Question

The complex ion $[Co(NH_3)_6]^{3+}$ is formed by $sp^3d^2$ hybridisation. Hence, the ion should possess

• A. octahedral geometry
• B. tetrahedral geometry
• C. square planar geometry
• D. tetragonal geometry

Answer 1

The Correct Option is A

The complex ion $[Co(NH_3)_6]^{3+}$ provides a classic example of an octahedral geometry caused by $sp^3d^2$ hybridisation. Let's analyze the reasoning behind this:

1. Electronic Configuration: The central atom is cobalt (Co), which in its 3+ oxidation state, has an electron configuration of [Ar] 3d^6\. Since it undergoes hybridization in this complex, it uses its 3d, 4s, and 4p orbitals.
2. Nature of the Complex: The ammonia ligands (NH₃) are neutral and donate electron pairs to the cobalt ion, leading to the formation of coordination bonds. The coordination number for cobalt in this complex is 6, as indicated by the six ammonia ligands.
3. Hybridization: The central cobalt ion undergoes $sp^3d^2$ hybridisation, using one 4s, three 4p, and two 3d orbitals to form six equivalent hybrid orbitals. These six hybrid orbitals account for the octahedral arrangement of ligands.
4. Geometry: The hybridization process results in an octahedral geometry where all six donor atoms from the ammonia ligands occupy vertices of an octahedron around the cobalt ion.
5. Conclusion: The presence of six hybrid orbitals aligns the ligand donor atoms in an octahedral geometry, thus confirming that $[Co(NH_3)_6]^{3+}$ exhibits octahedral geometry.

Therefore, the correct answer is octahedral geometry.