Question

ions
in an ion, the sum of the oxidation states is equal to the overall ionic charge. note that the sign of the oxidation states and the number of atoms associated with each oxidation state must be considered. in oh⁻, for example, the oxygen atom has an oxidation state of -2 and the hydrogen atom has an oxidation state of +1, for a total of (-2)+(+1)= -1.

part c
what is the oxidation state of an individual phosphorus atom in po₄³⁻?
express the oxidation state numerically (e.g., +1).
view available hint(s)

part d
what is the oxidation state of each individual carbon atom in c₂o₄²⁻?
express the oxidation state numerically (e.g., +1).
view available hint(s)

Explanation:

Step1: Set up equation for $PO_{4}^{3 - }$

Let the oxidation state of $P$ be $x$. Oxygen has an oxidation state of - 2 in most compounds. There are 4 oxygen atoms in $PO_{4}^{3 - }$. The overall charge of the ion is - 3. So, $x+4\times(-2)=-3$.

Step2: Solve for $x$ in $PO_{4}^{3 - }$ equation

$x - 8=-3$. Add 8 to both sides: $x=-3 + 8=+5$.

Step3: Set up equation for $C_{2}O_{4}^{2 - }$

Let the oxidation state of each $C$ atom be $y$. There are 2 carbon atoms and 4 oxygen atoms (with oxidation - state of - 2). The overall charge of the ion is - 2. So, $2y+4\times(-2)=-2$.

Step4: Solve for $y$ in $C_{2}O_{4}^{2 - }$ equation

$2y-8=-2$. Add 8 to both sides: $2y=-2 + 8 = 6$. Divide both sides by 2: $y = + 3$.

Answer:

Part C: +5
Part D: +3