Question

Which is more electronegative: a \[sp\] hybridized carbon or a \[s{{p}^{2}}\] hybridized carbon and why?

Answer 1

Hint: Electronegativity is defined as the relative tendency of an atom or species to attract electrons towards it and we know that, to attract the electron, positive charge is required that comes from the nucleus. So nearer an orbital is to the nucleus, higher electronegative it will be, as electrons in the orbital will feel more attraction force.

Complete answer:
In \[s{{p}^{3}}\] hybridized carbon four orbitals are involved in hybridized, among only one is s. if we calculate percentage of s character into \[s{{p}^{3}}\] hybridized orbitals, it will be equal to 25%.
In \[s{{p}^{2}}\] hybridized carbon three orbitals are involved in hybridized, among only one is s. if we calculate percentage of s character into \[s{{p}^{2}}\] hybridized orbitals, it will be equal to 33.33%.
In \[sp\] hybridized carbon two orbitals are involved in hybridization, among only one is s. If we calculate the percentage of s character into \[sp\] hybridized orbitals, it will be equal to 50%.

We can clearly see that \[sp\]hybridized orbital has a higher percentage of s character in it compared to \[s{{p}^{2}}\] hybridized orbitals. Higher percentage of s-character indicates that hybrid orbital is more closer to the nucleus and that’s why it has higher electronegativity in comparison to \[s{{p}^{2}}\] hybridized orbitals.

Order of electronegativity : $sp>s{{p}^{2}}>s{{p}^{3}}$.

Additional information: Because \[sp\] hybridized carbon is more electronegative than \[s{{p}^{2}}\] and \[s{{p}^{3}}\], that’s why terminal alkynes or 1-alkynes show acidic behavior.

Note: sp hybridized orbitals have the same shape and energy. Their shape is oval which is in between spherical shape and pear shape. sp- hybrid orbitals are collinear. That means they are aligned in a straight line in the opposite direction. That’s why the molecules that have sp hybridized central atoms possess linear geometry with bond angle of ${{180}^{0}}$.