Question

experimentally it was found that a metal oxide has formula M0.98O.Metal M is present as M2+ & M3+ in its oxide, fraction of the metal which exit as M3+ would be​

Answer 1

Answer:

Your answer is 0.04

Explanation:

Since the oxidation state of oxygen is −2. So, for M 0.980 to be neutral, the total oxidation state of M 0.98has to be +2.

Let the fraction of M 3+be x.

Then fraction of M 2+will be (0.98−x).

Now for the compound to be neutal,

3x+2(0.98−x)=2

3x+1.96−2x=2

x=2−1.96

x=0.04

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