Arrange the following metal complex/compounds in the increasing order of spin only magnetic moment. Presume all the three, high spin system.
(Atomic numbers Ce = 58, Gd = 64 and Eu = 63.)
(a) $$(NH_4)_2[Ce(NO_3)_6]$$
(b) $$Gd(NO_3)_3$$
(c) $$Eu(NO_3)_3$$

We need to find the number of unpaired electrons in each species to determine the spin-only magnetic moment $$\mu = \sqrt{n(n+2)}$$ BM, where $$n$$ is the number of unpaired electrons.

For $$(NH_4)_2[Ce(NO_3)_6]$$, cerium is in the +4 oxidation state. Cerium has atomic number 58 with electronic configuration $$[Xe]4f^1 5d^1 6s^2$$. Removing 4 electrons gives $$Ce^{4+}: [Xe]4f^0$$, which has $$n = 0$$ unpaired electrons. Therefore $$\mu = 0$$ BM.

For $$Gd(NO_3)_3$$, gadolinium is in the +3 oxidation state. Gadolinium has atomic number 64 with electronic configuration $$[Xe]4f^7 5d^1 6s^2$$. Removing 3 electrons gives $$Gd^{3+}: [Xe]4f^7$$, which has $$n = 7$$ unpaired electrons. Therefore $$\mu = \sqrt{7 \times 9} = \sqrt{63} \approx 7.94$$ BM.

For $$Eu(NO_3)_3$$, europium is in the +3 oxidation state. Europium has atomic number 63 with electronic configuration $$[Xe]4f^7 6s^2$$. Removing 3 electrons gives $$Eu^{3+}: [Xe]4f^6$$, which has $$n = 6$$ unpaired electrons (high spin). Therefore $$\mu = \sqrt{6 \times 8} = \sqrt{48} \approx 6.93$$ BM.

The increasing order of spin-only magnetic moment is: $$(a)$$ $$(NH_4)_2[Ce(NO_3)_6]$$ (0 BM) $$<$$ $$(c)$$ $$Eu(NO_3)_3$$ (6.93 BM) $$<$$ $$(b)$$ $$Gd(NO_3)_3$$ (7.94 BM), which gives the order $$(a) < (c) < (b)$$.