Question

How would you express tin ($IV$) oxide as a chemical formula?

Answer 1

Hint:In order to express the formula of a compound we must first know the symbols by which each of the constituent elements of the compound is represented.
Then, we find out the formal charge carried by each individual element in the compound, and finally we assign the chemical formula such that the charge on the compound is neutralized.

Complete step-by-step answer: We know that the metal oxides are the compounds which are made up of metals and oxygen. In the given question we will be considering the oxide of tin in its fourth oxidation state which is clearly indicated in the brackets. So, we know that the outermost orbital of the oxygen has six electrons. It is a known fact, but it can be proved by the outermost electronic configuration of the oxygen atom, which is \[1{{s}^{2}}2{{s}^{2}}2{{p}^{4}}\] as we can see that the outermost orbital having the principal quantum number of two, has six electrons. So, as a result, the oxygen needs two more electrons in order to attain a stable octet configuration consisting of eight electrons. So, after gaining two more electrons, consequently the oxidation state of this oxygen atom would become $-2$. And now we know that the tin oxide would have the metal tin combined with this oxygen ion. We know that the tin is in its $+4$ state as indicated by the question itself, we will be taking this charge of the tin and adding the charge of oxygen in it. Now we can see that $-2+4=+2$. So, the charge on the compound would be positive, so in order to balance this charge we will add another oxygen atom to this compound and it will have the same charge which is negative two. So, now the charge would be balanced and the compound will become neutral.
Finally we will take the chemical formulas of the individual atoms of tin and oxygen and combine them as discussed. The tin is represented by the symbol $Sn$, and as we already know the oxygen is represented by $O$. So, after combination the formula of tin ($IV$) oxide becomes $Sn{{O}_{2}}$, which is the required answer.

Note:The tin ($IV$) oxide is represented by the formula $Sn{{O}_{2}}$, because the oxidation state of oxygen is negative two, and the oxidation state of the tin as indicated in the question, is positive four.
So in order to balance out the unbalanced charge, we take two atoms of oxygen in the compound, and so the $+4$ charge is neutralized by two $-2$charges.