Oxidising agent

QuestionsCategory: ChemistryOxidising agent
Charles asked 4 months ago

From the principal oxidation states ofSulphur, identify the oxidation state of asulphur compound which is essentially areducing agent.

1 Answers
Edukamer answered 4 months ago

Sulfur is one of the elements that can form compounds with all other elements except the noble gases. It has different oxidation states ranging from −2 to +6.

Now, if we are talking of a sulfur compund that is a reducing agent, a good example will be sulfur dioxide (SO2) when it reacts with Halogens.

In a chemical reaction, the Sulfur dioxide is oxidised to a sulfuryl halide.

In our example, Sulfur in the SO2 is less electroagative than Oxygen. So, the electrons in the SO2 will be pulled more by Oxygen. This means here, S cannot have a negative oxidation number.

You don’t work out oxidation states by counting the numbers of electrons transferred. It would take far too long. Instead you learn some simple rules, and do some very simple sums!

1-) The oxidation state of an uncombined element is zero. That’s obviously so, because it hasn’t been either oxidised or reduced yet! This applies whatever the structure of the element – whether it is, for example, Xe or Cl2 or S8, or whether it has a giant structure like carbon or silicon.

2-) The sum of the oxidation states of all the atoms or ions in a neutral compound is zero.

3-) The sum of the oxidation states of all the atoms in an ion is equal to the charge on the ion.

4-) The more electronegative element in a substance is given a negative oxidation state. The less electronegative one is given a positive oxidation state. Remember that fluorine is the most electronegative element with oxygen second.

Some elements almost always have the same oxidation states in their compounds. For example, Oxygen will always have the oxidation state of -2 in all its compunds except in peroxides and F2O.

So, now coming back to our SO2, we have:

=> 2 Oxygens bonded with the S atom so the total oxidation number for the two Oxygen will be -2 x 2 = -4.

Taking in to account rule (2) above the oxidation number of S will be +4

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