Transition Elements

The strong oxidising agent is Co .
The number of unpaired electrons in Co ( ) is 4.
The spin-only magnetic moment is given by:
Substituting :

Manganese (Mn) has the highest oxidation state (+7, as in MnO₄^-) in the 3d series metals. It also exhibits multiple other oxidation states.

To determine the correct set of ions in an aqueous solution that have the same color, we need to consider the typical colors of transition metal ions. Transition metals often exhibit characteristic colors due to d-d transitions of electrons. Let's analyze each ion in the provided options:

1. , , :
   • : Pale violet or lavender color
   • : Typically green in aqueous solutions, but can appear violet as well depending on the ligands present.
   • : Often appears pale purple.
2. , , :
   • : Colorless because it's not a transition metal.
   • : Typically green.
   • : Usually yellow or brown in water.
3. , , :
   • : Colorless (not a transition metal).
   • : Typically blue.
   • : Pale pink.
4. , , :
   • : Colorless.
   • : Purple.
   • : Blue.

Based on the above reasoning, the correct answer is the first option, where , , and can appear in similar shades (pale violet, lavender, and purple). Therefore, the correct set is:

, ,

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 orbitals of the metal's cyanide complex, , using Crystal Field Theory.

Concept Used:

1. Standard Electrode Potential ( ): This value measures the tendency of a chemical species to be reduced. For the couple, a higher (more positive) value indicates a greater tendency for the ion to be reduced to the 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 ( ) 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 and a higher-energy set of two orbitals called .
4. Ligand Field Strength and Spin State: The cyanide ion ( ) is a strong-field ligand. It causes a large energy splitting ( ) between the and orbitals. This forces electrons to pair up in the lower-energy orbitals before occupying the higher-energy 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 couple among Mn, Cr, Co, and Fe. Let's list the standard reduction potentials for these metals:

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 . Let the oxidation state of Cobalt be .

So, the central metal ion is .

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 .

For the ion, we remove the two outermost electrons (from the 4s orbital):

This is a system.

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

The complex is an octahedral complex with a strong-field ligand ( ). Therefore, it will be a low-spin complex. We need to fill the 7 d-electrons into the and orbitals according to the low-spin configuration (pairing electrons in first).

The filling proceeds as follows:

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

The resulting electronic configuration is .

Step 5: State the final answer.

From the electronic configuration , we can see that there is 1 electron in the orbitals.

The number of electrons present in the orbital of the complex is 1.