Question

what is the oxidation number of the underlind element and how to rationalize the result KI3? Iodine is underlined ​

Answer 1

Answer:

(a) KI3  In KI3, the oxidation number (O.N.) of K is +1. Hence, the average oxidation number of I is  - 1/3  However, O.N. cannot be fractional. Therefore, we will have to consider the structure of KI3 to find the oxidation states. In a KI3 molecule, an atom of iodine forms a coordinate covalent bond with an iodine molecule.  Hence, in a KI3 molecule, the O.N. of the two I atoms forming the I2 molecule is 0, whereas the O.N. of the I atom forming the coordinate bond is –1.  (b) H2S4O6  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.  (c) Fe3O4  On taking the O.N. of O as –2, the O.N. of Fe is found to be .  +2(2/3)  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.  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 –2 in CH3COOH. Read more on Sarthaks.com - https://www.sarthaks.com/8792/what-are-oxidation-numbers-underlined-elements-each-following-rationalise-your-results

Explanation: