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Explain the following about the complexes [Fe(CN)_(6)]^(4-) and [Fe(H_(2)O)_(6)]^(2+): (i) What type of shapes do they have ? (ii) What type of hybridisation is involved in each case ? (iii) Which of them is outer orbital complex and which one is inner orbital complex ? (iv) Which of them is low spin complex and which one is high spin complex ? (v) Compare their magnetic behaviour. (vi) Why do they have different colours in dilute solution ? (vii) Write the electronic configuration of metal ion in each case in terms of t_(2g) and e_(g) orbitals. |
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Answer» Solution :(i) Both have octahedral shape. (ii) In `[Fe(CN)_(6)]^(4-)`, `CN^(-)` is a strong ligand. 3d ELECTRONS PAIR up. Hybridisation is `d^(2)sp^(3)`. In `[Fe(H_(2)O)_(6)]^(2+)` is an outer orbital complex while `[Fe(CN)_(6)]^(4-)` is an INNER orbital complex. (iv) In `[Fe(CN)_(6)]^(4-)`, after pairing up of 3d electrons, no unpaired electron is left while in `[Fe(H_(2)O)_(6)]^(2+)`, 4 unpaired electrons are present in the 3d SUBSHELL. Hence, `[Fe(CN)_(6)]^(4-)` is low spin complex while `[Fe(H_(2)O)_(6)]^(2+)` is high spin complex. (v) As `[Fe(CN)_(6)]^(4-)` has no unpaired electron, it is diamagnetic while `[Fe(H_(2)O)_(6)]^(2+)` has 4 unpaired electrons, it is paramagnetic. (vi) In both complexes, Fe is in +2 oxidation state and has the configuration `d^(6). CN^(-)` is a strong ligand while `H_(2)O` is a weak ligand. Hence, their crystal field splitting energies are different. Consequently, they absorb different wavelengths from white light and, therefore, the transmitted colours are also different. (vii) `[Fe(CN)_(6)]^(4-)=t_(2g)^(6)e_(g)^(0), [Fe(H_(2)O)_(6)]^(2+)=t_(2g)^(4)e_(g)^(2)`. |
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