Solveeit Logo

Question

Question: Explain the application of VBT to hexamine cobalt \[(III)\] ion \({{[Co{{(N{{H}_{3}})}_{6}}]}^{3+}}\...

Explain the application of VBT to hexamine cobalt (III)(III) ion [Co(NH3)6]3+{{[Co{{(N{{H}_{3}})}_{6}}]}^{3+}}

Explanation

Solution

Valence bond theory is a theory which is used to explain chemical bonding. It tells us about how the atomic orbitals combine to give individual chemical bonds. Strongest bond is formed when there is a maximum overlap of atomic orbitals.

Complete step by step answer:
As you can see that, [Co(NH3)6]3+{{[Co{{(N{{H}_{3}})}_{6}}]}^{3+}} , it is a metal amine complex, in which cobalt ion is attached with six ammonia ligands.
Valence bond theory says that a formation of covalent bond takes place between the overlapping of two atoms, having half – filled valence atomic orbitals, in which each atom contains one unpaired electron.
As you can see that, [Co(NH3)6]3+{{[Co{{(N{{H}_{3}})}_{6}}]}^{3+}} , coordination number of central metal ion is six. Coordination number six results in two possible hybridization, that are, d2sp3{{d}^{2}}s{{p}^{3}} and sp3d2s{{p}^{3}}{{d}^{2}} . Both the hybridization gives octahedral geometry. NH3N{{H}_{3}} is a strong ligand, [Co(NH3)6]3+{{[Co{{(N{{H}_{3}})}_{6}}]}^{3+}} is an inner orbital complex. In this complex, the six valence electrons of cobalt will occupy 3d3d orbital, whereas these ligands will occupy 3d,4s3d,4s and 4p4p orbitals, that gives d2sp3{{d}^{2}}s{{p}^{3}} hybridization.
As we know that, electronic configuration of CoCo is Ar3d74s2Ar\,3{{d}^{7}}\,4{{s}^{2}}
Electronic configuration of Co+3C{{o}^{+3}} is Ar3d6Ar\,3{{d}^{6}}
In presence of strong ligands NH3N{{H}_{3}} it will attract an inner dd orbital for bonding. As we discussed, the coordination number of Co+3C{{o}^{+3}} is six, so it will require six empty atomic orbitals to receive coordinated lone pairs of electrons. Hence, two 3d3d orbitals, three 4p4p orbitals, and one 4s4s orbital are combined to give d2sp3{{d}^{2}}s{{p}^{3}} hybridization.

So, the correct answer is Option A .

Note: 1.In [Co(NH3)6]3+{{[Co{{(N{{H}_{3}})}_{6}}]}^{3+}} , oxidation state of cobalt is +3+3 .
2.NH3N{{H}_{3}} is a strong ligand, which causes the pairing and cobalt undergoes d2sp3{{d}^{2}}s{{p}^{3}} hybridization.
3.Here, the electronic configuration of cobalt is d6{{d}^{6}} .
4.This molecule is diamagnetic because six pairs of electrons from each ammonia molecule occupy six hybrid orbitals.