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Considering the elements F, Cl, O and N the correct order of their chemical reactivity in terms of oxidizing property is:
A.$F $ > $Cl$ > $O$ > $N$
B.$F$ > $O$ > $Cl$ > $N$
C.$Cl$ > $F$ > $O$ > $N$
D.$O$ > $F$ > $N$ > $Cl$

Answer
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Hint: To solve this question, we must know that in a group, oxidizing power, i.e. the tendency to gain electrons, decreases from top to bottom as the size increases. But in a period, oxidizing power increases when we move left to right because the size decreases.

Complete answer:
- The oxidizing character of elements increases from left to right across a period because of the presence of vacant d orbitals in their valence shells. Thus, we get the decreasing order of oxidizing property as $F$ > $O$ > $N$.
- Again, the oxidizing character of elements decreases down a group. Thus, we get $F$ > $Cl$.
- Since Oxygen is more electronegative than Cl, O is a stronger oxidizing agent than Cl. Also, F is the best electronegative element. Hence, it is the best oxidizing agent.
- However, the oxidizing character of O is more than that of Cl i.e., $O$ > $Cl$.
- Hence, the correct order of chemical reactivity of F, Cl, O, and N in terms of their oxidizing property is $F$ > $O$ > $Cl$ > $N$.

Therefore, the answer to the question is (B) $F$ > $O$ > $Cl$ > $N$

Note:
As we go across a period of a periodic table, there is an increase in the tendency of accepting electrons. The atomic size decreases with a further increase in nuclear charge.