Total Number of N Co O Bond Angles in Co EDTA Complex

In the octahedral complex ion $[\text{Co}(\text{EDTA})]^{1-}$, the ethylenediaminetetraacetate ion ($\text{EDTA}^{4-}$) acts as a hexadentate ligand. It coordinates to the central $\text{Co}^{3+}$ ion through:

• $2$ Nitrogen ($\text{N}$) donor atoms
• $4$ Oxygen ($\text{O}$) donor atoms

The coordination number of the central cobalt ion is $6$, forming an octahedral geometry with a total of $6$ coordinating atoms: $N_1, N_2$ and $O_1, O_2, O_3, O_4$.

The total number of bond angles formed between any pair of donor atoms and the central metal atom in an octahedral complex of type $\text{MA}_2\text{B}_4$ (where $\text{A} = \text{N}$ and $\text{B} = \text{O}$) is given by: $\binom{6}{2} = \frac{6 \times 5}{2} = 15\text{ total bond angles}$

These $15$ bond angles can be categorized as:

1. $\text{N}-\text{Co}-\text{N}$ bond angles: $\binom{2}{2} = 1$

2. $\text{O}-\text{Co}-\text{O}$ bond angles: $\binom{4}{2} = \frac{4 \times 3}{2} = 6$

3. $\text{N}-\text{Co}-\text{O}$ bond angles: $\text{Number of N atoms} \times \text{Number of O atoms} = 2 \times 4 = 8$

Thus, the total number of $\text{N}-\text{Co}-\text{O}$ bond angles in $[\text{Co}(\text{EDTA})]^{1-}$ is $8$.