Question

NO2S4O6 ka oxidation number

Answer 1

Answer:

O.N of S is X

O.N of Na is +1

O.N of O is -2

Sum of all O.N is 2×+1+(4X)+(6)×−(2)

4X−10 is equal to 0

X equals to 10/4 which equals to 2.5

his compound must have sulfur atoms with mixed oxidation states. Since it is normal for sulfur to have oxidation states of -2, 0, +2, +4, and +6,  it is most likely that there are three sulfurs with a +2 oxidation state and one sulfur that is +4.

Explanation:

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