To identify the oxidation numbers of hydrogen and oxygen in compounds. Hydrogen

Question

To identify the oxidation numbers of hydrogen and oxygen in compounds.

Hydrogen has three possible oxidation numbers:

1 when bonded to a metal,

0 when in elemental form, H2, and

+1 when bonded to a nonmetal.

Oxygen has several possible oxidation numbers:

2 when bonded to a nonmetal (other than fluorine) or as the oxide ion, O2,

1 as the peroxide ion, O22,

1/2 as the superoxide ion, O2, and

0 when in elemental form, O2 or O3.

Disproportionation is a type of redox reaction in which the same element is both oxidized and reduced. For example, H2 can undergo disproportionation because some of the H atoms can be reduced from 0 to 1 and some can be oxidized from 0 to +1. However, hydrogen in the +1 state cannot undergo disproportionation because it can only be reduced, not oxidized. Similarly, hydrogen in the 1 state can only be oxidized, not reduced.

1. Classify each of the following binary compounds by the oxidation number of oxygen.

Drag each item to the appropriate bin.

-2

-1

-1/2

0

2. Which of the following forms of oxygen can undergo disproportionation?

Check all that apply.

Explanation / Answer

(1)

CaO

O.S = -2, Oxide

KO2

O.S = -1/2, superoxide

O2

O.S = 0, element

CO2

O.S = -2, oxide

K2O

O.S = -2 , oxide

CaO2

O.S = -1, peroxide

K2O2

O.S = -1, Peroxide

O3

O.S = 0, element

(2)

Peroxide can undergo dispropotionation reaction. Here O- can be oxidised to zero and can be reduced to -2.

Related Questions

Navigate

• Browse (All)
• Browse T
• Subjects

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.