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
Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.
${C_({graphite)}} + {O_2} \to C{O_2};\,\Delta H = - 393.4\,kJ/mole$
${C_({graphite})} \to {C_{{\rm({diamond})}}};\,\Delta H = $
$I.\, Al_2Cl_6\,\,\, II.\, B_2H_6\,\,\, III.\, Fe_2Cl_6\,\,\, IV.\, Si_2H_6$