Peroxides, hydrides and Roman numerals

By the end of this lesson

By the end you can handle the standard exceptions to the oxygen and hydrogen rules, work out an average oxidation number when it is fractional, and read or write a Roman numeral in a compound's name.

When oxygen does not take minus two

The rule "oxygen is $-2$ " comes from oxygen pulling electrons away from its partner. Sometimes that picture breaks.

In a peroxide, two oxygen atoms are bonded to each other: $\text{O}-\text{O}$ . Each oxygen now shares with another oxygen as well as with the other element, so it does not gain two whole electrons.

The result is an oxidation number of $-1$ for each oxygen in a peroxide.

Compound	Oxidation number of oxygen	Why
$\text{H}_2\text{O}$	$-2$	usual case
$\text{H}_2\text{O}_2$	$-1$	peroxide, O–O bond
$\text{Na}_2\text{O}_2$	$-1$	peroxide, O–O bond
$\text{OF}_2$	$+2$	fluorine is more electronegative, so oxygen loses

In $\text{OF}_2$ , fluorine is $-1$ by its own rule. Two fluorines give $-2$ in total, so oxygen must be $+2$ to make the neutral compound add to zero. Oxygen is the less electronegative partner here, which is rare.

Worked example: Worked example: hydrogen peroxide

Show that the oxidation number of oxygen in $\text{H}_2\text{O}_2$ is $-1$ .

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When hydrogen does not take plus one

Hydrogen is $+1$ when it is bonded to a more electronegative non-metal, which is most of the time.

With a reactive metal, the positions reverse. Sodium is much less electronegative than hydrogen, so in sodium hydride the shared electrons sit closer to hydrogen.

Hydrogen becomes $-1$ .

Compound	Oxidation number of hydrogen
$\text{HCl}$	$+1$
$\text{CH}_4$	$+1$
$\text{NaH}$	$-1$
$\text{CaH}_2$	$-1$

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