
What is the hybridisation of \[SO_4^{2 - }\] ion?
Answer
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Hint: To know the characteristic geometrical shapes of polyatomic molecules the concept of hybridisation was introduced. The atomic orbitals combine to form a new set of equivalent orbitals known as hybrid orbitals. Hybridisation is a process of intermixing of the orbitals of slightly different energies so that it redistributes the energies resulting in the formation of new orbitals of equivalent energies and shape.
Example- one \[2s\] and three \[2p\] orbitals of carbon hybridise to form four new \[s{p^3}\] hybrid orbitals.
Complete answer:
Sulphate(\[SO_4^{2 - }\]) is one of the most widely naturally available chemical compounds.
Sulphate is basically a chemical compound that is composed of sulphur and oxygen atoms. Sulphate forms salts with a variety of elements .
The number of sigma bond and lone pair electrons provides details regarding ion hybridisation.
\[sp\] hybridisation- Number of sigma bond and lone pair of electrons is 2.
\[s{p^2}\] hybridisation- Number of sigma bonds and lone pair of electrons is 3.
\[s{p^3}\] hybridisation- Number of sigma bonds and lone pair of electrons is 4.
According to VSEPR theory the \[SO_4^{2 - }\] ion has tetrahedral molecular geometry. The \[SO_4^{2 - }\] ion has four oxygen atoms arranged tetrahedrally around the sulphur atom. So, there are a total of four \[S - O\] sigma bonds and all four bonds are of equal lengths. Hence \[SO_4^{2 - }\] ion has \[s{p^3}\] hybridisation. The extra two electrons of sulphur form two double bonds to oxygen then they move over to their oxygen forming a net negative charge on each of the four oxygen atoms.
Note:
In each \[s{p^3}\] hybrid orbital there will be \[25\% \] s-character and \[75\% \] p-character. The angle between the \[s{p^3}\] hybrid orbital is \[{109.5^0}\]. The other examples for \[s{p^3}\] hybridisation is \[N{H_3}\], \[{H_2}O\],\[C{H_4}\],\[{C_2}{H_6}\] molecules etc. there are hybridisation involving d orbitals. They are \[ds{p^2}\],\[s{p^3}d\],\[s{p^3}{d^2}\] and \[{d^2}s{p^3}\] hybridisation. These are the hybridisations for higher molecules.
Example- one \[2s\] and three \[2p\] orbitals of carbon hybridise to form four new \[s{p^3}\] hybrid orbitals.
Complete answer:
Sulphate(\[SO_4^{2 - }\]) is one of the most widely naturally available chemical compounds.
Sulphate is basically a chemical compound that is composed of sulphur and oxygen atoms. Sulphate forms salts with a variety of elements .
The number of sigma bond and lone pair electrons provides details regarding ion hybridisation.
\[sp\] hybridisation- Number of sigma bond and lone pair of electrons is 2.
\[s{p^2}\] hybridisation- Number of sigma bonds and lone pair of electrons is 3.
\[s{p^3}\] hybridisation- Number of sigma bonds and lone pair of electrons is 4.
According to VSEPR theory the \[SO_4^{2 - }\] ion has tetrahedral molecular geometry. The \[SO_4^{2 - }\] ion has four oxygen atoms arranged tetrahedrally around the sulphur atom. So, there are a total of four \[S - O\] sigma bonds and all four bonds are of equal lengths. Hence \[SO_4^{2 - }\] ion has \[s{p^3}\] hybridisation. The extra two electrons of sulphur form two double bonds to oxygen then they move over to their oxygen forming a net negative charge on each of the four oxygen atoms.
Note:
In each \[s{p^3}\] hybrid orbital there will be \[25\% \] s-character and \[75\% \] p-character. The angle between the \[s{p^3}\] hybrid orbital is \[{109.5^0}\]. The other examples for \[s{p^3}\] hybridisation is \[N{H_3}\], \[{H_2}O\],\[C{H_4}\],\[{C_2}{H_6}\] molecules etc. there are hybridisation involving d orbitals. They are \[ds{p^2}\],\[s{p^3}d\],\[s{p^3}{d^2}\] and \[{d^2}s{p^3}\] hybridisation. These are the hybridisations for higher molecules.
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