Question

The correct order regarding the electronegativity of hybrid orbitals of carbon is:
A) $\text{ sp}>\text{ s}{{\text{p}}^{\text{2 }}}<\text{ s}{{\text{p}}^{\text{3 }}}\text{ }$
B) $\text{ sp}>\text{ s}{{\text{p}}^{\text{2 }}}>\text{ s}{{\text{p}}^{\text{3 }}}\text{ }$
C) $\text{ sp}<\text{ s}{{\text{p}}^{\text{2 }}}>\text{ s}{{\text{p}}^{\text{3 }}}\text{ }$
D) $\text{ sp}<\text{ s}{{\text{p}}^{\text{2 }}}<\text{ s}{{\text{p}}^{\text{3 }}}\text{ }$

Answer 1

The electronegativity of hybrid orbitals can be estimated from the percentage of s and p orbitals. S-orbitals are closest to the nucleus and thus have a higher affinity towards the electrons. The relation between the electronegativity and s-character is as below,
$\begin{aligned} & \text{ S-character }\propto \text{Electronegativity } \\\ & \frac{1}{\text{p-character}}\text{ }\propto \text{Electronegativity } \\\ \end{aligned}$

Complete Solution :
S orbital is a spherical orbital. It is closer to the nucleus and the s-orbital is tightly held by the nucleus than the p, d, and f orbital.
- Atomic orbitals of carbon atoms hybridized into three different sets of orbitals. The one s-orbital and three p-orbitals combine to produce $\text{ s}{{\text{p}}^{\text{3 }}}$ hybrid orbitals.in $\text{ s}{{\text{p}}^{\text{3 }}}$ hybrid orbitals, the percentage of s-character is $\text{ 25}{\scriptstyle{}^{0}/{}_{0}}\text{ }$ while p-orbitals has a total of $\text{ 75}{\scriptstyle{}^{0}/{}_{0}}\text{ }$ character. Due to the higher content of p-orbital, the $\text{ s}{{\text{p}}^{\text{3 }}}$hybrid orbitals cannot attract the electron pair effectively.
- The one s-orbital and two p-orbitals combine to produce $\text{ s}{{\text{p}}^{\text{2 }}}$ hybrid orbitals.in $\text{ s}{{\text{p}}^{\text{2 }}}$ hybrid orbitals, the percentage of s-character is $\text{ 33}\text{.33}{\scriptstyle{}^{0}/{}_{0}}\text{ }$ while p-orbitals has a total of $\text{ 66}\text{.66}{\scriptstyle{}^{0}/{}_{0}}\text{ }$ character. Due to the relatively higher content of p-orbital, the $\text{ s}{{\text{p}}^{\text{2 }}}$ hybrid orbitals are relatively more electronegative than $\text{ s}{{\text{p}}^{\text{3 }}}$a carbon atom. Here s-orbital has $\text{ 33}\text{.33}{\scriptstyle{}^{0}/{}_{0}}\text{ }$ the character which is more than that of $\text{ s}{{\text{p}}^{\text{3 }}}$.
- In the third set of hybrid orbitals one s-orbital and one p-orbitals combine to produce $\text{ sp }$ hybrid orbitals.in $\text{ sp }$ hybrid orbitals, the percentage of s-character is $\text{ 50}{\scriptstyle{}^{0}/{}_{0}}\text{ }$ while p-orbitals has a total of $\text{ 50}{\scriptstyle{}^{0}/{}_{0}}\text{ }$ character. Due to the higher s-orbital character, the $\text{ sp }$hybridized carbon atom has a higher tendency to attract the electron density. Therefore, $\text{ sp }$orbitals are more electronegative than $\text{ s}{{\text{p}}^{\text{2 }}}$ and $\text{ s}{{\text{p}}^{\text{3 }}}$ hybrid orbitals.
Now the correct order regarding the electronegativity of hybrid orbitals of carbon is as follows,
$\text{ sp}>\text{ s}{{\text{p}}^{\text{2 }}}>\text{ s}{{\text{p}}^{\text{3 }}}\text{ }$
So, the correct answer is “Option B”.

Note: Note that, the $\text{ sp }$ hybrid carbon atoms are more electronegative. Due to higher electronegativity, they can easily attract the electron from the bonding pair. Similarly, due to higher electronegativity, the $\text{ sp }$ hybridized orbitals have a greater tendency to lose a proton and are acidic in nature.