$$\mathop {2{K_2}Mn{O_4}}\limits^{ + 6} + \mathop {C{l_2}}\limits^0 \to 2K\mathop {Cl}\limits^{ - 1} + \mathop {2KMn{O_4}}\limits^{ + 7} $$
$$Mn \to $$  Increase in oxidation number hence oxidation.
$$C{l_2} \to $$  Decrease in oxidation number hence reduction.
$$C{l_2}$$  is acting as an oxidising agent.

$$\eqalign{ & S{O_2}\,{\text{is an oxidising agent}} \cr & 4{e^ - } + S{O_2} \to S \cr & {\text{Equivalent of}}\,S{O_2} = \frac{{64}}{4} = 16. \cr} $$

$$\eqalign{ & ClO_3^ - \to Cl_2^0 \cr & x - 6 = - 1\,\,x = 0 \cr & x = + 5\,\,\,\,x = 0\,\left( {x = {\text{oxidation number}}} \right) \cr & {\text{Equivalent mass}} \cr & = \frac{{{\text{Molecular mass}}}}{{{\text{Oxidation number}}}} \cr & = \frac{{84.45}}{5} \cr & = 16.89 \cr} $$

$$\eqalign{ & F{e_{0.94}}O:x \times 0.94 - 2 = 0 \cr & \therefore x = \frac{2}{{0.94}} = \frac{{200}}{{94}} \cr & \therefore {\text{Oxidation state of}}\,\,Fe = \frac{{200}}{{94}} \cr} $$
$$\left[ {Cr{{\left( {PP{h_3}} \right)}_3}{{\left( {CO} \right)}_3}} \right]:$$     $$x + \left( {0 \times 3} \right) + \left( {0 \times 3} \right) = 0,x = 0$$
$$\therefore {\text{Oxidation state of}}\,\,Cr = 0.$$

$${K_2}C{r_2}{O_7} + 7{H_2}S{O_4} + 6FeS{O_4} \to $$       $$3F{e_2}{\left( {S{O_4}} \right)_3} + C{r_2}{\left( {S{O_4}} \right)_3} + 7{H_2}O$$       $$ + {K_2}S{O_4}$$
$${\text{In ionic form}}\,:$$
$$2{K^ + } + C{r_2}O_7^{2 - } + 14{H^ + } + 7SO_4^{2 - }$$       $$ + 6F{e^{2 + }} + 6SO_4^{2 - } \to 6F{e^{3 + }} + 9SO_4^{2 - }$$       $$ + 2C{r^{3 + }} + 3SO_4^{2 - } + 2{K^ + } + SO_4^{2 - }$$       $$ + 7{H_2}O$$
$${\text{Eliminating the common species :}}$$
$$C{r_2}O_7^{2 - } + 14{H^ + } + 6F{e^{2 + }} \to $$       $$6F{e^{3 + }} + 2C{r^{3 + }} + 7{H_2}O$$

Since there is loss of electrons, hence it is oxidation reaction.

Since $$X$$ displaces $$Ni$$  from $$NiS{O_4}$$  solution, it means $$X$$ has higher oxidation potential or lower reduction potential than $$N{i^{2 + }}.$$  Since $$X$$ cannot displace $$Mn$$  from $$MnS{O_4}$$  solution, it means it has higher reduction potential than $$M{n^{2 + }}.$$  Lower the reduction potential, stronger is the reducing agent hence reducing power $$Mn > X > Ni.$$

In a metal displacement reaction, a metal in a compound is displaced by another metal in uncombined state.

$${I^ - }$$ is most easily oxidised due to lowest reduction potential among halides.

In (i) and (ii) both $$P$$ and $$S$$ are in highest oxidation state. In (iii) and (iv) ; $$P$$ has oxidation state of $$+4$$  which can be oxidized to $$+5$$  state, while in case of $$N{H_3}$$  nitrogen has oxidation state of $$–3$$  which can be oxidised.