IIT JEE , AIEEE Forum - Chemistry , Organic Chemistry

surbhihablani_1

explain the concept of ORBITALS and hybridisation.....

sjkalyan

hey it might be easy to explain u some stuff but u wont understan it by mere explanation

the orbital concept is really a vague though interesting one

u must be quite imaginative

to understand the essence of this concept its better u refer ncert XI chemistry book or some other standard book

if still u have problems browse teh net

there are beautiful pictures to support the theory

hey dont forget to rate me!!!

catch_arnnie

you should look for explanations in the books rather than askin here coz' we won't be able to make you understand properly

btw, you can always ask for your doubts & queries here...

neeraj_agarwal_1990

http://winter.group.shef.ac.uk/orbitron/AOs/4f/index.html

madhusudan.chavan

8892

Ground state electronic configuration of carbon is 1s² 2s² 2px¹y¹z⁰, indicating that it should be divalent. Bur carbon is tetravalent. Excited state configuration of carbon atom is 1s² 2s¹ 2px¹y¹z¹. In this case one of the bond will be different than rest of the three bonds. But all the four bonds in CH₄ are equivalent. Hence the concept of hybridization.

A phenomenon of mixing of two or more atomic orbitals of approx. equal energy to form equal number of new orbitals with equal energy, identical shape & size, and symmetrically oriented in space. The new orbitals are called hybrid orbitals.

Also refer to the study material.

Jyothi

Molecular Orbitals
Just as the valence electrons of atoms occupy atomic orbitals (AO), the shared electron pairs of covalently bonded atoms may be thought of as occupying molecular orbitals (MO). It is convenient to approximate molecular orbitals by combining or mixing two or more atomic orbitals. In general, this mixing of n atomic orbitals always generates n molecular orbitals. The hydrogen molecule provides a simple example of MO formation. In the following diagram, two 1s atomic orbitals combine to give a sigma (?) bonding (low energy) molecular orbital and a second higher energy MO referred to as an antibonding orbital. The bonding MO is occupied by two electrons of opposite spin, the result being a covalent bond.

The notation used for molecular orbitals parallels that used for atomic orbitals. Thus, s-orbitals have a spherical symmetry surrounding a single nucleus, whereas ?-orbitals have a cylindrical symmetry and encompass two (or more) nuclei. In the case of bonds between second period elements, p-orbitals or hybrid atomic orbitals having p-orbital character are used to form molecular orbitals. For example, the sigma molecular orbital that serves to bond two fluorine atoms together is generated by the overlap of p-orbitals (part A below), and two sp³ hybrid orbitals of carbon may combine to give a similar sigma orbital. When these bonding orbitals are occupied by a pair of electrons a covalent bond, the sigma bond results. Although we have ignored the remaining p-orbitals, their inclusion in a molecular orbital treatment does not lead to any additional bonding, as may be shown by activating the fluorine correlation diagram below.

Another type of MO (the ? orbital) may be formed from two p-orbitals by a lateral overlap, as shown in part A of the following diagram. Since bonds consisting of occupied ?-orbitals (pi-bonds) are weaker than sigma bonds, pi-bonding between two atoms occurs only when a sigma bond has already been established. Thus, pi-bonding is generally found only as a component of double and triple covalent bonds. Since carbon atoms involved in double bonds have only three bonding partners, they require only three hybrid orbitals to contribute to three sigma bonds. A mixing of the 2s-orbital with two of the 2p orbitals gives three sp² hybrid orbitals, leaving one of the p-orbitals unused. Two sp² hybridized carbon atoms are then joined together by sigma and pi-bonds (a double bond), as shown in part B.

The manner in which atomic orbitals overlap to form molecular orbitals is commonly illustrated by a correlation diagram. Two examples of such diagrams for the simple diatomic elements F₂ and N₂ will be drawn above when the appropriate button is clicked. The 1s and 2s atomic orbitals do not provide any overall bonding, since orbital overlap is minimal, and the resulting sigma bonding and antibonding components would cancel. In both these cases three 2p atomic orbitals combine to form a sigma and two pi-molecular orbitals, each as a bonding and antibonding pair. The overall bonding order depends on the number of antibonding orbitals that are occupied.

The subtle change in the energy of the ?_2p bonding orbital, relative to the two degenerate ?-bonding orbitals, is due to s-p hybridization that is unimportant to the present discussion. An impressive example of the advantages offered by the molecular orbital approach to bonding is found in the oxygen molecule.

A Chime model of the p and ? orbitals of a double bond may be examined by .

The p-orbitals in this model are represented by red and blue colored spheres, which represent different phases, defined by the mathematical wave equations for such orbitals.

Finally, in the case of carbon atoms with only two bonding partners only two hybrid orbitals are needed for the sigma bonds, and these sp hybrid orbitals are directed 180º from each other. Two p-orbitals remain unused on each sp hybridized atom, and these overlap to give two pi-bonds following the formation of a sigma bond (a triple bond), as shown below.

Jyothi Susan George

nitansh

In chemistry, a molecular orbital is a region in which an electron may be found in a molecule.^[1] In general, atomic orbitals combine to form molecular orbitals. The probability of finding an electron in a given region can be obtained by solving the Schrödinger wave equation.

Overview

In quantum chemistry, i.e. electronic structure theory, the molecular electronic states, which are the eigenstates of the electronic molecular Hamiltonian, are expanded (see configuration interaction expansion and basis) into linear combinations of anti-symmetrized products (Slater determinants) of one-electron functions. The spatial components of these one-electron functions are called molecular orbitals (MO). When one considers also their spin component, they are called molecular spin orbitals.

Most methods in computational chemistry today start by calculating the MOs of the system. A molecular orbital describes the behavior of one electron in the electric field generated by the nuclei and some average distribution of the other electrons. In the case of two electrons occupying the same orbital, the Pauli principle demands that they have opposite spin.

Qualitative discussion

For an imprecise, but qualitatively useful, discussion of the molecular structure, the molecular orbitals can be obtained from the "Linear combination of atomic orbitals molecular orbital method" ansatz (using eventually the concept of hybridized orbitals).

In this approach, the molecular orbitals are expressed as linear combinations of atomic orbitals, as if each atom were on its own.

The linear combination of atomic orbitals approximation for molecular orbitals was introduced in 1929 by Sir John Lennard-Jones. His ground-breaking paper showed how to derive the electronic structure of the fluorine and oxygen molecules from quantum principles. This qualitative approach to molecular orbital theory represents the dawn of modern quantum chemistry.

Some properties:

The number of molecular orbitals is equal to the number the atomic orbitals included in the linear expansion,
If the molecule has some symmetry, the degenerate atomic orbitals (with the same atomic energy) are grouped in linear combinations (called symmetry adapted atomic orbitals (SO)) which belong to the representation of the symmetry group, so the wave functions that describe the group is known as symmetry-adapted linear combinations (SALC).
The number of molecular orbitals belonging to one group representation is equal to the number of symmetry adapted atomic orbitals belonging to this representation,
Within a particular representation, the symmetry adapted atomic orbitals mix more if their atomic energy level are closer.

More quantitative approach

To obtain quantitative values for the molecular energy levels, one needs to have molecular orbitals which are such that the configuration interaction (CI) expansion converges fast towards the full CI limit. The most common method to obtain such functions is the Hartree-Fock method which expresses the molecular orbitals as eigenfunctions of the Fock operator. One usually solves this problem by expanding the molecular orbitals as linear combinations of gaussian functions centered on the atomic nuclei (see linear combination of atomic orbitals and basis set (chemistry)). The equation for the coefficients of these linear combinations is a generalized eigenvalue equation known as the Roothaan equations which are in fact a particular representation of the Hartree-Fock equation.

Simple accounts often suggest that experimental molecular orbital energies can be obtained by the methods of ultra-violet photoelectron spectroscopy for valence orbitals and X-ray photoelectron spectroscopy for core orbitals. This however is incorrect as these experiments measure the ionization energy, the difference in energy between the molecule and one of the ions resulting from the removal of one electron. Ionization energies are linked approximately to orbital energies by Koopmans' theorem. While the agreement between these two values can be close for some molecules, it can be very poor in other cases.

nitansh

sp³ hybrids

Hybridisation describes the bonding atoms from an atom's point of view. That is, for a tetrahedrally coordinated carbon (e.g. methane, CH₄), the carbon should have 4 orbitals with the correct symmetry to bond to the 4 hydrogen atoms. The problem with the existence of methane is now this: Carbon's ground-state configuration is 1s² 2s² 2p_x¹ 2p_y¹ or perhaps more easily read:

$C\quad \frac{\uparrow\downarrow}{1s}\; \frac{\uparrow\downarrow}{2s}\; \frac{\uparrow\,}{2p_x}\; \frac{\uparrow\,}{2p_y}\; \frac{\,\,}{2p_z}$

(Note: The 1s orbital is lower in energy than the 2s orbital, and the 2s orbital is lower in energy than the 2p orbitals)

The valence bond theory would predict, based on the existence of two half-filled p-type orbitals (the designations p_x p_y or p_z are meaningless at this point, as they do not fill in any particular order), that C forms two covalent bonds. CH₂. However, methylene is a very reactive molecule (see also: carbene) and cannot exist outside of a molecular system. Therefore, this theory alone cannot explain the existence of CH₄.

Furthermore, ground state orbitals cannot be used for bonding in CH₄. While exciting a 2s electron into a 2p orbital would theoretically allow for four bonds according to the valence bond theory, (which has been proved experimentally correct for systems like O₂) this would imply that the various bonds of CH₄ would have differing energies due to differing levels of orbital overlap. Once again, this has been experimentally disproved: any hydrogen can be removed from a carbon with equal ease.

To summarise, to explain the existence of CH₄ (and many other molecules) a method by which as many as 12 bonds (for transition metals) of equal strength (and therefore equal length) can be created was required.

The first step in hybridisation is the excitation of one (or more) electrons (we will have a look on the carbon atom in methane, for simplicity of the discussion):

$C^{*}\quad \frac{\uparrow\downarrow}{1s}\; \frac{\uparrow\,}{2s}\; \frac{\uparrow\,}{2p_x} \frac{\uparrow\,}{2p_y} \frac{\uparrow\,}{2p_z}$

The proton that forms the nucleus of a hydrogen atom attracts one of the valence electrons on carbon. This causes an excitation, moving a 2s electron into a 2p orbital. This, however, increases the influence of the carbon nucleus on the valence electrons by increasing the effective core potential (the amount of charge the nucleus exerts on a given electron = Charge of Core ? Charge of all electrons closer to the nucleus).

The combination of these forces creates new mathematical functions known as hybridised orbitals. In the case of carbon attempting to bond with four hydrogens, four orbitals are required. Therefore, the 2s orbital (core orbitals are almost never involved in bonding) mixes with the three 2p orbitals to form four sp³ hybrids (read as s-p-three). See graphical summary below.

becomes $C^{*}\quad \frac{\uparrow\downarrow}{1s}\; \frac{\uparrow\,}{sp^3}\; \frac{\uparrow\,}{sp^3} \frac{\uparrow\,}{sp^3} \frac{\uparrow\,}{sp^3}$

In CH₄, four sp³ hybridised orbitals are overlapped by hydrogen's 1s orbital, yielding four ? (sigma) bonds. The four bonds are of the same length and strength. This theory fits our requirements.

translates into

An alternative view is: View the carbon as the C^4? anion. In this case all the orbitals on the carbon are filled:

$C^{4-}\quad \frac{\uparrow\downarrow}{1s}\; \frac{\uparrow\downarrow}{2s}\; \frac{\uparrow\downarrow}{2p_x} \frac{\uparrow\downarrow}{2p_y} \frac{\uparrow\downarrow}{2p_z}$

If we now recombine these orbitals with the empty s-orbitals of 4 hydrogens (4 protons, H⁺) and allow maximum separation between the 4 hydrogens (i.e. tetrahedral surrounding of the carbon), we see that at any orientation of the p-orbitals, a single hydrogen has an overlap of 25% with the s-orbital of the C, and a total of 75% of overlap with the 3 p-orbitals (see that the relative percentages are the same as the character of the respective orbital in an sp³-hybridisation model, 25% s- and 75% p-character).

According to the orbital hybridization theory the valence electrons in methane should be equal in energy but its photoelectron spectrum ^[4] shows two bands, one at 12.7 eV (one electron pair) and one at 23 eV (three electron pairs). This apparent inconsistency can be explained when one considers additional orbital mixing taking place when the sp³ orbitals mix with the 4 hydrogen orbitals.

sp² hybrids

Other carbon based compounds and other molecules may be explained in a similar way as methane. Take, for example, ethene (C₂H₄). Ethene has a double bond between the carbons. The Lewis structure looks like this:

Ethene Lewis Structure. Each C bonded to two hydrogens and one double bond between them.

Carbon will sp² hybridise, because hybrid orbitals will form only ? bonds and one ? (pi) bond is required for the double bond between the carbons. The hydrogen-carbon bonds are all of equal strength and length, which agrees with experimental data.

In sp² hybridization the 2s orbital is mixed with only two of the three available 2p orbitals:

$C^{*}\quad \frac{\uparrow\downarrow}{1s}\; \frac{\uparrow\,}{sp^2}\; \frac{\uparrow\,}{sp^2} \frac{\uparrow\,}{sp^2} \frac{\uparrow\,}{p}$

forming a total of 3 sp² orbitals with one p-orbital remaining. In ethene the two carbon atoms form a ? bond by overlapping two sp² orbitals and each carbon atoms forms two covalent bonds with hydrogen by s?sp² overlap all with 120° angles. The ? bond between the carbon atoms perpendicular to the molecular plane is formed by 2p?2p overlap.

The amount of p-character is not restricted to integer values, i.e. hybridisations like sp^2.5 are also readily described. In this case the geometries are somewhat distorted from the ideally hybridised picture. For example, as stated in Bent's rule, a bond tends to have higher p-character when directed toward a more electronegative substituent.

sp hybrids

The chemical bonding in compounds such as alkynes with triple bonds is explained by sp hybridization.

$C^{*}\quad \frac{\uparrow\downarrow}{1s}\; \frac{\uparrow\,}{sp}\; \frac{\uparrow\,}{sp} \frac{\uparrow\,}{p} \frac{\uparrow\,}{p}$

In this model the 2s orbital mixes with only one of the three p-orbitals resulting in two sp orbitals and two remaining unchanged p orbitals. The chemical bonding in acetylene (C₂H₂) consists of sp?sp overlap between the two carbon atoms forming a ? bond and two additional ? bonds form by p?p overlap. Each carbon also bonds to hydrogen in a sigma s?sp overlap at 180° angles.

Hybridisation and molecule shape

Hybridisation, along with the VSEPR theory, helps to explain molecule shape:

AX₁ (eg, LiH): no hybridisation; trivially linear shape
AX₂ (eg, BeCl₂): sp hybridisation; linear or diagonal shape; bond angles are cos^?1(?1) = 180°
AX₃ (eg, BCl₃): sp² hybridisation; trigonal planar shape; bond angles are cos^?1(?1/2) = 120°
AX₄ (eg, CCl₄): sp³ hybridisation; tetrahedral shape; bond angles are cos^?1(?1/3) ? 109.5°
AX₅ (eg, PCl₅): sp³d hybridisation; trigonal bipyramidal shape
AX₆ (eg, SF₆): sp³d² hybridisation; octahedral (or square bipyramidal) shape

This holds if there are no lone electron pairs on the central atom. If there are, they should be counted in the X_i number, but bond angles become smaller due to increased repulsion. For example, in water (H₂O), the oxygen atom has two bonds with H and two lone electron pairs (as can be seen with the valence bond theory as well from the electronic configuration of oxygen), which means there are four such 'elements' on O. The model molecule is, then, AX₄: sp³ hybridization is utilized, and the electron arrangement of H₂O is tetrahedral. This agrees with the experimentally-determined shape for water, a non-linear, bent structure, with a bond angle of 104.5 degrees (the two lone-pairs are not visible).

In general, for an atom with s and p orbitals forming hybrids h_i and h_j with included angle

?

, the following holds: 1 +

?

_i

?

_j cos(

?

) = 0. The p-to-s ratio for hybrid i is

?

_i², and for hybrid j it is

?

_j². In the special case of equivalent hybrids on the same atom, again with included angle

?

, the equation reduces to just 1 +

?

² cos(

?

) = 0. For example, BH₃ has a trigonal planar geometry, three 120^o bond angles, three equivalent hybrids about the boron atom, and thus 1 +

?

² cos(

?

) = 0 becomes 1 +

?

² cos(120^o) = 0, giving

?

² = 2 for the p-to-s ratio. In other words, sp² hybrids, just as expected from the list above.

srini

Hi.....

hey surbhi......you are just in class 10 ..dont get tensed know......your inquistivity is appreciable......good............all the best

khinart

Concept of Hybridisation

It has already been pointed out that covalency of an element is equal to the number of half-filled orbitals present in the valence shell of its atoms. On applying this concept to carbon, we find that the valency of carbon sh

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