If two different non-axial d-orbitals having 'xz' nodal plane form `pi`-bond by overlapping each other, then internuclear axis will be :

Options

A. x
B. y
C. z
D. They don't form `pi`-bond

Explanation

To solve the problem, we need to analyze the given information about the non-axial d-orbitals and their nodal planes. Here’s a step-by-step solution:

### Step 1: Identify Non-Axial d-Orbitals
The non-axial d-orbitals are dxy, dyz, and dxz. These orbitals are characterized by their orientation in relation to the axes of the coordinate system.

**Hint:** Remember that non-axial d-orbitals do not point directly along the x, y, or z axes.

### Step 2: Understand the Nodal Plane
The question specifies that the d-orbitals have an 'xz' nodal plane. This means that the wave function of the orbitals has a node (a region of zero probability) in the xz plane, which affects their orientation and overlap.

**Hint:** Visualize the nodal plane as a flat surface where the probability of finding an electron is zero.

### Step 3: Draw the Orbitals
Draw the dxy and dyz orbitals to visualize their shapes and orientations. The dxy orbital lies in the xy plane, while the dyz orbital lies in the yz plane.

**Hint:** Sketching the orbitals can help you see how they relate to the xz nodal plane.

### Step 4: Analyze Overlap for Pi Bonding
For pi bonding to occur, the orbitals must overlap side by side. However, since the dxy and dyz orbitals are perpendicular to each other, they cannot overlap effectively to form a pi bond.

**Hint:** Remember that only orbitals aligned in the same plane can overlap to form pi bonds.

### Step 5: Conclude the Internuclear Axis
Since the dxy and dyz orbitals are perpendicular and cannot form a pi bond, there is no internuclear axis that allows for such bonding. Therefore, the answer to the question is that no pi bond formation occurs.

**Hint:** Consider the implications of perpendicular orbitals in terms of bonding capabilities.

### Final Answer
The internuclear axis will be such that no pi bond can be formed due to the perpendicular nature of the orbitals involved.