Question

The number of radial nodes and angular nodes for d-orbital can be represented as:
(a) $(n - 2)$radial nodes + 1 angular node = $(n - 1)$total nodes
(b)$(n - 1)$radial nodes + 1 angular node = $(n - 1)$ total nodes
(c) $(n - 3)$radial nodes + 2 angular nodes = $(n - l - 1)$total nodes
(d) $(n - 3)$radial nodes + 2 angular nodes = $(n - 1)$ total nodes

Answer 1

Hint: The nodes in an orbital are the points where the probability of finding an electron is zero. The total number of nodes is given by $(n - 1)$, the number of radial nodes by $(n - l - 1)$ and the number of angular nodes by $l$ . Here n is the principal quantum number and it denotes the shells while l is the azimuthal quantum number which describes the orbital angular momentum for an atomic orbital and describes the shape of the orbital.

Complete step by step answer:
1: As we know the total number of nodes is given by $(n - 1)$, where $n$ is the principal quantum number.
2: $l$ is the Azimuthal quantum number, which describes the shape of orbitals. The value of $l$ can be from 0 to $(n - 1)$ . The values describe a shape, so 0 is for s-orbital, 1 for p-orbital and 2 for d-orbital and so on.
3: Here, we have to calculate the number of radial and angular nodes. So,
Number of angular nodes = $l$
Here, for d-orbital, $l$ = 2.
∴ Number of angular nodes = 2
4: For radial nodes, the formula is: total nodes – angular nodes
$ = (n - 1) - l$
Putting the values of $n$ and $l$ in the formula, we get:
⸫Number of radial nodes $
= n - 2 - 1 \\
 = n - 3 \\
 $
Thus, the number of radial nodes is $(n - 3)$ .and as mentioned before, the total number of nodes is $(n - 1)$ .
The correct option is (d).

Note:
This answer can also be verified by adding the radial and angular nodes for d-orbital, which is:
Radial nodes + Angular nodes = Total nodes

=$(n - 3) + 2 = (n - 1)$