Electronegativity of carbon atoms depends upon their state of hybridisation. In which of the following compounds, the carbon marked with asterisk is most electronegative?

Solution :
The greater the $$s$$  character of the hybrid orbitals, the greater is the electronegativity. Thus, a carbon atom having an $$sp$$  hybrid orbital with $$50\% \,s$$  character is more electronegative than those possessing $$s{p^2}$$  or $$s{p^3}$$  hybridised orbitals .
The asterisk marked carbon in $$C{H_3} - C{H_2} - C \equiv \mathop C\limits^ * H$$     is $$sp$$  hybridised hence, it is most electronegative.