Question

Which of these statements about ${[Co{(CN)_6}]^{3 - }}$ is true?
A) ${[Co{(CN)_6}]^{3 - }}$ has four unpaired electrons and will be in a high-spin configuration
B) ${[Co{(CN)_6}]^{3 - }}$ has no unpaired electrons and will be in a high-spin configuration
C) ${[Co{(CN)_6}]^{3 - }}$ has no unpaired electrons and will be in low-spin configuration
D) ${[Co{(CN)_6}]^{3 - }}$ has four unpaired electrons and will be in a low-spin configuration

Answer 1

Hint: First, calculate the oxidation state central metal ion i.e., cobalt in the complex. Write the electronic configuration of cobalt in the complex according to crystal field theory. Now, the given ligand in the complex is CN which is a strong field ligand so there will be pairing of electrons.

Complete answer:
${[Co{(CN)_6}]^{3 - }}$ is a coordination complex which contains cobalt ($Co$) as central metal atom and six cyanide ($C{N^ - }$) ligands. Due to the presence of six ligands, the given complex will have octahedral geometry. Now, oxidation state of $Co$ (let $x$) in the given coordination complex will be calculated as follows:
Since, charge on $C{N^ - }$ is -1
Therefore, for ${[Co{(CN)_6}]^{3 - }}$ :
$x + 6( - 1) = - 3$
$\therefore x = + 3$
Hence, $Co$ in the complex is as $C{o^{3 + }}$.
Electronic configuration of $Co$: $3{d^7}4{s^2}$
Therefore, electronic configuration of $C{o^{3 + }}$: $3{d^6}4{s^0}$
Now, filling of these 6 electrons in the d-orbitals:
According to Crystal Field theory (CFT), in the presence of ligands in a definite geometry, the five degenerate d-orbitals split into two ${e_g}$ and three ${t_{2g}}$ orbitals. This phenomenon is known as crystal field splitting. The crystal field splitting, ${\Delta _o}$ also depends upon the field produced by the ligands. Here in the given complex, ligand is $C{N^ - }$ which is a strong field ligand according to the spectrochemical series. In the presence of strong field ligands, ${\Delta _o} > P$ (where P is the pairing energy) and the complexes form are low-spin complexes. When ${\Delta _o} > P$, there is pairing of the electrons in the orbitals.
So, filling of the ${d^6}$ electrons in the ${t_{2g}}$ and ${e_g}$ orbitals will be as follows:
 

Thus, there is pairing of electrons in ${t_{2g}}$ orbitals and hence, no unpaired electron is present.
Therefore, we concluded that ${[Co{(CN)_6}]^{3 - }}$has no unpaired electrons and will be in low-spin configuration.

Thus, option C is correct.

Note:
According to CFT, the energy of the two two ${e_g}$ orbitals increases by $\left( {\dfrac{3}{5}{\Delta _o}} \right)$ and that of the three ${t_{2g}}$ orbitals decreases by $\left( {\dfrac{2}{5}{\Delta _o}} \right)$ , where ${\Delta _o}$ is the crystal field splitting energy separation and the subscript o is for the octahedral geometry.