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What is the difference between d2sp3 and sp3d2 hybrid orbitals?
Wiki User
∙ 12y ago
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.
Wiki User
∙ 12y ago
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Explain d2sp3 hybridisation that takes place in potassium ferro cyanide?
as you now Fe configuration end by 4s2 3d5 we have to kind of hypredization depening on the lligand inter to Fe i will tak about d2sp3 just if the bonded ligand strong the electron in d5 will be compressed as follows 1 1 1 1 1 2 - - - d5 s2 3p empty will becom after intering a strong ligand 2 2 1- - 2 --- d5 with s2 3p empty 2 empty so the hypredization will be between 2 orbital from d with s orbital and 3 orbital of p producing d2sp3 and the complex called inner complex والله تعالى اعلم عبدالعزيز ابويامين
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.
Why cant ph5 phosphorus pentahydride exist?
This compound could possibly exist, but it would be very unstable towards decomposition to PH3 and H2. One could propose either an ionic (PH4+H-) or covalent structure for this compound. The covalent structure would involve an expanded valence shell involving d2sp3 hybrid orbitals. For phosphorus, valence shell expansion is usually favored by bonding to electronegative elements such as halogens. Since hydrogen is of similar electronegativity to phosphorus, this covalent structure would not be favored. Concerning the possible ionic structure, the hydrogen atoms bonded to PH4+ have some protic character. Therefore they would readily react with the hydride ion to form H2. If PH5 could be made, it would be stable only at very low temperatures under very high pressures.
What are the chemical properties of xenon tetrafluoride?
Xenon Tetrafluoride is a compound comprised of four (4) Fluoride atoms and one (1) Xenon atom. The structure in unique in the way that the central Xenon atom has two lone pairs of electrons in addition to the four single bonds connecting the fluorine atoms. The two lone pairs of the central atom, Xenon, cause the compound to assume an octahedral shape. More specifically, the compound assumes a square planar molecular geometry, with four (4) Xe-F bonds and the two lone pairs above and below, which, by the theory of valence shell electron pair repulsion, occupy the most space possible, creating bond angles of 90 degrees. The hybridization for the central Xenon atom is d2sp3, as there are six (6) areas: the Xe-F bonds and the two lone pairs. XeF4 t is non polar as it is symmetrical.
Explain d2sp3 hybridisation that takes place in potassium ferro cyanide?
as you now Fe configuration end by 4s2 3d5 we have to kind of hypredization depening on the lligand inter to Fe i will tak about d2sp3 just if the bonded ligand strong the electron in d5 will be compressed as follows 1 1 1 1 1 2 - - - d5 s2 3p empty will becom after intering a strong ligand 2 2 1- - 2 --- d5 with s2 3p empty 2 empty so the hypredization will be between 2 orbital from d with s orbital and 3 orbital of p producing d2sp3 and the complex called inner complex والله تعالى اعلم عبدالعزيز ابويامين
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.
Why cant ph5 phosphorus pentahydride exist?
This compound could possibly exist, but it would be very unstable towards decomposition to PH3 and H2. One could propose either an ionic (PH4+H-) or covalent structure for this compound. The covalent structure would involve an expanded valence shell involving d2sp3 hybrid orbitals. For phosphorus, valence shell expansion is usually favored by bonding to electronegative elements such as halogens. Since hydrogen is of similar electronegativity to phosphorus, this covalent structure would not be favored. Concerning the possible ionic structure, the hydrogen atoms bonded to PH4+ have some protic character. Therefore they would readily react with the hydride ion to form H2. If PH5 could be made, it would be stable only at very low temperatures under very high pressures.
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 compound comprised of four (4) Fluoride atoms and one (1) Xenon atom. The structure in unique in the way that the central Xenon atom has two lone pairs of electrons in addition to the four single bonds connecting the fluorine atoms. The two lone pairs of the central atom, Xenon, cause the compound to assume an octahedral shape. More specifically, the compound assumes a square planar molecular geometry, with four (4) Xe-F bonds and the two lone pairs above and below, which, by the theory of valence shell electron pair repulsion, occupy the most space possible, creating bond angles of 90 degrees. The hybridization for the central Xenon atom is d2sp3, as there are six (6) areas: the Xe-F bonds and the two lone pairs. XeF4 t is non polar as it is symmetrical.
