<\/p>\n<\/li>\n<\/ul>\n
We list here some examples of sp hybridization in detail:<\/p>\n
1)\u00c3\u201a\u00c2\u00a0\u00c3\u201a\u00c2\u00a0\u00c3\u201a\u00c2\u00a0 <\/b>Beryllium Chloride (BeCl2<\/sub>)<\/b><\/em><\/p>\n \u00c3\u00a2\u00e2\u201a\u00ac\u00cb\u0153Be\u00c3\u00a2\u00e2\u201a\u00ac\u00e2\u201e\u00a2 has electronic configuration as 1s2 2s2 in the ground state. Due to the absence of any unassociated electron, excitation takes place fostering its one 2s electron into the empty 2p orbital. Thus, in this phase, the electronic configuration is\u00c3\u201a\u00c2\u00a0 1s2 2s1 2p1.<\/p>\n BeCl2 is linear in shape. In case the beryllium atom formed bonds using pure orbitals, the shape of the molecule might have been angular. Hence for this reason, the below given sp hybridization was proposed.<\/p>\n During the sate of excitation, the atom undergoes sp hybridization involving the mixing of 2s and 2p orbitals. Thus two half filled ‘sp’ hybrid orbitals are formed, which are arranged linearly. These half filled Sp-orbitals form two \u00c3\u008f\u00c6\u2019 bonds with two ‘Cl’ atoms and hence this leads to the linear shape of Thus BeCl2 with the bond angle of 180o<\/sup>.<\/p>\n 2)\u00c3\u201a\u00c2\u00a0\u00c3\u201a\u00c2\u00a0\u00c3\u201a\u00c2\u00a0 <\/b>Acetylene (C2<\/sub>H2<\/sub>)<\/b><\/em><\/p>\n The ground state configuration of \u00c3\u00a2\u00e2\u201a\u00ac\u00cb\u0153C\u00c3\u00a2\u00e2\u201a\u00ac\u00e2\u201e\u00a2 being 1s2 2s2 2px12py1 has only two unassociated electrons. But carbon makes four bonds since its valency is four. For this 4 unpaired electrons are required. Hence it promotes its 2s electron into the empty 2pz<\/sub> orbital in the excited state.<\/p>\n Thus in the excited state, the electronic configuration\u00c3\u201a\u00c2\u00a0of carbon is 1s2<\/sup> 2s1<\/sup> \u00c3\u201a\u00c2\u00a0\u00c3\u201a\u00c2\u00a0\u00c3\u201a\u00c2\u00a0\u00c3\u201a\u00c2\u00a0\u00c3\u201a\u00c2\u00a0\u00c3\u201a\u00c2\u00a02px<\/sub>1<\/sup>2py<\/sub>1<\/sup>2pz<\/sub>1<\/sup>.\u00c3\u201a\u00c2\u00a0Each carbon atom undergoes ‘sp’ hybridization by using a 2s and one 2p orbitals in the excited state to give two half filled ‘sp’ orbitals, which are arranged linearly.<\/p>\n By the use of Sp-orbitals, a \u00c3\u008f\u00c6\u2019sp-sp bond is formed between the two carbon atoms. At the same time there are two unhybridized p orbitals i.e., 2py<\/sub> and 2pz<\/sub> on each carbon atom which being perpendicular to the Sp-orbitals form two \u00c3\u008f\u00e2\u201a\u00acp-p <\/sub>bonds between the two carbon atoms.\u00c3\u201a\u00c2\u00a0Hence this results in the formation of a triple bond between the carbon atoms. Each carbon also forms a \u00c3\u008f\u00c6\u2019sp-s <\/sub>bond with the hydrogen atom.<\/p>\n Result: Acetylene molecule has bond angle as 180 o and is linear in shape.<\/p>\n <\/p>\n sp2<\/sup> HYBRIDIZATIO<\/span><\/b>N<\/span><\/b><\/em><\/p>\n We now discuss some examples of sp2 hybridization<\/p>\n 1)\u00c3\u201a\u00c2\u00a0\u00c3\u201a\u00c2\u00a0\u00c3\u201a\u00c2\u00a0 <\/b>Boron trichloride (BCl3<\/sub>)<\/b><\/em><\/p>\n Ground state configuration being 1s2<\/sup> 2s2<\/sup> 2p1<\/sup>, it has only one unpaired electron. Since three unassociated electrons are required so there is promotion of one of 2s electron into the 2p sublevel by absorbing energy.<\/p>\n Thus the configuration becomes 1s2<\/sup> 2s2<\/sup> 2px<\/sub>1<\/sup>2py<\/sub>1<\/sup>.<\/p>\n However to account for the trigonal planar shape of this BCl3<\/sub> molecule, sp2<\/sup> hybridization before bond formation was advocated.<\/p>\n Boron experiences sp2<\/sup> hybridization by using a 2s and two 2p orbitals in the excitation state to give three half-filled sp2<\/sup> hybrid orbitals which are oriented in trigonal planar symmetry. It then forms three \u00c3\u008f\u00c6\u2019sp-p <\/sub>bonds with three chlorine atoms by using its half-filled sp2<\/sup> hybrid orbitals. For the \u00c3\u008f\u00c6\u2019-bond formation, the half-filled p-orbital are used.<\/p>\n Result: with bond angle 120o , the shape is trigonal.<\/p>\n 2)\u00c3\u201a\u00c2\u00a0\u00c3\u201a\u00c2\u00a0\u00c3\u201a\u00c2\u00a0 <\/b>Ethylene (C2<\/sub>H4<\/sub>)<\/b><\/em><\/p>\n In the state of excitation, each carbon atom experiences sp2<\/sup> hybridization by mixing 2s and two 2p orbitals to give three half-filled sp2<\/sup> hybrid orbitals oriented in trigonal planar symmetry.\u00c3\u201a\u00c2\u00a0Another unhybridized 2pz<\/sub> orbital is also present on each carbon perpendicular to the plane of sp2<\/sup> hybrid orbitals in half-filled state.<\/p>\n The carbon atoms form a \u00c3\u008f\u00c6\u2019sp<\/sub>2-sp<\/sub>2 bond with each other by using sp2<\/sup> hybrid orbitals.\u00c3\u201a\u00c2\u00a0Lateral overlapping between the unhybridized 2pz orbitals also leads to the formation of a <\/b>\u00c3\u008f\u00e2\u201a\u00acp-p<\/sub> bond. Thus there is a double bond (\u00c3\u008f\u00c6\u2019sp<\/sub>2-sp<\/sub>2 &\u00c3\u201a\u00c2\u00a0 \u00c3\u008f\u00e2\u201a\u00acp-p<\/sub>) between two carbon atoms.\u00c3\u201a\u00c2\u00a0Each carbon atom also forms two \u00c3\u008f\u00c6\u2019sp<\/sub>2-s <\/sub>bonds with two hydrogen atoms.<\/p>\n Result: Hence it is planar and has the \u00c3\u00a2\u00cb\u2020\u00c2\u00a0HCH & \u00c3\u00a2\u00cb\u2020\u00c2\u00a0HCC bond angles equal to 120o<\/sup><\/p>\n \u00c3\u201a\u00c2\u00a0SP3<\/sup> HYBRIDIZATION<\/b><\/em><\/p>\n 1) Methane (CH4<\/sub>)<\/b><\/em><\/b><\/p>\n In the state of excitation, the carbon atom undergoes sp3<\/sup> hybridization which involves the mixing of one \u00c3\u00a2\u00e2\u201a\u00ac\u00cb\u01532s\u00c3\u00a2\u00e2\u201a\u00ac\u00e2\u201e\u00a2 and three 2p orbitals to furnish four half-filled sp3<\/sup> hybrid orbitals, which are oriented in tetrahedral symmetry in space around the carbon atom.\u00c3\u201a\u00c2\u00a0Each one of these sp3<\/sup> hybrid orbitals forms a \u00c3\u008f\u00c6\u2019sp<\/sub>3-s <\/sub>bond with one hydrogen atom. Thus carbon forms four \u00c3\u008f\u00c6\u2019sp<\/sub>3-s <\/sub>bonds with four hydrogen atoms. Result: The shape is tetrahedral and it has bond angle of 109o<\/sup>28\u00c3\u00a2\u00e2\u201a\u00ac\u00e2\u201e\u00a2.<\/p>\n \u00c3\u201a\u00c2\u00a02) Ethane (C2<\/sub>H6<\/sub>)<\/b><\/b><\/p>\n The case is parallel to that of methane. It experiences sp3<\/sup> hybridization in the excited state which yields four sp3<\/sup> hybrid orbitals in tetrahedral geometry.<\/p>\n The overlapping of sp3 hybrid orbitals along the inter-nuclear axis leads to the formation of a \u00c3\u008f\u00c6\u2019sp<\/sub>3-sp<\/sub>3bond.\u00c3\u201a\u00c2\u00a0Each carbon atom also forms three \u00c3\u008f\u00c6\u2019sp<\/sub>3-s<\/sub> bonds with hydrogen atoms.<\/p>\n Result: Hence ethane has tetrahedral symmetry around each carbon with \u00c3\u00a2\u00cb\u2020\u00c2\u00a0HCH & \u00c3\u00a2\u00cb\u2020\u00c2\u00a0HCC bond angles equal to 109o<\/sup>28′.<\/p>\n 3) Ammonia (NH3<\/sub>)<\/b><\/b><\/p>\n The configuration in ground state being 1s2 2s2 2px12py12pz1 shows that there are three unassociated electrons in the 2p sublevel and hence the nitrogen atom can form three bonds with three hydrogen atoms. This results in the bond angle of 90o<\/sup> differing from the generally reported angle of 107o<\/sup>48′.<\/p>\n Hence it was advocated for the Nitrogen to undergo sp3<\/sup> hybridization of a 2s and three 2p orbitals to give four sp3<\/sup> orbitals, which henceforth are organized in a tetrahedral symmetry. Since it reduces the number of repulsions so the molecule is bound to become more stable. It includes one full filled and three half filled.<\/p>\n Nitrogen atom forms 3 \u00c3\u008f\u00c6\u2019sp<\/sub>3-s <\/sub>bonds with three hydrogen atoms by using three half-filled sp3 <\/sup>hybrid orbitals. A lone pair also exists on the nitrogen atom which requires more space than the bond pairs and belongs to full filled sp3<\/sup> hybrid orbital.<\/p>\n Nevertheless, \u00c3\u00a2\u00cb\u2020\u00c2\u00a0HNH bond angle does not equal the normal tetrahedral angle: 109o<\/sup>28′ which is also different from the reported angle of 107o<\/sup>48′. The reduction in the bond angle is caused by the repulsion of lone pair over the bond pairs.<\/p>\n Result: The ammonia is trigonal pyramidal and has a lone pair on nitrogen atom.<\/p>\n <\/p>\n 4) Water molecule (H2<\/sub>O)<\/b> Oxygen atom has two unpaired electrons which are free to combine with the hydrogen atoms. But the only restriction is that the resulting bond angle should be 90o<\/sup>. The electronic configuration of oxygen is 1s2 2s2 2px22py12pz1. The reported bond angles were 104o<\/sup>28′. To account this, sp3<\/sup> hybridization before the bond formation was proposed.<\/p>\n Formation of water molecule involves the sp3 hybridization of the oxygen atom by mixing of a 2s and three 2p orbitals to furnish four sp3 hybrid orbitals oriented in tetrahedral geometry. This comprises two half-filled and two completely filled. The half-filled hybrid orbitals are then used by the oxygen atoms to form two \u00c3\u008f\u00c6\u2019sp<\/sub>3-s <\/sub>bonds. The\u00c3\u201a\u00c2\u00a0reported bond angle is 104o<\/sup>28′ instead of regular tetrahedral angle: 109o<\/sup>28′. It is again due to repulsions caused by two lone pairs on the bond pairs.<\/p>\n<\/p>\n
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