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Question: The correct order regarding the electronegativity of hybrid orbitals of carbon is: A) \(\text{ sp}...

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

Explanation

Solution

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,
 S-character Electronegativity  1p-character Electronegativity  \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  sp\text{ s}{{\text{p}}^{\text{3 }}} hybrid orbitals.in  sp\text{ s}{{\text{p}}^{\text{3 }}} hybrid orbitals, the percentage of s-character is  250/0 \text{ 25}{\scriptstyle{}^{0}/{}_{0}}\text{ } while p-orbitals has a total of  750/0 \text{ 75}{\scriptstyle{}^{0}/{}_{0}}\text{ } character. Due to the higher content of p-orbital, the  sp\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  sp\text{ s}{{\text{p}}^{\text{2 }}} hybrid orbitals.in  sp\text{ s}{{\text{p}}^{\text{2 }}} hybrid orbitals, the percentage of s-character is  33.330/0 \text{ 33}\text{.33}{\scriptstyle{}^{0}/{}_{0}}\text{ } while p-orbitals has a total of  66.660/0 \text{ 66}\text{.66}{\scriptstyle{}^{0}/{}_{0}}\text{ } character. Due to the relatively higher content of p-orbital, the  sp\text{ s}{{\text{p}}^{\text{2 }}} hybrid orbitals are relatively more electronegative than  sp\text{ s}{{\text{p}}^{\text{3 }}}a carbon atom. Here s-orbital has  33.330/0 \text{ 33}\text{.33}{\scriptstyle{}^{0}/{}_{0}}\text{ } the character which is more than that of  sp\text{ s}{{\text{p}}^{\text{3 }}}.
- In the third set of hybrid orbitals one s-orbital and one p-orbitals combine to produce  sp \text{ sp } hybrid orbitals.in  sp \text{ sp } hybrid orbitals, the percentage of s-character is  500/0 \text{ 50}{\scriptstyle{}^{0}/{}_{0}}\text{ } while p-orbitals has a total of  500/0 \text{ 50}{\scriptstyle{}^{0}/{}_{0}}\text{ } character. Due to the higher s-orbital character, the  sp \text{ sp }hybridized carbon atom has a higher tendency to attract the electron density. Therefore,  sp \text{ sp }orbitals are more electronegative than  sp\text{ s}{{\text{p}}^{\text{2 }}} and  sp\text{ s}{{\text{p}}^{\text{3 }}} hybrid orbitals.
Now the correct order regarding the electronegativity of hybrid orbitals of carbon is as follows,
 sp> sp> sp \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  sp \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  sp \text{ sp } hybridized orbitals have a greater tendency to lose a proton and are acidic in nature.