Coordination Compounds 2 Question 3
3. The complex ion that will lose its crystal field stabilisation energy upon oxidation of its metal to $+3$ state is
Ignore pairing energy
(a) $\left[Co(\text { phen }) _3\right]^{2+}$
(b) $\left[Ni(\text { phen }) _3\right]^{2+}$
(c) $\left[\operatorname{Zn}(\text { phen }) _3\right]^{2+}$
(d) $\left[Fe(\text { phen }) _3\right]^{2+}$
(2019 Main, 12 April I)
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Answer:
Correct Answer: 3. (d)
Solution:
- Key Idea: Crystal field splitting occurs due to the presence of ligands in a definite geometry. In octahedral complexes the energy of two, $e _g$ orbitals will increase by $(0.6) \Delta _o$ and that of three $t _{2 g}$ will decrease by $(0.4) \Delta _0$.
The complex ion that will lose its crystal field stabilisation energy upon oxidation of its metal to $+3$ state is $\left[Fe(\text { phen }) _3\right]^{2+}$.
$ \left[Fe(\text { phen }) _3\right]^{2+} \xrightarrow{-e^{-}}\left[Fe(\text { phen }) _3\right]^{3+} $
In $\left[Fe(\text { phen }) _3\right]^{2+}$, electronic configuration of $Fe^{2+}$ is $3 d^{6} 4 s^{0}$. Phenanthrene is a strong field symmetrical bidentate ligand. The splitting of orbital in $Fe^{2+}$ is as follows:
CFSE $=6 \times-0.4 \Delta _o=-2.4 \Delta _o$.
The splitting of orbital and arrangement of electrons in $Fe^{3+}$ is as follows :
CFSE $=5 \times-0.4 \Delta _o=-2.0 \Delta _o$
$Fe^{2+}$ upon oxidation of its metal to $+3$ state lose its CFSE from $-2.4 \Delta _o$ to $-2.0 \Delta _o$.
BETA
