(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.)

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).

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.

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.

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.

(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.