Question

Identify the following species in which \(d^2sp^3\) hybridization is shown by the central atom.

• A. [Co(NH₃)₆]³⁺
• B. SF₆
• C. BrF₅
• D. [PtCl₄]^2-

Answer 1

The Correct Option is A

To identify the complex exhibiting \(d^2sp^3\) hybridization, we must first grasp the concept of hybridization and then apply it to each provided option.

Hybridization Explained: Hybridization is the atomic orbital combination process that yields new hybrid orbitals. The resultant hybridization type dictates the molecule's geometry.

1. [Co(NH₃)₆]³⁺

• Cobalt (Co) in this complex is in the +3 oxidation state.
• Cobalt's electronic configuration is [Ar] 3d⁷ 4s². For Co³⁺, this simplifies to [Ar] 3d⁶.
• The Co³⁺ ion undergoes \(d^2sp^3\) hybridization, utilizing its 3d, 4s, and 4p orbitals to construct the octahedral complex.

2. SF₆

• In SF₆, the sulfur atom is bonded to six fluorine atoms.
• Sulfur achieves six bonds via \(sp^3d^2\) hybridization, resulting in an octahedral geometry.

3. BrF₅

• In BrF₅, the central bromine atom is bonded to five fluorine atoms and possesses one lone pair.
• This molecule demonstrates \(sp^3d^2\) hybridization, leading to a square pyramidal shape.

4. [PtCl₄]^2-

• In this scenario, platinum (Pt) forms four coordinate bonds with chlorine atoms.
• The complex adopts \(dsp^2\) hybridization, yielding a square planar geometry, characteristic of Pt²⁺'s d⁸ electron count.

Therefore, [Co(NH₃)₆]³⁺ is the correct answer due to its formation of \(d^2sp^3\) hybridization.