Question

What is the formula for hexafluorocobaltate(III)? Is it high spin or low spin?

Answer 1

Hint: We know that transition metal compounds are also known as coordination compounds. Their nomenclature is done based on the rules given by IUPAC (International Union of Pure and Applied Chemistry). The nature of the bonding between the central metal atoms or ions and ligands in the coordination complex is explained by some theories known as Valence bond theory, Crystal field theory and Molecular orbital theory.

Complete answer:
In the IUPAC name of the given coordination complex, the "-ate" at the end indicates a negative overall charge to the complex. "Hexafluoro" represents six fluoro $ (F - ) $ ligands, and the III written after the name of the metal atom indicates the oxidation state of cobalt has a magnitude of 3.
Therefore, by balancing and comparing the charge, we can find out the charge on the coordination sphere as follows:
 $ 6( - 1) + 3 = x $
 $ x = - 3 $
Therefore, the formula o the given complex is:
 $ {[Co{F_6}]^{3 - }} $
Therefore, the oxidation state of cobalt in this complex is $ + 3 $ .
The electronic configuration of $ C{o^{3 + }} $ is:
 $ 1{s^2}2{s^2}2{p^6}3{s^2}3{p^6}3{d^6} $
We know that $ {F^ - } $ is a weak-field ligand (because it is a $ \pi $ donor), the crystal field splitting energy is small, and we say that it is an (octahedral) high spin complex.

Note:
The coordination compound is either paramagnetic or diamagnetic in nature.
Paramagnetic complex: It is a complex in which an unpaired electron is present is called the paramagnetic complex. Diamagnetic complex: It is a complex in which no unpaired electron is present is called diamagnetic complex in nature. The above given complex is paramagnetic in nature because electronic configuration of cobalt ion is $ 1{s^2}2{s^2}2{p^6}3{s^2}3{p^6}3{d^6} $ .