Q45E

Would you expect the complex \(\left( {Co{{(en)}_3}} \right)C{l_3}\) to have any unpaired electrons? Any isomers?

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Answer

\(\left( {Co{{(en)}_3}} \right)C{l_3}\) does not have unpaired electrons and it has an optical isomer.

1Step 1 of 3:Explanation

In \(\left( {{\rm{Co}}{{({\rm{en}})}_3}} \right){\rm{C}}{{\rm{l}}_3}\), the central atom is the transition metal cobalt which is attached with \({\bf{3}}\)ethylenediamine molecules.
Also, there are \(3{\rm{C}}{{\rm{l}}^ - }\)ions, which means the coordination sphere has positive charge \(\left( { + {\bf{3}}} \right)\)
Now, let us determine the oxidation state of \(Co\)as we use \(x\)for the oxidation state of \(Co\).
The general oxion state of ethylenediamine molecule is\({\bf{0}}\), so we can determine the oxidation state of \(Co\)like this:

\(\begin{aligned}{\underline{\phantom{xx}}}x + (3 \cdot 0) &= + 3\\x & = + 3\end{aligned}\)

2Step 2 of 3:Analysis

The atomic number of Co is 27 and its electronic configuration is\((Ar)3{d^7}4{s^2}\)
In oxidation state of \(C{o^{3 + }}\) the electronic configuration is: \((Ar)3{d^6}4{s^0}\) We know that ligand contributes a pair of electrons to the metal and since en is a strong field ligand hence, it causes larger splitting.
The magnitude of pairing energy \((P)\) is less than crystal field splitting energy in octahedral field: \({\bf{P}} < {\Delta _o}\)
It means the electron pairs donated by ligand go to the innermost orbitals.
From the crystal field diagram we can see that \(\left( {{\rm{Co}}{{({\rm{en}})}_3}} \right){\rm{C}}{{\rm{l}}_3}\) has \({\bf{0}}\) unpaired electrons.

3Step 3 of 3:Conclusion

This complex does not have any geometric isomers, but its mirror image is non-superimposable, which means it has an optical isomer.

Result:

\(\left( {Co{{(en)}_3}} \right)C{l_3}\) does not have unpaired electrons and it has an optical isomer.