Coordination Compounds 2 Question 4
####4. The crystal field stabilisation energy (CFSE) of $\left[Fe\left(H _2 O\right) _6\right] Cl _2$ and $K _2\left[NiCl _4\right]$, respectively, are
(2019 Main, 10 April II)
(a) $-0.4 \Delta _o$ and $-1.2 \Delta _t$
(b) $-0.4 \Delta _o$ and $-0.8 \Delta _t$
(c) $-2.4 \Delta _o$ and $-1.2 \Delta _t$
(d) $-0.6 \Delta _o$ and $-0.8 \Delta _t$
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Solution:
Key Idea Crystal field stabilisation energy (CFSE) for octahedral complexes $=(-0.4 x+0.6 y) \Delta _o$
where, $x=$ number of electrons occupying $t _{2 g}$ orbital.
$y=$ number of electrons occupying $e _g$ orbital.
CFSE for tetrahedral complexes
$=(-0.6 x+0.4 y) \Delta _t$
where, $x=$ number of electrons occupying $e$ orbital.
$y=$ number of electrons occupying $t$ orbital.
In $\left[Fe\left(H _2 O\right) _6\right] Cl _2, H _2 O$ is a weak field ligand, so it is a high spin (outer orbital) octahedral complex of $Fe^{2+}$.
$$ Fe^{2+}\left(3 d^{6}\right)=\begin{array}{|l|l|l|l|} \hline 1 & 1 & e _g \ \hline 1 L & 1 & 1 & t _{2 g} \end{array} $$
$\therefore CFSE=(-0.4 x+0.6 y) \Delta _o$
$$ =[-0.4 \times 4+0.6 \times 2] \Delta _o=-0.4 \Delta _o $$
In $K _2\left[NiCl _4\right], Cl^{-}$is a weak field ligand, so it is a high spin tetrahedral complex of $Ni^{2+}$.
$\therefore CFSE=(-0.6 \times 4+0.4 \times 4) \Delta _t=-0.8 \Delta _t$
