ci: C!: The central Sn atom is bonded to the three 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. Comment As this example illustrates, when a molecule exhibits resonance, any one of the resonance structures can be used to predict the molecular geometry. So, the molecular geometry is bent with an ideal bond angle of 120 (Table 9.2). Two of the domains are from bonds, and one is due to a nonbonding pair.

(Remember that a double bond counts as a single electron domain.) The arrangement of three electron domains is trigonal planar (Table 9.1). Thus, there are three electron domains about the central O atoms. In both resonance structures, the central O atom is bonded to the two outer O atoms and has one nonbonding pair. Solve (a) We can draw two resonance structures for : Because of resonance, the bonds between the central O atom and the outer O atoms are of equal length. We then obtain the molecular geometry (Notice that for Os, the A and B atoms are all oxygen atoms.) from the arrangement of the domains that are due to bonds. (b) Snci, SOLUTION Analyze We are given the molecular formulas of a molecule and a Plan To predict the molecular geometries, we draw their Lewis struc polyatomic ion, both conforming to the general formula AB and tures and count electron domains around the central atom to get the both having a central atom from the p block of the periodic table.

Transcribed image text: Sample Exercise 9.1 Using the VSEPR Model Use the VSEPR model to predict the molecular geometry of (a) 03.

Double bonds crystalmaker