Question

Which of the following species (there may be more than one) is/ are likely to have the structure shown below: (a) \(\mathrm{XeF}_{4}\), (b) \(\mathrm{BrF}_{4}{ }^{+},(\mathrm{c}) \mathrm{SiF}_{4}\) (d) \(\mathrm{TeCl}_{4}\), \((\mathrm{e}) \mathrm{HClO}_{4} ?\) (The colors shown do not reflect the identity of any element.) [Sections 22.3, 22.4, 22.6, and 22.10]

Short answer

None of the given species (\(\mathrm{XeF}_{4}\), \(\mathrm{BrF}_{4}^{+}\), \(\mathrm{SiF}_{4}\), \(\mathrm{TeCl}_{4}\), \(\mathrm{HClO}_{4}\)) match the structure provided in the problem because all of them result in an invalid configuration of non-bonding electrons.

Step-by-step solution

Identify the central atom

For each of the given species, identify the central atom. In this case, all the species have the central atom listed first. (a) Xe in \(\mathrm{XeF}_{4}\) (b) Br in \(\mathrm{BrF}_{4}^{+}\) (c) Si in \(\mathrm{SiF}_{4}\) (d) Te in \(\mathrm{TeCl}_{4}\) (e) Cl in \(\mathrm{HClO}_{4}\)

Determine the number of valence electrons

Find the number of valence electrons for the central atom in each species: (a) Xe: 8 valence electrons (b) Br: 7 valence electrons (since it has a +1 charge) (c) Si: 4 valence electrons (d) Te: 6 valence electrons (e) Cl: 7 valence electrons

Add up the total number of electrons used for bonding

Now that we have the valence electrons for each central atom, we also need to consider the number of electrons used in bonding to the surrounding atoms: (a) \(\mathrm{XeF}_{4}\): 4 bonds × 2 electrons/bond = 8 electrons (b) \(\mathrm{BrF}_{4}^{+}\): 4 bonds × 2 electrons/bond = 8 electrons (c) \(\mathrm{SiF}_{4}\): 4 bonds × 2 electrons/bond = 8 electrons (d) \(\mathrm{TeCl}_{4}\): 4 bonds × 2 electrons/bond = 8 electrons (e) \(\mathrm{HClO}_{4}\): 1 H-Cl bond × 2 electrons/bond + 4 Cl-O bonds × 2 electrons/bond = 10 electrons

Calculate the number of non-bonding electrons

Subtract the total number of bonding electrons from the valence electrons on the central atom to find the number of non-bonding electrons: (a) Xe in \(\mathrm{XeF}_{4}\): 8 - 8 = 0 non-bonding electrons (b) Br in \(\mathrm{BrF}_{4}^{+}\): 7 - 8 = -1 non-bonding electrons (not possible, so this structure doesn't match the given one) (c) Si in \(\mathrm{SiF}_{4}\): 4 - 8 = -4 non-bonding electrons (not possible, so this structure doesn't match the given one) (d) Te in \(\mathrm{TeCl}_{4}\): 6 - 8 = -2 non-bonding electrons (not possible, so this structure doesn't match the given one) (e) Cl in \(\mathrm{HClO}_{4}\): 7 - 10 = -3 non-bonding electrons (not possible, so this structure doesn't match the given one) From the above calculations, it seems that none of the given species match the structure provided in the problem, as none of them seem to have a valid configuration of the non-bonding electrons.

Key concepts

Valence Electrons

Valence electrons play a fundamental role in chemical bonding and molecular structure. They are the electrons located in the outermost shell of an atom and are involved in forming bonds with other atoms. The number of valence electrons an element has can be determined from its position in the periodic table. For instance, Group 1 elements have one valence electron, while Group 18 (the noble gases) have eight, which is generally considered a full set, leading to their characteristically inert nature.

To predict the molecular geometry and reactivity of an element or compound, it's imperative to know the valence electron count for the central atom. In the exercise provided, students are encouraged to identify these valence electrons as part of the step-by-step solution to understanding molecular geometry. An improvement for the solution would be to explain that for charged species, like ions, the valence electron count changes. Positive charges mean that electrons are lost, while negative charges indicate gained electrons.

Electron Pair Geometry

The electron pair geometry of a molecule is the three-dimensional arrangement of bonding and non-bonding electron pairs around the central atom. This concept arises from the VSEPR (Valence Shell Electron Pair Repulsion) theory which posits that electron pairs repel one another and will therefore adopt a spatial arrangement that minimizes this repulsion. The result is a distinctive geometric shape.

For instance, a molecule with four electron groups around a central atom, like methane (CH₄), will have a tetrahedral electron pair geometry. Should there be a lone pair of electrons, the molecular geometry changes accordingly. An important aspect often missing in solutions is that the distinction between electron pair geometry (based solely on repulsion) and the actual molecular shape (which considers only atoms) is critical for clear understanding.

Molecular Structure

While electron pair geometry concerns the arrangement of all electron pairs, molecular structure refers specifically to the spatial disposition of the atoms themselves. Non-bonding electron pairs, or lone pairs, affect the molecular structure but are not counted as part of the visible shape. It's essential to note that the presence of lone pairs leads to different observed geometries. For example, the angular bent shape of water (H₂O) is due to its two lone pairs forcing the bonded hydrogen atoms closer together.

Molecular structure has profound effects on a compound's properties, such as polarity, reactivity, and phase at room temperature. In the exercise given, the students aim to match a molecular structure with possible compounds - a task requiring the understanding of how bonding and non-bonding electrons shape a molecule. This understanding would be deepened with visualization aids such as molecular models or computer simulations to clearly illustrate the three-dimensional geometries.