assign oxidiation numbers to the underline elements each of the following species (a) k2Mno4 (b)H2S2O7
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(a) KMnO4
In KMnO4, the oxidation number of Mn can be found out by taking it as x
ON of K = +1
ON of O = -2
Since, KMnO4 is stable,
1+x-8 = 0 ------> x = +7
(b) H2S2O7
In H2S2O7, the oxidation number of S can be found out by taking it as x
ON of H = +1
ON of O = -2
Since H2S2O7 is stable,
2+2x-14 = 0 --------> 2x = 12 ------> x = +6
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