Electron Count and Oxidation State

A few examples of special relevance to homogeneous catalytic systems are given in Fig. 2.1, along with total electron counts. The rationales behind the schemes that are used to arrive at the electron counts are described in the following. 

RhCl(PPh3) 3 The chlorine radical (Cl ) accepts an electron from rhodium metal (electronic configuration Ad1,5s2) to give Cl and Rh+. The chloride ion then donates two electrons to the rhodium ion to form a dative or a coordinate bond. Each PPh3 donates a lone pair of electrons on the phosphorus atom to the rhodium ion. The total number of electrons around rhodium is therefore 8 + 2 + 3X2=16, and the oxidation state of rhodium is obviously 1 +. The other way of counting is to take the nine electrons of rhodium and add one electron for the chlorine radical and six for the three neutral phosphine ligands. This also gives the same electron count of 16. 

Co(CO)4 Since there is a net negative charge and CO is a neutral ligand, the formal oxidation state of cobalt is 1 —. The electron count is therefore 10 + 4X2=18. According to the covalent model, the electron count is also 9 + 4X2 + 1 = 18, but cobalt is assumed to be in a zero oxidation state, and one electron is added for the negative charge. 

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C.ii. Some Rules Concerning Formal Oxidation State and Electron Counting... 

Common oxidation states and /-electron counts for the transition metals are given in Table 25-1. Most /-transition metal ions have vacant d orbitals that can accept shares in electron pairs. Most act as Lewis acids by forming dative bonds with Lewis bases that donate one or more lone pairs of electrons in coordination compounds (coordination com-... 

Zirconocene dichloride is neutral. What is the oxidation state and electron count of zirconium ... 

Give (a) the oxidation states and electron counts of complexes 2.71-2.73 (b) the overall stoichiometric reaction between oxygen and RH, catalyzed by Fe of CytP j. ... 

Insertions and (3-eliminations are also the microscopic reverse of each other. In an insertion, an A=B 77 bond inserts into an M-X bond (M-X + A=B —> M-A-B-X). The M-X and A=B bonds are broken, and M-A and B-X bonds are formed. Insertion is usually preceded by coordination of the A=B 77 bond to the metal, so it is sometimes called migratory insertion. In an insertion, an M-X bond is replaced with an M-A bond, so there is no change in oxidation state, d electron count, or total electron count. However, a new a bond is formed at the expense of a 77 bond. The nature of the reaction requires that the new C-M and C-H bonds form to the same face of the A=B 77 bond, resulting in syn addition. The reaction of a borane (R2BH) with an alkene to give an alkylborane is a typical insertion reaction that you have probably seen before. 

Elimination is the microscopic reverse of insertion. Just as insertion does not, a / -elimination causes no change in the oxidation state, d electron count, or total electron count of the metal. By far the most common /3-elimination is the (3-hydride elimination, in which M-A-B-H -h> M-H + A=B. The /3-hydride elimination is the bane of the organometallic chemist s existence, as it causes many metal-alkyl bonds to be extremely labile. /3-Alkoxy and /3-halide eliminations are also known, as in the reaction of BrCH2CH2Br with Mg. 

In the oxidative addition of an A-B bond to a metal, new M-A and M-B bonds are formed as the A-B bond is cleaved (Eq. 2.1). The reverse reaction, reductive elimination, leads to the extrusion of an A-B molecule from a precursor M(A)(B) complex this is often the product forming step in a catalytic reaction. In the oxidative direction, we break the A-B bond and form an M-A and an M-B. Since A and B are always considered as le X-type (anionic) ligands, the oxidation state, the electron count, and coordination number of the metal all increase by two units during the reaction. The change of +2 in the formal oxidation state gives the reaction the name oxidative addition. These terms as well as the conceptual basis of organometallic chemistry are discussed in a previous work [4]. 

There is a second method for counting the electrons in a complex and deducing the metal s oxidation state and electronic configuration. This is the ionic model, in which one supposes that a complex is formed by a metal centre and by ligands which always act as Lewis bases, supplying one (or several) pairs of electrons. 

Similarly, for RhH(CO)(PPh3)3, structure 2.59, the rhodium oxidation state is 1+ because the hydrogen atom is assumed to carry, with some justification, a formal negative charge. The five ligands, H", CO, and three PPhj, each donate two electrons, and the electron count therefore is 8 + 5 x 2 = 18. If we do not assign an oxidation state then the hydrogen atom donates one electron, and rhodium is in the zero oxidation state. The electron count is 1+9 + 4x2 = 18. 

If alkylidene and the imido ligand are treated as dianions, and the two alkoxo groups as alkoxide, then tungsten is in the 6+ oxidation state. The electron count, 0 + 4 + 4 + 2x2 = 12, however, remains the same. 

There are two methods for counting electrons formal charge and oxidation state. These two counting methods actually represent two flipsides of the same coin. To calculate formal charge, we treat all bonds as covalent, regardless of whether they are or not ... 

C20-0040. Determine the oxidation states and d electron counts for the metal ions in the following... 

Probably the most common detachment step in late transition metal catalyzed processes is reductive elimination. In this transformation two groups, that are both attached to the same metal centre, will be released and form a covalent bond, with the concomitant formation of a metal whose formal oxidation state, coordination number and electron count are decreased by two. Figure 1-9 presents a general order of the ease of reductive elimination for the most common complexes. 

The formal oxidation state and d-electron count of the metal in the following complexes (a) (h6C6H6)2 Mo, (b) cp2ZrCl (OMe)... 

The useful complex (MeCN)2PdCl2 has palladium in the +2 oxidation state because of its tw chlorine atoms and the number of electrons is 8 for the Pd(II) oxidation state and another two eac from the four ligands making 16 in all. This complex does not fulfil the 18-electron rule and reactive. You would have got the same answer if you had counted ten for the palladium, two each fc the nitriles, and one each for the chlorines, but this is not so realistic. 

Figure 2.1 Formal oxidation states and valence electron counts of metal ions in some homogeneous catalysts.

Cp2Zr(CH3)(THF)]+ The zirconium oxidation state is 4+ and each Cp ligand donates six electrons. The ligand CTC donates two electrons. The solvent molecule, THF, also donates two electrons, and the total electron count is 12 + 0 + 2 + 2=16. With the covalent model zirconium is in the zero oxidation state and has four electrons Ad2,5s2) in the valence shell. Both Cp and CH3 are considered as radicals and therefore donate five and one electron, respectively. The valence electron count is therefore 4 + 2x5 + 1+ 2-1 = 16. Notice that because of the positive charge, we subtract one electron. 

What are the oxidation states and the valence electron counts of the metal ions in the complexes shown in Figs. 2.4, 2.6, and 2.11 ... 

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