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# Explain the $s{p^2}$ hybridisation?

Last updated date: 26th Feb 2024
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Hint: As we know that the atomic orbitals combine to form a new set of equivalent orbitals known as the hybrid orbitals and it helps in explaining the shape of molecules and equivalency of bonds. It can be classified into $sp,s{p^2},s{p^3}$ and so on which involves one s-orbital and one p-orbital, one s and two p-orbitals etc.

We also know that there are many different types of hybridisation depending upon the types of orbitals involved in mixing like $sp,s{p^2},s{p^3},s{p^3}d$ etc. let us talk about $s{p^2}$ hybridisation among these.
We can see that in $s{p^2}$ hybridisation, one s-orbital and two p- $({p_x}\;\& \;{p_y})$ orbitals of one atom hybridise to give three equivalent $s{p^2}$ hybrid orbitals. These three $s{p^2}$ hybrid orbitals are directed towards the three corners of an equilateral triangle with an angle of ${120^\circ }$ and give a triangular geometry to the molecule. We can represent it as:
$s{p^2}$ hybrid orbitals are larger in size than the $sp$-hybrid orbitals but slightly smaller than that of $s{p^3}$. Examples of $s{p^2}$ hybridisation includes the compounds of boron like $B{F_3},BC{l_3}$ and $B{H_3}$ as well as aluminium and carbon containing compounds such as $AlC{l_3}$ and $C{H_2} = C{H_2}$respectively.
Note:Always remember that $s{p^2}$ hybridisation involves the mixing of one s-orbital and two p-orbitals which includes the promotion of one electron in the s-orbital and the one to the any one of the p-orbital whose combination creates a three new hybrid orbitals of equivalent energy levels. The $s{p^2}$ hybridised orbitals possess a trigonal planar geometry of molecules and these hybrid orbitals are more effective in forming stable bonds than the pure atomic orbitals.