Try it on paper first. A hint, the key idea and the worked solution are here when you need them.
In ethane, ethene and ethyne, the carbon atoms are , and . Find the s-character of each kind of hybrid. The C-H bonds are about , and long, in no special order. Match each length to its molecule, and explain.
Hybridisation mixes atomic orbitals of nearly equal energy into the same number of equal hybrid orbitals. Two pairs on the central atom give sp (180°), three give sp2 (120°), four give sp3 (109.5°), five give sp3d (trigonal bipyramidal) and six give sp3d2 (octahedral). The s-character of each hybrid is 100% divided by the number of orbitals that mix. Unhybridised p orbitals form pi bonds.
Have you tried it?
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.
Step 1. s-character. : . : . : .
Step 2. The idea. More s-character keeps the hybrid closer to the nucleus. So the bond it makes is shorter.
Step 3. Matching. Ethyne () has the shortest C-H bond, . Ethene () has , and ethane () .
Answer: , and ; the C-H bond is shortest in ethyne and longest in ethane.
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