In $$2FeC{l_3} + {H_2}S \to 2FeC{l_2} + 2HCl + S$$
$$O.N.$$  of $$S$$  changes from $$–2$$  to $$0$$ ( hence oxidised ) $$O.N.$$  of $$Fe$$  changes from $$+3$$  to $$+2$$  hence reduced.

$${C_6}{H_{12}}{O_6}:6x + 12 + \left( { - 12} \right) = $$      $$0 \Rightarrow x = 0$$
$$CHC{l_3}:x + 1 + \left( { - 3} \right) = 0$$       $$ \Rightarrow x = + 2$$
$$C{H_3}C{H_3}:x + 3 + x + 3 = 0$$       $$ \Rightarrow x = - 3$$
$${\left( {COOH} \right)_2}:x + x + \left( { - 4 \times 2} \right) + \left( { + 2} \right)$$        $$ = 0 \Rightarrow x = + 3$$

$$\left( {Ni \to N{i^{2 + }}} \right)Ni.$$    loses electrons hence act as reducing agent.

$$\eqalign{ & {\text{Oxides :}}\,{N_2}O \to + 1,NO \to + 2 \cr & {N_2}{O_3} \to + 3,N{O_2} \to + 4,{N_2}{O_5} \to + 5 \cr} $$

A reagent which lowers the oxidation number of an element in a given substance is reducing agent or reductant.

$${I_2} + O{H^ - } \to IO_3^ - + {I^ - } + {H_2}O$$
Oxidation half : $${I_2} + O{H^ - } \to IO_3^ - + {H_2}O$$
Reduction half : $${I_2} \to {I^ - }$$
Oxidation half :
Adding $$O{H^ - },{I_2} + 12O{H^ - } \to 2IO_3^ - + 6{H_2}O$$
Adding electrons : $${I_2} + 12O{H^ - } \to 2IO_3^ - + 6{H_2}O + 10{e^ - }$$
Reduction half : Balancing $${e^ - },{I_2} + 2{e^ - } \to 2{I^ - }$$
$$\left. {{I_2} + 2{e^ - } \to 2{I^ - }} \right] \times 5$$
Adding both reactions
$$6{I_2} + 12O{H^ - } \to 2IO_3^ - + 10{I^ - } + 6{H_2}O$$
Dividing by $$2,3{I_2} + 6O{H^ - } \to IO_3^ - + 5{I^ - } + 3{H_2}O$$

$$Cr{O_2}C{l_2} \to x + \left( { - 4} \right) + \left( { - 2} \right)$$       $$ = 0 \Rightarrow x = + 6$$

All the statements are correct.

In the given reaction, oxidation state of $$Mg$$  is changing from $$0$$ to $$+2,$$  while in nitrogen it is changing from $$0$$ to $$– 3.$$  So oxidation of $$Mg$$  and reduction of nitrogen takes place.

$$\eqalign{ & C{H_3}Cl:x + 3 - 1 = 0 \Rightarrow x = - 2 \cr & CC{l_4}:x - 4 = 0 \Rightarrow x = + 4 \cr & CHC{l_3}:x + 1 - 3 = 0 \Rightarrow x = + 2 \cr & C{H_2}C{l_2}:x + 2 - 2 = 0 \Rightarrow x = 0 \cr} $$