In which of the following molecular shape `d_(z^(2))` orbital must not be involved in bonding ?

Options

A. Pentagonal planar
B. Trigonal planar
C. Linear
D. Square planar

Explanation

To determine in which molecular shape the \(d_{z^2}\) orbital is not involved in bonding, we will analyze the hybridization of each molecular shape provided in the options.

### Step-by-Step Solution:

1. **Identify the Molecular Shapes and Their Hybridizations:**
- **Pentagonal Planar:** The hybridization for this shape is \(sp^3d^3\).
- **Trigonal Planar:** The hybridization for this shape is \(sp^2\).
- **Linear:** The hybridization can vary, but common types include \(sp\), \(sp^2\), and \(sp^3d\).
- **Square Planar:** The hybridization for this shape is \(sp^3d^2\).

2. **Analyze Each Shape:**
- **Pentagonal Planar:**
- Hybridization: \(sp^3d^3\)
- Involves \(d_{z^2}\) orbital along with other \(d\) orbitals.
- **Conclusion:** \(d_{z^2}\) is involved.

- **Trigonal Planar:**
- Hybridization: \(sp^2\)
- Involves \(s\) and two \(p\) orbitals, with no \(d\) orbitals involved.
- **Conclusion:** \(d_{z^2}\) is NOT involved.

- **Linear:**
- Hybridization can be \(sp\), \(sp^2\), or \(sp^3d\).
- In the case of \(sp\) or \(sp^2\), \(d_{z^2}\) is not involved. However, in \(sp^3d\), \(d_{z^2}\) can be involved.
- **Conclusion:** Not a definitive answer; depends on specific hybridization.

- **Square Planar:**
- Hybridization: \(sp^3d^2\)
- Involves \(d_{x^2-y^2}\) and \(d_{z^2}\) orbitals.
- **Conclusion:** \(d_{z^2}\) is involved.

3. **Final Conclusion:**
- The only molecular shape where the \(d_{z^2}\) orbital is not involved in bonding is the **Trigonal Planar** shape with \(sp^2\) hybridization.

### Answer:
The \(d_{z^2}\) orbital must not be involved in bonding in the **Trigonal Planar** molecular shape.

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