Question

figure 2: the bef₂ molecule adopts a linear structure in which the two bonds are as far apart as possible, on opposite sides of the be atom.
part 2 — questions:

1. would you expect the bond angle in bef₂ to be larger or smaller than the bond angle in h₂s?
2. would you expect the bond angle in pcl₃ to be larger or smaller than the bond angle in pcl₄?

Explanation:

Question 1

To determine the bond angle comparison between \( \text{BeF}_2 \) and \( \text{H}_2\text{S} \), we analyze their molecular geometries using the VSEPR theory.

• \( \text{BeF}_2 \): Beryllium (Be) has 2 valence electrons, and it forms 2 single bonds with fluorine (F) atoms. There are no lone pairs on the central Be atom. According to VSEPR theory, the electron - pair geometry and molecular geometry are linear, with a bond angle of \( 180^\circ \).
• \( \text{H}_2\text{S} \): Sulfur (S) has 6 valence electrons. It forms 2 single bonds with hydrogen (H) atoms, leaving 2 lone pairs on the central S atom. The electron - pair geometry is tetrahedral (because there are 4 electron groups: 2 bonding pairs and 2 lone pairs), and the molecular geometry is bent. Lone pairs of electrons exert a greater repulsive force than bonding pairs. The repulsion between the lone pairs and the bonding pairs in \( \text{H}_2\text{S} \) compresses the bond angle between the two \( \text{H - S} \) bonds. The bond angle in \( \text{H}_2\text{S} \) is approximately \( 92^\circ \), which is much smaller than \( 180^\circ \).

So, the bond angle in \( \text{BeF}_2 \) is larger than that in \( \text{H}_2\text{S} \).

To compare the bond angles of \( \text{PCl}_3 \) and \( \text{PCl}_5 \), we again use the VSEPR theory.

• \( \text{PCl}_5 \): Phosphorus (P) has 5 valence electrons, and it forms 5 single bonds with chlorine (Cl) atoms. There are no lone pairs on the central P atom. The electron - pair geometry and molecular geometry are trigonal bipyramidal. In a trigonal bipyramidal geometry, the bond angles are \( 120^\circ \) (in the equatorial plane) and \( 90^\circ \) (between the equatorial and axial positions).
• \( \text{PCl}_3 \): Phosphorus (P) has 5 valence electrons. It forms 3 single bonds with chlorine (Cl) atoms, leaving 1 lone pair on the central P atom. The electron - pair geometry is tetrahedral (4 electron groups: 3 bonding pairs and 1 lone pair), and the molecular geometry is trigonal pyramidal. The lone pair on the central P atom in \( \text{PCl}_3 \) exerts a repulsive force on the bonding pairs. This repulsion compresses the bond angle between the \( \text{Cl - P - Cl} \) bonds. The bond angle in \( \text{PCl}_3 \) is approximately \( 100^\circ \).

When we compare the bond angles in the relevant parts of the structures, the bond angles in the trigonal bipyramidal \( \text{PCl}_5 \) (especially the equatorial bond angles of \( 120^\circ \)) are larger than the bond angle in the trigonal pyramidal \( \text{PCl}_3 \) (around \( 100^\circ \)). So, the bond angle in \( \text{PCl}_3 \) is smaller than the bond angle in \( \text{PCl}_5 \).

Answer:

The bond angle in \( \text{BeF}_2 \) is larger than the bond angle in \( \text{H}_2\text{S} \).

Question 2