Question

The d-orbitals involved in the hybridization to form square planar and trigonal bipyramidal geometries are, respectively,

• A. \( d_{z^2} \) and \( d_{z^2} \)
• B. \( d_{yz} \) and \( d_{z^2} \)
• C. \( d_{x^2 - y^2} \) and \( d_{z^2} \)
• D. \( d_{x^2 - y^2} \) and \( d_{yz} \)

Hint:

Square planar geometry typically involves \( d_{x^2 - y^2} \), while trigonal bipyramidal geometry involves \( d_{z^2} \).

Answer 1

The Correct Option is C

Step 1: Understanding the geometries.
For a square planar geometry, the \( d_{x^2 - y^2} \) orbital is involved in hybridization, while for trigonal bipyramidal geometry, the \( d_{z^2} \) orbital is involved.

Step 2: Analyzing the options.
(A) \( d_{z^2} \) and \( d_{z^2} \): Incorrect, two \( d_{z^2} \) orbitals are not involved in the hybridization for these geometries.
(B) \( d_{yz} \) and \( d_{z^2} \): Incorrect, \( d_{yz} \) is not involved in the hybridization for square planar or trigonal bipyramidal geometries.
(C) \( d_{x^2 - y^2} \) and \( d_{z^2} \): Correct, these orbitals are involved in the hybridization for square planar and trigonal bipyramidal geometries.
(D) \( d_{x^2 - y^2} \) and \( d_{yz} \): Incorrect, \( d_{yz} \) is not involved in the hybridization for these geometries.

Step 3: Conclusion.
The correct answer is (C) \( d_{x^2 - y^2} \) and \( d_{z^2} \).