What are organometallic compounds?

The compounds which contain at least one metal-carbon bond are termed as organometallic compounds. In organometallic compounds, the carbon should be part of an organic group. A metal bound with ligands is termed an organometallic compound. The ligands include PPH3, PF3, etc. The organometallic compounds could be differentiated by using the prefix organo. The coordination compound having one or more alkene ligands is termed transition metal alkene complex. Zeise salt is the first olefin (alkene) complex containing transition metal that is reported as an organometallic compound. The formula of Zeise salt is K[Pt(C2H4)Cl3].H2O. Similarly, ferrocene is the first sandwich compound containing transition metal, which is prepared by the Kealy group and the miller group.

Significance of organometallic compounds of d-block elements (transition metals)

The transition metals can form complexes since they have vacant d orbitals. They are also termed Coordination compounds. The organometallic compounds of the transition metals could be used as catalysts as well as reagents. For example, organocadmium, organozinc, organocopper, Grignard are used as the reagents for several reactions. Most of the complexes of organo transition metals are used as catalysts for many organic transformation reactions.

Metal carbonyl compounds

The binary metal carbonyl compounds which have only metal & carbon monoxide (CO) ligands could be produced by the reaction of reactive metal & CO or through the reduction of metal salt to zero valences followed by the reaction with high-pressure CO. The tetracarbonylnickel was initially synthesized via the reaction of nickel (Ni) metal with CO. There are also many of the metal carbonyl compounds having M-M bonds connecting three or more metals & terminal CO, µ-CO & µ₃-CO are coordinated to metal frames. An example of bridged metal carbonyl is Ru₃(CO)₁₂.

Organometallic chemistry of d-block elements (transition metals)

• Total valence electron, coordination number & the oxidation state of metal are some important terms used in organometallic chemistry.
• Total valence electron of the compound is the sum of the metal valence electron & the electrons donated by the ligands. The organometallic compounds containing a total of 18 valence electrons are more stable.

18 electron rule

This rule states that the organometallic compounds of the transition metals are thermodynamically more stable when their total valence electron is equal to 18.

The two methods which are used to count the number of electrons present in the organometallic compounds are,

• Oxidation state method.
• Neutral ligand method.

Oxidation state method

This method considers the ligand to donate electron pairs to the metal.

In this method, the total electron count could be calculated by using the charge of ligand & the oxidation state of the metal.

For example, consider the complex of Cr(CO)₆.

In this method, initially, the oxidation state of the metal should be determined. Here the oxidation state of Cr is zero. Therefore, it contains 6d electrons. The number of ligands donated by CO is two. Therefore,6+(6x2)=18. It obeys the 18 electron rule.

Neutral ligand method

In this method, all the ligands are considered neutral ligands & the oxidation state of the metal will not be accounted for. Here the number of d-electrons in the metal is considered in its zero oxidation state. Let we discuss the example of Cr(CO)₆ complex,

In this method, there is no requirement to calculate the oxidation state of the metal. Cr contains six d electrons in the oxidation state of zero & each CO gives two electrons. Therefore, 6+(6x2)=18. This transition metal complex obeys the 18 electron rule. Hence, it is highly stable.

Determination of M-M bond in complexes of transition metals   

In the complexes of transition metals containing 18 electrons, the number of M-M bonds could be calculated by using the formula.

Number of M-M bond =$\frac{18\mathrm{n}-\mathrm{TVE}}{2}$

Where,

n is the number of metals.

TVE is the total valence electrons.

Let us discuss the example of ${\mathrm{Mn}}_2{\left(\mathrm{CO}\right)}_{10}$

In this complex, the number of metal atoms is two.

The number of d electrons present in Mn in its zero oxidation state is seven. The number of electrons donated by CO is two. Hence, the total valence electron is $2\times 7+10\times 2=34$.

The number of M-M bonds could be calculated as follows:

$\mathrm{Number} \mathrm{of} \mathrm{M}-\mathrm{M} \mathrm{bond}=\frac{18\mathrm{n}-\mathrm{TVE}}{2}=\frac{18\times 3-34}{2}=1$ 

Hence, there is only one M-M bond is present in the complex.

Back donation

The metal carbonyl compounds contain CO coordinated with zero-valent metal. The normal coordination bonds can be produced by the electrons donated from the very basic ligands to the metal which produced the basis of the coordination theory of Werner. Since CO has less basicity and the M-C bonds are not stable. When the symmetry and shape of the metal d orbital and CO π orbital for C-O bond are fit for overlap, the bonding interaction between carbon & the metal is expected. The mechanism through which the electrons are provided from the filled metal d orbital to the empty CO π* orbital is termed as a back donation. Since the gathering of superfluous electrons on the low oxidation state metal atom is protected, back donation results in the stabilization of the metal carbonyl bond. The back donation could be depicted as follows:

Metallocene

Metallocenes are compounds consisting of two cyclopentadienyl anions connected to the metal center in the oxidation state II. The cyclopentadienyl ligand $\left(C_5{H}_5\right)$ could be abbreviated as Cp. If the H atoms of Cp ligands are replaced by methyl ($C{H}_3$ ) groups, then that ligand is termed as Cp star & it is depicted by Cp*.  An example of a metallocene is Ferrocene. Ferrocene is the stable sandwich compound where the Cp groups are joined with iron. The structure of ferrocene could be depicted as follows:

Context and Applications

This topic is significant for both undergraduate and postgraduate courses, especially for Bachelors and Masters in chemistry, Bachelors and Masters in Biochemistry.

Practice Problems

Question 1: The oxidation state of nickel & the number of M-M bonds in the complex of [Ni₂(CO)₆]^2- that is consistent with the 18 electron rule are ________.

1. Ni(IV), 2 bonds
2. Ni(-II), 1 bond
3. Ni(IV), 3 bonds
4. Ni(-I), 1 bond

Answer: Option 4 is correct.

Explanation:

The oxidation state of Ni could be calculated as follows,

$$\begin{gathered} \mathrm{Oxidation} \mathrm{state} = 2\mathrm{X}+6\left(0\right)=-2 \\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}2\mathrm{X} = -2 \\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em} \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\mathrm{X}=-1 \\ \mathrm{Total} \mathrm{valence} \mathrm{electron} = 2\times 10+6\times 2+2=34 \\ \mathrm{Number} \mathrm{of} \mathrm{metal}-\mathrm{metal} \mathrm{bond}=\frac{18\mathrm{n}-\mathrm{TVE}}{2} \\ =\frac{18\times 2-34}{2}=1 \end{gathered}$$ 

Hence, the oxidation state of Ni is -1 and the number of M-M bond present in the complex is one.

Question 2: Among the following, the naturally occurring organometallic compound is ________.

1. Chlorophyll
2. Vitamin B₁₂ coenzyme
3. Myoglobin
4. Cytochrome P-450.

Answer: Option 2 is correct.

Explanation: Vitamin B₁₂ coenzyme is the naturally occurring organometallic compound. Since it contains a metal-carbon bond (Cobalt-carbon bond).

Question 3: Among the following statement, which is not correct about ferrocene?

1. The Cp rings in ferrocene are staggered
2. The Cp rings in ferrocene are almost eclipsed
3. The decamethyl ferrocene is staggered in the case of solid state
4. Ferrocene could be nitrated by dil HNO₃

Answer: Option 1 is correct.

Explanation: Among the given statement, the incorrect one is ‘Cp rings in ferrocene are staggered’. Because the Cp rings in ferrocene are eclipsed.

Question 4: The binding modes of NO in 18 electron compounds [Ni(ƞ⁵-Cp)(NO)] and [Co(CO)₃(NO)] respectively are ________.

1. Linear and bent
2. Bent and linear
3. Bent and bent
4. Linear and linear

Answer: Option 4 is correct.

Explanation: The binding mode of NO could be determined by calculating the number of electrons donated by NO. If the number of electrons donated by NO is three, then the binding mode is linear. If the number of electrons donated by NO is one, then the binding mode is one.

In the case of $\left[\mathrm{Ni}\right({\eta }^5-\mathrm{Cp})(\mathrm{NO})$  the number of electrons donated by NO could be calculated as follows:

$$\begin{gathered} 9+6+\mathrm{X}=18 \\ \mathrm{X}=3 \end{gathered}$$

Hence, NO is in a linear mode of coordination.

Similarly, in the case of $\left[\mathrm{Co}{\left(\mathrm{CO}\right)}_3\right(\mathrm{NO})]$, the number of electrons donated by NO can be calculated as follows:

$$\begin{gathered} 10+5+\mathrm{X}=3 \\ \mathrm{X}=3 \end{gathered}$$

Hence, NO is linear.

Question 5: The hapticities of x and y of the arene moieties in the diamagnetic complex [(ƞ^x-C₆H₆)Ru(ƞ^y- C₆H₆) respectively are _________.

1. 4 & 4
2. 6 & 2
3. 6 & 6
4. 4 & 6

Answer: Option 4 is correct.

Explanation: For a stable complex, the total electron count is 18. The number of electrons donated by Ru in its zero oxidation state is 8. Hence, the hapticities of x and y could be calculated as follows:

x+8+y = 18

x+y=18-8=10

x+y=10= 4+6

Hence, the hapticities of x and y are 4 and 6.