The complex ion that will lose its crystal fi | vedclass

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Question

(D) The Crystal Field Stabilization Energy $(CFSE)$ for an octahedral complex is given by $CFSE = (-0.4n_{t2g} + 0.6n_{eg})\Delta_o$. For $[Fe(phen)_3

Vedclass · Accepted Answer

$[Fe(phen)_3]^{2+}$

Vedclass · Answer

(D) The Crystal Field Stabilization Energy $(CFSE)$ for an octahedral complex is given by $CFSE = (-0.4n_{t2g} + 0.6n_{eg})\Delta_o$. For $[Fe(phen)_3]^{2+}$,the central metal is $Fe^{2+}$ ($d^6$ configuration). Since $phen$ is a strong field ligand,it forms a low-spin complex with $t_{2g}^6 e_g^0$ configuration. $CFSE = (-0.4 	imes 6 + 0.6 	imes 0)\Delta_o = -2.4\Delta_o$. Upon oxidation to $Fe^{3+}$ ($d^5$ configuration),the complex becomes $[Fe(phen)_3]^{3+}$ with $t_{2g}^5 e_g^0$ configuration. $CFSE = (-0.4 	imes 5 + 0.6 	imes 0)\Delta_o = -2.0\Delta_o$. However,the question implies a significant loss or change in stability. Let's check $[Co(phen)_3]^{2+}$ $(d^7)$: $t_{2g}^6 e_g^1$ $(CFSE = -1.8\Delta_o)$. Oxidation to $Co^{3+}$ $(d^6)$: $t_{2g}^6 e_g^0$ $(CFSE = -2.4\Delta_o)$. Actually,for $Fe^{2+}$ $(d^6)$ to $Fe^{3+}$ $(d^5)$,the loss of an electron from the $t_{2g}$ orbital significantly reduces the stabilization energy. Thus,$[Fe(phen)_3]^{2+}$ is the correct answer.

The complex ion that will lose its crystal field stabilization energy upon oxidation of its metal to $+3$ state is

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