Question

\[s{p^3}\] hybridised orbital contains
A) \[\dfrac{1}{4}\] s-character
B) \[\dfrac{1}{2}\] s-character
C) \[\dfrac{2}{3}\] s-character
D) \[\dfrac{3}{4}\] s-character

Answer 1

Hint: The combination of atomic orbitals result to form hybrid orbitals. The hybrid orbitals are suitable for the bond formation following the predicted directions of VSEPR theory. Here, we have to calculate the s character in \[s{p^3}\] hybridization.

Complete Step by Step Answer:
The term \[s{p^3}\] hybridization defines the combination of one 2s orbital and three 2p orbitals. And this hybridization produces four hybrid orbits having identical properties. For an atom to be \[s{p^3}\] hybridised, an atom must have one number of s orbital and three numbers of p orbital.

The orbitals are defined as the regions where the electrons reside. And they are of four types, s, p, d and f. The azimuthal quantum number determines the values of s,p, d and f. The shape of the s orbital is spherical, the p orbital is polar in nature. The s orbital can accommodate two numbers of electrons, the p orbitals can hold six numbers of electrons, the d orbital has ten electrons and the f orbital can hold fourteen electrons.

In \[s{p^3}\]orbital, the s character has 25% or \[\dfrac{1}{4}\] s-character.
Therefore, the right answer is option A.

Additional Information:
In case of a molecule having \[s{p^2}\] hybridization, the percentage of s character is 33%. And in case of sp hybridization, the s-character is 50%.

Note: The amount of s or p character is determined by the orbital hybridization that can be used for the properties of molecules such as acidity or basicity with accuracy. The bonding in case of methane and alkanes is explained by the theory of hybridization.