Answer: C. Iron(III) is \(\text{Fe}^{3+}\); the ethanedioate ion is \(\text{C}_2\text{O}_4^{2-}\) (from \(\text{Na}_2\text{C}_2\text{O}_4\), where two Na⁺ balance the 2− charge).
Charge balance: the lowest common multiple of 3 and 2 is 6, so we need 2 \(\text{Fe}^{3+}\) (total +6) and 3 \(\text{C}_2\text{O}_4^{2-}\) (total −6). The formula is \(\text{Fe}_2(\text{C}_2\text{O}_4)_3\).
A and B use the wrong charge ratios. D inverts the ratio (3:2 instead of 2:3) — that would balance \(\text{Fe}^{2+}\) with a 3− anion, not iron(III).
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