JEE Advanced · 2018 · 20 May 2018 · Paper 2 · Coordination Compounds

The complex ${\left[\mathrm{Co}(\mathrm{en})({\mathrm{NH}}_3{)}_3({\mathrm{H}}_2\mathrm{O})\right]}^{3+}$ has two geometrical isomers (cis and trans). When bidentate 'en' is replaced by two cyanide ligands, ${\left[\mathrm{Co}({\mathrm{NH}}_3{)}_3(\mathrm{CN}{)}_2({\mathrm{H}}_2\mathrm{O})\right]}^{+}$ has three geometrical isomers. Since en and ${\mathrm{NH}}_3$ are strong field ligands, the complex is diamagnetic (not paramagnetic). ${\mathrm{NH}}_3$ is a stronger ligand than ${\mathrm{H}}_2\mathrm{O}$, so ${\left[\mathrm{Co}(\mathrm{en})({\mathrm{NH}}_3{)}_4\right]}^{3+}$ has a larger crystal field splitting and absorbs shorter wavelength than ${\left[\mathrm{Co}(\mathrm{en})({\mathrm{NH}}_3{)}_3({\mathrm{H}}_2\mathrm{O})\right]}^{3+}$, meaning the latter absorbs at longer wavelength.