Give the oxidation state of the metal, number of \(d\)electrons, and the number of unpaired electrons predicted for \(\left( {Co{{\left( {N{H_3}} \right)}_6}} \right)C{l_3}. \)

• In\(\left( {{\rm{Co}}{{\left( {{\rm{N}}{{\rm{H}}_3}} \right)}_6}} \right){\rm{C}}{{\rm{l}}_3}\), the central atom is the transition metal which is attached with \(6\)ammonia molecules.
• There are\(3{\rm{C}}{{\rm{l}}^ - }\)ions, which means the coordination sphere has positive charge\(\left( {{3^ + }} \right):{\left( {{\rm{Co}}{{\left( {{\rm{N}}{{\rm{H}}_3}} \right)}_6}} \right)^{3 + }}\)
• We can determine the Oxidation state of \(Co\)as we use \(x\)for the oxidation state of\(Co\).
• We know that general oxidation state of ammonia is\(0\), so we can determine the oxidation state of\(Co\)like this:

\(\begin{aligned}{}x + (6 \cdot 0) & = + 3\\x & = + 3\end{aligned}\)

• It means that the oxidation state of\(Co\)is\( + 3.\)
• If we look at the periodic table we can see that 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: \((Ar)3{d^6}4{s^2}\), Which means that the number of\(d\) -electrons in \(C{o^{3 + }}\)is \(6\).

• The ligand contributes a pair of electrons to the metal and since \({\rm{N}}{{\rm{H}}_3}\)is a strong field ligand hence, it causes larger splitting.
• The magnitude of pairing energy\(\left( P \right)\)is less than crystal field splitting energy in octahedral field:\(P < {\Delta _o}\), which means the electron pairs donated by ligand go to the innermost orbitals.
• From the crystal field diagram we can see that \(\left( {{\rm{Co}}{{\left( {{\rm{N}}{{\rm{H}}_3}} \right)}_6}} \right){\rm{C}}{{\rm{l}}_3}\)has 0 unpaired electrons.

Result

Oxidation state of \(Co\)is \( + 3.\)it has \(6\)-d-electrons, and the molecule does not have unpaired electrons.