According to VSEPR theory, what is the molecular geometry (shape) of interhalogen compound IF7 (iodine heptafluoride)?

Introduction / Context:
Valence Shell Electron Pair Repulsion (VSEPR) theory is used in chemistry to predict the shapes of molecules based on the repulsion between electron pairs around a central atom. Interhalogen compounds, where halogens bond with one another, provide interesting examples of unusual geometries. IF7, iodine heptafluoride, is a classic case often mentioned in textbooks. This question asks you to identify the correct molecular geometry of IF7 according to VSEPR theory, testing your knowledge of advanced molecular shapes.

Given Data / Assumptions:

• The compound is IF7, which means iodine bonded to seven fluorine atoms.
• Iodine is the central atom and forms seven I–F bonds.
• We assume that iodine in this compound has no lone pairs in its valence shell, just bonding pairs.
• The options include several possible geometries: pentagonal bipyramid, square pyramid, trigonal bipyramid, octahedral and tetrahedral.

Concept / Approach:
According to VSEPR theory, the shape of a molecule is determined by the arrangement of electron pairs around the central atom that minimises repulsion. Seven bonding pairs and no lone pairs around a central atom correspond to an AX7 type configuration. The geometry that minimises repulsions for seven electron pairs is a pentagonal bipyramidal arrangement, where five atoms form a planar pentagon around the central atom and two atoms occupy axial positions above and below that plane. Therefore, IF7 is predicted to have a pentagonal bipyramidal geometry, not square pyramidal, trigonal bipyramidal, octahedral or tetrahedral geometries, which correspond to other coordination numbers.

Step-by-Step Solution:
Step 1: Count the valence electrons on iodine. As a group 17 element, iodine has seven valence electrons. Step 2: In IF7, iodine forms seven covalent bonds with seven fluorine atoms, using all seven of its valence electrons in bonding. Step 3: Because each bond uses one electron from iodine, there are seven bonding pairs and no lone pairs on the central iodine atom. Step 4: In VSEPR notation, this gives an AX7 type, where A is the central atom and X represents a bonded atom. Step 5: For seven electron pairs, the geometry that minimises repulsions is pentagonal bipyramidal: five fluorine atoms arranged in a planar pentagon around iodine and two fluorine atoms along an axis perpendicular to that plane. Step 6: Other geometries listed correspond to different numbers of electron pairs: trigonal bipyramidal corresponds to AX5, octahedral to AX6 and tetrahedral to AX4. Step 7: Therefore, the correct molecular geometry for IF7 is pentagonal bipyramid.

Verification / Alternative check:
Structural studies, such as X ray diffraction, confirm that IF7 has seven equivalent I–F bonds and that five of these are in one plane forming a pentagon, with two additional fluorine atoms occupying axial positions. This matches the pentagonal bipyramidal arrangement predicted by VSEPR theory. Advanced inorganic chemistry texts specifically list IF7 as the standard example of a molecule with AX7 electron pair geometry. By comparing with other known geometries, such as PF5 (trigonal bipyramidal) and SF6 (octahedral), you can see that additional bonding pairs require expanding the coordination number to seven, which yields the pentagonal bipyramidal shape.

Why Other Options Are Wrong:
Option B, square pyramid, corresponds to AX5E type geometries, such as BrF5, where there are five bonds and one lone pair. Option C, trigonal bipyramid, is the geometry for AX5 molecules like PF5 with five bonding pairs. Option D, octahedral, describes AX6 molecules such as SF6, with six bonding pairs. Option E, tetrahedral, corresponds to AX4 molecules like CH4 with four bonding pairs. None of these geometries involve seven bonding pairs with no lone pairs on the central atom. Only option A, pentagonal bipyramid, matches the seven coordinate arrangement of IF7.

Common Pitfalls:
Students sometimes struggle with geometries beyond the most common shapes (linear, trigonal planar, tetrahedral, trigonal bipyramidal and octahedral) and may guess octahedral or trigonal bipyramidal out of habit. Another pitfall is miscounting electron pairs or forgetting to consider lone pairs on the central atom. To avoid these errors, systematically count valence electrons, determine the number of bonding and lone pairs and then match the AXn or AXnE notation with the known VSEPR geometries. Remembering that IF7 is the textbook example of a pentagonal bipyramidal molecule can also help fix this shape in memory.

Final Answer:
The molecular geometry of IF7 (iodine heptafluoride) according to VSEPR theory is Pentagonal bipyramid.