The d-and f-Block Elements

The electronic configurations are:

- **Cr$^{3+}$**: Cr is $[Ar]3d^54s^1$, so losing 3 electrons gives **$[Ar]3d^3$.
- Pm$^{3+}$: Pm is a lanthanoid; the stable trivalent ion is $[Xe]4f^4$.
- Cu$^{+}$**: Cu is $[Ar]3d^{10}4s^1$, so losing one electron gives **$[Ar]3d^{10}$.
- Ce$^{4+}$**: Ce is $[Xe]4f^15d^16s^2$; losing 4 electrons gives **$[Xe]4f^0$.
- Co$^{2+}$**: Co is $[Ar]3d^74s^2$, so losing 2 electrons gives **$[Ar]3d^7$.
- Lu$^{2+}$**: Lu is $[Xe]4f^{14}5d^16s^2$; losing 2 electrons gives **$[Xe]4f^{14}5d^1$.
- Mn$^{2+}$**: Mn is $[Ar]3d^54s^2$, so losing 2 electrons gives **$[Ar]3d^5$.
- Th$^{4+}$**: Th is $[Rn]6d^27s^2$; losing 4 electrons gives **$[Rn]$**.

The stability is explained by electronic configuration.

- **Mn$^{2+}$ has the configuration $3d^5$, which is a half-filled subshell** and therefore especially stable.
- If Mn$^{2+}$ is oxidised to Mn$^{3+}$, it becomes **$3d^4$, which is less stable.
- Fe$^{2+}$ has $3d^6$** configuration.
- Oxidation of Fe$^{2+}$ to Fe$^{3+}$ gives **$3d^5$, the stable half-filled configuration.

So, Mn$^{2+}$ resists oxidation, while Fe$^{2+}$ is more readily oxidised** because the product Fe$^{3+}$ is more stable.

In the first half of the first-row transition series, the **$+2$ state becomes more stable with increasing atomic number because:

- the increasing nuclear charge makes the ions more stable,
- electrons are added to the inner $3d$ subshell, so the $4s$ electrons** are removed first and the formation of $M^{2+}$ ions is easy,
- the stability of the resulting ions depends on electronic arrangements such as **$d^5$ or other relatively stable configurations.

As a result, in this part of the series the $+2$ oxidation state is commonly observed and becomes progressively more stable**.

The stability of oxidation states in the first transition series is decided only partly by electronic configuration.

- Certain configurations are especially stable, such as **$d^0$, $d^5$, and $d^{10}$.
- Therefore, oxidation states leading to these configurations are favoured. For example:
- Mn$^{2+}$ is stable because it is $3d^5$.
- Fe$^{3+}$ is relatively stable because it is also $3d^5$.
- Zn$^{2+}$ is stable because it is $3d^{10}$.
- Cr$^{2+}$ is reducing because it changes to Cr$^{3+}$, $d^3$, which is more stable.

But electronic configuration is not the only factor. Stability also depends on ionisation enthalpy, enthalpy of atomisation, hydration enthalpy, lattice energy, and ligand effects. For example:
- Cu$^{2+}$** is more stable in aqueous solution than Cu$^{+}$ because Cu$^{2+}$ has much greater hydration enthalpy.
- **Mn$^{3+}$** is much less stable than Fe$^{3+}$ even though both are transition ions, because the energy required to form Mn$^{3+}$ is high.

So, electronic configuration strongly influences oxidation-state stability, but the final stability is determined by several factors together.

The stable oxidation states are:

- **$3d^3$ → $+3$
- $3d^5$ → $+2$
- $3d^7$ → $+2$
- $3d^9$ → $+2$**

These correspond to the more common stable oxidation states discussed for the first-row transition elements.

The oxometal anions are:

- Chromate: $\mathrm{CrO_4^{2-}}$
- Dichromate: $\mathrm{Cr_2O_7^{2-}}$
- Vanadate: $\mathrm{VO_4^{3-}}$
- Permanganate: $\mathrm{MnO_4^-}$

In these ions, the metal shows an oxidation state equal to its group number.

The lanthanoid contraction is the regular decrease in atomic and ionic radii from lanthanum to lutetium as atomic number increases.

It happens because the added 4f electrons shield the nuclear charge very poorly. As a result, the effective nuclear attraction on the outer electrons increases, and the size decreases.

Consequences of lanthanoid contraction:
- The 4d and 5d transition elements become nearly the same in size, for example Zr (160 pm) and Hf (159 pm).
- Hence the second and third transition series show very similar properties.
- It explains the difficulty in separating some chemically similar elements.
- It influences the chemistry of elements that follow the lanthanoids.

The elements are called transition elements because their chemical properties are **transitional between those of $s$- and $p$-block elements.

Their main characteristics are:
- typical metallic properties such as lustre, malleability and conductivity,
- high melting and boiling points,
- variable oxidation states,
- formation of coloured ions,
- tendency to form complex compounds,
- catalytic activity,
- paramagnetism, and
- formation of interstitial compounds and alloys**.

Among the $d$-block elements, Zn, Cd and Hg are not regarded as transition elements because their ground-state and common oxidation-state configurations are **$d^{10}$**, i.e. the $d$ subshell is complete.

In transition elements, the outer electronic configuration is generally **$(n-1)d^{1-10}ns^{1-2}$. Thus, they have a partly filled $d$ subshell either in the atom or in one of their ions.

In non-transition elements**, the $d$ subshell is either absent or completely filled and not involved in the usual chemistry. Their valence electrons are mainly in the **$s$ and $p$ orbitals**.

Because of the close energy of $(n-1)d$ and $ns$ orbitals, transition elements can lose or use both types of electrons, which is why they show variable oxidation states and other special properties.

The lanthanoids mainly exhibit **$+3$ oxidation state. In addition, some of them also show $+2$ and $+4$ oxidation states in special cases.

So the oxidation states are $+2$, $+3$, and $+4$, with $+3$** being the most common.

If it refers to the book’s “Explain giving reasons” item on transition metals showing paramagnetism, high atomisation enthalpy, colour, and catalysis, then the reasons are:

- Paramagnetism: due to unpaired electrons. - High enthalpy of atomisation: due to strong metallic bonding from many unpaired electrons. - Coloured compounds: due to **$d$–$d$ electronic transitions. - Catalytic activity: due to variable oxidation states and complex formation**.

Interstitial compounds are formed when small atoms like H, C or N get trapped in the spaces between the metal atoms in a crystal lattice.

They are common with transition metals because these metals have metallic lattices with interstitial spaces that can accommodate such small atoms.

Their main properties are:
- high melting points,
- great hardness,
- metallic conductivity,
- chemical inertness, and
- often non-stoichiometric composition.

Examples include **TiC, Mn$_4$N, Fe$_3$H, VH$_{0.56}$ and TiH$_{1.7}$**.

The variability of oxidation states in transition metals is different from that in non-transition metals because:

- in transition metals, oxidation states often differ by one unit;
- in non-transition elements, oxidation states normally differ by two units.

This happens because transition metals have **incompletely filled $d$ orbitals**, so both $ns$ and $(n-1)d$ electrons can take part in bonding.

Examples:
- Transition metals: **V$^{2+}$, V$^{3+}$, V$^{4+}$, V$^{5+}$
- Non-transition elements: Sn$^{2+}$ and Sn$^{4+}$, Pb$^{2+}$ and Pb$^{4+}$

So transition metals show greater variability** in oxidation states.

Chromite ore, **$\mathrm{FeCr_2O_4}$, is first fused with sodium carbonate in the presence of air to form sodium chromate**:

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

The yellow solution of sodium chromate is filtered and acidified with sulphuric acid to give sodium dichromate. Then sodium dichromate is treated with potassium chloride to obtain potassium dichromate:

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

Effect of increasing pH: chromate and dichromate are interconvertible. On increasing pH, the equilibrium shifts from dichromate to chromate:

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

So a more alkaline solution favours chromate.

Potassium dichromate is a strong oxidising agent in acidic solution. It oxidises substances such as iodides, iron(II) salts and sulphides.

Ionic equations:

1. With iodide:
$$\mathrm{Cr_2O_7^{2-}}+14\mathrm{H^+}+6\mathrm{I^-}\rightarrow 2\mathrm{Cr^{3+}}+7\mathrm{H_2O}+3\mathrm{I_2}$$

2. With iron(II) solution:
$$\mathrm{Cr_2O_7^{2-}}+14\mathrm{H^+}+6\mathrm{Fe^{2+}}\rightarrow 2\mathrm{Cr^{3+}}+7\mathrm{H_2O}+6\mathrm{Fe^{3+}}$$

3. With hydrogen sulphide:
$$\mathrm{Cr_2O_7^{2-}}+14\mathrm{H^+}+3\mathrm{H_2S}\rightarrow 2\mathrm{Cr^{3+}}+7\mathrm{H_2O}+3\mathrm{S}$$

So acidified dichromate is reduced to **$\mathrm{Cr^{3+}}$** while the other species are oxidised.

Preparation of potassium permanganate:

1. **$\mathrm{MnO_2}$ is fused with KOH and an oxidising agent such as $\mathrm{O_2}$ / $\mathrm{KNO_3}$** to form potassium manganate:
$$2\mathrm{MnO_2}+4\mathrm{KOH}+\mathrm{O_2}\rightarrow 2\mathrm{K_2MnO_4}+2\mathrm{H_2O}$$
2. The manganate ion is then converted to permanganate by disproportionation in acidic/neutral medium or by electrolytic oxidation:
$$3\mathrm{MnO_4^{2-}}+4\mathrm{H^+}\rightarrow 2\mathrm{MnO_4^-}+\mathrm{MnO_2}+2\mathrm{H_2O}$$

Reactions of acidified permanganate:

(i) With iron(II):
$$5\mathrm{Fe^{2+}}+\mathrm{MnO_4^-}+8\mathrm{H^+}\rightarrow \mathrm{Mn^{2+}}+4\mathrm{H_2O}+5\mathrm{Fe^{3+}}$$

(ii) With sulphur dioxide / sulphite in acid medium:
$$5\mathrm{SO_3^{2-}}+2\mathrm{MnO_4^-}+6\mathrm{H^+}\rightarrow 2\mathrm{Mn^{2+}}+3\mathrm{H_2O}+5\mathrm{SO_4^{2-}}$$

(iii) With oxalic acid / oxalate:
$$5\mathrm{C_2O_4^{2-}}+2\mathrm{MnO_4^-}+16\mathrm{H^+}\rightarrow 2\mathrm{Mn^{2+}}+8\mathrm{H_2O}+10\mathrm{CO_2}$$

So acidified permanganate acts as a powerful oxidising agent.

The given values show that the standard electrode potentials are not regular.

This irregularity is explained by the irregular variation of ionisation enthalpies and the stability of certain electronic configurations:
- **$d^0$, $d^5$, and $d^{10}$ are especially stable.
- The first and second ionisation enthalpies and even sublimation enthalpies** vary irregularly.

So the observed $E^\circ$ values depend on a balance of these factors, not just on atomic number.

A solution is coloured when the ion has **partly filled $d$ orbitals and can undergo $d$–$d$ transitions.

- Ti$^{3+}$**: coloured — $3d^1$
- **V$^{3+}$**: coloured — $3d^2$
- **Cu$^{+}$**: colourless — $3d^{10}$
- **Sc$^{3+}$**: colourless — $3d^0$
- **Mn$^{2+}$**: coloured — $3d^5$
- **Fe$^{3+}$**: coloured — $3d^5$
- **Co$^{2+}$**: coloured — $3d^7$

So the coloured ions are **Ti$^{3+}$, V$^{3+}$, Mn$^{2+}$, Fe$^{3+}$ and Co$^{2+}$**.

The **$+2$ oxidation state is common throughout the first transition series, but its stability changes.

- In the first half** of the series, the $+2$ state becomes progressively more stable as atomic number increases.
- This is because the resulting **$M^{2+}$ ions are increasingly favoured electronically.
- Mn$^{2+}$ is particularly stable due to the half-filled $3d^5$ configuration.
- At the end of the series, Cu$^{2+}$ and Zn$^{2+}$ are also important; Zn$^{2+}$ is especially stable because it has the filled $3d^{10}$** configuration.

Thus, the $+2$ state is most stable when it leads to especially stable $d$-subshell arrangements.