electron count oxidation state coordination number

04/22/23 Counting Electrons 1

Electron CountOxidation 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!


• Every element has a certain number of valence orbitals:1 (1s) for H4 (ns, 3np) for main group elements9 (ns, 3np, 5(n-1)d) for transition metals

The basis of counting electrons

pxs py pz

dxzdxy dx2-y2dyz dz2


• Every orbital wants to be “used", i.e. contribute to binding an electron pair.

• Therefore, every element wants to be surroundedby 2/8/18 electrons.

• The strength of the preference for electron-precise structures depends on the position of the element in the periodic table.



• 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 -donor (nucleophile) and susceptible to electrophilic attack.



H2

Every H has 2 e. OK

CH4

H has 2 e, C 8. OK

NH3

N has 8 e. Nucleophile! OK

Use a localized (valence-bond) modelto count electrons

H H

C

H

H

H

H

N

H

H

H


C2H4

C has 8 e. OK

singlet CH2

C has only 6 e, and an empty pz orbital: extremely reactive ("singlet carbene"). Unstable. Sensitive to nucleophiles and electrophiles.

triplet CH2

C has only 6 e, is a "biradical" and extremely reactive ("triplet carbene"), but not especially for nucleophiles or electrophiles.

C

H

H

C

H

H

C

H

H

C

H

H


CH3+

C has only 6 e, and an empty pz orbital: extremely reactive. Unstable. Sensitive to nucleophiles.

CH3-

C has 8 e, but a lone pair. Sensitive to electrophiles.

Cl-

Cl has 8 e, 4 lone pairs. OK Somewhat sensitive to electrophiles.

C

H

H

H

C

H

H

H

Cl


BH3

B has only 6 e, not stable as monomer,forms B2H6:

B2H6

B has 8 e, all H's 2 (including the bridgingH!). 2-electron-3-center bonds! OK

AlCl3

Al has only 6 e, not stable as monomer,forms Al2Cl6:

Al2Cl6

Al has 8 e, all Cl's too (including thebridging Cl!). Regular2-electron-2-center bonds! OK

B

H

H

H

B

H

H

B

H

H

H

H

Al

Cl

Cl

Cl

Al

Cl

Cl

Cl

Al

ClCl

Cl


2 MeAlCl2 Me2Al2Cl4

2-electron-3-center bonds are a stopgap!

H3B·NH3

N-B: donor-acceptor bond (nucleophile NH3 has attacked electrophile BH3).

Organometallic chemists are "sloppy" and write . Writing or would be more correct (although the latter does not reflect the “real” charge distribution).

Al

Cl

Cl

Cl

Al

MeCl

Me

Al

Cl

Cl

Al

Cl

Cl

Me

Me

B

H

H

H

N

H

H

H

B NH

H

H

H

H

H B NH

H

H

H

H

H B NH

H

H

H

H

H


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 (hyperconjugation).Easy dissociation in PCl3 en Cl2.

HF2-

Write as FH·F-, mainly ion-dipole interaction.

?

P

Cl

Cl

Cl

Cl

Cl

P

Cl

Cl

Cl

Cl

Cl

?F H F F H F


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 atom4. Count 2 e for every dative bond from another

atom5. Add

How do you count?


B = 3- = 14H = 4tot = 8OK

Examples: counting electronsBH4

- B HH

H

H

H2CO C OH

H

C = 41=O= 22H = 2tot = 8OK

Pd =10- = 13Cl = 31NH3 = 2tot =16could have additional 2 e(Pd-Cl -bond?)

Cl Ru ClMe3P

PMe3

CH2

Ru = 82Cl = 22PMe3= 41CH2 = 2tot =16could have additional 2 e

PdCl3(NH3)-

Cl Pd NH3Cl

Cl-


Counting is not always trivial

Pd =102- = 23Cl = 31CH2= 1tot =16could have additional 2 e

Cl PdCl

Cl

NH3

Cl PdCl

Cl

NH3

-

is


• 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.

Remember, when counting:


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.

Oxidation States


1. Start with the formal charge on the metal2. Ignore dative bonds3. 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

Calculating theformal oxidation state


charge C =04C-Cl: C+-Cl- =+4tot =+4

Examples: oxidation statesCCl4

COCl2 charge C= 02C-Cl: C+-Cl- =+21C=O: C2+-O2- =+2tot =+4

AlCl4- charge Al= -14Al-Cl: Al+-Cl- = +4tot = +3

MnO4-

charge Mn = -14Mn=O: Mn2+-O2- = +8tot = +7

PdCl42- charge Pd= -24Pd-Cl: Pd+-Cl- = +4tot = +2


charge C = 03C-Cl: C+-Cl- =+3tot =+3trivalent carbon ?

Examples: oxidation states

charge Mg = 04Mg-Me: Mg+-Me- = +4tot = +4impossible, Mg has only 2 valence electrons!

charge Pt = -23Pt-Cl: Pt+-Cl- = +3tot = +1univalent Pt ?

C2Cl6Pt2Cl64-

MgMe4


Oxidation states are formal.However, they do give an indication whether a

structure or composition is reasonable (apart from the M-M complication).

The significance ofan oxidation state ?


For group n or n+10:– 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, -

acceptor ligands (CO) stabilize lower oxidation states– oxidation states usually change from m to m-2, m or m+2 in

reactions

Acceptable oxidation states


Simply the number of atoms directly bonded to the atom you are interested in, regardless of bond orders etc.

CH4: 4C2H4: 3C2H2: 2AlCl4

-: 4Me4Zn2-: 4OsO4: 4

Coordination number

B2H6: 4 (B)1 (terminal H)2 (bridging H)


For complexes with -system ligands, the whole ligand is usually counted as 1:

Cyclopentadienyl groups are sometimes counted as 3,because a single Cp group can replace 3 individual ligands:

Coordination Number

Cl PdCl

Cl-

ZrClCl C.N. 4

H

CoCOOC

COCO

CoOC CO


The most common coordination numbers for organometallic compounds are:

2-6 for main group metals4-6 for transition metalsCoordination numbers >6 are relatively rare. So are

very low coordination numbers (<4) together with a “too-low” electron count.

Coordination Number


C.N. "Normal" geometry2 linear or bent3 planar trigonal, pyramidal, "T-shaped"4 square planar, tetrahedral5 square pyramid, trigonal bipyramid6 octahedron

Coordination number and coordination geometry


Could WH6(PMe3)3 be ?Count W: 18 VE (OK), oxidation state 6 (OK), coordination number 9 (very high). Possible.

Protonation gives WH7(PMe3)3+.

Could that be ?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!

Illustration:protonation of WH6(PMe3)3

W PMe3Me3PMe3P HH H

HHH

W PMe3Me3PMe3P HH H

HHH

H

+


Give electron count and oxidation state for the following compounds. Draw conclusions about their (in)stability.

Me2Mg Pd(PMe3)4 MeReO3

ZnCl4 Pd(PMe3)3 OsO3(NPh)ZrCl4 ZnMe4

2- OsO4(pyridine)Co(CO)4

- Mn(CO)5- Cr(CO)6

V(CO)6- V(CO)6 Zr(CO)6

4+

PdCl(PMe3)3 RhCl2(PMe3)2

Ni(PMe3)Cl4 Ni(PMe3)Cl3 Ni(PMe3)2Cl2

Exercises

Cl PdMe3P

PMe3

BMe3

-

electron count oxidation state coordination number

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