Question 9.SE.1: Use the VSEPR model to predict the molecular geometry of (a)...

Analyze We are given the molecular formulas of a molecule and a polyatomic ion, both conforming to the general formula AB_n and both having a central atom from the p block of the periodic table. (Notice that for O_3, the A and B atoms are all oxygen atoms.)

Plan To predict the molecular geometries, we draw their Lewis structures and count electron domains around the central atom to get the electron-domain geometry. We then obtain the molecular geometry from the arrangement of the domains that are due to bonds.

Solve

(a) We can draw two resonance structures for O_3:

Because of resonance, the bonds between the central O atom and the outer O atoms are of equal length. In both resonance structures, the central O atom is bonded to the two outer O atoms and has one nonbonding pair. Thus, there are three electron domains about the central O atoms. (Remember that a double bond counts as a single electron domain.) The arrangement of three electron domains is trigonal planar (Table 9.1). Two of the domains are from bonds, and one is due to a nonbonding pair. So, the molecular geometry is bent with an ideal bond angle of 120° (Table 9.2).

Comment As this example illustrates, when a molecule exhibits resonance, any one of the resonance structures can be used to predict the molecular geometry.

(b) The Lewis structure for SnCl_3^- is:

\left[\begin{matrix} :\overset{..}{\underset{..}{Cl—}} & \overset{..}{Sn—} & \overset{..}{\underset{..}{Cl}}:\\ & | & \\ & :\underset{..}{Cl}: & \end{matrix} \right]^-

The central Sn atom is bonded to the three Cl atoms and has one nonbonding pair; thus, we have four electron domains, meaning a tetrahedral electron-domain geometry (Table 9.1) with one vertex occupied by a nonbonding pair of electrons. A tetrahedral electron-domain geometry with three bonding and one nonbonding domains leads to a trigonal-pyramidal molecular geometry (Table 9.2).