1. Find the oxidation state and electronic configuration of the metal ion in each compound

For lanthanoids, the magnetic moment depends on the number of unpaired electrons in the relevant ion.

(a) $(NH_4)_2[Ce(NO_3)_6]$

Let oxidation state of Ce be $x$.

Each nitrate ligand inside the complex is $NO_3^-$, so: $x + 6(-1) = -2$ $x - 6 = -2$ $x = +4$

So metal ion is $Ce^{4+}$.

Cerium: $Z=58$ Neutral Ce: $[Xe]4f^1 5d^1 6s^2$ Removing 4 electrons gives: $Ce^{4+} = [Xe]4f^0$ So number of unpaired electrons: $n = 0$

(b) $Gd(NO_3)_3$

Each nitrate is $-1$, so Gd is in $+3$ oxidation state: $Gd^{3+}$

Gadolinium: $Z=64$ Neutral Gd: $[Xe]4f^7 5d^1 6s^2$ Removing 3 electrons gives: $Gd^{3+} = [Xe]4f^7$ So number of unpaired electrons: $n = 7$

(c) $Eu(NO_3)_3$

Similarly, Eu is in $+3$ oxidation state: $Eu^{3+}$

Europium: $Z=63$ Neutral Eu: $[Xe]4f^7 6s^2$ Removing 3 electrons gives: $Eu^{3+} = [Xe]4f^6$ So number of unpaired electrons in $4f^6$ is: $n = 6$

2. Use spin-only magnetic moment formula

$\mu = \sqrt{n(n+2)}\; BM$

Now compare:

• For (a): $n=0$ $\mu = 0$
• For (c): $n=6$ $\mu = \sqrt{6(8)} = \sqrt{48}$
• For (b): $n=7$ $\mu = \sqrt{7(9)} = \sqrt{63}$

Thus, $0 < \sqrt{48} < \sqrt{63}$

So increasing order of spin-only magnetic moment is: $(a) < (c) < (b)$

3. Match with options

This corresponds to Option C.

4. Comparison with stored correct answer

Stored correct answer: C

Derived answer: C

So the derived answer agrees with the stored answer.