1.4 · Bonding & Lewis

Oxidation states

Electron bookkeeping that assumes every bond is ionic.

By the end you should be able to:

Assign oxidation states

Key idea

What an oxidation state means

The oxidation state (oxidation number) is the charge an atom would carry if every bond were fully ionic: each bonding pair is given entirely to the more electronegative atom, and a bond between two identical atoms is split evenly.

It is bookkeeping for redox chemistry: an increase in oxidation state is oxidation, a decrease is reduction.

Write the sign first for oxidation states (Cr is +6 in dichromate) and last for ionic charges ( $\ce{Cr^3+}$ ).

Method

Oxidation-state rules, in priority order

Work down the list; when two rules conflict, the higher one wins.

An atom in a free element is 0 ( $\ce{O2}$ , $\ce{Na}$ , $\ce{S8}$ ).
A monatomic ion equals its charge ( $\ce{Fe^3+}$ is +3, $\ce{Cl-}$ is −1).
F is −1 in all compounds.
Group 1 metals are +1 and group 2 metals are +2 in compounds.
H is +1 with nonmetals but −1 in metal hydrides ( $\ce{NaH}$ , $\ce{CaH2}$ ).
O is −2, except in peroxides (−1: $\ce{H2O2}$ , $\ce{Na2O2}$ ), superoxides ( $-\tfrac{1}{2}$ : $\ce{KO2}$ ) and $\ce{OF2}$ (+2).
The oxidation states add up to the overall charge (0 for a neutral compound); solve for the unknown atom.

Other halogens are usually −1, but positive when bonded to O or to a lighter halogen (Cl is +7 in $\ce{ClO4-}$ ).

Key idea

Oxidation state vs formal charge

Both hand out bonding electrons to atoms, but from opposite extremes:

	Formal charge	Oxidation state
Bonding electrons	split equally (pure covalent)	all to the more electronegative atom (pure ionic)
Used for	choosing Lewis structures	tracking redox reactions
C in $\ce{CO}$	−1	+2
O in $\ce{CO}$	+1	−2
N in $\ce{NH4+}$	+1	−3

Each set adds up to the overall charge, but neither is the real partial charge on the atom.

Common mistake

Oxidation-state traps

Wrong: O is −2 in $\ce{H2O2}$ , so H must be +2. Right: $\ce{H2O2}$ is a peroxide (O–O bond): O is −1 and H is +1.

Wrong: H is +1 in $\ce{NaH}$ . Right: in metal hydrides H is −1, because the metal is less electronegative (rule 4 outranks rule 5).

Wrong: oxidation states must be whole numbers. Right: averages can be fractional: Fe in $\ce{Fe3O4}$ is $+\tfrac{8}{3}$ and O in $\ce{KO2}$ is $-\tfrac{1}{2}$ .

Wrong: writing the oxidation state of Cr in $\ce{Cr2O7^2-}$ as +12. Right: +12 is the total for two Cr atoms; each Cr is +6.

Worked example

Worked example: Cr in Cr₂O₇²⁻

Let the oxidation state of Cr be $x$ . O is −2 (no O–O bonds), and the total must equal the ion's charge:

$2x + 7(-2) = -2 \;\Rightarrow\; 2x = 12 \;\Rightarrow\; x = +6$

Answer: each chromium is +6.

Worked example

Practice set with answers

Species	Atom	Working	Oxidation state
$\ce{MnO4-}$	Mn	$x - 8 = -1$	+7
$\ce{SO4^2-}$	S	$x - 8 = -2$	+6
$\ce{NH4+}$	N	$x + 4 = +1$	−3
$\ce{S2O3^2-}$	S (average)	$2x - 6 = -2$	+2
$\ce{OF2}$	O	$x - 2 = 0$	+2
$\ce{CH4}$	C	$x + 4 = 0$	−4
$\ce{CO2}$	C	$x - 4 = 0$	+4

Check yourself

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