Correct option is 12

The sum of the oxidation states of bromine in bromic acid (HBrO₃) and perbromic acid (HBrO₄) is +12.

In bromic acid, the oxidation state of bromine is +5, while in perbromic acid, it is +7. The sum of the oxidation states of bromine in both compounds is (+5) + (+7) = +12.

Correct option is 10

To find the sum of the oxidation states of the metals in ${\mathrm{Fe}(\mathrm{CO})}_5$, ${\mathrm{VO}}^{2+}$, and ${\mathrm{WO}}_3$, we first need to find the oxidation states of the metals in each compound.

1. ${\mathrm{Fe}(\mathrm{CO})}_5$: The compound $\mathrm{CO}$ is a neutral ligand, so it does not contribute to the oxidation state of the metal. Hence, the oxidation state of Fe in ${\mathrm{Fe}(\mathrm{CO})}_5$ is 0.




2. ${\mathrm{VO}}^{2+}$: The compound $\mathrm{O}$ has an oxidation state of -2, and since the overall charge of the compound is +2, the oxidation state of V must be +4 to balance this out.




3. ${\mathrm{WO}}_3$: The compound $\mathrm{O}$ has an oxidation state of -2, and since there are three oxygen atoms, the total oxidation state contributed by oxygen is -6. To balance this out, the oxidation state of W must be +6.

Therefore, the sum of the oxidation states of the metals in these three compounds is $0+4+6=10$.

Correct option is 6

In ammonium phosphomolybdate, the compound typically found in the reagent used for the quantitative determination of phosphorus, the molybdenum (Mo) atom is in the +6 oxidation state.

Correct option is 4

Hypophosphoric acid H₄P₂O₆

Oxidation state is +4

Correct option is 6

$C{r}_2{O}_7^{2-}+2O{H}^{-}$ ⇌ $2Cr{O}_4^{2-}+H_{2}O$

Let Oxidation state of Cr in $Cr{O}_4^{2-}$ is = x.

$\therefore$ x + (–2 × 4) = –2

$\Rightarrow$ x = 6