How to find the Oxidation Number for Xe in XeOF4 (Xenon oxytetrafluoride)

Wayne Breslyn
Wayne Breslyn
8.4 هزار بار بازدید - 6 سال پیش - To find the correct oxidation
To find the correct oxidation state of Xe in XeOF4 (Xenon oxytetrafluoride), and each element in the molecule, we use a few rules and some simple math.

First, since the XeOF4 molecule doesn’t have an overall charge (like NO3- or H3O+) we could say that the total of the oxidation numbers for XeOF4 will be zero since it is a neutral molecule.

We write the oxidation number (O.N.) for elements that we know and use these to figure out oxidation number for Xe.

----------
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 +2.

Here it is bonded to Xe so the oxidation number on Oxygen is -2. Using this information we can figure out the oxidation number for the element Xe in XeOF4.
6 سال پیش در تاریخ 1397/09/23 منتشر شده است.
8,447 بـار بازدید شده
... بیشتر