Question

Which of the following is/are examples of $\text{ }\\!\\!\sigma\\!\\!\text{ -bonded}$ organometallic compound?
(A) $A{{l}_{2}}{{\left( C{{H}_{3}} \right)}_{6}}$
(B) $Pb{{\left( C{{H}_{3}} \right)}_{4}}$
(C) $Zn{{\left( {{C}_{2}}{{H}_{5}} \right)}_{2}}$
(D) $Ferrocene$

Answer 1

Recollect what are organometallic compounds. Revise the formation of $\text{ }\\!\\!\sigma\\!\\!\text{ -bond}$ and pi-bonds. Find out in each of the given options which type of bond formation is taking place.

Complete step by step solution:
-Organometallic compounds are organic compounds containing a metal-carbon bond.
-$\text{ }\\!\\!\sigma\\!\\!\text{ -bond}$ is formed when the orbitals of two adjacent atoms overlap axially.
-pi-bond is formed when the orbitals of two adjacent atoms overlap laterally.
-Let’s have a look at the options given in this question.

(A) $A{{l}_{2}}{{\left( C{{H}_{3}} \right)}_{6}}$- Trimethylaluminum
Trimethylaluminum is an organometallic compound and it exists in a dimer state. Aluminium atoms form three $\text{ }\\!\\!\sigma\\!\\!\text{ -bonds}$ with three methyl groups to form trimethylaluminum. There is no pi-bond present in this compound.

(B) $Pb{{\left( C{{H}_{3}} \right)}_{4}}$- Tetramethyl lead
Tetramethyl lead is an organometallic compound. Here, lead forms four $\text{ }\\!\\!\sigma\\!\\!\text{ -bonds}$ with four methyl groups. There are no pi-bonds present in this compound.

(C) $Zn{{\left( {{C}_{2}}{{H}_{5}} \right)}_{2}}$- Diethyl zinc
Diethyl zinc is also an organometallic compound. Here, zinc forms two $\text{ }\\!\\!\sigma\\!\\!\text{ -bonds}$ with two ethyl groups. There are no pi-bonds present in this compound.

(D) Ferrocene
Ferrocene is an organometallic compound. Ferrocene contains one iron, Fe atom bound to two cyclopentadienyl rings. In this compound, Fe is pi-bonded to ligands.

-Therefore, the examples of $\text{ }\\!\\!\sigma\\!\\!\text{ -bonded}$ organometallic compounds are (A) $A{{l}_{2}}{{\left( C{{H}_{3}} \right)}_{6}}$, (B) $Pb{{\left( C{{H}_{3}} \right)}_{4}}$ and (C) $Zn{{\left( {{C}_{2}}{{H}_{5}} \right)}_{2}}$.

So, answer is options (A), (B) and (C).

Note: Carefully find out the type of bonds present in each compound and then come to a conclusion about which is the right answer. Remember $\text{ }\\!\\!\sigma\\!\\!\text{ -bonds}$ result due to axial overlapping of orbitals and pi-bonds result from lateral or side-ways overlapping of orbitals.