1. We need to count atoms with oxidation state $0$ in each molecule and then add them.

2. Molecule 1: $\mathrm{Br_3O_8}$

Let oxidation state of Br be $x$ on average: $3x + 8(-2)=0$ $3x-16=0 \Rightarrow x=\frac{16}{3}$

So bromine atoms are not all equivalent. This compound is better written as mixed anhydride corresponding to bromine oxyacids, containing:

• one Br in $+5$ state,
• two Br in $+\! \,?$

But no atom is in oxidation state $0$.

Thus, number of atoms in zero oxidation state in $\mathrm{Br_3O_8}$ is $0$

3. Molecule 2: $\mathrm{F_2O}$

In $\mathrm{F_2O}$, fluorine is always $-1$. Let oxidation state of oxygen be $x$: $x+2(-1)=0$ $x=+2$

So no atom has oxidation state $0$.

Thus, count = $0$

4. Molecule 3: $\mathrm{H_2S_4O_6}$ (tetrathionic acid)

Structure of tetrathionic acid is: $\mathrm{HO_3S-S-S-SO_3H}$

Now assign oxidation states:

• Terminal sulfur atoms bonded to $3$ oxygens each are in high oxidation state, actually $+5$ each.
• The two inner sulfur atoms are bonded only to sulfur atoms. In an S–S bond, electrons are shared equally, so each such sulfur has oxidation state $0$.

Check total: $2(+5)+2(0)+2(+1)+6(-2)=0$ $10+0+2-12=0$ Correct.

Thus, number of atoms in zero oxidation state in $\mathrm{H_2S_4O_6}$ is $2$

5. Molecule 4: $\mathrm{H_2S_5O_6}$ (pentathionic acid)

Structure: $\mathrm{HO_3S-S-S-S-SO_3H}$

Here:

• the two terminal sulfur atoms attached to $\mathrm{SO_3H}$ groups are in $+5$ state,
• the three inner sulfur atoms are bonded only to sulfur atoms, so each is in oxidation state $0$.

Check total: $2(+5)+3(0)+2(+1)+6(-2)=0$ $10+0+2-12=0$ Correct.

Thus, number of atoms in zero oxidation state in $\mathrm{H_2S_5O_6}$ is $3$

6. Molecule 5: $\mathrm{C_3O_2}$ (carbon suboxide)

Structure: $\mathrm{O=C=C=C=O}$

Now determine oxidation states of carbon atoms.

• Each terminal carbon is double bonded to oxygen. Since oxygen is more electronegative, each terminal carbon has oxidation state $+2$.
• The central carbon is bonded only to carbons, so its oxidation state is $0$.

Thus, number of atoms in zero oxidation state in $\mathrm{C_3O_2}$ is $1$

7. Total sum

Adding all zero oxidation state atoms: $0+0+2+3+1=6$

8. Final Answer

The required sum is: $\boxed{6}$

This matches the stored correct answer.