Question

(a) Explain why \(\mathrm{BrF}_{4}^{-}\) is square planar, whereas \(\mathrm{BF}_{4}^{-}\) is tetrahedral. (b) How would you expect the \(\mathrm{H}-\mathrm{X}-\mathrm{H}\) bond angle to vary in the series \(\mathrm{H}_{2} \mathrm{O}, \mathrm{H}_{2} \mathrm{~S}, \mathrm{H}_{2} \mathrm{Se}\) ? Explain. (Hint: The size of an electron pair domain depends in part on the electronegativity of the central atom.)

Short answer

(a) \(\mathrm{BrF}_{4}^{-}\) is square planar because there are 5 electron pair domains (4 bonding pairs and 1 lone pair) around the central Br atom, which adopts this geometry to minimize repulsion. In contrast, \(\mathrm{BF}_{4}^{-}\) is tetrahedral because it has 4 electron pair domains (4 bonding pairs) around the central B atom without any lone pairs. (b) The H-X-H bond angle in the series H2O, H2S, and H2Se decreases as the electronegativity of the central atom decreases. This leads to larger electron pair domain sizes and increased repulsion between lone pairs and bonding pairs, pushing the bonds closer together.

Step-by-step solution

Analyze \(\mathrm{BrF}_{4}^{-}\)

To analyze \(\mathrm{BrF}_{4}^{-}\), we first need to determine the number of valence electrons of all the atoms involved. Bromine has 7 valence electrons, each of the 4 fluorine atoms has 7, and there's an extra electron due to the -1 charge, giving us a total of 29 valence electrons. In the central atom (Br), we have a total of 5 electron pair domains (4 bonding pairs with F and 1 lone pair). According to the VSEPR theory, the molecular geometry that minimizes electron pair repulsion for 5 electron pair domains is the square pyramidal geometry. However, due to the presence of a lone pair, the molecule adopts a square planar geometry (lone pairs occupy more space than bonding pairs, and pushing the F atoms to be in the plane creates the necessary space).

Analyze \(\mathrm{BF}_{4}^{-}\)

To analyze \(\mathrm{BF}_{4}^{-}\), we first need to determine the number of valence electrons of all the atoms involved. Boron has 3 valence electrons, each of the 4 fluorine atoms has 7, and there's an extra electron due to the -1 charge, giving us a total of 32 valence electrons. In the central atom (B), we have a total of 4 electron pair domains (4 bonding pairs with F). According to the VSEPR theory, the molecular geometry that minimizes electron pair repulsion for 4 electron pair domains is the tetrahedral geometry. As there are no lone pairs, the molecule adopts this tetrahedral structure.

Compare \(\mathrm{BrF}_{4}^{-}\) and \(\mathrm{BF}_{4}^{-}\)

Given the analysis in Step 1 and Step 2, we have the following results: - \(\mathrm{BrF}_{4}^{-}\) has a square planar geometry due to the presence of a lone pair on the Br atom along with 4 bonding pairs. - \(\mathrm{BF}_{4}^{-}\) has a tetrahedral geometry due to the presence of 4 bonding pairs without any lone pairs.

Analyze the H-X-H bond angles

We are given H2O, H2S, and H2Se molecules, and asked to determine the H-X-H bond angle (X being O, S, or Se). In these molecules, the central atoms O, S and Se have two bonding pairs (H-X bonds) and two lone pairs. The general trend of electronegativity decreases from O to S to Se. As electronegativity decreases, the electron pair domain size increases, leading to a larger region of space occupied by lone pairs. As a result, the repulsion between lone pairs and bonding pairs will increase, pushing the bonding pairs closer together. This will cause the H-X-H bond angle to decrease. We can thus expect the bond angles to follow the trend: H2O > H2S > H2Se.

Summary

(a) \(\mathrm{BrF}_{4}^{-}\) is square planar because it has 4 bonding pairs and 1 lone pair, while \(\mathrm{BF}_{4}^{-}\) is tetrahedral due to its 4 bonding pairs without any lone pairs. (b) In the series H2O, H2S, and H2Se, the H-X-H bond angle decreases due to the decreasing electronegativity of the central atom, which results in larger electron pair domain size and increased repulsion between lone pairs and bonding pairs.

Key concepts

VSEPR Theory

The VSEPR (Valence Shell Electron Pair Repulsion) Theory is a model used to predict the shape of molecules. It suggests that electron pairs around a central atom will arrange themselves as far apart as possible to minimize repulsion. This helps determine the molecular geometry.
1. VSEPR focuses on electron pair interactions. These include bonding pairs, which are electrons shared between atoms, and lone pairs, which are pairs of electrons that are not shared.
2. The shape of a molecule can vary significantly based on these electron pair interactions.
For example, in the case of \(\mathrm{BrF}_4^-\), there are four bonding pairs and one lone pair around bromine. The lone pair influences the arrangement due to its larger repulsion, leading to a square planar shape.
In contrast, \(\mathrm{BF}_4^-\) has four bonding pairs and no lone pair, which naturally forms a tetrahedral geometry because all electron pairs are shared, minimizing repulsion evenly across the space.

Lone Pairs

Lone pairs are the electron pairs that do not participate in bonding. They hold great significance in determining molecular shape due to their strong repulsive force.

• Lone pairs require more space than bonding pairs because they are only attracted to one nucleus rather than two.
• In molecules like \(\mathrm{BrF}_4^-\), the presence of a lone pair on bromine causes distortion in shape, resulting in a square planar geometry. This occurs because the lone pair forces the bonding pairs into the same plane to reduce overall electron repulsion.
• When comparing \(\mathrm{BrF}_4^-\) with \(\mathrm{BF}_4^-\), the absence of lone pairs in \(\mathrm{BF}_4^-\) leads to a typical tetrahedral shape, as all electron clouds are uniformly spaced.

Lone pairs also influence bond angles, as seen in molecules like water, where the angle between bonds is smaller than in a regular tetrahedron due to the compression caused by lone pairs.

Bond Angles

Bond angles are the angles between two adjacent bonds at an atom. They are crucial for defining the geometry of the molecule and are directly influenced by the presence of lone pairs and the types of atoms involved.
When examining bond angles in molecules such as H\(_2\)O, H\(_2\)S, and H\(_2\)Se, one can see a trend. As we move from H\(_2\)O to H\(_2\)Se:

• The central atom's electronegativity decreases (oxygen to selenium).
• Larger electron pair domains form, increasing lone pair repulsions.

This leads to a decrease in the H-X-H bond angle as the central atoms become less electronegative.
In H\(_2\)O, the high electronegativity of oxygen means stronger attraction for bonding electrons, resulting in a more closed bond angle. However, in H\(_2\)Se, the weaker pull by selenium allows the lone pairs to push the bonded electrons closer together, further reducing the bond angle. This demonstrates how bond angles can vary based on changes in atomic characteristics and electron arrangements.