The probability of finding P_y electron is zero in — Atomic Structure Chemistry Question

💡 Solution & Explanation

### Step 1 - Geometry of p-Orbitals An atomic $p$-subshell ($l = 1$) contains three degenerate dumbbell-shaped orbitals oriented along the coordinate axes: * $p_x$: oriented along the $x$-axis * $p_y$: oriented along the $y$-axis * $p_z$: oriented along the $z$-axis ### Step 2 - Nodal Planes A **nodal plane** is a plane through the nucleus where the probability density $|\psi|^2 = 0$. Each $p$-orbital has exactly one nodal plane (since number of angular nodes $= l = 1$), always **perpendicular to the orbital's axis of orientation**. For the $p_y$ orbital: * Lobes lie along the $y$-axis * The nodal plane is perpendicular to the $y$-axis, passing through the origin * This plane is the **$xz$-plane** (defined by $y = 0$) $$\psi_{p_y}(x, 0, z) = 0 \implies |\psi_{p_y}(x, 0, z)|^2 = 0$$ ### Step 3 - Evaluation of Options * **Option (A) $xy$-plane:** Contains the $y$-axis → high electron density. Incorrect. * **Option (B) $yz$-plane:** Contains the $y$-axis → lobes lie within this plane. Incorrect. * **Option (C) $xz$-plane:** Perpendicular to $y$-axis → nodal plane, $|\psi|^2 = 0$. **Correct.** * **Option (D) $y$-axis:** Orbital lobes lie along this axis → maximum density. Incorrect. $$\text{Correct Option: } \boxed{\text{C}}$$