Calculate oxidation number of each element in KClO3 and CaCO3

Dear Student,

$$\begin{gathered} \left(\mathrm{i}\right) \mathrm{Oxidation} \mathrm{State} \mathrm{of} \mathrm{each} \mathrm{element} \mathrm{in} {\mathrm{KClO}}_3: \\ \mathrm{Oxidation} \mathrm{state} \mathrm{of} \mathbf{K}\mathbf{=}\mathbf{+}\mathbf{1} \\ \mathrm{Cl}=\mathrm{x} \\ \mathbf{O}\mathbf{=}\mathbf{-}\mathbf{2} \\ 1+ \mathrm{x} + 3 (-2) = 0 \\ \mathrm{x}-5 = 0 \\ \mathbf{x}\mathbf{=}\mathbf{+}\mathbf{5} \end{gathered}$$

$$\begin{gathered} \left(\mathrm{ii}\right) \mathrm{Oxidation} \mathrm{State} \mathrm{of} \mathrm{each} \mathrm{element} \mathrm{in} {\mathrm{CaCO}}_3: \\ \mathrm{Oxidation} \mathrm{state} \mathrm{of} \mathbf{Ca}\mathbf{ }\mathbf{=}\mathbf{+}\mathbf{2} \\ \mathrm{C}=\mathrm{x} \\ \mathbf{O}\mathbf{=}\mathbf{-}\mathbf{2} \\ 2 +\hspace{0.17em}\mathrm{x} + 3 (-2) = 0 \\ \mathrm{x}-4 = 0 \\ \mathbf{x}\mathbf{=}\mathbf{+}\mathbf{4} \end{gathered}$$

Hope this information clears your doubts about the topic.

Keep asking!!

Regards