Calculate the Oxiidation Number of atoms in the compounds
a) NaAuCl4
b) OF2
c) NH2OH
d) Ca(ClO2)2

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Solution

a) NaAuCl₄:
Na always exist in +1 oxidation state.
Cl always exist as -1 oxidation state.
So, let the oxidation state of Au=x
Hence, x+1-4=0 (This is equal to zero because there is net zero charge over the compound)
x=+3
b)OF₂:
F always have -1 oxidation state.
Hence oxidation state of oxygen=+2
c)NH₂OH:
oxidation state of oxygen=-2
Oxidation state of H= +1
Let oxidation state of nitrogen = x
x+2-2+1 = 0
x=-1
Oxidation state of N=-1
d) Ca(ClO₂)₂:
Oxidation state of Ca=+2
Oxidation state of oxygen=-2
ClO₂^-= Oxidation number of Cl + 2(Oxidation number of Oxygen)=-1
Hence, oxidation number of Cl = +3

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