Synergic Bonding in Metal-Carbonyl Complexes: 17 MCQs with answers

A two-way electron exchange between metal and ligand

Synergic bonding involves σ donation from metal to ligand and π back-donation from metal d-orbitals to ligand π* orbitals, so it is a two-way exchange.

Metal M to the carbon of CO

The σ bond arrow goes from M to C, meaning the metal donates σ electron density to the carbon of CO.

π 2px* and π 2py* antibonding orbitals

The diagram shows π back-bonds from metal d-orbitals into the π 2px* and π 2py* antibonding orbitals of CO.

CO, CN⁻, and NO

The source names CO, CN⁻, and NO as the classic synergic ligands because they have accessible π* orbitals.

π back-donation from the metal to CO

The arrows from metal d-orbitals into the π* orbitals of CO represent π back-donation.

True

Synergic bonding is defined by the mutual reinforcement of σ donation from metal to ligand and π back-donation from metal d-orbitals to ligand π* orbitals.

CO also accepts π electron density through its π* orbitals

Ignoring the π back-bond is a common mistake. CO acts as both a σ donor and a π acceptor in synergic bonding.

It becomes more electron-rich and a better π acceptor

σ donation makes the carbon more electron-rich, which strengthens its ability to accept π back-donation.

The metal becomes more electron-poor, so it donates σ density more readily

Back-donation removes electron density from the metal, making it more electron-poor and a stronger σ donor.

It weakens because π* antibonding orbitals fill with electron density

Electron density entering the π* antibonding orbitals reduces the C≡O bond order, weakening the bond.

They have completely filled d-orbitals

These ligands do not have d-orbitals; their key feature is having π* orbitals to accept electron density and lone pairs to donate.

They have partially filled d-orbitals that can donate π density

Transition metals have partially filled d-orbitals that enable π back-donation, making them central to synergic bonding.

π back-donation has increased, weakening the C≡O bond

A lower IR stretching frequency means a weaker C≡O bond, which results from increased π back-donation into the π* orbitals.

Both A and R are true, and R correctly explains A.

The mutual reinforcement of σ donation and π back-donation is exactly why the metal–CO bond is stronger than either interaction alone.

I and III only

Statement II is wrong because π back-donation goes from metal d-orbitals to ligand π* orbitals, not the reverse. Statements I and III are correct.

σ donation and π back-donation reinforce each other, strengthening the bond

The synergic effect—mutual reinforcement of σ donation and π back-donation—explains the stability of metal carbonyls.

It would prevent π back-donation, weakening the overall bond.

The correct option is B. If the metal's d-orbitals are fully occupied, there would be no available d-electrons to participate in π back-donation to the π* orbitals of the ligands. This lack of back-donation would weaken the overall synergic bond, as the mutual reinforcement of σ donation and π back-donation is essentia…