Oxidation state of S atom in H2SO3?

Answer 1

+5

Answer 2

IT has an oxidation number of +4.

Explanation: Let's assume the oxidation state of Sulfur to be 'x'. Now, hydrogen always has an oxidation state of +1 (Except when it is bonded with 1st group elements like Lithium, Sodium, etc- it has a valency of -1) and Oxygen always has an oxidation state of -2 (Except in peroxides and superoxides - it has an oxidation state of -1).

Now, the total charge on the compound is zero. (Since there is no indication on top of the compound). Thus, the sum of oxidation states of all the elements in the compound will be zero.

I.e., 2 atoms of Hydrogen plus 1 atom of Sulfur plus 3 atoms of Oxygen will be :

(+1x2) + (x) + (-2x3)=0

2 + x + (-6) = 0

(-4) + x = 0

x = 4.

Thus, the oxidation state of Sulfur in this compound is +4.

Answer 3

The Oxidation number of S here is to be calculated using the oxidation numbers of H and O, when the goal is that the complete compound will have an oxidation number of zero.

Oxidation numbers -

H = +1

O = -2

2x(+1)+(-2)x4= 2-8=(-6)

Since the combined oxidation numbers equal -6, to reach zero, S must be +6.

Answer 4

With a few exceptions that do not apply to this formula, oxygen is always assigned a formal oxidation number of -2 per atom, and hydrogen is assigned +1 per atom. The compound as a whole must always be electrically neutral. Therefore, in this molecule, the oxidation number of the single sulfur atom must be -[-2(4) +2] or +6.

Answer 5

4 because in this molecule 2+x+(

-6)=0

so we get x=4

so this is a oxidation state of h2so3

Answer 6

H = +1 oxidation state

S = +4 oxidation state

O = -2 oxidation state

Answer 7

+6 oxidation state

Answer 8

+6