Arrange the following metal complexes/compoun | vedclass

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(D) $_{58}Ce \rightarrow [Xe] 4f^1 5d^1 6s^2$. In the complex,$Ce^{4+} \rightarrow [Xe] 4f^0$. There are $0$ unpaired electrons,so $\mu_m = 0 \ B.M$.$

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$(a) < (c) < (b)$

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(D) $_{58}Ce \rightarrow [Xe] 4f^1 5d^1 6s^2$. In the complex,$Ce^{4+} \rightarrow [Xe] 4f^0$. There are $0$ unpaired electrons,so $\mu_m = 0 \ B.M$.$(b)$ $_{64}Gd^{3+} \rightarrow [Xe] 4f^7$. There are $7$ unpaired electrons,so $\mu_m = \sqrt{7(7+2)} = \sqrt{63} \ B.M$.$(c)$ $_{63}Eu^{3+} \rightarrow [Xe] 4f^6$. There are $6$ unpaired electrons,so $\mu_m = \sqrt{6(6+2)} = \sqrt{48} \ B.M$.Comparing the magnetic moments: $0 Thus,the increasing order is $(a) < (c) < (b)$.

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

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