Question

Confirm that the observed gas-phase structures of \(\mathrm{XeF}_{2}\), \(\mathrm{XeF}_{4}\) and \(\mathrm{XeF}_{6}\) are consistent with the VSEPR model.

Short answer

The VSEPR model predicts linear for \(\mathrm{XeF}_2\), square planar for \(\mathrm{XeF}_4\), and distorted octahedral for \(\mathrm{XeF}_6\), matching their observed structures.

Step-by-step solution

Determine the number of valence electrons for XeF2

The central atom Xenon (Xe) has 8 valence electrons, and each Fluorine (F) atom has 7 valence electrons. For \(\mathrm{XeF}_2\):- Xe contributes 8 electrons.- Two F atoms contribute 2 x 7 = 14 electrons.Total: \(8 + 14 = 22\) electrons.

Apply VSEPR to XeF2

According to VSEPR theory, we first determine the electron pair geometry. For 22 valence electrons, Xe uses 2 to form bonds with F atoms (4 bonding electrons), leaving 18 non-bonding electrons (9 pairs) as lone pairs on Xe. For \(\mathrm{XeF}_2\), 3 lone pairs and 2 bonding pairs create a trigonal bipyramidal electron pair geometry. To minimize repulsion, lone pairs are placed equatorially, resulting in a linear shape.

Determine the number of valence electrons for XeF4

For \(\mathrm{XeF}_4\):- Xe contributes 8 electrons.- Four F atoms contribute 4 x 7 = 28 electrons.Total: \(8 + 28 = 36\) electrons.

Apply VSEPR to XeF4

Xe uses 4 electrons for bonding with F (8 bonding electrons), leaving 28 non-bonding electrons (14 pairs) as lone pairs on Xe. For \(\mathrm{XeF}_4\), two lone pairs and four bonding pairs create an octahedral electron pair geometry. Placing lone pairs opposite each other minimizes repulsion, resulting in a square planar shape.

Determine the number of valence electrons for XeF6

For \(\mathrm{XeF}_6\):- Xe contributes 8 electrons.- Six F atoms contribute 6 x 7 = 42 electrons.Total: \(8 + 42 = 50\) electrons.

Apply VSEPR to XeF6

Xe uses 6 electrons to form bonds with F atoms (12 bonding electrons), leaving 38 non-bonding electrons (19 pairs) as lone pairs. For \(\mathrm{XeF}_6\), one lone pair and six bonding pairs result in a distorted octahedral or square pyramidal shape. Placing the lone pair to minimize repulsion gives a structure consistent with VSEPR predictions.

Key concepts

Valence Electrons

Valence electrons are the outermost electrons of an atom and play a crucial role in chemical bonding and molecular structure. They determine how an atom interacts and bonds with other atoms. When calculating the valence electrons of a molecule, consider the electrons in the outermost shell for each atom.

For example, Xenon (Xe) is the central atom in \({{\rm XeF}_{2}}\), \({{\rm XeF}_{4}}\), and \({{\rm XeF}_{6}}\). Xe has 8 valence electrons. Each fluorine (F) atom contributes 7 valence electrons. Thus, calculating the total valence electrons involves adding up the values from all atoms in the molecule.

• \({{\rm XeF}_{2}}\): Xe contributes 8, and 2 F atoms contribute 14, making a total of 22 valence electrons.
• \({{\rm XeF}_{4}}\): Xe plus 4 F atoms give a total of 36 valence electrons.
• \({{\rm XeF}_{6}}\): Xe plus 6 F atoms amount to 50 valence electrons.

Understanding this helps predict molecular geometry through models like the VSEPR by knowing how many electrons are available for bonding and lone pair formation.

Molecular Geometry

The geometry of a molecule describes the shape or three-dimensional arrangement of its atoms in space. This geometry is crucial, as it affects the molecule's chemical reactivity, polarity, and physical properties. The VSEPR (Valence Shell Electron Pair Repulsion) model is a common method used to predict molecular geometry.

According to VSEPR theory, electrons around a central atom arrange themselves to minimize repulsion due to their negative charge. This arrangement dictates the molecule's overall shape. For instance, in \({{\rm XeF}_{2}}\), five electron pairs (3 lone pairs + 2 bonding pairs) result in a trigonal bipyramidal arrangement, which simplifies to a linear structure when lone pairs are positioned to minimize repulsion.

• \({{\rm XeF}_{4}}\) displays an octahedral electron geometry. To reduce repulsion, lone pairs are placed opposite each other, creating a square planar shape.
• \({{\rm XeF}_{6}}\) starts with an octahedral geometry but transforms into a distorted octahedron or square pyramidal shape due to the presence of a lone pair.

Bonding and Lone Pairs in Molecules

Molecules have two types of electron pairs that determine their structure: bonding pairs and lone pairs. Bonding pairs are shared between atoms, forming covalent bonds and contributing directly to the molecular shape. In contrast, lone pairs are non-bonding electrons residing on the central atom, influencing geometry but not forming bonds.

Take \({{\rm XeF}_{2}}\) as an example: it features two bonding pairs (Xe to each F atom) and three lone pairs on Xe. The lone pairs opt for an equatorial position in a trigonal bipyramidal arrangement to minimize electron repulsion, leading to a linear molecular shape.

• For \({{\rm XeF}_{4}}\), four bonding pairs and two lone pairs on Xe achieve an octahedral configuration upon which the lone pairs take up axial positions, resulting in a square planar shape.
• In \({{\rm XeF}_{6}}\), six bonding pairs are joined by one lone pair on the central Xe atom, culminating in a distorted octahedral shape, as the lone pair distorts the symmetry.

Understanding bonding and lone pair arrangements is vital since they dictate the molecule's geometry and many of its chemical properties.