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(C) The magnetic moment $(\mu)$ is calculated using the formula $\mu = \sqrt{n(n+2)} 	ext{ BM}$,where $n$ is the number of unpaired electrons.$1$. Fo

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$Fe^{3+}$

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(C) The magnetic moment $(\mu)$ is calculated using the formula $\mu = \sqrt{n(n+2)} 	ext{ BM}$,where $n$ is the number of unpaired electrons.$1$. For $Mn^{6+}$ $([Ar]3d^1)$: $n = 1$,$\mu = \sqrt{1(1+2)} = 1.73 	ext{ BM}$.$2$. For $Ni^{2+}$ $([Ar]3d^8)$: $n = 2$,$\mu = \sqrt{2(2+2)} = 2.83 	ext{ BM}$.$3$. For $Fe^{3+}$ $([Ar]3d^5)$: $n = 5$,$\mu = \sqrt{5(5+2)} = 5.91 	ext{ BM}$.$4$. For $Ag^{+}$ $([Kr]4d^{10})$: $n = 0$,$\mu = 0 	ext{ BM}$.Thus,$Fe^{3+}$ has the maximum number of unpaired electrons $(n=5)$ and consequently shows the maximum magnetic moment.

Which of the following species shows the maximum magnetic moment?

Similar Questions

Which of the following compounds are colourless? ($TiF_6^{2-}, CoF_6^{3-}, Cu_2Cl_2$ and $NiCl_4^{2-}$) (Atomic numbers: $Ti = 22, Co = 27, Cu = 29, Ni = 28$)

Which among the following elements in their respective oxidation states develops the lowest spin-only magnetic moment? (Atomic numbers: $Cu=29, Fe=26, Ni=28, Co=27$)

In which of the following orders are the given complex ions arranged correctly with respect to their decreasing spin-only magnetic moment?
$(i)$ $[FeF_{6}]^{3-}$
$(ii)$ $[Co(NH_{3})_{6}]^{3+}$
$(iii)$ $[NiCl_{4}]^{2-}$
$(iv)$ $[Cu(NH_{3})_{4}]^{2+}$

$X$ is the number of geometrical isomers exhibited by $[Pt(NH_3)(H_2O)BrCl]$.
$Y$ is the number of optically inactive isomer$(s)$ exhibited by $[CrCl_2(ox)_2]^{3-}$.
$Z$ is the number of geometrical isomers exhibited by $[Co(NH_3)_3(NO_2)_3]$.
The value of $X+Y+Z$ is . . . . . . .

Which of the following complexes will not show optical isomerism?

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