Question

The complex ion which has the highest magnetic moment among the following is

• A. ${[Co{F}_6]}^{3-}$
• B. ${[Co{(N{H}_3)}_6]}^{3+}$
• C. ${[Ni{(N{H}_3)}_4]}^{2+}$
• D. ${[Ni{(CN)}_4]}^{2-}$
• E. ${[Fe{(CN)}_6]}^{4-}$

Answer 1

Answer: (A)

(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.