Problem 39 - Chemistry and Chemical Reactivity

Question

Antimony pentafluoride reacts with HF according to the equation $$2 \mathrm{HF}+\mathrm{SbF}_{5} \longrightarrow\left[\mathrm{H}_{2} \mathrm{F}\right]^{+}\left[\mathrm{SbF}_{6}\right]^{-}$$ (a) What is the hybridization of the Sb atom in the reactant and product? (b) Draw a Lewis structure for \(\mathrm{H}_{2} \mathrm{F}^{+}\). What is the geometry of \(\mathrm{H}_{2} \mathrm{F}^{+}\) ? What is the hybridization of \(\mathrm{F}\) in \(\mathrm{H}_{2} \mathrm{F}^{+} ?\)

Step-by-Step Solution

Answer

(a) Sb: \( \text{sp}^3\text{d} \) in reactant; \( \text{sp}^3\text{d}^2 \) in product. (b) \( \text{H}_2\text{F}^+ \) is bent, F is \( \text{sp}^3 \) hybridized.

Step 1: Determine the Hybridization of Sb in Reactant

The reactant \( \text{SbF}_5 \) consists of an antimony (Sb) atom bonded to five fluorine (F) atoms. Sb has 5 bonding pairs and no lone pairs. To determine the hybridization, use the formula: \( \text{Hybridization} = \text{number of bonding pairs} + \text{number of lone pairs} = 5 \). Thus, the hybridization is \( \text{sp}^3\text{d} \) (trigonal bipyramidal geometry).

Step 2: Determine the Hybridization of Sb in Product

In the product \( \left[\text{SbF}_6\right]^- \), the Sb atom is bonded to six fluorine atoms and has a formal charge of +1. To determine the hybridization, consider that there are no lone pairs: \( \text{Hybridization} = 6 \). Hence, the hybridization of Sb is \( \text{sp}^3\text{d}^2 \) (octahedral geometry).

Step 3: Draw Lewis Structure for \( \text{H}_2\text{F}^+ \)

In \( \text{H}_2\text{F}^+ \), the fluoride ion (F-) is bonded to two hydrogen ions (H+), creating a positive charge. The Lewis structure will show one fluoride atom centrally bonded to two hydrogen atoms, with two lone pairs on the fluoride atom.

Step 4: Determine Geometry of \( \text{H}_2\text{F}^+ \)

The molecule \( \text{H}_2\text{F}^+ \), with a central fluorine atom bonded to two hydrogens and holding two lone pairs, exhibits a bent or V-shaped geometry due to the lone pairs exerting repulsion.

Step 5: Determine the Hybridization of F in \( \text{H}_2\text{F}^+ \)

The fluoride in \( \text{H}_2\text{F}^+ \) forms two bonds and has two lone pairs. For hybridization calculation: \( 2\text{(bond pairs)} + 2\text{(lone pairs)} = 4 \), leading to \( \text{sp}^3 \) hybridization (tetrahedral basis).

Step 6: Summarize Findings

The Sb atom in \( \text{SbF}_5 \) is \( \text{sp}^3\text{d} \) hybridized, while in \( \left[\text{SbF}_6\right]^- \) it is \( \text{sp}^3\text{d}^2 \). The \( \text{H}_2\text{F}^+ \) is bent and the F is \( \text{sp}^3 \) hybridized.

Key Concepts

Lewis structure

Understanding the Lewis structure is an essential step in visualizing the arrangement of atoms within a molecule. It uses dots and lines to represent valence electrons and covalent bonds in a straightforward way. For example, in the molecule \(\text{H}_2\text{F}^+\), which is part of our given chemical reaction, we start by placing the fluoride (F) atom centrally. Then, each hydrogen (H) atom forms a single bond with F. This central F atom will typically have two lone pairs, illustrated by pairs of dots around it.

• Each single line represents a pair of shared electrons, denoting a covalent bond.
• The lone pairs (non-bonding valence electrons) are shown as pairs of dots.

These bonds and the charges balance the positive charge on the molecule, making it crucial to arrange them precisely to depict the correct chemical structure. Understanding this structure assists in predicting how the molecule might react or attach to others in a substance.

molecular geometry

Once we have a valid Lewis structure, we can predict the molecule's geometry. This means understanding the spatial arrangement of atoms in a molecule, defined by the bond angles and lengths. For \(\text{H}_2\text{F}^+\), the geometry is shaped by the central F atom bonded to two H atoms, with two lone electron pairs also present.The electron pair repulsion theory, VSEPR (Valence Shell Electron Pair Repulsion), helps us predict that the molecule takes on a bent or V-shaped geometry due to these lone pairs. Simple concepts to remember:

• The lone pairs take up more space than bonding pairs, often pushing bonded atoms closer together.
• Predicting this shape is crucial as it affects the molecule's physical and chemical properties.

In our example, this bent shape results in a bond angle that is slightly less than 109.5°, typical for molecules with a similar arrangement of bonds and lone pairs.

chemical bonding

Chemical bonding is the force that holds atoms together in molecules. It occurs when atoms share or transfer electrons to achieve more stable electron configurations. There are several types of chemical bonds, with covalent bonds being one of the most common in organic and inorganic chemistry.In the context of our exercise, both covalent and ionic interactions play a role. Here's how:

• Covalent bonds are illustrated in the \(\text{H}_2\text{F}^+\) molecule, where electrons are shared between the F and H atoms, forming a bond.
• Ionic characteristics arise when considering the \(\left[\text{SbF}_6\right]^-\) ion. This species involves ionic interactions between the positively charged \(\text{H}_2\text{F}^+\) and negatively charged \(\left[\text{SbF}_6\right]^-\) ions.

Recognizing how these bonds work helps explain reactions and interactions at the molecular level, providing insight into the mechanisms driving chemical processes.