Q38E

Question

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

Step-by-Step Solution

Verified
Answer

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

1Step 1 of 2: Given information


  • 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:

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

  • 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 \(6\).


2Step 2 0f 2 : Analysis


  • 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.