Valence Bond Theory PDF

MED-102 General Chemistry Bonding Theories – Valence Bond Theory  LOBs covered Discuss and apply valence bond theory Identify the types of hybrid orbitals employed by central atoms  Valence Bond Theory Explains WHY and HOW bonds are formed Valence Bond Theory –...

MED-102 General Chemistry Bonding Theories – Valence Bond Theory  LOBs covered Discuss and apply valence bond theory Identify the types of hybrid orbitals employed by central atoms  Valence Bond Theory Explains WHY and HOW bonds are formed Valence Bond Theory – Chemical bonds are formed by the overlap of atomic orbitals on adjacent atoms – Explains observed molecular geometry Valence Bond Theory Valence Bond Theory  Hybrid Atomic Orbitals Methane CH4 Carbon in the ground state can only form 2 covalent bonds We need 4 unpaired electrons  Hybrid Atomic Orbitals We can promote a 2s electron into the empty 2p orbital Now we can form 4 bonds, but not all the same  Hybrid Atomic Orbitals – Revision Slide WATCH: https://www.youtube.com/watch?v=vHXViZTxLXo According to the Valence Bond theory, we need unpaired electrons in the two orbitals that will overlap to form a covalent bond. The ground electronic state of C has only 2 unpaired electrons, but we need to make 4 bonds. Therefore, we will need 4 unpaired electrons. We do this by expending a bit of energy to promote a 2s electron into the empty 2p orbital. This produces 4 unpaired electrons, and therefore 4 single bonds can now form. These 4 bonds will not, however, be the same length and strength. We have one 2s electron and three 2p electrons. Thus, we will get three bonds that are the same, and one that is different. However, experimental evidence shows that all 4 bonds in CH4 are exactly the same. Therefore, we need to make all 4 orbitals exactly the same. We do this by hybridization.  Hybrid Atomic Orbitals We blend together the 2s orbital with the three 2p orbitals  Shape of sp3 hybrid orbitals There are 4 sp3 hybrid orbitals Shape of sp3 hybrid orbitals – Revision Slide In hybridization, we blend together the electron density of different atomic orbitals, forming an equal number of hybrid atomic orbitals. In the example above, we blend together one s orbital with three p orbitals, and end up with four sp3 hybrid orbitals. The sp3 hybrid orbitals have a big lobe and a small lobe. It is the big lobe that will be involved in overlapping and the formation of a covalent bond. The big lobe of an sp3 orbital is bigger in size than the lobe of an atomic 2p orbital. This leads to greater overlap and the formation of a stronger bond. When a chemical bond is formed, energy is released. The formation of stronger bonds with sp3 hybrid orbitals makes up for the energy used to promote the 2s electron to the 2p subshell. Methane CH4 Bonding  Methane CH4 Bonding – Revision Slide When we arrange the four sp3 orbitals to a common center (the atomic nucleus), placing the big lobes as far apart as possible, gives a tetrahedral geometry. Tetrahedral geometry is the experimentally observed geometry of the CH4 methane molecule. All bonds have the same length and strength, and the observed bond angles are 109.5o, the tetrahedral bond angle. Therefore, the use of sp3 hybrid orbitals solves two problems: (a) all bonds are identical, and (b) tetrahedral geometry. Ammonia NH3 Bonding Ammonia NH3 Bonding – Revision Slide The ground electronic state valence shell of the N atom is 2s2 2p3. In the left diagram above, the N atom does have 3 unpaired electrons, and can therefore form three covalent bonds. Clearly, there is no need for promotion. The question then is ‘why does the N atom undergo hybridization?’ The N atom undergoes hybridization because the sp3 lobes are bigger than the 2p lobes, and this leads to stronger bond formation, releasing more energy. Also, there is a need to undergo hybridization in order to explain the experimentally observed molecular geometry of the NH3 ammonia molecule, which is trigonal pyramidal, a subset of the tetrahedral electron-pair geometry. Water H2O bonding 5-Minute Break  Sigma () Bonds Cylindrical symmetry All single covalent bonds are  bonds  Pi () Bonds Involved in double and triple covalent bonds Double bond = 1 + 1 Triple bond = 1 + 2  Pi () Bond WATCH: https://www.youtube.com/watch?v=_qzdRPv4Ns4 Ethylene  Ethylene – Revision Slide WATCH: https://www.youtube.com/watch?v=G-D16aniBIs The double bond in ethylene is made up of 1 sigma and 1 pi bond. In ethylene, the C atoms undergo promotion and then sp2 hybridization. In doing so, one pure atomic 2p orbital is kept unhybridized. This is necessary because we need the 2p orbitals on both C atoms in order to form a pi bond. Therefore, each C atom in ethylene has the following electronic configuration: C 1s2 (2sp2)3 2p1 The sp2 orbitals form the three sigma chemical bonds, two with H atoms and one between the two C atoms. The 2p orbitals will join together to form a pi bond. Acetylene bonding Acetylene bonding – Revision Slide WATCH: https://www.youtube.com/watch?v=c7fatZwTEzo The triple bond in acetylene is made up of 1 sigma and 2 pi bonds. In acetylene, the C atoms undergo promotion and then sp hybridization. In the figure above, the sp orbitals are colored green. In doing so, two pure atomic 2p orbitals are kept unhybridized. This is necessary because we need the 2p orbitals on both C atoms in order to form the two pi bonds. Therefore, each C atom in acetylene has the following electronic configuration: C 1s2 (2sp)2 2p2 The sp orbitals form the two sigma chemical bonds, one with a H atom and one between the two C atoms. The 2p orbitals will join together to form two pi bonds. Hybrid Orbitals – an easy way to know There is a simple way to predict which hybrid orbitals are being used Ask the question: how many electron regions are on the central atom? Hybrid Orbitals – an easy way to know – Revision Slide Look at the atom of interest. Ask the question: ‘how many electron clouds are there on it?’ A double bond is ONE electron cloud. The same for a triple bond. A lone pair is also ONE electron cloud. We will need exactly the same total number of orbitals in our hybrid orbital. For example, in CH4 above left, the central C atom has four electron clouds around it. The same applies for the N atom in NH3, and the O atom in H2O. The hybrid orbital that is made up of 4 atomic orbitals is sp3. In acetone, bottom left, the central C atom has three electron clouds around it (the double bond counts as one). The hybrid orbital required there is sp2. In CO2, bottom right, the central C atom has two electron clouds, hence we need sp hybrid orbitals.  Summary for Revision Valence Bond theory explains how and why chemical bonds form. It also explains the observed molecular geometry of a central atom in a molecule. In Valence Bond theory, atomic orbitals come close together, and their lobes overlap, giving a common shared region, where the shared electrons exist. The greater the degree of overlap between the atomic orbitals, the stronger the bond formed. We need unpaired electrons to make chemical bonds. We can form as many bonds as there are unpaired electrons. In some cases, there is a need to promote electrons to higher energy orbitals to give the correct number of unpaired electrons. Following promotion, hybridization often happens. Hybridization leads to the formation of stronger bonds and gives the experimentally observed geometry. Sigma bonds are single covalent bonds. Pi bonds are involved in double and triple covalent bonds. Pi bond result from the overlap of adjacent p orbitals. A double bond is made up of a sigma and a pi bond. A triple bond contains one sigma and two pi bonds. An easy way to find the hybrid orbitals involved is to count the number of electron clouds (bond regions and lone pairs) on the atom of interest, and use a hybrid orbital that is made up of the same number of atomic orbitals.