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 1

Answer: (C)

11

Given principal quantum number, $n=3$

Subshells present are: $l=0,1,2$

as $3s,3p$ and $3d$-orbital, Number of nodal planes are:

1. $3s$-orbital: no nodal plane. It is spherically symmetrical orbital. Number of nodal planes =0

2. $3p$-orbital: It has $p_x,p_y,p_z$ orbitals where nodal planes are in yz, zx,xy-plane, respectively. Total nodal planes=3

3. 3d-orbital: It has $d_{xy},d_{yz},d_{yz},d_{x^2-y^2},d_{z^2}$ orbitals where nodal planes are in $(zx,yz),(xy,xz),(xy,zx),(twoz-inbetweenxy),none$ planes, respectively. Total nodal planes =8.

Thus the total number of nodal planes in $n=3$ are 0+3+8=11.