Question

Spin only magnetic moment of $[MnB{r}_6{]}^{4–}$is _____B.M. [round off to the closest integer]

• A. 6
• B. 6.00
• C. 6.0

Answer 1

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

$[MnB{r}_6{]}^{4-}$
$x–6=–4$
$x=+2$
Since, $B{r}^{-}$ is a weak field ligand, hence it will be a high spin complex.
$\therefore$ configuration of central metal atom $Mn$ will be , $t_{2g}^3{e}_g^2$
Hence, no. of unpaired $e^{-}$ will be $n=5$

$\mu =\sqrt{n(n+2)}$
$=\sqrt{35}\approx 6\text{B.M.}$