Q. The complex ion which has the highest magnetic moment among the following is
Solution:
(i) In $ {{[Co{{F}_{6}}]}^{3-}} $ ion, the oxidation state of cobalt is +3. $ C{{o}^{3+}} $ ion
Formation of $ {{[Co{{F}_{6}}]}^{3+}} $
Here, $ {{F}^{-}} $ ion provides a weak ligand field and is unable to pair up the unpaired electrons of the 3d-orbitals. Thus, it is highly paramagnetic. (ii) In $ {{[Co{{(N{{H}_{3}})}_{6}}]}^{3+}} $ ion, $ N{{H}_{3}} $ provides a strong field ligand and the electrons of metal are made to pair up, so the complex will be diamagnetic. Formation of $ {{[Co{{(N{{H}_{3}})}_{6}}]}^{3+}} $
(iii) Similarly in $ {{[Ni{{(N{{H}_{3}})}_{4}}]}^{2+}},N{{H}_{3}} $ is a strong field ligand and the complex is diamagnetic. $ N{{i}^{2+}} $ ion:
Formation of $ {{[Ni{{(N{{H}_{3}})}_{4}}]}^{2+}} $
(iv) $ {{[Ni{{(CN)}_{4}}]}^{2-}} $ has Ni in the + 2 oxidation state and $ ds{{p}^{2}} $ hybridisation. Thus it is also diamagnetic. (v) In $ {{[Fe{{(CN)}_{6}}]}^{4-}} $ ions, oxidation state of $ Fe $ is +2 and $ C{{N}^{-}} $ ligand is a strong field ligand. Thus the resulting complex ion involves $ {{d}^{2}}s{{p}^{3}} $ hybridisation and diamagnetic as it does not contain any unpaired electrons. $ F{{e}^{2+}} $ ion:
Formation of $ {{[Fe{{(CN)}_{6}}]}^{4+}} $
$ \because $ Magnetic moment $ =\sqrt{n(n+2)} $ where, $ n= $ no. of unpaired electrons $ \therefore $ $ {{[Co{{F}_{6}}]}^{3-}} $ has maximum magnetic moment.
