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What are the oxidation number of the underlined elements in each of the following and how do you rationalise your results? Kul(l)_(3) |
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Answer» SOLUTION :In `Kl_(3)`, the oxidation number of K is +1. Therefore, the average oxidation number of in it will be `-(1)/(3)`. SINCE, the average oxidation number is fractional, let us consider its structure as given below. `K^(+)[l-1larrl]^(-)` The structure reveals that a coordinate BOND is formed between `l_(2)` MOLECULE and `l^(-)` ion. Since, `l_(2)` is is a neutral molecule, the oxidation number of each I atom forming `l_(2)` molecule is zero. The oxidation number of l in `l^(-)` ion is -1. Thus, we have `K^(+)[overset(0)l-overset(0)llarroverset(-1)l]^(-)` |
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