Question

Which one of the following statements is incorrect?
(A) All halogens have weaker X-X bond than X-X’ bond, in interhalogens (except F-F bond in fluorine
(B) Radius ratio between iodine and fluorine (among halogens) is maximum
(C) Interhalogen compounds are more reactive than halogen compounds
(D) Among interhalogen compounds, maximum number of atoms are present in iodine fluoride

Answer 1

Hint: In the interhalogen compounds both the atoms will have differences in their electronegativities, which makes their bond polar. While between similar halogen atoms there is no difference in electronegativities.

Complete step by step answer:
-For (A): This statement says that X-X bonds are weaker than X-X’ bonds. We all know that this is not true. The interhalogen bonds are weaker bonds as compared to bonds between halogen atoms and thus interhalogen compounds are more reactive as compared to halogens. The reason for this behaviour of interhalogens is that the electronegativities of both the atoms in an interhalogen is different, making this bond polarised and so it can break easily. While the electronegativities of both the atoms in halogens is the same and so their bond is not polarised.
For example: ${I_2}$ bond is much stronger than I-Cl bond due to the difference in electronegativities of iodine and chlorine. So, the statement (A) given in the question is false.

-For (B): In a periodic table when we move from top to bottom the number of shells increases and so the size of the atoms in a group also increases. Since among halogens iodine (I) is at the lowermost end of the periodic table it has the largest size (maximum radius) and fluorine has the smallest size (shortest radius). So, their radius ratio will be the largest.
Thus this statement will be true.

-For (C): As we talked while discussing statement (A), the bond between the two atoms in an interhalogen compound is weaker than the bond between two similar halogen atoms due to difference in electronegativities of different halogen atoms. So, the interhalogen bond being weaker can be easily broken and thus interhalogens are more reactive than halogen compounds.
So, this statement is also true.

-For (D): It states that among all interhalogen compounds maximum numbers of atoms are found in iodine fluoride. This is true. Iodine fluorides are of 4 types: iodine monofluoride (IF), iodine trifluoride ($I{F_3}$), iodine pentafluoride ($I{F_5}$) and iodine heptafluoride ($I{F_7}$). Iodine monofluoride (IF) is a chocolate brown compound and is unstable by nature. So, at ${0^ \circ }C$ it decomposes and is disproportionate to form iodine pentafluoride and iodine molecules.
This statement is also true.
So, the correct answer is “Option A”. All halogens have weaker X-X bond than X-X’ bond, in interhalogens (except F-F bond in fluorine.

Note: Due to the high reactivity of interhalogen compounds, some of them like $I{F_5}$, ICl and $Br{F_3}$ act as good halogenating agents. Also $Br{F_5}$ is reactive enough to generate fluorine. Most of the interhalogens are covalent gases and all of them are diamagnetic. More is the difference between the electronegativities of both atoms in interhalogens, higher is the boiling point.