d- and f-Block Elements — Practice

Question 1

Zinc is in the d-block but is not a transition element. Justify this, citing three properties of zinc compounds.

Accepted Answer

A transition element must have a partly filled d subshell either in the free atom or in a common oxidation state. Zinc is $3d^{10}\,4s^2$ as an atom and $3d^{10}$ as $\mathrm{Zn^{2+}}$, so its d subshell is complete in every state it adopts. The consequences are that zinc compounds are colourless, since there is no vacancy for a d electron to be promoted into; zinc shows only the $+2$ state, since the filled d electrons are too stable to use; and $\mathrm{Zn^{2+}}$ is diamagnetic, having no unpaired electrons. Contrast silver, which qualifies through $\mathrm{Ag^{2+}}$ in $\mathrm{AgF_2}$.

Question 2

Explain why manganese has an unusually low melting point compared with chromium and iron on either side of it.

Accepted Answer

Melting point in the transition series depends on how many electrons enter the metallic bond, and both 4s and unpaired 3d electrons contribute. Manganese has a half-filled $3d^5$ configuration, unusually stable because all five electrons occupy separate orbitals with parallel spins and the exchange energy is maximised. Those electrons resist being delocalised into the metallic bond, so fewer electrons hold the lattice together than the position in the row suggests, and the melting point falls about 660 K below vanadium. Technetium repeats the anomaly directly below.

Question 3

Why does manganese show the widest range of oxidation states in the first transition series?

Accepted Answer

Because oxidation state range is set by how many electrons are energetically available, and manganese has exactly seven in its 4s and 3d subshells combined, more than any earlier element in the row. Since 4s and 3d are close in energy, any number of those seven can be used, giving every state from $+2$ to $+7$. After manganese the d electrons begin pairing, and paired electrons are harder to remove, so the maximum oxidation state falls again from iron onwards. The highest states appear only with oxygen or fluorine, as in $\mathrm{MnO_4^-}$.

Question 4

Calculate the spin-only magnetic moments of $\mathrm{Ti^{3+}}$, $\mathrm{Fe^{2+}}$ and $\mathrm{Mn^{2+}}$.

Accepted Answer

$\mathrm{Ti^{3+}}$ is $d^1$ with one unpaired electron, giving $\mu = \sqrt{1 	imes 3} = 1.73$ BM. $\mathrm{Fe^{2+}}$ is $d^6$, and in a weak field the sixth electron pairs while four remain unpaired, giving $\sqrt{4 	imes 6} = 4.90$ BM. $\mathrm{Mn^{2+}}$ is $d^5$ with all five unpaired, giving $\sqrt{5 	imes 7} = 5.92$ BM, the largest possible for a first-row ion. Remember to remove 4s electrons before 3d when writing the ion configurations.

Question 5

Explain why aqueous $\mathrm{Cu^{2+}}$ is blue while $\mathrm{Cu^+}$ is colourless, and why $\mathrm{Cu^+}$ disproportionates in water.

Accepted Answer

$\mathrm{Cu^{2+}}$ is $d^9$, so an electron can absorb visible light and be promoted into the vacancy in the upper d orbitals, giving a d-d transition and hence colour; $\mathrm{Cu^+}$ is $d^{10}$ with no vacancy, so it is colourless. Copper(I) is the cheaper ion to make, but in water $\mathrm{Cu^{2+}}$ carries twice the charge in a similar volume and is hydrated far more strongly, $-2121$ against $-593$ kJ mol$^{-1}$. That difference more than repays the second ionisation enthalpy, and the full cycle comes out at $-61$ kJ mol$^{-1}$, so $\mathrm{Cu^+}$ disproportionates into $\mathrm{Cu^{2+}}$ and copper metal.

Question 6

The permanganate ion is intensely purple yet manganese in it is $d^0$. Explain the colour.

Accepted Answer

Manganese is in the $+7$ state, so it has no d electrons at all and a d-d transition is impossible. The colour arises from ligand-to-metal charge transfer: an absorbed photon promotes an electron from a lone pair on one of the oxygen atoms into an empty orbital on manganese. Charge transfer transitions are symmetry-allowed and therefore far more intense than d-d transitions, which is why a very dilute permanganate solution is deeply coloured while a comparable copper solution is only faintly tinted. Dichromate's orange has the same origin.

Question 7

A yellow potassium chromate solution turns orange on acidification and yellow again on adding alkali. Explain, and state whether this is a redox change.

Accepted Answer

Chromate and dichromate exist in a pH-dependent equilibrium, $\mathrm{2CrO_4^{2-} + 2H^+ \rightleftharpoons Cr_2O_7^{2-} + H_2O}$. Adding acid raises the hydrogen ion concentration, so by Le Chatelier the system shifts right and orange dichromate forms; adding alkali removes hydrogen ions and it shifts back to yellow chromate. It is **not** a redox change, since chromium remains at $+6$ on both sides. It does change the oxidising power dramatically, from $+1.33$ V for dichromate to $-0.13$ V for chromate.

Question 8

Describe the preparation of potassium dichromate from chromite ore and give its structure.

Accepted Answer

Chromite ore, $\mathrm{FeCr_2O_4}$, is fused with sodium carbonate in the presence of air, which oxidises the chromium to $+6$ and gives sodium chromate. Acidifying the resulting solution converts chromate to dichromate. Adding potassium chloride then precipitates potassium dichromate, which is less soluble than the sodium salt, and it is purified by recrystallisation. Structurally the dichromate ion consists of two chromium-centred tetrahedra sharing a single corner oxygen, giving a bridging $\mathrm{Cr-O-Cr}$ linkage.

Question 9

Give the manganese-containing reduction product of permanganate in acidic, neutral and strongly alkaline media, with the number of electrons in each case.

Accepted Answer

In acidic solution manganese falls from $+7$ to $+2$, gaining five electrons and giving the almost colourless $\mathrm{Mn^{2+}}$ ion. In neutral or faintly alkaline solution it falls to $+4$, gaining three electrons and depositing brown $\mathrm{MnO_2}$. In strongly alkaline solution it gains only one electron to become the green manganate ion $\mathrm{MnO_4^{2-}}$, with manganese at $+6$. Because the electron count differs, the same volume of the same solution does three different amounts of oxidation, and the stoichiometry of any permanganate titration depends entirely on the medium.

Question 10

Why must a permanganate titration be acidified with dilute sulphuric acid rather than hydrochloric or nitric acid?

Accepted Answer

Because permanganate is a strong enough oxidising agent, at $E^\circ = 1.51$ V in acid, to oxidise chloride ions to chlorine. If hydrochloric acid were used, part of the permanganate would be consumed by the acid itself and the titre would be too high and meaningless. Nitric acid is avoided for the opposite reason: it is itself an oxidising agent and would attack the analyte. Sulphate cannot be oxidised further under these conditions, so dilute sulphuric acid is inert and safe.

Question 11

Explain lanthanoid contraction and give two of its consequences.

Accepted Answer

Across the lanthanoid series, 4f electrons are added and shield the nuclear charge very poorly, being diffuse and awkwardly shaped. Each added proton therefore pulls the outer shells inward more than each added f electron pushes them outward, so atomic and ionic radii decrease steadily across the fourteen elements. One consequence is that zirconium and hafnium have almost identical radii of 160 and 159 pm despite being a period apart, making them notoriously hard to separate. Another is that basicity falls across the series, so $\mathrm{La(OH)_3}$ is more basic than $\mathrm{Lu(OH)_3}$.

Question 12

State three ways in which the actinoids differ from the lanthanoids.

Accepted Answer

First, actinoids show a much wider range of oxidation states, from $+3$ up to $+7$, because their 5f orbitals are more spatially extended and closer in energy to 6d and 7s, so they take part in bonding readily; uranium alone shows $+3$, $+4$, $+5$ and $+6$, whereas lanthanoids are overwhelmingly $+3$ with exceptions only at $f^0$, $f^7$ and $f^{14}$. Second, all actinoids are radioactive, and those beyond uranium do not occur naturally in quantity, so their chemistry is known largely from trace amounts. Third, the actinoid contraction is larger, because 5f electrons shield even more poorly than 4f.

Practice — d- and f-Block Elements

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