Question

Which of the following diagram / statement is correct for the hybrid orbitals ?

• A. s + py ¾¾® z' z
• B. s + py + pZ ¾¾® sp2 orbitals lie in the xy plane
• C. ns + np3 + any d2 ¾¾® results octahedral geometry of sp3d 2 hybrid orbitals.
• D. None of these

Answer 1

Answer: (C)

C

Answer 2

The question asks us to identify the correct statement/diagram regarding hybrid orbitals. Let's analyze each option:

Option A: s + py hybridization forms two sp hybrid orbitals, which are linear and oriented along the y-axis. The given diagram/statement is incorrect.

Option B: s + py + pz hybridization (1s + 3p) results in sp3 hybridization, leading to four tetrahedral hybrid orbitals, not planar sp2 orbitals. The statement is incorrect.

Option C: sp3d2 hybridization involves one s, three p, and two d orbitals. This combination forms six sp3d2 hybrid orbitals that are directed towards the corners of an octahedron, resulting in an octahedral geometry. Although "any d2" is a slight simplification, it refers to the specific d orbitals ($d_{z^2}$ and $d_{x^2-y^2}$) required for octahedral symmetry. This statement accurately describes the outcome of sp3d2 hybridization.