Question

In page no117 of NCERT Chemistry book, it is given that the shape (a) is the most stable for CF3. However, isn't lone-pair lone-pair repulsion present in shape (a)? So, it should be the least stable one right? Since lone pair lone-pair repulsion is greater than lone-pair bond pair repulsion??
Please help me with my doubt...​

Answer 1

Answer:

Explanation:

When one or more of the groups is a lone pair of electrons (non-bonded electrons), the experimentally-observed geometry around an atom is slightly different than in the case where all groups are bonds. The actual bond angles are similar, but not exactly the same, as those predicted based on the total number of groups (the "parent" geometry). When there is a mixture of group types (lone pairs (E) and bonded groups (X)) there are three different types of angles to consider: bond angles between two bonded atoms (X-X angles), angles between a bonded atom and a lone pair (X-E angles), and angles between two lone pairs (E-E angles). Empirical evidence shows the following trend in the degree of bond angles in around atoms with a mixture of group types:

Trend in bond angles:

E-E >X-E >X-X

Using empirical evidence as a guide, we can predict that lone pairs repel other electron groups more strongly than bonded pairs. The molecular geometry of molecules with lone pairs of electrons are better predicted when we consider that electronic repulsion created by lone pairs is stronger than the repulsion from bonded groups. It is difficult to predict the exact bond angle based on this principle, but we can predict approximate angles, as described and summarized below in Table  3.2.1.1