Question

What is the number of angular and radial nodes of $4d$ orbital?
A) $3,1$
B) $1,2$
C) $3,0$
D) $2,1$

Answer 1

Hint:As we know that a node or nodal surface is basically the region where the probability of finding the electron is zero. These nodes are of two types: radial node which is a circular ring occurring as the principal quantum number increases and angular node representing planes passing through the nucleus.

Complete answer:
As we know that a node or a nodal surface is the place or region where electron finding probability is zero and they can be of two types: the first one is radial node which is a circular ring occurring as the principal quantum number increases and it can be calculated using the formula as:
$n - l - 1$ where $n$ is the principal quantum number and $l$ is azimuthal quantum number.
So, in $4d$ orbital we know that d- subshell has five d-orbitals and therefore the $l = 2$ and $n = 4$.
Therefore the radial node will be: $node = 4 - 2 - 1 = 1$.

Next we have an angular node that represents the planes passing through the nucleus or centre where the probability density of the electron is zero. Number of angular nodes is equivalent to the azimuthal quantum number which is $l$. And as we have already discussed that d-orbital possess $l = 2$.
Thus, we can say that the number of angular and radial nodes of $4d$ orbital are $2,1$.

Hence the correct answer is option (D).

Note: Remember that we can also verify and calculate the total number of nodes in a subshell using the formula $(n - 1)$, so $4d$ orbital will have a total number of nodes as $3$. So in conclusion we can say that angular nodes are present in all three planes and radial nodes are sections of these axes and it only depends on the distance from the origin or nucleus.