JEE Chemistry 2026 Question with Solution

Given: The complexes are [Co(NH$_3$)$_6$]$^{3+}$ (I), [Co(H$_2$O)$_6$]$^{3+}$ (II), [Co(CN)$_6$]$^{3-}$ (III), [Co(NH$_3$)$_5$(H$_2$O)]$^{3+}$ (IV), [CoF$_6$]$^{3-}$ (V).

Find: The correct order of wavelength of light absorbed.

Concept: The wavelength of light absorbed is inversely proportional to the crystal field splitting energy $\Delta_o$. Stronger field ligands produce larger splitting and therefore shorter absorbed wavelength.

The ligand field strength order is:

$$\text{CN}^- > \text{NH}_3 > \text{H}_2\text{O} > \text{F}^-$$

So, [Co(CN)$_6$]$^{3-}$ has the maximum splitting and absorbs the shortest wavelength, while [CoF$_6$]$^{3-}$ has the minimum splitting and absorbs the longest wavelength.

Also, [Co(NH$_3$)$_5$(H$_2$O)]$^{3+}$ lies between [Co(NH$_3$)$_6$]$^{3+}$ and [Co(H$_2$O)$_6$]$^{3+}$ in crystal field splitting.

Hence, the order of absorbed wavelength is:

$$\text{III} < \text{I} < \text{IV} < \text{II} < \text{V}$$

Therefore, the correct option is A.

Given: Absorption wavelength depends on octahedral crystal field splitting.

Find: Arrange the complexes in increasing order of absorbed wavelength.

For the same metal ion oxidation state, stronger ligands create larger $\Delta_o$. Since absorbed energy is proportional to $\Delta_o$ and energy is inversely proportional to wavelength, larger $\Delta_o$ means shorter wavelength.

Thus we compare ligands using the spectrochemical series:

$$\text{CN}^- > \text{NH}_3 > \text{H}_2\text{O} > \text{F}^-$$

Therefore:

1. [Co(CN)$_6$]$^{3-}$ gives the shortest wavelength.
2. [Co(NH$_3$)$_6$]$^{3+}$ comes next.
3. [Co(NH$_3$)$_5$(H$_2$O)]$^{3+}$ has one \text{H}_2\text{O replacing $\text{NH}_3$, so its splitting is slightly less than (I) and wavelength is slightly larger.
4. [Co(H$_2$O)$_6$]$^{3+}$ absorbs at still longer wavelength.
5. [CoF$_6$]$^{3-}$ absorbs at the longest wavelength.

So the final order is:

$$\text{III} < \text{I} < \text{IV} < \text{II} < \text{V}$$

Hence, A is correct.