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?

Short answer

(a) The ions \(\mathrm{BrF}_{4}^{-}\) and \(\mathrm{ClF}_{4}^{+}\) have more than an octet of electrons around the central atom. (b) The ion \(\mathrm{AlF}_{4}^{-}\) has the same electron-domain and molecular geometries (both tetrahedral). (c) The ion \(\mathrm{BrF}_{4}^{-}\) has an octahedral electron-domain geometry. (d) The ions \(\mathrm{PF}_{4}^{-}\) and \(\mathrm{ClF}_{4}^{+}\) exhibit a see-saw molecular geometry.

Step-by-step solution

Calculate the number of valence electrons for each ion

To determine the number of valence electrons around the central atom for each ion, we will sum the valence electrons contributed by each atom and add or subtract the charge of the ion. - \(\mathrm{PF}_{4}^{-}\): P has 5 valence electrons, each F has 7, and the ion has a -1 charge. Total valence electrons: \(5 + 4(7) - 1 = 33\). - \(\mathrm{BrF}_{4}^{-}\): Br has 7 valence electrons, each F has 7, and the ion has a -1 charge. Total valence electrons: \(7 + 4(7) - 1 = 41\). - \(\mathrm{ClF}_{4}^{+}\): Cl has 7 valence electrons, each F has 7, and the ion has a +1 charge. Total valence electrons: \(7 + 4(7) + 1 = 35\). - \(\mathrm{AlF}_{4}^{-}\): Al has 3 valence electrons, each F has 7, and the ion has a -1 charge. Total valence electrons: \(3 + 4(7) - 1 = 31\).

Count the number of electron pairs around the central atom

Now we will determine how many electron pairs surround the central atom for each ion. - \(\mathrm{PF}_{4}^{-}\): With 33 valence electrons, P uses 8 electrons in bonding with 4 F atoms. It leaves 25 electrons for the central P atom, meaning 12 lone pairs (an octet). - \(\mathrm{BrF}_{4}^{-}\): With 41 valence electrons, Br uses 8 electrons in bonding with 4 F atoms. It leaves 33 electrons for the central Br atom, meaning 16 lone pairs (more than an octet). - \(\mathrm{ClF}_{4}^{+}\): With 35 valence electrons, Cl uses 8 electrons in bonding with 4 F atoms. It leaves 27 electrons for the central Cl atom, meaning 13 lone pairs (more than an octet). - \(\mathrm{AlF}_{4}^{-}\): With 31 valence electrons, Al uses 8 electrons in bonding with 4 F atoms. It leaves 23 electrons for the central Al atom, meaning 11 lone pairs (an octet). #a) Answer#: The ions \(\mathrm{BrF}_{4}^{-}\) and \(\mathrm{ClF}_{4}^{+}\) have more than an octet of electrons around the central atom. #b) Identification of ions with the same electron-domain and molecular geometries#

Analyze electron-domain geometries

We will determine the electron-domain geometries of each ion by considering the number of electron domains around the central atom (lone pairs + bonding pairs). - \(\mathrm{PF}_{4}^{-}\): 4 bonding pairs + 1 lone pair = 5 electron domains. The electron-domain geometry is trigonal bipyramidal. - \(\mathrm{BrF}_{4}^{-}\): 4 bonding pairs + 2 lone pairs = 6 electron domains. The electron-domain geometry is octahedral. - \(\mathrm{ClF}_{4}^{+}\): 4 bonding pairs + 1 lone pair = 5 electron domains. The electron-domain geometry is trigonal bipyramidal. - \(\mathrm{AlF}_{4}^{-}\): 4 bonding pairs + 0 lone pairs = 4 electron domains. The electron-domain geometry is tetrahedral.

Analyze molecular geometries

Now we will determine the molecular geometries of each ion by considering the arrangement of atoms in space. - \(\mathrm{PF}_{4}^{-}\): With 4 bonding pairs and 1 lone pair in a trigonal bipyramidal arrangement, the molecular geometry is see-saw. - \(\mathrm{BrF}_{4}^{-}\): With 4 bonding pairs and 2 lone pairs in an octahedral arrangement, the molecular geometry is square planar. - \(\mathrm{ClF}_{4}^{+}\): With 4 bonding pairs and 1 lone pair in a trigonal bipyramidal arrangement, the molecular geometry is see-saw. - \(\mathrm{AlF}_{4}^{-}\): With 4 bonding pairs and 0 lone pairs in a tetrahedral arrangement, the molecular geometry is tetrahedral. #b) Answer#: The ion \(\mathrm{AlF}_{4}^{-}\) has the same electron-domain and molecular geometries (both tetrahedral). #c) Identification of ions with an octahedral electron-domain geometry# #c) Answer#: From our previous analysis, we can see that the ion \(\mathrm{BrF}_{4}^{-}\) has an octahedral electron-domain geometry. #d) Identification of ions with a see-saw molecular geometry# #d) Answer#: From our previous analysis, we can see that the ions \(\mathrm{PF}_{4}^{-}\) and \(\mathrm{ClF}_{4}^{+}\) exhibit a see-saw molecular geometry.

Key concepts

Electron-Domain Geometry

When studying molecular geometry, it's vital to understand the concept of electron-domain geometry. This term refers to the spatial arrangement of electron domains (regions of high electron density) around the central atom in a molecule or ion. Electron domains can be bonds (single, double, or triple bonds) or lone pairs of electrons.

• The electron-domain geometry determines the general shape of the molecule or ion.
• It includes all the electron regions, unlike molecular geometry, which considers only the atoms' arrangement.
• This concept helps predict the angles between bonds and the molecule’s overall shape.

In the exercise, we see different examples, such as octahedral geometry in extit{BrF}_4^{-} and trigonal bipyramidal in PF_4^{-}. These geometrical terms tell us how the electron pairs or domains are organized around the central atom to minimize repulsion and achieve a stable structure.

Valence Electrons

Valence electrons are the outermost electrons of an atom and play a key role in bonding and molecular structure. They are primarily responsible for the chemical properties of an element. In the context of molecular geometry, counting valence electrons helps us determine:

• The total number of electrons available for bonding in a particular ion or molecule.
• How these electrons are distributed among atoms and if they form bonds or remain as lone pairs.
• The eventual electron-domain geometry which influences molecular shape and properties.

To illustrate, consider extit{BrF}_4^{-}: Bromine has 7 valence electrons, plus additional 7 for each fluorine atom, supplemented with one extra electron to account for the negative charge, making a total of 41 valence electrons. Distributing these, while noting that 8 electrons are used per electron domain in bonding, helps deduce the arrangement configurations like octahedral or trigonal bipyramidal.

Trigonal Bipyramidal

The term trigonal bipyramidal describes a particular type of electron-domain geometry where five domains are arranged. In this geometry, there are three equatorial domains forming a triangle and two axial domains positioned perpendicular to the plane of the triangle.

• This arrangement offers more stability and minimized electron repulsion when there are five electron domains around a central atom.
• The angles between equatorial domains are 120°, while the angular distance from equatorial to axial domains is 90°.

For instance, the extit{PF}_4^{-} ion uses a trigonal bipyramidal electron-domain geometry because it consists of four bonding pairs and one lone pair. This form results in a see-saw molecular shape once lone pairs are considered, altering the arrangement slightly, yet maintaining the primary domain setup.

Octahedral Geometry

In octahedral geometry, six electron domains are symmetrically arranged around the central atom. This particular geometry minimizes repulsion between electrons by placing domains at 90-degree angles to each other.

• This results in a highly symmetrical shape often found in coordination compounds and complex ions.
• An octahedral shape is common when six bonds without lone pairs are present.

In the case discussed, extit{BrF}_4^{-} exhibits an octahedral electron-domain geometry as it comprises four F atom bonding pairs and two lone pairs around the bromine atom. These domains utilize the full symmetry of an octahedron, while the two lone pairs occupy positions that lead to a square planar molecular geometry, slightly differing from the electron-domain arrangement due to non-bonding pairs.