The number of angular and radial nodes for 4d orbital is respectively

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2 and 1

1 and 2

3 and 1

4 and 0

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This question was previously asked in

UPSC CAPF – 2018

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The correct answer is A) 2 and 1.

For an atomic orbital described by the principal quantum number *n* and azimuthal (angular momentum) quantum number *l*:
– The number of angular nodes is equal to *l*.
– The total number of nodes is equal to *n* – 1.
– The number of radial nodes is the total number of nodes minus the number of angular nodes, i.e., (*n* – 1) – *l*.
For a 4d orbital:
– Principal quantum number *n* = 4.
– For a d orbital, the azimuthal quantum number *l* = 2 (s=0, p=1, d=2, f=3).
– Number of angular nodes = *l* = 2.
– Number of radial nodes = (*n* – 1) – *l* = (4 – 1) – 2 = 3 – 2 = 1.
The question asks for the number of angular and radial nodes *respectively*.

Angular nodes are surfaces where the probability of finding the electron is zero, and their shape depends on the value of *l* (e.g., for p orbitals, the angular node is a plane; for d orbitals, there are two angular nodes, often planes or conical surfaces). Radial nodes are spherical surfaces where the radial probability density (probability per unit volume) is zero. The number of radial nodes depends on both *n* and *l*.

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