Question
How many nodal planes are there in the atomic orbitals for the principal quantum number $$n=3?$$
A.
10
B.
9
C.
11
D.
2
Answer :
11
Solution :
Shell with $$n = 3$$ has $$3s,3p\left( {{p_x},{p_y},{p_z}} \right)$$ and $$3d$$ $$\left( {{d_{xy}},{d_{xz}},{d_{yz}},{d_{{x^2} - {y^2}}},{\text{and}}\,\,{d_{{z^2}}}} \right)$$ 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.
$${d_{{z^3}}}$$ 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.