Survey
* Your assessment is very important for improving the workof artificial intelligence, which forms the content of this project
* Your assessment is very important for improving the workof artificial intelligence, which forms the content of this project
Electron Count Oxidation State Coordination Number • Basic tools for understanding structure and reactivity. • Doing them should be “automatic”. • Not always unambiguous don’t just follow the rules, understand them! 5/24/2017 Counting Electrons 1 The basis of counting electrons • Every element has a certain number of valence orbitals: 1 (1s) for H 4 (ns, 3np) for main group elements 9 (ns, 3np, 5(n-1)d) for transition metals s dxy 5/24/2017 px dxz py dyz Counting Electrons pz dx2-y2 dz2 2 The basis of counting electrons • Every orbital wants to be “used", i.e. contribute to binding an electron pair. • Therefore, every element wants to be surrounded by 2/8/18 electrons. • The strength of the preference for electron-precise structures depends on the position of the element in the periodic table. 5/24/2017 Counting Electrons 3 The basis of counting electrons • Too few electrons: An empty orbital makes the compound very electrophilic, i.e. susceptible to attack by nucleophiles. • Too many electrons: There are fewer covalent bonds than one would think (not enough orbitals available). An ionic model is required to explain part of the bonding. The "extra" bonds are relatively weak. • Metal-centered (unshared) electron pairs: Metal orbitals are fairly high in energy. A metal atom with a lone pair is a strong s-donor (nucleophile) and susceptible to electrophilic attack. 5/24/2017 Counting Electrons 4 Use a localized (valence-bond) model to count electrons H2 H H Every H has 2 e. OK H CH4 H has 2 e, C 8. OK H C H H H NH3 N N has 8 e. Nucleophile! OK 5/24/2017 H Counting Electrons H 5 C2H4 C has 8 e. OK H H C C H H singlet CH2 H C has only 6 e, and an empty pz orbital: extremely reactive ("singlet carbene"). Unstable. Sensitive to nucleophiles and electrophiles. H H triplet CH2 C C has only 6 e, is a "biradical" and extremely reactive ("triplet carbene"), but not especially for nucleophiles or electrophiles. 5/24/2017 C Counting Electrons H 6 CH3+ C has only 6 e, and an empty pz orbital: extremely reactive. Unstable. Sensitive to nucleophiles. H H C H CH3- H C has 8 e, but a lone pair. Sensitive to electrophiles. C H H ClCl has 8 e, 4 lone pairs. OK Somewhat sensitive to electrophiles. 5/24/2017 Counting Electrons Cl 7 BH3 H B has only 6 e, not stable as monomer, forms B2H6: H B H B2H6 H B has 8 e, all H's 2 (including the bridging H!). 2-electron-3-center bonds! OK H H B B H H H AlCl3 Al has only 6 e, not stable as monomer, forms Al2Cl6: Cl Cl Al Cl Al2Cl6 Al has 8 e, all Cl's too (including the bridging Cl!). Regular 2-electron-2-center bonds! OK 5/24/2017 Counting Electrons Cl Cl Al Cl Cl Al Cl Cl 8 2 MeAlCl2 Me2Al2Cl4 Me Me Cl Al Cl Cl Cl Al Cl Cl Al Me Al Cl Cl Me 2-electron-3-center bonds are a stopgap! H3B·NH3 N-B: donor-acceptor bond (nucleophile NH3 has attacked electrophile BH3). H Organometallic chemists are "sloppy" and write H H H H H H H H H . Writing or would be more H B N H B N H B N H H H H H H correct (although the latter does not reflect the “real” charge distribution). 5/24/2017 H H Counting Electrons H B H N H 9 PCl5 P would have 10 e, but only has 4 valence orbitals, so it cannot form more than 4 “net” P-Cl bonds. You can describe the bonding using ionic structures Cl (hyperconjugation). Cl Easy dissociation in PCl3 en Cl2. Cl Cl Cl P Cl P Cl Cl Cl ? Cl HF2Write as FH·F-, mainly ion-dipole interaction. F 5/24/2017 H F ? F H Counting Electrons F 10 How do you count? 1. Number of valence electrons (from periodic table) 2. Correct for charge, if any (only if it belongs to that atom!) 3. Count 1 e for every covalent bond to another atom 4. Count 2 e for every dative bond from another atom 5. Add 5/24/2017 Counting Electrons 11 Examples: counting electrons H BH4- H B H H B 4H tot OK = = = = 3 1 4 8 C O H C = 4 1=O = 2 2H tot OK - Cl Cl Pd NH3 Cl Pd =10 - =1 3Cl = 3 1NH3 = 2 tot =16 could have additional 2 e (Pd-Cl p-bond?) H H2CO PdCl3(NH3)- = 2 = 8 PMe3 Cl Ru Cl Me3P CH2 Ru = 8 2Cl = 2 2PMe3 = 4 1CH2 = 2 tot =16 could have additional 2 e 5/24/2017 Counting Electrons 12 Counting is not always trivial Cl Cl Pd Cl Cl is NH3 Cl Pd Cl NH3 Pd =10 2=2 3Cl = 3 1CH2 = 1 tot =16 could have additional 2 e 5/24/2017 Counting Electrons 13 Remember, when counting: • • • • Odd electron counts are rare. In reactions you nearly always go from even to even (or odd to odd), and from n to n-2, n or n+2. Electrons don’t just “appear” or “disappear”. The optimal count is 2/8/18 e. 16 e also occurs frequently, other counts are much more rare. 5/24/2017 Counting Electrons 14 Oxidation States Most elements have a clear preference for certain oxidation states. These are determined by (a.o.) electronegativity and the number of valence electrons: Li: nearly always +1. Has only 1 valence electron, so cannot go higher. Is very electropositive, so doesn’t want to go lower. Cl: nearly always -1. Already has 7 valence electrons, so cannot go lower. Is very electronegative, so doesn’t want to go higher. 5/24/2017 Counting Electrons 15 Calculating the formal oxidation state 1. Start with the formal charge on the metal 2. Ignore dative bonds 3. Ignore bonds between atoms of the same element (this one is a bit silly) 4. Assign every covalent electron pair to the most electronegative element in the bond: this produces + and – charges (usually + at the metal) 5. Add 5/24/2017 Counting Electrons 16 Examples: oxidation states CCl4 charge C= 0 4C-Cl: C+-Cl- =+4 tot =+4 COCl2 5/24/2017 AlCl4- charge Al = -1 4Al-Cl: Al+-Cl- = +4 tot = +3 charge C= 0 2C-Cl: C+-Cl- =+2 1C=O: C2+-O2- =+2 tot =+4 MnO4- PdCl42- charge Pd = -2 4Pd-Cl: Pd+-Cl- = +4 tot = +2 charge Mn = -1 4Mn=O: Mn2+-O2- = +8 tot = +7 Counting Electrons 17 Examples: oxidation states C2Cl6 charge C = 0 3C-Cl: C+-Cl- =+3 tot =+3 trivalent carbon ? MgMe4 5/24/2017 Pt2Cl64- charge Pt = -2 3Pt-Cl: Pt+-Cl- = +3 tot = +1 univalent Pt ? charge Mg = 0 4Mg-Me: Mg+-Me- = +4 tot = +4 impossible, Mg has only 2 valence electrons! Counting Electrons 18 The significance of an oxidation state ? Oxidation states are formal. However, they do give an indication whether a structure or composition is reasonable (apart from the M-M complication). 5/24/2017 Counting Electrons 19 Acceptable oxidation states For group n or n+10: – – – – – – – 5/24/2017 never >+n or <-n (except group 11: frequently +2 of +3) usually even for n even, odd for n odd usually 0 for metals usually +n for very electropositive metals usually 0-3 for 1st-row transition metals of groups 6-11, often higher for 2nd and 3rd row electronegative ligands (F,O) stabilize higher oxidation states, pacceptor ligands (CO) stabilize lower oxidation states oxidation states usually change from m to m-2, m or m+2 in reactions Counting Electrons 20 Coordination number Simply the number of atoms directly bonded to the atom you are interested in, regardless of bond orders etc. CH4: 4 B2H6: 4 (B) C2H4: 3 1 (terminal H) C2H2: 2 2 (bridging H) AlCl4 : 4 Me4Zn2-: 4 OsO4: 4 5/24/2017 Counting Electrons 21 Coordination Number For complexes with p-system ligands, the whole ligand is usually counted as 1: Cl Cl Pd Cl Zr Cl Cl C.N. 4 Cyclopentadienyl groups are sometimes counted as 3, because a single Cp group can replace 3 individual ligands: OC Co 5/24/2017 CO H CO CO Co OC CO Counting Electrons 22 Coordination Number The most common coordination numbers for organometallic compounds are: 2-6 for main group metals 4-6 for transition metals Coordination numbers >6 are relatively rare. So are very low coordination numbers (<4) together with a “too-low” electron count. 5/24/2017 Counting Electrons 23 Coordination number and coordination geometry C.N. 2 3 4 5 6 5/24/2017 "Normal" geometry linear or bent planar trigonal, pyramidal, "T-shaped" square planar, tetrahedral square pyramid, trigonal bipyramid octahedron Counting Electrons 24 Illustration: protonation of WH6(PMe3)3 Could WH6(PMe3)3 be H Me3P H H Me3P W PMe3 H H H ? Count W: 18 VE (OK), oxidation state 6 (OK), coordination number 9 (very high). Possible. Protonation gives WH7(PMe3)3+. + Could that be H Me3P H H Me3P W HPMe3 H H H ? Count W: 18 VE (OK), oxidation state 8 (too high), coordination number 10 (extremely high). W+ must form 7 covalent bonds using only 5 electrons. That will not work! 5/24/2017 Counting Electrons 25 Exercises Give electron count and oxidation state for the following compounds. Draw conclusions about their (in)stability. Me2Mg ZnCl4 ZrCl4 Co(CO)4V(CO)6PdCl(PMe3)3 Ni(PMe3)Cl4 5/24/2017 Pd(PMe3)4 Pd(PMe3)3 ZnMe42Mn(CO)5V(CO)6 RhCl2(PMe3)2 Ni(PMe3)Cl3 MeReO3 OsO3(NPh) OsO4(pyridine) Cr(CO)6 Zr(CO)64+ Cl Pd Me3P PMe3 Ni(PMe3)2Cl2 Counting Electrons - BMe3 26