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

In summary, BrF4- has a square planar geometry due to the presence of 6 electron pair domains and one lone pair, whereas BF4- adopts a tetrahedral geometry with 4 electron pair domains and no lone pairs. In the series H2O, H2S, and H2Se, the H-X-H bond angle decreases from H2O to H2Se due to the decrease in electronegativity and the increase in the size of the electron pair domains of the central atom.

Step-by-step solution

Part (a): VSEPR theory for BrF4- and BF4- ions

For the BrF4- ion, we have the following electron arrangement: 1. The central atom, Br, has 7 valence electrons. 2. There are 4 F atoms bonded to the central atom, which contributes 4 electrons. 3. The ion has a negative charge, so we add an extra electron. The total number of valence electrons for BrF4- is 7+4+1=12 electrons, which results in 6 electron pair domains. According to VSEPR theory, a molecule with 6 electron pair domains will adopt an octahedral geometry when all the domains are occupied by bonding pairs (lone pairs are not counted). Since BrF4- has one lone pair and one negative charge, one of the electron pair domains is occupied by the lone pair. This results in a square planar geometry for BrF4-. For the BF4- ion, we have the following electron pair distribution: 1. The central atom, B, has 3 valence electrons. 2. Four F atoms contribute 4 electrons in bonding pairs. So, the total number of valence electrons for BF4- is 3+4=7 electrons. As there are no lone pairs on the central atom, the BF4- ion has 4 bonding pairs of electrons around the central atom which will be distributed into 4 electron pair domains. According to VSEPR theory, a molecule with 4 electron pair domains will adopt a tetrahedral geometry when there are no lone pairs. Therefore, BF4- has a tetrahedral shape.

Part (b): H-X-H bond angle variation in H2O, H2S, and H2Se

In order to explain the variation of the H-X-H bond angle in H2O, H2S, and H2Se, we will consider the electronegativity of the central atom (X) and the size of the electron pair domains: 1. Electronegativity of central atoms: O (3.44) > S (2.58) > Se (2.55). 2. As the electronegativity of the central atom decreases, the electron pair domains' size increases, and the repulsion between the lone pairs and the bonding pairs in the central atom decreases. Because of this, the lone pairs are occupying larger volumes around the central atom, and the H-X-H bond angle shrinks as the electronegativity of the central atom decreases. Therefore, we can expect the H-X-H bond angle to decrease in the following order: H2O > H2S > H2Se. In conclusion, the bond angle decreases as we move down the group in the Periodic Table (from oxygen to selenium).

Key concepts

Molecular Geometry

Molecular geometry is the three-dimensional arrangement of atoms within a molecule. This arrangement is crucial in determining many properties of the molecule, such as polarity, reactivity, and biological activity.

When discussing molecular geometry, it often involves understanding how bonds and lone electron pairs are arranged around a central atom. The Valence Shell Electron Pair Repulsion (VSEPR) theory is commonly used to predict molecular geometry. This theory suggests that electron pairs around a central atom will spread out to minimize repulsion, effectively determining the shape of the molecule.

• For example, \( ext{BrF}_4^-\) is square planar because its electron pairs arrange themselves into six domains, forming an octahedral geometry at first, but one of these domains is a lone pair, leading to a square planar shape.
• On the other hand, \( ext{BF}_4^-\) forms a tetrahedral shape due to having only four bonding pairs and no lone pairs.

Electron Pair Repulsion

Electron pair repulsion is a key concept in predicting the structure of molecules. According to the VSEPR theory, the arrangement of electron pairs around a central atom is influenced by the tendency of these pairs to repel each other and arrange themselves as far apart as possible.

In molecules like \( ext{H}_2 ext{O}\), \( ext{H}_2 ext{S}\), or \( ext{H}_2 ext{Se}\), the bond angles change because the lone pairs on the central atom repel the bonding pairs. The more lone pairs on a central atom, the smaller the angle between the bonding pairs.

• In \( ext{H}_2 ext{O}\), where the central atom oxygen has higher electronegativity, the lone pair repulsion is significant, leading to a smaller bond angle.
• As we move to larger central atoms with lower electronegativity, like sulfur in \( ext{H}_2 ext{S}\) and selenium in \( ext{H}_2 ext{Se}\), this repulsion decreases, and the bond angles increase accordingly.

Bond Angles

Bond angles are the angles between adjacent lines representing bonds drawn from the central atom. Calculating bond angles is critical for understanding molecular geometry and the overall shape of the molecule.

In a perfect tetrahedral arrangement, like in \( ext{BF}_4^-\), the bond angles are 109.5^{ ext{o}}, reflecting no lone pair disturbances. However, in molecules with lone pairs like \( ext{BrF}_4^-\), deviations occur.

• The bond angle shrinks in molecules like \( ext{H}_2 ext{O}\) from the expected tetrahedral angle, due to lone pairs creating additional repulsive forces.
• As the central atom's electronegativity decreases down the group (e.g., \( ext{H}_2 ext{O} > ext{H}_2 ext{S} > ext{H}_2 ext{Se}\)), the bond angle decreases because lone pairs occupy more significant spaces, pushing bonding pairs closer.

Electronegativity

Electronegativity is the measure of the tendency of an atom to attract a bonding pair of electrons. It plays a vital role in determining the bond angles and geometry of a molecule.

The concept of electronegativity is significant when comparing the molecular geometry across similar compounds. For instance, in the series \( ext{H}_2 ext{O}\), \( ext{H}_2 ext{S}\), and \( ext{H}_2 ext{Se}\):

• Oxygen, with the highest electronegativity, holds its bonding pairs more tightly compared to sulfur and selenium, resulting in a smaller bond angle.
• As we proceed to elements with lower electronegativity within the group, such as sulfur and selenium, the bonding pairs are less tightly held, resulting in a gradual increase in bond angle.

Lone Pairs

Lone pairs are electron pairs located on an atom that are not involved in bonding. They are critical in determining molecular geometry since they exert repulsion on other electron pairs.

Consider how lone pairs affect the geometry of \( ext{BrF}_4^-\) and \( ext{BF}_4^-\):

• In \( ext{BrF}_4^-\), the lone pair occupies one of the positions in the octahedral arrangement, resulting in a square planar shape, diminishing the \( ext{90}^{ ext{o}}\) angles in a typical octahedral geometry into a \( ext{180}^{ ext{o}}\) effective planar arrangement.
• In \( ext{BF}_4^-\), no lone pairs are present, allowing for uniform repulsion among the bonding pairs and retaining a classic tetrahedral structure.