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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$(1)\,\,\begin{array}{*{20}{c}}
{\,\,\,\,\,\,C{H_3}} \\
| \\
{{C_2}{H_5}C{H_2}C - OC{H_3}} \\
| \\
{\,\,\,\,\,\,\,\,C{H_3}}
\end{array}$
$(2)\,\,\begin{array}{*{20}{c}}
{{C_2}{H_5}C{H_2}C = C{H_2}} \\
{\,\,\,\,|} \\
{\,\,\,\,\,\,\,\,\,\,C{H_3}}
\end{array}$
$(3)\,\,\begin{array}{*{20}{c}}
{{C_2}{H_5}CH = C - C{H_3}} \\
{\,\,\,\,\,\,\,\,\,\,\,|} \\
{\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,C{H_3}}
\end{array}$