Question

The common oxidation state for sulphur is:
A. +4
B. +6
C. Both A and B
D. None of these

Answer 1

Hint: Oxidation number or oxidation state is the charge present on the atom or on the molecule. But the oxidation number of an element is going to vary from one compound to another and depends on the chemical reaction in which it participates. Oxidation number will be positive or negative depending on the losing or gaining of electrons.

Complete step by step answer:
- In the question it is asked what the common oxidation state of sulphur is.
- Oxidation state also termed as oxidation number.
- Sulphur is an element which is present in the VI A group in the periodic table and belongs to p-block.
- The atomic number of sulphur is 16 means there are 16 electrons present in sulphur.
- The electronic configuration of sulphur is as follows.
$$1s^{2}2s^{2}2p^{6}3s^{2}3p^4$$
 - There are four electrons in the 3p orbital and 2 electrons in 3s orbital in the sulphur.
- To participate in the reaction with others sulphur initially loses these four electrons and will get +4 oxidation state.
- If other chemicals still need electrons to react with sulphur then the sulphur loses the two electrons from 3s orbital and will get an oxidation number of +6.
- Therefore sulphur will show two common oxidation states +4 and +6.
So, the correct option is C.

Note: If an element has positive oxidation number means the element lost the electrons during the reaction. If an element has negative oxidation number means the element accepts the electrons from other elements during the reaction.