How to find the Oxidation Numbers for (NH4)2S (Ammonium sulfide)

To find the correct oxidation state of in (NH4)2S (Ammonium sulfide), and each element in the compound, we use a few rules and some simple math. First, since the (NH4)2S doesn’t have an overall charge (like NO3- or H3O+) we could say that the total of the oxidation numbers for (NH4)2S will be zero since it is a neutral compound. We write the oxidation number (O.N.) for elements that we know and use these to figure out oxidation number for . --------- RESOURCES ---------- GENERAL RULES Free elements have an oxidation state of zero (e.g. Na, Fe, H2, O2, S8). In an ion the all Oxidation numbers must add up to the charge on the ion. In a neutral compound all Oxidation Numbers must add up to zero. Group 1 = +1 Group 2 = +2 Hydrogen with Non-Metals = +1 Hydrogen with Metals (or Boron) = -1 Fluorine = -1 Oxygen = -2 (except in H2O2 or with Fluorine) Group 17(7A) = -1 except with Oxygen and other halogens lower in the group ---------- We know that Oxygen usually is -2 with a few exceptions. When Oxygen is in a peroxide, like H2O2 (Hydrogen peroxide), it has a charge of -1. When it is bonded to Fluorine (F) it has an oxidation number of +1. Here it is bonded to element symbol so the oxidation number on Oxygen is -2. Using this information we can figure out the oxidation number for the element in (NH4)2S.