In which of the following compounds, the oxidation number of iodine is fractional?
Answer & explanation
Correct answer: option 2
The correct answer is option 2. \(I_3^–\).
To determine in which compound the oxidation number of iodine is fractional, let's analyze each compound:
1. \(IF_7\) (Iodine Heptafluoride):
Iodine can have a maximum oxidation state of +7 in \(IF_7\). Oxidation state of iodine in \(IF_7\) is \(+7\). It is not fractional.
2. \(I_3^–\) (Triiodide Ion):
The triiodide ion (\(I_3^-\)) consists of three iodine atoms with a total charge of \(-1\). Oxidation state of iodine in \(I_3^-\): Each iodine atom has an oxidation state of \(-\frac{1}{3}\). This is fractional.
3. \(IF_5\) (Iodine Pentafluoride):
Iodine in \(IF_5\) typically has an oxidation state of +5. Oxidation state of iodine in \(IF_5\) is \(+5\). It is not fractional.
4. \(IF_3\) (Iodine Trifluoride):
Iodine in \(IF_3\) typically has an oxidation state of +3. Oxidation state of iodine in \(IF_3\) is \(+3\). It is not fractional.
Based on the analysis, the compound in which the oxidation number of iodine is fractional is: (2) \(I_3^–\) (Triiodide Ion)
In \(I_3^-\), each iodine atom has an oxidation state of \(-\frac{1}{3}\), which is fractional.