Question

Write the formula to calculate EAN. Explain formation of [CO(NH\(_3\))\(_6\)]\(^{3+}\) complex ion with respect to : (i) Type of hybridisation (ii) Magnetic property

Hint:

The strength of the ligand determines whether a complex will be inner orbital (low-spin) or outer orbital (high-spin). Strong-field ligands like NH\(_3\) and CN\(^-\) force electrons to pair up, using inner d-orbitals for hybridisation. Weak-field ligands like H\(_2\)O and Cl\(^-\) do not, leading to the use of outer d-orbitals.

Answer 1

Formula for EAN (Effective Atomic Number)
\[ \text{EAN} = Z - X + Y \]
where:
- \( Z \) = Atomic number of the central metal atom.
- \( X \) = Number of electrons lost during the formation of the metal ion from its atom.
- \( Y \) = Number of electrons donated by the ligands.

Formation of [Co(NH\(_3\))\(_6\)]\(^{3+}\)
Step 1: Determine the oxidation state and electron configuration of Cobalt.
Cobalt (Co) has Z = 27. Its configuration is [Ar] 3d\(^7\) 4s\(^2\).
In the complex, NH\(_3\) is neutral, so the oxidation state of Co is +3.
Co\(^{3+}\) has the configuration [Ar] 3d\(^6\).

Step 2: Consider the ligand and its effect.
Ammonia (NH\(_3\)) is a strong-field ligand. It causes the pairing of the 3d electrons of Co\(^{3+}\). The six electrons in the 3d orbitals will pair up in three orbitals.
Step 3: Determine hybridisation and magnetic property.

[(i)] Type of hybridisation: For the formation of six coordinate bonds with six NH\(_3\) molecules, the Co\(^{3+}\) ion uses two empty 3d orbitals, one 4s orbital, and three 4p orbitals. This results in d\(^2\)sp\(^3\) hybridisation, forming an inner orbital or low-spin octahedral complex.
[(ii)] Magnetic property: Since all six 3d electrons are paired up due to the strong-field ligand, there are no unpaired electrons. Therefore, the complex is diamagnetic.