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Chapter 10: Problem 118

One possibility for the electron-group geometry for seven electron groups is pentagonal-bipyramidal, as found in the IF \(_{7}\) molecule. Write the VSEPR notation for this molecule. Sketch the structure of the molecule, labeling all the bond angles.

Short Answer

Expert verified
The VSEPR notation for the IF\(_7\) molecule is AX7. The molecule has a pentagonal-bipyramidal shape. Bond angles are \(90°\) from the equatorial to axial plane, \(72°\) in the equatorial plane, and \(180°\) in the axial plane.

Step by step solution

01

Identifying the Central Atom

In the IF\(_7\) molecule, iodine (I) is the central atom. Iodine has seven valence electrons. Each of the seven fluoride (F) atoms will share one electron with iodine to form a single bond.
02

Writing the VSEPR Notation

The VSEPR notation for this molecule is AX7 where A represents the central atom (iodine), X represents the surrounding atoms (fluoride), and the number (7) refers to the number of surrounding atoms.
03

Sketching the Structure

The molecular geometry of IF7 according to the VSEPR is a pentagonal bipyramid. The central iodine atom is surrounded by five fluorine atoms forming the base of the pentagonal pyramid. Two fluorine atoms are located above and below the iodine atom along the perpendicular axis forming the 'bi-pyramid'.
04

Labeling the Bond Angles

In a pentagonal-bipyramidal arrangement, the bond angle from the equatorial plane (formed by the five fluorine atoms in the pentagon) to the axial plane (formed by the central iodine atom and the two other fluorine atoms) is \(90°\). The bond angle in the pentagon (equatorial plane) is \(72°\) and the bond angle between the two fluorine atoms in the axial plane is \(180°\).

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