Question

Vp Patio tend to undergo positron emission. Question 5 What is the crystal field energy level diagram for the complex (Co(CN)613-? ad, dda 분 송순 스는 du du do (0) E (II) (IV) O I and IV only OIV only Oll only 0 Tonly Il and Ill only

Answer 1

In the complex $Co(CN)_{6}^{3-}$ , the metal atom, $Co^{3+}$ is . So, from the electron count of the complex and knowing that cyanide is a strong field ligand, we can conclude the answer (I) but certain things are important to discuss so that we can solve these problems with ease.

9

[ To calculate the charge of the central metal atom Co, assume its charge is 'x', now 6 cyanide ligands have '-1' charge each making a total of '-6' and the overall complex has a charge of '-3'. So, $x-6=-3,\Rightarrow x=+3$ . So, Co is in the oxidation state of '+3' in the complex. Now, Co has a configuration of
Co [Ar]3d" 4s
, so to make $Co^{3+}$ , we first have to remove the two 4s electrons and one 3d electron which makes it a
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complex.]

Now, $CN^{-}$ is a strong field pi-acceptor ligand (from spectrochemical series!) which means in the co, complex, it makes a higher energy barrier between the $t_{2g}$ and $e_{g}$ sets of orbitals. Now, to place the six electrons, we cannot use the $e_{g}$ as the energy of the complex cannot be compensated in that case, so we have to place all the six electrons in the $t_{2g}$ orbital only. So, the $e_{g}$ remains vacant and our configuration becomes the one that is depicted in .

Vp Patio tend to undergo positron emission. Question 5 What is the crystal field energy level diagram for the complex (Co(CN)613-? ad 부부 은 스는 d, du de . (0) (II) (II) 5 electrons, but we have 6 electrons in the complex! (IV) cyanide is a strong field ligand, so this configuration is not feasible! O I and IV only OIV only Oll only I only I and III only