Question

If a compound has octahedral geometry, what is the central atom’s hybridization?

• A. sp\(^2\)
• B. dsp\(^2\)
• C. sp\(^3\)d\(^2\)
• D. sp\(^3\)d\(^3\)

Hint:

In octahedral geometry, the hybridization of the central atom is \( sp^3d^2 \), using six orbitals.

Answer 1

The Correct Option is C

Step 1: Understanding octahedral geometry.
In an octahedral geometry, there are six bonding regions around the central atom. This requires the central atom to have six hybridized orbitals, which is achieved by using the \( sp^3d^2 \) hybridization.
Step 2: Conclusion.
The central atom in an octahedral compound undergoes \( sp^3d^2 \) hybridization, which corresponds to option (C).

Answer 2

Step 1: An octahedral shape has 6 bonding positions arranged around the central atom.

Step 2: To make 6 equivalent hybrid orbitals, the atom must mix one s, three p, and two d orbitals.

Step 3: That combination is called sp³d² hybridization — option (C).

Answer 3

Elimination approach: \( sp^2 \) hybridization only produces 3 orbitals, enough for a trigonal planar shape, not six positions. \( dsp^2 \) gives 4 orbitals arranged in a square plane, a 4-coordinate geometry, not 6. \( sp^3d^3 \) would need seven hybrid orbitals for a coordination number of 7, one too many for an octahedron. The only combination that yields exactly 6 equivalent orbitals is \( sp^3d^2 \), option (C).