Question

Predict whether \([CoF_6]^{3-}\) is diamagnetic or paramagnetic and why?
\([Atomic \ \ number : Co = 27]\)

Hint:

A complex is paramagnetic if it contains unpaired electrons in its metal's d-orbitals. If all electrons are paired, it is diamagnetic.

Answer 1

Paramagnetic or Diamagnetic Nature of \([CoF_6]^{3-}\):
To predict whether \([CoF_6]^{3-}\) is paramagnetic or diamagnetic, let's look at the electronic configuration of \(Co^{3+}\). The electronic configuration of \(Co\) is \([Ar]\, 3d^7\, 4s^2\). For \(Co^{3+}\), the electrons are removed from the 4s and 3d orbitals, leaving the configuration \(3d^6\). In the case of \([CoF_6]^{3-}\), fluorine is a weak field ligand and does not cause pairing of the electrons. Therefore, the 6 electrons in the \(3d\) orbitals remain unpaired, making \([CoF_6]^{3-}\) paramagnetic.


Step 1: Determine the electron configuration of \(Co^{3+}\)
\[ Co^{3+} : 3d^6 \]

Step 2: The presence of unpaired electrons in the \(3d\) orbitals means that the complex is paramagnetic.