Question

Consider this reaction:

Answer 1

Answer : The oxidation state of Mg in Mg(s) is (0).

Explanation :

Oxidation number or oxidation state : It represent the number of electrons lost or gained by the atoms of an element in a compound.

Oxidation numbers are generally written with the sign (+) and (-) first and then the magnitude.

Rules for Oxidation Numbers are :

The oxidation number of a free element is always zero.

The oxidation number of a monatomic ion equals the charge of the ion.

The oxidation number of  Hydrogen (H)  is +1, but it is -1 in when combined with less electronegative elements.

The oxidation number of  oxygen (O)  in compounds is usually -2.

The oxidation number of a Group 17 element in a binary compound is -1.

The sum of the oxidation numbers of all of the atoms in a neutral compound is zero.

The sum of the oxidation numbers in a polyatomic ion is equal to the charge of the ion.

The given chemical reaction is:

$Mg(s)+2H_2O(l)\rightarrow Mg(OH)_2(s)+H_2(g)$

In the given reaction, the oxidation state of Mg in Mg(s) is (0) because it is a free element and the oxidation state of Mg in $Mg(OH)_2$ is (+2).

Hence, the oxidation state of Mg in Mg(s) is (0).