$$\mathop {Ca}\limits^{ + 2} \mathop C\limits^{ + 4} \mathop {{O_3}}\limits^{ - 2} \to \mathop {CaO}\limits^{ + 2 - 2} + \mathop {C{O_2}}\limits^{ + 4 - 2} $$
Since there is no change in oxidation number of any species, it is not a redox reaction.
155.
In the balanced chemical reaction $$IO_3^ - + a{I^ - } + b{H^ - } \to c{H_2}O + d{I_2}$$
$$a, b, c$$ and $$d,$$ respectively, correspond to
(A) and (B) are neutralisation reactions. The oxidation state of $$Cu$$ is +2 in both reactant and product, and $$SO_4^{2 - }\,ion$$ does not change. In option (D) is a redox reaction.
$$\eqalign{
& Zn \to Z{n^{2 + }} + 2{e^ - }\left( {{\text{Oxidation}}} \right) \cr
& 2{H^ + } + 2{e^ - } \to {H_2}\left( {{\text{Reduction}}} \right) \cr} $$
159.
For the reaction : $${I^ - } + ClO_3^ - + {H_2}S{O_4} \to $$ $$C{l^ - } + HSO_4^ - + {I_2}$$
The incorrect statement in the balanced equation is
A
stoichiometric coefficient of $$HSO_4^ - $$ is 6
\[{{H}_{2}}S{{O}_{5}};\,H-O-\underset{\begin{smallmatrix}
\downarrow \\
O
\end{smallmatrix}}{\overset{\begin{smallmatrix}
O \\
\uparrow
\end{smallmatrix}}{\mathop{S}}}\,-O-O-H\]
Oxidation number of $$O$$ in peroxo linkage = - 1
Oxidation number of other $$O$$ atoms = - 2
$$ + 2 + x - 6 - 2 = 0\,\,{\text{or}}\,\,x = + 6$$