Dsp3 Hybridization

[Jules Altier]

AlsoI have a book that says that in compounds where the central atom is $\mathrmdsp^3$ hybridized, it's shape is square pyramidal, like that of $\ceBrF5,$ but the hybridization of $\ceBr$ in $\ceBrF5$ is $\mathrmsp^3d^2.$ Why is it so? Is my book wrong or am I missing something?

Formally, one may say that putting the d before s and p implies a lower-shell d-orbital. That would be the transition metal case above. And placing the d behind s and p would use d-orbitals from the same shell. That would be the main group case. Thus, the difference between the two (tl;dr) is:

@Jan While your answer is technically correct and is the method that would most likely be used in an upper division/graduate inorganic or physical chemistry course, I think the OP's question was from the perspective of general chemistry where things are kept a bit more simplified. For example, in the general chemistry textbook I use by Petrucci, et.al., they discuss the hybridization model for both main group and transition metal chemistry. They also point out that experimental data indicates difficulties with $\cesp^3d^2$ hybridization for $\ceSF6$ and suggest an alternative which involves 4 covalent bonds to the sulfur and 2 ionic bonds, i.e. $\ceSF4^2+(F^-)2$. No general chemistry textbook that I have seen (and I have seen quite a few) discusses 4-electron-3-center bonds because it becomes too complicated at the general chemistry level and not necessary for those students who are not chemistry majors. MO theory is discussed at the general chemistry level but only for diatomic molecules and ions from the second period, again because more atoms or higher periods makes it more complicated and it's just meant to give them a taste of what it is.

Hybridization is a process in which orbitals of different blocks on the periodic table, form their electrons together to create a new orbitals, with the same number of electrons. Chemist, Linus Pauling founded the idea of hybridization in the year 1931. It was invented when valence bond theory could not prove/predict molecular structure. He used a Methane molecule to prove his theory.

Hybridization was invented to explain the bonding anomalies for a covalent bonds. The difference between covalent and ionic bonds is that ionic compounds transfer electrons because they are more electronegative, whereas covalent bonds share electrons. Hybridization is not required for ionic because the bonding theory behind it is known. One element is positively charged, and the other is negatively charged.

Carbon has 4 lone pair available for bonding because hybridization cannot take place in an orbital that is full, such as 1s. As said early, 2s and 2p electrons will form and create 2sp3. If the carbon is taking place in a double bond, one p orbital will break off and 2sp2 will be formed. If the Carbon is taking place in a triple bond, two p orbitals will break off and sp will be formed. The same can be said for other elements that occupy higher orbitals.

The relationship between vsepr and hybridization is through the domain. The domain of the VSEPR notation and the domain of hybridized orbital must be equal. When a molecule has 2 domains, it has 2 electrons involved in the bonds. An sp hybridization involves 2 electrons and makes covalent bonds in 2 orbits. These 2 orbits are a hybrid of a 2 s orbital and a p orbital, and thus all AX2 hybrids are sp hybrids. Similarly, all derivatives of trigonal bipyramidal (AX5), such as see-saw (AX4E), T-shaped (AX3E2) or linear (AX2E3) have 5 domains and therefore 5 electrons involved in the bonds. The hybridization in this case expands beyond sp3 (which contains 4 orbits) to the 3d shell, and to obtain 5 orbits we get a dsp3 hybridization for all trigonal bipyramidal derivatives.

Draw Lewis structures and MO diagrams for \(\mathrmCN^+, \mathrmCN,\) and\(\mathrmCN^-\) . According to the Lewis model, which species is most stable? According to MO theory, which species is most stable? Do the two theories agree?

Molecular Orbital Theory Sketch the bonding and antibonding molecular orbitals that result from linear combinations of the 2\(p_x\) atomic orbitals in a homonuclear diatomic molecule. (The 2\(\beta_x\) orbitals are those whose lobes are oriented alongthe bonding axis.)

Valence Bond Theory The valence electron configurations of several atoms are shown below. How many bonds can each atom make without hybridization? $$\quad \text a. \mathrmB 2 \mathrms^2 2 p^1 \quad \text b. \mathrmN 2 \mathrms^2 2 \mathrmp^3 \quad \text c. \mathrmO 2\mathrms^2 2 \mathrmp^4$$

Molecular Orbital Theory Apply molecular orbital theory to predict if each molecule or ion exists in a relatively stable form. $$\quad \text a. \mathrmH_2^2- \quad \text b. \mathrmNe_2\quad \text c. \mathrmHe_2^2+ \quad \text d. \mathrmF_2^2-$$

In chemistry, orbital hybridisation (or hybridization) is the concept of mixing atomic orbitals to form new hybrid orbitals (with different energies, shapes, etc., than the component atomic orbitals) suitable for the pairing of electrons to form chemical bonds in valence bond theory. For example, in a carbon atom which forms four single bonds, the valence-shell s orbital combines with three valence-shell p orbitals to form four equivalent sp3 mixtures in a tetrahedral arrangement around the carbon to bond to four different atoms. Hybrid orbitals are useful in the explanation of molecular geometry and atomic bonding properties and are symmetrically disposed in space. Usually hybrid orbitals are formed by mixing atomic orbitals of comparable energies.[1]

Chemist Linus Pauling first developed the hybridisation theory in 1931 to explain the structure of simple molecules such as methane (CH4) using atomic orbitals.[2] Pauling pointed out that a carbon atom forms four bonds by using one s and three p orbitals, so that "it might be inferred" that a carbon atom would form three bonds at right angles (using p orbitals) and a fourth weaker bond using the s orbital in some arbitrary direction. In reality, methane has four C-H bonds of equivalent strength. The angle between any two bonds is the tetrahedral bond angle of 10928'[3] (around 109.5). Pauling supposed that in the presence of four hydrogen atoms, the s and p orbitals form four equivalent combinations which he called hybrid orbitals. Each hybrid is denoted sp3 to indicate its composition, and is directed along one of the four C-H bonds.[4] This concept was developed for such simple chemical systems, but the approach was later applied more widely, and today it is considered an effective heuristic for rationalizing the structures of organic compounds. It gives a simple orbital picture equivalent to Lewis structures.

Hybridisation theory is an integral part of organic chemistry, one of the most compelling examples being Baldwin's rules. For drawing reaction mechanisms sometimes a classical bonding picture is needed with two atoms sharing two electrons.[5] Hybridisation theory explains bonding in alkenes[6] and methane.[7] The amount of p character or s character, which is decided mainly by orbital hybridisation, can be used to reliably predict molecular properties such as acidity or basicity.[8]

Orbitals are a model representation of the behavior of electrons within molecules. In the case of simple hybridization, this approximation is based on atomic orbitals, similar to those obtained for the hydrogen atom, the only neutral atom for which the Schrdinger equation can be solved exactly. In heavier atoms, such as carbon, nitrogen, and oxygen, the atomic orbitals used are the 2s and 2p orbitals, similar to excited state orbitals for hydrogen.

Hybridisation describes the bonding of atoms from an atom's point of view. For a tetrahedrally coordinated carbon (e.g., methane CH4), the carbon should have 4 orbitals directed towards the 4 hydrogen atoms.

This diagram suggests that the carbon atom could use its two singly occupied p-type orbitals to form two covalent bonds with two hydrogen atoms in a methylene (CH2) molecule, with a hypothetical bond angle of 90 corresponding to the angle between two p orbitals on the same atom. However the true H-C-H angle in singlet methylene is about 102[9] which implies the presence of some orbital hybridisation.

The carbon atom can also bond to four hydrogen atoms in methane by an excitation (or promotion) of an electron from the doubly occupied 2s orbital to the empty 2p orbital, producing four singly occupied orbitals.

Quantum mechanically, the lowest energy is obtained if the four bonds are equivalent, which requires that they are formed from equivalent orbitals on the carbon. A set of four equivalent orbitals can be obtained that are linear combinations of the valence-shell (core orbitals are almost never involved in bonding) s and p wave functions,[10] which are the four sp3 hybrids.

Hybridisation helps to explain molecule shape, since the angles between bonds are approximately equal to the angles between hybrid orbitals. This is in contrast to valence shell electron-pair repulsion (VSEPR) theory, which can be used to predict molecular geometry based on empirical rules rather than on valence-bond or orbital theories.[11]

As the valence orbitals of transition metals are the five d, one s and three p orbitals with the corresponding 18-electron rule, spxdy hybridisation is used to model the shape of these molecules. These molecules tend to have multiple shapes corresponding to the same hybridization due to the different d-orbitals involved. A square planar complex has one unoccupied p-orbital and hence has 16 valence electrons.[13]

In some general chemistry textbooks, hybridization is presented for main group coordination number 5 and above using an "expanded octet" scheme with d-orbitals first proposed by Pauling. However, such a scheme is now considered to be incorrect in light of computational chemistry calculations.

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