Identify the homoleptic complex(es) that is/are low spin.
Solution
<p>(A) $[ \text{Fe(CN)}_5 \text{NO} ]^{2-}$: This is a heteroleptic complex with iron in the $\text{Fe}^{2+}$ oxidation state, resulting in a $3d^6$ electronic configuration. It forms a low spin complex with electron configuration $\text{t}_{2g}^6 \text{e}_g^0$ due to the strong field ligands (SFL).</p>
<p>(B) $[ \text{CoF}_6 ]^{3-}$: This homoleptic complex contains cobalt in the $\text{Co}^{3+}$ oxidation state, leading to a $3d^6$ electron configuration. It is a high spin complex with an $\text{sp}^3 \text{d}^2$ hybridization due to weak field ligands (WFL).</p>
<p>(C) $[ \text{Fe(CN)}_6 ]^{4-}$: This homoleptic complex has iron in the $\text{Fe}^{2+}$ state, giving a $3d^6$ configuration. It forms a low spin complex with hybridization $\text{d}^2 \text{sp}^3$ and electron configuration $\text{t}_{2g}^6 \text{e}_g^0$, due to the strong field nature of the cyanide ligands.</p>
<p>(D) $[ \text{Co(NH}_3 )_6 ]^{3+}$: This homoleptic complex contains cobalt in the $\text{Co}^{3+}$ state, resulting in a $3d^6$ electron configuration. It forms a low spin complex with hybridization $\text{d}^2 \text{sp}^3$ and configuration $\text{t}_{2g}^6 \text{e}_g^0$ due to the strong field ligands.</p>
<p>(E) $[ \text{Cr(H}_2\text{O})_6 ]^{2+}$: This homoleptic complex with chromium in the $\text{Cr}^{2+}$ state results in a $3d^4$ electronic configuration. It is a high spin complex with hybridization $\text{d}^2 \text{sp}^3$ and configuration $\text{t}_{2g}^3 \text{e}_g^1$ due to weak field ligands. </p>
<p>In summary, the homoleptic complexes that are low spin are:</p>
<p><p>(C) $[ \text{Fe(CN)}_6 ]^{4-}$</p></p>
<p><p>(D) $[ \text{Co(NH}_3 )_6 ]^{3+}$</p></p>
About this question
Subject: Chemistry · Chapter: Coordination Compounds · Topic: Ligands and Coordination Number
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