Determine the oxidation number of each element in the...

Determine the oxidation number of each element in the following ions or compounds,
(a) $\mathrm{PF}_{6}$
(b) $\mathrm{H}_{2} \mathrm{As} \mathrm{O}_{4}^{-}$
(c) $\mathrm{UO}^{2+}$
(d) $\mathrm{N}_{2} \mathrm{O}_{5}$
(e) $\mathrm{POCl}_{3}$
(f) $\mathrm{XeO}_{4}^{2-}$

Key Concepts

Summation of Oxidation Numbers in Ions and Molecules

In any chemical species, the sum of the oxidation numbers of all atoms must equal the overall charge of the molecule or ion. In a neutral molecule, the oxidation numbers sum to zero, whereas in a polyatomic ion they sum to the ion's overall charge. This rule is critical to systematically determine unknown oxidation numbers in complex compounds or ions.

Electronegativity and Electron Distribution

Electronegativity plays an implicit role in oxidation state assignments, as more electronegative atoms are assumed to attract electrons more strongly. This assumption helps determine how electrons are 'assigned' in a bond when applying oxidation number rules, even if the real electron distribution is covalent. The most electronegative atoms typically receive the negative oxidation states in compounds.

Oxidation State

The oxidation state (or oxidation number) is a formalism that represents the hypothetical charge an atom would have if all bonds were assumed to be completely ionic. It helps in bookkeeping the transfer of electrons in redox (oxidation-reduction) reactions and in understanding the electron distribution among atoms in a molecule or ion.

Standard Rules for Assigning Oxidation Numbers

A set of agreed-upon rules is used to assign oxidation numbers. For example, elements in their elemental form have an oxidation number of zero; in compounds, oxygen is usually assigned an oxidation number of -2 and hydrogen +1 (with specific exceptions). Additionally, when elements form compounds, certain elements like fluorine have fixed oxidation states (e.g., fluorine is always -1). These rules simplify the process of determining relative electron density in chemical species.

Transcript

00:01 Hello everyone, we're going to practice assigning oxidation numbers.

00:04 I went through these very brief rules on my last video, so we're just going to get right into writing our, or figuring out our oxidation numbers.

00:16 Our first substance is pf6, phosphorus, hexafluoride.

00:23 Looking at our rules, i've got to figure out my p and my f.

00:28 F is going to be negative 1.

00:30 And if f has a negative 1 charge, that means i have a net charge, an overall charge of negative 6.

00:38 Since i only have one phosphorus, the charge of my phosphorus has to be positive 6.

00:44 There's our first example.

00:47 Our second example, we are given h2, as, 04 with a negative 1 charge.

00:57 So i have hydrogen, arsenic, and oxygen.

01:03 The oxygen is going to equal negative 2.

01:08 So let's put that right here.

01:10 We get a negative 8 total.

01:16 And the sum of the hydrogen, the arsenic, and the oxygen has to equal the charge of our ion.