Question

The crystal field stabilisation energy $(CFSE)$ and the spin-only magnetic moment in Bohr Magneton $(BM)$ for the complex $K_3[Fe(CN{)}_6]$ are, respectively

• A. $0.0{\Delta }_0$ and$\sqrt{35}BM$
• B. $-2.0{\Delta }_0$ and$\sqrt{3}BM$
• C. $-0.4{\Delta }_0$ and$\sqrt{24}BM$
• D. $-2.4{\Delta }_0$ and $0BM$

Answer 1

Answer: (A)

The oxidation state of $Fe$ in $K_3[Fe(CN{)}_6]$ is $+3$.
Thus, the electronic configuration for $F{e}^{3+}$ will be $[Ar]3d^{5}4s^0$

$CFSE=-(0.4\times t_{2}g{e}^{-})+0.6\times e_g{e}^{-}$
$=-0.4\times 3+0.6\times 2$
$=-1.2+1.2=0.0{\Delta }_0$
Magnetic moment, $\mu =\sqrt{n\left(n+2\right)}$
In this case, $n=5$
$\mu =\sqrt{5\left(5+2\right)}=\sqrt{35}BM$