Question

Is the coordination compound $\mathbf{\left[}\mathbf{\text{Co}}{\mathbf{\left(}{\mathbf{\text{NH}}}_{\mathbf{\text{3}}}\mathbf{\right)}}_{\mathbf{\text{6}}}\mathbf{\right]}{\mathbf{\text{Cl}}}_{\mathbf{\text{2}}}$diamagneticor paramagnetic?

Short answer

The coordination complex $\left[\text{Co}{\left({\text{NH}}_{\text{3}}\right)}_{\text{6}}\right]{\text{Cl}}_{\text{2}}$ is paramagnetic.

Step-by-step solution

Magnetic property.

Coordination compounds have magnetic properties because of the presence of unpaired electrons.

Diamagnetism and Paramagnetism. 

The coordination compounds having no unpaired electron are called diamagnetic.

The coordination compounds having unpaired electrons are called paramagnetic.

The given coordination compound. 

The compound $\left[\text{Co}{\left({\text{NH}}_{\text{3}}\right)}_{\text{6}}\right]{\text{Cl}}_{\text{2}}$ has the central metal atom cobalt, which is present in the +2 oxidation state having the seven electrons in the d orbital:

${\text{CO}}^{\text{+2}}{\text{\hspace{0.17em}=\hspace{0.17em}d}}^{\text{7}}$

The cobalt metal in the +2 oxidation state has a weak field ammine ligand, which will cause the low splitting of d orbitals and form a high spin complex; that is, there are five electrons in ${\text{t}}_{\text{2g}}$orbital and two electrons in ${\text{e}}_{\text{g}}$orbitals.

Therefore, three unpaired electrons making the given coordination complex data-custom-editor="chemistry" $\left[\text{Co}{\left({\text{NH}}_{\text{3}}\right)}_{\text{6}}\right]{\text{Cl}}_{\text{2}}$paramagnetic.