Question

Only iodine forms heptafluoride \[\text{I}{{\text{F}}_{7}}\], but chlorine and bromine give pentafluorides. The reason for this is:
A. Low electron affinity of iodine
B. unusual pentagonal bipyramidal
C. That the larger iodine atom can accommodate more number of small fluorine atoms around it.
D. low chemical reactivity of $\text{I}{{\text{F}}_{7}}$

Answer 1

Hint: Atomic size trend differs in the period and group that is as we move down the group the size of the atoms increases due to the addition of a new shell whereas when we move from left to right along the period then the size of the atom decreases due to the increase in effective nuclear charge.

Complete step by step Answer:
-As we know that an atomic number of iodine, chlorine and bromine is 53, 17 and 35.
-So, we can say that the size of iodine is bigger than the chlorine and bromine because we know that moving down the group there will be decreases in the atomic size of the atom.
-Also, a new shell adds in each element as we move down the group.
-Due to which affinity to accommodate more bonds increases.
-That's why iodine makes heptafluoride and chlorine and bromine pentafluoride.
So, we can see that this matches with the option C.
Now, let's look at the other options.
-As due to the larger size of iodine the affinity to attract more electron pairs increase.
-So, option A is the incorrect answer.

-Pentagonal bipyramidal is a geometry of a molecule in which one central atom makes 7 bond pairs or ligands. It has hybridisation of $\text{s}{{\text{p}}^{3}}{{\text{d}}^{3}}$ which is the same as that of $\text{I}{{\text{F}}_{7}}$.
-So, it does not affect the bond formation or affinity to make bonds of the iodine.
-Hence, option B is an incorrect answer.

-$\text{I}{{\text{F}}_{7}}$ have less reactivity as compared to $\text{Cl}{{\text{F}}_{5}}\text{ and Br}{{\text{F}}_{5}}$.
-It is so because the bond between interhalogen compounds is weaker than the bond in pure halogens.
-But this statement is not the correct explanation. That's why option D is also an incorrect answer.

Therefore, option C is the correct answer.

Note: $\text{Cl}{{\text{F}}_{5}}\text{ and Br}{{\text{F}}_{5}}$has the hybridisation of $\text{s}{{\text{p}}^{3}}{{\text{d}}^{3}}$due to which they have square pyramidal geometry. The formula to calculate the hybridisation is 1/2 (V + M - C +A). Here, V is the valence electron of the central atom, M is the number of the monovalent atom, C is the total cationic charge and A is the total anionic charge.