The magnetic moment of Fe^{2+} is B.M. | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

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

Vedclass · Accepted Answer

$4.9$

Vedclass · Answer

(C) The magnetic moment $\mu$ is calculated using the formula $\mu = \sqrt{n(n + 2)} 	ext{ B.M. }$,where $n$ is the number of unpaired electrons.The atomic number of $Fe$ is $26$,and its electronic configuration is $[Ar] 3d^6 4s^2$.For $Fe^{2+}$,the configuration is $[Ar] 3d^6$.In the $3d$ subshell,there are $5$ orbitals. According to Hund's rule,the $6$ electrons occupy these orbitals as follows: one orbital has $2$ electrons (paired),and the remaining $4$ orbitals have $1$ electron each (unpaired).Thus,the number of unpaired electrons $n = 4$.Substituting $n = 4$ into the formula: $\mu = \sqrt{4(4 + 2)} = \sqrt{24} \approx 4.9 	ext{ B.M. }$

The magnetic moment of $Fe^{2+}$ is $B.M$.

Similar Questions

The number of geometrical isomers for octahedral $[CoCl_4(NH_3)_2]^-$,square planar $[AuBr_2Cl_2]^-$ and $[PtCl_2(en)]$ are

How many geometrical isomers are shown by the complex $[Pt(NH_3)(NH_2OH)(H_2O)(Py)]^+$?

The number of possible geometric isomers for a square planar complex $[Mabcd]$ is ............. ($M =$ central metal,$a, b, c, d =$ monodentate ligands).

The ratio of spin-only magnetic moment values $\mu_{eff}[Cr(CN)_6]^{3-} / \mu_{eff}[Cr(H_2O)_6]^{3+}$ is $.........$.

The spin magnetic moment of cobalt in the compound $Hg[Co(SCN)_4]$ is (in $B.M.$)

Vedclass Test Series

Exam Paper Generator

Online Exam Module