Question

Write Lewis structures for \(\mathrm{HCP}\) and \(\left[\mathrm{IOF}_{4}\right]^{-} .\) Use VSEPR theory to predict the electron-region and molecular geometries of these species, and note any differences between these geometries.

Short answer

HCP is linear; IOF4- is seesaw.

Step-by-step solution

Determine Valence Electrons for HCP

Calculate the total number of valence electrons for the molecule HCP. Hydrogen (H) has 1 electron, carbon (C) has 4 electrons, and phosphorus (P) has 5 electrons. Thus, HCP has a total of \(1 + 4 + 5 = 10\) valence electrons.

Draw Lewis Structure for HCP

The structure is linear with C as the central atom, forming a C−H bond and a C≡P triple bond. This uses all 10 electrons, with shared electrons in the bonds. This configuration ensures each atom achieves a stable electron configuration.

VSEPR Theory for HCP

With two bonding regions around the central carbon atom and no lone electron pairs, the electron-region geometry is linear. Consequently, the molecular geometry is also linear.

Determine Valence Electrons for IOF4-

Calculate the total number of valence electrons. Iodine (I) has 7 electrons, oxygen (O) has 6 electrons, and each fluorine (F) has 7 electrons. There are 4 fluorines, so we have a total of \(7 + 6 + 4 \times 7 + 1 = 40\) valence electrons, including the additional electron for the negative charge.

Draw Lewis Structure for IOF4-

Iodine is the central atom, forming single bonds with the four F atoms and a double bond with O. This accounts for 10 electrons in bonds (5 bonds × 2 electrons each). The remaining 30 electrons make 5 lone pairs around I, F, and O, with I having one lone pair (10 electrons), each F having 3 lone pairs (14 electrons), and O having 2 lone pairs (6 electrons).

VSEPR Theory for IOF4-

The central iodine atom in IOF4- has 5 electron regions (4 bonding and 1 lone pair). According to VSEPR theory, the electron-region geometry is trigonal bipyramidal. With one lone pair, the molecular shape is distorted from this geometry, leading to a seesaw molecular geometry.

Key concepts

VSEPR Theory

Lewis structures lay the groundwork for understanding molecules, but VSEPR Theory takes us a step further by predicting their shape. VSEPR, which stands for Valence Shell Electron Pair Repulsion, helps us visualize the three-dimensional arrangement of atoms in a molecule.
According to this theory, electron pairs in the valence shell of a central atom arrange themselves to be as far apart as possible. This arrangement minimizes repulsion between like charges. For example, the carbon atom in the linear HCP molecule aligns to minimize electron pair repulsions.
In the case of IOF4-, the iodine atom has five electron regions, including one lone pair. According to VSEPR Theory, these regions adopt a shape that helps balance repulsions, predicting an electron-domain geometry of trigonal bipyramidal, but altering the angles due to the lone pair, resulting in a seesaw molecular shape.

Valence Electrons

To accurately draw Lewis structures and predict molecular shapes using VSEPR, a fundamental step is determining the number of valence electrons. Valence electrons are those in the outermost electron shell that are involved in chemical bonding. They are critical in determining how atoms bond and form molecules.
For HCP, the valence electrons total to 10. This comes from adding 1 electron from hydrogen, 4 from carbon, and 5 from phosphorus. These electrons are distributed to fulfill the bonding requirements, achieving stable configurations for each atom.
In the case of IOF4-, it's essential to include the extra electron due to its negative charge. Thus, iodine, oxygen, and fluorines contribute a total of 40 valence electrons, accounting also for the additional one due to the negative charge. Accurately counting these electrons ensures correct Lewis structures and predictions of molecular geometries.

Molecular Geometry

Molecular geometry describes the precise 3D arrangement of atoms in a molecule, and it can differ from electron-domain geometry by the presence of lone pairs. Understanding and predicting molecular geometry is essential for grasping how molecules fit into and interact with their environment.
For HCP, the molecular geometry is straightforwardly linear, as dictated by the shared electron pairs forming bonds between the hydrogen, carbon, and phosphorus atoms.
The molecular geometry of IOF4- is more complex. Although the electron-domain geometry is trigonal bipyramidal, the presence of a lone pair on the iodine central atom distorts the geometry to what is known as a "seesaw" shape. This shape reflects the effort to maintain minimal electron pair repulsion, with lone pairs occupying equatorial positions whenever possible.

Electron-domain Geometry

Electron-domain geometry involves the spatial arrangement of all electron regions (bonding and non-bonding) around a central atom. Comprehending this concept helps predict the shape and orientation of molecules in space.
For molecules like HCP, where there are no lone pairs, the molecular and electron-domain geometries are the same—linear in this case, since there are two regions of electron density that align oppositely.
However, in IOF4-, electron-domain geometry takes into account the lone pair and the four bond pairs. This results in a trigonal bipyramidal geometry. The electron pairs, including the lone pairs on the central iodine, determine the electron-domain arrangement to minimize repulsion. When lone pairs are involved, they influence the overall molecular shape by bending and shaping the bond angles differently, as seen in the seesaw shape of IOF4-.