
Find the oxidation state of sulphur in .
A.
B.
C.
D.
Answer
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Hint: Oxidation state is also sometimes referred to as oxidation number, and is the total number of electrons that an atom gains or loses towards forming a chemical bond with another atom.
Complete step by step solution:
There are certain rules that one needs to keep in mind while calculating the oxidation states in a molecule.
i. Any free element has an oxidation number equal to zero.
ii. For monatomic ions, the oxidation number always has the same value as the net charge corresponding to the ion.
iii. The hydrogen atom has an oxidation state of . However, and when it is bonded to an element with less electronegativity than it shows an oxidation number of .
iv. Oxygen has an oxidation of in most of its compounds. However, when it is bonded in peroxides, the oxidation number corresponding to oxygen is .
v. All alkali metals or group elements have an oxidation state of in their compounds.
vi. All alkaline earth metals or group 2 elements show an oxidation state of in their compounds.
vii. In the compounds made up of two elements, a halogen has an oxidation number of assigned to them.
viii. In the case of neutral compounds, the sum of all the oxidation numbers of the constituent atoms totals to zero.
When polyatomic ions are considered, the sum of all the oxidation numbers of the atoms that constitute them is equal to the net charge of the polyatomic ion.
Now, let the oxidation state of the sulphur atom in be “X”.
Applying the above rules for molecule the equation becomes,
Where is the oxidation state for hydrogen and is the oxidation state of oxygen which are multiplied to the number of constituent atoms.
Further, solving for “X” we have,
Hence, the oxidation state for sulphur in is found to be (Option A).
Note: A student should take care of the fact that they are multiplying the oxidation number to the constituent number of atoms present in the given molecule, and correctly observing the electronegativities as well.
Complete step by step solution:
There are certain rules that one needs to keep in mind while calculating the oxidation states in a molecule.
i. Any free element has an oxidation number equal to zero.
ii. For monatomic ions, the oxidation number always has the same value as the net charge corresponding to the ion.
iii. The hydrogen atom has an oxidation state of
iv. Oxygen has an oxidation of
v. All alkali metals or group
vi. All alkaline earth metals or group 2 elements show an oxidation state of
vii. In the compounds made up of two elements, a halogen has an oxidation number of
viii. In the case of neutral compounds, the sum of all the oxidation numbers of the constituent atoms totals to zero.
When polyatomic ions are considered, the sum of all the oxidation numbers of the atoms that constitute them is equal to the net charge of the polyatomic ion.
Now, let the oxidation state of the sulphur atom in
Applying the above rules for
Where
Further, solving for “X” we have,
Hence, the oxidation state for sulphur in
Note: A student should take care of the fact that they are multiplying the oxidation number to the constituent number of atoms present in the given molecule, and correctly observing the electronegativities as well.
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