question_answer

Which of these statements about \[{{[Co{{(CN)}_{6}}]}^{3-}}\] is true?                                                                 

A)  \[{{[Co{{(CN)}_{6}}]}^{3-}}\]has no unpaired electrons and will be in a low-spin configuration.

B)  \[{{[Co{{(CN)}_{6}}]}^{3-}}\] has four unpaired electrons and will be in a low-spin configuration.

C)  \[{{[Co{{(CN)}_{6}}]}^{3-}}\] has four unpaired electrons and will be in a high-spin configuration.

D)  \[{{[Co{{(CN)}_{6}}]}^{3-}}\] has no unpaired electrons and will be in a high-spin configuration.

Correct Answer: A

Solution :

\[{{[Co{{(CN)}_{6}}]}^{3-}}\] \[C{{o}^{3+}}=1{{s}^{2}}2{{s}^{2}}2{{p}^{6}}3{{s}^{2}}3{{p}^{6}}3{{d}^{6}}\] \[C{{N}^{-}}\] is a strong field ligand and as it approaches the metal ion, the electrons must pair up. The splitting of the d-orbitals into two sets of orbitals in an octahedral \[{{[Co{{(CN)}_{6}}]}^{3-}}\] may be represented as Here, for \[{{d}^{6}}\] ions, three elections first enter orbitals with parallel spin put the remaining may pair up in \[{{t}_{2g}}\] orbital giving rise to low spin complex (strong ligand) field. \[\therefore \,{{[Co{{(CN)}_{6}}]}^{3-}}\] has no unpaired electron and will be in a low spin configuration.