Question

Find the oxidation number of Sulphur in $KAl{(S{O_4})_2}.12{H_2}O$

Answer 1

Hint: The charge that an atom seems to have when forming ionic bonds with other heteroatoms is defined as its oxidation number. A negative oxidation state is assigned to an atom with a higher electronegativity (even if it forms a covalent bond).

Complete answer:
Let the oxidation number of Sulphur be ‘ $x$ ’ .
We know,
Oxidation number of $K = + 1$
Oxidation number of $H = + 1$
Oxidation number of $O = - 2$
Oxidation number of $Al = + 3$
Therefore, the oxidation number of Sulphur will be :
$1( + 1) + 1( + 3) + 2(x) + 8( - 2) + 24( + 1) + 12( - 2) = 0$
$1 + 3 + 2x - 16 + 24 - 24 = 0$
From the above calculations we get,
$x = + 6$
Hence, the oxidation number of Sulphur will be $ + 6$.
The oxidation state or number of an atom or ion in a molecule/ion is determined by:
1.Calculating the constant oxidation state of other atoms/molecules/ions bonded to it.
2.Adding the total charge of a molecule or ion to the total oxidation state of the molecule or ion.

Additional Information: Neutral atoms or molecules have no net charge. As a result, their overall oxidation state is zero.
Oxidation number of all alkali metal ions is always = $ + 1$
Oxidation number of all alkaline earth metal ions is always = $ + 2$
Oxidation number of all boron family metal ions is always = $ + 3$
Oxidation number of hydrogen in protons (${H^ + }$) is $ + 1$ , and in hydride is $ - 1$.

Note:
The oxidation state is also known as the oxidation number. However, depending on whether we are considering the electronegativity of the atoms or not, these terms can have a different meaning. In coordination chemistry, the word oxidation number is widely employed.