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Q. How many nodal planes are there in the atomic orbitals for the principal quantum number $n = 3$ ?

Structure of Atom

Solution:

Shell with $n = 3$ has $3s$, $3p$ ($p_x$, $p_y$, $p_z$) and $3d$
($d_{xy}$, $d_{xz}$, $d_{yz}$, $d_{x^2-y^2}$, and $d_{z^2}$) orbitals.
$s$ has no nodal plane.
Each of $p_x$, $p_y$, $p_z$ has one nodal plane, which means a total of three nodal planes.
$d_{z^2}$ has no nodal plane.
Each of $d_{xy}$, $d_{xz}$, $d_{yz}$, $d_{x^2-y^2}$ has two nodal planes, which means a total of eight nodal planes.
Hence, for $n = 3$, a total of $11$ nodal planes are there.