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Question
Assign oxidation number atom in the following species.
\[\ce{Cr(OH)^Θ_4}\]
Solution
\[\ce{Cr(OH)^Θ_4}\]
Oxidation number of O = –2
Oxidation number of H = +1
\[\ce{Cr(OH)^Θ_4}\] is an ionic species.
∴ Sum of the oxidation numbers of all atoms = – 1
∴ Oxidation number of Cr + 4 × (Oxidation number of O) + 4 × (Oxidation number of H) = – 1
∴ Oxidation number of Cr + 4 × (–2) + 4 × (+1) = – 1
∴ Oxidation number of Cr – 8 + 4 = – 1
∴ Oxidation number of Cr – 4 = – 1
∴ Oxidation number of Cr = – 1 + 4
∴ Oxidation number of Cr in \[\ce{Cr(OH)^Θ_4}\] = +3
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