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 solve the problem of determining which complex exhibits \(d^2sp^3\) hybridization, we need to understand the concept of hybridization and apply it to each option. 

Concept of Hybridization: Hybridization is the process of combining atomic orbitals to form new hybrid orbitals. The type of hybridization helps determine the geometry of the molecule.

1. [Co(NH₃)₆]³⁺

• Cobalt (Co) in this complex is in the +3 oxidation state.
• The electronic configuration of Co is [Ar] 3d⁷ 4s². For Co³⁺, it becomes [Ar] 3d⁶.
• The Co³⁺ ion will undergo \(d^2sp^3\) hybridization, using its 3d, 4s, and 4p orbitals to form the octahedral complex.

2. SF₆

• In SF₆, the sulfur atom is surrounded by six fluorine atoms.
• Sulfur forms six bonds using \(sp^3d^2\) hybridization, resulting in an octahedral geometry.

3. BrF₅

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

4. [PtCl₄]^2-

• In this case, platinum (Pt) forms four coordinate bonds with chlorine atoms.
• The complex adopts a \(dsp^2\) hybridization, resulting in a square planar geometry, typical for d⁸ electron count of Pt²⁺.

Hence, the correct answer is [Co(NH₃)₆]³⁺ as it forms a d²sp³ hybridization.

Answer 2

\([ \text{Co(NH}_3)_6 ]^{3+}\) exhibits \(d^2sp^3\) hybridization as it forms an octahedral geometry. In this complex, the central cobalt ion uses two \(d\) orbitals, one \(s\) orbital, and three \(p\) orbitals to form six coordinate bonds with ammonia molecules.

For \( \text{BrF}_5 \), the central bromine atom exhibits \( sp^3d^2 \) hybridization, resulting in a square pyramidal structure.

In \([ \text{PtCl}_4 ]^{2-}\), platinum shows \( dsp^2 \) hybridization, resulting in a square planar geometry.

\( \text{SF}_6 \) involves \( sp^3d^2 \) hybridization, resulting in an octahedral geometry around the sulfur atom.

Thus, the correct species showing \(d^2sp^3\) hybridization is \([ \text{Co(NH}_3)_6 ]^{3+}\).

The correct option is (C) :[Co(NH₃)₆]³⁺