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The ligand called $\pi $-acid is:
(A) $CO$
(B) $N{H_3}$
(C) ${C_2}{O_4}^{ - 2}$
(D) Ethylene diamine

Last updated date: 29th May 2024
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Hint: The ligand which can accept d-orbital electrons of the metal to its empty $\pi$-antibonding orbitals alongside the formation of a coordinate covalent bond is called $\pi$-acids. The $\pi$-acid ligand gave in the option is also a greenhouse gas.

Complete step by step solution:
-The ligands that can do $\pi $-back bonding are also called $\pi$-acids. Let’s know something about $\pi $-back bonding in order to find in which ligand it occurs.
-$\pi $-back bonding involves back donation of the $\pi$-electrons. Here, electrons move from metal to the antibonding orbital having appropriate symmetry of the ligand. Only $\pi $-acceptor ligands can accept the electrons from metals.
-However, it has been found that it is a synergic process which involves the first donation of electrons from ligand to empty orbitals of metal. Then metal donates d-orbital electrons to empty $\pi $-orbitals of the ligand.
-Thus, we can say that it is necessary to have a $\pi$-orbital in order to accept electrons from metal in a back donation.
-Ammonia does not have any double bond between nitrogen and hydrogen. So, it is not a $\pi $-acid.
-The structure of ethylene diamine is ${H_2}N - C{H_2} - C{H_2} - N{H_2}$. So, it also does not involve any $\pi $-bond in its structure. Thus, it can also not act as a $\pi $-acid.
-The ligand ${C_2}{O_4}^{ - 2}$ do not have any empty $\pi $-antibonding orbital. So, it cannot accept electrons from metal as a back donation. So, it can also not act as a $\pi $-acid.
-Carbon monoxide $(CO)$ is a ligand which has an empty $\pi$-antibonding orbitals and so it can accept electrons from metal as a back donation. So, $CO$ can act as a $\pi $-acid.

Thus, the correct answer is (A).

Note: Remember that having $\pi $-orbitals is not just a criterion for the ligand to be considered as a $\pi $-acid. Actually, they should have an empty $\pi $-antibonding orbital that can accept electrons from the metal.