Assign an oxidation number to each atom in the products. 2na(s)+2h2o(l)→2naoh(aq)+h2(g) assign an oxidization number for: na , o , h (in naoh) , h (in h2) i'm not sure how to assign oxidization numbers, if someone could explain it to me that'd be great! i have an exam that involves oxidization numbers coming up, and i'm completely clueless on them! you in advance!

Answer : The oxidation number of Na, O and H in NaOH are, (+1), (-2) and (+1) respectively.

The oxidation number of H in  is zero.

Explanation :

Oxidation number : 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.

When the atoms are present in their elemental state then the oxidation number will be zero.

Rules for Oxidation Numbers :

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, but it is -1 in peroxides.

The oxidation number of a Group 1 element in a compound is +1.

The oxidation number of a Group 2 element in a compound is +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:

Now we have to determine the oxidation number of each atom in the products.

The given product is,

'Na' belongs to group 1 thus, the oxidation number is (+1)

The oxidation number of oxygen generally in all compound is (-2)

The oxidation number of hydrogen generally in all compound is (+1)

And as we know that, the sum of the oxidation numbers of all of the atoms in a neutral compound is zero.

(+1)+(-2)+(+1)=0

Thus, the oxidation number of Na, O and H in NaOH are, (+1), (-2) and (+1) respectively.

The given product is,

The oxidation number of hydrogen generally in all compound is (+1).

As we know that the oxidation number of atoms of elements at their standard states is always zero.

Thus, the oxidation number of H in  is zero.

lol for the pointsss