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Antimony pentafluoride reacts with HF according to the equation $$2\mathrm{HF}+\mathrm{SbF}_{3}\rightarrow\left[\mathrm{H}_{2}\mathrm{F}\right]^{+}\left[\mathrm{SbF}_{6}\right]^{-}$$ (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 \(\mathrm{H}_{2} \mathrm{F}^{+} ?\)

Short Answer

Expert verified
(a) Sb is \(sp^3\) in reactant, \(sp^3d^2\) in product. (b) \(\mathrm{H}_2\mathrm{F}^{+}\) is linear with \(sp\) hybridization for F.

Step by step solution

01

Determine Hybridization of Sb in Reactant

The compound \(\text{SbF}_3\) has three fluorine atoms bonded to antimony (Sb). Antimony uses its five valence electrons: three for bonding with fluorine and two as a lone pair. Therefore, the steric number is 3 (bonding pairs) + 1 (lone pair) = 4, suggesting \(sp^3\) hybridization for Sb.
02

Determine Hybridization of Sb in Product

In the product \(\left[\text{SbF}_6\right]^{-}\), antimony is surrounded by six fluorine atoms and does not have lone pairs. The steric number is 6 (six bonding pairs), indicating \(sp^3d^2\) hybridization for Sb.
03

Draw Lewis Structure for \(\mathrm{H}_{2} \\mathrm{F}^{+}\)

\(\mathrm{H}_{2} \mathrm{F}^{+}\) consists of two hydrogen atoms bonded to a central fluorine atom. Place a positive charge on the system, suggesting that it has one less electron. Connect two \(\text{H}\) atoms to \(\text{F}\), all single bonds, and place a positive charge to reflect one unaccounted electron.
04

Determine Geometry of \(\mathrm{H}_{2} \\mathrm{F}^{+}\)

The central fluorine atom in \(\mathrm{H}_{2} \mathrm{F}^{+}\) has two hydrogen atoms bonded to it, with no lone pairs on the fluorine. This arrangement implies a linear geometry for \(\mathrm{H}_{2} \mathrm{F}^{+}\).
05

Find Hybridization of \(F\) in \(\mathrm{H}_{2} \\mathrm{F}^{+}\)

With two sigma bonds and no lone pairs due to the positive charge, the fluorine atom in \(\mathrm{H}_{2} \mathrm{F}^{+}\) is \(sp\) hybridized, making it linear.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Lewis Structure
The Lewis structure is a simple way of representing molecules, showing how atoms are bonded. It uses dots to represent electrons and lines to indicate bonds between atoms. Understanding Lewis structures is crucial because it helps predict the shape, hybridization, and overall behavior of a molecule.

In the exercise, we are asked to draw the Lewis structure for the ion \(\mathrm{H}_{2} \mathrm{F}^{+}\). This molecule consists of a central fluorine atom bonded to two hydrogen atoms. Let's start by counting the total number of valence electrons. Normally, fluorine has 7 and hydrogen has 1. However, because \(\mathrm{H}_{2} \mathrm{F}^{+}\) has a positive charge, we subtract one electron, giving us a total of 8 electrons.

To construct the structure:
  • Place the fluorine atom in the center since it can form more bonds than hydrogen.
  • Attach each hydrogen to the fluorine with a single bond. This uses up 4 electrons.
  • Place the remaining 4 electrons as lone pairs on the central fluorine atom to satisfy the octet rule.
  • Add a positive charge to the molecule to account for the missing electron.
By considering the positive charge, we reflect an electron deficiency in the structure, which influences the molecule's properties and interactions with other compounds.
Molecular Geometry
Molecular geometry describes the three-dimensional arrangement of atoms within a molecule, which can affect physical and chemical properties. The VSEPR (Valence Shell Electron Pair Repulsion) theory helps us predict the shape of a molecule based on the number of bonds and lone pairs around the central atom.

In the given reaction, the geometry of \(\mathrm{H}_{2} \mathrm{F}^{+}\) is determined primarily by the lack of lone pairs on the central fluorine atom. Two hydrogen atoms are bonded to the fluorine atom. According to the VSEPR theory, when a central atom has two bonded atoms and no lone pairs, the geometry is linear.

Linear geometry indicates that the hydrogen and fluorine atoms form a straight line:
  • This geometry minimizes repulsion between the bonding pairs, allowing the molecule to maintain a stable shape.
  • It also affects the molecule's polarity because the dipole from the fluorine to hydrogen may not completely cancel out.
  • This linear arrangement allows the molecule to interact in specific ways with other molecules or ions in reactions.
Understanding the geometry of molecules like \(\mathrm{H}_{2} \mathrm{F}^{+}\) is paramount when predicting how they will interact in chemical processes.
Chemical Bonding
Chemical bonding is the attraction between atoms that enables the formation of chemical compounds. Hybridization is a key concept that helps explain the bonding and geometry of molecules. Hybridization involves the mixing of atomic orbitals to form new hybrid orbitals, which helps in the bonding process.

In the reaction given, we observe the hybridization of antimony (Sb) in different states:
  • For \(\text{SbF}_3\), where antimony bonds with three fluorine atoms, the steric number is 4 (including one lone pair), suggesting \(sp^3\) hybridization.
  • In the product \(\left[\text{SbF}_6\right]^{-}\), antimony forms bonds with six fluorine atoms. Here, the steric number is 6, indicating \(sp^3d^2\) hybridization.

  • Bonding in the reactant and product gives insight into the molecule's geometry and properties.
    • The electron pairs try to maintain the greatest possible distance from each other, which shapes the molecule into optimal geometry.
    • Hybrid orbitals allow molecules to form even more stable bonds by proper alignment of atomic orbitals during bond formation.
    This understanding of chemical bonding through hybridization explains not only the structure but also the function and reactivity of molecules in chemical reactions.

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Most popular questions from this chapter

Which of the homo-nuclear, diatomic molecules of the second-period elements (from \(\mathrm{Li}_{2}\) to \(\mathrm{Ne}_{2}\) ) are paramagnetic? Which have a bond order of \(1 ?\) Which have a bond order of \(2 ?\) Which diatomic molecule has the highest bond order?

Xenon forms well-characterized compounds (4 page 400 ). Two xenon-oxygen compounds are \(\mathrm{XeO}_{3}\) and \(\mathrm{XeO}_{4} .\) Draw the Lewis structures of these compounds, and give their electron-pair and molecular geometries. What are the hybrid orbital sets used by xenon in these two oxides?

Draw the two resonance structures that describe the bonding in the acetate ion. What is the hybridization of the carbon atom of the \(-\mathrm{CO}_{2}^{-}\) group? Select one of the two resonance structures and identify the orbitals that overlap to form the bonds between carbon and the three elements attached to it.

Bromine forms a number of oxides of varying stability. (a) One oxide has \(90.90 \%\) Br and \(9.10 \%\) O. Assuming its empirical and molecular formulas are the same, draw a Lewis structure of the molecule and specify the hybridization of the central atom (O). (b) Another oxide is unstable BrO. Assuming the molecular orbital diagram in Figure 9.18 applies to BrO, write its electron configuration (where Br uses \(4 s\) and \(4 p\) orbitals). What is the HOMO for the molecule?

Draw the Lewis structures of the acid \(\mathrm{HPO}_{2} \mathrm{F}_{2}\) and its anion \(\mathrm{PO}_{2} \mathrm{F}_{2}^{-}\). What is the molecular geometry and hybridization for the phosphorus atom in each species? (H is bonded to an O atom in the acid.)

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