In inorganic chemistry, what is the correct molecular geometry (shape) of iodine heptafluoride, IF7, according to the VSEPR model?

Introduction / Context:
Molecular geometry is an important topic in inorganic chemistry and is often analysed using the Valence Shell Electron Pair Repulsion (VSEPR) model. Iodine heptafluoride, IF7, is a classic example of a hypervalent molecule in which the central atom has more than eight electrons in its valence shell. This question asks you to recall or deduce the correct geometry of IF7, which is a commonly tested example in exams dealing with molecular shapes.

Given Data / Assumptions:
- The central atom in the molecule is iodine (I). - There are seven fluorine (F) atoms bonded to iodine, giving the formula IF7. - Options include trigonal bipyramidal, pentagonal bipyramidal, square pyramidal, and tetrahedral geometries. - We assume the use of the VSEPR model for predicting molecular shape.

Concept / Approach:
According to VSEPR theory, the shape of a molecule is determined by the repulsion between electron pairs around the central atom. In IF7, iodine uses an expanded valence shell to accommodate seven bonding pairs with no lone pairs. The arrangement that minimises repulsion among seven bonding pairs is a pentagonal bipyramidal geometry, where five fluorine atoms occupy the vertices of a pentagon in one plane and two fluorine atoms occupy axial positions above and below this plane. The correct approach is to count the electron domains and match them to the known geometries for seven domains.

Step-by-Step Solution:
Step 1: Determine the total number of electron pairs around the central iodine atom in IF7. There are seven I–F bonds and, in this structure, no lone pairs on iodine. Step 2: Recognise that seven bonding pairs with no lone pairs correspond to a coordination number of seven for the central atom. Step 3: Recall the standard VSEPR shapes: 5 electron pairs give trigonal bipyramidal, 6 give octahedral, and 7 give pentagonal bipyramidal geometry when there are no lone pairs. Step 4: Compare each option. Option a explicitly mentions pentagonal bipyramidal geometry with five fluorine atoms in a plane and two axial, which matches the VSEPR prediction. Step 5: Option b, trigonal bipyramidal, is appropriate for five bonding pairs, not seven, so it cannot describe IF7 correctly. Step 6: Option c, square pyramidal, is associated with five bonding pairs and one lone pair, such as in BrF5, not with seven bonds and no lone pairs. Step 7: Option d, tetrahedral, only accounts for four bonding pairs and is obviously not suitable for a heptafluoride.

Verification / Alternative check:
Textbooks in coordination chemistry and main group chemistry list IF7 as a classic pentagonal bipyramidal molecule. Structural studies support this geometry, showing five fluorine atoms arranged in a regular pentagon around iodine, with two more fluorine atoms aligned axially. This arrangement gives each bond approximately equal repulsion and stabilises the molecule. No alternative geometry among the options can accommodate seven equivalent bond pairs with minimal repulsion as effectively as the pentagonal bipyramidal shape.

Why Other Options Are Wrong:
Trigonal bipyramidal geometry is associated with five electron pairs, not seven, so option b is inconsistent with the formula IF7. Square pyramidal geometry normally corresponds to six electron domains (five bonds and one lone pair) and fits compounds like BrF5, not IF7. Tetrahedral geometry is limited to four bonding pairs and cannot describe a seven-coordinate molecule. Therefore, options b, c, and d do not match the coordination number and electron pair arrangement present in IF7.

Common Pitfalls:
Students sometimes incorrectly generalise that any molecule with more than four bonds must be trigonal bipyramidal or octahedral and forget that coordination numbers greater than six are possible for heavier elements. Another pitfall is mixing up examples: BrF5 is square pyramidal, SF6 is octahedral, and IF7 is pentagonal bipyramidal. Memorising a small set of standard examples with their shapes, and relating them back to the number of bonding pairs and lone pairs, helps avoid confusion in exam situations.

Final Answer:
The correct molecular geometry of IF7 is pentagonal bipyramidal, with five fluorine atoms in a plane and two in axial positions.