Question

Write the formula for a complex formed between cr3 and 2,2\'-bipyridine (c10h8n2), with a coordination number of 6.

Answer 1

Answer: The formula of the complex will be
`[Cr(C_(10)H_8N_2)_3]^(3+)\text{ or }[Cr(bpy)_3]^(3+)`

Step-by-step explanation:

Coordination number in a complex is defined as the number of ligands that are attached to the central metal atom.

• If the ligand is unidentate, it will occupy 1 place around the central metal atom.
• If the ligand is bidentate, it will occupy 2 places around the central metal atom.
• If the ligand is tridentate, it will occupy 3 places around the central metal atom.

We are given:

A metal ion having formula
`Cr^(3+)` and ligand attached to it is 2,2'-bipyridine (shortly named as 'bpy')

Coordination number of the complex is 6

We know that, 'bpy' is a bidentate ligand and does not carry any charge on it

Hence, the formula of the complex will be
`[Cr(C_(10)H_8N_2)_3]^(3+)\text{ or }[Cr(bpy)_3]^(3+)`

Answer 2

The reaction is:

Cr₃⁺ +3 C₁₀H₈N₂ → [Cr(C₁₀H₈N₂)₃]³⁺

2,2\'-bipyridine is bidentate ligand and neutral. It exhibits two N terminal, thus, it functions as bidentate ligand.

Therefore, three 2,2\'-bipyridine ligand fulfills the six coordination number of Cr₃⁺.

The formula for the complex is [Cr(C₁₀H₈N₂)₃]³⁺