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) (NH₄)₂[Ce(NO₃)₆]
(b) Gd(NO₃)₃ and
(c) Eu(NO₃)₃

Solution

Chemistry ma’am · 3 steps

Find each metal's oxidation state first, then count its 4f electrons, because only unpaired ones count.

1. Oxidation states. In (NH₄)₂[Ce(NO₃)₆], two NH₄⁺ give +2 and six NO₃⁻ give -6, so cerium is +4. In both nitrates, three NO₃⁻ make the metal +3.
2. Count f electrons. Ce is [Xe]4f¹ 5d¹ 6s², so Ce⁴⁺ is 4f⁰. Gd is [Xe]4f⁷ 5d¹ 6s², so Gd³⁺ is 4f⁷. Eu is [Xe]4f⁷ 6s², so Eu³⁺ is 4f⁶.

   μ = √n(n+2) BM

   Compound	Ion	Unpaired electrons n	μ (BM)
   (a) (NH₄)₂[Ce(NO₃)₆]	Ce⁴⁺, 4f⁰	0	0
   (c) Eu(NO₃)₃	Eu³⁺, 4f⁶	6	√48 = 6.93
   (b) Gd(NO₃)₃	Gd³⁺, 4f⁷	7	√63 = 7.94

3. So. The increasing order is (a) < (c) < (b), which is option (c).

The idea to keep. For lanthanides, remove the 6s and 5d electrons first, then the 4f. A half-filled 4f⁷ ion such as Gd³⁺ has the most unpaired electrons and the largest spin-only moment.

Answer from the NTA final answer key. Solution by PrepWiser.