Chemistry 6 (9 am section) Spring Covalent Bonding

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Beatrice Holland
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1 Chemistry 6 (9 am section) Spring 000 Covalent Bonding The stability of the bond in molecules such as H, O, N and F is associated with a sharing (equal) of the VALENCE ELECTRONS between the BONDED ATOMS. How do we know this? You recall that we described the behavior of an electron in an atom in terms of ATOMIC ORBITALS -- an A.O. is a solution to the Schrödinger equation. This suggests that we might try to describe the behavior of an electron in a MOLECULE in terms of MOLECULAR ORBITALS (M.O.). The MO's are obtained QUANTITATIVELY by solving the Schrödinger equation -- our approach will be mainly QUALITATIVE. We consider the simplest molecule H + ; this has TWO NUCLEI AND ONE ELECTRON. e - r A r B H A H B z For now we will write the MOLECULAR ORBITALS as ψ mol. We take the internuclear axis as the z-axis. What must ψ mol look like CLOSE to H A and when H B is FAR AWAY? ψ mol ψ 1sA = N e r A What must ψ mol look like CLOSE to H B and when H A is FAR AWAY? ψ mol ψ 1sB = N e r B (N is a constant.) This suggests that when H A and H B are close we should write ψ mol = c A ψ 1sA + c B ψ 1sB The MOLECULAR ORBITAL is written as a LINEAR COMBINATION OF ATOMIC ORBITALS (LCAO MO THEORY). 1

2 What possibilities exist for c A and c B? How would we find them? -- solve the Schrödinger equation. We will use a more qualitative approach. This involves examining the PROBABILITY DENSITY ψ mol ψ mol = c A ψ 1sA + c A c B ψ 1sA ψ 1sB + c B ψ 1sB Look at the Probability Density when the electron is very close to H A : ψ mol (evaluated at H A ) c A ψ1sa (evaluated at H A ) Look at the Probability Density when the electron is very close to H B : ψ mol (evaluated at H B ) c B ψ1sb (evaluated at H B ) Since the two ends of the molecule are IDENTICAL (there can t be any physical difference if we exchange the labels A and B) ψ mol (evaluated at H A ) = ψ mol (evaluated at H B ) i.e. But since c A ψ1sa (evaluated at H A ) = c B ψ1sb ψ 1sA (evaluated at H A ) = ψ 1sB (evaluated at H B ) (evaluated at H B ) therefore c A = c B Thus, either (i) c A = c B ; let c A = c B = c or (ii) c B = c A ; let c A = c', c B = c' Examine the POSITIVE COMBINATION first: ψ + = c ( ψ 1sA + ψ 1sB ) We will evaluate ψ + at points along the z-axis : ψ + = c ( ψ 1sA + ψ 1sA ψ 1sB + ψ 1sB ) The PRODUCT ψ 1sA ψ 1sB is referred to as the OVERLAP BETWEEN ψ 1sA and ψ 1sB. To recap, MOLECULAR ORBITAL THEORY is an approximation in which we attempt to describe the behaviour of electrons in molecules in terms of MOLECULAR ORBITALS (MO's) ψ i. We form such MO's by taking Linear Combinations of Atomic Orbitals (LCAO).

3 Example: for H + H A H B ψ 1 = ψ + = c ( ψ 1sA + ψ 1sB ) When we examine the Probability Density ψ +, we see that we get a BUILD-UP of ELECTRON DENSITY IN THE REGION BETWEEN THE TWO NUCLEI. ψ + is called a BONDING MOLECULAR ORBITAL. We can label MO's in a manner analogous to the scheme used for AO's. ψ + = ψ σ1s The symbol σ in the subscript (σ1s) indicates that the orbital is cylindrically symmetrical about the internuclear axis and DOES NOT POSSESS A NODAL PLANE that CONTAINS THE NUCLEI Examine ψ + = c ( ψ 1sA + ψ 1sA ψ 1sB + ψ 1sB ) The PRODUCT ψ 1sA ψ 1sB is referred to as the OVERLAP between the A.O.'s ψ 1sA and ψ 1sB. The BUILD-UP of Probability Density in the region between the nuclei is associated with the value of the product ψ 1sA ψ 1sB BETWEEN THE NUCLEI i.e. with the ψ 1sA ψ 1sB OVERLAP between the nuclei. As the overlap of ψ 1sA and ψ 1sB between the nuclei INCREASES, ψ + between the nuclei INCREASES and the BOND STRENGTH between H A and H B INCREASES. Now consider the other possible M.O. formed from the 1s A.O. s: ψ = ψ σ*1s = c' ( ψ 1sA ψ 1sB ) ( σ* 1s) This M.O. leads to an ELECTRON DEFICIENCY IN THE REGION BETWEEN THE NUCLEI. In fact, there is a NODE in the Probability Density on the plane perpendicular to the internuclear axis at the MID-POINT of the H A H B bond. The deficiency in the Probability Density between the nuclei is associated with the fact that in ψ the ψ 1sA ψ 1sB OVERLAP between the NUCLEI is NEGATIVE. We now examine the Energy Changes which occur when these M.O.'s are formed: 3

4 Energy ε ψ (= ε σ * 1s ) = Energy of the ANTIBONDING MO ψ ε 1sA Energy of 1s AO on atom A ε ψ + ε 1sB Energy of 1s AO on atom B (= ε σ1s ) = Energy of the BONDING MO ψ + The ENERGY LOWERING is proportional to the value of the ψ 1sA ψ 1sB PRODUCT between the nuclei, i.e. the ENERGY LOWERING is proportional to the ψ 1sA ψ 1sB OVERLAP BETWEEN THE NUCLEI NOTE: The relative values of the orbital energies shown above will, of course, vary depending on the nature of the nuclei A and B and on the nature of the diatomic molecule A B. Since the ψ 1sA ψ 1sB overlap (in ψ + ) will INCREASE as the H A H B separation DECREASES, the energy lowering of ψ σ1s shown will INCREASE as the INTERNUCLEAR SEPARATION DECREASES. However, as the internuclear separation DECREASES, the nuclear-nuclear repulsion energy will INCREASE. The EQUILIBRIUM SEPARATION in the diatomic H + ion reflects that H A H B distance at which the TOTAL POTENTIAL ENERGY HAS A MINIMUM. Predictions : H + : 1-electron system : (ψ σ1s ) 1 : r e (H + ) = 1.07 Å : D o (H + ) = 56 kj/mol H : -electron system : (ψ σ1s ) : r e (H ) = 0.74 Å : D o (H ) = 436 kj/mol The SHORTER BOND LENGTH OF 0.74 Å in H allows GREATER OVERLAP between ψ 1sA and ψ 1sB in the region between the nuclei: this is reflected in an INCREASED dissociation energy (D o ) value in H. He : 4-electron system : (ψ σ1s ) (ψ σ * 1s ) : electrons in BONDING MO'S and electrons in ANTIBONDING MO'S -- thus, we would NOT EXPECT ANY NET BONDING between the He atoms -- there is NO STABLE GROUND STATE He molecule. 4

5 Electronic Structure of Homonuclear Diatomic Molecules The conclusions from our discussions of Covalent Bonding in H + and in H may be summarized as follows: (i) The STABILITY of the BONDING MO is associated with a POSITIVE OVERLAP (in the region between the nuclei) of the AO's forming the MO (ii) NEGATIVE OVERLAP (in the region between the nuclei) of the AO's forming the MO leads to an ANTIBONDING MO Criteria for effective combination of Atomic Orbitals φ a and φ b there must be as LARGE A POSITIVE OVERLAP (in the region between the nuclei) between φ a and φ b as possible the ENERGIES of the AO's φ a and φ b in their respective atoms must be of COMPARABLE MAGNITUDE We can now extend the ideas we developed for H + to describe the electronic structure of homonuclear diatomic molecules formed from atoms in the second period (i.e. Li F). We must now consider how to form MO's from s AO's and p AO's. As our starting example we will consider O. What happens to the energy separation between the s and the p atomic orbitals as the atomic number Z increases? You recall from our consideration of Radial Probability Distribution Functions that an electron in a s orbital penetrates to the nucleus more strongly than does an electron in a p orbital. As a result Z eff (s) > Z eff (p) and E s < E p. Such considerations show that this orbital energy gap will INCREASE as Z INCREASES (i.e. as we go across the second period). 5

6 For the case of O, which s, p combinations satisfy the criteria outlined above? We label the two oxygen atoms "O a " and "O b ", and consider all the possible combinations. Combination Overlap Energy Match s Oa with s Ob Very Good Excellent p zoa with s Ob Very Good Poor p xoa with s Ob Zero Poor p yoa with s Ob Zero Poor p zoa with p zob Very Good Excellent p zoa with p xob Zero Excellent p zoa with p yob Zero Excellent p xoa with p xob Good Excellent p xoa with p yob Zero Excellent p yoa with p yob Good Excellent Once we know the possible combinations we next arrange the molecular orbitals in order of INCREASING ENERGY Factors determining Molecular Orbital Energies : 1) The type of AO from which the MO is constructed. For example; E σ1s, E σ*1s < E σs because E s << E s i.e. the 1s AO's are of much lower energy than the s AO's ) The Energy of a BONDING MO < the Energy of the ANTIBONDING MO partner For example: E σ1s < E σ*1s ; E πp < E π*p 3) The overlap between φ a and φ b : S φ a φ b For example: S s a s >> S b px a px ; S b pz a pz >> S b px a px b The sizes and energies of MO's vary with internuclear separation and with the type of nuclei in the molecule. Thus, it is difficult to give a single definitive order that is correct for every type of homonuclear diatomic molecule. 6

7 You may recall that the DIFFERENCE in energy between the s and p atomic orbitals INCREASES as we go across the second row from B to F. If the s -- p AO energy gap is LARGE, then the following Molecular Orbital ordering is correct: σ1s < σ 1s < σs < σ s < σp z < πp x, πp y < π p x, π p y < σ p z This ordering is correct for O and F. For homonuclear diatomic molecules containing the atoms B, C, and N, the following order is (probably) the correct one: σ1s < σ 1s < σs < σ s < πp x, πp y < σp z < π p x, π p y < σ p z We now build up the electronic structure of molecules following the AUFBAU principle the PAULI EXCLUSION principle HUND'S rule when required Consider the ground state electronic configuration of B -- B has a TOTAL of 10 electrons. The ground state electronic configuration is: (σ1s) (σ 1s) (σs) (σ s) (πp x ) (πp y ) Thus, B has TWO UNPAIRED ELECTRONS -- it is said to be PARAMAGNETIC We can define a BOND ORDER (or the number of effective bonds), N e, as BOND ORDER (N e ) = (# of bonding electrons # of antibonding electrons)/ For B, N e = (6 4)/ =1; i.e. B has a SINGLE bond between the B atoms. Consider the ground state electronic configuration of N -- N has a TOTAL of 14 electrons. The ground state electronic configuration is: (σ1s) (σ 1s) (σs) (σ s) (πp x ) (πp y ) (σp z ) For N, N e = (10 4)/ =3; ( 1 sigma(σ) bond, and pi (π) bonds) i.e. N has a TRIPLE bond between the N atoms -- ( 1 sigma(σ) bond, and pi (π) bonds) 7

8 Thus, N has ZERO UNPAIRED ELECTRONS -- it is said to be DIAMAGNETIC Consider the ground state electronic configuration of O -- O has a TOTAL of 16 electrons. The ground state electronic configuration is: (σ1s) (σ 1s) (σs) (σ s) (σp z ) (πp x ) (πp y ) (π p x ) (π p y ) For O, N e = (10 6)/ =; ( 1 sigma(σ) bond, and 1 pi (π) bond) i.e. O has a DOUBLE bond between the O atoms -- ( 1 sigma(σ) bond, and 1 pi (π) bond) Thus, O has TWO UNPAIRED ELECTRONS -- it is said to be PARAMAGNETIC (this will be demonstrated in lecture: liquid O is attracted to the pole pieces of a permanent magnet). Such ideas can be used to understand the equilibrium bond lengths (r e ) and the equilibrium bond dissociation energies (D o ) of the following homonuclear diatomic molecules: Molecule Bond Order D o (ev) r e (Å) H H He B C N O F

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