Sum of oxidation state (magnitude) and coordination number of cobalt in $$Na[Co(bpy)Cl_4]$$ is

Explanation: The coordination number is the total number of coordinate bonds formed between the metal ion and ligands.

The ligand $$\mathrm{bpy}$$ (2,2'-bipyridine) is a neutral bidentate ligand and forms two coordinate bonds.

Each chloride ligand $$\mathrm{Cl^-}$$ is monodentate and forms one coordinate bond.

Therefore:

$$\mathrm{CN = (1 \times 2) + (4 \times 1) = 6}$$

The compound is $$\mathrm{Na[Co(bpy)Cl_4]}$$.

Since sodium exists as $$\mathrm{Na^+}$$, the complex ion carries a charge of $$\mathrm{-1}$$:

$$\mathrm{[Co(bpy)Cl_4]^-}$$

Let the oxidation state of cobalt be $$\mathrm{x}$$.

The ligand $$\mathrm{bpy}$$ is neutral and each chloride contributes $$\mathrm{-1}$$ charge.

Thus:

$$\mathrm{x + (4 \times -1) = -1}$$

$$\mathrm{x - 4 = -1}$$

$$\mathrm{x = +3}$$

Hence, the oxidation state is $$\mathrm{3}$$ and the coordination number is $$\mathrm{6}$$.

Required sum:

$$\mathrm{3 + 6 = 9}$$

Correct Answer: $$\mathrm{9}$$