Question

Write the oxidation numbers of the individual ions within the given neutral compounds.
1.HPF4
2. KMnO4

Answer 1

Points to remember :

Oxidation no of some elements in combined state .

• F = -1
• H = 1 [ In ionic hydrides = -1 ]
• Alkali metals = +1  
• Alkaline earth metal = +2
• O = -2 [ In peroxide = -1 ]
• Halogens = -1
• Sulphur = - 2 [ In sulphides ]

The algebraic sum of oxidation numbers of all the elements in a compound is zero .

Solution :

(1) HPF 4

$\implies H + P + 4 (F) = 0$

here ,

• Oxidation number of H = + 1
• Oxidation number  of F = - 1

⇒ 1 + P - 4 = 0

⇒ P - 3 = 0

⇒ P = + 3

Oxidation number of P = + 3

(2) KMnO4

$\implies K + Mn + 4 (O) = 0$

here ,

• Oxidation number of K = + 1 [ Alkali metal ]
• Oxidation number  of O = - 2

⇒ 1 + Mn  + 4 x -2  = 0

⇒ 1 + Mn - 8  = 0

⇒ Mn - 7 = 0

⇒ Mn  = + 7

Oxidation number of Mn is + 7