Chapter 4: The d- and f-Block Elements (Inorganic Chemistry)

A transition element is one that has an incompletely filled d subshell in its atom or in any of its common oxidation states. Silver shows the +2 oxidation state (as in $\mathrm{AgF}{}_2$, AgO), where it has the configuration $4d^9$: an incomplete d subshell. So silver is a transition element.

In its +2 state silver is $4d^9$, an incompletely filled d subshell.

The enthalpy of atomisation depends on the strength of metallic bonding, which is greater when more unpaired electrons are available. Zinc ($3d^{10}4s^2$) has no unpaired electrons; its d electrons do not take part in metallic bonding. So the metallic bonds are weak and the enthalpy of atomisation is the lowest.

Zinc ($3d^{10}4s^2$) has no unpaired electrons, so its metallic bonding is weakest.

Manganese ($3d^{5}4s^2$). It has the largest number of electrons (seven: five 3d and two 4s) of similar energy available for bonding, so it shows every state from +2 to +7.

Manganese, because all seven of its 3d and 4s electrons can be used, giving +2 to +7.

$E^{\circ }$ depends on the energy needed to turn the solid metal into hydrated ions: atomisation + ionisation − hydration. Copper has a high enthalpy of atomisation and high ionisation enthalpies, and the hydration enthalpy of $\mathrm{Cu}{}^{2+}$ is not large enough to make up for them. The overall change is unfavourable, so $E^{\circ }$ is positive and copper does not liberate hydrogen from acids.

The high energy to atomise and ionise copper is not balanced by the hydration enthalpy of Cu²⁺.

Ionisation enthalpies rise only slowly and irregularly because added electrons go into the inner 3d subshell, which shields the 4s electrons, and because the 3d and 4s energies change relative to each other. The irregularities come from the extra stability of $d^0$, $d^5$ and $d^{10}$:

• Cr and Cu have unusually high second ionisation enthalpies: removing the second electron breaks the stable $d^5$ ($\mathrm{Cr}{}^{+}$) or $d^{10}$ ($\mathrm{Cu}{}^{+}$) arrangement.
• Mn has a relatively low second ionisation enthalpy, because $\mathrm{Mn}{}^{+}$ ($3d^{5}4s^1$) loses its 4s electron to become stable $3d^5$.
• Zn has a high first ionisation enthalpy because of its fully filled $3d^{10}4s^2$ configuration.

The extra stability of d⁰, d⁵ and d¹⁰ arrangements (e.g. high IE₂ for Cr and Cu, low IE₂ for Mn) makes the trend irregular.

Oxygen and fluorine are small and the most electronegative elements, so they can oxidise a metal to its highest state. Fluorine forms very strong ionic or covalent bonds (high lattice or bond energies); oxygen can also form multiple bonds (M=O), so fewer oxygen atoms are needed to stabilise a high state. Examples: $\mathrm{VF}{}_5$, $\mathrm{CrF}{}_6$, $\mathrm{Mn}{}_2\mathrm{O}{}_7$, $\mathrm{CrO}{}_3$.

O and F are small and highly electronegative; F forms strong bonds and O forms multiple bonds, which stabilise high oxidation states.

$\mathrm{Cr}{}^{2+}$. $E^{\circ }(\mathrm{Cr}{}^{3+}/\mathrm{Cr}{}^{2+})=-0.41$ V while $E^{\circ }(\mathrm{Fe}{}^{3+}/\mathrm{Fe}{}^{2+})=+0.77$ V, so $\mathrm{Cr}{}^{2+}$ is far more easily oxidised. On oxidation, $\mathrm{Cr}{}^{2+}$ ($d^4$) becomes $\mathrm{Cr}{}^{3+}$ ($d^3$, a stable half-filled $t_{2g}$ set), whereas $\mathrm{Fe}{}^{2+}$ ($d^6$) becomes $\mathrm{Fe}{}^{3+}$ ($d^5$), which is less favourable in water.

Cr²⁺: its E°(Cr³⁺/Cr²⁺) is negative (−0.41 V) and Cr³⁺ (t₂g³) is very stable.

Z = 27 is cobalt: Co is $[\mathrm{Ar}]3d^{7}4s^2$, so $\mathrm{Co}{}^{2+}$ is $3d^7$ with 3 unpaired electrons.

$$\mu =\sqrt{n(n+2)}=\sqrt{3\times 5}=\sqrt{15}\approx 3.87\text{BM}$$

About 3.87 BM.

In water, $\mathrm{Cu}{}^{+}$ disproportionates: $2\mathrm{Cu}{}^{+}(\mathrm{aq})\to \mathrm{Cu}{}^{2+}(\mathrm{aq})+\mathrm{Cu}(\mathrm{s})$. Although $\mathrm{Cu}{}^{+}$ has a $3d^{10}$ configuration, $\mathrm{Cu}{}^{2+}$ has a much larger (more negative) hydration enthalpy because of its smaller size and higher charge, and this more than makes up for the second ionisation enthalpy of copper.

Cu⁺ disproportionates to Cu²⁺ and Cu, because the high hydration enthalpy of Cu²⁺ outweighs copper's second ionisation enthalpy.

In actinoids the electrons are added to 5f orbitals, which are more diffuse and shield even more poorly than 4f orbitals. So the nuclear charge felt by the outer electrons increases more sharply from one element to the next, and the size decreases more.

5f electrons shield the outer electrons less effectively than 4f electrons.

• (i) $\mathrm{Cr}{}^{3+}$: $[\mathrm{Ar}]3d^3$
• (ii) $\mathrm{Pm}{}^{3+}$: $[\mathrm{Xe}]4f^4$
• (iii) $\mathrm{Cu}{}^{+}$: $[\mathrm{Ar}]3d^{10}$
• (iv) $\mathrm{Ce}{}^{4+}$: $[\mathrm{Xe}]$
• (v) $\mathrm{Co}{}^{2+}$: $[\mathrm{Ar}]3d^7$
• (vi) $\mathrm{Lu}{}^{2+}$: $[\mathrm{Xe}]4f^{14}5d^1$
• (vii) $\mathrm{Mn}{}^{2+}$: $[\mathrm{Ar}]3d^5$
• (viii) $\mathrm{Th}{}^{4+}$: $[\mathrm{Rn}]$

As listed above.

$\mathrm{Mn}{}^{2+}$ is $3d^5$, a stable half-filled subshell; turning it into $\mathrm{Mn}{}^{3+}$ ($3d^4$) means breaking this stable arrangement, so its third ionisation enthalpy is very high. $\mathrm{Fe}{}^{2+}$ is $3d^6$; losing one electron gives $\mathrm{Fe}{}^{3+}$, the stable $3d^5$ configuration, so it is oxidised easily.

Mn²⁺ is already the stable d⁵; Fe²⁺ (d⁶) becomes the stable d⁵ when oxidised.

In the first half (Sc to Mn), the +2 ions are $3d^1$ to $3d^5$. As the atomic number rises, the nuclear charge increases and the third ionisation enthalpy needed to go from $\mathrm{M}{}^{2+}$ to $\mathrm{M}{}^{3+}$ increases, reaching a maximum at Mn, where it would destroy the stable $3d^5$ arrangement. So the +2 state is increasingly preferred: $\mathrm{Ti}{}^{2+}$, $\mathrm{V}{}^{2+}$ and $\mathrm{Cr}{}^{2+}$ are strong reducing agents, while $\mathrm{Mn}{}^{2+}$ is very stable.

The third ionisation enthalpy (M²⁺ → M³⁺) rises with atomic number, peaking at Mn (d⁵), so +2 becomes more stable.

Electronic configuration is an important factor:

• Ions with $d^0$, $d^5$ or $d^{10}$ are especially stable: $\mathrm{Sc}{}^{3+}$, $\mathrm{Ti}{}^{4+}$, $\mathrm{V}{}^{5+}$ ($d^0$); $\mathrm{Mn}{}^{2+}$, $\mathrm{Fe}{}^{3+}$ ($d^5$); $\mathrm{Zn}{}^{2+}$ ($d^{10}$).
• Hence $\mathrm{Mn}{}^{2+}$ resists oxidation, $\mathrm{Fe}{}^{2+}$ is easily oxidised to $\mathrm{Fe}{}^{3+}$, and Sc shows only +3.
• A half-filled $t_{2g}$ set also stabilises $\mathrm{Cr}{}^{3+}$ ($d^3$) in water.

But configuration alone does not decide everything. Hydration enthalpy and lattice energy also matter: $\mathrm{Cu}{}^{2+}$ ($d^9$) is more stable in water than $\mathrm{Cu}{}^{+}$ ($d^{10}$) because of its much higher hydration enthalpy.

Configurations with d⁰, d⁵, d¹⁰ (and t₂g³) favour certain states (Sc³⁺, Mn²⁺, Fe³⁺, Zn²⁺, Cr³⁺), but hydration and lattice energies also matter (e.g. Cu²⁺ over Cu⁺).

• $3d^{3}4s^2$ (vanadium): +5 (also +2, +3, +4)
• $3d^{5}4s^2$ (manganese): +2 and +7; or $3d^{5}4s^1$ (chromium): +3 and +6
• $3d^{8}4s^2$ (nickel): +2
• $3d^4$: no element has this ground-state configuration ($3d^{4}4s^2$ does not occur; chromium is $3d^{5}4s^1$)

3d³: +5 (V); 3d⁵: +2, +7 (Mn) or +3, +6 (Cr); 3d⁸: +2 (Ni); 3d⁴: no such ground state.

• Vanadate, $\mathrm{VO}{}_3{}^{-}$ or $\mathrm{VO}{}_4{}^{3-}$: V +5 (group 5)
• Chromate $\mathrm{CrO}{}_4{}^{2-}$ and dichromate $\mathrm{Cr}{}_2\mathrm{O}{}_7{}^{2-}$: Cr +6 (group 6)
• Permanganate, $\mathrm{MnO}{}_4{}^{-}$: Mn +7 (group 7)

Vanadate (V +5), chromate and dichromate (Cr +6), permanganate (Mn +7).

Lanthanoid contraction is the steady decrease in the atomic and ionic radii of the lanthanoids from lanthanum to lutetium. It happens because each added 4f electron shields the others poorly from the increasing nuclear charge.

Consequences:

• The second (4d) and third (5d) transition series have almost the same atomic radii: Zr (160 pm) and Hf (159 pm). So pairs such as Zr/Hf, Nb/Ta and Mo/W have very similar properties and are hard to separate.
• The basic character of the hydroxides $\mathrm{Ln}(\mathrm{OH}){}_3$ decreases from La to Lu.
• The lanthanoids are very similar to each other, which makes their separation difficult.

The gradual decrease in size from La to Lu due to poor 4f shielding; it makes 4d and 5d elements nearly equal in size (Zr ≈ Hf), lowers basicity across the series, and makes lanthanoids hard to separate.

Characteristics: they are all metals with high melting and boiling points, high density and good conductivity; they show variable oxidation states; most form coloured ions and paramagnetic compounds; they form many complexes; they and their compounds are good catalysts; they form interstitial compounds and alloys.

Name: they lie between the s-block and p-block elements, and their properties show a gradual change (transition) from the highly reactive s-block metals to the p-block elements.

Not regarded as transition elements: Zn, Cd and Hg, which have a completely filled $d^{10}$ subshell both in the atom and in their common ions ($\mathrm{M}{}^{2+}$).

Metallic, variable oxidation states, coloured, paramagnetic, complexes, catalysts; named for their position between s- and p-blocks. Zn, Cd and Hg are not true transition elements (d¹⁰ in atom and ions).

Transition elements fill the penultimate (inner) d subshell: general configuration $(n-1)d^{1-10}n{s}^{1-2}$. In non-transition (s- and p-block) elements, the electrons are added to the outermost shell ($n{s}^{1-2}$ or $n{s}^{2}n{p}^{1-6}$), and any inner d subshell is either absent or completely filled.

Transition elements: $(n-1)d^{1-10}n{s}^{1-2}$ (inner d being filled); non-transition: only the outermost s or p subshell is being filled.

The common and most stable state is +3. Some also show +2 (Eu, Yb, Sm, Tm) and +4 (Ce, Pr, Tb, Dy), especially when that gives an empty, half-filled or full 4f subshell ($\mathrm{Ce}{}^{4+}$ $4f^0$, $\mathrm{Eu}{}^{2+}$ $4f^7$, $\mathrm{Tb}{}^{4+}$ $4f^7$, $\mathrm{Yb}{}^{2+}$ $4f^{14}$).

Mainly +3; some show +2 (Eu, Yb, Sm) and +4 (Ce, Pr, Tb).

(i) They have unpaired d electrons; each unpaired electron has a magnetic moment, so they are attracted by a magnetic field.

(ii) They have many unpaired electrons, so there is strong metallic bonding between atoms (involving both ns and (n−1)d electrons).

(iii) In their ions the d orbitals split in energy (in the presence of ligands). An electron absorbs visible light and jumps from a lower to a higher d orbital (d–d transition); the complementary colour is seen.

(iv) They show variable oxidation states and can form unstable intermediates, and they provide a large surface for reactants to be adsorbed, lowering the activation energy. Examples: Fe in the Haber process, $\mathrm{V}{}_2\mathrm{O}{}_5$ in the contact process, Ni in hydrogenation.

(i) Unpaired d electrons. (ii) Strong metallic bonding by many unpaired electrons. (iii) d–d transitions absorb visible light. (iv) Variable oxidation states and a large surface for adsorption.

Interstitial compounds form when small atoms such as H, C, N or B are trapped in the empty spaces (interstices) of a metal's crystal lattice, e.g. TiC, $\mathrm{Mn}{}_4\mathrm{N}$, $\mathrm{Fe}{}_3\mathrm{H}$, $\mathrm{VH}{}_{0.56}$. They are usually non-stoichiometric, very hard, have high melting points, keep metallic conductivity and are chemically inert.

They are common for transition metals because these metals have close-packed lattices with voids of the right size to hold small non-metal atoms.

Compounds with small atoms (H, C, N, B) held in the voids of a metal lattice; transition metal lattices have suitable voids.

In transition metals, the oxidation states usually differ by one unit, because the (n−1)d and ns electrons have similar energies and can be lost one by one: $\mathrm{Fe}{}^{2+}$/$\mathrm{Fe}{}^{3+}$, $\mathrm{Cu}{}^{+}$/$\mathrm{Cu}{}^{2+}$, Mn +2 to +7.

In non-transition (p-block) elements, the states usually differ by two units, mainly because of the inert pair effect: Sn +2/+4, Pb +2/+4, Tl +1/+3.

Transition metals: states differ by 1 (Fe²⁺/Fe³⁺, Mn +2 to +7). Non-transition metals: by 2 (Sn²⁺/Sn⁴⁺, Pb²⁺/Pb⁴⁺).

Step 1: fusion with sodium carbonate in air

$$4\mathrm{FeCr}{}_2\mathrm{O}{}_4+8\mathrm{Na}{}_2\mathrm{CO}{}_3+7\mathrm{O}{}_2\to 8\mathrm{Na}{}_2\mathrm{CrO}{}_4+2\mathrm{Fe}{}_2\mathrm{O}{}_3+8\mathrm{CO}{}_2$$

Step 2: acidification of the filtered sodium chromate solution with sulphuric acid:

$$2\mathrm{Na}{}_2\mathrm{CrO}{}_4+2\mathrm{H}{}^{+}\to \mathrm{Na}{}_2\mathrm{Cr}{}_2\mathrm{O}{}_7+2\mathrm{Na}{}^{+}+\mathrm{H}{}_2\mathrm{O}$$

Step 3: treatment with potassium chloride (potassium dichromate is less soluble and crystallises):

$$\mathrm{Na}{}_2\mathrm{Cr}{}_2\mathrm{O}{}_7+2\mathrm{KCl}\to \mathrm{K}{}_2\mathrm{Cr}{}_2\mathrm{O}{}_7+2\mathrm{NaCl}$$

Effect of increasing pH: in alkaline solution, orange dichromate changes to yellow chromate:

$$\mathrm{Cr}{}_2\mathrm{O}{}_7{}^{2-}+2\mathrm{OH}{}^{-}\to 2\mathrm{CrO}{}_4{}^{2-}+\mathrm{H}{}_2\mathrm{O}$$

Chromite is fused with Na₂CO₃ in air (Na₂CrO₄), acidified (Na₂Cr₂O₇), then treated with KCl (K₂Cr₂O₇). Raising the pH turns orange dichromate into yellow chromate.

In acid solution, dichromate is a strong oxidising agent ($E^{\circ }=1.33$ V), and Cr goes from +6 to +3 (orange to green):

$$\mathrm{Cr}{}_2\mathrm{O}{}_7{}^{2-}+14\mathrm{H}{}^{+}+6\mathrm{e}{}^{-}\to 2\mathrm{Cr}{}^{3+}+7\mathrm{H}{}_2\mathrm{O}$$

• (i) $\mathrm{Cr}{}_2\mathrm{O}{}_7{}^{2-}+14\mathrm{H}{}^{+}+6\mathrm{I}{}^{-}\to 2\mathrm{Cr}{}^{3+}+3\mathrm{I}{}_2+7\mathrm{H}{}_2\mathrm{O}$
• (ii) $\mathrm{Cr}{}_2\mathrm{O}{}_7{}^{2-}+14\mathrm{H}{}^{+}+6\mathrm{Fe}{}^{2+}\to 2\mathrm{Cr}{}^{3+}+6\mathrm{Fe}{}^{3+}+7\mathrm{H}{}_2\mathrm{O}$
• (iii) $\mathrm{Cr}{}_2\mathrm{O}{}_7{}^{2-}+8\mathrm{H}{}^{+}+3\mathrm{H}{}_2\mathrm{S}\to 2\mathrm{Cr}{}^{3+}+3\mathrm{S}+7\mathrm{H}{}_2\mathrm{O}$

In acid it takes 6 electrons (Cr⁺⁶ → Cr³⁺); it oxidises I⁻ to I₂, Fe²⁺ to Fe³⁺ and H₂S to S, as in the equations above.

Preparation: pyrolusite ($\mathrm{MnO}{}_2$) is fused with KOH in air (or with an oxidising agent like $\mathrm{KNO}{}_3$) to give dark green potassium manganate:

$$2\mathrm{MnO}{}_2+4\mathrm{KOH}+\mathrm{O}{}_2\to 2\mathrm{K}{}_2\mathrm{MnO}{}_4+2\mathrm{H}{}_2\mathrm{O}$$

The manganate disproportionates in neutral or acid solution to permanganate:

$$3\mathrm{MnO}{}_4{}^{2-}+4\mathrm{H}{}^{+}\to 2\mathrm{MnO}{}_4{}^{-}+\mathrm{MnO}{}_2+2\mathrm{H}{}_2\mathrm{O}$$

(Commercially, manganate is oxidised to permanganate electrolytically.)

Reactions in acid solution ($\mathrm{MnO}{}_4{}^{-}+8\mathrm{H}{}^{+}+5\mathrm{e}{}^{-}\to \mathrm{Mn}{}^{2+}+4\mathrm{H}{}_2\mathrm{O}$):

• (i) $\mathrm{MnO}{}_4{}^{-}+8\mathrm{H}{}^{+}+5\mathrm{Fe}{}^{2+}\to \mathrm{Mn}{}^{2+}+5\mathrm{Fe}{}^{3+}+4\mathrm{H}{}_2\mathrm{O}$
• (ii) $2\mathrm{MnO}{}_4{}^{-}+5\mathrm{SO}{}_2+2\mathrm{H}{}_2\mathrm{O}\to 2\mathrm{Mn}{}^{2+}+5\mathrm{SO}{}_4{}^{2-}+4\mathrm{H}{}^{+}$
• (iii) $2\mathrm{MnO}{}_4{}^{-}+16\mathrm{H}{}^{+}+5\mathrm{C}{}_2\mathrm{O}{}_4{}^{2-}\to 2\mathrm{Mn}{}^{2+}+10\mathrm{CO}{}_2+8\mathrm{H}{}_2\mathrm{O}$

MnO₂ is fused with KOH in air to K₂MnO₄, which is converted to KMnO₄ (disproportionation or electrolytic oxidation); acidified MnO₄⁻ oxidises Fe²⁺, SO₂ and oxalate as shown.

(i) The more positive $E^{\circ }(\mathrm{M}{}^{3+}/\mathrm{M}{}^{2+})$, the more easily $\mathrm{M}{}^{3+}$ is reduced, i.e. the less stable it is.

• $\mathrm{Cr}{}^{3+}$ (−0.4 V) is the most stable; it is not easily reduced.
• $\mathrm{Mn}{}^{3+}$ (+1.5 V) is very unstable; it is readily reduced to $\mathrm{Mn}{}^{2+}$ ($d^5$).
• $\mathrm{Fe}{}^{3+}$ (+0.8 V) is in between.

Stability: $\mathrm{Cr}{}^{3+}>\mathrm{Fe}{}^{3+}>\mathrm{Mn}{}^{3+}$.

(ii) The more negative $E^{\circ }(\mathrm{M}{}^{2+}/\mathrm{M})$, the more easily the metal is oxidised: Mn (−1.2 V) > Cr (−0.9 V) > Fe (−0.4 V). Iron is oxidised least easily of the three.

(i) Cr³⁺ > Fe³⁺ > Mn³⁺ in stability. (ii) Ease of oxidation of the metal: Mn > Cr > Fe, so iron is oxidised least easily.

An ion is coloured if it has unpaired d electrons (d–d transitions possible).

• $\mathrm{Ti}{}^{3+}$ ($3d^1$): coloured (purple)
• $\mathrm{V}{}^{3+}$ ($3d^2$): coloured (green)
• $\mathrm{Cu}{}^{+}$ ($3d^{10}$): colourless (no unpaired electron)
• $\mathrm{Sc}{}^{3+}$ ($3d^0$): colourless (no d electron)
• $\mathrm{Mn}{}^{2+}$ ($3d^5$): coloured (pale pink)
• $\mathrm{Fe}{}^{3+}$ ($3d^5$): coloured (yellow)
• $\mathrm{Co}{}^{2+}$ ($3d^7$): coloured (pink)

Coloured: Ti³⁺, V³⁺, Mn²⁺, Fe³⁺, Co²⁺ (unpaired d electrons). Colourless: Cu⁺ (d¹⁰), Sc³⁺ (d⁰).

• Sc does not form a stable +2 state.
• $\mathrm{Ti}{}^{2+}$, $\mathrm{V}{}^{2+}$ and $\mathrm{Cr}{}^{2+}$ exist but are strong reducing agents, easily oxidised to +3.
• $\mathrm{Mn}{}^{2+}$ ($3d^5$) is the most stable +2 ion.
• $\mathrm{Fe}{}^{2+}$ is fairly stable but is oxidised to $\mathrm{Fe}{}^{3+}$ ($d^5$) by air.
• $\mathrm{Co}{}^{2+}$ and $\mathrm{Ni}{}^{2+}$ are stable in water; $\mathrm{Cu}{}^{2+}$ is stable (more than $\mathrm{Cu}{}^{+}$); $\mathrm{Zn}{}^{2+}$ ($3d^{10}$) is very stable and is zinc's only state.

In general the +2 state becomes more stable across the series (the $E^{\circ }(\mathrm{M}{}^{2+}/\mathrm{M})$ values become less negative), with $\mathrm{Mn}{}^{2+}$ and $\mathrm{Zn}{}^{2+}$ especially stable.

+2 is unstable for the early elements (Ti, V, Cr) and becomes more stable across the series; Mn²⁺ (d⁵) and Zn²⁺ (d¹⁰) are especially stable.

Lanthanoids fill 4f, mostly +3, smaller contraction, less reactive. Actinoids fill 5f, many oxidation states (up to +7), larger contraction, more reactive and radioactive.

(i) $\mathrm{Cr}{}^{2+}$ ($d^4$) loses an electron to become $\mathrm{Cr}{}^{3+}$ ($d^3$, a stable half-filled $t_{2g}$ set), so it is reducing. $\mathrm{Mn}{}^{3+}$ ($d^4$) gains an electron to become $\mathrm{Mn}{}^{2+}$ ($d^5$, stable half-filled), so it is oxidising.

(ii) With strong-field ligands, $\mathrm{Co}{}^{3+}$ ($d^6$) forms low-spin complexes ($t_{2g}^6$) with a very large crystal field stabilisation energy. This extra stability more than makes up for the third ionisation enthalpy, so Co(II) is easily oxidised to Co(III) in complexes.

(iii) An ion with one d electron ($d^1$) easily loses it to reach the stable $d^0$ (noble gas) configuration, so such ions are unstable (e.g. $\mathrm{Ti}{}^{3+}$ is easily oxidised to $\mathrm{Ti}{}^{4+}$).

(i) Cr²⁺ → Cr³⁺ (t₂g³) and Mn³⁺ → Mn²⁺ (d⁵) are both favourable. (ii) Co³⁺ (d⁶) gains a large CFSE with strong ligands. (iii) d¹ ions easily lose the electron to become stable d⁰.

Disproportionation is a reaction in which the same element in one oxidation state is simultaneously oxidised and reduced.

• $2\mathrm{Cu}{}^{+}(\mathrm{aq})\to \mathrm{Cu}{}^{2+}(\mathrm{aq})+\mathrm{Cu}(\mathrm{s})$ (Cu +1 → +2 and 0)
• $3\mathrm{MnO}{}_4{}^{2-}+4\mathrm{H}{}^{+}\to 2\mathrm{MnO}{}_4{}^{-}+\mathrm{MnO}{}_2+2\mathrm{H}{}_2\mathrm{O}$ (Mn +6 → +7 and +4)

(Another: $3\mathrm{CrO}{}_4{}^{3-}+8\mathrm{H}{}^{+}\to 2\mathrm{CrO}{}_4{}^{2-}+\mathrm{Cr}{}^{3+}+4\mathrm{H}{}_2\mathrm{O}$, Cr +5 → +6 and +3.)

The same element is both oxidised and reduced; e.g. 2Cu⁺ → Cu²⁺ + Cu, and 3MnO₄²⁻ + 4H⁺ → 2MnO₄⁻ + MnO₂ + 2H₂O.

Copper. Its configuration is $3d^{10}4s^1$; losing the single 4s electron gives $\mathrm{Cu}{}^{+}$ with a stable, completely filled $3d^{10}$ subshell, and its first ionisation enthalpy is low.

Copper; Cu⁺ has the stable 3d¹⁰ configuration.

• $\mathrm{Mn}{}^{3+}$ ($3d^4$): 4 unpaired
• $\mathrm{Cr}{}^{3+}$ ($3d^3$): 3 unpaired
• $\mathrm{V}{}^{3+}$ ($3d^2$): 2 unpaired
• $\mathrm{Ti}{}^{3+}$ ($3d^1$): 1 unpaired

The most stable in water is $\mathrm{Cr}{}^{3+}$: its three electrons half-fill the lower-energy $t_{2g}$ orbitals in the octahedral aqua complex.

4, 3, 2, 1 unpaired electrons; Cr³⁺ is the most stable in water.

(i) MnO is basic but $\mathrm{Mn}{}_2\mathrm{O}{}_7$ is acidic; CrO is basic, $\mathrm{Cr}{}_2\mathrm{O}{}_3$ amphoteric and $\mathrm{CrO}{}_3$ acidic. In the low oxidation state the metal–oxygen bond is ionic and the oxide releases $\mathrm{O}{}^{2-}$ (basic). In a high oxidation state the small, highly charged metal ion forms covalent bonds and the oxide reacts with water to give an acid.

(ii) Examples: $\mathrm{VF}{}_5$, $\mathrm{CrF}{}_6$, $\mathrm{Mn}{}_2\mathrm{O}{}_7$, $\mathrm{CrO}{}_3$. Oxygen and fluorine are small and highly electronegative; fluorine forms strong bonds and oxygen can form multiple bonds, stabilising high oxidation states.

(iii) Examples: $\mathrm{MnO}{}_4{}^{-}$ (Mn +7), $\mathrm{CrO}{}_4{}^{2-}$ and $\mathrm{Cr}{}_2\mathrm{O}{}_7{}^{2-}$ (Cr +6), $\mathrm{VO}{}_4{}^{3-}$ (V +5). Oxygen's ability to form multiple bonds (π bonding) with the metal stabilises these very high oxidation states.

(i) Low states are ionic and basic, high states covalent and acidic (MnO basic, Mn₂O₇ acidic). (ii) O and F are small and highly electronegative (CrF₆, Mn₂O₇). (iii) Oxygen's multiple bonding stabilises high states in oxoanions (MnO₄⁻, Cr₂O₇²⁻).

(i) $\mathrm{K}{}_2\mathrm{Cr}{}_2\mathrm{O}{}_7$ from chromite:

• $4\mathrm{FeCr}{}_2\mathrm{O}{}_4+8\mathrm{Na}{}_2\mathrm{CO}{}_3+7\mathrm{O}{}_2\to 8\mathrm{Na}{}_2\mathrm{CrO}{}_4+2\mathrm{Fe}{}_2\mathrm{O}{}_3+8\mathrm{CO}{}_2$ (fused in air)
• $2\mathrm{Na}{}_2\mathrm{CrO}{}_4+2\mathrm{H}{}^{+}\to \mathrm{Na}{}_2\mathrm{Cr}{}_2\mathrm{O}{}_7+2\mathrm{Na}{}^{+}+\mathrm{H}{}_2\mathrm{O}$ (acidified)
• $\mathrm{Na}{}_2\mathrm{Cr}{}_2\mathrm{O}{}_7+2\mathrm{KCl}\to \mathrm{K}{}_2\mathrm{Cr}{}_2\mathrm{O}{}_7+2\mathrm{NaCl}$ (crystallised)

(ii) $\mathrm{KMnO}{}_4$ from pyrolusite:

• $2\mathrm{MnO}{}_2+4\mathrm{KOH}+\mathrm{O}{}_2\to 2\mathrm{K}{}_2\mathrm{MnO}{}_4+2\mathrm{H}{}_2\mathrm{O}$ (fused with KOH in air)
• $3\mathrm{MnO}{}_4{}^{2-}+4\mathrm{H}{}^{+}\to 2\mathrm{MnO}{}_4{}^{-}+\mathrm{MnO}{}_2+2\mathrm{H}{}_2\mathrm{O}$ (disproportionation), or electrolytic oxidation: $\mathrm{MnO}{}_4{}^{2-}\to \mathrm{MnO}{}_4{}^{-}+\mathrm{e}{}^{-}$

(i) Fusion with Na₂CO₃ in air → Na₂CrO₄ → acidify → Na₂Cr₂O₇ → KCl → K₂Cr₂O₇. (ii) Fusion with KOH in air → K₂MnO₄ → disproportionation or electrolytic oxidation → KMnO₄.

An alloy is a homogeneous solid mixture of two or more metals (or a metal and a non-metal), made by melting them together.

Misch metal contains about 95% of a mixture of lanthanoid metals and about 5% iron, with traces of S, C, Ca and Al. It is used in magnesium-based alloys for making bullets, shells and lighter flints.

A homogeneous mixture of metals; misch metal (≈95% lanthanoids + 5% Fe) is used in Mg alloys for bullets, shells and lighter flints.

Inner transition elements are the f-block elements, in which the last electron enters the (n−2)f subshell: the lanthanoids (Z = 58–71) and the actinoids (Z = 90–103).

From the list: 59 (Pr, a lanthanoid), 95 (Am) and 102 (No) (actinoids). The others (29 Cu, 74 W, 104 Rf) are d-block elements.

f-block elements (Z = 58–71 and 90–103); 59, 95 and 102.

Lanthanoids mostly show +3. Actinoids show a much wider and irregular range, because their 5f, 6d and 7s levels are close in energy:

• Th: +4; Pa: +5; U: up to +6; Np and Pu: up to +7
• After the middle of the series, +3 becomes the main state again (Am onwards).

Their radioactivity and short half-lives also make their chemistry hard to study.

Actinoids show many oxidation states (Th +4, Pa +5, U +6, Np/Pu +7) and then mainly +3, unlike the steady +3 of lanthanoids.

Lawrencium, Lr (Z = 103): $[\mathrm{Rn}]5f^{14}6d^{1}7s^2$. It shows the +3 oxidation state (losing the 6d and two 7s electrons gives the stable $5f^{14}$).

Lawrencium (Z = 103), [Rn]5f¹⁴6d¹7s²; oxidation state +3.

Ce (Z = 58): $[\mathrm{Xe}]4f^{1}5d^{1}6s^2$. Removing three electrons (two 6s and one 5d) gives $\mathrm{Ce}{}^{3+}$: $[\mathrm{Xe}]4f^1$, with 1 unpaired electron.

$$\mu =\sqrt{1\times (1+2)}=\sqrt{3}\approx 1.73\text{BM}$$

[Xe]4f¹; μ ≈ 1.73 BM.

• +4: Ce, Pr, Nd, Tb, Dy. $\mathrm{Ce}{}^{4+}$ is $4f^0$ and $\mathrm{Tb}{}^{4+}$ is $4f^7$.
• +2: Sm, Eu, Tm, Yb. $\mathrm{Eu}{}^{2+}$ is $4f^7$ and $\mathrm{Yb}{}^{2+}$ is $4f^{14}$.

These states are favoured when the ion reaches an empty ($f^0$), half-filled ($f^7$) or completely filled ($f^{14}$) 4f subshell, which is extra stable.

+4: Ce, Pr, Nd, Tb, Dy; +2: Sm, Eu, Tm, Yb. They tend to reach f⁰, f⁷ or f¹⁴.

• (i) Lanthanoids: $[\mathrm{Xe}]4f^{1-14}5d^{0-1}6s^2$ (4f filled). Actinoids: $[\mathrm{Rn}]5f^{1-14}6d^{0-1}7s^2$ (5f filled). The 5f orbitals are less deeply buried than 4f, so they take part in bonding more.
• (ii) Lanthanoids mainly +3 (occasionally +2, +4). Actinoids show a wide range, +3 to +7, because 5f, 6d and 7s energies are similar.
• (iii) Lanthanoids are reactive metals (similar to calcium) and form mainly ionic compounds. Actinoids are more reactive, especially when finely divided, react with boiling water and acids, and are all radioactive.

See the comparison above: 4f vs 5f filling, mostly +3 vs +3 to +7, and actinoids more reactive and radioactive.

• 61 (Pm): $[\mathrm{Xe}]4f^{5}6s^2$
• 91 (Pa): $[\mathrm{Rn}]5f^{2}6d^{1}7s^2$
• 101 (Md): $[\mathrm{Rn}]5f^{13}7s^2$
• 109 (Mt): $[\mathrm{Rn}]5f^{14}6d^{7}7s^2$

As listed above.

• (i) Configurations: generally similar down a group ($(n-1)d^{x}n{s}^{1-2}$), with some irregularities: Cr $3d^{5}4s^1$, Mo $4d^{5}5s^1$ but W $5d^{4}6s^2$; Ni $3d^{8}4s^2$, Pd $4d^{10}5s^0$, Pt $5d^{9}6s^1$.
• (ii) Oxidation states: in the first series the lower states (+2, +3) are more common and stable. In the second and third series the higher states are more stable: Mo(VI) and W(VI) are stable while Cr(VI) is strongly oxidising; Os and Ru reach +8.
• (iii) Ionisation enthalpies: the 5d series has higher values than the 3d and 4d series, because the 4f electrons shield poorly (lanthanoid contraction) and the nuclear charge is high.
• (iv) Atomic sizes: the 4d atoms are larger than the 3d atoms, but the 4d and 5d atoms of a group are almost the same size (e.g. Zr 160 pm, Hf 159 pm) because of lanthanoid contraction.

Similar configurations with some irregularities; heavier series favour higher oxidation states; 5d IEs are highest; 4d and 5d sizes are nearly equal because of lanthanoid contraction.

Water is a weak-field ligand, so these octahedral aqua ions are high-spin (electrons fill $t_{2g}$ and $e_g$ singly before pairing).

2, 3, 3, 5, 6, 5, 7, 8, 9 electrons; occupation as in the table (high spin).

The statement is true. Compared with the 4d and 5d metals, the first series elements:

• have smaller atomic radii and lower enthalpies of atomisation (weaker metallic bonding, lower melting points);
• show lower oxidation states more often (+2, +3), while heavier elements prefer higher states (Mo(VI), W(VI), Os(VIII));
• form mostly high-spin complexes with weak ligands, whereas heavier elements form mostly low-spin complexes;
• rarely form metal–metal bonds, which are common in heavier elements;
• have magnetic moments that fit the spin-only formula more closely.

True: the 3d metals are smaller, prefer lower oxidation states, form high-spin complexes and rarely M–M bonds, unlike the 4d and 5d metals.

Use $\mu =\sqrt{n(n+2)}$:

• $\mathrm{K}{}_4[\mathrm{Mn}(\mathrm{CN}){}_6]$: Mn²⁺, $3d^5$. 2.2 BM is close to 1.73 BM, i.e. 1 unpaired electron. $\mathrm{CN}{}^{-}$ is a strong-field ligand: low spin, $t_{2g}^5$, inner-orbital ($d^{2}s{p}^3$) octahedral complex.
• $[\mathrm{Fe}(\mathrm{H}{}_2\mathrm{O}){}_6]{}^{2+}$: Fe²⁺, $3d^6$. 5.3 BM is close to 4.90 BM, i.e. 4 unpaired electrons: high spin ($t_{2g}^4{e}_g^2$), outer-orbital ($s{p}^3{d}^2$) octahedral complex with the weak ligand water.
• $\mathrm{K}{}_2[\mathrm{MnCl}{}_4]$: Mn²⁺, $3d^5$. 5.9 BM matches 5 unpaired electrons: high spin, $s{p}^3$ tetrahedral complex.

[Mn(CN)₆]⁴⁻: 1 unpaired e⁻, low spin (d²sp³). [Fe(H₂O)₆]²⁺: 4 unpaired, high spin (sp³d²). [MnCl₄]²⁻: 5 unpaired, high spin (sp³, tetrahedral).