Redox Reactions

Given reaction is,

In this reaction,

Cl of $ \mathrm{KClO}_{3} $ (oxidation number $ =+5 $ ) is reduced to $ \mathrm{Cl}^{-} $and O of $ \mathrm{KClO}_{3} $

$ \mathrm{KClO}_{3} $ (oxidation number $ =-2 $ ) is oxidised to $ \mathrm{O}_{2} $.

The correct match is A-III, B-I, C-IV, D-II.

For $ \mathrm{Na}_{2} \mathrm{CO}_{3} $ the equivalent weight is $ \frac{M}{2} $ as it is diprotic base.

In acidic medium, the $ n $-factor for $ \mathrm{KMnO}_{4} $ is 5 . So, its equivalent weight is $ \frac{M}{5} $.

In neutral medium, the $ n $-factor for $ \mathrm{K}_{2} 2_{2} \mathrm{r}_{2} $ is 6 . So, its equivalent weight is $ \frac{M}{6} $. In neutural medium, the $ n $-factor for $ \mathrm{KMnO}_{4} $ is 3. So, its equivalent weight is $ \frac{M}{3} $.

20 Volume of $ \mathrm{H}_{2} \mathrm{O}_{2} $ solution refers to 1 L of $ \mathrm{H}_{2} \mathrm{O}_{2} $ solution that will liberale 20 L of $ \mathrm{O}_{2} $ at NTP.

$ \underset{68 \mathrm{~g}}{2 \mathrm{H}_{2} \mathrm{O}_{2}} \rightarrow 2 \mathrm{H}_{2} \mathrm{O}+\underset{22.4 \mathrm{~L} \text { at NTP }}{\mathrm{O}_{2}} $

22.4 L of $ \mathrm{O}_{2} $ at NTP produce from $ \mathrm{H}_{2} \mathrm{O}_{2}=68 \mathrm{~g} $

20 L of $ \mathrm{O}_{2} $ at NTP produce from $ \mathrm{H}_{2} \mathrm{O}_{2} $

$ =\frac{68}{22.4} \times 20=60.71 \mathrm{~g} $

Strength of $ \mathrm{H}_{2} \mathrm{O}_{2}=60.71 \mathrm{~g} / \mathrm{L} $

Equivalent weight of $ \mathrm{H}_{2} \mathrm{O}_{2}=\frac{68}{32} \times 8=17 $

$ \text { Normality }=\frac{\text { Strength }}{\text { Equivalent weight }}=\frac{60.71}{17} $

Normality $ \approx 3.570 \mathrm{~N} $

In the given three oxoanions, the oxidation state of $\mathrm{V}, \mathrm{Cr}$ and Mn in $\mathrm{VO}_2^{+}$, $\mathrm{Cr}_2 \mathrm{O}_7^{2-}$ and $\mathrm{MnO}_4^{-}$are $+5,+6$ and +7 respectively.

Hence, the correct increasing order is, $\mathrm{VO}_2^{+}<\mathrm{Cr}_2 \mathrm{O}_7^{2-}<\mathrm{MnO}_4^{-}$

Sodium is oxldised from 0 to +1. Water is reduced from +1 to 0.

Fluorine is reduced from 0 to -1. Water is oxidised from -2 to 0.

The following reactions are redox reactions.

Reaction (iii) is a double displacement reaction.

The process of acidifying chromate involves the following reaction:

$ 2 \mathrm{Na}_2 \mathrm{CrO}_4 + \mathrm{H}_2 \mathrm{SO}_4 \rightarrow \mathrm{Na}_2 \mathrm{Cr}_2 \mathrm{O}_7 + \mathrm{H}_2 \mathrm{O} + \mathrm{Na}_2 \mathrm{O}_4 $

This can be simplified to:

$ \mathrm{CrO}_4^{2-} \xrightarrow{\mathrm{H}^{+}} \mathrm{Cr}_2 \mathrm{O}_7^{2-} $

In this reaction, chromate ions ($\mathrm{CrO}_4^{2-}$) are converted into dichromate ions ($\mathrm{Cr}_2 \mathrm{O}_7^{2-}$).

In the compound $\mathrm{Na}_2 \mathrm{Cr}_2 \mathrm{O}_7$, sodium ($\mathrm{Na}$) is in the +1 oxidation state and oxygen ($\mathrm{O}$) is in the -2 oxidation state. To find the oxidation state of chromium ($\mathrm{Cr}$) in $\mathrm{Cr}_2 \mathrm{O}_7^{2-}$, we analyze the overall charge balance:

Total charge contributed by two sodium atoms: $2 \times (+1) = +2$.

Each oxygen atom contributes a -2 charge, and there are 7 oxygen atoms: $7 \times (-2) = -14$.

The net charge on the ion $\mathrm{Cr}_2 \mathrm{O}_7^{2-}$ is -2.

Setting up the equation for chromium:

$ 2 + 2x - 14 = 0 $

Solving for $x$ gives:

$ 2x = 12 \\ x = +6 $

Thus, the oxidation state of chromium in $\mathrm{Na}_2 \mathrm{Cr}_2 \mathrm{O}_7$ is +6.

Given, 100 mL of $0.1 \mathrm{M} \mathrm{Fe}^{2+}$ titrated with $\frac{1}{60} \mathrm{M} \mathrm{Cr}_2 \mathrm{O}_7^{2-}$ solution.

$ \text { The complete reaction is, } $

$ \begin{aligned} & \Rightarrow \quad 0.1 \mathrm{M} \quad \frac{1}{60} M =0.1 \mathrm{~N}=\frac{1}{10} \mathrm{~N}[\because N=M \times n \text {-factor }] \\ & \Rightarrow \quad V \mathrm{~mL} \quad 100 \mathrm{~mL}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,(say) \\ & \ \\ & \Rightarrow \quad V=\frac{1}{0.1} \times\left(100 \times \frac{1}{10}\right)=100 \mathrm{~mL} \end{aligned} $

So, $\quad V=0.1 \mathrm{~L}$

When hydrogen sulfide ($ \mathrm{H}_2 \mathrm{S} $) is passed through acidic potassium permanganate ($ \mathrm{KMnO}_4 $), sulfur is produced.

The balanced chemical equation for this reaction is:

$ \mathrm{KMnO}_4 + \mathrm{H}_2 \mathrm{S} + \mathrm{H}_2 \mathrm{SO}_4 \rightarrow \mathrm{H}_2 \mathrm{O} + \mathrm{K}_2 \mathrm{SO}_4 + \mathrm{MnSO}_4 + \mathrm{S} $

In this reaction, $ \mathrm{KMnO}_4 $ acts as an oxidizing agent, converting $ \mathrm{H}_2 \mathrm{S} $ into sulfur ($ \mathrm{S} $).

In a neutral medium, the permanganate ion oxidizes iodine to form iodate. Here's the reaction illustrating this process:

$ \begin{aligned} & \mathrm{KMnO}_4 \rightleftharpoons \mathrm{MnO}_4^{-}+\mathrm{K}^{+} \\ & 2 \mathrm{MnO}_4^{-}+\mathrm{H}_2 \mathrm{O}+\mathrm{I}^{-} \longrightarrow \\ & 2 \mathrm{MnO}_2+2 \mathrm{OH}^{-}+\mathrm{IO}_3^{-} \end{aligned} $

This reaction shows how permanganate ions in a neutral solution can facilitate the conversion of iodide ions to iodate ions, with manganese dioxide and hydroxide ions as additional products.

The complete reaction is as follows,

(a) $\mathrm{S}+2 \mathrm{H}_2 \mathrm{SO}_4 \longrightarrow 3 \mathrm{SO}_2+2 \mathrm{H}_2 \mathrm{O}$

Here, sulphur $(\mathrm{S})$ is oxidised to $\mathrm{SO}_2$

$\mathrm{H}_2 \mathrm{SO}_4$ is reduced to $\mathrm{SO}_2$.

Hence, both $X$ and $Y$ are $\mathrm{SO}_2$.

Let's analyze each statement step by step:

For Statement I (Decomposition of potassium chlorate):

Consider the reaction:

$2KClO_3 \rightarrow 2KCl + 3O_2$

In potassium chlorate (KClO₃), let's determine the oxidation state of chlorine.

For KClO₃, potassium (K) has an oxidation number of +1 and oxygen (O) is typically -2. Setting chlorine's oxidation number as x:

$+1 + x + 3(-2) = 0 \quad \Rightarrow \quad x = +5$

In potassium chloride (KCl), K is +1 and Cl must be -1 (to balance the compound).

Since chlorine goes from +5 to -1, it has gained electrons. This is a reduction. So Statement I is correct.

For Statement II (Reaction of sodium with oxygen to form Na₂O):

Consider the reaction:

$4Na + O_2 \rightarrow 2Na_2O$

Sodium (Na) starts with an oxidation state of 0 and in Na₂O, sodium is +1.

Oxygen (O₂) starts with 0 and in Na₂O, it is -2.

With sodium being oxidized (gain in oxidation number) and oxygen being reduced (loss in oxidation number), this clearly is a redox reaction. So Statement II is correct.

Since both statements are correct, the correct option is:

Option A

Both statements I and II are correct.

In a neutral medium, potassium permanganate oxidizes $ I^{-} $ to $ \text{IO}_3^{-} $ (iodate). The balanced chemical reaction for this process is as follows:

$ 2 \text{KMnO}_4 + \text{H}_2\text{O} + \text{KI} \longrightarrow 2 \text{MnO}_2 + 2 \text{KOH} + \text{KIO}_3 $

In this reaction, the permanganate ion acts as an oxidizing agent, facilitating the conversion of iodide ions into iodate ions.

Oxidation half reaction:

$ \mathrm{SO}_{2}(g)+2 \mathrm{H}_{2} \mathrm{O}(l) \longrightarrow \mathrm{HSO}_{4}^{-}(a q)+3 \mathrm{H}^{+}(a q)+2 e^{-} \quad \ldots \text { (i) } $

Reduction half reaction:

$ \mathrm{MnO}_{4}^{-}(a q)+8 \mathrm{H}^{+}+5 e^{-} \longrightarrow \mathrm{Mn}^{2+}(a q)+4 \mathrm{H}_{2} \mathrm{O}(l) \quad \ldots \text { (ii) } $

Multiply eq. (i) by 5 and (ii) by 2 and then add

$ \begin{aligned} 2 \mathrm{MnO}_{4}^{-}(a q)+5 \mathrm{SO}_{2}(g) & +2 \mathrm{H}_{2} \mathrm{O}(l)+\mathrm{H}^{+}(a q) \longrightarrow 2 \mathrm{Mn}^{+}(a q)+5 \mathrm{HSO}_{4}^{-}(a q) \end{aligned} $

Thus, the stoichiometric coefficient of $ \mathrm{SO}_{2} $ in the above balance equation is 5 .

To determine the volume of $ \frac{M}{8} \mathrm{KMnO_4} $ solution needed to completely react with $ 25.0 \, \mathrm{cm}^3 $ of $ \frac{M}{4} \mathrm{FeSO_4} $ in an acidic medium, we start by considering the balanced ionic equation:

$ \mathrm{MnO_4}^- + 5 \mathrm{Fe}^{2+} + 8 \mathrm{H}^+ \rightarrow \mathrm{Mn}^{2+} + 5 \mathrm{Fe}^{3+} + 4 \mathrm{H_2O} $

From this equation, it's evident that 1 mole of $ \mathrm{KMnO_4} $ reacts with 5 moles of $ \mathrm{FeSO_4} $.

For balancing the redox reaction, apply the molarity equation:

$ \frac{M_1 V_1}{n_1}(\mathrm{KMnO_4}) = \frac{M_2 V_2}{n_2}(\mathrm{FeSO_4}) $

Substitute the given values where:

$ M_1 = \frac{M}{8} $ (molarity of $\mathrm{KMnO_4}$)

$ M_2 = \frac{M}{4} $ (molarity of $\mathrm{FeSO_4}$)

$ V_2 = 25 \, \mathrm{cm}^3 $

$ n_1 = 1 $ (moles of $\mathrm{KMnO_4}$)

$ n_2 = 5 $ (moles of $\mathrm{FeSO_4}$)

$ \frac{\frac{1}{8} \times V_1}{1} = \frac{\frac{1}{4} \times 25}{5} $

Solving the equation for $ V_1 $:

$ \frac{1 \times V_1}{8} = \frac{1 \times 25}{4 \times 5} $

$ V_1 = \frac{1 \times 25 \times 8}{4 \times 5} $

$ V_1 = 10 \, \mathrm{cm}^3 \quad \text{or} \quad V_1 = 10.0 \, \mathrm{mL} $

Hence, the volume of $ \frac{M}{8} \mathrm{KMnO_4} $ solution required is 10.0 mL.

The correct answer is Option A: Both Statement I and Statement II are incorrect.

Explanation :

In redox titrations, the indicators used are sensitive to a change in oxidation potential, not pH. A redox titration (also called an oxidation-reduction titration) can accurately determine the concentration of an unknown analyte by measuring the amount of an oxidizing or reducing agent that it consumes. These indicators work by undergoing a definite color change at a particular electrode potential.

In contrast, in acid-base titrations, the indicators used are sensitive to a change in pH, not oxidation potential. Acid-base titrations are based on the neutralization reaction between the acid and the base. The point at which all the acid or base has reacted with the other component is called the equivalence point, and it often results in a sudden change in pH. This sudden change in pH can be detected using a pH-sensitive indicator.

The reaction given is a disproportionation reaction where iodine ($\mathrm{I}_2$) undergoes both oxidation and reduction. Disproportionation reactions are a type of redox reaction where an element is simultaneously oxidized and reduced.

In the process, iodine gets oxidized from 0 oxidation state (in $\mathrm{I}_2$) to +5 oxidation state (in $\mathrm{IO}_3^{-}$), and reduced from 0 oxidation state (in $\mathrm{I}_2$) to -1 oxidation state (in $\mathrm{I}^{-}$).

The n-factor is the total change in oxidation state per molecule that undergoes the redox reaction. In this case, the n-factor for $\mathrm{IO}_3^{-}$ is 5 (as iodine goes from 0 to +5) and for $\mathrm{I}^{-}$, it's 1 (as iodine goes from 0 to -1).

Now, to balance the redox reaction, the total increase in oxidation state (total oxidation) must equal the total decrease in oxidation state (total reduction). Hence, the molar ratio of $\mathrm{IO}_3^{-}$ to $\mathrm{I}^{-}$ must be 1:5.

So, the balanced reaction would be:

$\mathrm{IO}_3^{-}+6 \mathrm{H}^{+}+5 \mathrm{I}^{-} \rightarrow 3 \mathrm{I}_2+3 \mathrm{H}_2 \mathrm{O}$

This equation yields 3 moles of $\mathrm{I}_2$, but the original equation needs to produce 6 moles of $\mathrm{I}_2$, so the entire equation is multiplied by 2:

$2 \mathrm{IO}_3^{-}+12 \mathrm{H}^{+}+10 \mathrm{I}^{-} \rightarrow 6 \mathrm{I}_2+6 \mathrm{H}_2 \mathrm{O}$

This tells us that to get 6 moles of $\mathrm{I}_2$, you need 10 moles of $\mathrm{I}^{-}$. So, $x = 10$.

The structure of $\mathrm{C}_3 \mathrm{O}_2$ is

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

Assigning the oxidation number to each C -atom in molecule considering oxidation number of oxygen as -2

We get, oxidation state as

$ \mathrm{O}=\stackrel{+2}{\mathrm{C}}=\stackrel{0}{\mathrm{C}}=\stackrel{+2}{\mathrm{C}}=\mathrm{O} $

Thus, two C -atoms linked with O -atom has +2 oxidation state and carbon atom in the centre has 0 oxidation state.

The reaction (a) produce nitrogen

(II) oxide as one of the product.

$ \begin{aligned} 3 \mathrm{Cu}+8 \mathrm{HNO}_3 \text { dil. } \longrightarrow & 3 \mathrm{Cu}\left(\mathrm{NO}_3\right)_2 +2 \mathrm{NO}+4 \mathrm{H}_2 \mathrm{O} \end{aligned} $

Other reactions are as follows :

$ \begin{aligned} & \mathrm{Cu}+4 \mathrm{HNO}_3 \text { (conc.) } \longrightarrow \mathrm{Cu}\left(\mathrm{NO}_3\right)_2 +2 \mathrm{NO}_2+2 \mathrm{H}_2 \mathrm{O} \\ & 4 \mathrm{Zn}+10 \mathrm{HNO}_3 \text { (dil.) } \rightarrow 4 \mathrm{Zn}\left(\mathrm{NO}_3\right)_2 +5 \mathrm{H}_2 \mathrm{O}+\mathrm{N}_2 \mathrm{O} \\ & \mathrm{Zn}+4 \mathrm{HNO}_3 \text { (conc.) } \rightarrow \mathrm{Zn}\left(\mathrm{NO}_3\right)_2 +2 \mathrm{H}_2 \mathrm{O}+2 \mathrm{NO}_2 \end{aligned} $

$\mathrm{SO}_2$ is the strongest reducing agent. In $\mathrm{SO}_2$, sulphur have +4 oxidation state and it will attain +6 oxidation state and ion $\mathrm{SO}_4^{2-}$. That means it undergoes oxidation (going from lower oxidation state to higher oxidation state). $\mathrm{So}, \mathrm{SO}_2$ will oxidises itself and reduces others and act as reducing agent, while $\mathrm{TeO}_2, \mathrm{TeO}_3$ and $\mathrm{SO}_3$ acts as an oxidising agent.

In above reaction, oxidation and reduction takes place simultaneously that means it is a redox reaction but it is not a disproportionation reaction because same element is not simultaneously oxidised and reduced.

To determine which compound has the maximum equivalent weight, let's calculate the equivalent weights for each:

For $ \mathrm{Na}_2 \mathrm{C}_2 \mathrm{O}_4 \cdot 2 \mathrm{H}_2 \mathrm{O} $:

The formula for calculating equivalent weight is:

$ \text{Equivalent weight} = \frac{\text{Molar mass}}{\text{n-factor}} $

Here, n-factor is 2 because it involves a two-electron transfer process. The molar mass of $ \mathrm{Na}_2 \mathrm{C}_2 \mathrm{O}_4 \cdot 2 \mathrm{H}_2 \mathrm{O} $ is:

$ 2(23) + 2(12) + 4(16) + 2(18) = 170 $

$ \text{Equivalent weight} = \frac{170}{2} = 85 \, \mathrm{g} \, \mathrm{eq}^{-1} $

For $ \mathrm{KMnO}_4 / \mathrm{H}^{+} $:

This reaction has an n-factor of 5 due to the change in oxidation state from +7 to +2 for Mn. The molar mass is:

$ 39 + 55 + 4(16) = 158 $

$ \text{Equivalent weight} = \frac{158}{5} = 31.6 \, \mathrm{g} \, \mathrm{eq}^{-1} $

For $ \mathrm{KMnO}_4 / \mathrm{H}_2 \mathrm{O} $:

When Mn changes from +7 to +4, the n-factor is 3. The molar mass calculation remains the same:

$ 39 + 55 + 64 = 158 $

$ \text{Equivalent weight} = \frac{158}{3} = 52.6 \, \mathrm{g} \, \mathrm{eq}^{-1} $

For $ \mathrm{K}_2 \mathrm{Cr}_2 \mathrm{O}_7 / \mathrm{H}^{+} $:

The n-factor is 6 because chromium changes from an oxidation state of +6 to +3. The molar mass is:

$ 2(39) + 2(52) + 7(16) = 294 $

$ \text{Equivalent weight} = \frac{294}{6} = 49 \, \mathrm{g} \, \mathrm{eq}^{-1} $

The compound $ \mathrm{Na}_2 \mathrm{C}_2 \mathrm{O}_4 \cdot 2 \mathrm{H}_2 \mathrm{O} $ has the highest equivalent weight of 85 g eq$^{-1}$.

Disproportionation reactions are the reactions in which an element undergoes oxidation and reduction simultaneously forming two respective products. (One with higher oxidation state and one with lower.)

$\therefore \mathrm{A}, \mathrm{B}, \mathrm{D}$ only are disproportionation reactions.

Reaction of ferrous oxalate oxidised by $\mathrm{KMnO}_4$ is

Number of gram equivalent of $\mathrm{KMnO}_4$

$=$ Number of gram equivalent of ferrous oxalate

Number of moles $\times n$-factor

$=$ number of moles $\times n$-factor

Number of moles of $\mathrm{KMnO}_4 \times 2=1 \times 5$

Number of moles of $\mathrm{KMnO}_4=5 / 2$

Disproportionation reactions are those in which one element undergoes both oxidation and reduction both simultaneously.

In the reaction,

Oxygen in reactant $\mathrm{H}_2 \mathrm{O}_2$ is in -1 oxidation state. Oxygen in product $\mathrm{H}_2 \mathrm{O}$ is in -2 oxidation state and in other product $\mathrm{O}_2$, it is in 0 oxidation state. So, in this reaction oxygen undergoes oxidation and reduction both simultaneously.

In the reaction,

Chlorine in reactant $\mathrm{Cl}_2$ is in 0 oxidation state. Chlorine in product NaCl is in -1 oxidation state and in other product $\mathrm{NaOCl}, \mathrm{Na}$ is in +1 oxidation state, 0 is in -2 oxidation state and Cl is in +1 oxidation state. So, in this reaction chlorine undergoes oxidation and reduction both simultaneously.

When an atom is bonded to similar atoms, has zero oxidation state.

$\therefore$ Two middle sulphur atoms have zero oxidation state.

For other two sulphur atoms.

$\begin{aligned} 2 x+6 \times(-2) & =-2 \\ 2 x-12 & =-2 \\ 2 x & =-2+12 \\ 2 x & =+10 \\ x & =+5 \end{aligned}$

$\therefore$ The oxidation number on S-atoms in $\mathrm{S}_4 \mathrm{O}_6^{2-}$ is $+5,0,0+5$.

$\mathrm{Mg}+\mathrm{NO}_3^{-} \longrightarrow \mathrm{Mg}^{2+}+\mathrm{NH}_3$

The complete balanced equation of the given reaction is

$4 \mathrm{Mg}+\mathrm{NO}_3^{-}+9 \mathrm{H}^{\oplus} \longrightarrow 4 \mathrm{Mg}^{2+}+\mathrm{NH}_3+3 \mathrm{H}_2 \mathrm{O}$

Given, conc. of $\mathrm{NO}_3^{-}=0.1 \mathrm{~M}$

Volume of $\mathrm{NO}_3^{-}=100 \mathrm{~mL}$

We know that,

$\text { Molarity }=\frac{\text { Number of moles }}{\text { Volume of solution (in } \mathrm{L} \text { ) }}$

$\begin{aligned} & \text { Hence, number of moles }=\text { molarity } \times \text { volume of } \\ & \text { solution (in } \mathrm{L}) \\ & \begin{aligned} \text { So, number of moles } & =0.1 \times 100 \times 10^{-3} \\ & =10 \mathrm{~m} \mathrm{~mol} \\ & =10^{-2} \mathrm{~mol} \end{aligned} \end{aligned}$

From above Eq., 1 mole $\mathrm{NO}_3^{-}$ reacts with 4 moles Mg. So, $10^{-2}$ mole $\mathrm{NO}_3^{-}$ reacts with $4 \times 10^{-2}$ moles Mg. Now, to calculate the mass of Mg required, use the formula,

$\text { Number of moles }=\frac{\text { Given mass }}{\text { Molar mass }}$

Molar mass of $\mathrm{Mg}=24 \mathrm{~g} \mathrm{~mol}^{-1}$

On substituting the value,

$4 \times 10^{-2}=\frac{\text { Mass of } \mathrm{Mg}}{24}$

We will get,

Mass of $\mathrm{Mg}=24 \times 4 \times 10^{-2}$

Mass of $\mathrm{Mg}=0.96 \mathrm{~g}$

The given reaction is:

$\mathrm{SO}_3^{2-}(aq) + \mathrm{Br}_2(l) + \mathrm{H}_2O(l) \rightarrow \mathrm{SO}_4^{2-}(aq) + 2 \mathrm{Br}^-(aq) + 2 \mathrm{H}^+$

To determine the oxidation state of sulfur (S) in the product, we first identify that the sulfur-containing product is $\mathrm{SO}_4^{2-}$ (sulfate ion). Let's denote the oxidation state of sulfur as $ x $.

For the sulfate ion $\mathrm{SO}_4^{2-}$:

• Each oxygen (O) atom has an oxidation state of $-2$.


• There are four oxygen atoms, contributing a total of $ 4 \times (-2) = -8 $.

The net charge on the sulfate ion is $-2$. Using these data, we set up the following equation to solve for $ x $:

$ x + 4(-2) = -2 $

Simplifying this:

$ x - 8 = -2 $

Solving for $ x $:

$ x = -2 + 8 $

$ x = +6 $

Therefore, the oxidation state of sulfur in the sulfate ion $ \mathrm{SO}_4^{2-} $ is $ +6 $.

Phosphorus reacts with $\mathrm{HNO}_3$ to give $\mathrm{H}_3 \mathrm{PO}_4$ and $\mathrm{NO}_2$. In this reaction, phosphorus is oxidised to $\mathrm{H}_3 \mathrm{PO}_4$ and also N is reduced to $\mathrm{NO}_2$. The reaction is

$\mathrm{P}+5 \mathrm{HNO}_3 \longrightarrow \mathrm{H}_3 \mathrm{PO}_4+5 \mathrm{NO}_2+\mathrm{H}_2 \mathrm{O} $

Structure of Br$_3$O$_8$ is

Each terminal Br has three oxygen atoms attached to it. So, the oxidation state of Br -atom on terminals is +6 .

The bromine atom in the middle has two oxygen atoms attached to it. So, the oxidation state of Br atom in middle is +4.

Hence, the oxidation states of three Br -atoms in $\mathrm{Br}_3 \mathrm{O}_8$ molecule are $+6,+4,+6$.