Question

For radial probability curves, which of the following is/are correct?
(A) The number of maxima in 2s orbital is two
(B) The number of spherical or radial nodes is equal to \[n-l-1\]
(C) The number of angular nodes are $'l'$
(D). $3d_{z}^{2}$ has 3 angular nodes

Answer 1

Hint: Radial distribution curves give us the idea of how much electron density is present at a radial distance from the nucleus. Sum of radial and angular nodes are known as total nodes (n-1). Angular nodes are determined by an azimuthal quantum number.

Complete step by step solution:
Radial distribution curves give us the idea of how much electron density is present at a radial distance from the nucleus. At nodal point which is also known as radial point the value of radial probability density function $\left( 4\pi {{r}^{2}}{{\psi }^{2}} \right)$ becomes zero.
Nodes are of two types, one is a radial node which is a sphere but only at a fixed radius and also occurs as the increase in principal quantum number and. Whereas the other type of node is an angular node and they are typically flat at a fixed angle.
So, to find the number of radial nodes, You can use the formula that total no of nodes is the sum of angular nodes and radial nodes,
Written as Total no. of nodes = Angular nodes + radial nodes
So, radial nodes= Total no of nodes –angular nodes
Here total no of nodes is given by n-1,
And the number of angular nodes is given by $'l'$
Therefore it can be seen that radial nodes are given as $n-l-1$
Where n= principal quantum no,
 \['l'\]= azimuthal quantum no.
Now, we will find how much no of maxima an orbital can have. So, to find the maxima firstly we should know the radial nodes of that orbital.
Here in the question, we have to check the no of maxima in 2s orbital and for that,
Principal quantum no will be(n)= 2
Azimuthal quantum no will be = 0 (because for s orbital $'l'$ is 0)
Now put the values in the formula you will get,
Radial nodes = $n-l-1=2-0-1=1$
If the radial node is one the maxima will be 2. Therefore the no of maxima in 2s orbital is 2.
And the angular node for $3d_{z}^{2}$ is two and not three.

Thus the option A, B and C all are correct.

Note: To find the maximum you can also refer to the graph. For the 2s orbital, near the nucleus probability density is maximum. Angular nodes have directional nature and can show angles but radial nodes don’t have any idea about the angle or direction from the nucleus it only shows about the distance of electrons from the nucleus.