Question

Why is ${\left[\mathrm{Cr}{\left({\mathrm{NH}}_3\right)}_6\right]}^{3+}$ a high spin complex?

Answer 1

Crystal field theory

• According to the crystal field theory, the complex can be classified as high spin or low spin.
• The high spin complex is the compound with the greatest number of unpaired electrons.
• There are fewer unpaired electrons in the low spin complex.
• High spin complexes have low crystal field splitting energy
• Low spin complexes have high crystal field splitting energy.

${\left[\mathrm{Cr}{\left({\mathrm{NH}}_3\right)}_6\right]}^{\mathbf{3}\mathbf{+}}$ complex

• In the ${\left[\mathrm{Cr}{\left({\mathrm{NH}}_3\right)}_6\right]}^{3+}$ complex, the center metal atom chromium $\left(\mathrm{Cr}\right)$ is in $+3$ oxidation state.
• The electronic configuration is as follows:

${\mathrm{Cr}}^{3+}=\left[\mathrm{Ar}\right]3d^3$

• Only three of the five $3d$ orbitals are filled, leaving the remaining two for $4s$ or $4p$ orbitals.
• Ammonia $\left({\mathrm{NH}}_3\right)$ is a strong field ligand.
• Three $3d$ orbitals of a metal ion have three unpaired electrons, and all of these orbitals have the same energy.
• Thus, even in the presence of strong ligand, pairing will not occur.
• These $d^{2}s{p}^3$ hybrids form bonds with ligands ammonia $\left({\mathrm{NH}}_3\right)$ by accepting six lone pairs of electrons, one for each of the six ligands.

• As a result, ${\left[\mathrm{Cr}{\left({\mathrm{NH}}_3\right)}_6\right]}^{3+}$ is considered a high spin complex.