$ is readily oxidised:

$$\mathrm{Fe^{2+}\;(d^6) \longrightarrow Fe^{3+}\;(d^5,\;stable) + e^-}$$  — favourable.

Hence $$\mathrm{Mn^{2+}}$$ compounds are more stable towards oxidation to the $$+3$$ state than $$\mathrm{Fe^{2+}}$$ compounds.

In the first half of the first transition series (Sc, Ti, V, Cr, Mn) the $$+2$$ oxidation state is reached by the loss of the two $$4s$$ electrons. Its stability, relative to higher oxidation states, increases steadily with atomic number.

As we move from Sc to Mn the nuclear charge increases. The electrons being added enter the inner $$3d$$ subshell, where they shield one another rather poorly, so the effective nuclear charge experienced by the remaining electrons rises.

Consequently the $$3d$$ electrons become progressively more tightly bound to the nucleus. After the two $$4s$$ electrons have been removed to give the $$+2$$ state, the removal of a further electron from the $$3d$$ subshell — needed to reach the $$+3$$ or higher states — requires more and more energy (i.e. the third ionisation enthalpy increases steadily).

Because higher oxidation states thus become increasingly difficult to attain, the $$+2$$ state becomes more and more stable with increasing atomic number across the first half of the series.

To a large extent, the stability of an oxidation state of a transition element is decided by its electronic configuration. Configurations corresponding to empty ($$d^0$$), exactly half-filled ($$d^5$$) and completely-filled ($$d^{10}$$) $$d$$ subshells are especially stable.

Examples where the configuration favours stability:

• $$\mathrm{Sc^{3+}}$$ ($$d^0$$) and $$\mathrm{Ti^{4+}}$$ ($$d^0$$) — stable because of the empty $$d$$ subshell.
• $$\mathrm{Mn^{2+}}$$ ($$d^5$$) and $$\mathrm{Fe^{3+}}$$ ($$d^5$$) — stable because of the exactly half-filled $$d$$ subshell.
• $$\mathrm{Zn^{2+}}$$ ($$d^{10}$$) and $$\mathrm{Cu^+}$$ ($$d^{10}$$) — stable because of the completely-filled $$d$$ subshell.

However, electronic configuration is a major but not the sole deciding factor. Other factors — ionisation enthalpy, enthalpy of atomisation, hydration/lattice enthalpy and the nature of the element bonded to the metal — also influence which oxidation state is actually stable.

For instance, although $$\mathrm{Cu^+}$$ has the filled $$d^{10}$$ configuration, it is unstable in water and disproportionates ($$\mathrm{2Cu^+ \to Cu^{2+} + Cu}$$) because the high hydration enthalpy of $$\mathrm{Cu^{2+}}$$ outweighs the configurational stability of $$\mathrm{Cu^+}$$. Similarly, although $$\mathrm{Mn^{2+}}$$ ($$d^5$$) is configurationally very stable, manganese still shows the $$+7$$ state in $$\mathrm{MnO_4^-}$$ because oxygen stabilises that high state.

For a first-series transition metal atom the ground-state configuration is generally $$\mathrm{3d^x\,4s^2}$$. The stable oxidation states arise from loss of the $$4s^2$$ electrons together with a variable number of $$3d$$ electrons.

• $$3d^3$$: the atom is vanadium, $$\mathrm{V}$$ ($$\mathrm{[Ar]\,3d^3\,4s^2}$$). By losing $$4s^2$$ and the $$3d$$ electrons in steps it shows the oxidation states $$+2,\;+3,\;+4$$ and $$+5$$; the $$+5$$ state (use of all five valence electrons) is the characteristic stable one.
• $$3d^5$$: the atom is manganese, $$\mathrm{Mn}$$ ($$\mathrm{[Ar]\,3d^5\,4s^2}$$). It shows a wide range, $$+2$$ to $$+7$$; the most stable are $$+2$$ (half-filled $$d^5$$ ion) and $$+7$$ (maximum oxidation state).
• $$3d^8$$: the atom is nickel, $$\mathrm{Ni}$$ ($$\mathrm{[Ar]\,3d^8\,4s^2}$$). Its characteristic stable oxidation state is $$+2$$ (loss of the $$4s^2$$ electrons).
• $$3d^4$$: no atom of the first transition series actually has the ground-state configuration $$\mathrm{3d^4\,4s^2}$$. The element expected to be $$\mathrm{3d^4\,4s^2}$$ (chromium) is in fact $$\mathrm{3d^5\,4s^1}$$, because the half-filled $$3d^5$$ arrangement is more stable. Hence $$3d^4$$ does not occur as a ground-state atomic configuration — it is found only in ions such as $$\mathrm{Cr^{2+}}$$ and $$\mathrm{Mn^{3+}}$$.

An oxometal anion (oxoanion) is an anion in which the transition metal is surrounded by oxygen atoms. In some of these anions the metal is in its highest oxidation state, which equals its group number.

In each of these anions the oxidation state of the central metal is equal to its group number.

Lanthanoid contraction is the steady decrease in the atomic and ionic radii of the lanthanoid elements with increasing atomic number, i.e. as one moves from cerium ($$\mathrm{Ce}$$, $$Z=58$$) to lutetium ($$\mathrm{Lu}$$, $$Z=71$$).

Cause: as the atomic number increases, the additional electrons enter the inner $$4f$$ subshell. Because of their diffuse, multi-lobed shape, the $$4f$$ orbitals shield one another from the nucleus very imperfectly. Consequently the effective nuclear charge experienced by the outer electrons increases steadily, pulling them inward, so the size decreases regularly along the series.

Consequences of lanthanoid contraction:

• Similarity of the second and third transition series: the contraction almost exactly cancels the size increase expected on going from the second to the third transition series, so pairs such as $$\mathrm{Zr}$$ & $$\mathrm{Hf}$$ and $$\mathrm{Nb}$$ & $$\mathrm{Ta}$$ have nearly identical atomic radii and very similar chemical properties.
• Difficulty of separation of the lanthanoids: because their ionic sizes — and hence chemical properties — are so similar, the lanthanoids are very difficult to separate from one another.
• Variation in basic strength of hydroxides: since the size of $$\mathrm{Ln^{3+}}$$ decreases from $$\mathrm{La^{3+}}$$ to $$\mathrm{Lu^{3+}}$$, the covalent character of the hydroxides increases and their basic strength decreases — basicity falls steadily from $$\mathrm{La(OH)_3}$$ to $$\mathrm{Lu(OH)_3}$$.

Characteristics of the transition elements:

• They are all metals — hard, with high melting and boiling points and high densities.
• They have high enthalpies of atomisation (strong metallic bonding).
• They exhibit variable oxidation states.
• Most of their ions and compounds are coloured.
• They and many of their compounds are paramagnetic, owing to unpaired $$d$$ electrons.
• They and their compounds show good catalytic activity.
• They form complex (coordination) compounds, alloys and interstitial compounds.

Why they are called ‘transition’ elements: they are placed in the periodic table between the $$s$$-block (highly electropositive, reactive metals) and the $$p$$-block (largely non-metals). Their properties are intermediate — a ‘transition’ — between those of the two blocks. Their position also marks the transition from the filling of the $$ns$$ orbitals to the filling of the $$np$$ orbitals, the $$(n-1)d$$ orbitals being progressively filled in between.

$$d$$-block elements not regarded as transition elements: zinc, cadmium and mercury (group 12). Their atoms have the configuration $$(n-1)d^{10}\,ns^2$$, and their common ions $$\mathrm{M^{2+}}$$ have $$(n-1)d^{10}$$ — i.e. a completely filled $$d$$ subshell in both the atom and the ion. Having no partially filled $$d$$ subshell, $$\mathrm{Zn}$$, $$\mathrm{Cd}$$ and $$\mathrm{Hg}$$ are not true transition elements.

Transition elements: their general valence-shell configuration is $$(n-1)d^{1-10}\,ns^{1-2}$$. They have incompletely filled $$d$$ orbitals in the atomic state or in one of their ions. The differentiating (last-entering) electron goes into the $$(n-1)d$$ subshell — i.e. the penultimate shell.

Non-transition elements:

• $$s$$-block: general configuration $$ns^{1-2}$$ — the last electron enters the outermost $$ns$$ orbital.
• $$p$$-block: general configuration $$ns^2\,np^{1-6}$$ — the last electron enters the outermost $$np$$ orbital.

In the non-transition elements the $$d$$ orbitals are either absent, completely empty, or completely filled — never partially filled — and the differentiating electron enters the outermost ($$ns$$ or $$np$$) shell.

Thus the key difference is: in transition elements the last electron enters the inner $$(n-1)d$$ subshell, which remains incompletely filled, whereas in non-transition elements it enters the outermost $$ns$$ or $$np$$ subshell.

The lanthanoids exhibit the following oxidation states:

• $$+3$$ — the most common and characteristic oxidation state, shown by all the lanthanoids. It is the most stable state both in aqueous solution and in solid compounds.
• $$+2$$ — shown by a few elements, e.g. $$\mathrm{Sm}$$, $$\mathrm{Eu}$$ and $$\mathrm{Yb}$$ (for example, $$\mathrm{Eu^{2+}}$$ has the stable half-filled $$4f^7$$ configuration).
• $$+4$$ — shown by a few elements, e.g. $$\mathrm{Ce}$$, $$\mathrm{Pr}$$ and $$\mathrm{Tb}$$ (for example, $$\mathrm{Ce^{4+}}$$ has the stable empty $$4f^0$$ configuration).

The $$+2$$ and $$+4$$ states appear mainly when they lead towards a stable empty ($$4f^0$$), half-filled ($$4f^7$$) or completely-filled ($$4f^{14}$$) $$f$$ subshell; otherwise the $$+3$$ state dominates.

Interstitial compounds are compounds formed when small atoms of non-metals — such as hydrogen ($$\mathrm{H}$$), carbon ($$\mathrm{C}$$), nitrogen ($$\mathrm{N}$$) or boron ($$\mathrm{B}$$) — are trapped in the interstitial sites (the empty spaces or voids) of the crystal lattice of a metal.

Their chief characteristics:

• They are usually non-stoichiometric, i.e. their composition is not fixed — e.g. $$\mathrm{TiC}$$, $$\mathrm{Fe_3H}$$, $$\mathrm{Mn_4N}$$, $$\mathrm{VH_{0.56}}$$.
• They are very hard and have high melting points, often higher than those of the pure metal.
• They retain metallic conductivity.
• They are chemically inert.

Why they are well known for transition metals: transition metals crystallise in lattices that contain interstitial voids (holes) of a suitable size into which the small non-metal atoms can fit. The transition metal atoms are large enough to leave such voids, and the small atoms can enter them without much distortion of the lattice. The strong metal-metal and metal-non-metal bonding that results accounts for the hardness and high melting points. Because suitably-sized voids are a characteristic feature of transition-metal lattices, interstitial compounds are especially common among the transition metals.

Transition metals: the various oxidation states of a transition metal usually differ from one another by 1 unit. For example, iron shows $$+2$$ and $$+3$$ ($$\mathrm{Fe^{2+}, Fe^{3+}}$$), and vanadium shows $$+2,\;+3,\;+4$$ and $$+5$$. This is because, after losing the $$ns$$ electrons, a transition metal can lose a variable number of $$(n-1)d$$ electrons one at a time — the $$ns$$ and $$(n-1)d$$ orbitals are very close in energy, so successive $$d$$ electrons are removed in single steps. The lower oxidation states are generally ionic and the higher ones more covalent.

Non-transition metals (p-block): their oxidation states usually differ by 2 units. For example, tin shows $$+2$$ and $$+4$$ ($$\mathrm{Sn^{2+}, Sn^{4+}}$$), lead shows $$+2$$ and $$+4$$ ($$\mathrm{Pb^{2+}, Pb^{4+}}$$), and thallium shows $$+1$$ and $$+3$$. This is a consequence of the inert pair effect — the two $$ns$$ electrons tend to remain paired and are lost together (or retained together), so the oxidation states differ by 2.

In short, the variability in transition metals arises from the loss of $$(n-1)d$$ and $$ns$$ electrons (states differ by 1), whereas in non-transition metals it arises from the $$ns$$ and $$np$$ electrons (states differ by 2).

Preparation of potassium dichromate ($$\mathrm{K_2Cr_2O_7}$$) from iron chromite ore ($$\mathrm{FeCr_2O_4}$$):

Step 1 — Conversion of the ore to sodium chromate. The finely powdered chromite ore is fused with sodium carbonate (or sodium hydroxide) in the presence of air. Chromium is oxidised from the $$+3$$ to the $$+6$$ state, giving yellow sodium chromate:

$$\mathrm{4FeCr_2O_4 + 8Na_2CO_3 + 7O_2 \longrightarrow 8Na_2CrO_4 + 2Fe_2O_3 + 8CO_2}$$

Step 2 — Conversion of sodium chromate to sodium dichromate. The leached yellow solution of sodium chromate is acidified with sulphuric acid, giving orange sodium dichromate:

$$\mathrm{2Na_2CrO_4 + 2H^+ \longrightarrow Na_2Cr_2O_7 + 2Na^+ + H_2O}$$

Step 3 — Conversion of sodium dichromate to potassium dichromate. The sodium dichromate solution is treated with potassium chloride. Being much less soluble, potassium dichromate crystallises out on cooling as orange crystals:

$$\mathrm{Na_2Cr_2O_7 + 2KCl \longrightarrow K_2Cr_2O_7 + 2NaCl}$$

Effect of increasing pH: in solution the yellow chromate ion ($$\mathrm{CrO_4^{2-}}$$) and the orange dichromate ion ($$\mathrm{Cr_2O_7^{2-}}$$) exist in equilibrium, whose position depends on the pH:

$$\mathrm{2CrO_4^{2-} + 2H^+ \rightleftharpoons Cr_2O_7^{2-} + H_2O}$$

Increasing the pH makes the solution more alkaline (adding $$\mathrm{OH^-}$$, removing $$\mathrm{H^+}$$). By Le Chatelier's principle the equilibrium shifts to the left, so the orange dichromate is converted into the yellow chromate:

$$\mathrm{Cr_2O_7^{2-} + 2OH^- \longrightarrow 2CrO_4^{2-} + H_2O}$$

Thus on increasing the pH the solution changes colour from orange ($$\mathrm{Cr_2O_7^{2-}}$$) to yellow ($$\mathrm{CrO_4^{2-}}$$).

Preparation of potassium permanganate ($$\mathrm{KMnO_4}$$): it is prepared from the mineral pyrolusite, $$\mathrm{MnO_2}$$.

Step 1 — Conversion of $$\mathrm{MnO_2}$$ to potassium manganate. Pyrolusite is fused with potassium hydroxide in the presence of air (or an oxidising agent such as $$\mathrm{KNO_3}$$). Manganese is oxidised from the $$+4$$ to the $$+6$$ state, giving the dark green potassium manganate:

$$\mathrm{2MnO_2 + 4KOH + O_2 \longrightarrow 2K_2MnO_4 + 2H_2O}$$

Step 2 — Oxidation of manganate to permanganate. The green manganate (Mn in $$+6$$) is oxidised to the purple permanganate (Mn in $$+7$$). In industry this is done by electrolytic oxidation of the alkaline manganate solution at the anode:

$$\mathrm{MnO_4^{2-} \longrightarrow MnO_4^- + e^-}$$

(It may also be done chemically, e.g. by $$\mathrm{Cl_2}$$ or $$\mathrm{O_3}$$, or by acidification, in which case the manganate disproportionates: $$\mathrm{3MnO_4^{2-} + 4H^+ \to 2MnO_4^- + MnO_2 + 2H_2O}$$.)

Reactions of acidified $$\mathrm{KMnO_4}$$: in acidic medium permanganate is a powerful oxidising agent and is reduced to $$\mathrm{Mn^{2+}}$$:

$$\mathrm{MnO_4^- + 8H^+ + 5e^- \longrightarrow Mn^{2+} + 4H_2O}\qquad (E^{\circ} = +1.51\,\mathrm{V})$$

(i) With iron(II) ions: $$\mathrm{Fe^{2+}}$$ is oxidised to $$\mathrm{Fe^{3+}}$$  ($$\mathrm{5Fe^{2+} \to 5Fe^{3+} + 5e^-}$$):

$$\mathrm{MnO_4^- + 8H^+ + 5Fe^{2+} \longrightarrow Mn^{2+} + 4H_2O + 5Fe^{3+}}$$

(ii) With sulphur dioxide: $$\mathrm{SO_2}$$ is oxidised to sulphate  ($$\mathrm{SO_2 + 2H_2O \to SO_4^{2-} + 4H^+ + 2e^-}$$):

$$\mathrm{2MnO_4^- + 5SO_2 + 2H_2O \longrightarrow 2Mn^{2+} + 5SO_4^{2-} + 4H^+}$$

(iii) With oxalic acid: the oxalate ion is oxidised to carbon dioxide  ($$\mathrm{C_2O_4^{2-} \to 2CO_2 + 2e^-}$$); the reaction is carried out warm ($$\approx 333\,\mathrm{K}$$):

$$\mathrm{2MnO_4^- + 5C_2O_4^{2-} + 16H^+ \longrightarrow 2Mn^{2+} + 10CO_2 + 8H_2O}$$

A species is coloured in aqueous solution if it has partially filled (incomplete) $$d$$ orbitals — i.e. one or more unpaired $$d$$ electrons — which permit $$d$$-$$d$$ electronic transitions by absorption of visible light. A species with an empty ($$d^0$$) or completely-filled ($$d^{10}$$) $$d$$ subshell has no possible $$d$$-$$d$$ transition and is colourless.

Hence $$\mathrm{Ti^{3+}}$$, $$\mathrm{V^{3+}}$$, $$\mathrm{Mn^{2+}}$$, $$\mathrm{Fe^{3+}}$$ and $$\mathrm{Co^{2+}}$$ are coloured (they have unpaired $$d$$ electrons and undergo $$d$$-$$d$$ transitions), whereas $$\mathrm{Cu^{+}}$$ ($$d^{10}$$) and $$\mathrm{Sc^{3+}}$$ ($$d^0$$) are colourless.

The $$+2$$ oxidation state of a first-series transition metal is formed by losing the two $$4s$$ electrons. Its stability shows a definite trend across the series.

• At the beginning of the series (Sc, Ti) the $$+2$$ state is uncommon and unstable. Scandium does not form a stable $$+2$$ ion at all (it shows only $$+3$$); $$\mathrm{Ti^{2+}}$$ and $$\mathrm{V^{2+}}$$ are strongly reducing, since these elements prefer their higher oxidation states.
• Moving across the series the $$+2$$ state becomes more and more stable. As the nuclear charge increases, the $$3d$$ electrons are held more tightly, so removing a third electron to reach a higher state needs progressively more energy. This is reflected in the $$E^{\circ}(\mathrm{M^{2+}/M})$$ values becoming generally less negative across the series.
• $$\mathrm{Mn^{2+}}$$ ($$3d^5$$) is especially stable because of its exactly half-filled $$d$$ subshell, and $$\mathrm{Zn^{2+}}$$ ($$3d^{10}$$) is very stable because of its completely-filled $$d$$ subshell — indeed, for zinc the $$+2$$ state is the only state.

Thus, overall, the $$+2$$ oxidation state becomes progressively more stable from left to right across the first transition series, with extra stability shown by $$\mathrm{Mn^{2+}}$$ ($$d^5$$) and $$\mathrm{Zn^{2+}}$$ ($$d^{10}$$).

Disproportionation is a redox reaction in which a single species in an intermediate oxidation state is simultaneously oxidised (to a higher state) and reduced (to a lower state). It occurs when the intermediate oxidation state is less stable than the oxidation states lying on either side of it.

Example 1 — copper(I):

$$\mathrm{2Cu^+(aq) \longrightarrow Cu^{2+}(aq) + Cu(s)}$$

Here copper in the $$+1$$ state is oxidised to $$+2$$ (in $$\mathrm{Cu^{2+}}$$) and reduced to $$0$$ (in metallic $$\mathrm{Cu}$$).

Example 2 — manganate(VI):

$$\mathrm{3MnO_4^{2-} + 4H^+ \longrightarrow 2MnO_4^- + MnO_2 + 2H_2O}$$

Here manganese in the $$+6$$ state is oxidised to $$+7$$ (in $$\mathrm{MnO_4^-}$$) and reduced to $$+4$$ (in $$\mathrm{MnO_2}$$).

In the first transition series, copper exhibits the $$+1$$ oxidation state most frequently.

The ground-state configuration of copper ($$Z = 29$$) is $$\mathrm{[Ar]\,3d^{10}\,4s^1}$$. When copper loses its single $$4s$$ electron it forms $$\mathrm{Cu^+}$$:

$$\mathrm{Cu^+: [Ar]\,3d^{10}}$$

This ion has the stable, completely-filled $$3d^{10}$$ configuration. Because the loss of just one ($$4s$$) electron gives this especially stable arrangement, copper readily shows the $$+1$$ state — far more frequently than any other first-series transition metal.

To find the number of unpaired electrons, first write the $$d$$-electron configuration of each gaseous ion (the three electrons removed are the two $$4s$$ electrons plus the appropriate number of $$3d$$ electrons), then distribute the electrons over the five $$d$$ orbitals by Hund's rule.

Most stable in aqueous solution: $$\mathrm{Cr^{3+}}$$. It has the $$3d^3$$ configuration, in which all three electrons singly occupy the lower $$t_{2g}$$ set — a stable half-filled $$t_{2g}$$ arrangement. In the octahedral field of water molecules this gives the maximum crystal field stabilisation energy, making $$\mathrm{[Cr(H_2O)_6]^{3+}}$$ the most stable of these ions in aqueous solution.

An alloy is a homogeneous solid solution (or an intimate mixture) of two or more metals, or of a metal with one or more non-metals. Alloys are usually prepared by melting the components together and allowing the mixture to solidify. Because the metals involved often have similar atomic sizes, the atoms of one can readily replace those of another in the crystal lattice.

An important alloy containing lanthanoid metals — Misch metal. Misch metal consists of about 95% lanthanoid metals (mainly cerium), together with about 5% iron and traces of sulphur, carbon, calcium and aluminium.

Uses of Misch metal:

• It is used in cigarette and gas-lighter flints (it is pyrophoric, i.e. gives sparks on being struck).
• It is added to magnesium-based alloys used to produce bullets, shells and tracer ammunition.
• It is used to improve the quality of steel (as a deoxidiser and to remove sulphur).

Inner transition elements are the elements in which the last (differentiating) electron enters the $$(n-2)f$$ orbitals — i.e. the $$f$$-block elements. They comprise the lanthanoids ($$4f$$ series, $$Z = 58$$ to $$71$$) and the actinoids ($$5f$$ series, $$Z = 90$$ to $$103$$).

Examining the given atomic numbers:

Hence the atomic numbers 59 (praseodymium), 95 (americium) and 102 (nobelium) are those of inner transition elements.

The lanthanoids have a very uniform chemistry: nearly all of them show the single, dominant $$+3$$ oxidation state, with only a few showing $$+2$$ or $$+4$$ in addition. This regularity arises because the $$4f$$ orbitals lie deep inside the atom and take little part in bonding.

The actinoids, in contrast, show a much less regular (less smooth) chemistry, mainly because they exhibit a wide and variable range of oxidation states. In the actinoids the $$5f$$, $$6d$$ and $$7s$$ orbitals have comparable energies, so a variable number of electrons can take part in bonding.

Examples (oxidation states):

• Lanthanoids — almost all show $$+3$$ only (e.g. La, Pr, Nd, Gd are all essentially $$+3$$).
• Actinoids — they show $$+3, +4, +5, +6$$ and even $$+7$$. For example, uranium shows $$+3, +4, +5, +6$$; neptunium shows states from $$+3$$ up to $$+7$$; plutonium shows $$+3$$ to $$+7$$; americium shows $$+2$$ to $$+6$$.

Because of this irregular, multi-valent behaviour, the chemistry of the actinoids is not as smooth and uniform as that of the lanthanoids.

The actinoid series runs from thorium ($$Z = 90$$) to lawrencium ($$Z = 103$$). Hence the last element of the actinoid series is lawrencium, $$\mathrm{Lr}$$ ($$Z = 103$$).

Electronic configuration of lawrencium ($$Z = 103$$):

$$\mathrm{Lr: [Rn]\,5f^{14}\,6d^1\,7s^2}$$

Possible oxidation state: in $$\mathrm{Lr}$$ the $$5f$$ subshell is completely filled ($$5f^{14}$$) and very stable, so the $$5f$$ electrons are not available for bonding. Only the outer $$6d^1\,7s^2$$ electrons can be lost; their removal gives the stable $$\mathrm{[Rn]\,5f^{14}}$$ core. Therefore lawrencium exhibits only the $$+3$$ oxidation state.

Cerium has atomic number $$Z = 58$$ and the ground-state configuration

$$\mathrm{Ce: [Xe]\,4f^1\,5d^1\,6s^2}$$

To form the $$\mathrm{Ce^{3+}}$$ ion, three electrons are removed — the two $$6s$$ electrons and the single $$5d$$ electron — leaving only the one $$4f$$ electron:

$$\mathrm{Ce^{3+}: [Xe]\,4f^1}$$

By Hund's rule of maximum multiplicity, this single electron occupies one of the seven $$4f$$ orbitals on its own; it is therefore an unpaired electron. Hence the number of unpaired electrons is $$n = 1$$.

Spin-only magnetic moment:

$$\mu = \sqrt{n(n+2)} = \sqrt{1(1+2)} = \sqrt{3}$$

$$\mu \approx 1.73\;\mathrm{BM}$$

Lanthanoids that show the $$+4$$ oxidation state: cerium ($$\mathrm{Ce}$$), praseodymium ($$\mathrm{Pr}$$) and terbium ($$\mathrm{Tb}$$).

Lanthanoids that show the $$+2$$ oxidation state: samarium ($$\mathrm{Sm}$$), europium ($$\mathrm{Eu}$$), thulium ($$\mathrm{Tm}$$) and ytterbium ($$\mathrm{Yb}$$).

Correlation with electronic configuration: the normal oxidation state of every lanthanoid is $$+3$$. A $$+2$$ or $$+4$$ state appears only for those few elements whose ion thereby reaches a configuration that is exactly — or close to — one of the extra-stable arrangements: empty ($$4f^0$$), exactly half-filled ($$4f^7$$) or completely filled ($$4f^{14}$$).

A lanthanoid therefore forms a $$+4$$ ion by losing electrons until it reaches (or nearly reaches) $$4f^0$$ or $$4f^7$$, and a $$+2$$ ion by losing only its two $$6s$$ electrons when this leaves it at (or near) $$4f^7$$ or $$4f^{14}$$. For instance, $$\mathrm{Ce}$$ loses four electrons to give $$\mathrm{Ce^{4+}}$$ ($$4f^0$$), while $$\mathrm{Eu}$$ loses only the two $$6s$$ electrons to give $$\mathrm{Eu^{2+}}$$ ($$4f^7$$). Of these ions, $$\mathrm{Ce^{4+}}$$, $$\mathrm{Tb^{4+}}$$, $$\mathrm{Eu^{2+}}$$ and $$\mathrm{Yb^{2+}}$$ have exactly stable configurations and are the best characterised, whereas $$\mathrm{Pr^{4+}}$$, $$\mathrm{Sm^{2+}}$$ and $$\mathrm{Tm^{2+}}$$ are only near such configurations and are correspondingly less stable.

For each element, identify it from its atomic number and then write the configuration by filling the orbitals in order of increasing energy.

• $$Z = 61$$ — Promethium (Pm), a lanthanoid:  $$\mathrm{[Xe]\,4f^5\,6s^2}$$
• $$Z = 91$$ — Protactinium (Pa), an actinoid:  $$\mathrm{[Rn]\,5f^2\,6d^1\,7s^2}$$
• $$Z = 101$$ — Mendelevium (Md), an actinoid:  $$\mathrm{[Rn]\,5f^{13}\,7s^2}$$
• $$Z = 109$$ — Meitnerium (Mt), a $$d$$-block (transition) element:  $$\mathrm{[Rn]\,5f^{14}\,6d^7\,7s^2}$$

The number of $$3d$$ electrons in an ion is found by removing the appropriate electrons from the neutral atom (the $$4s$$ electrons are always removed before the $$3d$$ electrons).

In a hydrated (aqueous) ion the six water molecules form an octahedral field around the metal ion. This splits the five $$3d$$ orbitals into a lower triply-degenerate set, $$t_{2g}$$ (three orbitals), and an upper doubly-degenerate set, $$e_g$$ (two orbitals). Water is a weak-field ligand, so the splitting is small and the electrons fill the orbitals by Hund's rule (high-spin): the $$t_{2g}$$ and $$e_g$$ orbitals are first singly occupied as far as possible before any pairing takes place.

The statement is correct: the elements of the first transition series ($$3d$$) do differ in many properties from the heavier transition elements ($$4d$$ and $$5d$$). The chief differences are:

• Oxidation states: the first-series elements have their lower oxidation states as the more stable ones, whereas the heavier elements form much more stable higher oxidation states (e.g. $$\mathrm{Cr}$$ is stable as $$+3$$, but $$\mathrm{Mo}$$ and $$\mathrm{W}$$ are stable as $$+6$$).
• Atomic size and density: the heavier elements are larger; moreover the second and third series are almost equal in size because of the lanthanoid contraction, and the heavier elements have much higher densities and melting points.
• Enthalpy of atomisation: the heavier transition elements have higher enthalpies of atomisation and a much greater tendency to form metal-metal bonds and metal-atom clusters.
• Magnetic behaviour: first-series compounds are usually paramagnetic with moments close to the spin-only value, whereas low-spin compounds are far more common among the heavier elements (their ligand fields are stronger).
• Complex formation: the heavier elements form more stable complexes and use their $$d$$ orbitals more readily in covalent bonding.

Hence the first transition series shows several properties distinctly different from those of the heavier ($$4d$$ and $$5d$$) transition elements.

The spin-only magnetic moment is $$\mu = \sqrt{n(n+2)}\;\mathrm{BM}$$, where $$n$$ is the number of unpaired electrons. From the measured $$\mu$$ we can find $$n$$ and hence deduce the electronic arrangement and the nature of the ligand field.

$$\mathrm{K_4[Mn(CN)_6]}$$: here manganese is $$\mathrm{Mn^{2+}}$$ ($$3d^5$$). Solving $$\sqrt{n(n+2)} = 2.2$$ gives $$n(n+2) = 4.84$$, so $$n \approx 1$$. A $$d^5$$ ion with only one unpaired electron must be low-spin, configuration $$t_{2g}^5\,e_g^0$$. This indicates that $$\mathrm{CN^-}$$ is a strong-field ligand, forcing the electrons to pair up.

$$\mathrm{[Fe(H_2O)_6]^{2+}}$$: here iron is $$\mathrm{Fe^{2+}}$$ ($$3d^6$$). Solving $$\sqrt{n(n+2)} = 5.3$$ gives $$n(n+2) \approx 28$$, so $$n \approx 4$$. A $$d^6$$ ion with four unpaired electrons is high-spin, configuration $$t_{2g}^4\,e_g^2$$. This indicates that $$\mathrm{H_2O}$$ is a weak-field ligand (it causes no pairing beyond the minimum).

$$\mathrm{K_2[MnCl_4]}$$: here manganese is again $$\mathrm{Mn^{2+}}$$ ($$3d^5$$). Solving $$\sqrt{n(n+2)} = 5.9$$ gives $$n(n+2) \approx 35$$, so $$n = 5$$. All five $$d$$ electrons are unpaired, so the ion is high-spin $$d^5$$. This indicates that $$\mathrm{Cl^-}$$ is a weak-field ligand; the complex ion $$\mathrm{[MnCl_4]^{2-}}$$ is tetrahedral.

Conclusion: the magnetic moment values reveal the number of unpaired electrons, and hence whether a complex is low-spin or high-spin — which in turn tells us whether the ligand is strong-field ($$\mathrm{CN^-}$$) or weak-field ($$\mathrm{H_2O}$$, $$\mathrm{Cl^-}$$).