Question

The spin-only magnetic moment value of the
compound with strongest oxidizing ability among
$Mn{F}_4,Mn{F}_3$ and $Mn{F}_2$ is B.M. [nearest integer]

• A. (05.00)
• B. 5.0
• C. 5
• D. 5.00

Answer 1

Answer: (A), (B), (C), (D)

$Mn{F}_3$ has the strongest oxidising ability
$\left[\begin{array}{c}{E}_{M{n}^{3+}/M{n}^{2+}}^0\simeq 1.57V\\ \&E_{M{n}^{4+}/M{n}^{2+}}^0\simeq 1.2V\end{array}\right]$
In $M{n}^{3+}$ the configuration will be $3d^4$, i.e. 4 unpaired electrons.
So, spin only magnetic moment
$=\sqrt{4(4+2)}=\sqrt{24}B\cdot M\cdot$
$\simeq 5$