Question

What is the formula and Lewis structure of the coordination compound. Indicate the oxidation state and the d electron count of the central atom.

1. K₂ [Cr(CN)₂(NH₃)(O₂)(O)₂]

Answer 1

• Formula of the compound is given already, so I'm not providing that again. The formula of the coordination sphere i.e, the coordination part only is [Cr(CN)₂(NH₃)O₂(O)₂]^2-. Now, there are many ligands in the anionic part, lets calculate the charge occuring due to them:

LIGAND	CHARGE
CN-	-1
NH3	0
O - O	-2, i.e, O22-
O	-2

Now, since there are 2 K⁺ ion thus, charge on the anionic part is -2. Now, total charge occuring due to the ligands in the anionic part = 2x(-1) + 1x0 + 1x(-2) + 2x(-2) = -8

So, the Cr ion has to produce a charge +6 so that overall charge of anionic part becomes (+6 -8) = -2.

So, Cr is in (VI) state.

• We know, the electronic configuration of Cr is: [Ar] 3d⁵ 4s¹

so,Cr(VI) has electronic configuration: [Ar] 3d⁰ 4s⁰

So, d electron count of central atom is zero.

• The lewis structure is given below, I've shown only the coordination bonds with d²sp³ orbitals and p-orbitals of ligands, the bonds between Cr and O^2- however can not be shown in the same diagram because it will then be very much difficult to understand.

The reason why I coloured one orbital of Cr is to show it above the plane.

Due to 6 coordination number the structure is octahedral with a hybridization (according to VSEPR theory) d²sp³.