Question

Write the hybridization and magnetic character of $ {[Co{({C_2}{O_4})_3}]^{2 - }} $ . $ (At.no.ofCo = 27) $

Answer 1

Hint: To find the hybridization of any compound Concentrate on the electron pairs and other atoms linked directly to the concerned atom. This is crucial and one can directly get the state of hybridization and shape. And in case of magnetic behavior it depends on whether there is an attraction or repulsion by the pole of a magnet, matter is classified as being either paramagnetic or diamagnetic.

Complete answer:
We know, the atomic number of $ Co $ is $ 27 $ .
So, the electronic configuration of $ Co = [Ar]3{d^7}4{s^2} $ .
Now, we will calculate the oxidation state of $ Co $ .
So to find oxidation state of any compound we must follow these $ 5 $ steps
 $ - $ Find atoms without oxidation number rules
 $ - $ Find the known oxidation number for the other elements in the compound.
 $ - $ Multiply the number of each atom by its oxidation number.
 $ - $ Add the results together.
 $ - $ Calculate the unknown oxidation number based on the compound's charge.
So, the oxidation state of $ Co $ is
 $
  x + ( - 2) \times 3 = - 3 \\
  x - 6 = - 3 \\
  x = 13 + 6 \\
   = + 3 $
$ {[Co{({C_2}{O_4})_3}]^{2 - }} $ is an octahedral complex with $ 3 $ bidentate oxalate ligands binding to $ C{r^{3 + }} $ . The coordination number for $ Cr $ is $ 6 $ , oxidation state for $ Cr $ is $ + 3 $ , potassium is $ + 1 $ , and $ {C_2}{O_4} $ is $ - 2 $ .
Using valence bond theory we know that the oxalate ion is a strong field ligand, the $ 3 $ electrons singly occupy the $ 3 $ suborbital of $ 3d $ leaving $ 2 $ unoccupied orbital. $ 6 $ electron pairs from $ {C_2}{O_4} $ ligand occupy the $ 2 $ empty orbital of $ 3d,4s $ , and $ 3 $ of the $ 4p $ orbitals.
So, the hybridisation of $ {[Co{({C_2}{O_4})_3}]^{2 - }} $ is $ {d^2}s{p^3} $ .
As the hybridization of $ {[Co{({C_2}{O_4})_3}]^{2 - }} $ is $ {d^2}s{p^3} $ . So, it has no unpaired electron hence its magnetic behavior is diamagnetic.
The hybridization of $ {[Co{({C_2}{O_4})_3}]^{2 - }} $ is $ {d^2}s{p^3} $ and it is diamagnetic.

Note:
The valence bond theory describes chemical bonding. Valence bond theory states that the overlap of incompletely filled atomic orbitals leads to the formation of a chemical bond between two atoms. The unpaired electrons are shared and a hybrid orbital is formed.