(a) Why do transition metals generally form coloured compounds? (1 mark) (b) Draw the structures of permanganate ion. (1 mark) (c) Why $Cr^{3+}$ is reducing and $Mn^{3+}$ is oxidizing when both have $d^4$-configuration? (1 mark) (d) Give the reaction of potassium dichromate with

Question

(a) Why do transition metals generally form coloured compounds? (1 mark) (b) Draw the structures of permanganate ion. (1 mark) (c) Why $Cr^{3+}$ is reducing and $Mn^{3+}$ is oxidizing when both have $d^4$-configuration? (1 mark) (d) Give the reaction of potassium dichromate with (i) KI and (ii) $H_2S$ (in acidic medium). (2 marks)

**OR**

(e) Why do transition metals easily form alloys with other transition metals? (1 mark) (f) Why is silver (Atomic number = 47) considered to be a transition element whereas zinc (Atomic number = 30) is not? (1 mark) (g) Write the balanced chemical equation for the reaction of $KMnO_4$ with oxalic acid in acidic medium. (1 mark) (h) Write all the reactions occurring during the preparation of potassium permanganate from pyrolusite ore. (2 marks)

Accepted Answer

![Figure — Part (b) 'Draw the structures of permanganate ion' explicitly requires the drawn tetrahedral MnO4- structure,](/figures/the-d-and-f-block-elements/tetrahedral-permanganate-ion-purple-ball-and-stick-structure-sec-4-4-1.svg)

>

> [!TLDR]
> Transition-metal colour arises from d-d transitions; permanganate is tetrahedral $MnO_4^-$; $Cr^{2+}$/$Mn^{3+}$ redox behaviour follows from which configuration ($d^3$ or $d^5$) is more stable; and dichromate's oxidations of $I^-$ and $H_2S$ are balanced. *(Or: alloying, the Ag-vs-Zn transition-metal distinction, and $KMnO_4$'s balanced reaction with oxalic acid and its industrial preparation from pyrolusite are covered.)*
>
> **(a) Why transition metals form coloured compounds**
>
> Most transition-metal ions have **partially filled $d$-orbitals**. In a complex, the surrounding ligands split the five degenerate $d$-orbitals into two sets of different energy (crystal-field splitting, e.g. $t_{2g}$ and $e_g$ in an octahedral field). An electron can absorb a photon of visible light and jump from the lower-energy $d$-orbital set to the higher one (a **d–d transition**). Since only *specific* wavelengths of visible light are absorbed for this transition, the *transmitted/reflected* light — the complementary colour — is what we see as the colour of the compound. (Ions with an empty, $d^0$, or completely filled, $d^{10}$, configuration have no such transition available and are usually colourless/white, e.g. $Sc^{3+}$, $Zn^{2+}$.)
>
> **(b) Structure of the permanganate ion, $MnO_4^-$**
>
> $MnO_4^-$ is **tetrahedral**: manganese (oxidation state +7) sits at the centre with four oxygen atoms at the corners of a tetrahedron. All four Mn–O bonds are equivalent in length and bond order, due to resonance/delocalisation of $\pi$-electron density over all four Mn–O bonds ($p\pi$–$d\pi$ overlap between filled O $p$-orbitals and empty Mn $d$-orbitals), giving each bond partial double-bond character.
>
> **(c) Why $Cr^{2+}$ is reducing

Q.(a) Why do transition metals generally form coloured compounds? (1 mark)

Concept understanding — Colour of Transition Metal Ions

The intuition: a window that absorbs some colours

The precise mechanism: d–d transition

Why not all d-block ions are coloured

Q.(a) Why do transition metals generally form coloured compounds? (1 mark)

🎓Unlock everything free for 14 days

Unlock everything free for 14 days