Question
What are the oxidation number of the underlined elements in the following and how do you rationalise your results?
$KI_3$
$KI_3$
✓
Answer
In $KI_3$, the oxidation number (O.N.) of K is +1. Hence, the average oxidation number of I is $-\frac{1}{3}.$ However, O.N. cannot be fractional. Therefore, we will have to consider the structure of $KI_3$ to find the oxidation states. In a $KI_3$ molecule, an atom of iodine forms a coordinate covalent bond with an iodine molecule.
$\stackrel{{+1}}{\ \ \ \hbox{k}^+}\Big[\stackrel{{0}}{\hbox{I}}-\stackrel{{0}}{\hbox{I}}\leftarrow\stackrel{{-1}}{\hbox{I}}\Big]^-$
Hence, in a $KI_3$ molecule, the O.N. of the two I atoms forming the $I_2$ molecule is 0, whereas the O.N. of the I atom forming the coordinate bond is -1.
$\stackrel{{+1}}{\ \ \ \hbox{k}^+}\Big[\stackrel{{0}}{\hbox{I}}-\stackrel{{0}}{\hbox{I}}\leftarrow\stackrel{{-1}}{\hbox{I}}\Big]^-$
Hence, in a $KI_3$ molecule, the O.N. of the two I atoms forming the $I_2$ molecule is 0, whereas the O.N. of the I atom forming the coordinate bond is -1.
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