Properties and chemical reactions of mainly organic molecules can be simply described via considering the molecules to be shaped through sharing of electron pairs between the atoms. One other approach to configuration of molecules is the molecular orbital technique. Organic chemists contain for many years employed a bonding model that merges elements of molecular orbital theory by Lewis model of formation of covalent bond via electron sharing. This representation was proposed through Pauling in the year 1930 and is based on the model of orbital hybridization. This model utilizes the terminology of molecular orbital theory but treats the bonds between the atoms as although they are localized, as in the case of diatomic molecules. In other words, it is a kind of localized molecular orbital management of the bond.
We contain already deliberate that various kinds of orbital hybridization is probable depending on the number and nature of the orbitals involved. In this chapter, the hybridization involving s and p orbitals. Let us now learn each and every kind of hybridization involving s and p orbitals, in feature, to understand this idea and its utilize in describing the formation of molecules.
Hybridization is a theoretical idea that permits as realistic medelling of molecular structure as probable
The orbitals which undergo hybridization should not be energetically much dissimilar.
The number of hybrid orbitals created is forever equal to the number of atomic orbitals joined Sp3 is pronounced as s-p-3 and not sp cube.
The hybrid orbitals are attained via mathematical mixtures of atomic orbitals.
Let us consider the easiest organic compound, methane, containing the molecular formula CH4. We can remind that carbon has the electron configuration 1s22s22px12py1. Because only 2 unpaired electrons are there, one might expect that it should form only 2 bonds by 2 hydrogen atoms to form CH2. But in fact it forms 4 bonds through four hydrogen atoms to give CH4. Pauling recommended that this could be clarified via using orbital hybridization. In this process, atomic orbitals are combined to give up the new hybrid orbitals. In this case, in the 1st step one of the 2s electrons is promoted to the 2pz orbital, electron design can then be written as 2s12px12y1 2pz1. Bond formation through such pure atomic orbitals would direct to the situation where the bond formed via one 2s electron will be dissimilar from the bonds formed via three 2p electrons. But, in methane molecule, all the 4 bonds are equivalent. In order to illustrate this, the plan of orbital hybridization was invoked. In this procedure, one 2s and three 2p and three 2p orbitals on hybridization give up a set of 4 new equivalent orbitals. Such new orbitals are termed hybrid orbitals. Because they are formed via combining one s and three p orbitals, they are termed sp3 hybrid orbitals. All the 4 sp3 hybrid orbitals are of equivalent energy and each one of them has 25% s character and 75% p character.
Such sp3 hybrid orbitals are shown in Fig. we can see in Fig. that the
Fig: Hybridization of one 2s and three 2p orbitals to yield four sp3 hybrid orbitals. b) Four sp3hybrid orbitals directed towards the corners of a tetrahedron: small back lobes are not shown. c) Formation of methane molecule.
Sp3 hybrid orbital has 2 lobes of unequal size divided from each other via a node. This circumstance is alike to a p orbital but by the difference that here 1 lobe is extremely small and the other is extremely huge. In other terms, in sp3 hybrid orbitals, the electron density is focused in one direction that leads to superior overlap as compared to pure atomic orbitals. Therefore, the bonds formed via these orbitals will be stronger and more stable in comparison to those formed via using pure atomic orbital. The spatial orientation of such orbitals is attained through mathematical calculations and is shown in Fig. (b). this is in accordance through the VSEPR theory that we had studied. We can see in the shape that these orbitals are directed towards the corners of a tetrahedron and the bond angle between any 2 sp3 hybrid orbitals is 109.5o. In methane molecule, each of the four sp3 hybrid orbitals overlaps by 1s orbital of 4 hydrogen as shown in Fig (c).
Note that the bonds so shaped, for example: the C - H bonds, are σ (sigma) bonds. If as an alternative of combining by hydrogens, the hybrid orbital shapes a bond through same hybrid orbital of another carbon atom, then a C-C bond will consequence instead of the C - H bond. The C-C has a bond length of 154 pm and a bond energy of 348 kJ mol-1.
In a molecule as ethylene, wherever there is not sufficient hydrogen in the molecule to shape 6 C-H bonds, another kind of hybridization has to be consideration of.
In this sort of hybridization, as the name specifies, the 2s orbital of the carbon is hybridized by only 2 of the 3 accessible 2p orbitals, as revealed below.
Since 3 orbitals are hybridized, three equivalent sp2 hybrid orbitals are attained. We shall now clarify sp2 hybridization using ethylene as an instance.
According to the VSEPR theory, such orbitals are oriented in space making an angle of 120o through each other as shown in Fig. (a).
Note that the three sp2 hybrid orbitals are in one plane. The 3rd p orbital that is not used for hybridization is perpendicular to the sp2 hybrid orbitals and is show in red color in Fig. (a).
When two such sp2 hybridized carbon atoms shape a bond, the C-C bond formed is another time a σ bond. If the rest of the sp2 hybrid orbitals on each carbon atom overlap with 1s orbital of the 2 hydrogen atoms, then as shown in Fig.(b) the 2 unhybridised p orbitals on the 2 carbon atoms are equivalent to each other. Such p orbitals can overlap side ways to yield a second bond, recognized as π (pi) bond that is shown in Fig. (b). The C=C bond length for ethylene molecule so acquired is 134 pm. We can evaluate this value with C-C single bond length as specified before in case of ethane. You will study in detail, the compounds having sp2 hybridized carbon atoms these as alkenes and dienes.
(a) You are conscious that:
(i) δ bonds are shaped via the edge-on overlap of pure (s and p) or hybrid orbitals. The electron density in ∂ bonds is maximum along the internuclear axis.
(ii) π bonds are formed by sideways overlap of p orbitals. π bonds contain maximum electron density above and below the internuclear axis.
Fig: (a) sp2 hybrid orbitals. (b) Formation of a π bond in ethylene molecule.
Sp - Hybridization:
Let us now consider the 3rd kind of hybridization involving s and p orbitals in cases wherever a triple bond is stipulated. In carbon atom when 2s and merely one of the three 2p orbitals hybridize as given below, the hybridization is recognized as sp-hybridization. This leads to two new equivalent sp hybrid orbitails as shown in
Fig: (a) two sp hybrid orbitals. (b) Formation of one σ (sigma) bond, Formation of two Π bonds.
Acetylene is a linear molecule having cylindrically symmetrical Π electron density about the internuclear axis.
Such 2 orbitals are oriented in space at an angle of 180o according to the VSEPR theory. Let us learn sp-hybridization using acetylene as an instance. When one of the two sp hybrid orbitals on each carbon atom combines by another, a C-C sigma bond is shaped. The 2nd sp hybrid orbital on each carbon forms a sigma bond through 1s orbitals of 2 hydrogens, as given in Fig. (b).This leaves two p orbitals on each carbon atom that is not utilized in sp hybridization. These p orbitals are perpendicular to each other and as well to the sigma bond. Such p orbitals can overlap tangentially to provide rise to two Π bonds. Such a bond is termed a triple bond and we get the acetylene molecule, as given in Fig (b) The C = C bond in acetylene has a bond length of 120 pm and the H-C-C angle is 180o which shows that it is linear. Compounds having triple bond are termed alkynes. We can figure up the above information as given in Table.
Table: Bond characteristics and hybridization of carbon in simple molecules
From the data given in the Table we can terminate that
(i) In the hybrid orbitals, as the S character enhances, bond length reduces.
(ii) As the bond order increases, the bond length decreases before proceeding to the next division, answer the subsequent SAQ.
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