"d- and f-Block Elements" — Practice

Question 1

Explain why $	ext{Cr}_2	ext{O}_7^{2-}$ is orange and $	ext{MnO}_4^{-}$ purple, although chromium is $d^{0}$ and manganese is also $d^{0}$.

Accepted Answer

Both owe their colour to ligand-to-metal **charge transfer**, in which an electron moves from a filled oxygen orbital to an empty metal orbital.

The energy of that transition depends on how readily the metal accepts the electron, that is on its oxidising power.

Manganese(VII) is the stronger oxidant, so its transition is lower in energy and absorbs further into the visible, giving the deep purple.

Chromium(VI) absorbs at higher energy, in the blue, so the ion appears orange. Both are far more intense than any $d$-$d$ colour.

Question 2

Calculate the spin-only magnetic moments of $	ext{Ti}^{3+}$, $	ext{V}^{3+}$, $	ext{Cr}^{3+}$ and $	ext{Mn}^{2+}$, and state which is most paramagnetic.

Accepted Answer

$	ext{Ti}^{3+}$ is $3d^{1}$: $\mu=\sqrt3=1.73$ BM

$	ext{V}^{3+}$ is $3d^{2}$: $\mu=\sqrt8=2.83$ BM

$	ext{Cr}^{3+}$ is $3d^{3}$: $\mu=\sqrt{15}=3.87$ BM

$	ext{Mn}^{2+}$ is $3d^{5}$: $\mu=\sqrt{35}=5.92$ BM

Manganese(II) is the most paramagnetic, having the maximum possible five unpaired electrons for a $d$ set.

Question 3

Explain why $	ext{Sc}^{3+}$ solutions are colourless while $	ext{Ti}^{3+}$ solutions are violet.

Accepted Answer

$	ext{Sc}^{3+}$ is $3d^{0}$: the $d$ set is empty, so there is no electron to promote and no $d$-$d$ absorption is possible.

$	ext{Ti}^{3+}$ is $3d^{1}$: one electron can be promoted from the lower to the upper set of the split $d$ orbitals.

It absorbs in the green and yellow region, so the transmitted light appears violet.

Scandium also forms no charge-transfer band in the visible, since scandium(III) is not readily reduced, so its solutions remain colourless by both mechanisms.

Question 4

Explain why manganese shows oxidation states from $+2$ to $+7$ while zinc shows only $+2$.

Accepted Answer

Manganese is $3d^{5}4s^{2}$. All seven of those electrons lie close in energy and can in principle be removed, giving the full range from $+2$ to $+7$.

Zinc is $3d^{10}4s^{2}$. Losing the two $4s$ electrons gives the stable $3d^{10}$ configuration of $	ext{Zn}^{2+}$.

Removing a further electron would break into that filled set, requiring far more energy than any chemical process can supply.

Zinc is therefore fixed at $+2$, and by the same token shows no colour, no catalytic activity and no complex-ion chemistry of the transition-metal type.

Question 5

The third ionisation enthalpies of manganese and iron are $3248$ and $2957$ kJ mol$^{-1}$. Explain the reversal and its chemical consequence.

Accepted Answer

$	ext{Mn}^{2+}$ is $3d^{5}$, a half-filled set with maximum exchange stabilisation. Removing a third electron destroys that arrangement, so the enthalpy is exceptionally high.

$	ext{Fe}^{2+}$ is $3d^{6}$. Removing one electron **produces** the stable $3d^{5}$ arrangement, so the enthalpy is unusually low.

Chemically, iron(III) is common and stable in aqueous solution while manganese(III) is a strong oxidant that reverts readily to manganese(II).

One configuration explains both facts, read from the product side in one case and the reactant side in the other.

Question 6

Explain why the atomic radii of the first transition series barely change from vanadium to nickel, and why copper and zinc are slightly larger.

Accepted Answer

The added electrons enter the inner $3d$ shell, which screens the outer $4s$ electrons quite effectively.

The rise in nuclear charge is therefore largely cancelled, and the radius stays nearly constant across the middle of the series.

Towards the end the $d$ set becomes crowded and electron-electron repulsion within it grows appreciable.

That repulsion outweighs the small remaining increase in effective nuclear charge, so copper and zinc are slightly larger than nickel.

Question 7

Explain why copper does not displace hydrogen from dilute hydrochloric acid but does dissolve in dilute nitric acid.

Accepted Answer

$E^{\circ}\left(	ext{Cu}^{2+}/	ext{Cu}ight)=+0.34$ V, so oxidising copper to its ion is unfavourable relative to the hydrogen electrode at $0$ V.

Hydrochloric acid offers only the proton as oxidant, which is too weak, so no reaction occurs.

Nitric acid supplies the nitrate ion, whose reduction potential is around $+0.96$ V, comfortably above copper's.

Copper therefore dissolves, but the gas evolved is nitric oxide or nitrogen dioxide rather than hydrogen.

Question 8

Write the equations for the preparation of potassium permanganate from pyrolusite, and explain why the industrial route uses electrolysis for the second step.

Accepted Answer

Fusion with alkali and an oxidant gives the green manganate:

$2	ext{MnO}_2+4	ext{KOH}+	ext{O}_2ightarrow2	ext{K}_2	ext{MnO}_4+2	ext{H}_2	ext{O}$

Acidifying causes disproportionation:

$3	ext{MnO}_4^{2-}+4	ext{H}^+ightarrow2	ext{MnO}_4^-+	ext{MnO}_2+2	ext{H}_2	ext{O}$

That route converts only two of every three manganese atoms to permanganate, wasting a third as manganese dioxide. Electrolytic oxidation of the manganate converts **all** of it, which is why industry prefers it.

Question 9

A solution of potassium dichromate is treated with sodium hydroxide and then reacidified. Describe the colour changes, name the species present at each stage, and state the oxidation state of chromium throughout.

Accepted Answer

Initially orange, containing $	ext{Cr}_2	ext{O}_7^{2-}$.

Adding base removes hydrogen ions and shifts the equilibrium $2	ext{CrO}_4^{2-}+2	ext{H}^+ightleftharpoons	ext{Cr}_2	ext{O}_7^{2-}+	ext{H}_2	ext{O}$ to the left, giving yellow $	ext{CrO}_4^{2-}$.

Reacidifying restores the orange dichromate.

Chromium is at $+6$ in **both** species. No electrons are transferred at any point, which is why the change is reversible simply by adjusting pH.

Question 10

Explain why potassium dichromate is used as a primary standard in volumetric analysis while potassium permanganate is not.

Accepted Answer

Dichromate has an n-factor of $6$ in every medium, since it is always reduced to chromium(III). Permanganate's varies between $5$, $3$ and $1$ depending on acidity.

Dichromate can be obtained pure and dry and is stable indefinitely; permanganate solutions decompose slowly in light and are contaminated by manganese dioxide.

Dichromate solutions can therefore be made up by direct weighing, whereas permanganate must be standardised before each use.

The cost is that dichromate needs an external indicator, since its own colour change is less sharp.

Question 11

Explain why almost all lanthanoids show the $+3$ oxidation state, and why cerium and europium are exceptions in opposite directions.

Accepted Answer

The $4f$ orbitals are buried beneath the filled $5s$ and $5p$ shells and take almost no part in bonding, so the three outermost electrons are lost and $+3$ results across the series.

Cerium reaches $+4$ because doing so empties the $f$ shell entirely, giving $4f^{0}$.

Europium drops to $+2$ because that leaves a half-filled $4f^{7}$ shell.

Both revert to $+3$ in aqueous solution, so cerium(IV) is a strong oxidant and europium(II) a strong reductant — opposite reagents from the same underlying cause.

Question 12

Give three ways in which actinoid chemistry differs from lanthanoid chemistry, and explain the underlying reason.

Accepted Answer

**Oxidation states:** actinoids show $+3$ through $+6$ and beyond, while lanthanoids are almost uniformly $+3$.

**Radioactivity:** every actinoid is radioactive, whereas only promethium is among the lanthanoids.

**Contraction:** the actinoid contraction is larger, since $5f$ orbitals shield even more poorly than $4f$.

The common cause is that $5f$ orbitals are less buried than $4f$ and lie closer in energy to the valence shell, so they can participate in bonding rather than remaining chemically inert.

Practice — "d- and f-Block Elements"

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