Question

A transition metal (M) among Mn, Cr, Co and Fe has the highest standard electrode potential \( (M^{3+} / M^{2+}) \). It forms a metal complex of the type \( [M(CN)_6]^{4-} \). The number of electrons present in the \( e_g \) orbital of the complex is ________.

Hint:

In coordination chemistry, the number of electrons in the \( e \)-orbital of a complex depends on its d-orbital splitting. This splitting occurs when ligands interact with the central metal ion, influencing the number of available electrons in the \( e \)-orbital.

Answer 1

Correct Answer: 1

This problem requires two steps. Firstly, identify a specific transition metal (M) from a provided list by examining its standard electrode potential. Secondly, utilize Crystal Field Theory to determine the number of electrons occupying the \( e_g \) orbitals within the metal's cyanide complex, \( [\text{M(CN)}_6]^{4-} \).

Concepts Applied:

1. Standard Electrode Potential (\( E^\circ \)): A measure of a species' tendency to undergo reduction. For the \( M^{3+}/M^{2+} \) couple, a more positive \( E^\circ \) signifies a greater propensity for \( M^{3+} \) to reduce to \( M^{2+} \).
2. Oxidation State Determination: The sum of oxidation states for the central metal ion and its ligands equals the complex ion's overall charge. The cyanide ligand (\( \text{CN}^- \)) carries a -1 charge.
3. Crystal Field Theory (CFT): In octahedral complexes, the five d-orbitals split into two energy levels: the lower \( t_{2g \) (three orbitals) and the higher \( e_g \) (two orbitals).
4. Ligand Field Strength and Spin State: Cyanide (\( \text{CN}^- \)) is a strong-field ligand, inducing a significant energy gap (\( \Delta_o \)) between \( t_{2g \) and \( e_g \) orbitals. This results in low-spin complexes where electrons preferentially pair in \( t_{2g \) before occupying \( e_g \).

Solution Procedure:

Step 1: Metal Identification.

The transition metal (M) is the one exhibiting the highest standard electrode potential for the \( M^{3+}/M^{2+} \) couple among Mn, Cr, Co, and Fe. Standard reduction potentials are:

• \( E^\circ(\text{Mn}^{3+}/\text{Mn}^{2+}) = +1.51 \, \text{V} \)
• \( E^\circ(\text{Cr}^{3+}/\text{Cr}^{2+}) = -0.41 \, \text{V} \)
• \( E^\circ(\text{Co}^{3+}/\text{Co}^{2+}) = +1.82 \, \text{V} \)
• \( E^\circ(\text{Fe}^{3+}/\text{Fe}^{2+}) = +0.77 \, \text{V} \)

Cobalt (Co) possesses the highest standard electrode potential. Thus, M = Co.

Step 2: Oxidation State of Cobalt.

For the complex \( [\text{Co(CN)}_6]^{4-} \), let the oxidation state of Cobalt be \( x \).

\[x + 6 \times (\text{charge of CN}^-) = \text{overall charge}\]\[x + 6 \times (-1) = -4\]\[x - 6 = -4\]\[x = +2\]

The central metal ion is \( \text{Co}^{2+} \).

Step 3: Electronic Configuration of Cobalt Ion.

Cobalt (Co) has atomic number 27. Its neutral atom configuration is \( [\text{Ar}] \, 3d^7 4s^2 \).

Removing two electrons from the 4s orbital yields the \( \text{Co}^{2+} \) ion:

\[\text{Co}^{2+}: [\text{Ar}] \, 3d^7\]

This is a \( d^7 \) configuration.

Step 4: Electron Distribution via Crystal Field Theory.

The complex \( [\text{Co(CN)}_6]^{4-} \) is octahedral and features a strong-field ligand (\( \text{CN}^- \)), resulting in a low-spin configuration. We fill the 7 d-electrons into the \( t_{2g \) and \( e_g \) orbitals prioritizing pairing in \( t_{2g \):

1. The first three electrons occupy the \( t_{2g \) orbitals individually.
2. The subsequent three electrons pair up in the \( t_{2g \) orbitals, completing the \( t_{2g \) set with 6 electrons.
3. The final electron fills one of the \( e_g \) orbitals.

The resulting configuration is \( (t_{2g})^6 (e_g)^1 \).

Step 5: Final Result.

Based on the \( (t_{2g})^6 (e_g)^1 \) configuration, there is 1 electron in the \( e_g \) orbitals.

The number of electrons in the \( e_g \) orbital of the complex is 1.