Chlorine trifluoride has three bonds and two lone pairs, so the five electron pairs sit at the…
Contributed by StemPy LabTry it on paper first. A hint, the key idea and the worked solution are here when you need them.
Chlorine trifluoride has three bonds and two lone pairs, so the five electron pairs sit at the corners of a trigonal bipyramid. Each corner is either axial, on the axis, or equatorial, in the plane. A pair at 90° to another repels it strongly, and pairs at 120° or 180° repel much less. Count the 90° interactions for the three ways of placing the two lone pairs: both equatorial, one axial and one equatorial, and both axial. For each way give the number of lone pair-lone pair interactions at 90° and the number of lone pair-bond pair interactions at 90°. Use the order lone pair-lone pair lone pair-bond pair bond pair-bond pair. Say which way has the lowest repulsion.
Need a reminder? The key ideaShow
For the central atom, lone pairs , and the steric number is the number of bonded atoms plus the lone pairs. In a trigonal bipyramid a lone pair prefers an equatorial place, where it has fewer neighbours at 90°. The order of repulsion is lone pair-lone pair, then lone pair-bond pair, then bond pair-bond pair.
The worked solution
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Work it out on paper first. It is fine to get stuck or make a mistake: that is how we learn. When you are ready, go through the solution one step at a time, and check each step against your own.
Plan. List the neighbours at 90° of each place, then put in the lone pairs.
Step 1. Both equatorial. Each equatorial lone pair has the two axial places at 90°. Both are bond pairs. That gives lone pair-bond pair interactions. The two lone pairs are at 120° and not at 90°, so there are 0 lone pair-lone pair interactions at 90°.
Step 2. One axial and one equatorial. The axial lone pair has three equatorial places at 90°. One of them is the other lone pair, and two are bond pairs. The equatorial lone pair has the two axial places at 90°: one is the lone pair already counted and one is a bond pair. In all: 1 lone pair-lone pair and lone pair-bond pair interactions.
Step 3. Both axial. Each axial lone pair has three equatorial bond pairs at 90°: lone pair-bond pair and no lone pair-lone pair, as the two are at 180°.
Step 4. The best way. Both equatorial has 4 lone pair-bond pair and no lone pair-lone pair at 90°. The way with one axial has a lone pair-lone pair interaction at 90°, which is the strongest repulsion. Both axial has 6 lone pair-bond pair interactions. So both equatorial is the best.
Answer: (0, 4) for both equatorial, (1, 3) for one axial and (0, 6) for both axial. Both equatorial wins, and the shape is T-shaped.
Check. The same way gives a F-Cl-F angle of about 87.5°, which is slightly under the ideal 90° because of the lone pairs.
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