The correct order regarding the electronegativity of hybrid orbitals of carbon is
- sp < sp2 < sp3
- sp > sp2 < sp3
- sp > sp2 > sp3
- sp < sp2 > sp3
Answer
(3) sp > sp2 > sp3
Solution
s-character: sp = 50%, sp2 = 33.3%, sp3 = 25%.
As per NCERT, the greater the s-character of a hybrid orbital, the closer its electrons are to the nucleus, so the more electronegative it is (this is why ethyne's hydrogen is acidic).
∴ Electronegativity: sp > sp2 > sp3
Answer: (3)