Problem 87 - Chemistry The Central Science

Question

Consider the following \(\mathrm{XF}_{4}\) ions: \(\mathrm{PF}_{4}^{-}, \mathrm{BrF}_{4}^{-}, \mathrm{ClF}_{4}^{+},\) and \(\mathrm{AlF}_{4}^{-}\) (a) Which of the ions have more than an octet of electrons around the central atom? (b) For which of the ions will the electron-domain and molecular geometries be the same? (c) Which of the ions will have an octahedral electron-domain geometry (d) Which of the ions will exhibit a see-saw molecular geometry?

Step-by-Step Solution

Answer

a) \(\mathrm{BrF}_{4}^{-}\) and \(\mathrm{ClF}_{4}^{+}\) have more than an octet of electrons around the central atom. b) All ions have the same electron-domain and molecular geometries. c) Only \(\mathrm{BrF}_{4}^{-}\) has an octahedral electron-domain geometry. d) None of the ions exhibit a see-saw molecular geometry.

Step 1: Valence electrons

For each ion, count the total number of valence electrons: - \(\mathrm{PF}_{4}^{-}\): P has 5 valence electrons and each F has 7. Add 1 electron for the negative charge: 5 + (4 × 7) + 1 = 33 electrons. - \(\mathrm{BrF}_{4}^{-}\): Br has 7 valence electrons and each F has 7. Add 1 electron for the negative charge: 7 + (4 × 7) + 1 = 36 electrons. - \(\mathrm{ClF}_{4}^{+}\): Cl has 7 valence electrons and each F has 7. Subtract 1 electron for the positive charge: 7 + (4 × 7) - 1 = 34 electrons. - \(\mathrm{AlF}_{4}^{-}\): Al has 3 valence electrons and each F has 7. Add 1 electron for the negative charge: 3 + (4 × 7) + 1 = 32 electrons.

Step 2: Electron-domain geometry

Determine electron-domain geometry for each ion using VSEPR Theory: - \(\mathrm{PF}_{4}^{-}\): With 33 electrons, the electron-domain geometry is trigonal bipyramidal. - \(\mathrm{BrF}_{4}^{-}\): With 36 electrons, the electron-domain geometry is octahedral. - \(\mathrm{ClF}_{4}^{+}\): With 34 electrons, the electron-domain geometry is square planar. - \(\mathrm{AlF}_{4}^{-}\): With 32 electrons, the electron-domain geometry is tetrahedral.

Step 3: Matching electron-domain and molecular geometries

Identify ions with matching electron-domain and molecular geometries: - For \(\mathrm{PF}_{4}^{-}\), \(\mathrm{BrF}_{4}^{-}\), \(\mathrm{ClF}_{4}^{+}\), and \(\mathrm{AlF}_{4}^{-}\), the molecular geometries match the electron-domain geometries, so the answer is all ions.

Step 4: Octahedral electron-domain geometry

Identify ions with an octahedral electron-domain geometry: - Only \(\mathrm{BrF}_{4}^{-}\) has an octahedral electron-domain geometry.

Step 5: See-saw molecular geometry

Identify ions with a see-saw molecular geometry: - None of the given ions exhibit a see-saw molecular geometry. #Results#: a) \(\mathrm{BrF}_{4}^{-}\) and \(\mathrm{ClF}_{4}^{+}\) have more than an octet of electrons around the central atom. b) All ions have the same electron-domain and molecular geometries. c) Only \(\mathrm{BrF}_{4}^{-}\) has an octahedral electron-domain geometry. d) None of the ions exhibit a see-saw molecular geometry.

Key Concepts

Electron-Domain Geometry

In chemistry, understanding the electron-domain geometry of molecules is crucial as it gives insight into their spatial arrangement. This concept is based on the VSEPR (Valence Shell Electron Pair Repulsion) Theory, which predicts the shape of a molecule based on the repulsion between electron pairs. The core idea is that electron pairs around a central atom will arrange themselves to minimize repulsion, thereby determining the geometry.

Take, for example, the ion \( \mathrm{BrF}_4^- \). It has 36 valence electrons, which results in an octahedral electron-domain geometry. The presence of extra lone pairs affects the arrangement, as these pairs also repel bonded electrons, contributing to the structure. In contrast, \( \mathrm{AlF}_4^- \) with 32 electrons adopts a tetrahedral electron-domain geometry, as it naturally supports the symmetrical distribution of these pairs.

Identifying electron-domain geometry involves understanding the number of electron pairs, both bonding and non-bonding, around the central atom. This geometry serves as the foundation for predicting the molecule's actual shape.

Octet Rule

The octet rule is a principle in chemistry that atoms tend to bond in such a way that each atom has eight electrons in its valence shell, achieving a state similar to that of the noble gases. While this rule applies well to main-group elements, there are notable exceptions, especially involving elements capable of holding more than eight electrons due to available d-orbitals.

For instance, in the ions \( \mathrm{BrF}_{4}^{-} \) and \( \mathrm{ClF}_{4}^{+} \), the central atoms, bromine and chlorine, respectively, exceed the octet rule. \( \mathrm{BrF}_{4}^{-} \) accumulates 36 valence electrons, while \( \mathrm{ClF}_{4}^{+} \) totals 34. These additional electrons allow the central atoms to form more bonds without strictly adhering to the octet rule.

Understanding when the octet rule does not apply is key, especially for elements located in period 3 and beyond of the periodic table, where d-orbitals become available for bonding.

Molecular Geometry

Molecular geometry defines the three-dimensional arrangement of atoms within a molecule and relates directly to electron-domain geometry but delves into the actual placement of atoms rather than electron pairs alone.

For any given ion, such as \( \mathrm{ClF}_4^+ \), the molecule's geometry may adjust slightly to accommodate lone pairs versus bonding pairs. Here, \( \mathrm{ClF}_4^+ \) achieves a square planar geometry, which is consistent with its electron-domain configuration. This pedagogical match between electron-domain and molecular geometry allows for a straightforward prediction of molecular shape using VSEPR theory.

In all ions considered, their electron-domain geometries align with the molecular geometries because the number of surrounding atoms and lone pairs cooperate to maintain expected shapes. While the electron-domain geometry considers all electron densities, the molecular geometry focuses on the atoms' position, creating clarity in complex structures.