Show that the probability of finding a \(3 d_{x z}\) electron in the \(x y\) plane is zero.

Short Answer

Expert verified
Due to the shape and orientation of the \(3 d_{x z}\) orbital, the probability of finding an electron in the \(x y\) plane is zero, as it is a nodal plane, meaning the wavefunction, and thus the electron density, equals zero in this plane.

Step by step solution

01

Understanding electron orbitals

It's crucial to note that one of the main tenets of quantum mechanics is that the location of particles cannot be pinpointed precisely (principle of uncertainty). Instead, we use probability densities to specify where particles are likely to be found. For electrons in an atom, these are described by the solutions to the Schrödinger equation called wavefunctions \(\psi(r, \theta, \phi)\). For every set of quantum numbers, there corresponds a certain wavefunction.
02

Understanding d-orbitals

A 3d orbital refers to an atom's electron orbitals where the principal quantum number \(n = 3\), orbital quantum number \(l = 2\), and magnetic quantum number \(m = -2,-1,0,1,2\). These orbitals have two nodal planes where the probability of finding an electron is zero. In the \(d_{x z}\) orbital, these are the yz and xy planes. Instead, the electrons are most likely found in lobes lying in the xz plane.
03

Asserting zero probability in certain planes

The solution to the Schrödinger equation, the wavefunction, gives the probability density for the electron. The absolute square of the wavefunction \(\psi(r, \theta, \phi)\) gives the probability. For the \(d_{x z}\) orbital, the wavefunction includes a part that changes the sign when \(x\) or \(z\) changes to \(-x\) or \(-z\). Thereby, the wavefunction is forced to pass through-zero at the xy-plane, thus making it a nodal plane.

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