Question

For each of the following, write the Lewis structure(s), predict the molecular structure (including bond angles), and give the expected hybridization of the central atom. a. \(\mathrm{KrF}_{2}\) c. \(\mathrm{XeO}_{2} \mathrm{F}_{2}\) b. \(\mathrm{KrF}_{4}\) d. \(\mathrm{XeO}_{2} \mathrm{F}_{4}\)

Short answer

a. \(\mathrm{KrF}_{2}\): Linear, 180°, sp³d c. \(\mathrm{XeO}_{2} \mathrm{F}_{2}\): Square planar, 90° and 180°, sp³d² b. \(\mathrm{KrF}_{4}\): Square planar, 90° and 180°, sp³d² d. \(\mathrm{XeO}_{2} \mathrm{F}_{4}\): Octahedral, 90° and 180°, sp³d³

Step-by-step solution

a. \(\mathrm{KrF}_{2}\)

Step 1: Draw the Lewis structure Kr = 8 valence electrons F = 7 valence electrons Total = 8 + (7 x 2) = 22 valence electrons - Kr is the central atom - Place 2 single bonds between Kr and the two F atoms - Distribute the remaining 18 electrons among Kr and F atoms as lone pairs - Fulfill octet rule for all the atoms Step 2: Predict the molecular structure and bond angles The molecular structure is linear, with Kr in the center and the two F atoms on the sides. Bond angle: 180° Step 3: Determine hybridization of the central atom Kr has 2 sigma bonds and 3 lone pairs, implying that its hybridization is sp³d (s + p's + d's = 5).

c. \(\mathrm{XeO}_{2} \mathrm{F}_{2}\)

Step 1: Draw the Lewis structure Xe = 8 valence electrons O = 6 valence electrons F = 7 valence electrons Total = 8 + (6 x 2) + (7 x 2) = 34 valence electrons - Xe is the central atom - Place 2 single bonds between Xe and the two F atoms, and 2 double bonds between Xe and the two O atoms - Distribute the remaining 18 electrons among Xe, O, and F atoms as lone pairs - Fulfill octet rule for all the atoms Step 2: Predict the molecular structure and bond angles The molecular structure is square planar Bond angles: 90° and 180° Step 3: Determine hybridization of the central atom Xe has 4 sigma bonds and 2 lone pairs, implying that its hybridization is sp³d² (s + p's + d's = 6).

b. \(\mathrm{KrF}_{4}\)

Step 1: Draw the Lewis structure Kr = 8 valence electrons F = 7 valence electrons Total = 8 + (7 x 4) = 36 valence electrons - Kr is the central atom - Place 4 single bonds between Kr and the four F atoms - Distribute the remaining 32 electrons among Kr and F atoms as lone pairs - Fulfill octet rule for all the atoms Step 2: Predict the molecular structure and bond angles The molecular structure is square planar Bond angles: 90° and 180° Step 3: Determine hybridization of the central atom Kr has 4 sigma bonds and 2 lone pairs, implying that its hybridization is sp³d² (s + p's + d's = 6).

d. \(\mathrm{XeO}_{2} \mathrm{F}_{4}\)

Step 1: Draw the Lewis structure Xe = 8 valence electrons O = 6 valence electrons F = 7 valence electrons Total = 8 + (6 x 2) + (7 x 4) = 50 valence electrons - Xe is the central atom - Place 4 single bonds between Xe and the four F atoms, and 2 double bonds between Xe and the two O atoms - Distribute the remaining 34 electrons among Xe, O, and F atoms as lone pairs - Fulfill octet rule for all the atoms Step 2: Predict the molecular structure and bond angles The molecular structure is octahedral Bond angles: 90° and 180° Step 3: Determine hybridization of the central atom Xe has 6 sigma bonds and no lone pairs, implying that its hybridization is sp³d³ (s + p's + d's = 7).

Key concepts

Molecular Structure

Understanding the molecular structure is crucial for comprehending how molecules interact with each other and how they participate in chemical reactions. Molecular structures describe the three-dimensional arrangement of atoms within a molecule.

• Linear structure: In this arrangement, two atoms or groups are connected to a central atom in a straight line. An example from the exercise is KrF₂, where the central krypton atom forms a linear structure with the two fluorine atoms, resulting in bond angles of 180°.
• Square planar structure: This structure involves a central atom surrounded by other atoms or groups in a square on a single plane. Both KrF₄ and
  XeO₂F₂ are examples of square planar geometry. Here, the bond angles are typically 90° and 180°.
• Octahedral structure: An octahedral structure features a central atom with six surrounding atoms or groups positioned at the corners of an octahedron. For instance,
  XeO₂F₄ has an octahedral structure where the bond angles are 90° and 180°.

Understanding the geometry is not just about knowing the shape but also predicting the distribution of charges, the potential reactivity of the molecule, and how it might interact with light or other molecules.

Bond Angles

Bond angles are essential for predicting the shape and physical properties of molecules. These angles are formed by the intersection of two covalent bonds at the atom that forms the vertex of the angle. Knowing the bond angles helps predict the molecular geometry, which is vital for defining the spatial arrangement of the molecule's atoms.

• In a linear molecular geometry, such as in KrF₂, the bond angles are 180° as the molecule is spread out in a straight line.
• In square planar molecules, like KrF₄ and
  XeO₂F₂, the bond angles are typically 90° between adjacent bonds and 180° across the square plane.
• For octahedral molecules such as
  XeO₂F₄, the central atom is 90° from each bonding pair (since it is equidistant to all surrounding atoms) and 180° from the atoms across from each other along the same axis.

The bond angles are fundamental for understanding how molecules interact, as they affect properties like polarity and intermolecular forces.

Hybridization of Central Atom

Hybridization is the concept of mixing atomic orbitals to form new hybrid orbitals that can host paired or unpaired electrons and are responsible for the bonding with other atoms. The type of hybridization gives insight into both the bond angles and the molecular geometry.

• KrF₂ has a central krypton atom with sp³d hybridization, which correlates with a linear molecular structure and a bond angle of 180°.
• KrF₄ and XeO₂F₂ both have a sp³d² hybridization of the central atom, indicating a square planar geometry with 90° and 180° bond angles.
• For XeO₂F₄, the central xenon atom's sp³d³ hybridization is indicative of an octahedral structure with 90° and 180° bond angles as well.

Identifying the hybridization of the central atom is not only important for understanding the molecule's shape but also for predicting chemical reactivity and properties such as the ability to form particular types of chemical bonds.