Question

he number of spherical nodes in $4s$ Orbital:
A. $4$
B. $\infty$
C. $2$
D. $3$

Answer 1

Hint:
A spherical node is known as a radial node. Radial nodes are given by $(n - l - 1)$, where $n - 1$ is the total number of nodes, and ($l$ ) is the number of angular nodes. ($n$ Is the principal quantum number, and ( $l$ ) is the angular momentum quantum number.

Complete step by step solution:
Nodes are the points where electron density is zero. There are two types of nodes for a given orbital-
1. Radial node
2. Angular node.
An Angular node is also called a nodal plane. It is a plane that passes through the nucleus. Angular nodes are equal to the azimuthal quantum number $\left( l \right).$ The total number of nodes present in this orbital is equal to$(n - 1)$ . The quantum number $\ell$ determines the number of angular nodes
Radial node is also called a nodal region. Radial node is a spherical surface where the probability of finding an electron is zero. The number of radial nodes increases with the principal quantum number (n).A radial node that occurs when the radial wave function for an atomic orbital is equal to zero or changes sign.
Number of spherical nodes is given by the formulae $= {\text{ (}}n - l - 1)$
where $n$ Is the principal quantum number and $\left( l \right)$ is azimuthal quantum number.
Since here the orbital is S-orbital so it’s azimuthal quantum number $\left( l \right)$= 0
Here $n$=4 and $l$=0 ,
So putting these values in the formula we get, $= {\text{ (}}n - l - 1)$ for 4s.
= $(4 - 0 - 1) = 3$
Number of spherical nodes $= {\text{ }}3$

**So the option (d) is correct.

Note: **
In general, a ($np$) orbital has $\left( {n{\text{ }} - {\text{ }}2} \right)$ radial nodes, so the $4p$-orbital has $\left( {4{\text{ }} - {\text{ }}2} \right){\text{ }} = {\text{ }}2$ radial nodes. The higher $p-$orbitals $\left( {5p,{\text{ }}6p,{\text{ }}and{\text{ }}7p} \right)$ are more complex since they have more spherical nodes.