What is the hybridization of the central atom in #"BCl"_3# ?

1 Answer

An easy way to figure out the answer is to write out the following string:

#"s p p p d d d d d"#

Look at the number of sigma bonds and lone pairs on the center atom. Counting from left to right, tick off the number you just counted on the above string.

So, #BCl_3# has three bonds on the center atom and no lone pairs. So I tick off three increments on the string:

#cancel("s") cancel("p") cancel("p") "p d d d d d"#

which would be #sp^2#.