What are the oxidation numbers of the underlined elements in each
of the following and how do you rationalise your results?
(a) KI3 (b)
In KI3, the oxidation
number (O.N.) of K is +1. Hence, the average oxidation number of
I is . However, O.N.
cannot be fractional. Therefore, we will have to consider the
structure of KI3to find the
In a KI3molecule, an
atom of iodine forms a coordinate covalent bond with an iodine
Hence, in a
KI3molecule, the O.N. of the
two I atoms forming the
I2molecule is 0, whereas the
O.N. of the I atom forming the coordinate bond is
However, O.N. cannot be fractional. Hence, S must be
present in different oxidation states in the molecule.
The O.N. of two of the four S atoms is +5 and the O.N. of
the other two S atoms is 0.
On taking the O.N. of O as â€“2, the O.N.
of Fe is found to be. However, O.N.
cannot be fractional.
Here, one of the three Fe atoms exhibits the O.N. of +2 and
the other two Fe atoms exhibit the O.N. of +3.
2 (x) + 6 (+1) + 1 (-2) = 0
or, 2x + 4 = 0
or, x = -2
Hence, the O.N. of C is â€“2.
2 (x) + 4 (+1) + 2 (-2) = 0
or, 2x = 0
or, x = 0
However, 0 is average O.N. of C. The two carbon atoms
present in this molecule are present in different environments.
Hence, they cannot have the same oxidation number. Thus, C
exhibits the oxidation states of +2 and â€“2 in
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