The change in hybridisation affects the electronegativity of carbon. The greater the s-character of the hybrid orbitals, the closer its electrons are to the nucleus, and thus, the greater is its electronegativity. The s-character in $sp$, $sp^2$, and $sp^3$ hybrid orbitals is 50%, 33.33%, and 25%, respectively. Therefore, a carbon atom having an $sp$ hybrid orbital is more electronegative than that possessing $sp^2$ or $sp^3$ hybridised orbitals. The correct decreasing order of electronegativity is $sp > sp^2 > sp^3$.