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Which of the following hybridizations has maximum s-characters?
A.\[s{p^3}\]
B.\[s{p^2}\]
C. sp
D. None of these

Answer
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Hint: Hybridization is the mix of two atomic orbitals to create a new type of hybridized orbitals.
One s-orbital is involved in \[s{p^3}\], \[s{p^2}\], and sp hybridization.

Complete step by step solution:Here in this question, we have to find out the maximum s-character from the given hybridization.
S-character is nothing but the percentage of the s-orbital present in the hybrid orbitals.
Let us first know about hybridization.
It is a mix of two atomic orbitals to create a new type of hybridized orbitals.
The new hybrid orbital possesses a percentage of both orbitals.
For example, in sp hybridization, there is a percentage of s-orbital and a percentage of p-orbital as well.
This is known as the s-character and p-character respectively.
A. \[s{p^3}\]
Here one s-orbital and three p-orbitals are hybridized.
So, the s-character is\[25\% \] and the p-character is\[75\% \].
B. \[s{p^2}\]
Here one s-orbital and two p-orbitals are hybridized.
So, the s-character is\[33.33\% \] and the p-character is\[66.67\% \].
C. sp
Here one s-orbital and one p-orbital are hybridized.
So, the s-character is\[50\% \] and the p-character is\[50\% .\]
So, the maximum s-character is present in the sp hybridization.

So, option C is correct.

Note: Bonding molecular orbital is created by the mix of orbitals maintaining similar symmetry but an antibonding molecular orbital is constructed by the unsymmetrical orbitals.
The energy of the antibonding MO is greater than the combining atomic orbitals and it causes the molecule to be less stable.