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Good question. A lot of people use them interchangeably, and assume they are the same. They are not the same. With sp3d2, the s, p and d orbitals which are hybridized all come from the same energy level, for instance, it has been taught that when sulfur combines with six fluorine atoms to make SF6 that the 3s, 3p and two 3d orbitals hybridize to make the sp3d2 hybrid orbital set.
But d2sp3 is different. In this case the d-orbitals come from the n-1 energy level. Transition metals may exhibit d2sp3 hybridization where the d orbitals are from the 3d and the s and p orbitals are the 4s and 3d.
The bottom line is this, in sp3d2 hybridization all of the orbitals have the same principal quantum number. In d2sp3, the principle quantum number of the d orbitals is one less than the principal quantum numbers of the s and p orbitals. We see d2sp3 hybridization in the transitions metals and sp3d2 hybridization in the nonmetals.
There is one more issue. Chemists today are finding out that in compounds like SF6 there is no involvement of d-orbitals. In other words, there is no sp3d2 hybridization in SF6. A more likely explanation involves what is called "3-center, 4-electron" bonding in which three orbitals overlap axially (in a straight line) and contain a total of 4 electrons. This means that the 3 unhybridized p-orbitals of sulfur are all that is needed to make the six bonds with fluorine atoms.
Now you can be the first in your class to point out that there really isn't any sp3d2 hybridization at all.
Oh, dude, it's like asking the difference between a hot dog and a sausage. So, d2sp3 and sp3d2 hybrid orbitals are like different ways to mix up your atomic orbitals for bonding. In d2sp3, you start with a d orbital and mix it with two s orbitals and three p orbitals, while in sp3d2, you mix an s orbital with three p orbitals and two d orbitals. It's just like rearranging your furniture - same pieces, different layout.
The main difference between d2sp3 and sp3d2 hybrid orbitals is the order in which the atomic orbitals are combined. In d2sp3 hybridization, first d orbitals are combined with s and p orbitals, while in sp3d2 hybridization, s and p orbitals are combined first with d orbitals. This leads to different geometric arrangements of the resulting hybrid orbitals.
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Explain d2sp3 hybridisation that takes place in potassium ferro cyanide?
In potassium ferrocyanide, the central iron atom undergoes d^2sp^3 hybridization. This means that the 3d, 4s, and three 4p orbitals of iron combine to form five hybrid orbitals. These hybrid orbitals are then used to form bonds with surrounding atoms, such as carbon and nitrogen in the ferrocyanide complex.
Why cant ph5 phosphorus pentahydride exist?
Phosphorus pentahydride, PH5, cannot exist because phosphorus typically forms stable compounds with a maximum coordination number of five due to its electron configuration. In PH5, the phosphorus atom would need to have a coordination number of six, which is energetically unfavorable. Furthermore, there are no known stable compounds of phosphorus in the +5 oxidation state with hydrogen.
Why is d3 orbital more stable in water than d5?
Lets first take the case of the d3 compound. The no.of orbitals in the 3d shell is 5. If three electrons occupy three orbitals then there are two free orbitals.Therefore According to Valence bond theory the six water ligands will use the two inner d orbitals the outer s and the p orbitals to form an inner orbital complex with hybrisation d2sp3. In the second case we have the d5 compund. Since there are five electrons in the d subshell the five electrons singly occupy all the five d orbitals. Here 's where the concept of the weak ligand comes in. Since water is a weak ligand it cannot force pairing of the unpaired d electrons to make room for an inner orbital complex. Thus it has to use the outer d orbital to form an outer orbital complex with hybridisation of sp3d2. Since the Inner orbital (low spin) complex is more stable than the outer orbital (high spin) complex. Thus d3 configuration is more stable than d5 configuration in aqueous medium.
What are the chemical properties of xenon tetrafluoride?
Xenon tetrafluoride is a strong fluorinating agent, meaning it can readily donate fluorine atoms to other compounds. It is highly reactive due to the electronegativity of fluorine and the large size of xenon, which results in a weak xenon-fluorine bond. Xenon tetrafluoride is a colorless solid at room temperature and is considered to be a powerful oxidizing agent.
Explain d2sp3 hybridisation that takes place in potassium ferro cyanide?
In potassium ferrocyanide, the central iron atom undergoes d^2sp^3 hybridization. This means that the 3d, 4s, and three 4p orbitals of iron combine to form five hybrid orbitals. These hybrid orbitals are then used to form bonds with surrounding atoms, such as carbon and nitrogen in the ferrocyanide complex.
Why cant ph5 phosphorus pentahydride exist?
Phosphorus pentahydride, PH5, cannot exist because phosphorus typically forms stable compounds with a maximum coordination number of five due to its electron configuration. In PH5, the phosphorus atom would need to have a coordination number of six, which is energetically unfavorable. Furthermore, there are no known stable compounds of phosphorus in the +5 oxidation state with hydrogen.
Why is d3 orbital more stable in water than d5?
Lets first take the case of the d3 compound. The no.of orbitals in the 3d shell is 5. If three electrons occupy three orbitals then there are two free orbitals.Therefore According to Valence bond theory the six water ligands will use the two inner d orbitals the outer s and the p orbitals to form an inner orbital complex with hybrisation d2sp3. In the second case we have the d5 compund. Since there are five electrons in the d subshell the five electrons singly occupy all the five d orbitals. Here 's where the concept of the weak ligand comes in. Since water is a weak ligand it cannot force pairing of the unpaired d electrons to make room for an inner orbital complex. Thus it has to use the outer d orbital to form an outer orbital complex with hybridisation of sp3d2. Since the Inner orbital (low spin) complex is more stable than the outer orbital (high spin) complex. Thus d3 configuration is more stable than d5 configuration in aqueous medium.
Is ICl4- a square planar and polar?
Yes, it is square planar. The central iodine atom exceeds the octet rule by bonding with all four chlorine atoms and having two lone pairs. A central atom with six electron pairs (d2sp3 hybridization) and two lone electron pairs by definition is square planar (see VSEPR theory for more information). Because of its symmetrical geometry, it will have no dipole moment.
What are the chemical properties of xenon tetrafluoride?
Xenon tetrafluoride is a strong fluorinating agent, meaning it can readily donate fluorine atoms to other compounds. It is highly reactive due to the electronegativity of fluorine and the large size of xenon, which results in a weak xenon-fluorine bond. Xenon tetrafluoride is a colorless solid at room temperature and is considered to be a powerful oxidizing agent.
