assign oxidation numbers to
PF-6(aq)
Is that PF6^-?
F is in group VIIA (or 17) and has an oxidation state of -1. That makes -6 total for F. To leave -1 on the ion, P must have an oxidation state of +5. Check my work.
Here is a site that contains some very simple rules for assigning oxidation states.
(Broken Link Removed)
To assign oxidation numbers to each element in the PF6- ion, you need to follow some rules:
1. The oxidation number of an element in its elemental form is always zero. Therefore, the oxidation number of P (phosphorus) is 0.
2. The sum of the oxidation numbers in a compound must equal the charge of the compound. The charge of the PF6- ion is -1.
3. Fluorine (F) usually has an oxidation number of -1, except when it reacts with more electronegative elements like oxygen or fluorine itself.
Based on these rules, we can work out the oxidation numbers:
- Let the oxidation number of P be "x."
- The overall charge of the PF6- ion is -1.
- There are six fluorine atoms, each with an oxidation number of -1, totaling -6.
So, the sum of the oxidation numbers must be equal to the charge:
x + (-1) x 6 = -1
Simplifying the equation, we get:
x - 6 = -1
Adding 6 to both sides:
x = 5
Therefore, the oxidation number of phosphorus (P) in the PF6- ion is +5.