MCQ
Angular part of wave function for an orbital is = ${\left( {\frac{{15}}{{4\pi }}} \right)^{\frac{1}{2}}}\,\sin \theta \,\cos \theta \,\sin \phi $ where $\theta $ = angle from $z-$ axis Angular node $(s)$ are
  • A
    $XY$ plane only
  • B
    $YZ$ plane only
  • $XY$ & $XZ$ plane only
  • D
    $XY$ , $YZ$ & $ZX$ plane

Answer

Correct option: C.
$XY$ & $XZ$ plane only
c
$\psi_{(\theta, \phi)}=\left(\frac{15}{4 \pi}\right)^{1 / 2} \sin \theta \cos \theta \sin \phi$

It is angular part of $d y z$ oxbital

How we identified that it is $d_{y z}$ See Power of $\theta$. Here it is $2 \Rightarrow d$ orbital

$\text { For } y z, \text { it is } \frac{\sin \theta \sin \phi \cos \theta}{y} \Rightarrow d_{y z} \text { orbital }$

xy and $x z$ are the nodal plane

(nodal plane is the plane where probability of finding electron is minimum)

$\Rightarrow$ C option

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