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

The problem is a two-part question. First, we must identify a specific transition metal (M) from a given list based on its standard electrode potential. Second, we need to determine the number of electrons in the \( e_g \) orbitals of the metal's cyanide complex, \( [\text{M(CN)}_6]^{4-} \), using Crystal Field Theory.

Concept Used:

1. Standard Electrode Potential (\( E^\circ \)): This value measures the tendency of a chemical species to be reduced. For the \( M^{3+}/M^{2+} \) couple, a higher (more positive) \( E^\circ \) value indicates a greater tendency for the \( M^{3+} \) ion to be reduced to the \( M^{2+} \) ion.
2. Oxidation State Calculation: The sum of the oxidation states of the central metal ion and the ligands in a complex ion equals the overall charge of the ion. The cyanide ligand (\( \text{CN}^- \)) has a charge of -1.
3. Crystal Field Theory (CFT): In an octahedral complex, the five d-orbitals of the central metal ion split into two sets: a lower-energy set of three orbitals called \( t_{2g} \) and a higher-energy set of two orbitals called \( e_g \).
4. Ligand Field Strength and Spin State: The cyanide ion (\( \text{CN}^- \)) is a strong-field ligand. It causes a large energy splitting (\( \Delta_o \)) between the \( t_{2g} \) and \( e_g \) orbitals. This forces electrons to pair up in the lower-energy \( t_{2g} \) orbitals before occupying the higher-energy \( e_g \) orbitals, resulting in a low-spin complex.

Step-by-Step Solution:

Step 1: Identify the transition metal (M).

We are given that M has the highest standard electrode potential for the \( M^{3+}/M^{2+} \) couple among Mn, Cr, Co, and Fe. Let's list the standard reduction potentials for these metals:

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

Comparing the values, Cobalt (Co) has the highest standard electrode potential. Therefore, M = Co.

Step 2: Determine the oxidation state of Cobalt in the complex.

The complex is given as \( [\text{Co(CN)}_6]^{4-} \). Let the oxidation state of Cobalt be \( x \).

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

So, the central metal ion is \( \text{Co}^{2+} \).

Step 3: Determine the electronic configuration of the central metal ion.

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

For the \( \text{Co}^{2+} \) ion, we remove the two outermost electrons (from the 4s orbital):

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

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

Step 4: Apply Crystal Field Theory to find the electron distribution in the \( d \)-orbitals.

The complex \( [\text{Co(CN)}_6]^{4-} \) is an octahedral complex with a strong-field ligand (\( \text{CN}^- \)). Therefore, it will be a low-spin complex. We need to fill the 7 d-electrons into the \( t_{2g} \) and \( e_g \) orbitals according to the low-spin configuration (pairing electrons in \( t_{2g} \) first).

The filling proceeds as follows:

1. The first three electrons occupy the \( t_{2g} \) orbitals singly.
2. The next three electrons pair up with the electrons already in the \( t_{2g} \) orbitals. This fills the \( t_{2g} \) level with 6 electrons.
3. The seventh and final electron must go into the higher-energy \( e_g \) orbitals.

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

Step 5: State the final answer.

From the electronic configuration \( (t_{2g})^6 (e_g)^1 \), we can see that there is 1 electron in the \( e_g \) orbitals.

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