Predict the powerful oxidizing agent in the 3 rd period: (a) sulphur (b) sodium (c) chlorine (d) bromine

Short Answer

Expert verified
Answer: (c) Chlorine

Step by step solution

01

Understanding oxidizing agents

An oxidizing agent is a substance that can accept electrons from another substance. The strength of an oxidizing agent depends on its ability to accept electrons, which is usually determined by its electronegativity and the change in oxidation state during the reaction.
02

Determining the electronegativity and oxidation state of given elements

Let's determine the electronegativity and common oxidation states of the given elements. You can find these values in a periodic table. (a) Sulphur: Electronegativity = 2.58, Oxidation states = -2, +4, +6 (b) Sodium: Electronegativity = 0.93, Oxidation states = +1 (c) Chlorine: Electronegativity = 3.16, Oxidation states = -1, +1, +3, +5, +7 (d) Bromine: Electronegativity = 2.96, Oxidation states = -1, +1, +3, +4, +5
03

Comparing the oxidizing abilities of given elements

A strong oxidizing agent will have higher electronegativity and a greater increase in its oxidation state during the reaction. The greater electronegativity makes it easier for the atom to attract electrons, while the increase in oxidation state ensures that it gains electrons. After comparing the periods, we can see that chlorine has the highest electronegativity among the given elements and can undergo the greatest increase in oxidation state (from -1 to +7). Thus, it has the strongest oxidizing ability.
04

Concluding which element is the powerful oxidizing agent

Based on the comparison, it is clear that chlorine (\(Cl\)) is the most powerful oxidizing agent among the given elements in the 3rd period. The correct option is (c) chlorine.

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