Why $Zn^{2+}$ is colourless whereas $Mn^{2+}$ is violet in colour?

**OR**

Explain why $Cr^{2+}$ is a good reducing agent and $Mn^{3+}$ is a good oxidising agent even though both ions have the same $d^4$ configuration.

Question

Why $Zn^{2+}$ is colourless whereas $Mn^{2+}$ is violet in colour?

**OR**

Explain why $Cr^{2+}$ is a good reducing agent and $Mn^{3+}$ is a good oxidising agent even though both ions have the same $d^4$ configuration.

Accepted Answer

>

> [!TLDR]
> Colour in transition-metal ions comes from d–d electronic transitions, which require partially filled d-orbitals; $Zn^{2+}$ has none available ($3d^{10}$) while $Mn^{2+}$ does ($3d^5$). In the alternative question, both $Cr^{2+}$ and $Mn^{3+}$ are $d^4$, but each is driven toward a more stable configuration by opposite electron-transfer directions.
>
> **Why $Zn^{2+}$ is colourless**
>
> Colour in transition-metal compounds typically arises from **d–d transitions**: an electron absorbs a photon of visible light and jumps from a lower-energy $d$ orbital to a higher-energy one (split by the ligand field). This is only possible if the $d$ subshell is **partially filled** — a completely empty or completely full $d$ subshell has no such transition available. $Zn$ has the configuration $[Ar]3d^{10}4s^2$, so $Zn^{2+}$ is $[Ar]3d^{10}$ — a **completely filled** $d$ subshell. With no vacant $d$ orbital for an electron to be promoted into, no d–d transition (and hence no visible-light absorption of this type) is possible, so $Zn^{2+}$ salts and solutions are colourless.
>
> **Why $Mn^{2+}$ is coloured**
>
> $Mn$ is $[Ar]3d^54s^2$, so $Mn^{2+}$ is $[Ar]3d^5$ — a **partially filled** (half-filled) $d$ subshell with unpaired electrons. This allows d–d transitions to occur, giving $Mn^{2+}$ compounds colour. (In practice, because the $3d^5$ configuration is high-spin and any transition must flip an electron's spin, these transitions are **spin-forbidden**, so they are unusually weak — $Mn^{2+}$(aq) is characteristically only a very pale pink rather than an intensely coloured ion — but the underlying principle asked here, that a partially filled $d$ subshell permits colour while a filled one does not, is exactly why $Mn^{2+}$ shows colour and $Zn^{2+}$ shows none.)
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> **Alternative (Or): $Cr^{2+}$ as reductant vs $Mn^{3+}$ as oxidant, both $d^4$**
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>

> [!TLDR]
> Both ions have $d^4$ configurations, but electron loss from $Cr^{2+}$ reaches the extra-stable half-f

Q.Why $Zn^{2+}$ is colourless whereas $Mn^{2+}$ is violet in colour? OR Explain why $Cr^{2+}$ is a good reducing agent and $Mn^{3+}$ is a good oxidising agent even though both ions have the same $d^4$ configuration.

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.Why Zn2+Zn^{2+}Zn2+ is colourless whereas Mn2+Mn^{2+}Mn2+ is violet in colour? OR Explain why Cr2+Cr^{2+}Cr2+ is a good reducing agent and Mn3+Mn^{3+}Mn3+ is a good oxidising agent even though both ions have the same d4d^4d4 configuration.

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Q.Why $Zn^{2+}$ is colourless whereas $Mn^{2+}$ is violet in colour? OR Explain why $Cr^{2+}$ is a good reducing agent and $Mn^{3+}$ is a good oxidising agent even though both ions have the same $d^4$ configuration.

Unlock everything free for 14 days