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There are rules to assigning conventional oxidation numbers to some elements. For those unspecified, you can solve them. This is how you solve it.

*S8. Since this is in elemental form, its oxidation number is assigned as 0.

*H2S. H is assigned with +1. Since the compound is neutral, the overall charge is 0. So,
2(+1) + x = 0
x = -2
The charge of S here is -2.

*SO₂. O is assigned with (-2). Using the same procedure,
x + 2(-2) = 0
x = +2
The charge of S here is +2.

*H₂SO₃.
2(+1) + x + 3(-2) = 0
x = +4.
The charge of S here is +4.

*K₂SO₄. K is assigned with +1. 
2(+1) + x + 4(-2) = 0
x = +6
The charge of S here is +6.

The S with the highest oxidation number is the one in K₂SO₄.
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Sulfur has highest oxidation number for [tex]\rm \bold {K_2SO_4}[/tex].

Oxidation number:

The total number of electrons that an atom either gains or loses in order to form a chemical bond with another atom.

The oxidation number of each atom can be calculated by subtracting the sum of lone pairs and electrons it gains from bonds from the number of valence electrons.

For elemental sulfur [tex]\rm \bold {S_8}[/tex], have no charge, hence have no oxidation number.

For [tex]\rm \bold {H_2S}[/tex]. H has +1 charge and compound has no charge. Hence Oxidation number of S is -2.

= 0- 2 (+1)

=-2

We can calculate oxidation number of sulfur for other compounds,

Sulfur has +2 Oxidation number for [tex]\rm \bold {SO_2}[/tex]

Sulfur has +4 Oxidation number for [tex]\rm \bold {H_2SO_3}[/tex]

Sulfur has +6 Oxidation number for [tex]\rm \bold {K_2SO_4}[/tex]

So, we can conclude that Sulfur has highest oxidation number for [tex]\rm \bold {K_2SO_4}[/tex].

To know more about Oxidation number, refer to the link:

https://brainly.com/question/15167411?referrer=searchResults

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