Question

How to write a balanced chemical equation for the reaction of molecular oxygen with radium metal to form oxide?

Answer 1

i) In case of a chemical reaction, balancing of the reaction requires identification of the elements and the compounds which are involved in the reaction and the symbols by which they are represented.
ii) After that, the reactant and products are written in a raw format also called skeletal equation which is not balanced.
iii) And then the number of species are counted on both the sides and eventually coefficients are added to make the number of each atom equal on both the sides.

Complete step-by-step answer:
In order to write a balanced chemical equation we must follow a systematic way. At first we will identify all the elements which are involved in the reaction by taking their chemical symbols. The radium is represented by $Ra$ and the oxygen is represented by the symbol $O$. Now we know that oxygen exists in diatomic form, in the environment.

We also know that the radium is an element on the second group, and so it loses two electrons from its outermost orbital to attain a noble gas configuration. So, it has an oxidation state $+2$. Now, we know that oxygen is a group sixteen element, and consequently it gains two electrons to attain a stable octet configuration. As a result it has an oxidation state of $-2$.

Now, when the two ions combine, after the reaction of radium with oxygen molecule, the molecule of oxygen breaks, and it combines with the radium, to form its oxide. The unbalanced equation can be written as,
$Ra+{{O}_{2}}\to RaO$

Here we can see that the atom of radium combines with a molecule of oxygen in order to form its oxide. Now we are supposed to balance this chemical equation. We will simply count the number of atoms of each element in both the sides and see if they are equal. If in case they are unequal we will try to balance the numbers by using appropriate coefficients.

The number of radium is same on both the sides but the number of oxygen atoms is two on the reactant side, so we add a coefficient $2$ on the product side which makes the number of radium atoms two on the product side, so we balance the number of radium at reactant side by adding the same coefficient. And the equation becomes,
$2Ra+{{O}_{2}}\to 2RaO$
Which is the complete balanced chemical equation of the given reaction.

Note: Balancing a chemical equation requires making the number of involved atoms in the molecules on both the reactant as well as the product side. By adding the appropriate coefficient we can balance the numbers, and after in case of the redox reactions, the oxidation state or the charges are also balanced by adding electrons and water molecules on the required sides.