Shapes of certain interhalogen compounds are stated below. Which one of them is not correctly stated?

To solve this question, we need to evaluate the given shapes of interhalogen compounds and identify which one is incorrectly stated. Let's analyze each option step by step.

Starting with option A: IF$$_7$$. Iodine heptafluoride (IF$$_7$$) has iodine as the central atom bonded to seven fluorine atoms. Iodine has seven valence electrons, and each bond uses one electron from iodine. Since there are seven bonds and no lone pairs on iodine, the total number of electron pairs around iodine is seven. For seven electron pairs, the geometry is pentagonal bipyramidal. Therefore, this statement is correct.

Next, option B: BrF$$_5$$. Bromine pentafluoride (BrF$$_5$$) has bromine as the central atom bonded to five fluorine atoms. Bromine has seven valence electrons. Five electrons are used in bonding with fluorine atoms, leaving two electrons, which form one lone pair. Thus, bromine has five bonding pairs and one lone pair, making a total of six electron pairs. For six electron pairs, the electron geometry is octahedral. However, the molecular geometry is determined by the arrangement of atoms, not lone pairs. With one lone pair occupying an axial position, the five fluorine atoms arrange in a square pyramidal shape. Trigonal bipyramidal geometry is for five electron pairs (e.g., PF$$_5$$), not six. Therefore, stating BrF$$_5$$ as trigonal bipyramidal is incorrect.

Now, option C: BrF$$_3$$. Bromine trifluoride (BrF$$_3$$) has bromine bonded to three fluorine atoms. Bromine has seven valence electrons. Three electrons are used in bonding, leaving four electrons, which form two lone pairs. So, bromine has three bonding pairs and two lone pairs, totaling five electron pairs. The electron geometry is trigonal bipyramidal. The two lone pairs occupy equatorial positions to minimize repulsion, resulting in a planar T-shaped molecular geometry. Hence, this statement is correct.

Finally, option D: ICl$$_3$$. Iodine trichloride (ICl$$_3$$) is not typically monomeric. Iodine has seven valence electrons. In a hypothetical monomer, iodine would have three bonds and two lone pairs (five electron pairs), leading to a T-shaped geometry. However, ICl$$_3$$ exists as a dimer, I$$_2$$Cl$$_6$$, where two iodine atoms are bridged by chlorine atoms. This dimeric structure is planar, similar to Al$$_2$$Cl$$_6$$. Therefore, describing it as planar dimeric is correct.

After evaluating all options, option B is not correctly stated because BrF$$_5$$ has a square pyramidal geometry, not trigonal bipyramidal. Hence, the correct answer is Option B.