CHE 201; DR GERVAS ASSEY; LECTURE 08

ELECTONIC COUNTING IN ORGANOMETALLIC CHEMISTRY

This lecture comprises of the following parts:

1.     The eighteen electron rule

2.     Electron counting scheme for common ligand

3.     Tutorial questions.

1.     ELECTRONIC COUNTING IN ORGANOMETALLIC CHEMISTRY.

Ø Just as in the main group chemistry requires valence shell of 8 electrons (octet rule), the organometallic chemistry is based on a total valence electron count of 18 on the central metal atom.....
.

                Why 18 electron?

The rule is called eighteen electron rule because in TM we start to fill electrons after the basis eighteen electron of the first orbital. Therefore the number 18 represents a filled valence shell for a transition metal (1s² 2s² 2p⁶ 3s² 3p⁶) or (s²p⁶d¹⁰). Just like octet rule (s²p⁶) which represent complete valence configuration for the main group chemistry.

Ø Several schemes exist for counting electrons in organometallic compounds. But the following main steps are involved.

i.                   Calculate the oxidation state of the central metal first, by considering the charge of the ligand as given in the scheme table.

ii.                 Count the electron in the central metal (don’t forget the oxidation state obtained above). Consider just those in 4s and 3d.

For example if the complex is Cr(CO)₆

The oxidation state of Cr is zero (because CO is neutral ligand)

Cr = [Ar] 4s¹ 3d⁵

Hence Cr = 6 electrons

iii.              Count the electron in the ligand, by considering the contribution of each ligand.

6CO = 2e^- x 6 = 12 electrons

iv.              Then the total number of electron is given by summing the electron in the central metal and those contributed by the ligand.

6 + 12 = 18 electrons

Hence the total number of electron in Cr(CO)₆ = 18 electrons.

I hope it was helpful

2.     ELECTRON COUNTING SCHEME FOR COMMON LIGANDS.

LIDANDS	METHOD A	METHOD B
H	2(H-)	1
Cl, Br, I	2(X-)	1
OH, OR	2(OH-  OR-)	1
NO (bent(M-N-O)	2(NO-)	1
NO (Linear(M-N-O)	2(NO+)	3
CO,PR3	2	2
NH3, H2O	2	2
=CRR’(Carbene)	2	2
H2C=CH2(ethylene)	2	2
CNR	2	2
=O, =S	4(O2- , S2-)	2
3-C3H5(-allyl)	2(C3H+5)	3
CR(carbyne)	3	3
Ethylene diamine(en)	4(2 per N)	4
Bipyridine (bipy)	4(2 per N)	4
Butadiene	4	4
Ƞ5-C5H5 (cyclopentadienyl)	6(C5H-5)	5
Ƞ6-(C6H6) benzene	6	6
	6(N3-)	3
Ƞ7-C7H7 (cycloheptatrienyl)	6(C7H+7)	7
CN	2(CN-)	1
CH3, CR3	2(CH-3, CR-3)	1

Example 01

Count electrons in (Ƞ⁵-C₅H₅)Fe(CO)₂Cl.

Solution:

Ø Calculating the oxidation state of Fe (the central metal),

(-1 x 1) + Fe + (0 x 2) + (-1) = 0 (because the complex is neutral)

Hence the oxidation state of Fe in the complex is +2

Ø Counting electron of Fe in the complex

Since Fe = [Ar] 4s² 3d⁶

Then Fe²⁺ = [Ar] 3d⁶

Hence there are 6 electrons

Ø -Ƞ⁵-C₅H₅ = 6electrons (from the scheme table)

-2(CO) = 2 X 2 = 4 electrons

-For chlorine, Cl = 2 electrons

Hence on summing up, the total number of electorn obtained is 18 electrons.

Example 02

Count electrons in [Mn(CO)]⁺

Solution

-Oxidation state of Mn is calculated as follows

Mn + (0) = +1(singly positive charged)

Hence Mn has +1 oxidation state.

-counting electron in Mn⁺¹

   Mn = [Ar] 4s² 3d⁵

    Mn⁺¹ = [Ar] 4s¹ 3d⁵ = 6 electrons

-for the ligand

       6C0 = 12 electrons

Hence the total number of electron is 18 electrons.

                3.0 TUTORIAL QUESTIONS

1. Determine the valence electron counts for the transition metals in the following

^{a)        2-}

^{b)       +}

^c)         Fe

^d)       ₈  ( has a single Co  -  Co bond )

2. Identify the first row transition metal for the following 18 – electro species

^{a)        -}

^b)       ₅

^c)         ₃

^d)       ₂      Assume single  bond.