Q.26 The number of radial node(s) for the valence orbital of U(III) ion is _____ (in
integer).
(Given: Atomic number of U is 92)
Solution Overview
Uranium (atomic number 92) has the ground-state electron configuration [Rn] 5f³ 6d¹ 7s². For the U(III) ion, three electrons are removed from the outermost orbitals (7s² and 6d¹), yielding [Rn] 5f³, where the valence orbital is 5f.
Radial nodes in an orbital are calculated using the formula: number of radial nodes = n – l – 1, where n is the principal quantum number and l is the azimuthal quantum number.
For the 5f valence orbital, n = 5 and l = 3 (f orbitals), so radial nodes = 5 – 3 – 1 = 1.
Answer: 1
Detailed Solution
Neutral U loses electrons sequentially from 7s (highest energy in actinides) then 6d, leaving 5f electrons as valence for U³⁺. The 5f orbital defines chemical reactivity in this ion, consistent with f-block actinide behavior.
Radial Node Formula Application
- Principal quantum number (n) = 5
- Azimuthal quantum number (l) = 3
- Radial nodes = 5 – 3 – 1 = 1
This matches GATE Life Sciences 2023 data, confirming 1 radial node.
Step-by-Step Calculation
- Neutral U: [Rn] 5f³ 6d¹ 7s²
- U(III): Remove 7s² + 6d¹ → [Rn] 5f³
- Valence orbital: 5f (n=5, l=3)
- Nodes calculation:
• Radial nodes = 1
• Angular nodes = l = 3
• Total nodes = n-1 = 4
Radial Nodes Formula Explained
Radial nodes represent spherical regions of zero electron probability. Use:
Radial nodes = n – l – 1
For 5f orbital: n=5, l=3 → 5-3-1 = 1 radial node
Node Types Summary
| Node Type | Formula | 5f Orbital Value |
|---|---|---|
| Radial Nodes | n – l – 1 | 1 |
| Angular Nodes | l | 3 |
| Total Nodes | n – 1 | 4 |
Exam Relevance
This solves GATE XL 2023 Q.26 precisely. Essential for CSIR NET Life Sciences and GATE Life Sciences preparation.


