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# Why is ${\left[\mathrm{Cr}{\left({\mathrm{NH}}_{3}\right)}_{6}\right]}^{3+}$ a high spin complex?

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## Crystal field theoryAccording 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 energyLow spin complexes have high crystal field splitting energy.${\left[\mathrm{Cr}{\left({\mathrm{NH}}_{3}\right)}_{6}\right]}^{\mathbf{3}\mathbf{+}}$ complexIn 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:${\mathrm{Cr}}^{3+}=\left[\mathrm{Ar}\right]3{d}^{3}$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.

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