JEE Main · 2022 · 27 Jun 2022 · Shift 2 · Coordination Compounds

For each complex, determine the number of unpaired electrons (n):
(A) $[{\mathrm{FeF}}_6{]}^{3-}$: ${\mathrm{Fe}}^{3+}$ is $3{\mathrm{d}}^5$, F⁻ is weak field, high spin → n=5
(B) $[\mathrm{Fe}(\mathrm{CN}{)}_6{]}^{3-}$: ${\mathrm{Fe}}^{3+}$ is $3{\mathrm{d}}^5$, CN⁻ is strong field, low spin → n=1
(C) $[{\mathrm{MnCl}}_6{]}^{3-}$: ${\mathrm{Mn}}^{3+}$ is $3{\mathrm{d}}^4$, Cl⁻ is weak field, high spin → n=4
(D) $[\mathrm{Mn}(\mathrm{CN}{)}_6{]}^{3-}$: ${\mathrm{Mn}}^{3+}$ is $3{\mathrm{d}}^4$, CN⁻ is strong field, low spin → n=2
Magnetic moment $\mu \propto \sqrt{n(n+2)}$, so the order is B < D < C < A.