2(b) (i) Why transition metal complexes have higher measured lattice energy as compared to the normal metals and explain the reason for the hump. [2+1] (ii) If the CFSE of [Co(H2O)6]2+ is -0.8 ?o, what spin state is it in?
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These complexes involve the central metal ion and the surrounding ligands, which can be ions or molecules. The formation of these complexes involves the d-orbitals of the transition metal, which can accommodate more electrons than the s-orbitals of normal metals. Show more…
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The Crystal Field Stabilization Energy (CFSE) of $\left[\mathrm{CoF}_{3}\left(\mathrm{H}_{2} \mathrm{O}\right)_{3}\right]\left(\Delta_{0}<\mathrm{P}\right)$ is : (a) $-0.8 \Delta_{0}+2 \mathrm{P}$ (b) $-0.4 \Delta_{0}$ (c) $-0.8 \Delta_{0}$ (d) $-0.4 \Delta_{0}+\mathrm{P}$
Co-ordination Compounds
Topic 2 : Bonding, Stability and Application of Coordination Compounds
Which one of the following high spin complexes has the largest CFSE (Crystal Field Stabilization Energy)? (a) $\left[\mathrm{Cr}\left(\mathrm{H}_{2} \mathrm{O}\right)_{6}\right]^{2^{+}}$ (b) $\left[\mathrm{Cr}\left(\mathrm{H}_{2} \mathrm{O}\right)_{6}\right]^{3+}$ (c) $\left[\mathrm{Mn}\left(\mathrm{H}_{2} \mathrm{O}\right)\right]^{2+}$ (d) $\left[\mathrm{Mn}\left(\mathrm{H}_{2} \mathrm{O}\right)_{6}\right]^{3^{+}}$
Generally speaking, for a given metal and ligand, the stability of a coordination compound is greater for the metal in the $3+$ rather than in the $2+$ oxidation state. Furthermore, for a given ligand the complexes of the bivalent metal ions of the first transition series tend to increase in stability in the order $\mathrm{Mn}(\mathrm{II})<\mathrm{Fe}(\mathrm{II})<\mathrm{Co}(\mathrm{II})<$ Ni(II) < Cu(II). Explain how these two observations are consistent with one another and also consistent with a crystal-field picture of coordination compounds.
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