After much heated discussion on a hazy Wednesday afternoon at North Spine MacDonald,our two Chemistry gods, Nathanal Wong and Kang Jing have suggested that we should approach this question by looking at how the ligands approach the central metal ion. The other 3  immortals namely Sophia, Amber and me, Alenson, agreed unanimously. The axis in which the ligands approach will encounter greater electrostatic forces of repulsion between the electrons in the respective d orbital of the metal ion and the lone pair on the ligands. As a result, those d- orbital which bisects the angle of approach will encounter greater electrostatic forces of attraction and will be of higher energy level than d-orbitals which are perpendicular to the angle of approach.

For example in the case of linear geometry where the electrons approach from 1 and 6, which is the z axis, dz2 orbitals encounter greatest electrostatic forces of attraction and hence has the highest energy level since the ligands approach directly at the z axis, in which the dz2 lies on.

The same applies for dxz and dyz orbital. However, unlike dz2 orbitals, these two orbitals do not lie entirely on the z-axis, but only partially on the z plane. Hence they will encounter lesser electrostatic forces of repulsion than dz2 but greater than dxy and d x2-y2 orbitals which are perpendicular away from the z axis.

That’s all folks =) Have a great week ahead!

Love Chemistry!

Date: 4th September 2015

Location: Intense discussion over whatsapp

Members present: 5/5 full attendance

This question gave our team a headache. The first challenge was to figure out the structure of the azide ion. We were unsure if it was single bond, double bond, or triple bond. Upon some heated discussion and research on the internet, we decided on it being double bond. We also understood that the confusion was due to the ion having several resonance forms.

The next challenge was to justify the presence of two lone pairs on each terminal N atom despite it being only sp hybridised. Eventually, we came to a conclusion that one lone pair resides in the sp orbital, while the other lone pair resides in the unhybridised p orbital, thereby giving rise to a linear geometry.

The final challenge was to determine which atom has the greatest electron density. We were pretty sure that it was the terminal N atom, however was struggling to find a solid explanation for it. Upon discussion, we have various inputs, such as considering the HOMO of the compound, as well as the electron distortion due to the formal charges.

Date: 25th August 2015 / 28th August 2015

Location: During the group discussion in tutorial class followed by final discussion on Whatsapp

Members present: 5/5 full attendance

This question was more straightforward and we started off by drawing out the MO diagram for all 3 compounds, taking into consideration s-p mixing for N and C. Once the MO diagram was done, we could then calculate the bond order and derive from there the expected bond strength and bond length.

Date: 19th August 2015

Location: Whatsapp discussion due to clashing of schedules to arrange a physical meetup

Members present: 5/5 full attendance

We brainstormed different approach to the question.

1) The electron was promoted from the 1s antibonding MO to the 2s bonding MO after the collision.

2) Inelastic collision between the two atoms and the kinetic energy loss is being absorbed by an electron to promote to higher energy level

However, after further discussion, we agreed that one helium atom was excited before the collision, and thus giving rise to a molecule with a bond order of 1 after working out a solution.