Calculating the oxidation numbers of an atom – Examples
Finding oxidation numbers on all atoms in platinum oxide, [PtO4]2+
Platinum oxide ion PtO42+
Platinum oxide ion is polyatomic and has a total charge of +2
Oxidation number of oxygen (O):
In this ion, oxygen has an oxidation number of −2 (see ‘Rules for calculations of oxidation numbers’) Oxidation number of platinum (Pt):
To calculate oxidation number of atom Pt, let x be the oxidation number of Pt, then
x + 4 × (oxidation number of oxygen) = total charge of platinum oxide
∴ x + 4(−2) = +2
⇒ x = +10
Therefore, Pt in Platinum oxide ion has an oxidation number of +10
Oxidation number of oxygen (O):
In this ion, oxygen has an oxidation number of −2 (see ‘Rules for calculations of oxidation numbers’) Oxidation number of platinum (Pt):
To calculate oxidation number of atom Pt, let x be the oxidation number of Pt, then
x + 4 × (oxidation number of oxygen) = total charge of platinum oxide
∴ x + 4(−2) = +2
⇒ x = +10
Therefore, Pt in Platinum oxide ion has an oxidation number of +10
Finding oxidation numbers on all atoms in iridium tetroxide [IrO4]+
Iridium tetroxide ion IrO4+
Ion Iridium tetroxide is polyatomic and has a total charge of +1
Oxidation number of oxygen (O):
In this ion, oxygen has an oxidation number of -2 (see ‘Rules for calculations of oxidation numbers’) Oxidation number of Iridium (Ir):
Let x be the oxidation number of Ir, then
x + 4 × (oxidation number of oxygen) = total charge of iridium tetroxide
∴ x + 4(−2) = +1
⇒ x = +9
Therefore, Ir has an oxidation number of +9
Oxidation number of oxygen (O):
In this ion, oxygen has an oxidation number of -2 (see ‘Rules for calculations of oxidation numbers’) Oxidation number of Iridium (Ir):
Let x be the oxidation number of Ir, then
x + 4 × (oxidation number of oxygen) = total charge of iridium tetroxide
∴ x + 4(−2) = +1
⇒ x = +9
Therefore, Ir has an oxidation number of +9
Finding oxidation states on all atoms in alumino-boron carbide Al3BC
Alumino-boron Carbide Al3BC
In nature Al3BC exists in crystal structure format only. To determine the oxidation state (or oxidation number) of atoms in Al3BC, it must be treated as a hypothetical neutral compound, that is the sum of oxidation numbers on its atoms is zero. (see ‘Rules for calculations of oxidation numbers’)
As electronegativity of aluminium (Al), boron (B), and carbon (C) are in the following ascending order Al < B < C (see ‘The Periodic Table’), treating all bonds in the compound as if they were fully ionic, each Al atom would lose all 3 covalent electrons, and C atom would gain 4 electrons to completely fill its valence orbital (8 valence electrons). Therefore, it is reasonable to deduce that oxidation state of C is −4 and of Al is +3.To calculate oxidation state of boron, let x be the oxidation state of boron, then
x + 3 × (oxidation state of aluminium) + 1 × (oxidation state of carbon) = total charge of compound alumino-boron carbide Al3BC
∴ x + 3(+3) + (-4) = 0
⇒ x = -5
Therefore the oxidation state of boron in alumino-boron carbide is −5, which is a rare oxidation state for boron.