IIT-JEE 2010 · Paper 1, Chemistry question 24

Shape of $\ce{BrF5}$. Br has 5 bond pairs and 1 lone pair: octahedral electron geometry, square-pyramidal shape.

Ideal angles. In an ideal square pyramid, the apical F makes 90° with each of the 4 basal F, and adjacent basal F atoms are 90° apart (4 more): 8 angles of 90°.

Real angles. The lone pair below the base repels the basal bonds, pushing them up towards the apical F; the F(apical)-Br-F(basal) angles become about 85°, and the basal F-Br-F angles also fall slightly below 90° (the basal F atoms lie above the Br). So, strictly, no angle is exactly 90°: the answer is 0.

The official key accepts 0 (actual geometry) or 8 (idealised geometry).

Answer: 0 (key: 0 or 8)