Question

The oxidation number of Chlorine in $NaCl{{O}_{3}}$ is:

Answer 1

Hint: Usually the oxidation number of an element is the number of valence electrons. In ionic compounds it is simply the number of electrons gained or lost by an element and chlorine generally shows oxidation state of -1.

Complete step by step solution:

Firstly, let’s have a clear idea what Oxidation and reduction is and how redox reactions occur.

Reduction: It is a process in which an atom gains an electron and therefore decreases (or reduces
its oxidation number). In other words, the positive character of the species is reduced.
Oxidation: It is a process in which an atom loses an electron and therefore increases its oxidation
number. In other words, the positive character of the species is increased.
So basically, redox reactions are reactions in which one species is reduced while the other is
oxidized in which the oxidation states of the species changes ultimately.

Now, let’s come to our current problem.

The oxidation state of O in $NaCl{{O}_{3}}$ is -2 and the oxidation state of Na in $NaCl{{O}_{3}}$ is +1.

The equation to find the oxidation state of is:
$1(Oxidation\,state\,of\,Na)\,+\,1(Oxidation\,state\,of\,Cl)\,+\,3(Oxidation\,state\,of\,O)\,=\,0$

On rearranging the above equation for the oxidation state of Cl.
$\,1(Oxidation\,state\,of\,Cl)=\,-[1(Oxidation\,state\,of\,Na)\,+3(Oxidation\,state\,of\,O)]$

Finally, we will substitute the oxidation state of O i.e. +1 and the oxidation state of Na i.e. +1 in the equation.

& \,Oxidation\,state\,of\,Cl=\,-[1(+1)\,+3(-2)] \\\ & \,\,\,\,\,\,\,\,\,\,\,\,\,\,=-(1-6) \\\ & \,\,\,\,\,\,\,\,\,\,\,\,\,\,=-(-5) \\\ & \,\,\,\,\,\,\,\,\,\,\,\,\,\,=+5 \\\ \end{aligned}$$ Therefore, we have found the oxidation state of Cl in $$NaCl{{O}_{3}}$$ is +5. Note: The reaction by which Sodium chlorate ($$NaCl{{O}_{3}}$$) is formed is a special type of redox reaction called disproportionation reaction, where oxidation and reduction of an element occurs simultaneously.