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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$\begin{array}{*{20}{c}}
{C{H_3} - CH - CH - C{H_2} - C{H_3}} \\
{\,\,\,|\,\,\,\,\,\,\,\,\,\,\,\,\,\,|\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,} \\
{\,\,CN\,\,\,\,\,\,\,\,\,CH = C{H_2}\,\,}
\end{array}$
$(A)$ $Rb$ and $Cs$
$(B)$ $Na$ and $K$
$(C)$ $Ar$ and $Kr$
$(D)$ $I$ and $At$
Choose the correct answer from the options given below.