Simplified absorption spectra of three complexes [(i), (ii) and (iii)] of M^n+ ion are provided belo

Q: Why does a stronger ligand absorb light of a lower wavelength?

Stronger ligands cause a larger splitting ($\Delta_0$) between the d-orbitals. To excite an electron across this larger energy gap, a photon with higher energy is required. Since energy is inversely proportional to wavelength ($E = \frac{hc}{\lambda}$), higher energy corresponds to a lower wavelength.

Q: How do I remember the spectrochemical series?

Use mnemonics! A general rule of thumb is that halogens are weak field ligands, oxygen donors are intermediate, and nitrogen or carbon donors are strong field ligands.

Q: Does the metal ion affect the absorption spectrum?

Absolutely! The oxidation state and the nature of the metal ion also significantly affect the magnitude of $\Delta_0$. However, in this specific problem, the metal ion ($M^{n+}$) is kept constant across all three complexes, allowing us to isolate the effect of the ligands.

Q: What if two ligands have very similar strengths?

If two ligands have similar strengths, their absorption peaks would be very close to each other on the spectrum. In such cases, you might need more precise experimental data or consider other subtle factors like pairing energy to distinguish them.

Q: Why are the peaks in the graph broad instead of sharp lines?

In solution, complexes are dynamic. Molecular vibrations and interactions with solvent molecules mean that the energy gap isn't a single, rigid value. Instead, a range of closely related energies can be absorbed, leading to the broad absorption bands you see in the graph.

Question

Simplified absorption spectra of three complexes [(i), (ii) and (iii)] of $M^{n+}$ ion are provided below; their $\lambda_{\max}$ values are marked as $A, B$ and $C$ respectively. The correct match between the complexes and their $\lambda_{\max}$ values is

(i) $[M(NCS)_6]^{(-6+n)}$
(ii) $[MF_6]^{(-6+n)}$
(iii) $[M(NH_3)_6]^{n+}$

Options:

(A) A-(iii), B-(i), C-(ii)

(B) A-(ii), B-(i), C-(iii)

(C) A-(ii), B-(iii), C-(i)

(D) A-(i), B-(ii), C-(iii)

Sigma Percentile · Accepted Answer

### Decoding Absorption Spectra in Coordination Complexes Have you ever wondered why coordination complexes exhibit such brilliant and diverse colors? The secret lies in the intricate dance between light, energy, and the nature of the ligands surrounding the central metal ion. In this problem, we are presented with the absorption spectra of three different complexes and tasked with matching them to their respective ligands. Let's embark on this fascinating journey through Crystal Field Theory. ### Analyzing the Setup Imagine you are looking at the provided absorption graph. We see three distinct peaks labeled **A**, **B**, and **C**. The x-axis represents the wavelength ($\lambda$) of the absorbed light. By simply observing the positions of these peaks, we can establish a clear relationship: $$\lambda_{\max}^A < \lambda_{\max}^B < \lambda_{\max}^C$$ This tells us that complex A absorbs light of the shortest wavelength, while complex C absorbs light of the longest wavelength. ### The Master Equation To connect this visual data to the chemical properties of the complexes, we must invoke Planck's famous equation, which relates the energy of a photon to its wavelength: $$\Delta E = \frac{hc}{\lambda_{\max}}$$ In the context of coordination compounds, this energy difference ($\Delta E$) corresponds to the Crystal Field Splitting Energy ($\Delta_0$). Because energy and wavelength are inversely proportional, the order of splitting energies is the exact opposite of the wavelength order: $$\Delta_0^A > \Delta_0^B > \Delta_0^C$$ This is a crucial realization! It means that the complex responsible for peak A experiences the greatest splitting of its d-orbitals. ### The Spectrochemical Series Now, what causes this splitting? It's the ligands! According to Crystal Field Theory, the magnitude of $\Delta_0$ is directly proportional to the strength of the ligand field. Stronger ligands cause a larger energy gap between the $t_{2g}$ and $e_g$ orbitals. We are given three ligands to evaluate: $NCS^-$, $F^-$, and $NH_3$. To rank them, we rely on the **spectrochemical series**, a vital tool for any chemistry student. The series dictates the following order of ligand strength: $$F^- < NCS^- < NH_3$$ ### Final Calculation With all the pieces of the puzzle in hand, we can now make the final connections. Ammonia ($NH_3$) is the strongest ligand among the three. Therefore, it will produce the largest splitting energy ($\Delta_0$), which corresponds to the absorption of the highest energy and shortest wavelength light. This perfectly matches **Peak A**. Conversely, the fluoride ion ($F^-$) is the weakest ligand. It will cause the smallest splitting, absorbing the lowest energy and longest wavelength light. This corresponds to **Peak C**. By elimination, the thiocyanate ion ($NCS^-$), which has intermediate strength, corresponds to **Peak B**. Summarizing our findings: - **A** $\rightarrow$ (iii) $[M(NH_3)_6]^{n+}$ - **B** $\rightarrow$ (i) $[M(NCS)_6]^{(-6+n)}$ - **C** $\rightarrow$ (ii) $[MF_6]^{(-6+n)}$ This elegant interplay of physics and chemistry leads us directly to the correct option, **(a)**.

Select Answer:

Visualized Solution

$Visualizing the Absorption Spectra$

$Relating Wavelength to Energy$

$Energy and Ligand Strength$

$Identifying the Ligands$

$Spectrochemical Series$

$Final Matching$

$The Way Forward$

The Sigma Insight: Bonding and Crystal field

Decoding Absorption Spectra in Coordination Complexes

Analyzing the Setup

The Master Equation

The Spectrochemical Series

Final Calculation

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