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Question: Write the hybridization and magnetic character of \( {[Co{({C_2}{O_4})_3}]^{2 - }} \) . \( (At.no.of...

Write the hybridization and magnetic character of [Co(C2O4)3]2{[Co{({C_2}{O_4})_3}]^{2 - }} . (At.no.ofCo=27)(At.no.ofCo = 27)

Explanation

Solution

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 CoCo is 2727 .
So, the electronic configuration of Co=[Ar]3d74s2Co = [Ar]3{d^7}4{s^2} .
Now, we will calculate the oxidation state of CoCo .
So to find oxidation state of any compound we must follow these 55 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 CoCo is
x+(2)×3=3 x6=3 x=13+6 =+3 x + ( - 2) \times 3 = - 3 \\\ x - 6 = - 3 \\\ x = 13 + 6 \\\ = + 3
[Co(C2O4)3]2{[Co{({C_2}{O_4})_3}]^{2 - }} is an octahedral complex with 33 bidentate oxalate ligands binding to Cr3+C{r^{3 + }} . The coordination number for CrCr is 66 , oxidation state for CrCr is +3+ 3 , potassium is +1+ 1 , and C2O4{C_2}{O_4} is 2- 2 .
Using valence bond theory we know that the oxalate ion is a strong field ligand, the 33 electrons singly occupy the 33 suborbital of 3d3d leaving 22 unoccupied orbital. 66 electron pairs from C2O4{C_2}{O_4} ligand occupy the 22 empty orbital of 3d,4s3d,4s , and 33 of the 4p4p orbitals.
So, the hybridisation of [Co(C2O4)3]2{[Co{({C_2}{O_4})_3}]^{2 - }} is d2sp3{d^2}s{p^3} .
As the hybridization of [Co(C2O4)3]2{[Co{({C_2}{O_4})_3}]^{2 - }} is d2sp3{d^2}s{p^3} . So, it has no unpaired electron hence its magnetic behavior is diamagnetic.
The hybridization of [Co(C2O4)3]2{[Co{({C_2}{O_4})_3}]^{2 - }} is d2sp3{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.