A certain orbital has n = 4 and m L = -3. The number of radial nodes in this orbital is ____________. (Round off to the Nearest Integer).

Solution

Chemistry ma’am · 3 steps

Let mₗ = -3 tell you l first. Once you know it is a 4f orbital, the node count follows at once.

1. Find l. mₗ runs from -l to +l, so mₗ = -3 needs l ≥ 3. For n = 4, l can be at most n - 1 = 3. So l = 3 is the only choice, and this is a 4f orbital.

   radial nodes = n - l - 1

2. Put in numbers. Radial nodes = 4 - 3 - 1 = 0. All n - 1 = 3 nodes of this orbital are angular, because the number of angular nodes equals l = 3.
3. So. The orbital has 0 radial nodes.

The common slip: 3. Using the total node count n - 1 instead of n - l - 1 gives 3. Those 3 nodes are all angular, not radial.

The idea to keep. Whenever mₗ is given, use it to fix the smallest allowed l. Then split the n - 1 total nodes into l angular nodes and n - l - 1 radial nodes.

Answer from the NTA final answer key. Solution by PrepWiser.