Question

What is the oxidation number of ${\text{Cl}}$ in the ${\text{ClO}}_{\text{3}}^{\text{ - }}$ ion?
A.$ - 1$
B.$ + 5$
C.$ - 5$
D.$ + 7$

Answer 1

Hint: The oxidation state of an atom depends upon its electronic configuration.Put the value for oxygen and find out the oxidation state of ${\text{Cl}}$ in the compound.

Complete step by step answer:
The oxidation state of an atom depends upon its electronic configuration. Mostly, in the case of non-metals, they tend to accept extra electrons to completely fill their valence shell and reach a stable noble gas configuration. They mostly form anions. In the case of metals, they tend to lose electrons from their valence shell to reach a stable configuration. They form cations. Elements of group 17 or halogens, they only require one extra electron to reach noble gas configuration, but sometimes, they act as metals and lose electrons instead to form compounds with more than one atom. This mainly occurs due to the high electron density of halogens and their reactivity due to which the electrons feel repulsion from other electrons.
In the question, the ion ${\text{ClO}}_{\text{3}}^{\text{ - }}$ is present with a net charge of $ - 1$. So, this means the oxidation state of chlorine and oxygen will add up to $ - 1$. In the compound, oxygen being the more electronegative element, gets assigned $ - 2$ oxidation state.
So, if we assume the oxidation state of chlorine as $x$, then: $x + 3 \times ( - 2) = - 1$.
Solving we get,
$ \Rightarrow x = - 1 + 6 = + 5$
So, the oxidation state of ${\text{Cl}}$ in ${\text{ClO}}_{\text{3}}^{\text{ - }}$ is $ + 5$.

$\therefore $ The correct option is option B, i.e. $ + 5$.

Note:
Chlorine is also an electronegative element, but oxygen is more electronegative and smaller in size than chlorine. As a result, electrons in oxygen experience more force of attraction with the nucleus as compared to chlorine and thus, extra electrons are added more easily in oxygen. This is why oxygen attains a negative charge in the compound.