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 $0\,BM$

Answer 1

Answer: (A)

The oxidation state of $Fe$ in $K_3 [Fe(CN)_6]$ is $+3$.
Thus, the electronic configuration for $Fe^{3+}$ will be $[Ar]3d^54s^0$

$CFSE = - (0.4 \times t_2g e^{-} ) + 0.6 \times e_g e^{-}$
$= - 0.4\times3 + 0.6\times 2$
$= -1.2+ 1.2= 0.0\Delta_0$
Magnetic moment, $\mu=\sqrt{n\left(n+2\right)} $
In this case, $n = 5$
$\mu=\sqrt{5\left(5+2\right)}=\sqrt{35}\, BM$