CAUTION: The first row of main group elements (n = 2) almost always obey the octet rule, but heavier main group elements often do not.

Transcription

1 2. Molecular Structure and Bonding Suggested Reading: Chapter 2, sections , Review of Lewis dot structures: Lewis proposed that a covalent bond is formed when two neighboring atoms share an electron pair. Single bond (A:B) Double bond (A::B) Triple bond (A:::B) An unshared pair of valence electrons is a lone pair. (A:) Lewis found that many molecules obey the octet rule. Each atom shares electrons with neighboring atoms to achieve a total of eight valence electrons. CAUTION: The first row of main group elements (n = 2) almost always obey the octet rule, but heavier main group elements often do not. Quick rules for drawing Lewis dot structures: 1. Account for all the valence electrons of each atom present. 2. Create bonding pairs and lone pairs of electrons such that atoms in the molecules obey the octet rule.if it is not possible to do so for all the atoms in a molecule, then satisfy the octet rule for the n = 2 row of elements first, then the n = 3, etc. 3. If possible, find a distribution of electrons that does not lead to charge separation (see ahead). Let s do one example go home and try others on your own! 55

2 2.1. Formal Charges and Oxidation States Formal charge is the charge an atom would have if electron pairs were shared equally. Lewis structures with low formal charges typically have the lowest energy! Formal charge = [# of valence electrons in free atom] [# of unshared electrons in atom] [# of bonds to the atom] Purely fictitious number obtained by homolytic cleavage (each bonding partner gets one electron) of every bond in a given molecule. Useful to assess relative importance of a resonance structure. Structures with minimal formal charges and negative charges on more electronegative atoms more realistic/important. an extreme example: Oxidation State: Determined by heterolytic bond cleavage with the more electronegative partner getting both electrons. e.g. nitric acid (oxidation states in red; formal charges in black) This is one possible resonance structure. It is not possible to draw a Lewis dot structure that obeys the octet rule for all first row atoms without charge separation. Note: the oxidation states of the oxygen atoms are still all the same The sum of the oxidation numbers of all the atoms in the species is equal to its total charge. 56

3 Some rules for determination of oxidation number Oxidation State Atoms in their elemental form 0 Fluorine (in all compounds except F 2 ) -1 Atoms of Group 1 +1 Atoms of Group 2 +2 Hydrogen +1 in combination with nonmetals -1 in combination with metals Oxygen - 1 / 3 in ozonides (O - 3 ) - 1 / 2 in superoxides (O - 2 ) -1 in peroxides (O 2-2 ) -2 unless combined with F Halogens (other than F) -1 unless combined O or more electronegative halogens Oxidation states help identify redox (i.e., electron transfer) processes. e.g., Chlorine, when dissolved in water, undergoes rapid disproportionataion to hypochlorous acid and hydrochloric acid, with an equilibrium constant of 5 x 10-4 mol 2 L -2. The reverse process is called comproportionation. e.g., synthesis of elemental sulfur from SO 2 and H 2 S 2 H 2 S (g) + SO 2(g) 3/8 S 8(s) + 2 H 2 O (g) S(-2) S(+4) S(0) 57

4 Maximum valences and typical oxidation states: Maximum oxidation states in bold. Negative oxidation states are possible the maximum is defined by the difference of the # of electrons in the free atom and a full octet shell. For the main group elements oxidation states typically change in units of 2 as shown in the table. Other oxidation states are very rare, highly reactive, and unstable against oxidation or reduction. Group # XeO 4 8 IF 7 7 SO HNO CCl 4, CH 4 ±4 4 4 BH 3, NH 3 3 ±3 3 MgO, H 2 Se 2 2 ±2 2 NaH 1 1 ±1 He Observation #1 Stability of highest oxidation state: Decreases from top to bottom within a group, e.g. P +V >> As +V >> Sb +V S +VI >> Se +VI >> Te +VI This is not true for the most electronegative elements N and O that typically only occur in maximum oxidation states of +3, +2. i.e., They do not realize their maximum oxidation state (exception HNO 3 ). Fluorine as the most electronegative element occurs (by definition) only in oxidation state 1. Observation #2 The inert pair effect: For the main group elements of the 6 th period to the right of the transition metals, the highest oxidation state is often unstable/very oxidizing. Tl +III, Pb +IV and Bi +V are relatively unstable and highly oxidizing. i.e., They prefer to be in oxidation state +1, +2, and +3 respectively. E.g. TlI 3 and PbI 4 are unknown, as they would instantaneously decompose ( into what?) 58

5 Explanations a) (The simple one) The insertion of the lanthanides (i.e., occupied f-orbitals) results in very high Z* (effective nuclear charge) for the 6s 2 electrons, which in turn stabilizes the 6s orbital. i.e., Makes it energetically more difficult to realize the higher oxidation state. b) (The better one) The inertness of the 6s 2 electron pair is due to a relativistic contraction: Relativistic effects most directly concern electrons in the 1s orbital (no node at nucleus). As nuclear charge increases, so must the radial velocity of core electrons in order to avoid capture by the nucleus (simple Bohr Model). In fact, one type of naturally occurring nuclear reaction is Electron Capture (EC) β-decay wherein an orbital electron is captured by a heavy nucleus, changing a proton into a neutron, and changing the element identity. As the nuclear charge approaches Z = , the radial velocity of the core electrons approaches the speed of light, e.g. for mercury (element 80) the velocity of the electron is 80/ times the speed of light. According to Einstein s theory: (v = radial velocity, c = speed of light, m 0 = rest mass). i.e. the mass of the electron at 0.58 c is 1.2 times its rest mass! The Bohr radius of an electron is inversely proportional to its mass Therefore: The 1s-orbitals of the heavier elements contract. The other s-orbitals (including 6s orbital) must also contract in order to remain orthogonal to the 1s orbital. The [noble gas shell]ns 2 electrons are held more tightly in heavier elements. More difficult to remove these electrons Inert pair! The effect is also present, but much weaker for the p-orbitals. For the d- and f-orbitals the effect is reversed, i.e. they undergo a relativistic expansion. Why could that be? Think about the effects of penetration and shielding! 59

6 2.2. The VSEPR model and molecular geometry (Valence Shell Electron Pair Repulsion) (see: J. Chem Ed., 1963, 40, ; J. Chem. Ed., 2004, 81, ) VSEPR Simple rules for determining molecular structures 1. Valence shell electron pairs in a Lewis structure keep as far apart as possible. 2. Lone pairs repel more strongly than bond pairs, therefore angles between lone pairs and bond pairs are larger than angles between bond pairs. 3. The repulsion exerted by a bond pair decreases with increasing electronegativity of the ligand, therefore angles involving this bond pair are decreased. 4. Multiple bonds repel more strongly than single bonds, therefore angles involving multiple bonds are increased. Repulsive forces between electron pairs arise mainly from the operation of the Pauli principle: The wavefunction for a many-electron molecule must be antisymmetric with respect to electron interchange. I.e., electrons with the same spin have a zero probability of being found in the same location, and an increasing probability of being found at an increasing distance apart. (Q. What is another repulsive force between electron pairs?) VSEPR predicts stereochemically active electron pairs VSEPR predicts the effect of π-bonds on structures Let s do some examples together: SO 4 2- XeF 4 SF 4 Question. Why are multiple bonds more common for small atoms with large electronegativities (e.g., C, N, O)? Suggested Textbook Exercises

7 2.3 Bond Strength and Bond Enthalpies The strength of a bond is measured by its dissociation enthalpy, ΔH θ (A-B), the standard reaction enthalpy for the process AB(g) A(g) + B(g) The mean bond enthalpy B is the average bond dissociation enthalpy measured over a series of A-B bonds in a variety of molecules (that contain A-B bonds). Obviously, there can be great variations in ΔH θ (A-B) in different molecules, so B is not always a good estimate. Trend: For an element E that has no lone pairs, the E-X bond enthalpy decreases down a group. B (kj/mol) B (kj/mol) B (kj/mol) C C 347 C H 412 C Cl 327 Si Si 222 Si H 328 Si Cl 391 Ge Ge 188 Ge H 289 Ge Cl 342 Smaller atoms form stronger bonds because the shared electrons are closer to each of the atomic nuclei. The larger values of Si Cl and Ge Cl are attributed to large Δχ (see ahead). 61

8 Trend: For an element E that has lone pairs, the E-X bond enthalpy typically increases between periods 2 and 3 and then decreases down the group. B (kj/mol) B (kj/mol) B (kj/mol) N N 165 N Cl 194 N H 388 P P 200 P Cl 319 P H 322 As As 180 As Cl 317 As H 247 The relative weakness of single bonds between Period 2 elements that have lone pairs is often attributed to the strong repulsion between lone pairs on neighboring atoms as they are brought into close proximity (smaller atoms means shorter bond distances). Note that the trend is not observed for E H because H doesn t have lone pairs. Bond Enthalpies can be used to rationalize: Catenation: Extensive catentation (formation of chain compounds) is observed for Carbon: CH 3 CH 3, CH 3 (CH 2 ) n CH 3 Sulfur: S n n = 5-8 are common But not Nitrogen: H 2 NNH 2 (hydrazine) and N 3 are known, but longer chains are not. Disproportionation of subvalent compounds: Compounds in which fewer bonds are formed than valence rules suggest are often unstable with respect to disproportionation. 3 PH 2 (g) 2 PH 3 (g) + 1 / 4 P 4 (s) P(-2) P(-3) P(0) The strong P P bond drives this. Note that entropy would favour the other direction. To be more accurate, one should actually consider the Gibbs energy. 62

9 Recall: Pauling Electronegativity derived from bond energies Therefore, electronegativity can be used to estimate bond energies/types and to make qualitative assessments regarding bond polarity. A Ketelaar triangle classifies binary compounds based on the difference in electronegativities of the two elements and their average electronegativities. E.g., Predict the type of bonding that is likely to dominate in S 2 Cl 2. (use Pauling ENs from p.31 of your textbook) χ Cl = 3.16 Δχ = = 0.58 χ S = 2.58 χ mean = ( ) / 2 = 2.87 Covalent! Suggested Textbook Exercises SIDE NOTE: Sulfur monochloride (S 2 Cl 2 ) is a yellow liquid. Although it reacts with water, it can still be handled in air. It may appear to smoke if the laboratory environment is quite humid. 2 S 2 Cl H 2 O SO HCl + 3/8 S 8 Because it is a precursor to the preparation of mustard gas (a chemical warfare agent), its sale and distribution is tightly controlled. S 2 Cl C 2 H 4 (ClC 2 H 4 ) 2 S + 1/8 S 8 63