For the octahedral complex ion (Co(NH3)6]3+

For the octahedral complex ion (Co(NH3)6]3+

a) Draw a molecular orbital (MO) energy level diagram.

b) Calculate the number of unparied electrons.

c) Indicate weather it is high spin or low spin complex.

Breifly explain the synergic bond in the Cr(CO)6. Illustrate your answer with drawing.

The Correct Answer and Explanation is :

Let’s break down the two questions with their answers:

1. Octahedral Complex: [Co(NH₃)₆]³⁺

For the complex ion ([Co(NH₃)₆]^{3+}), cobalt is in the +3 oxidation state. The electron configuration of Co³⁺ is ( [Ar] 3d^6 ), since Co is in the transition metals group, and the electrons are removed first from the 4s orbital followed by 3d orbitals.

a) Molecular Orbital (MO) Energy Level Diagram:

The MO diagram for an octahedral complex is typically based on the d-orbitals, which split in energy when placed in a ligand field. In an octahedral crystal field, the d-orbitals split into two sets:

• ( e_g ) (higher energy): ( d_{z^2} ) and ( d_{x^2 – y^2} )
• ( t_{2g} ) (lower energy): ( d_{xy} ), ( d_{xz} ), and ( d_{yz} )

For ( [Co(NH₃)₆]^{3+} ), since the complex is in the +3 oxidation state, it has 6 electrons in the ( 3d^6 ) configuration. The electron filling follows the rule of Hund’s rule (to minimize energy), meaning electrons will occupy the lower energy orbitals first before filling the higher ones.

b) Number of Unpaired Electrons:

In this case, the ( t_{2g} ) orbitals will be filled first, followed by the ( e_g ) orbitals. The electrons will be arranged as follows:

• The ( t_{2g} ) orbitals can hold 6 electrons.
• For the ( 3d^6 ) configuration, there will be 2 electrons in the ( t_{2g} ) orbitals, and 4 electrons will be in the ( e_g ) orbitals.

In a weak-field ligand such as ( NH₃ ), the electrons are less likely to pair up, so we can expect some unpaired electrons. The number of unpaired electrons is 4 (2 unpaired electrons from the ( e_g ) orbitals and 2 from the ( t_{2g} ) orbitals).

c) High Spin or Low Spin Complex:

Since ( NH₃ ) is a weak field ligand, it does not cause strong splitting of the d-orbitals. Therefore, the complex is high spin, meaning there will be more unpaired electrons.

2. Synergic Bond in ( Cr(CO)_6 ):

In the complex ( Cr(CO)_6 ), the chromium is in the zero oxidation state, with an electron configuration of ( [Ar] 3d^5 4s^1 ).

The CO ligand acts as a π-acceptor ligand, meaning it can accept electron density into its empty ( \pi^* ) anti-bonding orbitals. This is in contrast to regular ( \pi )-donor ligands, which donate electron density. The synergic bonding in ( Cr(CO)_6 ) occurs due to the combination of two effects:

• The ( CO ) ligand donates electron density from its lone pair into the metal’s empty ( d )-orbitals, which is a ( \pi )-donor interaction.
• The metal atom simultaneously donates electron density into the empty ( \pi^* )-orbitals of the ( CO ) ligands, which is a ( \pi )-acceptor interaction.

This mutual electron donation and acceptance strengthen the metal-ligand bond, creating a synergic bond.

Drawing:

I’ll now generate the molecular orbital diagram to illustrate the bonding in ( Cr(CO)_6 ).

Here is the molecular orbital diagram illustrating the synergic bonding in the ( Cr(CO)_6 ) complex. It shows the interaction between chromium (Cr) and the CO ligands, with chromium donating electrons into the anti-bonding ( \pi^* ) orbitals of CO, while CO donates electrons into the chromium’s ( d )-orbitals. This mutual electron donation and acceptance strengthen the metal-ligand bond.