Question

Calculate the difference in the net stabilization energies (in $kJmo{l}^{-1})$ of $F{e}^{2+}$ octahedral complexes with $C{N}^{-}$ ligands (${\Delta }_0=25000c{m}^{-1})$ and with $H_{2}O$ ligands $({\Delta }_0=10000c{m}^{-1})$. Given that the pairing energy, $E_p=15000c{m}^{-1}$ for $F{e}^{2+}$ ion in both the complexes. (Given: h = $6\times {10}^{-34}$ Jsec, Avogadro's Number = $6\times {10}^{23}$, Speed of light =$3\times {10}^{8}mse{c}^{-1}$)

Answer 1

$\left[Fe\right(CN{)}_6{]}^{4-}$ complex is a low spin complex with configuration $t_{2g}^6{e}_g^0$.
While $\left[Fe\right(H_{2}O{)}_6{]}^{2+}$ complex is a high spin complex with configuration $t_{2g}^4{e}_g^2$
For $\left[Fe\right(CN{)}_6{]}^{4-}$,

CFSE=$\frac{-2}{5}\times 6\times {\Delta }_0+3E_p=(\frac{-2}{5}\times 6\times 25000+3\times 15000)c{m}^{-1}=-15000c{m}^{-1}$
For $\left[Fe\right(H_{2}O{)}_6{]}^{2+}$,
CFSE=$(\frac{-2}{5}\times 4+\frac{3}{5}\times 2)\times {\Delta }_0+15000=(-0.4\times 10000+15000)c{m}^{-1}=11000c{m}^{-1}$
So, difference in CFSE = $11000-(-15000)=26000c{m}^{-1}$
$\Rightarrow E=\frac{hc}{\lambda }=hc\overline{v}=(6\times {10}^{-34})\times (3\times {10}^8)\times (26000\times {10}^2)\times 6\times {10}^{23}\times {10}^{-3}kJ/mol$
= 280.80 kJ/mol