The angular part of the wave function for an | vedclass

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Question

(C) The given angular wave function is $\psi_{(	heta, \phi)} = \left( \frac{15}{4\pi} ight)^{1/2} \sin 	heta \cos 	heta \sin \phi$. This correspo

Vedclass · Accepted Answer

$XY$ and $XZ$ planes only

Vedclass · Answer

(C) The given angular wave function is $\psi_{(	heta, \phi)} = \left( \frac{15}{4\pi} ight)^{1/2} \sin 	heta \cos 	heta \sin \phi$. This corresponds to the $d_{yz}$ orbital. In spherical coordinates,$y = r \sin 	heta \sin \phi$ and $z = r \cos 	heta$. The product $yz$ is proportional to $\sin 	heta \sin \phi \cos 	heta$,which matches the angular part provided. The nodal planes for a $d_{yz}$ orbital are the $XY$ plane $(z=0)$ and the $XZ$ plane $(y=0)$,where the probability of finding the electron is zero. Therefore,the correct option is $C$.

The angular part of the wave function for an orbital is given by $\psi_{(\theta, \phi)} = \left( \frac{15}{4\pi} \right)^{1/2} \sin \theta \cos \theta \sin \phi$,where $\theta$ is the angle from the $z$-axis. The angular node$(s)$ are:

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