Question

Which of the following orbitals does not have the angular node?
A. ${{p}_{x}}-orbital$
B. ${{d}_{z}}^{2}-orbital$
C. ${{p}_{y}}-orbital$
D. $1s-orbital$

Answer 1

Hint: To find out the correct option, use the formula: the number of angular nodes equal to the azimuthal quantum number (generally represented by ‘l’). Different orbits have different azimuthal quantum numbers.

Complete Solution :
Nodes are the points where the electron density is zero. There are two types of nodes for a given orbital, which are angular nodes and spherical or radial nodes. Angular nodes are also called nodal planes and they are found in p, d and f-orbital. The angular nodes are equal to the azimuthal quantum number (denoted by ‘l’).
So, here let us look at what is the angular node of the given option.

- In the first option i.e. ${{p}_{x}}-orbital$, as we know the azimuthal quantum number of p-orbital is $1$. So, the angular node will be $1$.

- In option two i.e. ${{d}_{z}}^{2}-orbital$, the azimuthal quantum number of d-orbital is $2$. Therefore, the angular nodes will be $2$.

- In the third option i.e. ${{p}_{y}}-orbital$, the azimuthal quantum number of p-orbital is $1$. Thus, the number of angular nodes will be $1$.

- In the last option i.e. $1s-orbital$, the azimuthal quantum number of s-orbital is $0$. Therefore, the s-orbital will have no angular nodes.
Therefore, $1s-orbital$ will not have any angular nodes.
So, the correct answer is “Option D”.

Note: Radial or spherical nodes are found in $2s,3s,3p,4p,5d$ orbitals. The formula used to find out the spherical nodes is $n-l-1$, where n is the principal quantum number and l us the azimuthal quantum number. The total number of nodes of any orbital is given by $n-1$.