MCAT Chemistry Question — Bond Angles

Similar bond angles generally – but not always – imply that the molecules have similar VSEPR geometries. However, we also need to consider factors including the presence of lone pairs and the overall Lewis structures. Let’s look at the shapes of each pair individually to find the right answer.

A. tetrahedral, linear. CH4, like a typical tetrahedral molecule, has approximate bond angles of 109.5 degrees. In contrast, a linear arrangement implies that bond angles are 180 degrees. These are too distant to be correct.

B. trigonal planar, trigonal pyramidal. These may seem similar initially! However, trigonal planar bond angles are 120 degrees, while a trigonal pyramidal arrangement resembles a tetrahedral structure with one substituent replaced by a lone pair. As a result, the associated angles are slightly smaller than 109.5 (107.8, if you want to be accurate).

C. bent, bent. This, too, it tempting! But H2O possesses two lone pairs on its central atom. As a result, its bond angles are even smaller than 107.8. In fact, they’re around 104.5 degrees. This differs significantly from the 120-degree angles expected in SO2.

D. tetrahedral, bent. Finally, here’s our correct choice! Tetrahedral bond angles are typically 109.5 degrees, while that between the O-H bonds of water is 104.5. This is only a 5-degree difference (smaller, if you consult some textbooks, though we don’t really care about the exact details). The key is that, although their shapes differ, the presence of different numbers of lone pairs cause these angles to be nearly identical.