Question

The correct order of increasing oxidizing power is:
A. $C{{l}_{2}}$ < $B{{r}_{2}}$ < ${{F}_{2}}$ < ${{I}_{2}}$
B. ${{F}_{2}}$ < $B{{r}_{2}}$ < $C{{l}_{2}}$ < ${{I}_{2}}$
C. ${{F}_{2}}$ > $B{{r}_{2}}$ > $C{{l}_{2}}$ > ${{I}_{2}}$
D. ${{I}_{2}}$ < $B{{r}_{2}}$ < $C{{l}_{2}}$ < ${{F}_{2}}$

Answer 1

Hint: Think about what it means when a substance is oxidized. How is oxidation related to the number of electrons that are present in an atom or molecule? Consider the trends of the periodic table to determine the answer to this question.

Complete step by step answer:
When any substance is oxidized, it is considered that a net loss of electrons occurs concerning those atoms. So, if the oxidation number of any element is increasing after the reaction, then that element is said to be oxidized.
Now let us look at how easily halogens can oxidize other elements. This will determine their oxidizing power. Since they have to oxidize the other element, they themselves will get reduced and gain electrons. So, if we find the order in which the tendency to gain electrons increases, we will get the oxidizing power of those halogens.
Consider the trend in atomic radius when we go down a group. As we move down the group, shells are added to the atom and the atomic radius increases. As the atomic radius increases appreciably, the effective nuclear charge on each electron (the force with which the nucleus attracts the electron) reduces due to the shielding effect caused by the electrons in the smaller shells. So, as we move down a group, the tendency to lose electrons will increase. The increasing trend in atomic radius is: ${{I}_{2}}$ > $B{{r}_{2}}$ > $C{{l}_{2}}$ > ${{F}_{2}}$

The element with the lowest oxidizing power will be the one that loses electrons easily since it needs to abstract electrons in order to be a good oxidizing agent. The tendency to lose electrons will decrease in the manner: ${{I}_{2}}$ > $B{{r}_{2}}$ > $C{{l}_{2}}$ > ${{F}_{2}}$

So, we can conclude that iodine has the least oxidizing power and fluorine has the most. Hence, the correct answer to this question is . ${{I}_{2}}$ < $B{{r}_{2}}$ < $C{{l}_{2}}$ < ${{F}_{2}}$’
So, the correct answer is “Option D”.

Note: Remember that oxidation means losing electrons and reduction means gaining electrons. Use the mnemonic OIL RIG to remember this. Here, Oxygen Is Losing (electrons) and Reduction Is Gaining (electrons). Also remember that when the oxidizing power is asked, it means how easily a substance can oxidize other substances (reduce itself) and not how easily it can get oxidized.