To find the oxidation state of the metal ion in Co(ox)₃₄, we first need to understand what ‘ox’ represents. In this context, ‘ox’ refers to the oxalate ion (C₂O₄²⁻). Since there are four oxalate ions, we need to account for their total charge.
The oxalate ion, C₂O₄²⁻, carries a charge of -2. Therefore, for four oxalate ions, the total contribution to the charge is:
- 4 (C₂O₄²⁻) = 4 * -2 = -8
Let x be the oxidation state of the cobalt ion (Co). The overall charge of the complex Co(ox)₃₄ is neutral (0). Hence, we can set up the equation:
x + (-8) = 0
Simplifying this, we find:
- x – 8 = 0
- x = +8
Therefore, the oxidation state of the cobalt ion in Co(ox)₃₄ is +8.