
What is the spatial orientation of $ s{p^3}{d^3} $ hybrid orbitals?
Answer
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Hint : The $ 3 - $ dimensional orientation of orbitals in space is called the spatial orientation of orbitals. The geometry of $ s{p^3}{d^3} $ hybridization is pentagonal bipyramidal. In this geometry, five bonds are in the plane and one bond lies above the plane and one bond below the plane.
Complete Step By Step Answer:
The spatial arrangement of $ s{p^3}{d^3} $ hybrid orbitals is said to be pentagonal bipyramidal. Let us consider iodine heptafluoride $ (I{F_7}) $ to understand this. The electronic configuration of iodine in its ground state is $ [Kr]4{d^{10}}5{s^2}5{p^5} $ . The formation of iodine heptafluoride needs seven unpaired electrons. The iodine atom will promote three electrons into the empty $ 5d $ orbitals. One electron from the $ 5s $ orbital and two electrons from $ 5p $ orbital are promoted to the empty $ 5d $ orbitals. This is called the third excited state. In this third excited state, the iodine atom will undergo $ s{p^3}{d^3} $ hybridization and it gives seven half filled $ s{p^3}{d^3} $ hybrid orbitals in the pentagonal bipyramidal geometry. These will further form $ 7{\sigma _{s{p^3}{d^3} - p}} $ bonds with the fluorine atoms.
Therefore, the spatial orientation of $ s{p^3}{d^3} $ hybrid orbitals is pentagonal bipyramidal.
Note :
Hybridization is the process of mixing atomic orbitals into new hybrid orbitals which have different shapes and energies than the parent atomic orbitals, suitable for pairing of electrons to form chemical bonds in the valence bond theory. The orientation of hybrid orbitals is in such a way that there is minimum repulsion between those orbitals. The process of hybridization takes place when atomic orbitals form a new atomic orbital. Note that the new orbital formed can hold exactly the same number of electrons as the old atomic orbitals.
Complete Step By Step Answer:
The spatial arrangement of $ s{p^3}{d^3} $ hybrid orbitals is said to be pentagonal bipyramidal. Let us consider iodine heptafluoride $ (I{F_7}) $ to understand this. The electronic configuration of iodine in its ground state is $ [Kr]4{d^{10}}5{s^2}5{p^5} $ . The formation of iodine heptafluoride needs seven unpaired electrons. The iodine atom will promote three electrons into the empty $ 5d $ orbitals. One electron from the $ 5s $ orbital and two electrons from $ 5p $ orbital are promoted to the empty $ 5d $ orbitals. This is called the third excited state. In this third excited state, the iodine atom will undergo $ s{p^3}{d^3} $ hybridization and it gives seven half filled $ s{p^3}{d^3} $ hybrid orbitals in the pentagonal bipyramidal geometry. These will further form $ 7{\sigma _{s{p^3}{d^3} - p}} $ bonds with the fluorine atoms.
Therefore, the spatial orientation of $ s{p^3}{d^3} $ hybrid orbitals is pentagonal bipyramidal.
Note :
Hybridization is the process of mixing atomic orbitals into new hybrid orbitals which have different shapes and energies than the parent atomic orbitals, suitable for pairing of electrons to form chemical bonds in the valence bond theory. The orientation of hybrid orbitals is in such a way that there is minimum repulsion between those orbitals. The process of hybridization takes place when atomic orbitals form a new atomic orbital. Note that the new orbital formed can hold exactly the same number of electrons as the old atomic orbitals.
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