In Chapter 10, we have talked about the chemical bonds of diatomic molecules, e.
ID: 1062255 • Letter: I
Question
In Chapter 10, we have talked about the chemical bonds of diatomic molecules, e.g. H2. For polyatomic molecules with more than two atoms, the formation of chemical bonds usually involves the formation of hybrid orbitals. In quantum mechanics, bond hybridization is defined as the process that several atomic orbitals from the same atom and with similar energies to be linearly combined. Hybridization yields hybrid orbitals with the same number as the atomic orbitals combined. The configurations of hybrid orbitals are responsible for the shapes of polyatomic molecules and the bond orders of multiple bonds.
3. In Chapter 10, we have talked about the chemical bonds of diatomic molecules, eg. H2. For olyatomic molecules with more than two atoms, the formation of chemical bonds usually involves the formation of hybrid orbitals. In quantum mechanics, bond hybridization is defined as the process that several atomic orbitals from the same atom and with similar energies to be linearly combined. Hybridization yields hybrid orbitals with the same number as the atomic orbitals combined. The configurations of hybrid orbitals are responsible for the shapes of polyatomic molecules and the bond orders of multiple bonds. Consider the sp-hybrid orbitals. Two sp-hybrid orbitals are generated from the linear combination of one s orbital and one p orbital. The corresponding wave functions of the two sp-hybrid orbitals are (w. +w.) 180 sp,1 sp,2 The two sp-hybrid orbitals are on the same line and form a bond angle of 180°. This is why carbon dioxide molecules are linear Based on the above information, answer the following question. (a) Which combination shown below do you think will most likely to form the sp-hybrid orbitals in carbon dioxide? Explain your choice [C-carbon; O-oxygen] A. one C 1sorbital and one C 2p orbital B. one C 2s orbital and one O 2p orbital C. one C 2s orbital and one C 2p orbital D. one C 2s orbital and one C 3p orbital (b) Given that s and p orbitals of the same atom are always orthonormal, verify that the two sp-hybrid orbitals are also orthonormal. Show your proof. (c) Given the energy of a s orbital is Es and that of a p orbital is Ep. Express the energy of each sp-hybrid orbitals in terms of Es and Ep. Do they have the same energy? (This is a general conclusion for all other hybrid orbitals.) d What advantage (related to the total energy of formed molecules) does the hybridization process have?Explanation / Answer
(a) In CO2 (O=C=O), carbon, C is the central atom and binds to two oxygen, O atoms in a linear fashion. The O-C-O bond angle in CO2 is 180 and both the O-C-O bonds have equal lengths and bond energies. Since the central C atom must form equivalent bonds to the two O atoms, hence the C atom must invoke hybridization. The outer (valence) orbitals on C are involved in hybridization. Since the bond angle in CO2 is 180, the C atom uses one s and one p orbital to form two sp hybrid orbitals. The valence shell in C houses the 2s and 2p orbitals and hence the correct option is (c).
We ruled out (a) and (b) because these involve p orbital on O and (d) because the 2s orbital is much lower in energy than 3p and cannot participate in hybridization with 3p. An important criterion for hybridization is that the participating orbitals should have similar energies. Moreover, the 3s and 3p orbitals in C are vacant.
Ans: (c) one C 2s orbital and one C 2p orbital
(b) Given that s and p orbitals on a particular atom are orthonormal, we can write,
sp = ps = 0
We evaluate the integral below as:
sp,1sp,2d (d represents the volume element) = [1/2*(s + p)][1/2*(s – p).d = ½*(s2 – sp + ps – p2)d = ½*(s2 – p2)d = ½*(1 – 1)d = 0 (ans).
Since the integral is 0, the two hybrid orbitals, sp,1 and sp,2 are orthonormal to each other (ans).
(c) Given the energy of the s and p orbitals as Es and Ep, the energy of the hybrid orbital (in simple qualitative terms, without going into quantum mechanical calculations) is
<E> = ½*(Es + Ep)
The hybrid orbitals will have average energy of the two orbitals; this is a general conclusion for all hybrid orbitals. Hybrid orbitals in general have the mean energy of the participating orbitals.
The two hybrid orbitals will have the same energy. The two hybrid orbitals are degenerate.
(d) Hybridization leads to lowering of energy of the molecule. If the 2s and 2p orbitals on C were to bind to two 2p orbitals on O to form C-O single bonds in CO2, the system would have had a higher energy (since the 2p orbitals on both C and O are more or less of equivalent energy and appreciably higher than the energy of the 2s orbital on C). The sp hybrid orbitals have intermediate energy (as compared to 2s and 2p orbitals on C) and bonding between the hybrid orbitals on C and 2p orbitals on O leads to an overall lowering of energy of CO2 and hence higher stabilization as compared to individual C and O atoms.