A) \[{{[Co{{({{H}_{2}}O)}_{6}}]}^{3+}}\]
B) \[{{[Co{{(N{{H}_{3}})}_{6}}]}^{3+}}\]
C) \[{{[Co{{(CN)}_{6}}]}^{3-}}\]
D) \[{{[Co{{({{C}_{2}}{{O}_{4}})}_{3}}]}^{3-}}\]
Correct Answer: C
Solution :
| [c] Key Idea: The magnitude of\[{{\Delta }_{oct}},\](the orbital splitting energy) is decided by the nature of ligand. Strong field ligand has highest \[{{\Delta }_{oct}}.\] The increasing field strength is as |
| \[{{I}^{-}}<B{{r}^{-}}<C{{l}^{-}}<{{F}^{-}}<O{{H}^{-}}<{{H}_{2}}O<{{C}_{2}}O_{4}^{2-}\] |
| \[<N{{H}_{3}}<en<NO_{2}^{-}<C{{N}^{-}}\] |
| The \[C{{N}^{-}}\] is the strongest ligand among these, hence the magnitude of \[{{\Delta }_{oct}}\] will be maximum in\[{{[Co{{(CN)}_{6}}]}^{3-}}.\] |
| Note: The magnitude of \[{{\Delta }_{oct}}\] is also decided by oxidation state of the metal ion. Thus, greater the ionic charge on the central metal ion, the greater the value of \[{{\Delta }_{oct}}\]. But here the oxidatinon state of Co metal ion is same in all complexes. |
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