Determining the Number of Radial Nodes in Atomic Orbitals

Total nodes equal n - 1. They are the sum of radial (spherical) nodes, given by n - l - 1, and angular (planar/conical) nodes, given by l.

The number of atomic orbitals from the following having 5 radial nodes is 7s, 7p, 6s, 8p, 8d

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What is the difference between total nodes and radial nodes?

Total nodes equal $n - 1$ . They are the sum of radial (spherical) nodes, given by $n - l - 1$ , and angular (planar/conical) nodes, given by $l$ .

Accepted answer: 3

There are 3 atomic orbitals ( $7 p$ , $6 s$ , and $8 d$ ) that have exactly 5 radial nodes.

Given: Given atomic orbitals: $7 s, 7 p, 6 s, 8 p, 8 d$ . Target condition: number of radial nodes $= 5$ .
Strategise: Use the formula for radial nodes: $Radial Nodes = n - l - 1$ , where $l$ is the principal quantum number and $l$ is the azimuthal quantum number ( $s \to 0, p \to 1, d \to 2$ ). Calculate the radial nodes for each orbital and count those equal to $5$ .
Execute: Calculate radial nodes for each orbital: - For $7 s$ : $n = 7, l = 0 ⟹ 7 - 0 - 1 = 6$ - For $7 p$ : $n = 7, l = 1 ⟹ 7 - 1 - 1 = 5$ - For $6 s$ : $n = 6, l = 0 ⟹ 6 - 0 - 1 = 5$ - For $8 p$ : $n = 8, l = 1 ⟹ 8 - 1 - 1 = 6$ - For $8 d$ : $n = 8, l = 2 ⟹ 8 - 2 - 1 = 5$ Orbitals with 5 radial nodes are $7 p$ , $6 s$ , and $8 d$ . The total count is $3$ .
Verify: Verify total nodes ( $n - 1$ ): $7 s : 6 + 0 = 6$ (total $7 - 1 = 6$ ) $7 p : 5 + 1 = 6$ (total $7 - 1 = 6$ ) $6 s : 5 + 0 = 5$ (total $6 - 1 = 5$ ) $8 p : 6 + 1 = 7$ (total $8 - 1 = 7$ ) $8 d : 5 + 2 = 7$ (total $8 - 1 = 7$ ) Calculations match. Exactly 3 orbitals have 5 radial nodes.