Chemistry · Structure Of Atom · NEET
A node is a region where the probability of finding the electron is zero. In simple words, the electron is never found there. The wave function (ψ) becomes zero at a node. There are two types: radial nodes (spherical shells where ψ = 0) and angular nodes (flat planes or cones where ψ = 0).
Only three, and they come from NCERT (Unit 2, p.59). Angular nodes = l. Radial nodes = n − l − 1. Total nodes = n − 1. Notice that angular + radial = l + (n − l − 1) = n − 1 = total. So once you know two, you get the third by subtraction.
The number in front is n (principal quantum number). The letter gives l: s → l = 0, p → l = 1, d → l = 2, f → l = 3. Example: 3d means n = 3 and l = 2. So angular nodes = 2, radial nodes = 3 − 2 − 1 = 0, total nodes = 3 − 1 = 2.
A radial node depends only on distance r from the nucleus. It is a spherical shell where ψ = 0, so it looks like a hollow ball layer. An angular node depends on direction (angle). It is a flat plane (or cone) passing through the nucleus where ψ = 0. Angular nodes decide the SHAPE of the orbital; radial nodes decide how many times ψ crosses zero as you move outward.
For 2p: n = 2, l = 1. Radial nodes = n − l − 1 = 2 − 1 − 1 = 0. Angular nodes = l = 1. So 2p has no spherical node but one planar (nodal plane) node that gives the dumbbell shape its two lobes. Total nodes = n − 1 = 1, which matches (0 radial + 1 angular).
Count how many times the curve crosses the zero line (the x-axis), not touching at the very start or at infinity. Each crossing is one radial node. For an s orbital, radial nodes = n − 1, because s has l = 0, so all its nodes are radial. Example: a 3s curve crosses zero twice → 2 radial nodes.
An orbital having 3 angular nodes and 3 total nodes is:
Consider schematic plots of the orbital wavefunction ψr against distance r from the nucleus (A: keeps decreasing, no crossing; B: crosses zero once; C: crosses zero twice; D: many oscillations). The figure representing two radial nodes in the orbital is:
Which one is a wrong statement?
Try the real previous-year questions from this chapter — each with the answer and a full solution.
No. A node must be a full surface (a whole spherical shell or a whole plane/cone) where ψ = 0 everywhere on it. The value being zero only at the nucleus or only at infinite distance is not counted as a node.
For 4d: n = 4, l = 2. Total nodes = n − 1 = 3. Angular nodes = l = 2. Radial nodes = n − l − 1 = 4 − 2 − 1 = 1. So 1 radial + 2 angular = 3 total.
For p and d orbitals the angular nodes are usually flat nodal planes. For some orbitals like d(z²) the angular nodes are cones, not flat planes. So it is safer to say angular nodes, and their number always equals l.
Because they test one clean formula set (angular = l, radial = n − l − 1, total = n − 1) that links quantum numbers to orbital shape. They are fast, scoring marks if you memorise the three formulas, and they often combine with reading a ψ vs r graph.
