The oxidation number of each sulphur atom in the following compounds are given below

(a) $\mathrm{Na}_2 \mathrm{~S}_2 \mathrm{O}_3$ Let us consider the structure of $\mathrm{Na}_2 \mathrm{~S}_2 \mathrm{O}_3$.

There is a coordinate bond between two sulphur atoms. The oxidation number of acceptor S -atom is -2 . Let, the oxidation number of other S -atom be $x$.

$$\mathrm{\mathop {2( + 1)}\limits_{For\,Na} + \mathop {3 \times ( - 2)}\limits_{For\,O - atoms} + x + \mathop {1( - 2) = 0}\limits_{For\,coordinate\,S - atom}}$$

$x=+6$

Therefore, the two sulphur atoms in $\mathrm{Na}_2 \mathrm{~S}_2 \mathrm{O}_3$ have -2 and +6 oxidation number.

(b) $\mathrm{Na}_2 \mathrm{~S}_4 \mathrm{O}_6$ Let us consider the structure of $\mathrm{Na}_2 \mathrm{~S}_4 \mathrm{O}_6$.

In this structure, two central sulphur atoms have zero oxidation number because electron pair forming the S-S bond remain in the centre. Let, the oxidation number of (remaining S-atoms) S-atom be $x$.

$$\mathrm{\mathop {2( + 1)}\limits_{For\,Na} + \mathop {6\,( - 2)}\limits_{For\,O} + 2x + \mathop {2(0) = 0}\limits_{}}$$

$2-12+2 x=0$ or $x=+\frac{10}{2}=+5$

Therefore, the two central S-atoms have zero oxidation state and two terminal S-atoms have +5 oxidation state each.

(c) $\mathrm{Na}_2 \mathrm{SO}_3$ Let the oxidation number of S in $\mathrm{Na}_2 \mathrm{SO}_3$ be $x$.

$$2(+1)+x+3(-2)=0 \text { or } x=+4$$

(d) $\mathrm{Na}_2 \mathrm{SO}_4$ Let the oxidation number of S be $x$.

$$2(+1)+x+4(-2)=0 \text { or } x=+6$$

Oxidation number method

(a)

Balance increase and decrease in oxidation number.

$$6 \mathrm{Fe}^{2+}+\mathrm{H}^{+}+\mathrm{Cr}_2 \mathrm{O}_7^{2-} \longrightarrow 2 \mathrm{Cr}^{3+}+6 \mathrm{Fe}^{3+}+\mathrm{H}_2 \mathrm{O}$$

Balance charge by multiplying $\mathrm{H}^{+}$by 14.

$$6 \mathrm{Fe}^{2+}+14 \mathrm{H}^{+}+\mathrm{Cr}_2 \mathrm{O}_7^{2-} \longrightarrow 2 \mathrm{Cr}^{3+}+6 \mathrm{Fe}^{3+}+\mathrm{H}_2 \mathrm{O}$$

Balance H and O -atoms by multiplying $\mathrm{H}_2 \mathrm{O}$ by 7 .

$$6 \mathrm{Fe}^{2+}+14 \mathrm{H}^{+}+\mathrm{Cr}_2 \mathrm{O}_7^{2-} \longrightarrow 2 \mathrm{Cr}^{3+}+6 \mathrm{Fe}^{3+}+7 \mathrm{H}_2 \mathrm{O}$$

This represents a balanced redox reaction.

(b)

Balance increase and decrease in oxidation number

$$\mathrm{I}_2+10 \mathrm{NO}_3^{-} \longrightarrow 10 \mathrm{NO}_2+2 \mathrm{IO}_3^{-}$$

Balance charge by writing $8 \mathrm{H}^{+}$in LHS of the equation.

$$\mathrm{I}_2+10 \mathrm{NO}_3^{-}+8 \mathrm{H}^{+} \longrightarrow 10 \mathrm{NO}_2+2 \mathrm{IO}_3^{-}$$

Balance H -atoms by writing $4 \mathrm{H}_2 \mathrm{O}$ in RHS of the equation.

$$\mathrm{I}_2+10 \mathrm{NO}_3^{-}+8 \mathrm{H}^{+} \longrightarrow 10 \mathrm{NO}_2+2 \mathrm{IO}_3^{-}+4 \mathrm{H}_2 \mathrm{O}$$

Oxygen atoms are automatically balanced. This represents a balanced redox reaction.

(c)

(Multiply $\mathrm{S}_2 \mathrm{O}_3^{2-}$ by 2 because there are 4 S -atoms in $\mathrm{S}_4 \mathrm{O}_6^{2-}$ ion.)

Increase and decrease in oxidation number is already balanced. Charge and oxygen atoms are also balanced.

This represents a balanced redox reaction.

(d)

Increase and decrease in oxidation number is already balanced.

Add $4 \mathrm{H}^{+}$ towards LHS of the equation to balance charge.

$$\mathrm{MnO}_2+\mathrm{C}_2 \mathrm{O}_4^{2-}+4 \mathrm{H}^{+} \longrightarrow \mathrm{Mn}^{2+}+2 \mathrm{CO}_2$$

Add $2 \mathrm{H}_2 \mathrm{O}$ towards RHS of the equation to balance H -atoms

$$\mathrm{MnO}_2+\mathrm{C}_2 \mathrm{O}_4^{2-}+4 \mathrm{H}^{+} \longrightarrow \mathrm{Mn}^{2+}+2 \mathrm{CO}_2+2 \mathrm{H}_2 \mathrm{O} $$

This represents a balanced redox reaction.

(a) Writing the O.N. on each atom above its symbol, then

Here, the $\mathrm{O} . \mathrm{N}$. of Cl increases from -1 in HCl to O in $\mathrm{Cl}_2$, therefore, $\mathrm{Cl}^{-}$is oxidised and hence HCl acts as the reducing agent.

The O.N. of N decreases from +5 in $\mathrm{HNO}_3$ to +3 in NOCl , therefore, $\mathrm{HNO}_3$ acts as the oxidising agent.

Thus, this reaction is a redox reaction.

(b) Writing the O.N. of each atom above its symbol, we have,

Here, the O.N. of none of the atoms undergo a change, therefore, this reaction is not a redox reaction.

(c)

Here, O.N. of Fe decreases from +3 in $\mathrm{Fe}_2 \mathrm{O}_3$ to 0 in Fe , therefore, $\mathrm{Fe}_2 \mathrm{O}_3$ acts as an oxidising agent. Further, O.N. of C increases from +2 in CO to +4 in $\mathrm{CO}_2$, therefore, CO acts as a reducing agent.

Thus, this reaction is an example of redox reaction.

(d) Writing the O.N. of each atom above its symbol, then

Here, O.N. of none of the atoms undergo a change, therefore, this reaction is not a redox reaction.

(e) Writing the O.N. of each atom above its symbol, then

Here, O.N. of N increases from -3 to 0 in $\mathrm{N}_2$, therefore, $\mathrm{NH}_3$ acts as a reducing agent. Further, $\mathrm{O} . \mathrm{N}$. of O decreases from O in $\mathrm{O}_2$ to -2 in $\mathrm{H}_2 \mathrm{O}$, therefore, $\mathrm{O}_2$ acts as a oxidising agent. Thus, this reaction is a redox reaction.

Write the O. N. of all atoms above their respective symbols.

O. N. decreases by, 3 per Cr-atom

Divide the given equation into two half reactions

Reduction half reaction: $\mathrm{Cr}_2 \mathrm{O}_7 \rightarrow \mathrm{Cr}^{3+}$

Oxidation half reaction: $\mathrm{I}^{-} \rightarrow \mathrm{I}_2$

To balance reduction half reaction.

$$\mathrm{Cr}_2 \mathrm{O}_7^{2-}+14 \mathrm{H}^{+}+6 \mathrm{e}^{-} \longrightarrow 2 \mathrm{Cr}^{3+}+7 \mathrm{H}_2 \mathrm{O}$$

To balance oxidation half reaction

$$2 \mathrm{I}^{-} \longrightarrow \mathrm{I}_2+2 \mathrm{e}^{-}$$

To balance the reaction by electrons gained and lost

$$\mathrm{Cr}_2 \mathrm{O}_7^{2-}+14 \mathrm{H}^{+}+6 \mathrm{e}^{-} \longrightarrow 2 \mathrm{Cr}^{3+}+7 \mathrm{H}_2 \mathrm{O}$$

$6 \mathrm{I}^{-} \longrightarrow 3 \mathrm{I}_2+6 \mathrm{e}^{-}$

$\mathrm{Cr}_2 \mathrm{O}_7^{2-}+14 \mathrm{H}^{+}+6 \mathrm{I}^{-} \longrightarrow 2 \mathrm{Cr}^{3+}+3 \mathrm{I}_2+7 \mathrm{H}_2 \mathrm{O}$

This gives the final balanced ionic equations. (b) Write the skeletal equation of the given reaction

$$\mathrm{Cr}_2 \mathrm{O}_7^{2-}(a q)+\mathrm{Fe}^{2+}(a q) \longrightarrow \mathrm{Cr}^{3+}(a q)+\mathrm{Fe}^{3+}(a q)$$

Write the O. N. of all the elements above their respective symbols.

Divide the given equation into two half reactions Oxidation half reaction:

$\mathrm{Fe}^{2+}(a q) \rightarrow \mathrm{Fe}^{3+}(a q)$ reduction half reaction : $\mathrm{Cr}_2 \mathrm{O}_7^{2-}(\mathrm{aq}) \rightarrow \mathrm{Cr}^{3+}(a q)$

To balance oxidation half reaction

$$\mathrm{Fe}^{2+}(a q) \rightarrow \mathrm{Fe}^{3+}(a q)+\mathrm{e}^{-}$$

To balance reduction half reaction

$$\mathrm{Cr}_2 \mathrm{O}_7^{2-}(a q)+6 \mathrm{e}^{-} \longrightarrow 2 \mathrm{Cr}^{3+}(a q)$$

Balance charge by adding $\mathrm{H}^{+}$ ions.

$$\mathrm{Cr}_2 \mathrm{O}_7^{2-}(\mathrm{aq})+14 \mathrm{H}^{+}(\mathrm{aq})+6 \mathrm{e}^{-} \longrightarrow 2 \mathrm{Cr}^{3+}(a q)$$

Balance O atoms by adding $\mathrm{H}_2 \mathrm{O}$ molecules

$$\mathrm{Cr}_2 \mathrm{O}_7^{2-}(a q)+14 \mathrm{H}^{+}(a q)+6 \mathrm{e}^{-} \longrightarrow 2 \mathrm{Cr}^{3+}(a q)+7 \mathrm{H}_2 \mathrm{O}(l)$$

To balance the reaction

$$\begin{gathered} 6 \mathrm{Fe}^{2+}(a q) \longrightarrow 6 \mathrm{Fe}^{3+}(a q)+6 \mathrm{e}^{-} \\ \mathrm{Cr}_2 \mathrm{O}_7^{2-}(a q)+14 \mathrm{H}^{+}(a q)+6 \mathrm{e}^{-} \longrightarrow 2 \mathrm{Cr}^{3+}(a q)+7 \mathrm{H}_2 \mathrm{O}(l) \end{gathered}$$

(c) Write the O.N. of all atoms above their respective symbols.

Divide the skeleton equation into two half-reactions.

Reduction half reaction: $\mathrm{MnO}_4^{-} \longrightarrow \mathrm{Mn}^{2+}$

Oxidation half reaction : $\mathrm{SO}_3^{2-} \longrightarrow \mathrm{SO}_4^{2-}$

To balance reduction half reaction

$$\mathrm{MnO}_4^{-}+8 \mathrm{H}^{+}+5 \mathrm{e}^{-} \longrightarrow \mathrm{Mn}^{2+}+4 \mathrm{H}_2 \mathrm{O}$$

To balance oxidation half reaction

$$\mathrm{SO}_3^{2-} \longrightarrow \mathrm{SO}_4^{2-}+2 e^{-}$$

Balance charge by adding $\mathrm{H}^{+}$ions.

$$\mathrm{SO}_3^{2-} \longrightarrow \mathrm{SO}_4^{2-}+2 \mathrm{H}^{+}+2 \mathrm{e}^{-}$$

Balance O -atoms by adding $\mathrm{H}_2 \mathrm{O}$ molecules

$$\mathrm{SO}_3^{2-}+\mathrm{H}_2 \mathrm{O} \longrightarrow \mathrm{SO}_4^{2-}+2 \mathrm{H}^{+}+2 \mathrm{e}^{-}$$

To balance the reaction

$$\begin{aligned} & 2 \mathrm{MnO}_4^{-}+16 \mathrm{H}^{+}+10 \mathrm{e}^{-} \longrightarrow 2 \mathrm{Mn}^{2+}+8 \mathrm{H}_2 \mathrm{O} \\ & 5 \mathrm{SO}_3^{2-}+5 \mathrm{H}_2 \mathrm{O} \longrightarrow 5 \mathrm{SO}_4^{2-}+10 \mathrm{H}^{+}+10 e^{-} \end{aligned}$$

$2 \mathrm{MnO}_4^{-}+5 \mathrm{SO}_3^{2-}+6 \mathrm{H}^{+} \longrightarrow 2 \mathrm{Mn}^{2+}+5 \mathrm{SO}_4^{2-}+3 \mathrm{H}_2 \mathrm{O}$

This represents the correct balanced redox equation.

(d) Write the O. N. of all the atoms above their respective symbols.

(d) Write the O.N. of all the atoms above their respective symbols.

Divide skeleton equation into two half reactions

Reduction half reaction $\mathrm{MnO}_4^{-} \rightarrow \mathrm{Mn}^{2+}$

Oxidation half reaction $\mathrm{Br}^{-} \rightarrow \mathrm{Br}_2$

To balance reduction half reaction

$$\mathrm{MnO}_4^{-}+8 \mathrm{H}^{+}+5 \mathrm{e}^{-} \longrightarrow \mathrm{Mn}^{2+}+4 \mathrm{H}_2 \mathrm{O}$$

To balance oxidation half reaction

$$2 \mathrm{Br}^{-} \longrightarrow \mathrm{Br}_2+2 e^{-}$$

To balance the reaction

$$\begin{aligned} 2 \mathrm{MnO}_4^{-}+16 \mathrm{H}^{+}+10 e^{-} & \longrightarrow 2 \mathrm{Mn}^{2+}+8 \mathrm{H}_2 \mathrm{O} \\ 10 \mathrm{Br}^{-} & \longrightarrow 5 \mathrm{Br}_2+10 e^{-} \end{aligned}$$

$$2 \mathrm{MnO}_4^{-}+10 \mathrm{Br}^{-}+16 \mathrm{H}^{+} \longrightarrow 2 \mathrm{Mn}^{2+}+5 \mathrm{Br}_2+8 \mathrm{H}_2 \mathrm{O}$$

This represents the correct balanced ionic equation.

$$\mathrm{A.\to(4)\quad B.\to(5)\quad C.\to(3)\quad D.\to(1)}$$

Suppose that $x$ be the oxidation states of central atoms.

A. Oxidation number of Cr in $\mathrm{Cr}_2 \mathrm{O}_7^{2-}$

$$\begin{aligned} 2 x+7(-2) & =-2 \\ 2 x-14 & =-2 \\ 2 x & =+12 \\ x & =+6 \end{aligned}$$

$$\begin{aligned} &\text { B. Oxidation number of } \mathrm{Mn} \text { in } \mathrm{MnO}_4^{-}\\ &\begin{aligned} x+4(-2) & =-1 \\ x-8 & =-1 \\ x & =+7 \end{aligned} \end{aligned}$$

$$\begin{aligned} &\text { C. Oxidation number of } \mathrm{V} \text { in } \mathrm{VO}_3^{-}\\ &\begin{aligned} x+3(-2) & =-1 \\ x-6 & =-1 \\ x & =+5 \end{aligned} \end{aligned}$$

$$\begin{aligned} &\text { D. Oxidation number of } \mathrm{Fe} \text { in } \mathrm{FeF}_6^{3-}\\ &\begin{aligned} x+6(-1) & =-3 \\ x-6 & =-3 \\ \text { or }\quad x & =+3 \end{aligned}\\ \end{aligned}$$