Animated Solution for Chemistry - Redox Reactions: Consider the following molecules : ${\text{Br}}_3{\text{O}}_8$, ${\text{F}}_2\text{O}$, ${\text{H}}_2{\text{S}}_4{\text{O}}_6$, ${\text{H}}_2{\text{S}}_5{\text{O}}_6$, and ${\text{C}}_3{\text{O}}_2$. Count the number of atoms existing in their zero oxidation state in each molecule. Their sum is______.

The concept of oxidation states is a fundamental pillar of redox chemistry, allowing us to track electron flow and understand the charge distribution within molecules. In this problem, we are tasked with identifying atoms that possess an oxidation state of exactly zero across a variety of fascinating chemical structures. Let's embark on a structural journey to uncover these hidden zeros.

Before we dive into the molecules, we must recall a critical rule of assigning oxidation states: bonds between identical atoms (homoatomic bonds) do not contribute to the oxidation state of either atom. Because there is no electronegativity difference between two identical atoms, the shared electrons are considered to be divided equally. This principle is the key to solving our problem.

Let's start with tribromooctoxide, ${\text{Br}}_3{\text{O}}_8$. Its structure consists of a linear chain of three bromine atoms. The terminal bromine atoms are each double-bonded to three oxygen atoms. Since oxygen is more electronegative, it pulls electron density away, giving each terminal bromine an oxidation state of $+6$. The central bromine is double-bonded to two oxygen atoms, resulting in an oxidation state of $+4$. None of the atoms here have a zero oxidation state.

Next, we examine oxygen difluoride, ${\text{F}}_2\text{O}$. Fluorine is the undisputed champion of electronegativity, so it always takes a $-1$ oxidation state in its compounds. With two fluorine atoms pulling electron density, the central oxygen atom is forced into a $+2$ oxidation state. Again, we find zero atoms with a zero oxidation state.

A Chain of Zeros

Now, the story gets interesting with the polythionic acids. Consider tetrathionic acid, ${\text{H}}_2{\text{S}}_4{\text{O}}_6$. Its structure features a continuous chain of four sulfur atoms: $\text{HO}-\text{S}(=\text{O}{)}_2-\text{S}-\text{S}-\text{S}(=\text{O}{)}_2-\text{OH}$. The terminal sulfur atoms are bonded to highly electronegative oxygen atoms, giving them a $+5$ oxidation state. However, the two sulfur atoms in the middle are bonded only to other sulfur atoms. Thanks to our golden rule, these homoatomic bonds contribute nothing to their oxidation state. Thus, these two middle sulfur atoms have an oxidation state of exactly $0$.

Similarly, pentathionic acid, ${\text{H}}_2{\text{S}}_5{\text{O}}_6$, extends this chain to five sulfur atoms. The terminal sulfurs remain at $+5$, but now we have three sulfur atoms sandwiched in the middle, bonded exclusively to other sulfurs. Consequently, all three of these central sulfur atoms boast a $0$ oxidation state.

The Final Piece

Finally, we look at carbon suboxide, ${\text{C}}_3{\text{O}}_2$. This molecule has a striking linear structure: $\text{O}=\text{C}=\text{C}=\text{C}=\text{O}$. The terminal carbon atoms are double-bonded to oxygen, assigning them an oxidation state of $+2$. The central carbon, however, is double-bonded to two other carbon atoms. Since it is bonded only to identical atoms, its oxidation state is $0$. This gives us one more atom to add to our tally.

Summing up our findings, we have $0$ atoms from ${\text{Br}}_3{\text{O}}_8$, $0$ from ${\text{F}}_2\text{O}$, $2$ from ${\text{H}}_2{\text{S}}_4{\text{O}}_6$, $3$ from ${\text{H}}_2{\text{S}}_5{\text{O}}_6$, and $1$ from ${\text{C}}_3{\text{O}}_2$.

This problem beautifully illustrates how a deep understanding of molecular structure and electronegativity rules is essential for mastering oxidation states.